Cálculo de Forças Internas e Momentos em Estruturas

Problem 11 Determine the resultant internal normal force acting on the cross section through point A in each column In a segment BC weighs 300 kgm and segment CD weighs 400kgm In b the column has a mass of 200 kgm a Given g 981 m s2 wBC 300 kg m LBC 3m wCA 400 kg m FB 5kN LCA 12m FC 3kN Solution Σ Fy 0 FA wBC g LBC wCA g LCA FB 2FC 0 FA wBC g LBC wCA g LCA FB 2FC FA 245 kN Ans b Given g 981 m s2 w 200 kg m L 3m F1 6kN FB 8kN F2 45kN Solution Σ Fy 0 FA w L g FB 2F1 2F2 0 FA w L g FB 2F1 2F2 FA 3489 kN Ans Problem 12 Determine the resultant internal torque acting on the cross sections through points C and D of the shaft The shaft is fixed at B Given TA 250N m TCD 400N m TDB 300N m Solution Equations of equilibrium TA TC 0 TC TA TC 250N m Ans TA TCD TD 0 TD TCD TA TD 150N m Ans Problem 13 Determine the resultant internal torque acting on the cross sections through points B and C Given TD 500N m TBC 350N m TAB 600N m Solution Equations of equilibrium Σ Mx 0 TB TBC TD 0 TB TBC TD TB 150N m Ans Σ Mx 0 TC TD 0 TC TD TC 500N m Ans Problem 14 A force of 80 N is supported by the bracket as shown Determine the resultant internal loadings acting on the section through point A Given P 80N θ 30deg φ 45deg a 03m b 01m Solution Equations of equilibrium ΣFx0 NA P cos φ θ 0 NA P cos φ θ NA 7727 N Ans ΣFy0 VA P sin φ θ 0 VA P sin φ θ VA 2071 N Ans ΣΜA0 MA P cos φ a cos θ P sin φ b asin θ 0 MA P cos φ a cos θ P sin φ b asin θ MA 0555 N m Ans Note Negative sign indicates that MA acts in the opposite direction to that shown on FBD Problem 15 Determine the resultant internal loadings acting on the cross section through point D of member AB Given ME 70N m a 005m b 03m Solution Segment AB Support Reactions ΣΜA0 ME By 2 a b 0 By ME 2a b By 175 N At B Bx By 150 200 Bx 13125 N Segment DB NB Bx VB By ΣFx0 ND NB 0 ND NB ND 13125 N Ans ΣFy0 VD VB 0 VD VB VD 175 N Ans ΣΜD0 MD ME By a b 0 MD ME By a b MD 875 N m Ans Problem 16 The beam AB is pin supported at A and supported by a cable BC Determine the resultant internal loadings acting on the cross section at point D Given P 5000N a 08m b 12m c 06m d 16m e 06m Solution θ atan b d θ 3687 deg φ atan a b d θ φ 1447 deg Member AB ΣΜA0 FBC sin φ a b P b 0 FBC P b sin φ a b FBC 1201 kN Segment BD ΣFx0 ND FBC cos φ P cos θ 0 ND FBC cos φ P cos θ ND 1563 kN Ans ΣFy0 VD FBC sin φ P sin θ 0 VD FBC sin φ P sin θ VD 0kN Ans ΣΜD0 FBC sin φ P sin θ d c sin θ MD 0 MD FBC sin φ P sin θ d c sin θ MD 0kN m Ans Note Member AB is the twoforce member Therefore the shear force and moment are zero Problem 17 Solve Prob 16 for the resultant internal loadings acting at point E Given P 5000N a 08m b 12m c 06m d 16m e 06m Solution θ atan b d θ 3687 deg φ atan a b d θ φ 1447 deg Member AB ΣΜA0 FBC sin φ a b P b 0 FBC P b sin φ a b FBC 1201 kN Segment BE ΣFx0 NE FBC cos φ P cos θ 0 NE FBC cos φ P cos θ NE 1563 kN Ans ΣFy0 VE FBC sin φ P sin θ 0 VE FBC sin φ P sin θ VE 0kN Ans ΣΜE0 FBC sin φ P sin θ e ME 0 ME FBC sin φ P sin θ e ME 0kN m Ans Note Member AB is the twoforce member Therefore the shear force and moment are zero Problem 18 The boom DF of the jib crane and the column DE have a uniform weight of 750 Nm If the hoist and load weigh 1500 N determine the resultant internal loadings in the crane on cross sections through points A B and C Given P 1500N w 750 N m a 21m b 15m c 06m d 24m e 09m Solution Equations of Equilibrium For point A ΣFx0 NA 0 Ans VA w e P 0 ΣFy0 VA w e P VA 217 kN Ans ΣΜA0 MA w e 05 e P e 0 MA w e 05 e P e MA 1654 kN m Ans Note Negative sign indicates that MA acts in the opposite direction to that shown on FBD Equations of Equilibrium For point B ΣFx0 NB 0 Ans VB w d e P 0 ΣFy0 VB w d e P VB 398 kN Ans ΣΜB0 MB w d e 05 d e P d e 0 MB w d e 05 d e P d e MB 9034 kN m Ans Note Negative sign indicates that MB acts in the opposite direction to that shown on FBD Equations of Equilibrium For point C ΣFx0 VC 0 Ans NC w b c d e P 0 ΣFy0 NC w b c d e P NC 555 kN Ans ΣΜB0 MC w c d e 05 c d e P c d e 0 MC w c d e 05 c d e P c d e MC 11554 kN m Ans Note Negative sign indicates that NC and MC acts in the opposite direction to that shown on FBD Problem 19 The force F 400 N acts on the gear tooth Determine the resultant internal loadings on the root of the tooth ie at the centroid point A of section aa Given P 400N θ 30deg φ 45deg a 4mm b 575mm Solution α φ θ Equations of equilibrium For section a a ΣFx0 VA P cos α 0 VA P cos α VA 38637 N Ans ΣFy0 NA sin α 0 NA P sin α NA 10353 N Ans ΣΜA0 MA P sin α a P cos α b 0 MA P sin α a P cos α b MA 1808 N m Ans Problem 110 The beam supports the distributed load shown Determine the resultant internal loadings on the cross section through point C Assume the reactions at the supports A and B are vertical Given w1 45 kN m w2 60 kN m a 18m b 18m c 24m d 135m e 135m Solution L1 a b c L2 d e Support Reactions ΣΜA0 By L1 w1 L1 05 L1 05w2 L2 L1 L2 3 0 By w1 L1 05 05w2 L2 1 L2 3 L1 By 2282 kN ΣFy0 Ay By w1 L1 05w2 L2 0 Ay By w1 L1 05w2 L2 Ay 1229 kN Equations of Equilibrium For point C ΣFx0 NC 0 Ans ΣFy0 Ay w1 a b VC 0 VC Ay w1 a b VC 392 kN Ans ΣΜC0 MC w1 a b 05 a b Ay a b 0 MC w1 a b 05 a b Ay a b MC 1507 kN m Ans Note Negative sign indicates that VC acts in the opposite direction to that shown on FBD Problem 111 The beam supports the distributed load shown Determine the resultant internal loadings on the cross sections through points D and E Assume the reactions at the supports A and B are vertical Given w1 45 kN m w2 60 kN m a 18m b 18m c 24m d 135m e 135m Solution L1 a b c L2 d e Support Reactions ΣΜA0 By L1 w1 L1 05 L1 05w2 L2 L1 L2 3 0 By w1 L1 05 05w2 L2 1 L2 3 L1 By 2282 kN ΣFy0 Ay By w1 L1 05w2 L2 0 Ay By w1 L1 05w2 L2 Ay 1229 kN Equations of Equilibrium For point D ΣFx0 ND 0 Ans ΣFy0 Ay w1 a VD 0 VD Ay w1 a VD 418 kN Ans ΣΜD0 MD w1 a 05 a Ay a 0 MD w1 a 05 a Ay a MD 14823 kN m Ans Equations of Equilibrium For point E ΣFx0 NE 0 Ans ΣFy0 VE 05w2 05 e 0 VE 05w2 05 e VE 203 kN Ans ΣΜD0 ME 05w2 05 e e 3 0 ME 05 w2 05 e e 3 ME 0911 kN m Ans Note Negative sign indicates that ME acts in the opposite direction to that shown on FBD Problem 112 Determine the resultant internal loadings acting on a section aa and b section bb Each section is located through the centroid point C Given w 9 kN m θ 45deg a 12m b 24m Solution L a b Support Reactions ΣΜA0 Bx L sin θ w L 05 L 0 Bx w L 05 L L sin θ Bx 2291 kN ΣFy0 Ay w L sin θ 0 Ay w L sin θ Ay 2291 kN ΣFx0 Bx w L cos θ Ax 0 Ax w L cos θ Bx Ax 0 kN a Equations of equilibrium For Section a a ΣFx0 NC Ay sin θ 0 NC Ay sin θ ΣFy0 VC Ay cos θ w a 0 VC Ay cos θ w a VC 54 kN Ans ΣΜA0 MC w a 05 a Aycos θ a 0 MC w a 05 a Aycos θ a MC 1296 kN m Ans b Equations of equilibrium For Section b b ΣFx0 NC w a cos θ 0 NC w a cos θ NC 764 kN Ans ΣFy0 VC w a sin θ Ay 0 VC w a sin θ Ay VC 1527 kN Ans ΣΜA0 MC w a 05 a Aycos θ a 0 MC w a 05 a Aycos θ a MC 1296 kN m Ans NC 162 kN Ans Problem 113 Determine the resultant internal normal and shear forces in the member at a section aa and b section bb each of which passes through point A Take θ 60 degree The 650N load is applied along the centroidal axis of the member Given P 650N θ 60deg a Equations of equilibrium For Section a a ΣFy0 P Naa 0 Naa P Naa 650N Ans ΣFx0 Vaa 0 Ans b Equations of equilibrium For Section b b ΣFy0 Vbb P cos 90deg θ 0 Vbb P cos 90deg θ Vbb 56292 N Ans ΣFx0 Nbb P sin 90deg θ 0 Nbb P sin 90deg θ Nbb 325N Ans Problem 114 Determine the resultant internal normal and shear forces in the member at section bb each as a function of θ Plot these results for 0o θ 90o The 650N load is applied along the centroidal axis of the member Given P 650N θ 0 Equations of equilibrium For Section b b ΣFx0 Nbb P cos θ 0 Nbb P cos θ Ans ΣFy0 Vbb P cos θ 0 Vbb P cos θ Ans Problem 115 The 4000N load is being hoisted at a constant speed using the motor M which has a weight of 450 N Determine the resultant internal loadings acting on the cross section through point B in the beam The beam has a weight of 600 Nm and is fixed to the wall at A Given W1 4000N w 600 N m W2 450N a 12m b 12m c 09m d 09m e 12m f 045m r 0075m Solution Tension in rope T W1 2 T 200 kN Equations of Equilibrium For point B ΣFx0 NB T 0 NB T NB 2 kN Ans ΣFy0 VB w e W1 0 VB w e W1 VB 472 kN Ans ΣΜB0 MB w e 05 e W1 e r T f 0 MB w e 05 e W1 e r T f MB 4632 kN m Ans Problem 116 Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob 115 Given W1 4000N w 600 N m W2 450N a 12m b 12m c 09m d 09m e 12m f 045m r 0075m Solution Tension in rope T W1 2 T 200 kN Equations of Equilibrium For point C LC d e ΣFx0 NC T 0 NC T NC 2 kN Ans ΣFy0 VC w LC W1 0 VC w LC W1 VC 526 kN Ans ΣΜC0 MC w LC 05 LC W1 LC r T f 0 MC w LC 05 LC W1 LC r T f MC 9123 kN m Ans Equations of Equilibrium For point D LD b c d e ΣFx0 ND 0 ND 0kN Ans ΣFy0 VD w LD W1 W2 0 VD w LD W1 W2 VD 697 kN Ans ΣΜC0 MD w LD 05 LD W1 LD r W2 b 0 MD w LD 05 LD W1 LD r W2 b MD 22932 kN m Ans Problem 117 Determine the resultant internal loadings acting on the cross section at point B Given w 900 kN m a 1m b 4m Solution L a b Equations of Equilibrium For point B ΣFx0 NB 0 NB 0kN Ans ΣFy0 VB 05 w b L b 0 VB 05 w b L b VB 1440kN Ans ΣΜB0 MB 05 w b L b b 3 0 MB 05 w b L b b 3 MB 1920 kN m Ans Problem 118 The beam supports the distributed load shown Determine the resultant internal loadings acting on the cross section through point C Assume the reactions at the supports A and B are vertical Given w1 05 kN m a 3m w2 15 kN m Solution L 3 a w w2 w1 Support Reactions ΣΜA0 By L w1 L 05 L 05 w L 2L 3 0 By w1 L 05 05 w L 2 3 By 525 kN ΣFy0 Ay By w1 L 05 w L 0 Ay By w1 L 05 w L Ay 375 kN Equations of Equilibrium For point C ΣFx0 NC 0 NC 0kN Ans ΣFy0 VC w1 a 05 w a L a Ay 0 VC w1 a 05 w a L a Ay VC 175 kN Ans ΣΜC0 MC w1 a 05 a 05 w a L a a 3 Ay a 0 MC w1 a 05 a 05 w a L a a 3 Ay a MC 85 kN m Ans Problem 119 Determine the resultant internal loadings acting on the cross section through point D in Prob 118 Given w1 05 kN m a 3m w2 15 kN m Solution L 3 a w w2 w1 Support Reactions ΣΜA0 By L w1 L 05 L 05 w L 2L 3 0 By w1 L 05 05 w L 2 3 By 525 kN ΣFy0 Ay By w1 L 05 w L 0 Ay By w1 L 05 w L Ay 375 kN Equations of Equilibrium For point D ΣFx0 ND 0 ND 0kN Ans ΣFy0 VD w1 2a 05 w 2a L 2a Ay 0 VD w1 2a 05 w 2 a L 2a Ay VD 125 kN Ans ΣΜD0 MD w1 2a a 05 w 2 a L 2a 2a 3 Ay 2a 0 MD w1 2a a 05 w 2 a L 2a 2a 3 Ay 2a MD 95 kN m Ans Problem 120 The wishbone construction of the power pole supports the three lines each exerting a force of 4 kN on the bracing struts If the struts are pin connected at A B and C determine the resultant internal loadings at cross sections through points D E and F Given P 4kN a 12m b 18m Solution Support Reactions FBD a and b Given ΣΜA0 By a Bx 05 b P a 0 1 ΣΜC0 Bx 05 b P a By a P a 0 2 Solving 1 and 2 Initial guess Bx 1kN By 2kN Bx By Find Bx By Bx By 267 2 kN From FBD a ΣFx0 Bx Ax 0 Ax Bx Ax 267 kN ΣFy0 Ay P By 0 Ay P By Ay 6kN From FBD b ΣFx0 Cx Bx 0 Cx Bx Cx 267 kN ΣFy0 Cy By P P 0 Cy 2P By Cy 6kN Equations of Equilibrium For point D FBD c ΣFx0 VD 0 VD 0kN Ans ΣFy0 ND 0 ND 0kN Ans ΣΜD0 MD 0 MD 0kN m Ans For point E FBD d ΣFx0 VF Ax Cx 0 VF Ax Cx VF 0kN Ans ΣFy0 NF Ay Cy 0 NF Ay Cy NF 12kN Ans ΣΜF0 MF Ax Cx 05 b 0 MF Ax Cx 05 b MF 48 kN m Ans ΣFx0 Ax VE 0 VE Ax ΣFy0 NE Ay 0 NE Ay ΣΜE0 ME Ax 05 b 0 ME Ax 05 b ME 24 kN m Ans For point F FBD e VE 267 kN Ans NE 6kN Ans Problem 121 The drum lifter suspends the 25kN drum The linkage is pin connected to the plate at A and B The gripping action on the drum chime is such that only horizontal and vertical forces are exerted on the drum at G and H Determine the resultant internal loadings on the cross section through point I Given P 25 kN θ 60deg a 200mm b 125mm c 75mm d 125mm e 125mm f 50mm Solution Equations of Equilibrium Memeber Ac and BD are twoforce members ΣFy0 P 2 F sin θ 0 1 F P 2 sin θ 2 F 1443 kN Equations of Equilibrium For point I ΣFx0 VI F cos θ 0 VI F cos θ VI 0722 kN Ans ΣFy0 NI F sin θ 0 NI F sin θ NI 125 kN Ans ΣΜI0 MI F cos θ a 0 MI F cos θ a MI 0144 kN m Ans Problem 122 Determine the resultant internal loadings on the cross sections through points K and J on the drum lifter in Prob 121 Given P 25 kN θ 60deg a 200mm b 125mm c 75mm d 125mm e 125mm f 50mm Solution Equations of Equilibrium Memeber Ac and BD are twoforce members ΣFy0 P 2 F sin θ 0 1 F P 2 sin θ 2 F 1443 kN Equations of Equilibrium For point J ΣFy0 VI 0 VI 0kN Ans ΣFx0 NI F 0 NI F NI 1443 kN Ans ΣΜJ0 MJ 0 MJ 0kN m Ans Note Negative sign indicates that NJ acts in the opposite direction to that shown on FBD Support Reactions For Member DFH ΣΜH0 FEF c F cos θ a b c F sin θ f 0 FEF F cos θ a b c c F sin θ f c FEF 3016 kN Equations of Equilibrium For point K ΣFx0 NK FEF 0 NK FEF NK 3016 kN Ans ΣFy0 VK 0 VK 0kN Ans ΣΜK0 MK 0 MK 0kN m Ans Problem 123 The pipe has a mass of 12 kgm If it is fixed to the wall at A determine the resultant internal loadings acting on the cross section at B Neglect the weight of the wrench CD Given g 981 m s2 ρ 12 kg m P 60N a 0150m b 0400m c 0200m d 0300m Solution w ρ g ΣFx0 NBx 0N Ans ΣFy0 VBy 0N Ans ΣFz0 VBz P P w b c 0 VBz P P w b c VBz 706 N Ans ΣΜx0 TBx P b P b w b 05 b 0 TBx P b P b w b 05 b TBx 942 N m Ans ΣΜy0 MBy P 2a w b c w c 05 c 0 MBy P 2 a w b c w c 05 c MBy 623 N m Ans ΣΜz0 MBz 0N m Ans Problem 124 The main beam AB supports the load on the wing of the airplane The loads consist of the wheel reaction of 175 kN at C the 6kN weight of fuel in the tank of the wing having a center of gravity at D and the 2kN weight of the wing having a center of gravity at E If it is fixed to the fuselage at A determine the resultant internal loadings on the beam at this point Assume that the wing does not transfer any of the loads to the fuselage except through the beam Given PC 175kN PE 2kN PD 6kN a 18m b 12m e 03m c 06m d 045m Solution ΣFx0 VAx 0kN Ans ΣFy0 NAy 0kN Ans ΣFz0 VAz PD PE PC 0 VAz PD PE PC VAz 167 kN Ans ΣΜx0 MAx PD a PE a b c PC a b MAx 507 kN m Ans ΣΜy0 TAy PD d PE e 0 TAy PD d PE e TAy 21 kN m Ans ΣΜz0 MAz 0kN m Ans MAx PD a PE a b c PC a b 0 Problem 125 Determine the resultant internal loadings acting on the cross section through point B of the signpost The post is fixed to the ground and a uniform pressure of 50 Nm2 acts perpendicular to the face of the sign Given a 4m d 2m p 50 N m2 b 6m e 3m c 3m Solution P p c d e ΣFx0 VBx P 0 VBx P VBx 750N Ans ΣFy0 VBy 0N Ans ΣFz0 NBz 0N Ans ΣΜx0 MBx 0N m Ans ΣΜy0 MBy P b 05 c 0 MBy P b 05 c MBy 5625N m Ans ΣΜz0 TBz P e 05 d e 0 TBz P e 05 d e TBz 375N m Ans Problem 126 The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft Determine the resultant internal loadings acting on the cross section through point D The 400N forces act in the z direction and the 200N and 80N forces act in the y direction The journal bearings at A and B exert only y and z components of force on the shaft Given P1z 400N P2y 200N P3y 80N a 03m b 04m c 03m d 04m Solution L a b c d Support Reactions ΣΜz0 2P3y d 2P2y c d Ay L 0 Ay 2P3y d L 2P2y c d L Ay 24571 N ΣFy0 Ay By 2 P2y 2 P3y 0 By Ay 2 P2y 2 P3y By 31429 N ΣΜy0 2P1z b c d Az L 0 Az 2P1z b c d L Az 62857 N ΣFz0 Bz Az 2 P1z 0 Bz Az 2 P1z Bz 17143 N Equations of Equilibrium For point D ΣFx0 NDx 0N Ans ΣFy0 VDy By 2 P3y 0 VDy By 2 P3y VDy 1543 N Ans ΣFz0 VDz Bz 0 VDz Bz VDz 1714 N Ans ΣΜx0 TDx 0N m Ans ΣΜy0 MDy Bz d 05 c 0 MDy Bz d 05 c MDy 9429 N m Ans ΣΜz0 MDz By d 05 c 2 P3y 05 c 0 MDz By d 05 c 2 P3y 05 c MDz 14886 N m Ans Problem 127 The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft Determine the resultant internal loadings acting on the cross section through point C The 400N forces act in the z direction and the 200N and 80N forces act in the y direction The journal bearings at A and B exert only y and z components of force on the shaft Given P1z 400N P2y 200N P3y 80N a 03m b 04m c 03m d 04m Solution L a b c d Support Reactions ΣΜz0 2P3y d 2P2y c d Ay L 0 Ay 2P3y d L 2P2y c d L Ay 24571 N ΣFy0 Ay By 2 P2y 2 P3y 0 By Ay 2 P2y 2 P3y By 31429 N ΣΜy0 2P1z b c d Az L 0 Az 2P1z b c d L Az 62857 N ΣFz0 Bz Az 2 P1z 0 Bz Az 2 P1z Bz 17143 N Equations of Equilibrium For point C ΣFx0 NCx 0N Ans ΣFy0 VCy Ay 0 VCy Ay VCy 2457 N Ans ΣFz0 VCz Az 2P1z 0 VCz Az 2P1z VCz 1714 N Ans ΣΜx0 TCx 0N m Ans ΣΜy0 MCy Az a 05 b 2 P1z 05 b 0 MCy Az a 05 b 2 P1z 05 b MCy 15429 N m Ans ΣΜz0 MCz Ay a 05 b 0 MCz Ay a 05 b MCz 12286 N m Ans Problem 128 Determine the resultant internal loadings acting on the cross section of the frame at points F and G The contact at E is smooth Given a 12m b 15m c 09m P 400N d 09m e 12m θ 30deg Solution L d2 e2 Member DEF ΣΜD0 NE b P a b 0 NE P a b b NE 720N Member BCE ΣΜB0 FAC e L d NE sin θ c d 0 FAC L e d NE sin θ c d FAC 900N ΣFx0 Bx FAC d L NE cos θ 0 Bx FAC d L NE cos θ Bx 8354 N ΣFy0 By FAC e L NE sin θ 0 By FAC e L NE sin θ By 360N Equations of Equilibrium For point F ΣFy0 NF 0 NF 0N Ans ΣFx0 VF P 0 VF P VF 400N Ans ΣΜF0 MF P 05 a 0 MF P 05 a MF 240N m Ans Equations of Equilibrium For point G ΣFx0 Bx NG 0 NG Bx NG 8354 N Ans ΣFy0 VG By 0 VG By VG 360N Ans ΣΜG0 MG By 05 d 0 MG By 05 d MG 162N m Ans Problem 129 The bolt shank is subjected to a tension of 400 N Determine the resultant internal loadings acting on the cross section at point C Given P 400N r 150mm θ 90deg Solution Equations of Equilibrium For segment AC ΣFx0 NC P 0 NC P NC 400N Ans ΣFy0 VC 0 VC 0N Ans ΣΜG0 MC P r 0 MC P r MC 60 N m Ans Problem 130 The pipe has a mass of 12 kgm If it is fixed to the wall at A determine the resultant internal loadings acting on the cross section through B Given P 750N MC 800N m ρ 12 kg m g 981 m s2 a 1m b 2m c 2m Solution Py 4 5 P Pz 3 5 P Equations of Equilibrium For point B ΣFx0 VBx 0kip Ans ΣFy0 NBy Py 0 NBy Py Ans NBy 600 N Ans ΣFz0 VBz Pz ρ g c ρ g b 0 VBz Pz ρ g c ρ g b VBz 9209 N Ans ΣΜx0 MBx Pz b ρ g c b ρ g b 05 b 0 MBx Pz b ρ g c b ρ g b 05 b MBx 16063 N m Ans ΣΜy0 TBy 0N m Ans ΣΜz0 MBy Mc 0 MBy MC MBy 800 N m Ans Problem 131 The curved rod has a radius r and is fixed to the wall at B Determine the resultant internal loadings acting on the cross section through A which is located at an angle θ from the horizontal Solution P kN θ deg Equations of Equilibrium For point A ΣFx0 NA P cos θ 0 NA P cos θ Ans ΣFy0 VA P sin θ 0 VA P sin θ Ans ΣΜA0 MA P r 1 cos θ 0 MA P r 1 cos θ r Ans Problem 132 The curved rod AD of radius r has a weight per length of w If it lies in the horizontal plane determine the resultant internal loadings acting on the cross section through point B Hint The distance from the centroid C of segment AB to point O is CO 09745r Given θ 225deg r m a 09745r w kN m2 Solution Equations of Equilibrium For point B ΣFz0 VB π 4 r w 0 VB 0785 w r Ans ΣFx0 NB 0 Ans ΣΜx0 TB π 4 r w 009968r 0 TB 00783w r2 Ans ΣΜy0 MB π 4 r w 037293r 0 MB 0293 w r2 Ans Problem 133 A differential element taken from a curved bar is shown in the figure Show that dNdθ V dVdθ N dMdθ T and dTdθ M Solution Problem 134 The column is subjected to an axial force of 8 kN which is applied through the centroid of the crosssectional area Determine the average normal stress acting at section aa Show this distribution of stress acting over the areas cross section Given P 8kN b 150mm d 140mm t 10mm Solution A 2 b t d t A 440000 mm2 σ P A σ 182 MPa Ans Problem 135 The anchor shackle supports a cable force of 30 kN If the pin has a diameter of 6 mm determine the average shear stress in the pin Given P 30kN d 6mm Solution ΣFy0 2 V P 0 V 05P V 1500N A π d2 4 A 282743 mm2 τavg V A τavg 5305 MPa Ans Problem 136 While running the foot of a 75kg man is momentarily subjected to a force which is 5 times his weight Determine the average normal stress developed in the tibia T of his leg at the mid section aa The cross section can be assumed circular having an outer diameter of 45 mm and an inner diameter of 25 mm Assume the fibula F does not support a load Given g 981 m s2 M 75kg do 45mm di 25mm Solution A π 4 do 2 di 2 A 10995574 mm2 σ 5M g A σ 3345 MPa Ans Problem 137 The thrust bearing is subjected to the loads shown Determine the average normal stress developed on cross sections through points B C and D Sketch the results on a differential volume element located at each section Units Used kPa 103Pa Given P 500N Q 200N dB 65mm dC 140mm dD 100mm Solution AB π dB 2 4 AB 33183 mm2 σB P AB σB 1507 kPa Ans AC π dC 2 4 AC 153938 mm2 σC P AC σC 325 kPa Ans AD π dD 2 4 AD 78540 mm2 σD Q AD σD 255 kPa Ans Problem 138 The small block has a thickness of 5 mm If the stress distribution at the support developed by the load varies as shown determine the force F applied to the block and the distance d to where it is applied Given a 60mm b 120mm t 5mm σ1 0MPa σ2 40MPa σ3 60MPa Solution F A σ d F 05 σ2 a t σ2 b t 05 σ3 σ2 b t F 3600 kN Ans Require F d A x σ d d 05 σ2 a t 2a 3 σ2 b t a 05 b 05 σ3 σ2 b t a 2 b 3 F d 110mm Ans Problem 139 The lever is held to the fixed shaft using a tapered pin AB which has a mean diameter of 6 mm If a couple is applied to the lever determine the average shear stress in the pin between the pin and lever Given a 250mm b 12mm d 6mm P 20N Solution ΣΜO0 V b P 2a 0 V P 2a b V 83333 N A π d2 4 A 282743 mm2 τavg V A τavg 2947 MPa Ans Problem 140 The cinder block has the dimensions shown If the material fails when the average normal stress reaches 0840 MPa determine the largest centrally applied vertical load P it can support Given σallow 0840MPa ao 150mm ai 100mm bo 2 1 2 3 2 mm bi 2 1 3 mm Solution A ao bo ai bi A 1300mm2 Pallow σallow A Pallow 1092 kN Ans Problem 141 The cinder block has the dimensions shown If it is subjected to a centrally applied force of P 4 kN determine the average normal stress in the material Show the result acting on a differential volume element of the material Given P 4kN ao 150mm ai 100mm bo 2 1 2 3 2 mm bi 2 1 3 mm Solution A ao bo ai bi A 1300mm2 σ P A σ 308 MPa Ans Problem 142 The 250N lamp is supported by three steel rods connected by a ring at A Determine which rod is subjected to the greater average normal stress and compute its value Take θ 30 The diameter of each rod is given in the figure Given W 250N θ 30deg φ 45deg dB 9mm dC 6mm dD 75mm Solution Initial guess FAC 1N FAD 1N Given ΣFx0 FAC cos θ FAD cos φ 0 1 ΣFy0 FAC sin θ FAD sin φ W 0 2 Solving 1 and 2 FAC FAD Find FAC FAD FAC FAD 18301 22414 N Rod AB AAB π dB 2 4 AAB 6361725 mm2 σAB W AAB σAB 393 MPa Rod AD AAD π dD 2 4 AAD 4417865 mm2 σAD FAD AAD σAD 5074 MPa Rod AC AAC π dC 2 4 AAC 2827433 mm2 σAC FAC AAC σAC 6473 MPa Ans Problem 143 Solve Prob 142 for θ 45 Given W 250N θ 45deg φ 45deg dB 9mm dC 6mm dD 75mm Solution Initial guess FAC 1N FAD 1N Given ΣFx0 FAC cos θ FAD cos φ 0 1 ΣFy0 FAC sin θ FAD sin φ W 0 2 Solving 1 and 2 FAC FAD Find FAC FAD FAC FAD 17678 17678 N Rod AB AAB π dB 2 4 AAB 6361725 mm2 σAB W AAB σAB 393 MPa Rod AD AAD π dD 2 4 AAD 4417865 mm2 σAD FAD AAD σAD 4001 MPa Rod AC AAC π dC 2 4 AAC 2827433 mm2 σAC FAC AAC σAC 6252 MPa Ans Problem 144 The 250N lamp is supported by three steel rods connected by a ring at A Determine the angle of orientation θ of AC such that the average normal stress in rod AC is twice the average normal stress in rod AD What is the magnitude of stress in each rod The diameter of each rod is given in the figure Given W 250N φ 45deg dB 9mm dC 6mm dD 75mm Solution Rod AB AAB π dB 2 4 AAB 6361725 mm2 Rod AD AAD π dD 2 4 AAD 4417865 mm2 Rod AC AAC π dC 2 4 AAC 2827433 mm2 Since σAC 2σAD Therefore FAC AAC 2 FAD AAD Initial guess FAC 1N FAD 2N θ 30deg Given FAC AAC 2 FAD AAD 1 ΣFx0 FAC cos θ FAD cos φ 0 2 ΣFy0 FAC sin θ FAD sin φ W 0 3 Solving 1 2 and 3 FAC FAD θ Find FAC FAD θ FAC FAD 18038 14092 N θ 5647 deg σAB W AAB σAB 393 MPa Ans σAD FAD AAD σAD 319 MPa Ans σAC FAC AAC σAC 638 MPa Ans Problem 145 The shaft is subjected to the axial force of 30 kN If the shaft passes through the 53mm diameter hole in the fixed support A determine the bearing stress acting on the collar C Also what is the average shear stress acting along the inside surface of the collar where it is fixed connected to the 52mm diameter shaft Given P 30kN dhole 53mm dshaft 52mm dcollar 60mm hcollar 10mm Solution Bearing Stress Ab π 4 dcollar 2 dhole 2 σb P Ab σb 483 MPa Ans Average Shear Stress As π dshaft hcollar τavg P As τavg 184 MPa Ans Problem 146 The two steel members are joined together using a 60 scarf weld Determine the average normal and average shear stress resisted in the plane of the weld Given P 8kN θ 60deg b 25mm h 30mm Solution Equations of Equilibrium ΣFx0 N P sin θ 0 N P sin θ N 6928 kN ΣFy0 V P cos θ 0 V P cos θ V 4kN A h b sin θ σ N A σ 8MPa Ans τavg V A τavg 462 MPa Ans Problem 147 The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N Determine the average normal stress developed in the bolt BC if the bolt has a diameter of 8 mm Assume A is a pin Given P 775N a 40mm b 30mm d 8mm θ 20deg Solution Support Reaction ΣFA0 P a FBC cos θ a b 0 FBC P a a b cos θ FBC 47128 N Average Normal Stress ABC π d2 4 σ FBC ABC σ 938 MPa Ans Problem 148 The board is subjected to a tensile force of 425 N Determine the average normal and average shear stress developed in the wood fibers that are oriented along section aa at 15 with the axis of the board Given P 425N θ 15deg b 25mm h 75mm Solution Equations of Equilibrium ΣFx0 V P cos θ 0 V P cos θ V 410518 N ΣFy0 N P sin θ 0 N P sin θ N 1N Average Normal Stress A h b sin θ σ N A σ 00152 MPa Ans τavg V A τavg 00567 MPa Ans Problem 149 The open square butt joint is used to transmit a force of 250 kN from one plate to the other Determine the average normal and average shear stress components that this loading creates on the face of the weld section AB Given P 250kN θ 30deg b 150mm h 50mm Solution Equations of Equilibrium ΣFx0 V P sin θ 0 V P sin θ V 125kN ΣFy0 N P cos θ 0 N P cos θ N 216506 kN Average Normal and Shear Stress A h b sin 2θ σ N A σ 25MPa Ans τavg V A τavg 14434 MPa Ans Problem 150 The specimen failed in a tension test at an angle of 52 when the axial load was 100 kN If the diameter of the specimen is 12 mm determine the average normal and average shear stress acting on the area of the inclined failure plane Also what is the average normal stress acting on the cross section when failure occurs Given P 100kN d 12mm θ 52deg Solution Equations of Equilibrium ΣFx0 V P cos θ 0 V P cos θ V 61566 kN ΣFy0 N P sin θ 0 N P sin θ N 78801 kN Inclined plane A π 4 d2 sin θ σ N A σ 54905 MPa Ans τavg V A τavg 42896 MPa Ans Cross section A πd2 4 σ P A σ 88419 MPa Ans τavg 0 τavg 0MPa Ans Problem 151 A tension specimen having a crosssectional area A is subjected to an axial force P Determine the maximum average shear stress in the specimen and indicate the orientation θ of a section on which it occurs Solution Equations of Equilibrium ΣFy0 V P cos θ 0 V P cos θ Inclined plane Aincl A sin θ τ V Aincl τ P cos θ sin θ A τ P sin 2θ 2A dτ dθ P cos 2θ A dτ dθ 0 cos 2θ 0 2θ 90deg θ 45deg Ans τmax P sin 90 2A τmax P 2A Ans Problem 152 The joint is subjected to the axial member force of 5 kN Determine the average normal stress acting on sections AB and BC Assume the member is smooth and is 50mm thick Given P 5kN θ 45deg φ 60deg dAB 40mm dBC 50mm t 50mm Solution α 90deg φ α 3000 deg AAB t dAB ABC t dBC ΣFx0 NAB cos α P cos θ 0 NAB P cos θ cos α NAB 4082 kN ΣFy0 NAB sin α P sin θ NBC 0 NBC NAB sin α P sin θ NBC 1494 kN σAB NAB AAB σAB 2041 MPa Ans σBC NBC ABC σBC 0598 MPa Ans Problem 153 The yoke is subjected to the force and couple moment Determine the average shear stress in the bolt acting on the cross sections through A and B The bolt has a diameter of 6 mm Hint The couple moment is resisted by a set of couple forces developed in the shank of the bolt Given P 25kN M 120N m ho 62mm hi 50mm d 6mm θ 60deg Solution As a force on bolt shank is zero then τA 0 Ans Equations od Equilibrium ΣFz0 P 2Fz 0 Fz 05P Fz 125 kN ΣMz0 M Fx hi 0 Fx M hi Fx 24 kN Average Shear Stress A πd2 4 The bolt shank subjected to a shear force of VB Fx 2 Fz 2 τB VB A τB 9571 MPa Ans Problem 154 The two members used in the construction of an aircraft fuselage are joined together using a 30 fishmouth weld Determine the average normal and average shear stress on the plane of each weld Assume each inclined plane supports a horizontal force of 2 kN Given P 40kN b 375mm hhalf 25m θ 30deg Solution Equations of Equilibrium ΣFx0 V 05P cos θ 0 V 05P cos θ V 1732 kN ΣFy0 N 05P sin θ 0 N 05P sin θ N 1kN Average Normal and Shear Stress A hhalf b sin θ σ N A σ 53333 Pa Ans τavg V A τavg 92376 Pa Ans Problem 155 The row of staples AB contained in the stapler is glued together so that the maximum shear stress the glue can withstand is τ max 84 kPa Determine the minimum force F that must be placed on the plunger in order to shear off a staple from its row and allow it to exit undeformed through the groove at C The outer dimensions of the staple are shown in the figure It has a thickness of 125 mm Assume all the other parts are rigid and neglect friction Given τmax 0084MPa a 125mm b 75mm t 125mm Solution Average Shear Stress A a b a 2t b t τmax V A V τmax A V 263 N Fmin V Fmin 263 N Ans Problem 156 Rods AB and BC have diameters of 4mm and 6 mm respectively If the load of 8 kN is applied to the ring at B determine the average normal stress in each rod if θ 60 Given W 8kN θ 60deg dA 4mm dC 6mm Solution Rod AB AAB π dA 2 4 Rod BC ABC π dC 2 4 ΣFy0 FBC sin θ W 0 FBC W sin θ FBC 9238 kN ΣFx0 FBC cos θ FAB 0 FAB FBC cos θ FAB 4619 kN σAB FAB AAB σAB 3676 MPa Ans σBC FBC ABC σBC 3267 MPa Ans Problem 157 Rods AB and BC have diameters of 4 mm and 6 mm respectively If the vertical load of 8 kN is applied to the ring at B determine the angle θ of rod BC so that the average normal stress in each rod is equivalent What is this stress Given W 8kN dA 4mm dC 6mm Solution Rod AB AAB π dA 2 4 Rod BC ABC π dC 2 4 ΣFy0 FBC sin θ W 0 ΣFx0 FBC cos θ FAB 0 Since FAB σ AAB FBC σ ABC Initial guess σ 100MPa θ 50deg Given σ ABC sin θ W 0 1 σ ABC cos θ σ AAB 0 2 Solving 1 and 2 σ θ Find σ θ θ 6361 deg Ans σ 31585 MPa Ans Problem 158 The bars of the truss each have a crosssectional area of 780 mm2 Determine the average normal stress in each member due to the loading P 40 kN State whether the stress is tensile or compressive Given P 40kN a 09m b 12m A 780mm2 Solution c a2 b2 c 15 m h b c v a c Joint A ΣFy0 v FAB P 0 FAB P v FAB 66667 kN ΣFx0 h FAB FAE 0 FAE h FAB FAE 53333 kN σAB FAB A σAB 8547 MPa T Ans σAE FAE A σAE 68376 MPa C Ans Joint E ΣFy0 FEB 075P 0 FEB 075P FEB 30kN ΣFx0 FED FAE 0 FED FAE FED 53333 kN σEB FEB A σEB 38462 MPa T Ans σED FED A σED 68376 MPa C Ans Joint B ΣFy0 v FBD v FAB FEB 0 FBD FAB FEB v FBD 116667 kN ΣFx0 FBC h FAB h FBD 0 FBC h FAB h FBD FBC 146667 kN σBC FBC A σBC 188034 MPa T Ans σBD FBD A σBD 149573 MPa C Ans Problem 159 The bars of the truss each have a crosssectional area of 780 mm2 If the maximum average normal stress in any bar is not to exceed 140 MPa determine the maximum magnitude P of the loads that can be applied to the truss σallow 140MPa Given a 09m b 12m A 780mm2 Solution c a2 b2 c 15 m h b c v a c For comparison purpose set P 1kN Joint A ΣFy0 v FAB P 0 FAB P v FAB 1667 kN ΣFx0 h FAB FAE 0 FAE h FAB FAE 1333 kN σAB FAB A σAB 2137 MPa T σAE FAE A σAE 1709 MPa C Joint E ΣFy0 FEB 075P 0 FEB 075P FEB 075 kN ΣFx0 FED FAE 0 FED FAE FED 1333 kN σEB FEB A σEB 0962 MPa T σED FED A σED 1709 MPa C Joint B ΣFy0 v FBD v FAB FEB 0 FBD FAB FEB v FBD 2917 kN ΣFx0 FBC h FAB h FBD 0 FBC h FAB h FBD FBC 3667 kN σBC FBC A σBC 4701 MPa T σBD FBD A σBD 3739 MPa C Since the crosssectional areas are the same the highest stress occurs in the member BC which has the greatest force Fmax max FAB FAE FEB FED FBD FBC Fmax 3667 kN Pallow P Fmax σallow A Pallow 2978 kN Ans Problem 160 The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p 650 Pa Determine the average shear stress which the glue exerts on the sides of the tube needed to hold the cap in place Given p 650Pa a 25mm di 35mm do 40mm Solution Ap π di 2 4 As π do a P a FBC cos θ a b 0 P p Ap P 0625 N Average Shear Stress τavg P As τavg 1991 Pa Ans Problem 161 The crimping tool is used to crimp the end of the wire E If a force of 100 N is applied to the handles determine the average shear stress in the pin at A The pin is subjected to double shear and has a diameter of 5 mm Only a vertical force is exerted on the wire Given P 100N a 375mm b 50mm c 25mm d 125mm dpin 5mm Solution From FBD a ΣFx0 Bx 0 Bx 0N ΣΜD0 P d By c 0 By P d c By 500N From FBD b ΣFx0 Ax 0 Ax 0N ΣΜE0 Ay a By a b 0 Ay By a b a Ay 116667 N Average Shear Stress Apin π dpin 2 4 VA 05 Ay VA 583333 N τavg VA Apin τavg 29709 MPa Ans Problem 162 Solve Prob 161 for pin B The pin is subjected to double shear and has a diameter of 5 mm Given a 375mm b 50mm c 25mm d 125mm dpin 5mm P 100N Solution From FBD a ΣFx0 Bx 0 Bx 0N ΣΜD0 P d By c 0 By P d c By 500N Average Shear Stress Pin B is subjected to doule shear Apin π dpin 2 4 VB 05 By VB 250N τavg VB Apin τavg 12732 MPa Ans Problem 163 The railcar docklight is supported by the 3mmdiameter pin at A If the lamp weighs 20 N and the extension arm AB has a weight of 8 Nm determine the average shear stress in the pin needed to support the lamp Hint The shear force in the pin is caused by the couple moment required for equilibrium at A Given w 8 N m P 20N a 900mm h 32mm dpin 3mm Solution From FBD a ΣFx0 Bx 0 ΣΜA0 V h w a 05a P a 0 V w a 05 a h P a h V 66375 N Average Shear Stress Apin π dpin 2 4 τavg V Apin τavg 93901 MPa Ans Problem 164 The twomember frame is subjected to the distributed loading shown Determine the average normal stress and average shear stress acting at sections aa and bb Member CB has a square cross section of 35 mm on each side Take w 8 kNm Given w 8 kN m a 3m b 4m A 00352 m2 Solution c a2 b2 c 5m h a c v b c Member AB ΣMA0 By a w a 05a 0 By 05w a By 12kN ΣFy0 v FAB By 0 FAB By v FAB 15kN Section aa σaa FAB A σaa 1224 MPa Ans τaa 0 τaa 0MPa Ans Section bb ΣFx0 N FAB h 0 N FAB h N 9kN ΣFy0 V FAB v 0 V FAB v V 12kN Abb A h σbb N Abb σbb 441 MPa Ans τbb V Abb τbb 588 MPa Ans Problem 165 Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN Determine the average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member B which has a thickness of 30 mm Given P 5kN θ 60deg φ 30deg drod 10mm h 40mm t 30mm Solution AB t h Arod π 4 drod 2 ΣFx0 P cos θ FB 0 FB P cos θ FB 25 kN ΣFy0 Fc P sin θ 0 FC P sin θ FC 433 kN Average Normal Stress σB FB AB σB 2083 MPa Ans σC FC Arod σC 55133 MPa Ans Problem 166 Consider the general problem of a bar made from m segments each having a constant crosssectional area Am and length Lm If there are n loads on the bar as shown write a computer program that can be used to determine the average normal stress at any specified location x Show an application of the program using the values L1 12 m d1 06 m P1 2 kN A1 1875 mm2 L2 06 m d2 18 m P2 15 kN A2 625 mm2 Problem 167 The beam is supported by a pin at A and a short link BC If P 15 kN determine the average shear stress developed in the pins at A B and C All pins are in double shear as shown and each has a diameter of 18 mm Given P 15kN a 05m b 1m c 15m d 15m e 05m θ 30deg dpin 18mm Solution L a b c d e Support Reactions ΣΜA0 By L P L a 4P c d e 4P d e 2P e 0 By P L a L 4 P c d e L 4 P d e L 2 P e L By 825 kN ΣFy0 By P 4 P 4 P 2 P Ay 0 Ay By P 4 P 4P 2 P Ay 825 kN FBC By sin θ FBC 165kN Ax FBC cos θ Ax 14289 kN Average Shear Stress Apin π dpin 2 4 For Pins B and C τBandC 05FBC Apin τBandC 3242 MPa Ans For Pin A FA Ax 2 Ay 2 FA 165kN τA 05FA Apin τA 3242 MPa Ans Problem 168 The beam is supported by a pin at A and a short link BC Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa All pins are in double shear as shown and each has a diameter of 18 mm Given τallow 80MPa a 05m b 1m c 15m d 15m e 05m θ 30deg dpin 18mm Solution L a b c d e For comparison purpose set P 1kN Support Reactions ΣΜA0 By L P L a 4P c d e 4P d e 2P e 0 By P L a L 4 P c d e L 4 P d e L 2 P e L By 55 kN ΣFy0 By P 4 P 4 P 2 P Ay 0 Ay By P 4 P 4P 2 P Ay 55 kN FBC By sin θ FBC 11kN Ax FBC cos θ Ax 953 kN FA Ax 2 Ay 2 FA 11kN Require Fmax max FBC FA Fmax 11kN Apin π dpin 2 4 Pallow P Fmax τallow 2Apin Pallow 370 kN Ans Problem 169 The frame is subjected to the load of 1 kN Determine the average shear stress in the bolt at A as a function of the bar angle θ Plot this function 0 θ 90o and indicate the values of θ for which this stress is a minimum The bolt has a diameter of 6 mm and is subjected to single shear Given P 1kN dbolt 6mm a 06m b 045m c 015m Solution Support Reactions ΣΜC0 FAB cos θ c FAB sin θ a P a b 0 FAB P a b cos θ c sin θ a Average Shear Stress Pin B is subjected to doule shear τ FAB Abolt Abolt π dbolt 2 4 τ 4P a b cos θ c sin θ a π dbolt 2 dτ dθ 4P a b π dbolt 2 sin θ c cos θ a cos θ c sin θ a 2 dτ dθ 0 sin θ c cos θ a 0 tan θ a c θ atan a c θ 7596 deg Ans Problem 170 The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the beam 1ft x 12ft If the hoist is rated to support a maximum of 75 kN determine the maximum average normal stress in the 18mmdiameter tie rod BC and the maximum average shear stress in the 16mmdiameter pin at B Given P 75kN xmax 36m a 3m θ 30deg drod 18mm dpin 16mm Solution Support Reactions ΣΜC0 FBC sin θ a P x 0 FBC P x sin θ a Maximum FBC occurs when x xmax Therefore FBC P xmax sin θ a FBC 1800 kN Arod π drod 2 4 Apin π dpin 2 4 τpin 05 FBC Apin τpin 44762 MPa Ans σrod FBC Arod σrod 70736 MPa Ans Problem 171 The bar has a crosssectional area A and is subjected to the axial load P Determine the average normal and average shear stresses acting over the shaded section which is oriented at θ from the horizontal Plot the variation of these stresses as a function of θ 0o θ 90o Solution Equations of Equilibrium ΣFx0 V P cos θ 0 V P cos θ ΣFy0 N P sin θ 0 N P sin θ Inclined plane Aθ A sin θ σ N A σ P A sin θ 2 Ans τavg V A τavg P 2A sin 2θ Ans Problem 172 The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC If the cable has a diameter of 15 mm plot the average normal stress in the cable as a function of the boom position θ for 0o θ 90o Given W 3kN a 1m do 15mm Solution Angle B φB 05 90deg θ φB 45deg 05θ Support Reactions ΣΜA0 FBC sin φB a W 05a cos θ 0 FBC 05W cos θ sin 45deg 05θ Average Normal Stress σBC FAB ABC ABC π do 2 4 σBC 2W π do 2 cos θ sin 45deg 05θ Ans Problem 173 The bar has a crosssectional area of 400 106 m2 If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads as shown determine the average normal stress in the bar as a function of for 0 x 05m Given P1 3kN P2 6kN w 8 kN m A 400 10 6 m2 a 05m b 075m Solution L a b ΣFx0 N P1 P2 w L x 0 N P1 P2 w L x Average Normal Stress σ N A σ P1 P2 w L x A Ans Problem 174 The bar has a crosssectional area of 400 106 m2 If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads as shown determine the average normal stress in the bar as a function of for 05m x 125m Given P1 3kN P2 6kN w 8 kN m A 400 10 6 m2 a 05m b 075m Solution L a b ΣFx0 N P1 w L x 0 N P1 w L x Average Normal Stress σ N A σ P1 w L x A Ans Problem 175 The column is made of concrete having a density of 230 Mgm3 At its top B it is subjected to an axial compressive force of 15 kN Determine the average normal stress in the column as a function of the distance z measured from its base Note The result will be useful only for finding the average normal stress at a section removed from the ends of the column because of localized deformation at the ends Given P 3kN ρ 23 103 kg m3 g 981 m s2 r 180mm h 075m Solution A π r2 w ρ g A ΣFz0 N P w h z 0 N P w h z Average Normal Stress σ N A σ P w h z A Ans Problem 176 The twomember frame is subjected to the distributed loading shown Determine the largest intensity of the uniform loading that can be applied to the frame without causing either the average normal stress or the average shear stress at section bb to exceed σ 15 MPa and τ 16 MPa respectively Member CB has a square crosssection of 30 mm on each side Given σallow 15MPa τallow 16MPa a 4m b 3m A 00302 m2 Solution c a2 b2 c 5m v b c h a c Set wo 1 kN m Member AB ΣMA0 By b wo b 05b 0 By 05wo b By 15 kN Section bb By FBC h FBC By h FBC 188 kN ΣFx0 FBC h Vbb 0 Vbb FBC h Vbb 15 kN ΣFy0 Nbb FBC v 0 Nbb FBC v Nbb 1125 kN Abb A v σbb Nbb Abb σbb 075 MPa τbb Vbb Abb τbb 1MPa Assume failure due to normal stress wallow wo σallow σbb wallow 2000 kN m Assume failure due to shear stress wallow wo τallow τbb wallow 1600 kN m Ans Controls Problem 177 The pedestal supports a load P at its center If the material has a mass density ρ determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant The cross section is circular Solution Require σ P w1 A σ P w1 dw A dA P dA w1 dA A dw dw dA P w1 A dw dA σ 1 dA π r dr 2 πr2 dA 2πr dr dw πr2 ρ g dz From Eq1 πr2 ρ g dz 2πr dr σ r ρ g dz 2dr σ ρ g 2σ 0 z z 1 d r1 r r 1 r d ρ g z 2σ ln r r1 r r1 e ρ g 2σ z However σ P π r1 2 Ans r r1 e π r1 2 ρ g 2P z Problem 178 The radius of the pedestal is defined by r 05e008y2 m where y is given in meters If the material has a density of 25 Mgm3 determine the average normal stress at the support Given ro 05m h 3m g 981 m s2 r ro e 008 y2 m ρ 25 103 kg m3 yunit 1m Solution dr π e 008 y2 dy Ao πro 2 Ao 07854 m2 dV π r2 dy dV πro 2 e 008 y2 2 dy V 0 3 y πro 2 e 008 y2 2 yunit d V 1584 m3 W ρ g V W 38835 kN σ W Ao σ 004945 MPa Ans Problem 179 The uniform bar having a crosssectional area of A and mass per unit length of m is pinned at its center If it is rotating in the horizontal plane at a constant angular rate of ω determine the average normal stress in the bar as a function of x Solution Equation of Motion ΣFxMaNMω r2 N m L 2 x ω x 1 2 L 2 x N m ω 8 L2 4 x2 Average Normal Stress σ N A σ m ω 8A L2 4 x2 Ans Problem 180 Member B is subjected to a compressive force of 4 kN If A and B are both made of wood and are 10mm thick determine to the nearest multiples of 5mm the smallest dimension h of the support so that the average shear stress does not exceed τallow 21 MPa Given P 4kN t 10mm τallow 21MPa a 300mm b 125mm Solution c a2 b2 c 325mm h a c v b c V P v V 154 kN τallow V t h h V t τallow h 7326 mm Use h 75mm h 75mm Ans Problem 181 The joint is fastened together using two bolts Determine the required diameter of the bolts if the failure shear stress for the bolts is τfail 350 MPa Use a factor of safety for shear of FS 25 Given P 80kN τfail 350MPa γ 25 Solution τallow τfail γ τallow 140MPa Vbolt 05 P 2 Vbolt 20kN Abolt Vbolt τallow π 4 d2 Vbolt τallow d 4 π Vbolt τallow d 1349 mm Ans Problem 182 The rods AB and CD are made of steel having a failure tensile stress of σfail 510 MPa Using a factor of safety of FS 175 for tension determine their smallest diameter so that they can support the load shown The beam is assumed to be pin connected at A and C Given P1 4kN P2 6kN P3 5kN a 2m b 2m c 3m d 3m γ 175 σfail 510MPa Solution L a b c d Support Reactions ΣΜA0 FCD L P1 a P2 a b P3 a b c 0 FCD P1 a L P2 a b L P3 a b c L FCD 670 kN ΣΜC0 FAB L P1 b c d P2 c d P3 d 0 FAB P1 b c d L P2 c d L P3 d L FAB 830 kN Average Normal Stress Design of rod sizes σallow σfail γ σallow 29143 MPa For Rod AB Abolt FAB σallow π 4 dAB 2 FAB σallow dAB 4 π FAB σallow dAB 602 mm Ans For Rod CD Abolt FCD σallow π 4 dCD 2 FCD σallow dCD 4 π FCD σallow dCD 541 mm Ans Problem 183 The lever is attached to the shaft A using a key that has a width d and length of 25 mm If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle determine the dimension d if the allowable shear stress for the key is τallow 35 MPa Given P 200N τallow 35MPa L 500mm a 20mm b 25mm Solution ΣΜA0 Faa a P L 0 Faa P L a Faa 5000N For the key Aaa Faa τallow b d Faa τallow d 1 b Faa τallow d 571 mm Ans Problem 184 The fillet weld size a is determined by computing the average shear stress along the shaded plane which has the smallest cross section Determine the smallest size a of the two welds if the force applied to the plate is P 100 kN The allowable shear stress for the weld material is τallow 100 MPa Given P 100kN τallow 100MPa L 100mm θ 45deg Solution Shear Plane in the Weld Aweld L a sin θ Aweld 05P τallow L a sin θ 05P τallow a 1 L sin θ 05P τallow a 7071 mm Ans Problem 185 The fillet weld size a 8 mm If the joint is assumed to fail by shear on both sides of the block along the shaded plane which is the smallest cross section determine the largest force P that can be applied to the plate The allowable shear stress for the weld material is τallow 100 MPa Given a 8mm L 100mm θ 45deg τallow 100MPa Solution Shear Plane in the Weld Aweld L a sin θ P τallow 2Aweld P τallow 2 L a sin θ P 11314 kN Ans Problem 186 The eye bolt is used to support the load of 25 kN Determine its diameter d to the nearest multiples of 5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washer will not penetrate or shear through it The allowable normal stress for the bolt is σallow 150 MPa and the allowable shear stress for the supporting material is τallow 35 MPa Given P 25kN dwasher 25mm σallow 150MPa τallow 35MPa Solution Allowable Normal Stress Design of bolt size Abolt P σallow π 4 d2 P σallow d 4 π P σallow d 14567 mm Use d 15mm d 15mm Ans Allowable Shear Stress Design of support thickness Asupport P τallow π dwasher h P τallow h 1 π dwasher P τallow h 9095 mm Use d 10mm d 10mm Ans Problem 187 The frame is subjected to the load of 8 kN Determine the required diameter of the pins at A and B if the allowable shear stress for the material is τallow 42 MPa Pin A is subjected to double shear whereas pin B is subjected to single shear Given P 8kN τallow 42MPa a 15m b 15m c 15m d 06m Solution θBC 45deg Support Reactions From FBD a ΣΜD0 FBC sin θBC c P c d 0 FBC P c d sin θBC c FBC 15839 kN From FBD b ΣΜA0 Dy a b P c d 0 Dy P c d a b Dy 56 kN ΣFx0 Ax P 0 Ax P Ax 8kN ΣFy0 Dy Ay 0 Ay Dy FA Ax 2 Ay 2 FA 977 kN Apin 05FA τallow π 4 d2 05FA τallow d 4 π 05FA τallow d 12166 mm Ans For pin B Pin A is subjected to single shear and FB FBC Apin FB τallow π 4 d2 FB τallow d 4 π FB τallow d 21913 mm Ans Problem 188 The two steel wires AB and AC are used to support the load If both wires have an allowable tensile stress of σallow 200 MPa determine the required diameter of each wire if the applied load is P 5 kN Given P 5kN σallow 200MPa a 4m b 3m θ 60deg Solution c a2 b2 h a c v b c At joint A Initial guess FAB 1kN FAC 2kN Given ΣFx0 FAC h FAB sin θ 0 1 ΣFy0 FAC v FAB cos θ P 0 2 Solving 1 and 2 FAB FAC Find FAB FAC FAB FAC 43496 47086 kN For wire AB AAB FAB σallow π 4 dAB 2 FAB σallow dAB 4 π FAB σallow dAB 526 mm Ans For wire AC AAC FAC σallow π 4 dAC 2 FAC σallow dAC 4 π FAC σallow dAC 548 mm Ans Problem 189 The two steel wires AB and AC are used to support the load If both wires have an allowable tensile stress of σallow 180 MPa and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm determine the greatest force P that can be applied to the chain before one of the wires fails Given σallow 180MPa a 4m b 3m θ 60deg dAB 6mm dAC 4mm Solution c a2 b2 h a c v b c Assume failure of AB FAB AAB σallow FAB π 4 dAB 2 σallow FAB 509 kN At joint A Initial guess P1 1kN FAC 2kN Given ΣFx0 FAC h FAB sin θ 0 1 ΣFy0 FAC v FAB cos θ P1 0 2 Solving 1 and 2 P1 FAC Find P1 FAC P1 FAC 58503 55094 kN Assume failure of AC FAC AAC σallow FAC π 4 dAC 2 σallow FAC 226 kN At joint A Initial guess P2 1kN FAB 2kN Given ΣFx0 FAC h FAB sin θ 0 1 ΣFy0 FAC v FAB cos θ P2 0 2 Solving 1 and 2 FAB P2 Find FAB P2 FAB P2 20895 24019 kN Chosoe the smallest value P min P1 P2 P 240 kN Ans Problem 190 The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stress of σallow 168 MPa Determine the greatest load that can be supported without causing the cable to fail when θ 30 and φ 45 Neglect the size of the winch Given σallow 168MPa do 6mm θ 30deg φ 45deg Solution For the cable Tcable Acable σallow Tcable π 4 do 2 σallow Tcable 47501 kN At joint B Initial guess FAB 1kN W 2kN Given ΣFx0 Tcable cos θ FAB cos φ 0 1 ΣFy0 W FAB sin φ Tcable sin θ 0 2 Solving 1 and 2 FAB W Find FAB W FAB W 5818 1739 kN Ans Problem 191 The boom is supported by the winch cable that has an allowable normal stress of σallow 168 MPa If it is required that it be able to slowly lift 25 kN from θ 20 to θ 50 determine the smallest diameter of the cable to the nearest multiples of 5mm The boom AB has a length of 6 m Neglect the size of the winch Set d 36 m Given σallow 168MPa W 25kN d 36m a 6m Solution Maximum tension in canle occurs when θ 20deg sin θ a sin ψ d ψ asin d a sin θ ψ 11842 deg At joint B Initial guess FAB 1kN Tcable 2kN Given φ θ ψ ΣFx0 Tcable cos θ FAB cos φ 0 1 ΣFy0 W FAB sin φ Tcable sin θ 0 2 Solving 1 and 2 FAB Tcable Find FAB Tcable FAB Tcable 114478 103491 kN For the cable Acable P σallow π 4 do 2 Tcable σallow do 4 π Tcable σallow do 28006 mm Use do 30mm do 30mm Ans Problem 192 The frame is subjected to the distributed loading of 2 kNm Determine the required diameter of the pins at A and B if the allowable shear stress for the material is τallow 100 MPa Both pins are subjected to double shear Given w 2 kN m τallow 100MPa r 3m Solution Member AB is atwoforce member θ 45deg Support Reactions ΣΜA0 FBC sin θ r w r 05r 0 FBC 05w r sin θ FBC 4243 kN ΣFy0 Ay FBC sin θ w r 0 Ay FBC sin θ w r Ay 3kN ΣFx0 Ax FBC cos θ 0 Ax FBC cos θ Ax 3kN Average Shear Stress Pin A and pin B are subjected to double shear FA Ax 2 Ay 2 FA 4243 kN FB FBC FB 4243 kN Since both subjected to the same shear force V 05 FA and V 05FB Apin V τallow π 4 dpin 2 V τallow dpin 4 π V τallow dpin 520 mm Ans Problem 193 Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required to support is P 150 kN The allowable tensile stress bearing stress and shear stress is σtallow 175 MPa σballow 275 MPa and τallow 115 MPa Given P 150kN σtallow 175MPa τallow 115MPa σballow 275MPa d2 30mm Solution Allowable Normal Stress Design of end cap outer diameter A P σtallow π 4 d1 2 d2 2 P σtallow d1 4 π P σtallow d2 2 d1 4462 mm Ans Allowable Bearing Stress Design of circular shaft diameter A P σballow π 4 d3 2 P σballow d3 4 π P σballow d3 2635 mm Ans Allowable Shear Stress Design of end cap thickness A P τallow π d3 t P τallow t 1 π d3 P τallow t 1575 mm Ans Problem 194 If the allowable bearing stress for the material under the supports at A and B is σballow 28 MPa determine the size of square bearing plates A and B required to support the loading Take P 75 kN Dimension the plates to the nearest multiples of 10mm The reactions at the supports are vertical Given σballow 28MPa P 75 kN P1 10kN P2 10kN P3 15kN P4 10kN a 15m b 25m Solution L 3 a b Support Reactions ΣΜA0 By 3a P2 a P3 2a P4 3a P L 0 By P2 a 3a P3 2a 3a P4 3a 3a P L 3a By 35kN ΣΜB0 Ay 3 a P1 3 a P2 2 a P3 a P b 0 Ay P1 3a 3a P2 2a 3a P3 a 3a P b 3a Ay 175 kN For Plate A AplateA Ay σballow aA 2 Ay σballow aA Ay σballow aA 79057 mm Use aA x aA plate aA 80mm Ans For Plate B AplateB By σballow aB 2 By σballow aB By σballow aB 111803 mm Use aB x aB plate aB 120mm Ans Rb 35kN b is in subscript Problem 195 If the allowable bearing stress for the material under the supports at A and B is σballow 28 MPa determine the maximum load P that can be applied to the beam The bearing plates A and B have square cross sections of 50mm x 50mm and 100mm x 100mm respectively Given σballow 28MPa P1 10kN P2 10kN P3 15kN P4 10kN a 15m b 25m aA 50mm aB 100mm Solution L 3 a b Support Reactions ΣΜA0 By 3a P2 a P3 2a P4 3 a P L 0 By P2 a 3a P3 2 a 3a P4 3 a 3a P L 3a ΣΜB0 Ay 3 a P1 3 a P2 2 a P3 a P b 0 Ay P1 3a 3a P2 2a 3a P3 a 3a P b 3a For Plate A Ay aA 2 σballow aA 2 σballow P1 3a 3a P2 2a 3a P3 a 3a P b 3a P P1 3a b P2 2a b P3 a b aA 2 σballow 3a b P 26400 kN Pcase1 P For Plate B By aB 2 σballow aB 2 σballow P2 a 3a P3 2 a 3a P4 3 a 3a P L 3a P P2 a L P3 2 a L P4 3 a L aB 2 σballow 3a L P 3000 kN Pcase2 P Pallow min Pcase1 Pcase2 Pallow 3kN Ans Problem 196 Determine the required crosssectional area of member BC and the diameter of the pins at A and B if the allowable normal stress is σallow 21 MPa and the allowable shear stress is τallow 28 MPa Given σallow 21MPa τallow 28MPa P 75kip θ 60deg a 06m b 12m c 06m Solution L a b c Support Reactions ΣΜA0 By L P a P a b 0 By P a L P a b L FBC By sin θ FBC 38523 kN By 33362 kN Bx FBC cos θ Bx 19261 kN ΣFy0 By P P Ay 0 Ay By P P Ay 33362 kN ΣFx0 Bx Ax 0 Ax Bx Ax 19261 kN FA Ax 2 Ay 2 FA 38523 kN Member BC ABC FBC σallow ABC 1834416 mm2 Ans Pin A AA FA τallow π 4 dA 2 FA τallow dA 4 π FA τallow dA 41854 mm Ans Pin B AB 05FBC τallow π 4 dB 2 05FBC τallow dB 4 π 05FBC τallow dB 29595 mm Ans Problem 197 The assembly consists of three disks A B and C that are used to support the load of 140 kN Determine the smallest diameter d1 of the top disk the diameter d2 within the support space and the diameter d3 of the hole in the bottom disk The allowable bearing stress for the material is τallowb 350 MPa and allowable shear stress is τallow 125 MPa Given P 140kN τallow 125MPa σballow 350MPa hB 20mm hC 10mm Solution Allowable Shear Stress Assume shear failure dor disk C A P τallow π d2 hC P τallow d2 1 π hC P τallow d2 3565 mm Ans Allowable Bearing Stress Assume bearing failure dor disk C A P σballow π 4 d2 2 d3 2 P σballow d3 d2 2 4 π P σballow d3 2760 mm Ans Allowable Bearing Stress Assume bearing failure dor disk B A P σballow π 4 d1 2 P σballow d1 4 π P σballow d1 2257 mm Since d3 d1 disk B might fail due to shear τ P A τ P π d1 hB τ 9873 MPa τallow OK Therefore d1 2257 mm Ans Problem 198 Strips A and B are to be glued together using the two strips C and D Determine the required thickness t of C and D so that all strips will fail simultaneously The width of strips A and B is 15 times that of strips C and D Given P 40N t 30mm bA 15m bB 15m bC 1m bD 1m Solution Average Normal Stress Requires σA σB σB σC σC σD N bA t 05N bC tC tC 05 bA t bC tC 225 mm Ans Problem 199 If the allowable bearing stress for the material under the supports at A and B is σballow 28 MPa determine the size of square bearing plates A and B required to support the loading Dimension the plates to the nearest multiples of 10mm The reactions at the supports are vertical Take P 75 kN Given σballow 28MPa P 75kN w 10 kN m a 45m b 225m Solution L a b Support Reactions ΣΜA0 By a w a 05 a P L 0 By w 05 a P L a By 3375 kN ΣΜB0 Ay a w a 05 a P L 0 Ay w 05 a P b a Ay 1875 kN Allowable Bearing Stress Design of bearing plates For Plate A Area Ay σballow aA 2 Ay σballow aA Ay σballow aA 81832 mm Use aA x aA plate aA 90mm aA 90mm Ans For Plate B Area By σballow aB 2 By σballow aB By σballow aB 109789 mm Use aB x aB plate aB 110mm aB 11000 mm Ans Problem 1100 If the allowable bearing stress for the material under the supports at A and B is σballow 28 MPa determine the maximum load P that can be applied to the beam The bearing plates A and B have square cross sections of 50mm x 50mm and 100mm x 100mm respectively Given σballow 28MPa w 10 kN m a 45m b 225m aA 50mm aB 100mm Solution L a b Support Reactions ΣΜA0 By a w a 05 a P L 0 P By a L w a L 05 a ΣΜB0 Ay a w a 05 a P b 0 P Ay a b w a b 05 a Allowable Bearing Stress Assume failure of material occurs under plate A Ay aA 2 σballow P aA 2 σballow a b w a 05a b P 31kN Pcase1 P Assume failure of material occurs under plate B By aB 2 σballow P By a L w a L 05 a P 367 kN Pcase2 P Pallow min Pcase1 Pcase2 Pallow 367 kN Ans Problem 1101 The hanger assembly is used to support a distributed loading of w 12 kNm Determine the average shear stress in the 10mmdiameter bolt at A and the average tensile stress in rod AB which has a diameter of 12 mm If the yield shear stress for the bolt is τy 175 MPa and the yield tensile stress for the rod is σy 266 MPa determine the factor of safety with respect to yielding in each case Given τy 175MPa w 12 kN m σy 266MPa a 12m b 06m e 09m do 10mm drod 12mm Solution c a2 e2 h a c v e c Support Reactions L a b ΣΜC0 FAB v a w L 05 L 0 FAB w L a v 05 L FAB 27kN For bolt A Bolt A is subjected to double shear and V 05FAB V 135 kN A π 4 do 2 τ V A τ 17189 MPa Ans FS τy τ FS 102 Ans For rod AB N FAB N 27kN A π 4 drod 2 σ N A σ 23873 MPa Ans FS σy σ FS 111 Ans Problem 1102 Determine the intensity w of the maximum distributed load that can be supported by the hanger assembly so that an allowable shear stress of τallow 95 MPa is not exceeded in the 10mmdiameter bolts at A and B and an allowable tensile stress of σallow 155 MPa is not exceeded in the 12mmdiameter rod AB Given τallow 95MPa σallow 155MPa a 12m b 06m e 09m do 10mm drod 12mm Solution c a2 e2 h a c v e c Support Reactions L a b ΣΜC0 FAB v a w L 05 L 0 FAB w L a v 05 L Assume failure of pin A or B V 05FAB V τallow A A π 4 do 2 05 w L a v 05 L τallow π 4 do 2 w a v 05L 2 τallow π 4 do 2 w 6632 kN m controls Ans Assuming failure of rod AB N FAB N σallow A A π 4 drod 2 w L a v 05 L σallow π 4 drod 2 w a v 05L2 σallow π 4 drod 2 w 7791 kN m Problem 1103 The bar is supported by the pin If the allowable tensile stress for the bar is σtallow 150 MPa and the allowable shear stress for the pin is τallow 85 MPa determine the diameter of the pin for which the load P will be a maximum What is this maximum load Assume the hole in the bar has the same diameter d as the pin Take t 6 mm and w 50 mm Given τallow 85MPa σtallow 150MPa t 6mm w 50mm Solution Given Allowable Normal Stress The effective crosssectional area Ae for the bar must be considered here by taking into account the reduction in crosssectional area introduced by the hole Here effective area Ae is equal to w d t and σallow equals to P Ae σtallow P w d t 1 Allowable Shear Stress The pin is subjected to double shear and therefore the allowable τ equals to 05P Apin and the area Apin is equal to π4 d2 τallow 2 π P d2 2 Solving 1 and 2 Initial guess d 20mm P 10kN P d Find P d P 3123 kN Ans d 1529 mm Ans Problem 1104 The bar is connected to the support using a pin having a diameter of d 25 mm If the allowable tensile stress for the bar is σtallow 140 MPa and the allowable bearing stress between the pin and the bar is σballow 210 MPA determine the dimensions w and t such that the gross area of the cross section is wt 1250 mm2 and the load P is a maximum What is this maximum load Assume the hole in the bar has the same diameter as the pin Given σtallow 140MPa σballow 210MPa A 1250mm2 d 25mm Solution A w t Given Allowable Normal Stress The effective crosssectional area Ae for the bar must be considered here by taking into account the reduction in crosssectional area introduced by the hole Here effective area Ae is equal to w d t that is A d t and σallow equals to P Ae σtallow P A d t 1 Allowable Bearing Stress The projected area Ab is equal to d t and σallow equals to P Ab σballow P d t 2 Solving 1 and 2 Initial guess t 05in P 1kip P t Find P t P 10500 kN Ans t 2000 mm Ans And w A t w 6250 mm Ans Problem 1105 The compound wooden beam is connected together by a bolt at B Assuming that the connections at A B C and D exert only vertical forces on the beam determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is σtallow 150 MPa and the allowable bearing stress for the wood is σballow 28 MPa Assume that the hole in the washers has the same diameter as the bolt Given P1 3kN P2 15kN P3 2kN σtallow 150MPa a 2m σballow 28MPa b 15m Solution From FBD a Given ΣΜD0 Cy 4 b By 3b P2 2 b P3 b 0 1 From FBD b ΣΜA0 By 2 a b Cy 2 a P1 a 0 2 Solving 1 and 2 Initial guess By 1kN Cy 2kN By Cy Find By Cy By Cy 44 455 kN For bolt Arae By σtallow π 4 dB 2 By σtallow dB 4 π By σtallow dB 611 mm Ans For washer Area By σballow π 4 dw 2 dB 2 By σballow dw dB 2 4 π By σballow dw 1541 mm Ans Problem 1106 The bar is held in equilibrium by the pin supports at A and B Note that the support at A has a single leaf and therefore it involves single shear in the pin and the support at B has a double leaf and therefore it involves double shear The allowable shear stress for both pins is τallow 150 MPa If a uniform distributed load of w 8 kNm is placed on the bar determine its minimum allowable position x from B Pins A and B each have a diameter of 8 mm Neglect any axial force in the bar Given τallow 150MPa do 8mm a 2m w 8 kN m b 2m Solution ΣΜA0 By a w b x a x 05 b x 0 1 ΣΜB0 Ay a w b x x 05 b x 0 2 Assume failure of pin A Arae Ay τallow π 4 do 2 Ay τallow Ay π 4 do 2 τallow Ay 75398 kN Substitute value of force A into Eq 2 Given Ay a w b x1 x1 05 b x1 0 2 Initial guess x1 03m x1 Find x1 x1 0480 m xcase1 x1 Assume failure of pin B Arae 05By τallow π 4 do 2 05By τallow By 2 π 4 do 2 τallow By 150796 kN Substitute value of force A into Eq 1 Given By a w b x2 a x2 05 b x2 0 1 Initial guess x2 03m x2 Find x2 x2 0909 m xcase2 x2 Choose the larger x value x max xcase1 xcase2 x 0909 m Ans Problem 1107 The bar is held in equilibrium by the pin supports at A and B Note that the support at A has a single leaf and therefore it involves single shear in the pin and the support at B has a double leaf and therefore it involves double shear The allowable shear stress for both pins is τallow 125 MPa If x 1 m determine the maximum distributed load w the bar will support Pins A and B each have a diameter of 8 mm Neglect any axial force in the bar Given τallow 125MPa x 1m do 8mm a 2m b 2m Solution wo kN m Given ΣΜA0 Bw a wo b x a x 05 b x 0 1 ΣΜB0 Aw a wo b x x 05 b x 0 2 Initial guess Bw 1kN Aw 1kN Solving 1 and 2 Bw Aw Find Bw Aw Bw Aw 175 075 kN For pin A Ay w1 Aw wo Arae Ay τallow π 4 do 2 w1 τallow Aw wo w1 π 4 do 2 τallow wo Aw w1 8378 kN m For pin B By w Bw wo Arae 05By τallow π 2 do 2 w2 τallow Bw wo w2 π 2 do 2 τallow wo Bw w2 7181 kN m The smalleer w controls w min w1 w2 w 7181 kN m Ans Problem 1108 The bar is held in equilibrium by the pin supports at A and B Note that the support at A has a single leaf and therefore it involves single shear in the pin and the support at B has a double leaf and therefore it involves double shear The allowable shear stress for both pins is τallow 125 MPa If x 1 m and w 12 kNm determine the smallest required diameter of pins A and B Neglect any axial force in the bar Given τallow 125MPa x 1m do 8mm a 2m b 2m Solution w 12 kN m Given ΣΜA0 By a w b x a x 05 b x 0 1 ΣΜB0 Ay a w b x x 05 b x 0 2 Initial guess By 1kN Ay 1kN Solving 1 and 2 By Ay Find By Ay By Ay 21 9 kN For pin A Arae Ay τallow π 4 dA 2 Ay τallow dA 4 π Ay τallow dA 957 mm Ans For pin B Arae 05By τallow π 2 dB 2 By τallow dB 2 π By τallow dB 1034 mm Ans Problem 1109 The pin is subjected to double shear since it is used to connect the three links together Due to wear the load is distributed over the top and bottom of the pin as shown on the freebody diagram Determine the diameter d of the pin if the allowable shear stress is τallow 70 MPa and the load P 40 kN Also determine the load intensities w1 and w2 Given τallow 70MPa P 40kN a 375mm b 25mm Solution Pin ΣFy0 P w1 a 0 w1 P a w1 106667 kN m Ans Link ΣFy0 P 2 05w2 b 0 w2 P b w2 160000 kN m Ans Shear Stress Area 05P τallow π 2 d2 P τallow d 2 π P τallow d 19073 mm Ans Problem 1110 The pin is subjected to double shear since it is used to connect the three links together Due to wear the load is distributed over the top and bottom of the pin as shown on the freebody diagram Determine the maximum load P the connection can support if the allowable shear stress for the material is τallow 56 MPa and the diameter of the pin is 125 mm Also determine the load intensities w1 and w2 Given τallow 56MPa d 125mm a 375mm b 25mm Solution Shear Stress Area 05P τallow π 2 d2 P τallow P π 2 d2 τallow P 137445 kN Ans Pin ΣFy0 P w1 a 0 w1 P a w1 36652 kN m Ans Link ΣFy0 P 2 05w2 b 0 w2 P b w2 54978 kN m Ans Problem 1111 The cotter is used to hold the two rods together Determine the smallest thickness t of the cotter and the smallest diameter d of the rods All parts are made of steel for which the failure tensile stress is σfail 500 MPa and the failure shear stress is τfail 375 MPa Use a factor of safety of FSt 250 in tension and FSs 175 in shear Given σfail 500MPa τfail 375MPa P 30kN d2 40mm h 10mm FSt 250 FSs 175 Solution Allowable Normal Stress Design of rod size σallow σfail FSt σallow 200MPa Area P σallow π 4 d2 P σallow d 4 π P σallow d 1382 mm Ans Allowable Shear Stress Design of cotter size τallow τfail FSs τallow 21429 MPa Area 05P τallow h t 05P τallow t 1 h 05P τallow t 7mm Ans Problem 1112 The long bolt passes through the 30mmthick plate If the force in the bolt shank is 8 kN determine the average normal stress in the shank the average shear stress along the cylindrical area of the plate defined by the section lines aa and the average shear stress in the bolt head along the cylindrical area defined by the section lines bb Given P 8kN dshank 7mm daa 18mm dbb 7mm hhead 8mm hplate 30mm Solution Average Normal Stress Ashank π 4 dshank 2 σt P Ashank σt 2079 MPa Ans Average Shear Stresses Aaa π daa hplate τaavg P Aaa τaavg 472 MPa Ans Abb π dbb hhead τbavg P Abb τbavg 4547 MPa Ans Problem 1113 The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN Determine the average normal and shear stress acting on the plane through section aa Show the results on a ifferential volume element located on the plane Given P 6kN d 150mm θ 30deg Solution φ 90deg θ Equations of Equilibrium ΣFx0 Vaa P cos φ 0 Vaa P cos φ Vaa 3kN ΣFy0 Naa P sin φ 0 Naa P sin φ Naa 5196 kN At inclined plane A d2 sin φ σaa Naa A σaa 02 MPa Ans τaa Vaa A τaa 0115 MPa Ans Problem 1114 Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame Given w 25 kN m a 12m b 09m c 15m e 045m Solution d a2 b2 d 15 m v a d h b d Support Reactions L b c ΣΜA0 By b w L 05 L 0 By w L 05 L b By 8kN ΣFy0 Ay By w L 0 Ay By w L Ay 2kN By FBC v FBC By v FBC 10kN ΣFx0 FBC h Ax 0 Ax FBC h Ax 6kN Segment AD ΣFx0 ND Ax 0 ND Ax ND 6kN Ans ΣFy0 Ay w e VD 0 VD Ay w e VD 313 kN Ans ΣΜD0 MD w e 05 e Ay e 0 MD w e 05 e Ay e MD 1153 kN m Ans Segment CE ΣFx0 NE FBC 0 ND FBC ND 10 kN Ans ΣFy0 VD 0 VD 0kN Ans ΣΜE0 MD 0 MD 0kN m Ans Problem 1115 The circular punch B exerts a force of 2 kN on the top of the plate A Determine the average shear stress in the plate due to this loading Given P 2kN dpunch 4mm hplate 2mm Solution Average Shear Stresses Aaa π dpunch hplate τaavg P Aaa τaavg 7958 MPa Ans Problem 1116 The cable has a specific weight γ weightvolume and crosssectional area A If the sag s is small so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis determine the average normal stress in the cable at its lowest point C Solution Equations of Equilibrium ΣMA0 T s γAL 2 L 4 0 γ γ A L 8 s Average Normal Stresses σ T A σ γ L 8 s Ans Problem 1117 The beam AB is pin supported at A and supported by a cable BC A separate cable CG is used to hold up the frame If AB weighs 20 kNm and the column FC has a weight of 30 kNm determine the resultant internal loadings acting on cross sections located at points D and E Neglect the thickness of both the beam and column in the calculation Given wb 20 kN m L 36m d 18m wc 30 kN m H 48m e 12m a 36m b 36m c 12m Solution Beam AB Lc L2 c2 vb c Lc hb L Lc ΣΜA0 By L wAB L 05 L 0 By wb L 05 L L By 793664 m s2 lb Ay By wb L Ay 36 kN ΣFy0 Ay By wb L 0 By FBC vb FBC By vb FBC 1138 kN ΣFx0 FBC h Ax 0 Ax FBC hb Ax 108 kN Segment AD ΣFx0 ND Ax 0 ND Ax ND 108 kN Ans ΣFy0 Ay wb d VD 0 VD Ay wb d VD 0kN Ans ΣΜD0 MD wb d 05 d Ay d 0 MD wb d 05 d Ay d MD 324 kN m Ans Member CG Hb H2 b2 vc H Hb hc b Hb Column FC ΣΜC0 Fx H Ax c 0 Fx Ax c H Fx 27 kN ΣFx0 FBC hb Ax Fx FCG hc 0 ΣFy0 FCG FBC hb Ax Fx hc FCG 45 kN Fy By wc H FBC vb FCG vc 0 Fy By wc H FBC vb FCG vc Fy 252 kN Segment FE ΣFx0 VE Fx 0 VE Fx VE 27 kN Ans ΣFy0 NE wc e Fy 0 NE wc e Fy NE 216 kN Ans ΣΜE0 ME Fy e 0 ME Fy e ME 3024 kN m Ans Problem 1118 The 3Mg concrete pipe is suspended by the three wires If BD and CD have a diameter of 10 mm and AD has a diameter of 7 mm determine the average normal stress in each wire Given M 3000kg g 981 m s2 h 2m r 1m α 120deg dBD 10mm dCD 10mm dAD 7mm Solution W M g W 2943 kN θ 05α θ 60deg L r2 h2 v h L Equations of Equilibrium ΣΜx0 2 F r cos θ FAD r 0 FBD FCD F ΣΜy0 FBD r sin θ FCD r sin θ 0 FAD F ΣFz0 3 F v W 0 F W 3v F 1097 kN Allowable Normal Stress σBD σCD σ1 σ1 F Area σ1 F π 4 dBD 2 σ1 13965 MPa Ans σAD σ2 σ2 F Area σ2 F π 4 dAD 2 σ2 285MPa Ans Problem 1119 The yokeandrod connection is subjected to a tensile force of 5 kN Determine the average normal stress in each rod and the average shear stress in the pin A between the members Given P 5kN dpin 25mm d30 30mm d40 40mm Solution Average Normal Stress A40 π 4 d40 2 σ40 P A40 σ40 3979 MPa Ans A30 π 4 d30 2 σ30 P A30 σ30 7074 MPa Ans Average Shear Stresses Apin π 4 dpin 2 τavg 05P Apin τavg 5093 MPa Ans Problem 21 An airfilled rubber ball has a diameter of 150 mm If the air pressure within it is increased until the balls diameter becomes 175 mm determine the average normal strain in the rubber Given d0 150mm d 175mm Solution ε πd πd0 πd0 ε 01667 mm mm Ans Problem 22 A thin strip of rubber has an unstretched length of 375 mm If it is stretched around a pipe having an outer diameter of 125 mm determine the average normal strain in the strip Given L0 375mm Solution L π 125 mm ε πL πL0 πL0 ε 00472 mm mm Ans Problem 23 The rigid beam is supported by a pin at A and wires BD and CE If the load P on the beam causes the end C to be displaced 10 mm downward determine the normal strain developed in wires CE and BD Given a 3m LCE 4m b 4m LBD 4m LCE 10mm Solution LBD a a b LCE LBD 42857 mm εCE LCE LCE εCE 000250 mm mm Ans εBD LBD LBD εBD 000107 mm mm Ans Problem 24 The center portion of the rubber balloon has a diameter of d 100 mm If the air pressure within it causes the balloons diameter to become d 125 mm determine the average normal strain in the rubber Given d0 100mm d 125mm Solution ε πd πd0 πd0 ε 02500 mm mm Ans Problem 25 The rigid beam is supported by a pin at A and wires BD and CE If the load P on the beam is displace 10 mm downward determine the normal strain developed in wires CE and BD Given a 3m b 2m c 2m LCE 4m LBD 3m tip 10mm Solution LBD a LCE a b LCE a b tip a b c LCE a b a b c tip LCE 71429 mm LBD a a b c tip LBD 42857 mm Average Normal Strain εCE LCE LCE εCE 000179 mm mm Ans εBD LBD LBD εBD 000143 mm mm Ans Problem 26 The rigid beam is supported by a pin at A and wires BD and CE If the maximum allowable normal strain in each wire is εmax 0002 mmmm determine the maximum vertical displacement of the load P Given a 3m b 2m c 2m LCE 4m LBD 3m εallow 0002 mm mm Solution LBD a LCE a b LCE a b tip a b c Average ElongationVertical Displacement LBD LBD εallow LBD 600 mm tip a b c a LBD tip 1400 mm LCE LCE εallow LCE 800 mm tip a b c a b LCE tip 1120 mm Controls Ans Problem 27 The two wires are connected together at A If the force P causes point A to be displaced horizontally 2 mm determine the normal strain developed in each wire Given a 300mm θ 30deg A 2mm Solution Consider the triangle CAA φA 180deg θ φA 150deg LCA a2 A 2 2 a A cos φA LCA 301734 mm CA LCA a a CA 000578 mm mm Ans Problem 28 Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB If a force is applied to the end D of the member and causes it to rotate by θ 03 determine the normal strain in the cable Originally the cable is unstretched Given a 400mm b 300mm c 300mm θ 03deg Solution LAB a2 b2 LAB 500mm Consider the triangle ACB φC 90deg θ φC 903 deg LAB a2 b2 2 a b cos φC LAB 501255 mm εAB LAB LAB LAB εAB 000251 mm mm Ans Problem 29 Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB If a force is applied to the end D of the member and causes a normal strain in the cable of 00035 mmmm determine the displacement of point D Originally the cable is unstretched Given a 400mm b 300mm c 300mm εAB 00035 mm mm Solution LAB a2 b2 LAB 500mm LAB LAB 1 εAB LAB 501750 mm Consider the triangle ACB φC 90deg θ LAB a2 b2 2 a b cos φC φC acos a2 b2 LAB 2 2 a b φC 90419 deg θ φC 90deg θ 041852 deg θ 000730 rad D b c θ D 4383 mm Ans Problem 210 The wire AB is unstretched when θ 45 If a vertical load is applied to bar AC which causes θ 47 determine the normal strain in the wire Given θ 45deg θ 2deg Solution LAB L2 L2 LAB 2L LCB 2L 2 L2 LCB 5L From the triangle CAB φA 180deg θ φA 13500 deg sin φB L sin φA LCB φB asin L sin φA 5L φB 18435 deg From the triangle CAB φB φB θ φB 20435 deg sin φB L sin 180deg φA LCB φA 180deg asin 5 L sin φB L φA 128674 deg φC 180deg φB φA φC 30891 deg sin φB L sin φC LAB LAB sin φC sin φB L LAB 2L εAB LAB LAB LAB εAB 003977 Ans Problem 211 If a load applied to bar AC causes point A to be displaced to the left by an amount L determine the normal strain in wire AB Originally θ 45 Given θ 45deg Solution εAC L L LAB L2 L2 LAB 2L LCB 2L 2 L2 LCB 5L From the triangle AAB φA 180deg θ φA 13500 deg LAB L2 LAB 2 2 L LAB cos φA LAB L2 2 L2 2 L L εAB LAB LAB LAB εAB L2 2 L2 2 L L 2L 2L εAB 1 2 L L 2 1 L L 1 Neglecting the higherorder terms εAB 1 L L 05 1 εAB 1 1 2 L L 1 Binomial expansion εAB 1 2 L L Ans Alternatively εAB LAB LAB LAB εAB L sin θ 2L εAB 1 2 L L Ans Problem 212 The piece of plastic is originally rectangular Determine the shear strain γxy at corners A and B if the plastic distorts as shown by the dashed lines Given a 400mm b 300mm Ax 3mm Ay 2mm Cx 2mm Cy 2mm Bx 5mm By 4mm Solution Geometry For small angles α Cx b Cy α 000662252 rad β By Cy a Bx Cx β 000496278 rad ψ Bx Ax b By Ay ψ 000662252 rad θ Ay a Ax θ 000496278 rad Shear Strain γxyB β ψ γxyB 11585 10 3 rad Ans γxyA θ ψ γxyA 11585 10 3 rad Ans Problem 213 The piece of plastic is originally rectangular Determine the shear strain γxy at corners D and C if the plastic distorts as shown by the dashed lines Given a 400mm b 300mm Ax 3mm Ay 2mm Cx 2mm Cy 2mm Bx 5mm By 4mm Solution Geometry For small angles α Cx b Cy α 000662252 rad β By Cy a Bx Cx β 000496278 rad ψ Bx Ax b By Ay ψ 000662252 rad θ Ay a Ax θ 000496278 rad Shear Strain γxyD α θ γxyD 11585 10 3 rad Ans γxyC α β γxyC 11585 10 3 rad Ans Problem 214 The piece of plastic is originally rectangular Determine the average normal strain that occurs along th diagonals AC and DB Given a 400mm b 300mm Ax 3mm Ay 2mm Cx 2mm Cy 2mm Bx 5mm By 4mm Solution Geometry LAC a2 b2 LAC 500mm LDB a2 b2 LDB 500mm LAC a Ax Cx 2 b Cy Ay 2 LAC 5008 mm LDB a Bx 2 b By 2 LDB 5064 mm Average Normal Strain εAC LAC LAC LAC εAC 1601 10 3 mm mm Ans εBD LDB LDB LDB εBD 12800 10 3 mm mm Ans Problem 215 The guy wire AB of a building frame is originally unstretched Due to an earthquake the two columns of the frame tilt θ 2 Determine the approximate normal strain in the wire when the frame is in thi position Assume the columns are rigid and rotate about their lower supports Given a 4m b 3m c 1m θ 2deg Solution θ θ 180 π θ 003490659 rad Geometry The vertical dosplacement is negligible Ax c θ Ax 34907 mm Bx b c θ Bx 139626 mm LAB a2 b2 LAB 5000mm LAB a Bx Ax 2 b2 LAB 508416 mm Average Normal Strain εAB LAB LAB LAB εAB 16833 10 3 mm mm Ans Problem 216 The corners of the square plate are given the displacements indicated Determine the shear strain along the edges of the plate at A and B Given ax 250mm ay 250mm v 5mm h 75mm Solution At A tan θA 2 ax h ay v θA 2 atan ax h ay v θA 152056 rad γntA π 2 θA γntA 005024 rad Ans At B tan φB 2 ay v ax h φB 2 atan ay v ax h φB 162104 rad γntB φB π 2 γntB 005024 rad Ans Problem 217 The corners of the square plate are given the displacements indicated Determine the average normal strains along side AB and diagonals AC and DB Given ax 250mm ay 250mm v 5mm h 75mm Solution For AB LAB ax 2 ay 2 LAB ax h 2 ay v 2 εAB LAB LAB LAB LAB 35355339 mm LAB 35189665 mm εAB 4686 10 3 mm mm Ans For AC LAC 2 ay LAC 500mm LAC 2 ay v LAC 510mm εAC LAC LAC LAC εAC 20000 10 3 mm mm Ans For DB LDB 2 ax LDB 500mm LDB 2 ax h LDB 485mm εDB LDB LDB LDB εDB 30000 10 3 mm mm Ans Problem 218 The square deforms into the position shown by the dashed lines Determine the average normal strain along each diagonal AB and CD Side DB remains horizontal Given a 50mm b 50mm Bx 3 mm Cx 8mm Cy 0mm θA 915deg LAD 53mm Solution For AB By LAD cos θA 90deg b By 29818 mm LAB a2 b2 LAB 707107 mm LAB a Bx 2 b By 2 LAB 708243 mm εAB LAB LAB LAB εAB 1606 10 3 mm mm Ans For CD Dy By Dy 29818 mm LCD a2 b2 LCD 707107 mm LCD a Cx 2 b Dy 2 2 a Cx b Dy cos θA LCD 795736 mm εCD LCD LCD LCD εCD 125340 10 3 mm mm Ans Problem 219 The square deforms into the position shown by the dashed lines Determine the shear strain at each of its corners A B C and D Side DB remains horizontal Given a 50mm b 50mm Bx 3 mm Cx 8mm Cy 0mm θA 915deg LAD 53mm Solution θA 1597 rad Geometry By LAD cos θA 90deg b By 29818 mm Dy By Dy 29818 mm Dx LAD sin θA 90deg Dx 13874 mm In triangle CBD LCB Cx Bx 2 b By 2 LCB 541117 mm LDB a Bx Dx LDB 483874 mm LCD a Cx Dx 2 b Dy 2 LCD 79586 mm cos θB LCB 2 LDB 2 LCD 2 2 LCB LDB θB acos LCB 2 LDB 2 LCD 2 2 LCB LDB θB 101729 deg θB 17755 rad θD 180deg θA θD 88500 deg θD 15446 rad θC 180deg θB θC 78271 deg θC 13661 rad Shear Strain γxyA 05π θA γxyA 26180 10 3 rad Ans γxyB 05π θB γxyB 204710 10 3 rad Ans γxyC 05π θC γxyC 204710 10 3 rad Ans γxyD 05π θD γxyD 26180 10 3 rad Ans Problem 220 The block is deformed into the position shown by the dashed lines Determine the average normal strain along line AB Given xBA 70 30 mm yBA 100mm xBA 55 30 mm yBA 1102 152 mm Solution For AB LAB xBA 2 yBA 2 LAB 1077033 mm LAB xBA 2 yBA 2 LAB 1118034 mm εAB LAB LAB LAB εAB 38068 10 3 mm mm Ans Problem 221 A thin wire lying along the x axis is strained such that each point on the wire is displaced x k x2 along the x axis If k is constant what is the normal strain at any point P along the wire Given x k x2 Solution ε x d x d ε 2 k x Ans Problem 222 The rectangular plate is subjected to the deformation shown by the dashed line Determine the averag shear strain γxy of the plate Given a 150mm b 200mm a 0mm b 3 mm Solution θ atan b a θ 1146 deg θ 199973 10 3 rad Shear Strain γxy θ γxy 19997 10 3 rad Ans Problem 223 The rectangular plate is subjected to the deformation shown by the dashed lines Determine the averag shear strain γxy of the plate Given a 200mm a 3mm b 150mm b 0mm Solution θ atan a b θ 1146 deg θ 199973 10 3 rad Shear Strain γxy θ γxy 19997 10 3 rad Ans Problem 224 The rectangular plate is subjected to the deformation shown by the dashed lines Determine the average normal strains along the diagonal AC and side AB Given a 200mm b 150mm Ax 3 mm Ay 0mm Bx 0mm By 0mm Cx 0mm Cy 0mm Dx 3 mm Dy 0mm Solution Geometry LAC a2 b2 LAC 250mm LAB b LAB 150mm LAC a Cx Ax 2 b Cy Ay 2 LAC 25241 mm LAB Ax 2 b2 LAB 15003 mm Average Normal Strain εAC LAC LAC LAC εAC 9626 10 3 mm mm Ans εAB LAB LAB LAB εAB 199980 10 6 mm mm Ans Problem 225 The piece of rubber is originally rectangular Determine the average shear strain γxy if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines Given a 300mm b 400mm Ax 0mm Ay 0mm Bx 0mm By 2mm Dx 3mm Dy 0mm Solution θAB atan By a θAB 038197 deg θAB 66666 10 3 rad θAD atan Dx b θAD 042971 deg θAD 74999 10 3 rad Shear Strain γxyA θAB θAD γxyA 14166 10 3 rad Ans Problem 226 The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines Determine the average normal strain along the diagonal DB and side AD Given a 300mm b 400mm Ax 0mm Ay 0mm Bx 0mm By 2mm Dx 3mm Dy 0mm Solution Geometry LDB a2 b2 LDB 500mm LAD b LAD 400mm LDB a Bx Dx 2 b Dy By 2 LDB 4966 mm LAD Dx 2 b Dy 2 LAD 40001 mm Average Normal Strain εBD LDB LDB LDB εBD 6797 10 3 mm mm Ans εAD LAD LAD LAD εAD 28125 10 6 mm mm Ans Problem 227 The material distorts into the dashed position shown Determine a the average normal strains εx and εy the shear strain γxy at A and b the average normal strain along line BE Given a 80mm Bx 0mm Ex 80mm b 125mm By 100mm Ey 50mm Ax 0mm Cx 10mm Dx 15mm Ay 0mm Cy 0mm Dy 0mm Solution xAC Cx Ax xAC 1000 mm yAC Cy Ay yAC 000 mm θAC atan Cx b θAC 798300 10 3 rad Since there is no deformation occuring along the y and xaxis εxA yAC εxA 0 Ans εyA xAC 2 b2 b b εyA 000319 Ans γxyA θAC γxyA 79830 10 3 rad Ans Geometry Bx Cx By b Bx By b Cx Bx 8mm By 0mm Ex Dx Ey b Ex Ey b Dx Ex 6mm Ey 0mm LBE Ex Bx 2 Ey By 2 LBE 9434 mm LBE a Ex Bx 2 Ey Ey By 2 LBE 9265 mm εBE LBE LBE LBE εBE 17913 10 3 mm mm Ans Note Negative sign indicates shortening of BE Problem 228 The material distorts into the dashed position shown Determine the average normal strain that occurs along the diagonals AD and CF Given a 80mm Bx 0mm Ex 80mm b 125mm By 100mm Ey 50mm Ax 0mm Cx 10mm Dx 15mm Ay 0mm Cy 0mm Dy 0mm Solution xAC Cx Ax xAC 1000 mm yAC Cy Ay yAC 000 mm θAC atan Cx b θAC 798300 10 3 rad Geometry LAD a2 b2 LAD 14841 mm LCF a2 b2 LCF 14841 mm LAD a Dx Ax 2 b Dy Ay 2 LAD 15700 mm LCF a Cx 2 b Cy 2 LCF 14327 mm Average Normal Strain εAD LAD LAD LAD εAD 57914 10 3 mm mm Ans εCF LCF LCF LCF εCF 34653 10 3 mm mm Ans Problem 229 The block is deformed into the position shown by the dashed lines Determine the shear strain at corners C and D Given a 100mm Ax 15 mm b 100mm Bx 15 mm LCA 110mm Solution Geometry θC asin Ax LCA θC 784 deg θC 01368 rad θD asin Bx LCA θD 784 deg θD 01368 rad Shear Strain γxyC θC γxyC 136790 10 3 rad Ans γxyD θD γxyD 136790 10 3 rad Ans Problem 230 The bar is originally 30 mm long when it is flat If it is subjected to a shear strain defined by γxy 00 x where x is in millimeters determine the displacement y at the end of its bottom edge It is distorte into the shape shown where no elongation of the bar occurs in the x direction Given L 300mm γxy 002 x unit 1mm Solution dy dx tan γxy dy dx tan 002 x 0 y 1 y d 0 L x tan 002 x unit d y 0 30 x tan 002x unit d y 960 mm Ans Problem 231 The curved pipe has an original radius of 06 m If it is heated nonuniformly so that the normal strain along its length is ε 005 cos θ determine the increase in length of the pipe Given r 06m ε 005 cos θ Solution L L ε d L 0 90deg rθ 005 cos θ d L 0 90deg θ 005 r cos θ d L 3000 mm Ans Problem 232 Solve Prob 231 if ε 008 sin θ Given r 06m ε 008 sin θ Solution L L ε d L 0 90deg rθ 008 sin θ d L 0 90deg θ 008 r sin θ d L 00480 m Ans Problem 233 A thin wire is wrapped along a surface having the form y 002 x2 where x and y are in mm Origina the end B is at x 250 mm If the wire undergoes a normal strain along its length of ε 00002x determine the change in length of the wire Hint For the curve y f x ds 1 dy dx 2 dx Given Bx 250mm ε 00002 x unit 1mm Solution y 002x2 dy dx 004x ds 1 dy dx 2 dx ds 1 004x 2 dx L s ε d L 0 Bx x 00002x 1 004x 2 d L unit 0 250 x 00002x 1 004x 2 d L 42252 mm Ans Problem 234 The fiber AB has a length L and orientation If its ends A and B undergo very small displacements uA and vB respectively determine the normal strain in the fiber when it is in position AB Solution Geometry LAB L cos θ uA 2 L sin θ vB 2 LAB L2 uA 2 vB 2 2L vB sin θ uA cos θ Average Normal Strain εAB LAB L L εAB 1 uA 2 vB 2 L2 2 vB sin θ uA cos θ L 1 Neglecting higherorder terms uA 2 and vB 2 εAB 1 2 vB sin θ uA cos θ L 1 Using the binomial theorem εAB 1 1 2 2 vB sin θ uA cos θ L 1 εAB vB sin θ L uA cos θ L Ans Problem 235 If the normal strain is defined in reference to the final length that is ε n p p s s s lim instead of in reference to the original length Eq22 show that the difference in these strains is represented as a secondorder term namely εn εn εn εn Solution εn S S S εn εn S S S S S S εn εn S2 2 S S S2 S S εn εn S S 2 S S εn εn S S S S S S εn εn εn εn QED 0 1 10 4 2 10 4 3 10 4 4 10 4 5 10 4 6 10 4 7 10 4 2 4 6 8 10 12 14 16 σ Y x ε x δ 00000 00150 00300 00500 00650 00850 01000 01125 01250 01550 01750 01875 Problem 31 A concrete cylinder having a diameter of 150 mm and gauge length of 300 mm is tested in compression The results of the test are reported in the table as load versus contraction Draw the stressstrain diagram using scales of 10 mm 2 MPa and 10 mm 01103 mmmm From the diagram determine approximately the modulus of elasticity Given d 150 L 300 Solution A π 4 d2 σ 103P A ε δ L σPA MPa εδL mmmm x min ε 000005 max ε Regression curve Coeff loess ε σ 15 σ 000 141 269 467 580 722 849 976 1089 1316 1415 1500 ε 0000000 0000050 0000100 0000167 0000217 0000283 0000333 0000375 0000417 0000517 0000583 0000625 Y x interp Coeff ε σ x Modulus of Elasticity From the stressstrain diagram ε 00003 0 mm mm σ 80 0 MPa Eapprox σ ε Eapprox 2667 GPa Ans Contraction mm P 00 250 475 825 1025 1275 1500 1725 1925 2325 2500 2650 Load kN Problem 32 Data taken from a stressstrain test for a ceramic are given in the table The curve is linear between the origin and the first point Plot the diagram and determine the modulus of elasticity and the modulus of resilience Unit used MJ 106 J σ 00 2324 3185 3458 3605 3738 ε 00000 00006 00010 00014 00018 00022 Solution Regression curve x min ε 000005 max ε Coeff loess ε σ 09 Y x interp Coeff ε σ x 0 5 10 4 0001 00015 0002 00025 40 80 120 160 200 240 280 320 360 400 σ Y x ε x Modulus of Elasticity From the stressstrain diagram ε 00006 0 mm mm σ 2324 0 MPa Eapprox σ ε Eapprox 3873 GPa Ans Modulus of Resilience The modulus of resilience is equal to the area under the initial linear portion of the curve ε 00006 0 mm mm σ 2324 0 103 kN m2 ur 1 2 ε σ ur 00697 MJ m3 Ans σPA MPa εδL mmmm Problem 33 Data taken from a stressstrain test for a ceramic are given in the table The curve is linear between the origin and the first point Plot the diagram and determine approximately the modulus of toughness The rupture stress is σr 3738 MPa Unit used MJ 106 J σ 00 2324 3185 3458 3605 3738 ε 00000 00006 00010 00014 00018 00022 Solution Regression curve x min ε 000005 max ε Coeff loess ε σ 09 Y x interp Coeff ε σ x 0 5 10 4 0001 00015 0002 00025 40 80 120 160 200 240 280 320 360 400 σ Y x ε x Modulus of Resilience The modulus of resilience is equal to the area under the curve A1 1 2 2324 00004 00010 A2 3185 00022 00010 A3 1 2 3738 3185 00022 00010 A4 1 2 3185 2324 00010 00006 Atotal A1 A2 A3 A4 ut Atotal 106 J m3 ut 0595 MJ m3 Ans σPA MPA εδL mmmm Problem 34 A tension test was performed on a steel specimen having an original diameter of 13 mm and gauge length of 50 mm The data is listed in the table Plot the stressstrain diagram and determine approximately the modulus of elasticity the yield stress the ultimate stress and the rupture stress Use a scale of 10 mm 209 MPa and 10 mm 005 mmmm Redraw the elastic region using the same stress scale but a strain scale of 10 mm 0001 mmmm Given d 125 L 50 P 00 75 230 400 550 590 590 600 830 1000 1075 975 925 δ 00000 00125 00375 00625 00875 01250 02000 05000 10000 25000 70000 100000 115000 Solution A π 4 d2 σ 103 P A ε δ L σPA MPa εδL mmmm σ 000 6112 18742 32595 44818 48078 48078 48892 67634 81487 87599 79450 75376 ε 000000 000025 000075 000125 000175 000250 000400 001000 002000 005000 014000 020000 023000 Curve Fit Use F x x x03 x06 Fit linfit ε σ F Fit 4189696 2285571 596538 x min ε 000005 max ε Y x F x Fit 0 0001 0002 0003 0004 0005 200 400 600 800 1000 σ ε 0 005 01 015 02 025 200 400 600 800 1000 σ Y x ε x Modulus of Elasticity From the stressstrain diagram Load kN Elongation mm From the stressstrain diagram σY 448MPa Ans ε 000125 0 mm mm σult 890MPa Ans σ 326 0 MPa σR 7538MPa Ans Eapprox σ ε Eapprox 2608 GPa Ans Problem 35 The stressstrain diagram for a steel alloy having an original diameter of 12mm and a gauge length of 50 mm is given in the figure Determine approximately the modulus of elasticity for the material the load on the specimen that causes yielding and the ultimate load the specimen will support Given d 12mm L 50mm Solution A π 4 d2 Modulus of Elasticity From the stressstrain diagram ε 0001 0 mm mm σ 290 0 MPa E σ ε E 290GPa Ans From the stressstrain diagram σY 290MPa Yield Load PY σY A PY 3280 kN Ans From th stressstrain diagram σu 550MPa Ultimate Load Pu σu A Pu 6220 kN Ans Problem 36 The stressstrain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure If the specimen is loaded until it is stressed to 500MPa determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded Given d 12mm L 50mm σmax 500MPa Solution A π 4 d2 Modulus of Elasticity From the stressstrain diagram ε 0001 0 mm mm σ 290 0 MPa E σ ε E 290GPa Ans Elastic Recovery rRe σmax E rRe 000172 mm mm AmountRe rRe L AmountRe 008621 mm Ans From the stressstrain diagram εmax 008 mm mm Permanent set Permanent elongation rps εmax rRe rps 007828 mm mm Amountps rps L Amountps 391379 mm Ans Problem 37 The stressstrain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure Determine approximately the modulus of resilience and the modulus of toughness for the material Unit used MJ 106 J Given d 12mm L 50mm Solution Modulus of Resilience The modulus of resilience is equal to the area under the initial linear portion of the curve ε 0001 0 mm mm σ 290 0 MPa ur 1 2 ε σ ur 0145 MPa Ans Modulus of Toughness The modulus of toughness is equal to the area under the curve and could be approximated by counting the number of sqaures the total number of squares is n 33 εsq 004 mm mm σsq 100MPa ut n εsq σsq ut 132MPa Ans Problem 38 The stressstrain diagram for a steel bar is shown in the figure Determine approximately the modulus of elasticity the proportional limit the ultimate stress and the modulus of resilience If the bar is loaded until it is stressed to 450 MPa determine the amount of elastic strain recovery and the permanent set or strain in the bar when it is unloaded Unit used kJ 103J Given σmax 450MPa Solution Modulus of Elasticity From th stressstrain diagram ε 00015 0 mm mm σ 325 0 MPa E σ ε E 21667 GPa Ans Modulus of Resilience The modulus of resilience is equal to the area under the initial linear portion of the curve ε 00015 0 mm mm σ 325 0 MPa ur 1 2 ε σ ur 24375 kJ m3 Ans Elastic Recovery rRe σmax E rRe 000208 mm mm Ans From th stressstrain diagram εmax 00750 mm mm Permanent set Permanent elongation rps εmax rRe rps 007292 mm mm Ans Problem 39 The σε diagram for elastic fibers that make up human skin and muscle is shown Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience Modulus of Elasticity From the stressstrain diagram ε 200 0 mm mm σ 77 0 MPa E σ ε E 385 MPa Ans Modulus of Resilience The modulus of resilience is equal to the area under the initial linear portion of the curve ε 200 0 mm mm σ 77 0 MPa ur 1 2 ε σ ur 7700 MPa Ans Modulus of Toughness The modulus of toughness is equal to the area under the curve ε0 0 ε1 200 ε2 225 σ0 0 σ1 77MPa σ2 385MPa A1 ur A2 05 σ1 σ2 ε2 ε1 ut A1 A2 ut 13475 MPa Ans Problem 310 An A36 steel bar has a length of 1250 mm and crosssectional area of 430 mm2 Determine the length of the bar if it is subjected to an axial tension of 25 kN The material has linearelastic behavior Given A 430mm2 L0 1250mm P 25kN σY 250MPa Est 200GPa Solution Normal Stress σ P A σ 58140 MPa less than yield stress σy Hence Hooks law is still valid Normal Strain ε σ Est ε 2906977 10 6 mm mm Thus δL ε L0 δL 036337 mm L L0 δL L 1250363 mm Ans Problem 311 The stressstrain diagram for polyethylene which is used to sheath coaxial cables is determined from testing a specimen that has a gauge length of 250 mm If a load P on the specimen develops a strain of ε 0024 mmmm determine the approximate length of the specimen measured between the gauge points when the load is removed Assume the specimen recovers elastically Given L0 250mm Solution Modulus of Elasticity From th stressstrain diagram ε 0004 0 mm mm σ 140 0 MPa E σ ε E 350000 MPa Elastic Recovery From the stressstrain diagram εmax 0024 mm mm σmax 26MPa rRe σmax E rRe 000743 mm mm Permanent set rps εmax rRe rps 001657 mm mm Permanent elongation L rps L0 L 414286 mm L L0 L L 254143 mm Ans Problem 312 Fiberglass has a stressstrain diagram as shown If a 50mmdiameter bar of length 2 m made from this material is subjected to an axial tensile load of 60 kN determine its elongation Given L0 2m d 50mm P 60kN unit 1Pa Solution A π 4 d2 σ P A σ 30558 106 Pa σ σ unit Given σ 300 106 ε05 ε σ2 9 1016 ε 0010375 mm mm L ε L0 L 2075 mm Ans Problem 313 The change in weight of an airplane is determined from reading the strain gauge A mounted in the planes aluminum wheel strut Before the plane is loaded the strain gauge reading in a strut is ε1 000100 mmmm whereas after loading ε2 000243 mmmm Determine the change in the force on the strut if the crosssectional area of the strut is 2200 mm2 Eal 70 GPa Given A 2200mm2 Eal 70 GPa ε1 000100 mm mm ε2 000243 mm mm Solution Stressstrain Relationship Applying Hooks law σ Eε σ1 Eal ε1 σ1 7000 MPa σ2 Eal ε2 σ2 17010 MPa Normal Force Applying equation σ P A P1 A σ1 P1 15400 kN P2 A σ2 P2 37422 kN Thus P P2 P1 P 22022 kN Ans Problem 314 A specimen is originally 300 mm long has a diameter of 12 mm and is subjected to a force of 25 kN When the force is increased to 9 kN the specimen elongates 225 mm Determine the modulus of elasticity for the material if it remains elastic Given d 12mm L0 300mm P1 25kN P2 9kN L 225mm Solution A π 4 d2 Normal Force Applying equation σ P A σ1 P1 A σ1 22105 MPa σ2 P2 A σ2 79577 MPa Thus σ σ2 σ1 σ 57473 MPa ε L L0 E σ ε E 7663 MPa Ans Problem 315 A structural member in a nuclear reactor is made from a zirconium alloy If an axial load of 20 kN is to be supported by the member determine its required crosssectional area Use a factor of safety of 3 with respect to yielding What is the load on the member if it is 1m long and its elongation is 05 mm Ezr 100 GPa σY 400 MPa The material has elastic behavior Given P 20kN L0 1m L 05mm Ezr 100 GPa σY 400 MPa FoS 3 Solution σallow σY FoS σallow 13333 MPa Areq P σallow Areq 150mm2 Ans A Areq ε L L0 ε 00005 mm mm σ Ezr ε σ 50MPa P σ A P 75 kN Ans Problem 316 The pole is supported by a pin at C and an A36 steel guy wire AB If the wire has a diameter of 5 mm determine how much it stretches when a horizontal force of 15 kN acts on the pole Given P 15kN Est 200 GPa a 12m b 1m d 5mm θ 30deg Solution A π 4 d2 L a b Support Reactions ΣΜB0 Cx L P b 0 Cx P b L Cx 682 kN ΣFx0 Cx P Bx 0 Bx Cx P FAB Bx sin θ FAB 1636 kN Bx 818 kN LAB L cos θ LAB 254 m σAB FAB A σAB 833393 MPa εAB σAB Est εAB 00041670 mm mm δAB εAB LAB δAB 10586 mm Ans Problem 317 By adding plasticizers to polyvinyl chloride it is possible to reduce its stiffness The stressstrain diagrams for three types of this material showing this effect are given below Specify the type that should be used in the manufacture of a rod having a length of 125 mm and a diameter of 50 mm that is required to support at least an axial load of 100 kN and also be able to stretch at most 6 mm Given d 50mm L0 125mm P 100kN δL 6mm Solution A π 4 d2 Normal Stress σ P A σ 50930 MPa Normal Strain ε δL L0 ε 0048000 mm mm From the stressstrain diagramthe copolymer will satisfy both stress and strain requirements Ans Problem 318 The steel wires AB and AC support the 200kg mass If the allowable axial stress for the wires is σallow 130 MPa determine the required diameter of each wire Also what is the new length of wire AB after the load is applied Take the unstretched length of AB to be 750 mm Est 200 GPa Given g 981 m s2 m 200kg θ 60deg v 4 5 h 3 5 L0 750mm Est 200GPa σallow 130MPa Solution W m g Axial force in steel wires AB and AC Initial guess FAC 1N FAB 2N Given ΣFx0 FAB cos θ FAC h 0 1 ΣFy0 FAB sin θ FAC v W 0 2 Solving 1 and 2 FAC FAB Find FAC FAB FAC FAB 106675 128010 N Wire AB FAB σallow π 4 dAB 2 dAB 4 π FAB σallow dAB 354 mm Ans Wire AC FAC σallow π 4 dAC 2 dAC 4 π FAC σallow dAC 323 mm Ans Stressstrain Relationship Applying Hooks law σ Eε εAB σallow Est εAB 0000650 mm mm Thus LAB L0 1 εAB LAB 750487 mm Ans Problem 319 The two bars are made of polystyrene which has the stressstrain diagram shown If the crosssectional area of bar AB is 950 mm2 and BC is 2500 mm2 determine the largest force P that can be supported before any member ruptures Assume that buckling does not occur Ans Given a 12m b 09m AAB 950mm2 ABC 2500mm2 Solution c a2 b2 h a c v b c ΣFy0 FAB v P 0 1 ΣFx0 FAB h FBC 0 2 Solving Eqs1 and 2 FAB 5 3P 1a FBC 4 3P 2a Assume tension failure of BC From the stressstrain diagram σRt 35MPa FBC ABC σRt FBC 8750 kN From Eq2a P 075 FBC P 6563 kN Pcase1 P Assume compression failure of AB From the stressstrain diagram σRc 175MPa FAB AAB σRc FAB 16625 kN From Eq1a P 060 FAB P 9975 kN Pcase2 P Chosoe the smallest value P min Pcase1 Pcase2 P 6563 kN Problem 320 The two bars are made of polystyrene which has the stressstrain diagram shown Determine the crosssectional area of each bar so that the bars rupture simultaneously when the load P 15 kN Assume that buckling does not occur Ans Given a 12m b 09m P 15kN Solution c a2 b2 h a c v b c ΣFy0 FAB v P 0 1 ΣFx0 FAB h FBC 0 2 Solving Eqs1 and 2 FAB 5 3P FAB 2500 kN FBC 4 3P FBC 2000 kN For member BC From the stressstrain diagram σRt 35MPa ABC FBC σRt ABC 57143 mm2 Ans For member AB From the stressstrain diagram σRc 175MPa AAB FAB σRc AAB 14286 mm2 Problem 321 The stressstrain diagram for a polyester resin is given in the figure If the rigid beam is supported by a strut AB and post CD both made from this material and subjected to a load of P 80 kN determine the angle of tilt of the beam when the load is applied The diameter of the strut is 40 mm and the diameter of the post is 80 mm Given P 80kN LAB 2m LCD 05m LAC 15m dAB 40mm dCD 80mm Solution Ay P 2 Cy P 2 Support Reactions FAB Ay FCD Cy AreaAB π 4 dAB 2 AreaCD π 4 dCD 2 From the stressstrain diagram E 322MPa 001 E 322000 MPa σAB FAB AreaAB σAB 31831 MPa εAB σAB E εAB 00098854 mm mm δAB εAB LAB δAB 19771 mm σCD FCD AreaCD σCD 7958 MPa εCD σCD E εCD 00024714 mm mm δCD εCD LCD δCD 1236 mm Angle of tilt α tan α δAB δCD LAC α atan δAB δCD LAC α 0708 deg Ans Problem 322 The stressstrain diagram for a polyester resin is given in the figure If the rigid beam is supported by a strut AB and post CD made from this material determine the largest load P that can be applied to the beam before it ruptures The diameter of the strut is 12 mm and the diameter of the post is 40 mm Given LAB 2m LCD 05m LAC 15m dAB 12mm dCD 40mm Solution Support Reactions Ay P 2 Cy P 2 FAB Ay FCD Cy AreaAB π 4 dAB 2 AreaCD π 4 dCD 2 For rupture of strut AB From the stressstrain diagram σRt 500MPa FAB AreaAB σRt P 2AreaAB σRt P 1131 kN Ans Controls For rupture of post CD From the stressstrain diagram σRc 950MPa FCD AreaCD σRc P 2AreaCD σRc P 23876 kN Problem 323 The beam is supported by a pin at C and an A36 steel guy wire AB If the wire has a diameter of 5 mm determine how much it stretches when a distributed load of w 15 kNm acts on the pipe The material remains elastic Given L 3m w 15 kN m dAB 5mm θ 30deg Est 200 GPa Solution Support Reactions ΣΜC0 FAB sin θ L w L 05 L 0 FAB w L 2sin θ AreaAB π 4 dAB 2 LAB L cos θ σAB FAB AreaAB σAB 229183 MPa εAB σAB Est εAB 00011459 mm mm δAB εAB LAB δAB 3970 mm Ans Problem 324 The beam is supported by a pin at C and an A36 steel guy wire AB If the wire has a diameter of 5 mm determine the distributed load w if the end B is displaced 18 mm downward Given L 3m δBy 18mm dAB 5mm θB 30deg Est 200 GPa Solution Consider triangle BBC L sin θC δBy θC asin δBy L Consider triangle ABC θC 90deg θC LAC L tan θB LAB LAC 2 L2 2 LAC L cos θC LAB 34731 m LAB L cos θB AreaAB π 4 dAB 2 δAB LAB LAB δAB 89883 mm εAB δAB LAB εAB 00025947 mm mm σAB εAB Est σAB 518942 MPa FAB σAB AreaAB FAB 1019 kN Support Reactions ΣΜC0 FAB sin θB L w L 05 L 0 w 2sin θB L FAB w 340 kN m Ans Problem 325 Direct tension indicators are sometimes used instead of torque wrenches to insure that a bolt has a prescribed tension when used for connections If a nut on the bolt is tightened so that the six heads of the indicator that were originally 3 mm high are crushed 03 mm leaving a contact area on each head of 15 mm2 determine the tension in the bolt shank The material has the stressstrain diagram shown Given h 3mm δh 03mm Area 15mm2 number 6 unit 1MPa Solution Stressstrain Relationship ε δh h ε 01000 mm mm From the stressstrain diagram σ 450 600 450 ε 00015 03 00015 σ 450 600 450 ε 00015 03 00015 unit σ 49950 MPa Axial Force For each head P σ Area P 0749 kN Thus the tension in the bolt is T number P T 450 kN Ans Problem 326 The acrylic plastic rod is 200 mm long and 15 mm in diameter If an axial load of 300 N is applied to it determine the change in its length and the change in its diameter Ep 270 GPa νp 04 Given P 300N L 200mm d 15mm EP 270 GPa ν 04 Solution A π 4 d2 σ P A σ 1698 MPa εlong σ EP εlong 00006288 mm mm δ εlong L δ 0126 mm Ans εlat ν εlong εlat 00002515 mm mm d εlat d d 0003773 mm Ans Problem 327 The block is made of titanium Ti6A14V and is subjected to a compression of 15 mm along the y axis and its shape is given a tilt of θ 897Determine εy εx and γxy Given Lx 125mm Ly 100mm δy 15 mm θ 897deg ν 036 Solution Normal Strain εy δy Ly εy 001500 mm mm Ans Poissons Ratio The lateral and longitudinal strain can be related using Poissons ratio εx ν εy εx 000540 mm mm Ans Shear Strain β 180deg θ β 9030 deg β 158 rad Thus γxy π 2 β γxy 000524 rad Ans Problem 328 A short cylindrical block of bronze C86100 having an original diameter of 38 mm and a length of 75 mm is placed in a compression machine and squeezed until its length becomes 745 mm Deterinme the new diameter of the block Given L 75mm L 745mm d 38mm ν 034 Solution A π 4 d2 εlong L L L εlong 00066667 mm mm εlat ν εlong εlat 00022667 mm mm d εlat d d 0086133 mm d d d d 380861 mm Ans Problem 329 The elastic portion of the stressstrain diagram for a steel alloy is shown in the figure The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm When the applied load on the specimen is 50 kN the diameter is 1299265 mm Determine Poissons ratio for the material Given d 13mm d 1299265mm L 50mm P 50kN Solution Normal Stress A π 4 d2 σ P A σ 376698 MPa Normal Strain From the stressstrain diagram the modulus of elasticity is E 400MPa 0002 E 200GPa Applying Hooks law σ Eε εlong σ E εlong 00018835 mm mm εlat d d d εlat 00005654 mm mm Poissons Ratio The lateral and longitudinal strain can be related using Poissons ratio ν εlat εlong ν 030018 Ans Problem 330 The elastic portion of the stressstrain diagram for a steel alloy is shown in the figure The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm If a load of P 20 kN is applied to the specimen determine its diameter and gauge length Take ν 04 Given L 50mm d 13mm P 20kN ν 04 Solution Normal Stress A π 4 d2 σ P A σ 150679 MPa Normal Strain From the stressstrain diagram the modulus of elasticity is E 400MPa 0002 E 200GPa Applying Hooks law σ Eε εlong σ E εlong 00007534 mm mm Thus δL εlong L δL 0037670 mm L L δL L 500377 mm Ans Poissons Ratio The lateral and longitudinal strain can be related using Poissons ratio εlat ν εlong εlat 000030136 mm mm d εlat d d 0003918 mm d d d d 1299608 mm Ans Problem 331 The shear stressstrain diagram for a steel alloy is shown in the figure If a bolt having a diameter of 6 mm is made of this material and used in the lap joint determine the modulus of elasticity E and the force P required to cause the material to yield Take ν 03 Given d 6mm ν 03 Solution Modulus of Rigidity From the stressstrain diagram G 350MPa 0004 G 87500MPa Modulus of Elasticity G E 2 1 ν E 2G 1 ν E 227500MPa Ans Yielding Stress From the stressstrain diagram τY 350MPa A π 4 d2 τY P A P τY A P 9896 kN Ans Problem 332 The brake pads for a bicycle tire are made of rubber If a frictional force of 50 N is applied to each side of the tires determine the average shear strain in the rubber Each pad has crosssectional dimensions of 20 mm and 50 mm Gr 020 MPa Given a 20mm b 50mm V 50N G 020MPa Solution A a b A 100000 mm2 Average Shear Stress The shear force is V τ V A τ 0050 MPa Shear Stressstrain Relationship Applying Hookes law for shear τ G γ γ τ G γ 0250 rad Ans Problem 333 The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm Both the plug and the sleeve are 50 mm long Determine the axial pressure p that must be applied to the top of the plug to cause it to contact the sides of the sleeve Also how far must the plug be compressed downward in order to do this The plug is made from a material for which E 5 MPa ν 045 Given d 30mm d 32mm L 50mm E 5MPa ν 045 Solution εlat d d d εlat 00666667 mm mm ν εlat εlong εlong εlat ν εlong 014815 mm mm Applying Hooks law σ Eε p E εlong p 0741 MPa Ans δL εlong L δL 741 mm Ans Problem 334 The rubber block is subjected to an elongation of 075 mm along the x axis and its vertical faces are given a tilt so that θ 893 Deterinme the strains εx εy and γxy Take νr 05 Given Lx 100mm Ly 75mm δx 075mm θ 893deg ν 05 Solution Normal Strain εx δx Lx εx 000750 mm mm Ans Poissons Ratio The lateral and longitudinal strain can be related using Poissons ratio εy ν εx εy 000375 mm mm Ans Shear Strain θ 8930 deg θ 156 rad Thus γxy π 2 θ γxy 001222 rad Ans Problem 335 The elastic portion of the tension stressstrain diagram for an aluinmum alloy is shown in the figure The specimen used for the test has a gauge length of 50 mm and a diameter of 125 mm When the applied load is 45 kN the new diameter of the specimen is 1248375 mm Compute shear modulus Gal for the aluinmum Given d 125mm d 1248375mm L 50mm P 45kN Solution Normal Stress A π 4 d2 σ P A σ 366693 MPa Normal Strain From the stressstrain diagram the modulus of elasticity is E 500MPa 000614 E 8143322 MPa Applying Hooks law σ Eε εlong σ E εlong 00045030 mm mm εlat d d d εlat 00013000 mm mm Poissons Ratio The lateral and longitudinal strain can be related using Poissons ratio ν εlat εlong ν 028870 G E 2 1 ν G 3160 GPa Ans Problem 336 The elastic portion of the tension stressstrain diagram for an aluinmum alloy is shown in the figure The specimen used for the test has a gauge length of 50 mm and a diameter of 125 mm If the applied load is 50 kN deterinme the new diameter of the specimen The shear modulus is Gal 28 GPa Given d 125mm L 50mm P 50kN G 28 GPa Solution Normal Stress A π 4 d2 σ P A σ 40744 MPa Normal Strain From the stressstrain diagram the modulus of elasticity is E 500MPa 000614 E 8143322 MPa Applying Hooks law σ Eε εlong σ E εlong 00050033 mm mm Poissons Ratio G E 2 1 ν Thus ν E 2 G 1 ν 0454 The lateral and longitudinal strain can be related using Poissons ratio εlat ν εlong εlat 000227233 mm mm d εlat d d 0028404 mm d d d d 124716 mm Ans Problem 337 The head H is connected to the cylinder of a compressor using six steel bolts If the clamping force in each bolt is 4 kN deterinme the normal strain in the bolts Each bolt has a diameter of 5 mm If σY 280 MPa and Est 200 GPa what is the strain in each bolt when the nut is unscrewed so that the clamping force is released Given d 5mm σY 28MPa P 4kN E 200 GPa Solution Normal Stress A π 4 d2 σ P A σ 2037183 MPa σY 280 MPa Normal Strain Since σ σY Hooks law is still valid ε σ E ε 00010186 mm mm Ans If the nut is unscrewed the load is zero Therefore the strain ε 0 Ans Problem 338 The rigid pipe is supported by a pin at C and an A36 steel guy wire AB If the wire has a diameter of 5 mm determine how much it stretches when a load of P 15 kN acts on the pipe The material remains elastic Given P 15kN Est 200 GPa LBC 24m d 5mm θ 60deg Solution Support Reactions ΣΜC0 FAB cos θ LAB P LAB 0 FAB P cos θ FAB 3kN Normal Stress Area π 4 d2 σAB FAB Area σAB 152789 MPa Normal Strain LAB LBC sin θ LAB 2771 m Applying Hooks law σ Eε εAB σAB Est εAB 00007639 mm mm Thus δLAB εAB LAB δLAB 21171 mm Ans Problem 339 The rigid pipe is supported by a pin at C and an A36 guy wire AB If the wire has a diameter of 5 mm determine the load P if the end B is displaced 25 mm to the right Est 200 GPa Given L 24m d 5mm θ 60deg δBx 25mm Est 200 GPa Solution Consider triangle BBC L sin θC δBx θC asin δBx L Consider triangle ABC θC 90deg θC LAC L cot θ LAB LAC 2 L2 2 LAC L cos θC LAB 27725 m Normal Strain LAB L sin θ εAB LAB LAB LAB εAB 00004510 mm mm Normal Stress Area π 4 d2 Applying Hooks law σ Eε σAB εAB Est σAB 90191 MPa Thus FAB σAB Area FAB 1771 kN Support Reactions ΣΜC0 FAB cos θ L P L 0 P FAB cos θ P 0885 kN Ans Problem 340 While undergoing a tension test a copperalloy specimen having a gauge length of 50 mm is subjected to a strain of 040 mmmm when the stress is 490 MPa If σY 315 MPa when εY 00025 mmmm determine the distance between the gauge points when the load is released Given L0 50mm ε1 040 mm mm σ1 490 MPa εY 00025 mm mm σY 315MPa Solution Modulus of Elasticity E σY εY E 12600 GPa Elastic Recovery rRe σ1 E rRe 00038889 mm mm Permanent set rps ε1 rRe rps 039611 mm mm Permanent elongation L rps L0 L 19806 mm L L0 L L 69806 mm Ans Problem 341 The 8mmdiameter bolt is made of an aluminum alloy It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm If the original lengths of the bolt and sleeve are 80 mm and 50 mm respectively determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN Assume the material at A is rigid Eal 70 GPa Emg 45 GPa Given Lb 80mm db 8mm Ls 30mm dso 20mm dsi 12mm P 8kN Eal 70GPa Emg 45GPa Solution Ab π 4 db 2 As π 4 dso 2 dsi 2 Normal Stress σb P Ab σb 15915 MPa σs P As σs 3979 MPa Normal Strain εb σb Eal εb 0002274 mm mm Ans εs σs Emg εs 0000884 mm mm Ans Problem 342 A tension test was performed on a steel specimen having an original diameter of 125mm and a gauge length of 50 mm The data is listed in the table Plot the stressstrain diagram and determine approximately the modulus of elasticity the ultimate stress and the rupture stress Use a scale of 20 mm 50 MPa and 20 mm 005 mmmm Redraw the linearelastic region using the same stress scale but a strain scale of 20 mm 0001 mmmm Given d 125 L 50 P 00 111 319 378 409 436 534 623 645 623 588 δ 00000 00175 00600 01020 01650 02490 10160 30480 63500 88900 119380 Solution A π 4 d2 σ P 103 A ε δ L σPA MPa εδL mmmm Use F x x x03 x06 σ 00000 904509 2599446 3080221 3332832 3552848 4351423 5076661 5255933 5076661 4791455 ε 0000000 0000350 0001200 0002040 0003300 0004980 0020320 0060960 0127000 0177800 0238760 Fit 107325 2123113 2190474 x min ε 0005 max ε Y x F x Fit 0 0001 0002 0003 0004 0005 50 100 150 200 250 300 350 400 450 500 550 σ ε 0 005 01 015 02 025 50 100 150 200 250 300 350 400 450 500 550 σ Y x ε x Load kN Elongation mm Fit linfit ε σ F Modulus of Elasticity From th stressstrain diagram From th stressstrain diagram σult 530MPa Ans ε 00005 0 mm mm σR 479MPa Ans σ 125 0 MPa Eapprox σ ε Eapprox 250GPa Ans Problem 343 A tension test was performed on a steel specimen having an original diameter of 125 mm and a gauge length of 50 mm Using the data listed in the table plot the stressstrain diagram and determine approximately the modulus of toughness Use a scale of 20 mm 50 MPa and 20 mm 005 mmmm Given d 125 L 50 P 00 111 319 378 409 436 534 623 645 623 588 δ 00000 00175 00600 01020 01650 02490 10160 30480 63500 88900 119380 Solution A π 4 d2 σ P 103 A ε δ L σPA MPa εδL mmmm Use F x x x03 x06 σ 00000 904509 2599446 3080221 3332832 3552848 4351423 5076661 5255933 5076661 4791455 ε 0000000 0000350 0001200 0002040 0003300 0004980 0020320 0060960 0127000 0177800 0238760 Fit linfit ε σ F Fit 107325 2123113 2190474 x min ε 0005 max ε Y x F x Fit 0 005 01 015 02 025 50 100 150 200 250 300 350 400 450 500 550 σ Y x ε x Modulus of Toughness The modulus of toughness is equal to the area under the curve and could be approximated by counting the number of sqaures the total number of squares is n 1885 εsq 0025 mm mm σsq 25MPa ut n εsq σsq ut 11781 MPa Ans Elongation mm Load kN Problem 344 An 8mmdiameter brass rod has a modulus of elasticity of Ebr 100 GPa If it is 3 m long and subjected to an axial load of 2 kN determine its elongation What is its elongation under the same load if its diameter is 6 mm Given P 2kN L 3m d1 8mm d2 6mm Ebr 100 GPa Solution Case 1 A1 π 4 d1 2 σ1 P A1 σ1 39789 MPa ε1 σ1 Ebr ε1 00003979 mm mm δ1 ε1 L δ1 1194 mm Ans Case 2 A2 π 4 d2 2 σ2 P A2 σ2 70736 MPa ε2 σ2 Ebr ε2 00007074 mm mm δ2 ε2 L δ2 2122 mm Ans Problem 41 The ship is pushed through the water using an A36 steel propeller shaft that is 8 m long measured from the propeller to the thrust bearing D at the engine If it has an outer diameter of 400 mm and a wall thickness of 50 mm determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN The bearings at B and C are journal bearings Given F 5kN L 8m do 400mm t 50mm E 200 GPa Solution di do 2t A π 4 do 2 di 2 Internal Force As shown on FBD P F Displacement δA P L A E δA 000364 mm Ans Note Negative sign indicates that end A moves towards end D Problem 42 The A36 steel column is used to support the symmetric loads from the two floors of a building Determine the vertical displacement of its top A if P1 200 kN P2 310 kN and the column has a crosssectional area of 14625 mm2 Given L 36m A 14625mm2 P1 200kN P2 310kN E 200 GPa Solution Internal Force As shown on FBD Displacement δAB 2 P1 L A E δBC 2 P1 P2 L A E δA δAB δBC δA 174769 mm Ans Note Negative sign indicates that end A moves towards end C Problem 43 The A36 steel column is used to support the symmetric loads from the two floors of a building Determine the loads P1 and P2 if A moves downward 3 mm and B moves downward 225 mm when the loads are applied The column has a crosssectional area of 14625 mm2 Given L 36m A 14625mm2 δA 3 mm δB 225 mm E 200 GPa Solution Internal Force As shown on FBD Displacement Initial guess P1 1kN P2 2kN Given For AB δA δB 2 P1 L A E 1 For BC δB 2 P1 P2 L A E 2 Solving 1 and 2 P1 P2 Find P1 P2 P1 P2 30469 60938 kN Ans Problem 44 The copper shaft is subjected to the axial loads shown Determine the displacement of end A with respect to end D if the diameters of each segment are dAB 20 mm dBC 25 mm and dCD 12 mm Take Ecu 126 GPa Given LAB 2m dAB 20mm LBC 375m dBC 25mm LCD 25m dCD 12mm E 126 GPa PA 40 kN PB 25kN PC 10kN PD 30 kN Solution Internal Force As shown on FBD Displacement AAB π 4 dAB 2 δAB PA LAB E AAB ABC π 4 dBC 2 δBC PA 2PB LBC E ABC ACD π 4 dCD 2 δCD PA 2PB 2PC LCD E ACD δAD δAB δBC δCD δAD 38483 mm Ans Note The positive sign indicates that end A moves away from end D Problem 45 The A36 steel rod is subjected to the loading shown If the crosssectional area of the rod is 60 mm2 Determine the displacement of B and A Neglect the size of the couplings at B C and D Given LAB 050m LBC 150m LCD 075m PA 8kN PB 2kN PC 33kN θC 60deg hB 4 5 vB 3 5 A 60mm2 E 200 GPa Solution Internal Force As shown on FBD Displacement δAB PA LAB E A δBC PA 2PB vB LBC E A δCD PA 2PB vB 2PC sin θC LCD E A δB δBC δCD δB 2307 mm Ans δA δAB δBC δCD δA 2641 mm Ans Problem 46 The assembly consists of an A36 steel rod CB and a 6061T6 aluminum rod BA each having a diameter of 25 mm Determine the applied loads P1 and P2 if A is displaced 2 mm to the right and B is displaced 05 mm to the left when the loads are applied The unstretched length of each segment is shown in the figure Neglect the size of the connections at B and C and assume that they are rigid Given LBA 12m LCB 06m δA 2mm δB 05 mm d 25mm ECB 200 GPa EBA 689 GPa Solution Internal Force As shown on FBD Displacement A π 4 d2 Initial guess P1 1kN P2 2kN Given For BA δA δB P1 LBA A EBA 1 For CB δB P1 P2 LCB A ECB 2 Solving 1 and 2 P1 P2 Find P1 P2 P1 P2 7046 15227 kN Ans Problem 47 The 15mmdiameter A36 steel shaft AC is supported by a rigid collar which is fixed to the shaft at B If it is subjected to an axial load of 80 kN at its end determine the uniform pressure distribution p on the collar required for equilibriumAlso what is the elongation on segment BC and segment BA Given LAB 200mm LBC 500mm d 15mm PC 80kN E 200 GPa rc 35mm Solution Acollar π 4 2rc 2 d2 Equations of equilibrium ΣFy0 p Acollar PC 0 p PC Acollar p 2179 MPa Ans Internal Force As shown on FBD Displacement PBC PC PBA 0 A π 4 d2 δBC PBC LBC E A δBC 1132 mm Ans δBA PBA LAB E A δBA 0mm Ans Problem 48 The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC Determine the vertical displacement of the 25kN load if the members were horizontal when the load was originally applied Each wire has a crosssectional area of 16 mm2 Given LDE 09m LCF 09m P 25kN LDH 03m LHC 06m A 16mm2 LAH 054m LBG 15m E 193 GPa LAI 09m LIB 03m Solution LDC LDH LHC LAB LAI LIB Internal Forces in the wires From FBD b ΣΜA0 FBG LAB P LAI 0 ΣFy0 FAH FBG P 0 From FBD a FBG P LAI LAB FBG 1875 kN ΣΜD0 FCF LDC FAH LDH 0 FAH P FBG FAH 0625 kN ΣFy0 FCF FDE FAH 0 FCF FAH LDH LDC FCF 02083 kN FDE FAH FCF FDE 04167 kN Displacement δD FDE LDE E A δD 012143782 mm δC FCF LCF E A δC 006071891 mm δH δC LHC LDC δD δC δH 010119819 mm δA δH FAH LAH E A δB FBG LBG E A δI δA LAI LAB δB δA δI 0736 mm Ans LHC 06m LDE 09m LCF 09m Given P 25kN LDH 03m Solution A 16mm2 LAH 054m LBG 15m E 193 GPa LAI 09m LIB 03m Ans LDC LDH LHC LAB LAI LIB Internal Forces in the wires From FBD b ΣΜA0 FBG LAB P LAI 0 ΣFy0 FAH FBG P 0 From FBD a FBG P LAI LAB FBG 1875 kN ΣΜD0 FCF LDC FAH LDH 0 FAH P FBG FAH 0625 kN ΣFy0 FCF FDE FAH 0 FCF FAH LDH LDC FCF 02083 kN FDE FAH FCF FDE 04167 kN Displacement δD FDE LDE E A δD 012143782 mm δC FCF LCF E A δC 006071891 mm δH δC LHC LDC δD δC δH 010119819 mm tan αDC δD δC LDC αDC atan δD δC LDC αDC 00039 deg Ans δA δH FAH LAH E A δA 021049223 mm δB FBG LBG E A δB 091078368 mm tan βAB δB δA LAB βAB atan δB δA LAB βAB 00334 deg Problem 49 The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC Determine the angle of tilt of each member after the 25kN load is applied The members were originally horizontal and each wire has a crosssectional area of 16 mm2 Problem 410 The bar has a crosssectional area of 1800 mm2 and E 250 GPa Determine the displacement of its end A when it is subjected to the distributed loading Given L 15m E 250 GPa A 1800mm2 w 500 x 1 3 N m Solution Internal Force As shown on FBD Px 0 x w x x d Px 0 x x 500 x 1 3 x d Px 1500 4 x 4 3 Displacement unit 1N m δA 0 L x Px E A d L 150 m δA unit A E 0 48 x 1500 4 x 4 3 d δA 2990 mm Ans Problem 411 The assembly consists of three titanium Ti6A14V rods and a rigid bar AC The crosssectional area of each rod is given in the figure If a force of 30 kN is applied to the ring F determine the horizontal displacement of point F Given LCD 12m LEC 06m LEF 03m LAB 18m LAE 03m P 30kN E 120 GPa ACD 600mm2 AAB 900mm2 AEF 1200mm2 Solution LAC LAE LEC Internal Forces in the rods From FBD FCD P LAE LAC ΣΜA0 FCD LAC P LAE 0 FCD 1000 kN ΣFy0 FCD FAB P 0 FAB P FCD FAB 2000 kN Displacement δC FCD LCD E ACD δC 01667 mm δA FAB LAB E AAB δA 03333 mm δE δC LEC LAC δA δC δE 02778 mm δFE P LEF E AEF δFE 00625 mm δF δE δFE δF 0340278 mm Ans Problem 412 The assembly consists of three titanium Ti6A14V rods and a rigid bar ACThe crosssectional area of each rod is given in the figure If a force of 30 kN is applied to the ring F determine the angle of tilt of bar AC Given LCD 12m LEC 06m LEF 03m LAB 18m LAE 03m P 30kN E 120 GPa ACD 600mm2 AAB 900mm2 AEF 1200mm2 Solution LAC LAE LEC Internal Forces in the rods From FBD FCD P LAE LAC ΣΜA0 FCD LAC P LAE 0 FCD 1000 kN ΣFy0 FCD FAB P 0 FAB P FCD FAB 2000 kN Displacement δC FCD LCD E ACD δC 01667 mm δA FAB LAB E AAB δA 03333 mm tan αAC δA δC LAC αAC atan δA δC LAC αAC 001061 deg Ans Problem 413 A springsupported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k 60 kNm three 304 stainless steel rods AB and CD which have a diameter of 5 mm and EF which has a diameter of 12 mm and a rigid beam GH If the pipe and the fluid it carries have a total weight of 4 kN determine the displacement of the pipe when it is attached to the support Given LAB 075m LCD 075m LEF 075m dAB 5mm dCD 5mm dEF 12mm LGE 025m LEH 025m k 60 kN m P 4kN E 193 GPa Solution LGH LGE LEH Internal Forces in the rods From FBD a ΣΜA0 FCD LGH P LGE 0 ΣFy0 FCD FAB P 0 FCD P LGE LGH FCD 200 kN FAB P FCD FAB 200 kN From FBD b ΣFy0 FEF FCD FAB 0 FEF FCD FAB FEF 400 kN Displacement AAB π 4 dAB 2 ACD π 4 dCD 2 AEF π 4 dEF 2 δE FEF LEF E AEF δE 013744 mm δD δE FCD k δD 3347077 mm Due to symmetry δB δD δCD FCD LCD E ACD δCD 039583 mm Due to symmetry δBA δCD δtotal δD δCD δtotal 338666 mm Ans Problem 414 A springsupported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k 60 kNm three 304 stainless steel rods AB and CD which have a diameter of 5 mm and EF which has a diameter of 12 mm and a rigid beam GH If the pipe is displaced 82 mm when it is filled with fluid determine the weight of the fluid Given LAB 075m LCD 075m LEF 075m dAB 5mm dCD 5mm dEF 12mm LGE 025m LEH 025m k 60 kN m δtotal 82mm E 193 GPa Solution LGH LGE LEH Internal Forces in the rods Iniatlly set P 1kN From FBD a ΣΜA0 FCD LGH P LGE 0 ΣFy0 FCD FAB P 0 FCD P LGE LGH FCD 050 kN FAB P FCD FAB 050 kN From FBD b ΣFy0 FEF FCD FAB 0 FEF FCD FAB FEF 100 kN Displacement AAB π 4 dAB 2 ACD π 4 dCD 2 AEF π 4 dEF 2 δE FEF LEF E AEF δE 003436 mm δD δE FCD k δD 836769 mm Due to symmetry δB δD δCD FCD LCD E ACD δCD 009896 mm Due to symmetry δBA δCD δtotal δD δCD δtotal 84666 mm W P δtotal δtotal W P δtotal δtotal W 969 kN Ans Problem 415 The assembly consists of three titanium rods and a rigid bar ACThe crosssectional area of each rod is given in the figure If a vertical force P 20 kN is applied to the ring F determine the vertical displacement of point F Eti 350 GPa Given LBA 2m LDC 2m LEF 15m LAE 05m LEC 075m ABA 60mm2 ADC 45mm2 AEF 75mm2 P 20kN E 350 GPa Solution LAC LAE LEC Internal Forces in the rods From FBD FDC P LAE LAC ΣΜA0 FDC LAC P LAE 0 FDC 800 kN ΣFy0 FDC FBA P 0 FBA P FDC FBA 1200 kN Displacement δC FDC LDC E ADC δC 10159 mm δA FBA LBA E ABA δA 11429 mm δE δC LEC LAC δA δC δE 10921 mm δFE P LEF E AEF δFE 11429 mm δF δE δFE δF 22349 mm Ans Problem 416 The linkage is made of three pinconnected A36 steel members each having a crosssectional area of 500 mm2 If a vertical force of P 250 kN is applied to the end B of member AB determine the vertical displacement of point B Given a 15m b 2m LAD a2 b2 LAD 25 m LAB 3m LAC a2 b2 LAC 25 m E 200 GPa A 500mm2 P 250kN Solution c a2 b2 h a c v b c θ atan a b θ 36869898 deg Equation of equilibrium For AB ΣFy0 FAD FAC v P 0 Since FAD FAC FAC P 2v FAC 15625 kN Displacement δAC FAC LAC E A δAC 39063 mm δBA P LAB E A δBA 75 mm Consider triangle AAC LAC LAC δAC LAA δA θA 180deg θ sin φA LAC sin θA LAC φA asin LAC sin θA LAC φA 3680289 deg θC 180deg θA φA θC 00670094 deg sin θC δA sin θA LAC δA LAC sin θC sin θA δA 48731 mm δB δA δBA δB 1237 mm Ans Problem 417 The linkage is made of three pinconnected A36 steel members each having a crosssectional area of 500 mm2 Determine the magnitude of the force P needed to displace point B 25 mm downward Given a 15m b 2m LAD a2 b2 LAD 25 m LAB 3m LAC a2 b2 LAC 25 m E 200 GPa A 500mm2 δB 25mm Solution c a2 b2 h a c v b c θ atan a b θ 36869898 deg Equation of equilibrium For AB ΣFy0 FAD FAC v P 0 Since FAD FAC FAC P 2v FAD P 2v Displacement δAC FAC LAC E A δAC P LAC 2v E A δBA P LAB E A δB δA δBA δA δB P LAB E A Consider triangle AAC LAC LAC δAC LAA δA θA 180deg θ θA 143130102 deg LAC 2 δA 2 LAC 2 2 δA LAC cos θA Initial guess P 1kN Given LAC P LAC 2v E A 2 δB P LAB E A 2 LAC 2 2 δB P LAB E A LAC cos θA Solving P Find P P 5047 kN Ans Problem 418 Consider the general problem of a bar made from m segments each having a constant crosssectional area Am and length Lm If there are n loads on the bar as shown write a computer program that can be used to determine the displacement of the bar at any specified location x Show an application of the program using the values L1 12 m d1 06 m P1 2 kN A1 1875 mm2 L2 06 m d2 18 m P2 15 kN A2 625 mm2 Problem 419 The rigid bar is supported by the pinconnected rod CB that has a crosssectional area of 14 mm2 and is made from 6061T6 aluminum Determine the vertical deflection of the bar at D when the distributed load is applied Given L 4m w 300 N m a 15m b 2m A 14mm2 E 689 GPa Solution LBC a2 b2 v a LBC h b LBC Support Reactions ΣΜA0 FBC v b w L 05 L 0 FBC w L2 2 v b FBC 2kN Displacement δBC FBC LBC E A δBC 5183 mm Consider triangle ABC LBC LBC δBC LBC 2505183 m LBC 2 a2 b2 2 a b cos θA θA acos a2 b2 LBC 2 2 a b θA 9024775 deg φ θA 90deg φ 024775 deg φ 00043241 rad δD φ L δD 1730 mm Ans Problem 420 The rigid beam is supported at its ends by two A36 steel tie rods If the allowable stress for the steel is σallow 115 MPa the load w 50 kNm and x 12 m determine the diameter of each rod so that the beam remains in the horizontal position when it is loaded Given LCD 18m LAC 24m LAB 18m x 12m w 50 kN m σallow 115 MPa Solution Internal Forces in the rods From FBD FCD w x2 2 LAC ΣΜA0 FCD LAC w x 05x 0 FCD 1500 kN ΣFy0 FCD FAB w x 0 FAB w x FCD FAB 4500 kN Displacement To maintain the rigid beam in the horizontal position the elongation of both rods AB and CD must be the same δA δC AAB π 4 dAB 2 ACD π 4 dCD 2 δA FAB LAB E AAB δC FCD LCD E ACD Thus FAB LAB E π 4 dAB 2 FCD LCD E π 4 dCD 2 dCD dAB FCD LCD FAB LAB dAB 3 dCD Allowable Normal Stress Assume failure of rods AB σallow FAB π 4 dAB 2 dAB 4 FAB π σallow dAB 22321 mm dCD dAB 3 dCD 12887 mm Assume failure of rods CD σallow FCD π 4 dCD 2 dCD 4 FCD π σallow dCD 12887 mm Ans dAB 3 dCD dAB 22321 mm Ans Problem 421 The rigid beam is supported at its ends by two A36 steel tie rods The rods have diameters dAB 12 mm and dCD 75 mm If the allowable stress for the steel is σallow 115 MPa determine the intensity of the distributed load w and its length x on the beam so that the beam remains in the horizontal position when it is loaded Given LCD 18m LAC 24m LAB 18m dAB 12mm dCD 75mm σallow 115 MPa Solution AAB π 4 dAB 2 ACD π 4 dCD 2 Allowable Normal Stress Assume failure of rods AB σallow FAB AAB FAB σallow π 4 dAB 2 FAB 13006 kN Displacement To maintain the rigid beam in the horizontal position the elongation of both rods AB and CD must be the same δA δC δA FAB LAB E AAB δC FCD LCD E ACD Thus FAB LAB E π 4 dAB 2 FCD LCD E π 4 dCD 2 FCD FAB dCD 2 dAB 2 FCD dCD 2 dAB 2 FAB FCD 5081 kN Internal Forces in the rods Given From FBD ΣΜA0 FCD LAC w x 05x 0 ΣFy0 FCD FAB w x 0 Initial guess w 1 kN m x 1m Solving w x Find w x w 1341 kN m Ans x 135 m Allowable Normal Stress Assume failure of rods CD σallow FCD ACD FCD σallow π 4 dCD 2 FCD 5081 kN Displacement To maintain the rigid beam in the horizontal position the elongation of both rods AB and CD must be the same δA δC δA FAB LAB E AAB δC FCD LCD E ACD Thus FAB LAB E π 4 dAB 2 FCD LCD E π 4 dCD 2 FCD FAB dCD 2 dAB 2 FAB dAB 2 dCD 2 FCD FAB 13006 kN The results are the same as assuming failure of rod AB Therefore rods AB and CD fail simultaneously Ans Problem 422 The post is made of Douglas fir and has a diameter of 60 mm If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is uniformly distributed along its sides of w 4 kNm determine the force F at its bottom needed for equilibrium Also what is the displacement of the top of the post A with respect to its bottom B Neglect the weight of the post Given L 2m d 60mm P 20kN w 4 kN m E 131 GPa Solution Equation od Equilibrium For entire post FBD a ΣFy0 P w L F 0 F P w L F 12kN Internal Force FBD b ΣFy0 P w y Fy 0 Fy P w y Displacement A π 4 d2 unit 1kN δAB 0 L y Fy A E d δAB unit A E 0 L y P w y 1 kN d δAB 0864 mm Ans Note Negative sign indicates that ens A moves towards end B Problem 423 The post is made of Douglas fir and has a diameter of 60 mm If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w 0 at y 0 to w 3 kNm at y 2 m determine the force F at its bottom needed for equilibrium Also what is the displacement of the top of the post A with respect to its bottom B Neglect the weight of the post Given L 2m d 60mm P 20kN wo 3 kN m E 131 GPa Solution Equation od Equilibrium For entire post FBD a ΣFy0 P 05wo L F 0 F P 05wo L F 17kN Internal Force FBD b ΣFy0 P 05 wo y L y Fy 0 Fy P wo 2 L y2 Displacement A π 4 d2 unit 1kN δAB 0 L y Fy A E d δAB unit A E 0 L y P wo 2 L y2 1 kN d δAB 1026 mm Ans Note Negative sign indicates that ens A moves towards end B Problem 424 The rod has a slight taper and length L It is suspended from the ceiling and supports a load P at its end Show that the displacement of its end due to this load is δ PLπEr2r1 Neglect the weight of the material The modulus of elasticity is E Solution Geometry L xo r2 xo r1 xo L r1 r2 r1 Thus rx r1 r2 r1 L x rx r1 L r2 r1 x L Ax π rx 2 Ax π L2 r1 L r2 r1 x 2 Displacement δ 0 L x P Ax E d δ P L2 π E 0 L x 1 r1 L r2 r1 x 2 d L 0 δ P L2 π E 1 r2 r1 r1 L r2 r1 x δ P L2 π E r2 r1 1 r1 L r2 r1 L 1 r1 L δ P L2 π E r2 r1 1 r2 L 1 r1 L δ P L2 π E r2 r1 r2 r1 r1 r2 L δ P L π E r1 r2 Ans Problem 425 Solve Prob 424 by including both P and the weight of the material considering its specific weight to be γ weight per volume Solution Internal Force FBD b ΣFy0 Px wx P 0 Px Wx P Geometry L xo r2 xo r1 xo L r1 r2 r1 Thus rx r1 r2 r1 L x rx r1 L r2 r1 x L Ax π rx 2 Ax π L2 r1 L r2 r1 x 2 Wx γ 3 Ax x0 x γ π 3 r1 2 x0 Wx γ π 3 L2 r1 L r2 r1 x 2 L r1 r2 r1 x γ π 3 r1 2 L r1 r2 r1 Wx γ π 3 L2 r2 r1 r1 L r2 r1 x 3 r1 3 L3 Displacement δ1 γ 3 E r2 r1 0 L x r1 L r2 r1 x 3 r1 3 L3 r1 L r2 r1 x 2 d δ1 0 L x Wx Ax E d δ1 γ 3 E r2 r1 0 L x r1 L r2 r1 x d r1 3 L3 0 L x 1 r1 L r2 r1 x 2 d Using the result of Prob424 for the 2nd Integral we have δ1 γ 3 E r2 r1 r1 L2 r2 r1 L2 2 r1 3 L3 1 L r1 r2 δ1 γ L2 r2 r1 6 E r2 r1 γ L2 r1 2 3 E r2 r1 r2 Using the result of Prob424 for the 2nd Integral we have δ δ1 0 L x P Ax E d δ γ L2 r2 r1 6 E r2 r1 γ L2 r1 2 3 E r2 r1 r2 P L π E r1 r2 Ans Problem 426 The support is made by cutting off the two opposite sides of a sphere that has a radius r0 If the original height of the support is r0 2 determine how far it shortens when it supports a load PThe modulus of elasticity is E Solution Geometry r ro cos θ y ro sin θ Ay π r2 dy ro cos θ dθ Ay π ro 2 cos θ 2 h 025 ro h ro sin θo Displacement 025 ro ro sin θo δ 2 0 h y P Ay E d δ 0 θo θ 2P ro cos θ π ro 2 cos θ 2 E d θo asin 025 θo 1448 deg δ 2P π ro E 0 1448o θ 1 cos θ d 48o 14 0 δ 2 P π ro E ln sec θ tan θ δ 0511 P π ro E Ans Alternatively Geometry Ay π x2 Ay π ro 2 y2 Displacement δ 0 025 ro θ 2 P π ro 2 y2 E d δ 2 0 h y P Ay E d 25 or 0 0 δ 2 P π E 1 2 ro ln ro y ro y δ 0511 P π ro E Ans Problem 427 The ball is truncated at its ends and is used to support the bearing load P If the modulus of elasticity for the material is E determine the decrease in its height when the load is applied Solution Geometry x2 r2 y2 Ay π x2 Ay π r2 y2 Displacement δ 2 0 h y P Ay E d r 2 2 r2 h2 δ 0 0866 r θ 2 P π r2 y2 E d h 3 2 r 866 r 0 0 h 0866 r δ 2 P π E 1 2 r ln r y r y δ 263 P π ro E Ans Problem 428 Determine the elongation of the aluminum strap when it is subjected to an axial force of 30 kN Eal 70 GPa Given L1 250mm L2 800mm d1 15mm d2 50mm t 6mm P 30kN E 70 GPa Solution Displacement δ 2 P L1 E t d2 d1 ln d2 d1 P L2 E t d2 δ 2371 mm Ans Problem 429 The casting is made of a material that has a specific weight γ and modulus of elasticity E If it is formed into a pyramid having the dimensions shown determine how far its end is displaced due to gravity when it is suspended in the vertical position Solution Internal Force FBD b ΣFy0 Pz 1 3γ A z 0 Pz 1 3γ A z Displacement δ 0 L y Pz A E d δ 0 L y γ A z 3A E d δ γ 3 E 0 L z y d δ γ L2 6 E Ans Problem 430 The pedestal is made in a shape that has a radius defined by the function r 22 y12 m where y is in meter If the modulus of elasticity for the material is E 100 MPa determine the displacement of its top when it supports the 5kN load Given H 4m do 1m d1 05m P 5kN E 100 106 Pa r 2 2 y05 Solution Ay 4 π 2 y05 2 Displacement Ay π r2 unit 1m H 4 δ 0 H y P Ay E d δ P E 1 unit 0 H y 2 y05 2 4 π d δ 01804 mm Ans Problem 431 The column is constructed from highstrength concrete and six A36 steel reinforcing rods If it is subjected to an axial force of 150 kN determine the average normal stress in the concrete and in each rod Each rod has a diameter of 20 mm Given L 12m rconc 100mm dst 20mm P 150kN Est 200 GPa Econc 29 GPa Solution Ast 6 π 4 dst 2 Aconc π rconc 2 Ast Compatibility δst δconc Given Pst L Ast Est Pconc L Aconc Econc 1 Equations of equilibrium ΣFy0 Pst Pconc P 0 2 Initial guess Pconc 1kN Pst 2kN Solving 1 and 2 Pconc Pst Find Pconc Pst Pconc Pst 104152 45848 kN Average Normal Stress σst Pst Ast σst 24323 MPa Ans σconc Pconc Aconc σconc 3527 MPa Ans Problem 432 The column is constructed from highstrength concrete and six A36 steel reinforcing rods If it is subjected to an axial force of 150 kN determine the required diameter of each rod so that onefourth of the load is carried by the concrete and threefourths by the steel Given L 12m rconc 100mm P 150kN Pst 075P Pconc 025P Est 200 GPa Econc 29 GPa Solution Pst 11250 kN Pconc 3750 kN Compatibility δst δconc Pst L Ast Est Pconc L Aconc Econc Ast 6 π 4 dst 2 Aconc π rconc 2 Ast Thus π rconc 2 Ast Ast Est Pst Pconc Econc π rconc 2 Ast Est Pst Pconc Econc 1 Ast π rconc 2 Est Pst Pconc Econc 1 Ast 95232948 mm2 dst 2 2 3 π Ast dst 2 3 π Ast dst 4495 mm Ans Problem 433 The A36 steel pipe has a 6061T6 aluminum core It is subjected to a tensile force of 200 kN Determine the average normal stress in the aluminum and the steel due to this loading The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm Given L 400mm do 80mm P 200kN di 70mm Est 200GPa Eal 689GPa Solution Aal π 4 di 2 Ast π 4 do 2 di 2 Compatibility δst δal Given Pst L Ast Est Pal L Aal Eal 1 Equations of equilibrium ΣFx0 Pst Pal P 0 2 Initial guess Pal 1kN Pst 2kN Solving 1 and 2 Pal Pst Find Pal Pst Pal Pst 105899 94101 kN Average Normal Stress σst Pst Ast σst 7988 MPa Ans σal Pal Aal σal 2752 MPa Ans Problem 434 The concrete column is reinforced using four steel reinforcing rodseach having a diameter of 18 mm Determine the stress in the concrete and the steel if the column is subjected to an axial load of 800 kN Est 200 GPa Ec 25 GPa Given P 800kN bc 300mm dst 18mm Est 200 GPa Ec 25 GPa Solution Ast 4 π 4 dst 2 Ac bc 2 Ast Set L 1m Compatibility δst δconc Given Pst L Ast Est Pc L Ac Ec 1 Equations of equilibrium ΣFy0 Pst Pc P 0 2 Initial guess Pc 1kN Pst 2kN Solving 1 and 2 Pc Pst Find Pc Pst Pc Pst 732928 67072 kN Average Normal Stress σst Pst Ast σst 6589 MPa Ans σc Pc Ac σc 824 MPa Ans Problem 435 The column is constructed from highstrength concrete and four A36 steel reinforcing rods If it is subjected to an axial force of 800 kN determine the required diameter of each rod so that onefourth of the load is carried by the steel and threefourths by the concrete Est 200 GPa Ec 25 GPa Given P 800kN Ec 25 GPa bc 300mm Pc 075P Est 200 GPa Pst 025P Solution Pst 200kN Pc 600kN Compatibility δst δc Pst L Ast Est Pc L Ac Ec Ast 4 π 4 dst 2 Ac bc 2 Ast Thus bc 2 Ast Ast Est Pst Pc Ec bc 2 Ast Est Pst Pc Ec 1 Ast bc 2 Est Pst Pc Ec 1 Ast 3600mm2 dst 2 1 π Ast dst 1 π Ast dst 3385 mm Ans Problem 436 The A36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm If it fits snugly between the fixed walls before it is loaded determine the reaction at the walls when it is subjected to the load shown Given LAB 300mm ro 20mm LBC 700mm ri 15mm P 8kN Est 200GPa Solution A π ro 2 ri 2 L LAB LBC By superposition C δC 0 2P LAB A E FC L A E 0 FC 2 P LAB L FC 480 kN Ans Equations of equilibrium ΣFx0 FA FC 2P 0 FA 2P FC FA 1120 kN Ans Problem 437 The 304 stainless steel post A has a diameter of d 50 mm and is surrounded by a red brass C83400 tube B Both rest on the rigid surface If a force of 25 kN is applied to the rigid cap determine the average normal stress developed in the post and the tube Given L 200mm dst 50mm ro 75mm t 12mm P 25kN Est 193 GPa Ebr 101 GPa Solution ri ro t Ast π 4 dst 2 Abr π ro 2 ri 2 Compatibility δst δbr Given Pst L Ast Est Pbr L Abr Ebr 1 Equations of equilibrium ΣFy0 Pst Pbr P 0 2 Initial guess Pbr 1kN Pst 2kN Solving 1 and 2 Pbr Pst Find Pbr Pst Pbr Pst 14525 10475 kN Average Normal Stress σst Pst Ast σst 5335 MPa Ans σbr Pbr Abr σbr 2792 MPa Ans Problem 438 The 304 stainless steel post A is surrounded by a red brass C83400 tube B Both rest on the rigid surface If a force of 25 kN is applied to the rigid cap determine the required diameter d of the steel post so that the load is shared equally between the post and tube Given L 200mm ro 75mm t 12mm P 25kN Est 193 GPa Ebr 101 GPa Pst 05P Pbr 05P Solution ri ro t Pst 125 kN Pbr 125 kN Compatibility δst δbr Pst L Ast Est Pbr L Abr Ebr Ast π 4 dst 2 Abr π ro 2 ri 2 Thus π ro 2 ri 2 Ast Est Pst Pbr Ebr Ast π ro 2 ri 2 Est Pst Pbr Ebr Ast 272254 mm2 dst 2 4 π Ast dst 4 π Ast dst 5888 mm Ans Problem 439 The load of 75 kN is to be supported by the two vertical steel wires for which σY 500 MPa If originally wire AB is 1250 mm long and wire AC is 12525 mm long determine the force developed in each wire after the load is suspended Each wire has a crosssectional area of 125 mm2 Given LAB 1250mm LAC 12525mm A 125mm2 W 75kN E 200 GPa EY 70500 MPa Solution L LAC LAB L 250 mm Compatibility δAC L δAB Given TAC LAC A E L TAB LAB A E 1 Equations of equilibrium ΣFy0 TAC TAB W 0 2 Initial guess TAC 1kN TAB 2kN Solving 1 and 2 TAC TAB Find TAC TAB TAC TAB 1249 6251 kN Ans Check Average Normal Stress σAC TAC A σAC 999 MPa σY 500 MPa σAB TAB A σAB 5001 MPa σY 500 MPa Problem 440 The load of 4 kN is to be supported by the two vertical steel wires for which σY 560 MPa If originally wire AB is 1250 mm long and wire AC is 12525 mm long determine the crosssectional area of AB if the load is to be shared equally between both wires Wire AC has a crosssectional area of 13 mm2 Given W 4kN LAB 1250mm AAC 13mm2 TAC 05W LAC 12525mm EY 560 MPa TAB 05W E 200 GPa Solution L LAC LAB L 250 mm TAC 2kN TAB 2kN Compatibility δAC L δAB TAC LAC AAC E L TAB LAB AAB E Thus AAB TAB LAB TAC LAC AAC L E AAB 360911 mm2 Ans Check Average Normal Stress σAC TAC AAC σAC 153846 MPa σY 560 MPa σAB TAB AAB σAB 55415 MPa σY 560 MPa Problem 441 The support consists of a solid red brass C83400 post surrounded by a 304 stainless steel tube Before the load is applied the gap between these two parts is 1 mm Given the dimensions shown determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials Given Lbr 0251m Ebr 101 GPa σYbr 70 MPa Lst 0250m Est 193 GPa dbr 60mm di 80mm t 10mm Solution L Lbr Lst L 1mm do di 2t Abr π 4 dbr 2 Ast π 4 do 2 di 2 Compatibility δbr L δst Fbr Lbr Abr Ebr L Fst Lst Ast Est 1 Equations of equilibrium ΣFy0 Fst Fbr P 0 2 Assume brass yields then Fbr σYbr Abr Fbr 19792 kN εbr σYbr Ebr εbr 000069307 mm mm δbr εbr Lbr δbr 017 mm L 1 mm Thus only the brass is loaded P Fbr P 19792 kN Ans Problem 442 Two A36 steel wires are used to support the 325kN 325kg engine Originally AB is 800 mm long and AB is 8002 mm long Determine the force supported by each wire when the engine is suspended from them Each wire has a crosssectional area of 625 mm2 Given LAB 800mm A 625mm2 LAB 8002mm W 325kN E 29 103 ksi Solution L LAB LAB L 0200 mm Compatibility δAB L δAB Given 1 TAB LAB A E L TAB LAB A E Equations of equilibrium ΣFy0 TAB TAB W 0 2 Initial guess TAB 1kN TAB 2kN Solving 1 and 2 TAB TAB Find TAB TAB TAB TAB 1469 1781 kN Ans Problem 443 The bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter of 40 mm andan outer diameter of 50 mm The bolt and sleeve are made of A36 steel and are secured to the rigid brackets as shown If the bolt length is 220 mm and the sleeve length is 200 mm determine the tension in the bolt when a force of 50 kN is applied to the brackets Given Lb 220mm db 20mm do 50mm Ls 200mm P 25kN di 40mm E 200GPa Solution Ab π 4 db 2 As π 4 do 2 di 2 Compatibility δb δs Given 1 Pb Lb Ab E Ps Ls As E Equations of equilibrium ΣFx0 Pb Ps 2P 0 2 Initial guess Pb 1kN Ps 2kN Solving 1 and 2 Pb Ps Find Pb Ps Pb Ps 14388 35612 kN Problem 444 The specimen represents a filamentreinforced matrix system made from plastic matrix and glass fiber If there are n fibers each having a crosssectional area of Af and modulus of Ef embedded in a matrix having a crosssectional area of Am and modulus of Em determine the stress in the matrix and each fiber when the force P is imposed on the specimen Solution Compatibility δm δf Pm L Am Em Pf L n Af Ef Pm Am Em n Af Ef Pf 1 Equations of equilibrium ΣFy0 P Pm Pf 0 2 Solving 1 and 2 Pm Am Em n Af Ef Am Em P Pf n Af Ef n Af Ef Am Em P Average Normal Stress σm Pm Am σm Em n Af Ef Am Em P Ans σf Pf n Af σf Ef n Af Ef Am Em P Ans Problem 445 The distributed loading is supported by the three suspender bars AB and EF are made from aluminum andCD is made from steel If each bar has a crosssectional area of 450 mm2 determine the maximum intensity w of the distributed loading so that an allowable stress of σallowst 180 MPa in the steel and σallowal 94 MPa in the aluminum is not exceeded Est 200 GPa Eal 70 GPa Given L 2m b 15m A 450mm2 Est 200 GPa Eal 70 GPa σalallow 94MPa σstallow 180MPa Solution Compatibility δA δC FAB L Eal A FCD L Est A FAB Eal Est FCD 1 Equations of equilibrium ΣΜC0 FEF b FAB b 0 FAB FEF ΣFy0 FAB FCD FEF w 2b 0 2 FAB w 2b FCD 2 Assume failure of AB and EF FAB σalallow A FAB 4230 kN From 1 FCD Est Eal FAB FCD 12086 kN From 2 w FAB b FCD 2 b w 6849 kN m Assume failure of CD FCD σstallow A FCD 8100 kN From 1 FAB Eal Est FCD FAB 2835 kN From 2 w FAB b FCD 2 b w 4590 kN m Controls Ans Problem 446 The rigid link is supported by a pin at A a steel wire BC having an unstretched length of 200 mm and crosssectional area of 225 mm2 and a short aluminum block having an unloaded length of 50 mm and crosssectional area of 40 mm2 If the link is subjected to the vertical load shown determine the average normal stress in the wire and the block Est 200 GPa Eal 70 GPa Given Lk 200mm Ak 225mm2 eD 50mm AD 40mm2 P 450N a 100mm b 150mm c 150mm Est 200GPa Eal 70GPa Solution Compatibility θBC θAD δBC b δD c Given 1 FBC Lk Ak Est b FD eD AD Eal c Equations of equilibrium ΣΜA0 FBC b FD c P a b 0 2 Initial guess FBC 1kN FD 2kN Solving 1 and 2 FBC FD Find FBC FD FBC FD 214968 535032 N Average Normal Stress σBC FBC Ak σBC 9554 MPa Ans σD FD AD σD 13376 MPa Ans Problem 447 The rigid link is supported by a pin at A a steel wire BC having an unstretched length of 200 mm and crosssectional area of 225 mm2 and a short aluminum block having an unloaded length of 50 mm and crosssectional area of 40 mm2 If the link is subjected to the vertical load shown determine the rotation of the link about the pin A Report the answer in radians Est 200 GPa Eal 70 GPa Given Lk 200mm Ak 225mm2 eD 50mm AD 40mm2 P 450N a 100mm b 150mm c 150mm Est 200GPa Eal 70GPa Solution Compatibility θBC θAD δBC b δD c Given 1 FBC Lk Ak Est b FD eD AD Eal c Equations of equilibrium ΣΜA0 FBC b FD c P a b 0 2 Initial guess FBC 1kN FD 2kN Solving 1 and 2 FBC FD Find FBC FD FBC FD 214968 535032 N Displacement δD FD eD AD Eal δD 0009554 mm θAD δD c θAD 6369 10 6 rad Ans Problem 448 The three A36 steel wires each have a diameter of 2 mm and unloaded lengths of LAC 160 m and LAB LAD 200 m Determine the force in each wire after the 150kg mass is suspended from the ring at A Given LAC 160m a 3 d 2mm LAB 200m b 4 M 150kg LAD 200m c 5 g 981 m s2 Solution h a c v b c A π 4 d2 W M g Compatibility In triangle AAB since the displacement δA is very small cos θA v δAC cos θA δAD δAC v δAD FAC LAC A E v FAD LAD A E 1 Equations of equilibrium ΣFx0 FAB h FAD h 0 FAB FAD ΣFy0 FAB v FAC FAD v W 0 Given 2 FAD v W FAC 2 From 1 FAC LAC v FAD LAD 3 Initial guess FAD 1N FAC 2N Solving 2 and 3 FAD FAC Find FAD FAC FAD FAC 4651 7268 N Ans FAB FAD Problem 449 The A36 steel wires AB and AD each have a diameter of 2 mm and the unloaded lengths of each wire are LAC 160 m and LAB LAD 200 m Determine the required diameter of wire AC so that each wire is subjected to the same force caused by the 150kg mass suspended from the ring at A Given LAC 160m a 3 dAB 2mm LAB 200m b 4 dAD 2mm LAD 200m c 5 g 981 m s2 M 150kg Solution h a c v b c W M g AAB π 4 dAB 2 AAD π 4 dAD 2 Equations of equilibrium Since each wire is required to carry the same amount of load Hence FAB F FAC F FAD F Compatibility In triangle AAB since the displacement δA is very small cos θA v δAC cos θA δAD δAC v δAD FAC LAC AAC E v FAD LAD AAD E AAC LAC LAD v AAD AAC 2010619 mm2 AAC π 4 dAC 2 dAC 4 π AAC dAC 160 mm Ans Problem 450 The three suspender bars are made of the same material and have equal crosssectional areas A Determinethe average normal stress in each bar if the rigid beam ACE is subjected to the force P Solution Compatibility δC δE d δA δE 2d 2δC δA δE 2 FCD L A E FAB L A E FEF L A E 2FCD FAB FEF 1 Equations of equilibrium ΣΜA0 FCD d FEF 2 d P d 2 0 FCD 2FEF 05P 2 ΣFy0 FAB FCD FEF P 0 3 Solving 1 2 and 3 FAB 7 12P FCD 1 3P FEF 1 12P σAB 7P 12A σAB P 3A σAB P 12A Ans Problem 451 The assembly consists of an A36 steel bolt and a C83400 red brass tube If the nut is drawn up snug against the tube so that L 75 mm then turned an additional amount so that it advances 002 mm on the bolt determine the force in the bolt and the tube The bolt has a diameter of 7 mm and the tube has a crosssectional area of 100 mm2 Given L 75mm Ebr 101 GPa L 002mm Est 200 GPa dst 7mm Abr 100mm2 Solution Ast π 4 dst 2 Kst Est Ast Kbr Ebr Abr Equations of equilibrium Since no external load is applie the force acting on the tube and the bolt is the same Compatibility L δst δbr L P L Kst P L Kbr L P L Kst Kbr Kst Kbr P L L Kst Kbr Kst Kbr P 1165 kN Ans Problem 452 The assembly consists of an A36 steel bolt and a C83400 red brass tube The nut is drawn up snug against the tube so that L 75 mm Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield The bolt has a diameter of 7 mm and the tube has a crosssectional area of 100 mm2 Given L 75mm Ebr 101 GPa dst 7mm Est 200 GPa Abr 100mm2 σYst 250MPa σYbr 70MPa Solution Ast π 4 dst 2 Kst Est Ast Kbr Ebr Abr Allowable Normal Stress Pst σYst Ast Pst 9621 kN Pbr σYbr Abr Pbr 7000 kN Since Pst Pbr by comparison the brass will yield first P min Pst Pbr P 700 kN Compatibility L δst δbr L P L Kst P L Kbr L P L Kst P L Kbr L 0120 mm Ans Problem 453 The 10mmdiameter steel bolt is surrounded by a bronze sleeve The outer diameter of this sleeve is 20 mm and its inner diameter is 10 mm If the bolt is subjected to a compressive force of P 20 kN determine the average normal stress in the steel and the bronze Est 200 GPa Ebr 100 GPa Given ds 10mm do 20mm Est 200GPa P 20kN di 10mm Ebr 100GPa Solution Ast π 4 ds 2 Abr π 4 do 2 di 2 Compatibility δb δs Pbr L Abr Ebr Pst L Ast Est Given Pbr Abr Ebr Pst Ast Est 1 Equations of equilibrium ΣFy0 Pbr Pst P 0 2 Initial guess Pbr 1kN Pst 2kN Solving 1 and 2 Pbr Pst Find Pbr Pst Pbr Pst 12000 8000 kN σst Pst Ast σst 10186 MPa Ans σbr Pbr Abr σbr 5093 MPa Ans Problem 454 The 10mmdiameter steel bolt is surrounded by a bronze sleeveThe outer diameter of this sleeve is 20 mmand its inner diameter is 10 mm If the yield stress for the steel is σYst 640 MPa and for the bronze σYbr 520 MPa determine the magnitude of the largest elastic load P that can be applied to the assembly Est 200 GPa Ebr 100 GPa Given do 20mm Est 200GPa σYst 640MPa di 10mm Ebr 100GPa σYbr 520MPa ds 10mm Solution Ast π 4 ds 2 Abr π 4 do 2 di 2 Equations of equilibrium ΣFy0 Pbr Pst P 0 1 Compatibility δb δs Pbr L Abr Ebr Pst L Ast Est Pbr Abr Ebr Pst Ast Est 2 Assume failure of bolt then Pst σYst Ast Pst 50265 kN From 2 Pbr Abr Ebr Ast Est Pst Pbr 75398 kN From 1 P Pbr Pst P 12566 kN Controls Ans Assume failure of sleeve then Pbr σYbr Abr Pbr 122522 kN From 2 Pst Ast Est Abr Ebr Pbr Pst 81681 kN From 1 P Pbr Pst P 20420 kN Problem 455 The rigid member is held in the position shown by three A36 steel tie rods Each rod has an unstretched length of 075 m and a crosssectional area of 125 mm2 Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn The lead of the screw is 15 mm Neglect the size of the turnbuckle and assume that it is rigid Note The lead would cause the rod when unloaded to shorten 15 mm when the turnbuckle is rotated one revolution Given L 075m b 05m A 125mm2 L 15mm E 200 GPa Solution Equations of equilibrium ΣΜE0 TCD b TAB b 0 TCD TAB 1 ΣFy0 TAB TCD TEF 0 1 TEF 2TAB 2 Compatibility Rod EF shortens 15mm causing AB and CD to elongate Thus L δA δC L TAB L E A TEF L E A L TAB L E A 2TAB L E A TAB E A 3L L TAB 16667 kN Ans TCD TAB TCD 16667 kN Ans From 1 TEF 2TAB TEF 33333 kN Ans From 2 Problem 456 The bar is pinned at A and supported by two aluminum rods each having a diameter of 25 mm and a modulus of elasticity Eal 70 GPa If the bar is assumed to be rigid and initially vertical determine the displacement of the end B when the force of 10 kN is applied Given LCD 06m LEF 03m d 25mm a 03m P 10kN Eal 70GPa Solution A π 4 d2 Compatibility δE 3 a δC a Given FEF LEF 3A Eal FCD LCD A Eal FEF LEF 3FCD LCD 1 Equations of equilibrium ΣΜA0 FEF 3 a FCD a P 2a 0 3 FEF FCD 2P 0 2 Initial guess FEF 1kN FCD 2kN Solving 1 and 2 FEF FCD Find FEF FCD FEF FCD 631579 105263 kN Displacement δE FEF LEF A Eal δE 0055142 mm δB 4 a δE 3a δB 4 3δE δB 0073522 mm Ans Problem 457 The bar is pinned at A and supported by two aluminum rods each having a diameter of 25 mm and a modulus of elasticity Eal 70 GPa If the bar is assumed to be rigid and initially vertical determine the force in each rod when the 10kN load is applied Given LCD 06m LEF 03m d 25mm a 03m P 10kN Eal 70GPa Solution A π 4 d2 Compatibility δE 3 a δC a Given FEF LEF 3A Eal FCD LCD A Eal FEF LEF 3FCD LCD 1 Equations of equilibrium ΣΜA0 FEF 3 a FCD a P 2a 0 3 FEF FCD 2P 0 2 Initial guess FEF 1kN FCD 2kN Solving 1 and 2 FEF FCD Find FEF FCD FEF FCD 6316 1053 kN Problem 458 The assembly consists of two posts made from material 1 having a modulus of elasticity of E1 and each a crosssectional area A1 and a material 2 having a modulus of elasticity E2 and crosssectional area A2 If a central load P is applied to the rigid cap determine the force in each material Solution FAB F1 FEF F2 Equations of equilibrium ΣΜE0 FCD d FAB d 0 FCD FAB FCD F1 ΣFy0 FAB FCD FEF P 0 2F1 F2 P 1 Compatibility δA δB δB δC F1 L A1 E1 F2 L A2 E2 F1 A1 E1 A2 E2 F2 2 Solving 1 and 2 F1 A1 E1 2A1 E1 A2 E2 P Ans F2 A2 E2 2A1 E1 A2 E2 P Ans Problem 459 The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticity E1 ofand each a crosssectional area A1 and a central post EF made from material 2 having a modulus of elasticity E2 and a crosssectional area A2 If posts AB and CD are to be replaced by those having a material 2 determine the required crosssectional area of these new posts so that both assemblies deform the same amount when loaded Solution FAB F1 FEF F2 Equations of equilibrium ΣΜE0 FCD d FAB d 0 FCD FAB FCD F1 ΣFy0 FAB FCD FEF P 0 2F1 F2 P 1 Compatibility δA δB δB δC δA δ F1 L A1 E1 F2 L A2 E2 F1 A1 E1 A2 E2 F2 2 Solving 1 and 2 F1 A1 E1 2A1 E1 A2 E2 P F2 A2 E2 2A1 E1 A2 E2 P When material 1 has been replaced by material 2 for two side posts then Equations of equilibrium 1 becomes 2 F1 F2 P 1 Compatibility 2 becomes F1 A1 A2 F2 2 Solving 1 and 2 F1 A1 2A1 A2 P F2 A2 2A1 A2 P Requires δB δB F2 L A2 E2 F2 L A2 E2 A2 E2 2A1 E1 A2 E2 P A2 2A1 A2 P A1 E1 E2 A1 Ans Problem 460 The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticity E1 ofand each a crosssectional area A1 and a central post EF made from material 2 having a modulus of elasticity E2 and a crosssectional area A2 If post EF is to be replaced by one having a material 1 determine the required crosssectional area of this new post so that both assemblies deform the same amount when loaded Solution FAB F1 FEF F2 Equations of equilibrium ΣΜE0 FCD d FAB d 0 FCD FAB FCD F1 ΣFy0 FAB FCD FEF P 0 2F1 F2 P 1 Compatibility δA δB δB δC δA δ F1 L A1 E1 F2 L A2 E2 F2 A2 E2 A1 E1 F1 2 Solving 1 and 2 F1 A1 E1 2A1 E1 A2 E2 P F2 A2 E2 2A1 E1 A2 E2 P When material 2 has been replaced by material 1 for central post then Equations of equilibrium 1 becomes 2 F1 F2 P 1 Compatibility 2 becomes F2 A2 A1 F1 2 Solving 1 and 2 F1 A1 2A1 A2 P F2 A2 2A1 A2 P Requires δA δA F1 L A1 E1 F1 L A1 E1 A1 E1 2A1 E1 A2 E2 P A1 2A1 A2 P A2 E2 E1 A2 Ans Problem 461 The bracket is held to the wall using three A36 steel bolts at B C and D Each bolt has a diameter of 125 mm and an unstretched length of 50 mm If a force of 4 kN is placed on the bracket as shown determine the force developed in each bolt For the calculation assume that the bolts carry no shear rather the vertical force of 4 kN is supported by the toe at A Also assume that the wall and bracket are rigid A greatly exaggerated deformation of the bolts is shown Given a 125mm b 25mm c 50mm d 125mm e 50mm L 50mm P 4kN E 200GPa Solution LAD a b c LAC a b A π 4 d2 LAB a Compatibility δD LAD δC LAC FD L LAD A E FC L LAC A E δB LAB δC LAC FB L LAB A E FC L LAC A E Given FD LAD LAC FC 1 FB LAB LAC FC 2 Equations of equilibrium ΣΜA0 FD LAD FC LAC FB LAB P e 0 3 Initial guess FB 10kN FC 20kN FD 30kN Solving 1 2 and 3 FB FC FD Find FB FC FD FB FC FD 02712 08136 18983 kN Ans Problem 462 The bracket is held to the wall using three A36 steel bolts at B C and D Each bolt has a diameter of 125 mm and an unstretched length of 50 mm If a force of 4 kN is placed on the bracket as shown determine how far s the top bracket at bolt D moves away from the wall For the calculation assume that the bolts carry no shear rather the vertical force of 4 kN is supported by the toe at A Also assume that the wall and bracket are rigid A greatly exaggerated deformation of the bolts is shown Given a 125mm b 25mm c 50mm d 125mm e 50mm L 50mm P 4kN E 200GPa Solution LAD a b c LAC a b LAB a A π 4 d2 Compatibility δD LAD δC LAC FD L LAD A E FC L LAC A E δB LAB δC LAC FB L LAB A E FC L LAC A E Given FD LAD LAC FC 1 FB LAB LAC FC 2 Equations of equilibrium ΣΜA0 FD LAD FC LAC FB LAB P e 0 3 Initial guess FB 10kN FC 20kN FD 30kN Solving 1 2 and 3 FB FC FD Find FB FC FD FB FC FD 02712 08136 18983 kN Displacement δD FD L A E δD 0003867 mm Ans Problem 463 The rigid bar is supported by the two short white spruce wooden posts and a spring If each of the posts has an unloaded length of 1 m and a crosssectional area of 600 mm2 and the spring has a stiffness of k 2 MNm and an unstretched length of 102 m determine the force in each post after the load is applied to the bar Given L 1m Lk 102m b 2m w 50 kN m k 2 103 kN m A 600mm2 E 965GPa Solution Lk Lk L Lk 002 m Compatibility δA Lk δk 1 Equations of equilibrium ΣΜC0 FB b FA b 0 FA FB ΣFy0 FA FB Fsp w b 0 Given 2FA Fsp w b 0 2 1 From 1 FA L E A Lk Fsp k Initial guess FA 1kN Fsp 2kN Solving 2 and 1 FA Fsp Find FA Fsp FA Fsp 2558 4884 kN Ans FB FA FB 2558 kN Ans Problem 464 The rigid bar is supported by the two short white spruce wooden posts and a spring If each of the posts has an unloaded length of 1 m and a crosssectional area of 600 mm2 and the spring has a stiffness of k 2 MNm and an unstretched length of 102 m determine the vertical displacement of A and B after the load is applied to the bar Given L 1m Lk 102m b 2m w 50 kN m k 2 103 kN m A 600mm2 E 965GPa Solution Lk Lk L Lk 002 m Compatibility δA Lk δk 1 Equations of equilibrium ΣΜC0 FB b FA b 0 FA FB ΣFy0 FA FB Fsp w b 0 Given 2FA Fsp w b 0 2 1 From 1 FA L E A Lk Fsp k Initial guess FA 1kN Fsp 2kN Solving 2 and 1 FA Fsp Find FA Fsp FA Fsp 2558 4884 kN FB FA FB 2558 kN Displacement δA FA L A E δA 4418 mm Ans δB FB L A E δB 4418 mm Ans Problem 465 The wheel is subjected to a force of 18 kN from the axle Determine the force in each of the three spokes Assume the rim is rigid and the spokes are made of the same material and each has the same crosssectional area Given L 04m θ 120deg P 18kN Solution φ 180deg θ Compatibility δAB cos φ δAC FAB L E A cos φ FAC L E A 1 Equations of equilibrium At A ΣFx0 FAC sin φ FAD sin b 0 FAC FAD ΣFy0 FAB FAC cos φ FAD cos φ P 0 Given FAB 2 FAC cos φ P 0 2 From 1 FAB cos φ FAC 1 Initial guess FAB 1kN FAC 2kN Solving 2 and 1 FAB FAC Find FAB FAC FAB FAC 12 6 kN Ans FAD FAC FAD 6kN Ans Problem 466 The post is made from 6061T6 aluminum and has a diameter of 50 mm It is fixed supported at A and B and at its center C there is a coiled spring attached to the rigid collar If the spring is originally uncompressed determine the reactions at A and B when the force P 40 kN is applied to the collar Given LCA 025m P 40kN d 50mm LBC 025m E 689GPa k 200 103 kN m Solution A π 4 d2 kCA E A LCA kBC E A LBC Equations of equilibrium ΣFy0 FA P Fk FB 0 1 Compatibility Considera a combined force F FBFk acted at free end B B δP δF B 0 Given 0 P kBC FB Fk kCA FB Fk kBC k 2 Also spBC Fk k FB Fk kBC k 3 Initial guess FB 1kN Fk 2kN Solving 1 and 2 FB Fk Find FB Fk FB Fk 1688 624 kN Ans From 1 FA P Fk FB FA 1688 kN Ans Problem 467 The post is made from 6061T6 aluminum and has a diameter of 50 mm It is fixed supported at A and B and at its center C there is a coiled spring attached to the rigid collar If the spring is originally uncompressed determine the compression in the spring when the load of P 50 kN is applied to the collar Given LCA 025m P 50kN d 50mm LBC 025m E 689GPa k 200 103 kN m Solution A π 4 d2 kCA E A LCA kBC E A LBC Equations of equilibrium ΣFy0 FA P Fk FB 0 1 Compatibility Considera a combined force F FBFk acted at free end B B δP δF B 0 Given 0 P kBC FB Fk kCA FB Fk kBC k 2 Also spBC Fk k FB Fk kBC k 3 Initial guess FB 1kN Fk 2kN Solving 1 and 2 FB Fk Find FB Fk FB Fk 21101 7799 kN Thus sp Fk k sp 003899 mm Ans Problem 468 The rigid bar supports the uniform distributed load of 90 kNm Determine the force in each cable if each cable has a crosssectional area of 36 mm2 and E 200 GPa Given a 1m b 2m A 36mm2 E 200GPa w 90 kN m Solution c a2 b2 v b c h a c LAC 4a2 b2 L 3a LBC c LDC c Equations of equilibrium ΣΜA0 TBC v a TDC v L w L 05L 0 1 Compatibility In triangle ABC LBC 2 a2 LAC 2 2 a LAC cos θA In triangle ADC LDC 2 L2 LAC 2 2 L LAC cos θA Thus eliminating cos θA LDC 2 L LBC 2 a L a LAC 2 1 L 1 a LDC 2 3LBC 2 6a2 2LAC 2 3LBC 2 LDC 2 2a2 2b2 3LBC 2 LDC 2 2c2 But LBC LBC BC LDC LDC DC LBC c BC LDC c DC Thus 3 c BC 2 c DC 2 2c2 Neglect squares of s since small strain occurs 3 c2 2cBC c2 2cDC 2c2 3 c2 2cBC c2 2cDC 2c2 3BC DC 0 3 TBC c E A TDC c E A 0 TDC 3TBC 2 Substituting 2 into 1 TBC v a 3TBC v L w L 05L 0 TBC 9a c 20 b w TBC 452804 kN Ans From 2 TDC 3TBC TDC 1358411 kN Ans Problem 469 The rigid bar is originally horizontal and is supported by two cables each having a crosssectional area of 36 mm2 and E 200 GPa Determine the slight rotation of the bar when the uniform load is applied Given a 1m b 2m A 36mm2 E 200GPa w 90 kN m Solution c a2 b2 v b c h a c LAC 4a2 b2 L 3a LBC c LDC c Equations of equilibrium ΣΜA0 TBC v a TDC v L w L 05L 0 1 Compatibility See solution of Prob 468 TDC 3TBC 2 Solving 1 and 2 TDC 271682 kN DC TDC c E A DC 84374919 mm Geometry tan θA b 2a θA atan b 2a θA 45000 deg In triangle ADC LDC 2 L2 LAC 2 2 L LAC cos θA LDC DC 2 L2 LAC 2 2 L LAC cos θA θA acos L2 LAC 2 LDC DC 2 2 L LAC θA 45180 deg θ θA θA θ 0180 deg Ans Problem 470 The electrical switch closes when the linkage rods CD and AB heat up causing the rigid arm BDE both to translate and rotate until contact is made at F Originally BDE is vertical and the temperature is 20C If AB is made of bronze C86100 and CD is made of aluminum 6061T6 determine the gap s required so that the switch will close when the temperature becomes 110C Unit used C deg Given LBD 400mm LDE 200mm L 300mm T1 20C T2 110C αcu 170 10 6 1 C αst 240 10 6 1 C Solution T T2 T1 T 90C Thermal Expansion δAB αcu T L δAB 04590 mm δCD αst T L δCD 06480 mm Geomotry θ δCD δAB LBD s δAB θ LBD LDE s 07425 mm Ans Problem 471 A steel surveyors tape is to be used to measure the length of a line The tape has a rectangular cross section of 125 mm by 5 mm and a length of 30 m when T1 20C and the tension or pull on the tape is 100 N Determine the true length of the line if the tape shows the reading to be 139 m when used with a pull of 175 N at T2 40C The ground on which it is placed is flat αst 17106C Est 200 GPa Unit used C deg Given a 5mm b 125mm T1 20C P1 100N L1 30m T2 40C P2 175N L2 139m αst 17 10 6 1 C Est 200GPa Solution A a b T T2 T1 T 20C Thermal Expansion δT αst T L2 δT 472600 mm δT 00473 m δP P2 P1 L2 A Est δP 83400 mm δP 00083 m L L2 δT δP L 139056 m Ans Problem 472 The assembly has the diameters and material makeup indicated If it fits securely between its fixed supports when the temperature is T1 20C determine the average normal stress in each material when the temperature reaches T2 40C Unit used C deg Given T1 20C T2 40C L1 12m d1 300mm L2 18m d2 200mm L3 09m d3 100mm α1 23 10 6 1 C α2 17 10 6 1 C α3 17 10 6 1 C E1 731GPa E2 103GPa E3 193GPa Solution T T2 T1 T 20C A1 π 4 d1 2 A2 π 4 d2 2 A3 π 4 d3 2 Equations of equilibrium ΣFx0 FA FD 0 FA FD Let FAF Then FA FAB FBC FCD FD F Thermal Expansion δT α1 T L1 α2 T L2 α3 T L3 δT 1470000 mm Elongation negative sign indicates shortening δF F L1 A1 E1 F L2 A2 E2 F L3 A3 E3 Compatibility 0 δT δF Given 0 δT F L1 A1 E1 F L2 A2 E2 F L3 A3 E3 1 Initial guess F 1kN Solving 1 F Find F F 106349 kN Average Normal Stress σal F A1 σal 1505 MPa Ans σbr F A2 σbr 3385 MPa Ans σst F A3 σst 13541 MPa Ans Problem 473 A highstrength concrete driveway slab has a length of 6 m when its temperature is 10C If there is a gap of 3 mm on one side before it touches its fixed abutment determine the temperature required to close the gap What is the compressive stress in the concrete if the temperature becomes 60C Unit used C deg Given α 11 10 6 1 C L 6m T1 10C gap 3mm T2 60C E 29GPa Solution Require gap α T T1 L T1 gap α L T1 T1 5545 C Ans For T T2 T1 T 50C gap δT δF gap α T L F L A E F A α T E gap E L σ F A σ α T E gap E L σ 145 MPa Ans Problem 474 A 18mlong steam pipe is made of steel with σY 280 MPa It is connected directly to two turbines A and B as shownThe pipe has an outer diameter of 100 mm and a wall thickness of 6 mmThe connection was made at T1 20C If the turbines points of attachment are assumed rigid determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 135C Unit used C deg Given L 18m T1 20C T2 135C σY 280MPa do 100mm t 6mm E 200GPa α 12 10 6 1 C Solution di do 2t A π 4 do 2 di 2 T T2 T1 T 115C Compatibility 0 δT δF 0 α T L F L E A F α T E A F 48903 kN Ans Check stress σ F A σ 27600 MPa σY 280 MPa ok Problem 475 A 18mlong steam pipe is made of steel with σY 280 MPa It is connected directly to two turbines A and B as shownThe pipe has an outer diameter of 100 mm and a wall thickness of 6 mm The connection was made at T1 20C If the turbines points of attachment are assumed to have a stiffness of k 16 MNm determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 135C Unit used C deg Given L 18m T1 20C T2 135C σY 280MPa do 100mm t 6mm E 200GPa α 12 10 6 1 C k 16 103 kN mm Solution di do 2t A π 4 do 2 di 2 T T2 T1 T 115C Compatibility 2x δT δF 2x α T L k x L E A x α T E A L k L 2E A x 0029830 mm F k x F 47729 kN Ans Check stress σ F A σ 26937 MPa σY 280 MPa ok Problem 476 The 12mlong A36 steel rails on a train track are laid with a small gap δ between them to allow forthermal expansion Determine the required gap so that the rails just touch one another when the temperature is increased from T1 30C to T2 30C Using this gap what would be the axial force in the rails if the temperature were to rise to T3 40C The crosssectional area of each rail is 3200 mm2 Unit used C deg Given L 12m A 3200mm2 T1 30 C T2 30C T3 40C E 200GPa α 12 10 6 1 C Solution Require gap α T2 T1 L gap 8640 mm Ans For T T3 T1 T 70C Compatibility gap δT δF gap α T L F L E A F α T L gap E A L F 7680 kN Ans Problem 477 The two circular rod segments one of aluminum and the other of copper are fixed to the rigid walls such that there is a gap of 02 mm between them when T1 15C What larger temperature T2 is required in order to just close the gap Each rod has a diameter of 30 mm αal 24106C Eal 70 GPa αcu 17106C Ecu 126 GPa Determine the average normal stress in each rod if T2 95C Unit used C deg Given Lcu 100mm Lal 200mm d 30mm T1 15C T2 95C gap 02mm Ecu 126GPa αcu 17 10 6 1 C Eal 70GPa αal 24 10 6 1 C Solution A π 4 d2 Require gap αcu T T1 Lcu αal T T1 Lal T T1 gap αcu Lcu αal Lal T 4577 C Ans For T T2 T1 T 80C Compatibility δcu δ1T δ1F δal δ2T δ2F gap δcu δal gap αcu T Lcu F Lcu Ecu A αal T Lal F Lal Eal A F αcu T Lcu αal T Lal gap Lcu Ecu A Lal Eal A F 61958 kN Ans Average Normal Stress σ F A σ 8765 MPa Ans Problem 478 The two circular rod segments one of aluminum and the other of copper are fixed to the rigid walls such that there is a gap of 02 mm between them when T1 15C Each rod has a diameter of 30 mm αal 24106C Eal 70 GPa αcu 17106C Ecu 126 GPa Determine the average normal stress in each rod if T2 150C and also calculate the new length of the aluminum segment Unit used C deg Given Lcu 100mm Lal 200mm d 30mm T1 15C T2 150C gap 02mm Ecu 126GPa αcu 17 10 6 1 C Eal 70GPa αal 24 10 6 1 C Solution A π 4 d2 For T T2 T1 T 135C Compatibility δcu δ1T δ1F δal δ2T δ2F gap δcu δal gap αcu T Lcu F Lcu Ecu A αal T Lal F Lal Eal A F αcu T Lcu αal T Lal gap Lcu Ecu A Lal Eal A F 131176 kN Average Normal Stress σ F A σ 18558 MPa Ans Displacement δal αal T Lal F Lal Eal A δal 0117783 mm Lal Lal δal Lal 200117783 mm Ans Problem 479 Two bars each made of a different material are connected and placed between two walls when the temperature is T1 10C Determine the force exerted on the rigid supports when the temperature becomes T2 20C The material properties and crosssectional area of each bar are given in the figure Unit used C deg Given T1 10C T2 20C L 300mm Ast 200mm2 Abr 450mm2 αst 12 10 6 1 C αbr 21 10 6 1 C Est 200GPa Ebr 100GPa Solution T T2 T1 T 10C Equations of equilibrium ΣFx0 FA FC 0 FA FC Let FAF Then FA FAB FBC FC F Compatibility 0 δT δF 0 αst T L αbr T L F L Ast Est F L Abr Ebr F αst αbr T 1 Ast Est 1 Abr Ebr F 6988 kN Ans Problem 480 The center rod CD of the assembly is heated from T1 30C to T2 180C using electrical resistance heating At the lower temperature T1 the gap between C and the rigid bar is 07 mm Determine the force in rods AB and EF caused by the increase in temperature Rods AB and EF are made of steel and each has a crosssectional area of 125 mm2 CD is made of aluminum and has a crosssectional area of and 375 mm2 Est 200 GPa Eal 70 GPa αst 12106C αal 23106C Unit used C deg Given T1 30C T2 180C gap 07mm Ast 125mm2 Aal 375mm2 αst 12 10 6 1 C αal 23 10 6 1 C Est 200GPa Eal 70GPa Lst 300mm Lal 240mm Solution T T2 T1 T 150C Equations of equilibrium ΣΜC0 FAB b FEF b 0 FAB FEF Let FAB Fst Then FAB FEF Fst ΣFy0 FAB FEF Fal 0 Fal 2Fst 1 Compatibility δst δal gap Fst Lst Est Ast αal T Lal Fal Lal Eal Aal gap 2 Substituting 1 into 2 Fst αal T Lal gap Lst Ast Est 2Lal Aal Eal Fst 4226 kN Ans From 1 Fal 2Fst Fal 8453 kN Ans Problem 481 The center rod CD of the assembly is heated from T1 30C to T2 180C using electrical resistance heating Also the two end rods AB and EF are heated from T1 30C to T2 50C At the lower temperature T1 the gap between C and the rigid bar is 07 mm Determine the force in rods AB and EF caused by the increase in temperature Rods AB and EF are made of steel and each has a crosssectional area of 125 mm2 CD is made of aluminum and has a crosssectional area of and 375 mm2 Est 200 GPa Eal 70 GPa αst 12106C αal 23106C Unit used C deg Given T1 30C T2 180C T2 50C Ast 125mm2 Aal 375mm2 αst 12 10 6 1 C αal 23 10 6 1 C Est 200GPa Eal 70GPa gap 07mm Lst 300mm Lal 240mm Solution T T2 T1 T 150C T T2 T1 T 20C Equations of equilibrium ΣΜC0 FAB b FEF b 0 FAB FEF Let FAB Fst Then FAB FEF Fst ΣFy0 FAB FEF Fal 0 Fal 2Fst 1 Compatibility δst δal gap αst T Lst Fst Lst Est Ast αal T Lal Fal Lal Eal Aal gap 2 Substituting 1 into 2 Fst αal T Lal gap αst T Lst Lst Ast Est 2Lal Aal Eal Fst 1849 kN Ans From 1 Fal 2Fst Fal 3698 kN Ans Problem 482 The pipe is made of A36 steel and is connected to the collars at A and BWhen the temperature is 15 C there is no axial load in the pipe If hot gas traveling through the pipe causes its temperature to rise by T 20 30xC where x is in meter determine the average normal stress in the pipe The inner diameter is 50 mm the wall thickness is 4 mm Unit used C deg Given L 24m di 50m T1 15C t 4mm T 20 30x C E 200GPa α 12 10 6 1 C Solution do di 2t A π 4 do 2 di 2 Compatibility 0 δT δF 0 0 L x α T d F L E A unit 1m C Lx L m F α E A L unit 0 Lx x 20 30x d F 84452766 kN Average Normal Stress σ F A σ 13440 MPa Ans Problem 483 The bronze 86100 pipe has an inner radius of 125 mm and a wall thickness of 5 mm If the gas flowing through it changes the temperature of the pipe uniformly from TA 60C at A to TB 15C at B determine the axial force it exerts on the walls The pipe was fitted between the walls when T 15C Unit used C deg Given L 24m ri 125mm TA 60C t 5mm TB 15C To 15C E 103GPa α 17 10 6 1 C Solution ro ri t A π ro 2 ri 2 Tx 60 TB TA L x C Tx 60 1875x C T Tx To T 45 1875x C Compatibility 0 δT δF 0 0 L x α T d F L E A unit 1m C Lx L m F α E A L unit 0 Lx x 45 1875x d F 18566 kN Ans Problem 484 The rigid block has a weight of 400 kN and is to be supported by posts A and B which are made of A36 steel and the post C which is made of C83400 red brass If all the posts have the same original length before they are loaded determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 10C Each post has a crosssectional area of 5000 mm2 Unit used C deg Given T 10C A 5000mm2 b 1m Est 200GPa Ebr 101GPa W 400kN αbr 18 10 6 1 C Solution Set L 1m Equations of equilibrium ΣΜC0 FA b FB b 0 FA FB Let FA Fst Then FA FB Fst ΣFy0 FA FB Fbr W 0 Given Fbr W 2Fst 1 Compatibility δst δbr Fst L Est A Fbr L Ebr A αbr T L 2 Initial guess Fst 1kN Fbr 2kN Solving 1 and 2 Fst Fbr Find Fst Fbr Fst Fbr 123393 153214 kN Ans Average Normal Stress σst Fst A σst 2468 MPa Ans σbr Fbr A σbr 3064 MPa Ans Problem 485 The bar has a crosssectional area A length L modulus of elasticity E and coefficient of thermal expansion α The temperature of the bar changes uniformly along its length from TA at A to TB at B so that at any point x along the bar T TA x TB TAL Determine the force the bar exerts on the rigid walls Initially no axial force is in the bar Solution Compatibility 0 δT δF 1 Tx TA TB TA L x T Tx TA T TB TA x L d T TB TA L x However d δT α T dx d δT α TB TA x L dx δT 0 L x α TB TA x L d δT α L 2 TB TA From 1 0 α L 2 TB TA F L E A F α E A 2 TB TA Ans Problem 486 The rod is made of A36 steel and has a diameter of 6 mm If the springs are compressed 12 mm when the temperature of the rod is T 10C determine the force in the rod when its temperature is T 75C Unit used C deg Given L 12m d 6mm k 200 N mm T1 10C T2 75C xo 12mm E 200GPa α 12 10 6 1 C Solution A π 4 d2 For T T2 T1 T 65C F k x xo Compatibility 2x δT δF 2x α T L F L E A x α T L 2 k x xo L 2E A x α T E A L k xo L 2E A k L x 020892 mm F k x xo F 2442 kN Ans Problem 487 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P 8 kN Given P 8kN d 20mm w 40mm h 20mm t 5mm r 10mm Solution For the fillet A h t w h 2 r h 05 From Fig 424 K 14 σmax K σavg σmax K P A σmax 112MPa Ans For the hole Ao w d t ro d 2 ro w 025 From Fig 425 Ko 2375 σmax K σavg σmax Ko P Ao σmax 190MPa Ans Problem 488 If the allowable normal stress for the bar is σallow 120 MPa determine the maximum axial force P that can be applied to the bar Given d 20mm t 5mm r 10mm w 40mm h 20mm σallow 120MPa Solution Assume failure of the fillet A h t w h 2 r h 05 From Fig 424 K 14 σallow σmax σmax K σavg σallow K P A P σallow A K P 857 kN Assume failure of the hole Ao w d t ro d 2 ro w 025 From Fig 425 Ko 2375 σallow σmax σmax K σavg σmax Ko P Ao P σallow Ao Ko P 505 kN Ans Controls Problem 489 The steel bar has the dimensions shown Determine the maximum axial force P that can be applied so as not to exceed an allowable tensile stress of σallow 150 MPa Given d 24mm t 20mm r 15mm w 60mm h 30mm σallow 150MPa Solution Assume failure occurs at the fillet A h t w h 2 r h 05 From Fig 424 K 14 σallow σmax σmax K σavg σallow K P A P σallow A K P 6429 kN Assume failure occrs at the hole Ao w d t ro d 2 ro w 02 From Fig 425 Ko 245 σallow σmax σmax K σavg σmax Ko P Ao P σallow Ao Ko P 4408 kN Ans Controls Problem 490 Determine the maximum axial force P that can be applied to the bar The bar is made from steel and has an allowable stress of σallow 147 MPa Given d 15mm t 4mm r 5mm w 375mm h 25mm σallow 147MPa Solution Assume failure of the fillet A h t w h 15 r h 02 From Fig 424 K 173 σallow σmax σmax K σavg σallow K P A P σallow A K P 8497 kN Assume failure of the hole Ao w d t ro d 2 ro w 02 From Fig 425 Ko 245 σallow σmax σmax K σavg σmax Ko P Ao P σallow Ao Ko P 54 kN Ans Controls Problem 491 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P 8 kN Given d 15mm t 4mm r 5mm w 375mm h 25mm P 8kN Solution For the fillet A h t w h 15 r h 02 From Fig 424 K 173 σmax K σavg σmax K P A σmax 1384 MPa For the hole Ao w d t ro d 2 ro w 020 From Fig 425 Ko 245 σmax K σavg σmax Ko P Ao σmax 21778 MPa Ans Controls Problem 492 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P 8 kN Given P 8kN d 12mm w 60mm h 30mm t 5mm r 15mm Solution At the fillet A h t w h 2 r h 05 From Fig 424 K 14 σmax K σavg σmax K P A σmax 7467 MPa At the hole Ao w d t ro d 2 ro w 01 From Fig 425 Ko 265 σmax K σavg σmax Ko P Ao σmax 8833 MPa Ans Controls Problem 493 The resulting stress distribution along section AB for the bar is shown From this distribution determine the approximate resultant axial force P applied to the bar Also what is the stressconcentration factor for this geometry Given h 80mm t 10mm u 5MPa v 20mm Solution P A σ d P Volume under curve Number of sqaures n 19 P n u v t P 1900 kN Ans σavg P h t σavg 2375 MPa Ans From the Figure σmax 30MPa K σmax σavg K 126 Ans Problem 494 The resulting stress distribution along section AB for the bar is shown From this distribution determine the approximate resultant axial force P applied to the barAlso what is the stressconcentration factor for this geometry Given h 80mm t 20mm u 18MPa v 20mm P A σ d P Volume under curve Number of sqaures n 10 P n u v t P 7200 kN Ans σavg P h t σavg 4500 MPa Ans From the Figure σmax 72MPa K σmax σavg K 160 Ans Solution Problem 495 The A36 steel plate has a thickness of 12 mm If there are shoulder fillets at B and C and σallow 150 MPa determine the maximum axial load P that it can support Compute its elongation neglecting the effect of the fillets Given t 12mm r 30mm w 120mm h 60mm L1 200mm L2 800mm σallow 150MPa E 200GPa Solution A1 h t A2 w t Maximum Normal Stress at fillet w h 2 r h 05 From Fig 423 K 14 σallow σmax σmax K σavg σallow K P A1 P σallow A1 K P 7714 kN Ans Displacement δ δ1 δ2 δ1 δ 2 P L1 E A1 P L2 E A2 δ 0429 mm Ans Problem 496 The 1500kN weight is slowly set on the top of a post made of 2014T6 aluminum with an A36 steel core If both materials can be considered elastic perfectly plastic determine the stress in each material Given ro 50mm Est 200 GPa ri 25mm Eal 731 GPa rs 25mm σYst 250MPa P 1500kN σYal 414MPa Solution Ast π rs 2 Aal π ro 2 ri 2 Equations of equilibrium Given ΣFy0 Pal Pst P 0 1 Compatibility δst δal Pal L Aal Eal Pst L Ast Est Pal Aal Eal Pst Ast Est 2 Initial guess Pal 1kN Pst 2kN Solving 1 and 2 Pal Pst Find Pal Pst Pal Pst 7845218 7154782 kN Average Normal stress σal Pal Aal σal 13318 MPa σYal 414MPa ok σst Pst Ast σst 36439 MPa σYst 250MPa Thererfore the steel core yields and so the elastic analysis is invalid Plastic Analysis The stress in the steel is σst σYst σst 250MPa Ans Pst σYst Ast Pst 49087 kN From 1 Pal P Pst Pal 100913 kN σal Pal Aal σal 17131 MPa σYal 414MPa ok Ans Problem 497 The 10mmdiameter shank of the steel bolt has a bronze sleeve bonded to it The outer diameter of this sleeve is 20 mm If the yield stress for the steel is σYst 640 MPa and for the bronze σYbr 520 MPa determine the magnitude of the largest elastic load P that can be applied to the assembly Est 200 GPa Ebr 100 GPa Given do 20mm Est 200GPa σYst 640MPa di 10mm Ebr 100GPa σYbr 520MPa ds 10mm Solution Ast π 4 ds 2 Abr π 4 do 2 di 2 Equations of equilibrium ΣFy0 Pbr Pst P 0 1 Compatibility δb δs Pbr L Abr Ebr Pst L Ast Est Pbr Abr Ebr Pst Ast Est 2 Assume yielding of bolt then Pst σYst Ast Pst 50265 kN From 2 Pbr Abr Ebr Ast Est Pst Pbr 75398 kN From 1 P Pbr Pst P 12566 kN Controls Ans Assume yielding of sleeve then Pbr σYbr Abr Pbr 122522 kN From 2 Pst Ast Est Abr Ebr Pbr Pst 81681 kN From 1 P Pbr Pst P 20420 kN Problem 498 The weight is suspended from steel and aluminum wires each having the same initial length of 3 m and crosssectional area of 4 mm2 If the materials can be assumed to be elastic perfectly plastic with σYst 120 MPa and σYal 70 MPa determine the force in each wire if the weight is a 600 N and b 720 N Eal 70 GPa Est 200 GPa Given L 3m Est 200GPa σYst 120MPa A 4mm2 Eal 70GPa σYal 70MPa Solution a W 600N Equations of equilibrium Given ΣFy0 Pal Pst W 0 1 Compatibility δst δal Pal L A Eal Pst L A Est Pal A Eal Pst A Est 2 Initial guess Pal 1N Pst 2N Solving 1 and 2 Pal Pst Find Pal Pst Pal Pst 15556 44444 N Ans Average Normal stress σal Pal A σal 3889 MPa σYal 70MPa ok σst Pst A σst 11111 MPa σYst 120MPa ok The average normal stress for both wires do not exceed their respective yield stress Thererfore the elastic analysis is valid for both wires Solution b W 720N Equations of equilibrium Given ΣFy0 Pal Pst W 0 1 Compatibility δst δal Pal L A Eal Pst L A Est Pal A Eal Pst A Est 2 Initial guess Pal 1N Pst 2N Solving 1 and 2 Pal Pst Find Pal Pst Pal Pst 18667 53333 N Average Normal stress σal Pal A σal 4667 MPa σYal 70MPa ok σst Pst A σst 13333 MPa σYst 120MPa Thererfore the steel wire yields and so the elastic analysis is invalid Plastic Analysis The stress in the steel is σst σYst σst 120MPa Pst σYst A Pst 48000 N Ans From 1 Pal W Pst Pal 24000 N Ans σal Pal A σal 60MPa σYal 70MPa ok Problem 499 The bar has a crosssectional area of 625 mm2 If a force of P 225 kN is applied at B and then removed determine the residual stress in sections AB and BC σY 210 MPa Given L 1m LAB 075L LBC 025L A 625mm2 σY 210MPa P 225kN Solution By superposition C δC 0 P LAB A E FC L A E 0 FC P LAB L FC 16875 kN Equations of equilibrium ΣFy0 FA FC P 0 FA P FC FA 5625 kN 1 Average Normal stress σAB FA A σAB 90MPa σY 210MPa ok σBC FC A σBC 270MPa σY 210MPa Thererfore the segment BC yields and so the elastic analysis is invalid Plastic Analysis The stress in the BC is σBC σY σBC 210MPa FC σY A FC 13125 kN From 1 FA P FC FA 9375 kN σAB FA A σAB 150MPa σY 210MPa ok A reversal of force of 45kip applied results in a reversed FC270 MPa and FA90 MPa which produce σAB 90 MPa T σBC 270 MPa C Hence σAB σAB σAB σAB 60MPa T Ans σBC σBC σBC σBC 60MPa T Ans Problem 4100 The bar has a crosssectional area of 300 mm2 and is made of a material that has a stressstrain diagram that can be approximated by the two line segments shown Determine the elongation of the bar due to the applied loading Given LAB 15m LBC 06m A 300mm2 PB 40kN PC 25kN σ1 140MPa σ2 280MPa ε2 0021 mm mm ε1 0001 mm mm Solution PBC PC PAB PC PB Average Normal stress and Strain For segment BC σBC PBC A σBC 8333 MPa 140MPa εBC σBC σ1 ε1 εBC 000060 mm mm Average Normal stress and Strain For segment AB σAB PAB A σAB 21667 MPa 140MPa εAB ε1 σAB σ1 σ2 σ1 ε2 ε1 εAB 001195 mm mm Elongation δAB εAB LAB δAB 1792857 mm δBC εBC LBC δBC 035714 mm δTotal δAB δBC δTotal 18286 mm Ans Problem 4101 The rigid bar is supported by a pin at A and two steel wires each having a diameter of 4 mm If the yield stress for the wires is σY 530 MPa and Est 200 GPa determine the intensity of the distributed load w that can be placed on the beam and will just cause wire EB to yield What is the displacement of point G for this case For the calculation assume that the steel is elastic perfectly plastic Given a 400mm b 250mm c 150mm d 4mm Lwire 800mm Est 200GPa σY 530MPa Solution L a b c A π 4 d2 Plastic Analysis Wire CD will yield first followed by wire BE When both wores yield FBE σY A FBE 6660 kN FCD σY A FCD 6660 kN Equations of equilibrium ΣΜA0 FBE a FCD a b w L 05L 0 w FBE 2a L2 FCD 2 a b L2 w 2185 kN m Ans Displacement When wire BE achieves yield stress the corresponding yield strain is εY σY Est εY 0002650 mm mm δBE εY Lwire δBE 2120 mm Geometry δBE a δG L δG L a δBE δG 424 mm Ans Problem 4102 The rigid bar is supported by a pin at A and two steel wires each having a diameter of 4 mm If the yield stress for the wires is σY 530 MPa and Est 200 GPa determine a the intensity of the distributed load w that can be placed on the beam that will cause only one of the wires to start to yield and b the smallest intensity of the distributed load that will cause both wires to yield For the calculation assume that the steel is elastic perfectly plastic Given a 400mm b 250mm c 150mm d 4mm Lwire 800mm Est 200GPa σY 530MPa Solution L a b c A π 4 d2 Equations of equilibrium ΣΜA0 FBE a FCD a b w L 05L 0 1 a By observation wire CD will yield first Then FCD σY A FCD 6660 kN Geometry δBE a δCD a b δBE a a b δCD FBE Lwire Est A a a b FCD Lwire Est A FBE a a b FCD FBE 40986 kN From 1 w FBE 2a L2 FCD 2 a b L2 w 1865 kN m Ans b When both wires yield FBE σY A FBE 6660 kN FCD σY A FCD 6660 kN From 1 w FBE 2a L2 FCD 2 a b L2 w 2185 kN m Ans Problem 4103 The rigid beam is supported by the three posts AB and C of equal length Posts A and C have a diameter of 75 mm and are made of aluminum for which Eal 70 GPa and σYal 20 MPa Post B has a diameter of 20 mm and is made of brass for which Ebr 100 GPa and σYbr 590 MPa Determine the smallest magnitude of P so that a only rods A and C yield and b all the posts yield Given dA 75mm dB 20mm dC 75mm Eal 70GPa σYal 20MPa L 4m Ebr 100GPa σYbr 590MPa Solution AA π 4 dA 2 AB π 4 dB 2 AC π 4 dC 2 Equations of equilibrium ΣΜB0 FC L P 05L FA L P 05L 0 FA FC Let FA Fal Then FA FC Fal ΣFy0 FA FC Fbr 2P 0 2Fal Fbr 2P 0 1 a Post A and C will yield Fal σYal AA Fal 88357 kN εal σYal Eal εal 00002857 mm mm Compatibility condition δal δbr Fal Lpost Eal AA Fbr Lpost Ebr AB Fbr Ebr AB Eal AA Fal Fbr 8976 kN σbr Fbr AB σbr 2857 MPa σY ok From 1 P Fal 05Fbr P 9285 kN Ans b All the posts yield Then Fal σYal AA Fal 88357 kN Fbr σYbr AB Fbr 185354 kN From 1 P Fal 05Fbr P 1810 kN Ans Problem 4104 The rigid beam is supported by the three posts AB and C Posts A and C have a diameter of 60 mm and are made of aluminum for which Eal 70 GPa and σYal 20 MPa Post B is made of brass for which Ebr 100 GPa and σYbr 590 MPa If P 130 kN determine the largest diameter of post B so that all the posts yield at the same time Given dA 60mm dC 60mm P 130kN Eal 70GPa σYal 20MPa L 4m Ebr 100GPa σYbr 590MPa Solution AA π 4 dA 2 AC π 4 dC 2 Equations of equilibrium ΣΜB0 FC L P 05L FA L P 05L 0 FA FC Let FA Fal Then FA FC Fal ΣFy0 FA FC Fbr 2P 0 2Fal Fbr 2P 0 1 When all the posts yield Fal σYal AA Fal 56549 kN From 1 Fbr 2P 2Fal Fbr 14690 kN Also Fbr σYbr AB AB π 4 dB 2 Fbr σYbr π 4 dB 2 dB 4Fbr π σYbr dB 17805 mm Ans Problem 4105 The rigid beam is supported by three A36 steel wires each having a length of 12 m The cross sectional area of AB and EF is 10 mm2 and the crosssectional area of CD is 4 mm2 Determine the largest distributed load w that can be supported by the beam before any of the wires begin to yield If the steel is assumed to be elastic perfectly plastic determine how far the beam is displaced downward just before all the wires begin to yield Given L 12m b 15m AAB 10mm2 AEF 10mm2 ACD 4mm2 E 200 GPa σY 250MPa Solution Compatibility Beam remains horizontal after the displacement since the loading and the system are symmetrical δAB δCD FAB L E AAB FCD L E ACD FCD ACD AAB FAB 1 Equations of equilibrium ΣΜC0 FEF b FAB b 0 FAB FEF ΣFy0 FAB FCD FEF w 2b 0 2 FAB w 2b FCD 2 Plastic Analysis Assume wires AB and EF yield FAB σY AAB FAB 2500 kN From 1 FCD ACD AAB FAB FCD 1000 kN From 2 w FAB b FCD 2 b w 20000 kN m Plastic Analysis Assume failure of CD FCD σY ACD FCD 1000 kN From 1 FAB AAB ACD FCD FAB 2500 kN From 2 w FAB b FCD 2 b w 20000 kN m The three wires AB CD and EF yield simultaneously Hence w 200 kN m Ans Displacement δ FCD L E ACD δ 1500 mm Ans Problem 4106 A material has a stressstrain diagram that can be described by the curve σ cε 12 Determine the deflection δ of the end of a rod made from this material if it has a length L crosssectional area A and a specific weight γ Solution Stressstrain relationship σ c ε σ2 c2 ε However σ P A and ε dδ dx Thus P2 A2 c2 dδ dx dδ dx P2 c2 A2 Since P γ A x dδ dx γ2 c2 x2 Displacement δ 0 L x γ2 c2 x2 d δ γ2 c2 0 L x x2 d δ γ2 L3 3 c2 Ans Problem 4107 Solve Prob 4106 if the stressstrain diagram is defined by σ cε 32 Solution Stressstrain relationship σ c ε3 σ2 c2 ε3 However σ P A and ε dδ dx Thus P2 A2 c2 dδ dx 3 dδ dx P c A 2 3 Since P γ A x dδ dx γ c 2 3 x 2 3 Displacement δ 0 L x γ c 2 3 x 2 3 d δ γ c 2 3 0 L x x 2 3 d δ 3 5 γ c 2 3 L 5 3 Ans Problem 4108 The bar having a diameter of 50 mm is fixed connected at its ends and supports the axial load P If the material is elastic perfectly plastic as shown by the stressstrain diagram determine the smallest load P needed to cause segment AC to yield If this load is released determine the permanent displacement of point C Given LAC 06m LCB 09m d 50mm σY 140MPa εY 0001 mm mm Solution A π 4 d2 When P is increased the segment AC will become plastic first then CB will become plastic Thus FA σY A FA 274889 kN FB σY A FB 274889 kN Equations of equilibrium ΣFx0 FA FB P 0 1 From the figure E σY εY E 140 103 MPa P FA FB P 54978 kN Ans The deflection of point C is δC εY LCB δC 0900 mm Consider the reverse of P on the bar FA 15FB 2 FA LAC A E FB LCB A E Equations of equilibrium ΣFx0 FA FB P 0 1 Substituting 2 into 1 FA 06P FA 32987 kN FB 04P FB 21991 kN The deflection of point C is δC FB LCB A E δC 072000 mm Hence δ δC δC δ 0180 mm Ans Problem 4109 Determine the elongation of the bar in Prob 4108 when both the load P and the supports are removed Given LAC 06m LCB 09m d 50mm σY 140MPa εY 0001 mm mm Solution A π 4 d2 L LAC LCB When P is increased the segment AC will become plastic first then CB will become plastic Thus FA σY A FA 274889 kN FB σY A FB 274889 kN Equations of equilibrium ΣFx0 FA FB P 0 1 From the figure E σY εY E 140 103 MPa P FA FB P 54978 kN Ans The deflection of point C is δC εY LCB δC 0900 mm Consider the reverse of P on the bar FA 15FB 2 FA LAC A E FB LCB A E Equations of equilibrium ΣFx0 FA FB P 0 1 Substituting 2 into 1 FA 06P FA 32987 kN FB 04P FB 21991 kN The resultant reactions are FA FA FA FA 54978 kN FB FB FB FB 54978 kN When the supports are removed the elongation will be δC FA L A E δC 0300 mm Ans Problem 4110 A 6mmdiameter steel rivet having a temperature of 800C is secured between two plates such that at this temperature it is 50 mm long and exerts a clamping force of 125 kN between the plates Determine the approximate clamping force between the plates when the rivet cools to 5C For the calculation assume that the heads of the rivet and the plates are rigid Take αst 14106C Est 200 GPa Is the result a conservative estimate of the actual answer Why or why not Unit used C deg Given T1 800C T2 5C P 1250kN L 50mm d 6mm αst 14 10 6 1 C Est 200GPa Solution T T1 T2 T 795C A π 4 d2 By superposition 0 δT δF 0 αst T L FT L A Est FT αst T A Est FT 62939 kN F P FT F 64189 kN Ans Yes Because as the rivet cools the plates and the rivet head will also deform Consequently the force FT on the rivets will not be as great Problem 4111 Determine the maximum axial force P that can be applied to the steel plate The allowable stress is σallow 150 MPa Given d 24mm t 6mm r 6mm w 120mm h 60mm σallow 150MPa Solution Assume failure of the fillet A h t w h 2 r h 01 From Fig 424 K 24 σallow σmax σmax K σavg σallow K P A P σallow A K P 225 kN Controls Ans Assume failure of the hole Ao w d t ro d 2 ro w 01 From Fig 425 Ko 265 σallow σmax σmax K σavg σmax Ko P Ao P σallow Ao Ko P 32604 kN Problem 4112 The rigid link is supported by a pin at A and two A36 steel wires each having an unstretched length of 300 mm and crosssectional area of 78 mm2 Determine the force developed in the wires when the link supports the vertical load of 175 kN Given L 300mm A 78mm2 P 175kN a 150mm b 100mm c 125mm Solution Compatibility δB b δC b c FB L b A E FC L b c A E Given FB b FC b c 1 Equations of equilibrium ΣΜA0 FC b c FB b P a 0 2 Initial guess FC 1kN FB 2kN Solving 1 and 2 FC FB Find FC FB FC FB 0974 0433 kN Ans Problem 4113 The force P is applied to the bar which is composed of an elastic perfectly plastic material Construct a graph to show how the force in each section AB and BC ordinate varies as P abscissa is increased The bar has crosssectional areas of 625 mm2 in region AB and 2500 mm2 in region BC and σY 210 MPa Given LAB 150mm AAB 625mm2 LBC 50mm ABC 2500mm2 σY 210MPa Solution Equations of equilibrium ΣFx0 P FA FC 0 1 Elastic behavior 0 C δC 0 P LAB E AAB FC LBC E ABC FC LAB E AAB 0 6P FC 05 6 FC 12 13P 2 Substituting 2 into 1 FA 1 13P 3 By comparison segment BC will yield first Hence FC σY ABC FC 525kN From 2 P 13 12 FC P 56875 kN From 3 FA 1 13 P FA 4375 kN When segment AB yields FA σY AAB FA 13125 kN FC σY ABC FC 525kN From 1 P FA FC P 65625 kN Problem 4114 The 2014T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A and B when T1 25C If the temperature becomes T2 20C and an axial force of P 80 kN is applied to the rigid collar as shown determine the reactions at A and B Unit used C deg Given LAC 125mm T1 25C d 12mm LCB 200mm T2 20 C P 80N E 731 GPa α 23 10 6 1 C Solution T T2 T1 T 45 C L LAC LCB A π 4 d2 By superposition 0 B T δB 0 P LAC A E α T L FB L A E FB P LAC L α T A E FB 8526 kN Ans Equations of equilibrium ΣFx0 FA P FB 0 FA P FB FA 8606 kN Ans Problem 4115 The 2014T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A and B when T1 40C Determine the force P that must be applied to the collar so that when T 0C the reaction at B is zero Unit used C deg Given LAC 125mm T1 40C d 12mm LCB 200mm T2 0C FB 0kN E 731 GPa α 23 10 6 1 C Solution T T2 T1 T 40 C L LAC LCB A π 4 d2 By superposition 0 B T δB 0 P LAC A E α T L FB L A E P L LAC α T A E FB P 19776 kN Ans Problem 4116 The A36 steel column having a crosssectional area of 11250 mm2 is encased in highstrength concrete as shown If an axial force of 300 kN is applied to the column determine the average compressive stress in the concrete and in the steel How far does the column shorten It has an original length of 24 m Given ac 225mm bc 400mm Ast 11250mm2 L 24m P 300kN Est 200 GPa Ec 29 GPa Solution Ac ac bc Ast Compatibility δst δconc Given Pst L Ast Est Pc L Ac Ec 1 Equations of equilibrium ΣFy0 Pst Pc P 0 2 Initial guess Pc 1kN Pst 2kN Solving 1 and 2 Pc Pst Find Pc Pst Pc Pst 151117 148883 kN Average Normal Stress σst Pst Ast σst 1323 MPa Ans σc Pc Ac σc 192 MPa Ans Displacement Either the concrete or steel can be used for the calculation δ Pst L Ast Est δ 015881 mm Ans Problem 4117 The A36 steel column is encased in highstrength concrete as shown If an axial force of 300 kN is applied to the column determine the required area of the steel so that the force is shared equally between the steel and concrete How far does the column shorten It has an original length of 24 m Given ac 225mm bc 400mm L 24m P 300kN Est 200 GPa Ec 29 GPa Pc 05P Pst 05P Solution Ac ac bc Ast Compatibility δst δconc Pst L Ast Est Pc L Ac Ec Ast Ec Ec Est ac bc Ast 1139738 mm2 Ans Displacement Either the concrete or steel can be used for the calculation δ Pst L Ast Est δ 015793 mm Ans Problem 4118 The assembly consists of a 30mmdiameter aluminum bar ABC with fixed collar at B and a 10mmdiameter steel rod CD Determine the displacement of point D when the assembly is loaded as shown Neglect the size of the collar at B and the connection at C Est 200 GPa Eal 70 GPa Given LAB 300mm dAB 30mm LBC 500mm dBC 30mm Eal 70GPa LCD 700mm dCD 10mm Est 200GPa PB 8 kN PD 20kN Solution Internal Force As shown on FBD Displacement AAB π 4 dAB 2 δAB PD PB LAB Eal AAB ABC π 4 dBC 2 δBC PD LBC Eal ABC ACD π 4 dCD 2 δCD PD LCD Est ACD δD δAB δBC δCD δD 1166 mm Ans Problem 4119 The joint is made from three A36 steel plates that are bonded together at their seams Determine the displacement of end A with respect to end B when the joint is subjected to the axial loads shown Each plate has a thickness of 5 mm Given b 100mm t 5mm LAB 600mm AAB t b LBC 200mm ABC 3t b LCD 800mm ACD 2t b P 46kN E 200GPa Solution Internal Force As shown on FBD δAD δAB δBC δCD δAD P LAB E AAB P LBC E ABC P LCD E ACD δAD 0491 mm Ans Problem 51 A shaft is made of a steel alloy having an allowable shear stress of τallow 84 MPa If the diameter of the shaft is 375 mm determine the maximum torque T that can be transmitted What would be the maximum torque T if a 25mmdiameter hole is bored through the shaft Sketch the shearstress distribution along a radial line in each case Given do 375mm di 25mm τallow 84MPa Solution c do 2 a Allowable shear srtess Aplying the torsion formula τallow T c J J π 2 do 2 4 T τallow J c T 087 kN m Ans b Allowable shear srtess Aplying the torsion formula τallow T c J J π 2 do 2 4 di 2 4 T τallow J c T 087 kN m Ans Shera stress at ρ 05 di τρ T ρ J τρ 6978 MPa Problem 52 The solid shaft of radius r is subjected to a torque T Determine the radius r of the inner core of the shaft that resists onehalf of the applied torque T2 Solve the problem two ways a by using the torsion formula b by finding the resultant of the shearstress distribution Problem 53 The solid shaft of radius r is subjected to a torque T Determine the radius r of the inner core of the shaft that resists onequarter of the applied torque T4 Solve the problem two ways a by using the torsion formula b by finding the resultant of the shearstress distribution Problem 54 The tube is subjected to a torque of 750 Nm Determine the amount of this torque that is resisted by the gray shaded section Solve the problem two ways a by using the torsion formula b by finding the resultant of the shearstress distribution Given ro 100mm ri 25mm T 750N m rc 75mm Solution c ro a Aplying the torsion formula J π 2 ro 4 ri 4 J π 2 ro 4 rc 4 τmax T c J τmax T c J τmax 0479 MPa T τmax J c T 0515 kN m Ans b Integartion Method dT ρ τ dA dA 2π ρ dρ τ ρ c τmax dT ρ ρ c τmax 2π ρ dρ dT 2π c τmax ρ3 dρ T 2π c τmax rc ro ρ ρ3 d T 0515 kN m Ans Problem 55 The solid 30mmdiameter shaft is used to transmit the torques applied to the gears Determine the absolute maximum shear stress on the shaft Given a 300mm TA 300 N m b 400mm TC 500N m c 500mm TD 200N m do 30mm TB 400 N m Solution c do 2 Internal Torque As shown in the torque diagram Allowable shear srtess Aplying the torsion formula From the torque diagram Tmax TB J π 2 do 2 4 τmax Tmax c J τmax 7545 MPa Ans x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c T1 x1 TA 1 N m T2 x2 TA TC 1 N m T3 x3 TA TC TD 1 N m 0 02 04 06 08 1 500 0 500 Distance m Torque Nm T1 x1 T2 x2 T3 x3 x1 x2 x3 Problem 56 The solid 32mmdiameter shaft is used to transmit the torques applied to the gears If it is supported by smooth bearings at A and B which do not resist torque determine the shear stress developed in the shaft at points C and D Indicate the shear stress on volume elements located at these points Given do 32mm T1 185N m T2 260 N m T3 75N m Solution c do 2 J π 2 do 2 4 TC T1 TD T1 T2 τC TC c J τC 2875 MPa Ans τD TD c J τD 1166 MPa Ans Problem 57 The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm If it is subjected to the applied torques as shown determine the absolute maximum shear stress developed in the shaft The smooth bearings at A and B do not resist torque Given do 32mm di 25mm T1 185N m T2 260 N m T3 75N m Solution c do 2 J π 2 do 2 4 di 2 4 Tmax T1 τmax Tmax c J τmax 4582 MPa Ans Problem 58 The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm If it is subjected to the applied torques as shown plot the shearstress distribution acting along a radial line lying within region EA of the shaft The smooth bearings at A and B do not resist torque Given do 32mm di 25mm T1 185N m T2 260 N m T3 75N m Solution c do 2 J π 2 do 2 4 di 2 4 TEA T1 τmax TEA c J τmax 4582 MPa Ans Shera stress at ρ 05 di τρ TEA ρ J τρ 3580 MPa Ans Problem 59 The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B The smaller pipe has an outer diameter of 1875 mm and an inner diameter of 17 mm whereas the larger pipe has an outer diameter of 25 mm and an inner diameter of 215 mm If the pipe is tightly secured to the wall at C determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench Given d1o 1875mm d1i 17mm F 75N d2o 25mm d2i 215mm aL 150mm aR 200mm Solution T F aL F aR T 2625 N m Segment AB c1 d1o 2 J1 π 2 d1o 2 4 d1i 2 4 τAB T c1 J1 τAB 6255 MPa Ans Segment BC c2 d2o 2 J2 π 2 d2o 2 4 d2i 2 4 τBC T c2 J2 τBC 1889 MPa Ans Problem 510 The link acts as part of the elevator control for a small airplane If the attached aluminum tube has an inner diameter of 25 mm and a wall thickness of 5 mm determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables Also sketch the shearstress distribution over the cross section Given di 25mm t 5mm P 600N a 75mm Solution do di 2t c do 2 J π 2 do 2 4 di 2 4 T P 2a T 9000 N m τmax T c J τmax 1445 MPa Ans Shera stress at ρ 05 di τρ T ρ J τρ 1032 MPa Ans Problem 511 The shaft consists of three concentric tubes each made from the same material and having the inner and outer radii shown If a torque of T 800 Nm is applied to the rigid disk fixed to its end determin the maximum shear stress in the shaft Given ri1 20mm ro1 25mm ri2 26mm ro2 30mm ri3 32mm ro3 38mm T 800N m L 2m Solution cmax ro3 J1 π 2 ro1 4 ri1 4 J2 π 2 ro2 4 ri2 4 J3 π 2 ro3 4 ri3 4 J J1 J2 J3 τmax T cmax J τmax 1194 MPa Ans Problem 512 The solid shaft is fixed to the support at C and subjected to the torsional loadings shown Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points Given ro 35mm ρA 35mm ρB 20mm TD 800N m TB 300 N m Solution J π 2 ro 4 TBA TD TB TAB TD τA TBA ρA J τB TAB ρB J τA 742 MPa τB 679 MPa Ans Problem 513 A steel tube having an outer diameter of 625 mm is used to transmit 3 kW when turning at 27 revmin Determine the inner diameter d of the tube to the nearest multiples of 5mm if the allowable shear stress is τallow 70 MPa Unit used rpm 2π 60 rad s Given do 625mm ω 27 rpm P 3kW τallow 70MPa Solution ω 283 rad s T P ω T 106103 N m Max stress c do 2 τallow T c J J π 2 do 2 4 di 2 4 di 2 do 2 4 2 π T c τallow 025 di 5683 mm Use di 60mm Ans Problem 514 The solid aluminum shaft has a diameter of 50 mm and an allowable shear stress of τallow 6 MPa Determine the largest torque T1 that can be applied to the shaft if it is also subjected to the other torsional loadings It is required that T1 act in the direction shown Also determine the maximum she stress within regions CD and DE Given do 50mm τallow 6MPa TA 68N m TC 49N m TD 35N m Solution J π 32 do 4 c do 2 Assume failure at region BC TBC T1 TA Aplying the torsion formula τallow TBC c J τallow T1 TA c J T1 τallow J c TA T1 21526 N m Ans Internal Torque Maximum torque occurs ithin region BC as indicated on the torque diagram Maximum shear srtesses at Other Regions TCD TA T1 TC TCD 9826 N m τmaxCD TCD c J τmaxCD 400 MPa Ans TDE TCD TD TDE 6326 N m τmaxDE TDE c J τmaxDE 258 MPa Ans Let a 1m x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a T1 x1 TA 1 N m T2 x2 TA T1 1 N m T3 x3 TA T1 TC 1 N m T4 x3 TA T1 TC TD 1 N m 0 1 2 3 4 200 100 0 100 Distance m Torque Nm T1 x1 T2 x2 T3 x3 T4 x4 x1 x2 x3 x4 Problem 515 The solid aluminum shaft has a diameter of 50 mm Determine the absolute maximum shear stress in the shaft and sketch the shearstress distribution along a radial line of the shaft where the shear stress is maximum Set T1 20 Nm Given do 50mm T1 20 N m TA 68N m TC 49N m TD 35N m Solution J π 32 do 4 c do 2 Maximum Torque Maximum torque occurs ithin region DE as indicated on the torque diagram TDE TA T1 TC TD TDE 13200 N m Tmax TDE Aplying the torsion formula τmax Tmax c J τmax 538 MPa Ans Let a 1m x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a T1 x1 TA 1 N m T2 x2 TA T1 1 N m T3 x3 TA T1 TC 1 N m T4 x3 TA T1 TC TD 1 N m 0 1 2 3 4 0 50 100 150 Distance m Torque Nm T1 x1 T2 x2 T3 x3 T4 x4 x1 x2 x3 x4 Problem 516 The motor delivers a torque of 50 Nm to the shaft AB This torque is transmitted to shaft CD using the gears at E and F Determine the equilibrium torque T on shaft CD and the maximum shear stress each shaft The bearings B C and D allow free rotation of the shafts Given dAB 30mm RE 50mm TAB 50N m dCD 35mm RF 125mm Solution Equilibrium ΣΜE0 TAB F RE 0 F TAB RE ΣΜF0 T F RF 0 T F RF T 125N m Ans Internal Torque As shown in FBD JAB π 32 dAB 4 cAB 05dAB JCD π 32 dCD 4 cCD 05dCD Maximum shear srtesses τmaxAB TAB cAB JAB τmaxAB 943 MPa Ans τmaxCD T cCD JCD τmaxCD 1485 MPa Ans Problem 517 If the applied torque on shaft CD is T 75 Nm determine the absolute maximum shear stress in eac shaft The bearings B C and D allow free rotation of the shafts and the motor holds the shafts fixed from rotating Given dEA 30mm RE 50mm TAB 50N m dCD 35mm RF 125mm T 75N m Solution Equilibrium ΣΜF0 T F RF 0 F T RF ΣΜE0 TAB F RE TA 0 TA TAB F RE TA 20N m Ans Internal Torque As shown in FBD TEA TAB TA TEA 3000 N m JEA π 32 dEA 4 cEA 05dEA JCD π 32 dCD 4 cCD 05dCD Maximum shear srtesses τmaxEA TEA cEA JEA τmaxEA 566 MPa Ans τmaxCD T cCD JCD τmaxCD 891 MPa Ans Problem 518 The copper pipe has an outer diameter of 625 mm and an inner diameter of 575 mm If it is tightly secured to the wall at C and a uniformly distributed torque is applied to it as shown determine the shear stress developed at points A and B These points lie on the pipes outer surface Sketch the shear stress on volume elements located at A and B Given do 625mm di 575mm q 625 N m m LOA 300mm LAB 225mm LBC 100mm Solution Internal Torque As shown on FBD TA q LOA TA 18750 N m TB q LOA LAB TB 32813 N m Max shear stress τ T c J c do 2 J π 2 do 2 4 di 2 4 τA TA c J τA 1379 MPa Ans τB TB c J τB 2414 MPa Ans Problem 519 The copper pipe has an outer diameter of 625 mm and an inner diameter of 575 mm If it is tightly secured to the wall at C and it is subjected to the uniformly distributed torque along its entire length determine the absolute maximum shear stress in the pipe Discuss the validity of this result Given do 625mm di 575mm q 625 N m m LOA 300mm LAB 225mm LBC 100mm Solution Internal Torque The maximum torque occurs at the support C TC q LOA LAB LBC TC 39063 N m Max shear stress τ T c J c do 2 J π 2 do 2 4 di 2 4 τC TC c J τC 2873 MPa Ans According to SaintVenants principle application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading Problem 520 The 60mmdiameter solid shaft is subjected to the distributed and concentrated torsional loadings shown Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points Given L1 03m L2 04m T1 400N m T2 600 N m do 60mm q 2 kN m m Solution Internal Torque As shown on FBD TA T1 TA 40000 N m TB T1 T2 q L2 TB 60000 N m Maximum shear stress τ T c J c do 2 J π 2 do 2 4 τA TA c J τA 943 MPa Ans τB TB c J τB 1415 MPa Ans Problem 521 The 60mm diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown Determine the absolute maximum and minimum shear stresses in the shaft and specify their locations measured from the fixed end Given L1 03m L2 04m T1 400N m T2 600 N m do 60mm q 2 kN m m Solution Internal Torque TC T1 T2 q 2L2 TC 140000 N m The maximum torque occurs at the fixed support C Tmax TC Tmax 140000 N m The minimum torque occurs in segment loaded with q Tmin 0 xo 2L2 TC q 2L2 xo TC q xo 0700 m Shear stress τ T c J c do 2 J π 32 do 4 τabsmin Tmin c J τabsmin 000 MPa Ans τabsmax Tmax c J τabsmax 3301 MPa Ans According to SaintVenants principle application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading Therefore the absolute τmax is not valid x1 0 001 2 L2 2L2 x2 2L2 101 2L2 2L2 2L1 Ta x1 TC q x1 1 N m Tb x2 TC 2q L2 T2 1 N m 0 02 04 06 08 1 12 1500 1000 500 0 Distance m Torque Nm Ta x1 Tb x2 x1 x2 Problem 522 The solid shaft is subjected to the distributed and concentrated torsional loadings shown Determine the required diameter d of the shaft if the allowable shear stress for the material is τallow 175 MPa Given L1 03m L2 04m T1 400N m T2 600 N m τallow 175MPa q 2 kN m m Solution Internal Torque TC T1 T2 q 2L2 TC 140000 N m The maximum torque occurs at the fixed support C Tmax TC Tmax 140000 N m Allowable Shear stress τ T c J c do 2 J π 32 do 4 τallow 16Tmax π do 3 do 3 16Tmax π τallow do 3441 mm Ans According to SaintVenants principle application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading Therefore the above analysis is not valid x1 0 001 2 L2 2L2 x2 2L2 101 2L2 2L2 2L1 Ta x1 TC q x1 1 N m Tb x2 TC 2q L2 T2 1 N m 0 02 04 06 08 1 12 1500 1000 500 0 Distance m Torque Nm Ta x1 Tb x2 x1 x2 Problem 523 The steel shafts are connected together using a fillet weld as shown Determine the average shear stress in the weld along section aa if the torque applied to the shafts is T 60 Nm Note The critical section where the weld fails is along section aa Given do 50mm a 12mm T 60N m θa 45deg Solution The maen radius of weld is rweld do a 2 rweld 3100 mm Shear stress V T rweld Aweld 2π rweld a sin θa τavg V Aweld τavg 117 MPa Ans Problem 524 The rod has a diameter of 12 mm and a weight of 80 Nm Determine the maximum torsional stress in the rod at a section located at A due to the rods weight Given d 12mm w 80 N m LA 03m Lx 09m Ly 09m Lz 03m Solution Equilibrium Σ Mx 0 TA w Ly 05Ly w Lz Ly 0 TA w Ly 05 Ly w Lz Ly TA 5400 N m Max shear stress τ T c J c d 2 J π 2 d 2 4 τA TA c J τA 15915 MPa Ans Problem 525 Solve Prob 524 for the maximum torsional stress at B Given d 12mm w 80 N m LA 03m Lx 09m Ly 09m Lz 03m Solution Equilibrium Σ Mx 0 TB w Ly 05Ly w Lz Ly 0 TB w Ly 05 Ly w Lz Ly TB 5400 N m Max shear stress τ T c J c d 2 J π 2 d 2 4 τB TB c J τB 15915 MPa Ans Problem 526 Consider the general problem of a circular shaft made from m segments each having a radius of cm If there are n torques on the shaft as shown write a computer program that can be used to determine the maximum shearing stress at any specified location x along the shaft Show an application of the program using the values L1 06 m c1 50 mm L2 12 m c2 25 mm T1 1200 Nm d1 0 T2 900 Nm d2 15 m Problem 527 The wooden post which is half buried in the ground is subjected to a torsional moment of 50 Nm that causes the post to rotate at constant angular velocityThis moment is resisted by a linear distribution of torque developed by soil friction which varies from zero at the ground to t0 Nmm at its base Determine the equilibrium value for t0 and then calculate the shear stress at points A and B which lie on the outer surface of the post Given do 100mm L 075m LAB 05 m T 50N m Solution Equilibrium Σ Mz 0 05to L T 0 to 2 T L to 13333 N m m Internal Torque As shown on FBD TA T TA 5000 N m TB T 05to LAB L LAB TB 2778 N m Maximum Shear Stress τ T c J c do 2 J π 32 do 4 τA TA c J τA 0255 MPa Ans τB TB c J τB 0141 MPa Ans Problem 528 A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft If the ring is held fixed and a torque T is applied to the shaft determine the maximum shear stress in the rubber Solution Shear stress τ F A F T r A 2π r h τ T 2π r2 h Shear stress is maximm when r is the smallest ie r ri Hence τmax T 2π ri 2 h Ans Problem 529 The shaft has a diameter of 80 mm and due to friction at its surface within the hole it is subjected to a variable torque described by the function t 25x ex2 Nmm where x is in meters Determine the minimum torque T0 needed to overcome friction and cause it to twist Also determine the absolute maximum stress in the shaft Given do 80mm L 2m t 25x ex2 N m m Solution unit N m Equilibrium Σ Mz 0 To unit 0 L m x 25x ex2 d To 66998 N m Ans Maximum Shear Stress τ T c J c do 2 J π 32 do 4 τabsmax To c J τabsmax 6664 MPa Ans Problem 530 The solid shaft has a linear taper from rA at one end to rB at the other Derive an equation that gives th maximum shear stress in the shaft at a location x along the shafts axis Problem 531 When drilling a well at constant angular velocity the bottom end of the drill pipe encounters a torsion resistance TA Also soil along the sides of the pipe creates a distributed frictional torque along its length varying uniformly from zero at the surface B to tA at A Determine the minimum torque TB tha must be supplied by the drive unit to overcome the resisting torques and compute the maximum shea stress in the pipe The pipe has an outer radius ro and an inner radius ri Problem 532 The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow 56 MPa If the outer diameter of the shaft is 625 mm and the engine delivers 165 kW to the shaft when is turning at 1140 revmin determine the minimum required thickness of the shafts wall Unit used rpm 2π 60 rad s Given do 625mm ω 1140 rpm P 165kW τallow 56MPa Solution ω 11938 rad s T P ω T 138214 J Max shear stress c do 2 τallow T c J J π 2 do 2 4 di 2 4 di 2 do 2 4 2 π T c τallow 025 di 5216 mm t do di 2 t 517 mm Ans Problem 533 The drive shaft AB of an automobile is to be designed as a thinwalled tube The engine delivers 125 kW when the shaft is turning at 1500 revmin Determine the minimum thickness of the shafts wall if the shafts outer diameter is 625 mmThe material has an allowable shear stress of τallow 50 MPa Unit used rpm 2π 60 rad s Given do 625mm ω 1500 rpm P 125kW τallow 50MPa Solution ω 15708 rad s T P ω T 79577 J Max shear stress c do 2 τallow T c J J π 2 do 2 4 di 2 4 di 2 do 2 4 2 π T c τallow 025 di 5650 mm t do di 2 t 2998 mm Ans Problem 534 The gear motor can develop 100 W when it turns at 300 revmin If the shaft has a diameter of 12 m determine the maximum shear stress that will be developed in the shaft Unit used rpm 2π 60 rad s Given d 12mm ω 300 rpm P 100W Solution ω 3142 rad s T P ω T 3183 N m Max shear stress c d 2 J π 2 d 2 4 τmax T c J τmax 9382 MPa Ans Problem 535 The gear motor can develop 100 W when it turns at 80 revmin If the allowable shear stress for the shaft is τallow 28 MPa determine the smallest diameter of the shaft to the nearest multiples of 5mm that can be used Unit used rpm 2π 60 rad s Given P 100W ω 80 rpm τallow 28MPa Solution ω 838 rad s T P ω T 11937 N m Max shear stress c d 2 J π 2 d 2 4 τallow T c J π 2 d 2 4 T d 2 τallow d 16T π τallow 1 3 d 1295 mm Use d 15mm Ans Problem 536 The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow 42 MPa If the outer diameter is 75 mm and the engine delivers 145 kW to the shaft when it is turning at 1250 revmin determine the minimum required thickness of the shafts wall Unit used rpm 2π 60 rad s Given do 75mm ω 1250 rpm P 145kW τallow 42MPa Solution ω 13090 rad s T P ω T 1107718 N m Max shear stress c do 2 τallow T c J J π 2 do 2 4 di 2 4 di 2 do 2 4 2 π T c τallow 025 di 6815 mm t do di 2 t 3427 mm Ans Problem 537 The 25kW reducer motor can turn at 330 revmin If the shaft has a diameter of 20 mm determine the maximum shear stress that will be developed in the shaft Unit used rpm 2π 60 rad s Given d 20mm ω 330 rpm P 25kW Solution ω 3456 rad s T P ω T 72343 N m Max shear stress c d 2 J π 2 d 2 4 τmax T c J τmax 46055 MPa Ans Problem 538 The 25kW reducer motor can turn at 330 revmin If the allowable shear stress for the shaft is τallow 56 MPa determine the smallest diameter of the shaft to the nearest multiples of 5mm that can be used Unit used rpm 2π 60 rad s Given d 20mm ω 330 rpm P 25kW τallow 56MPa Solution ω 3456 rad s T P ω T 72343 N m Max shear stress c d 2 J π 2 d 2 4 τallow T c J π 2 d 2 4 T d 2 τallow d 16T π τallow 1 3 d 1874 mm Use d 20mm Ans Problem 539 The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E It is coupled to a motor at C which delivers 3 kW of power to the shaft while it is turning at 50 revs If gears A and B remove 1 kW and 2 kW respectively determine the maximum shear stress developed i the shaft within regions AB and BC The shaft is free to turn in its support bearings D and E Unit used rpm 2π 60 rad s Given do 25mm ω 50 60 rpm PA 1 kW Po 3kW PB 2 kW Solution ω 31416 rad s TC Po ω TC 9549 N m TA PA Po TC TA 3183 N m Maximum Shear Stress c do 2 J π 2 do 2 4 τABmax TA c J τABmax 1038 MPa Ans τBCmax TC c J τBCmax 3113 MPa Ans Problem 540 A ship has a propeller drive shaft that is turning at 1500 revmin while developing 1500 kW If it is 2 m long and has a diameter of 100 mm determine the maximum shear stress in the shaft caused by torsion Unit used rpm 2π 60 rad s Given d 100mm ω 1500 rpm P 1500kW L 24m Solution ω 15708 rad s T P ω T 9549297 N m Max shear stress c d 2 J π 2 d 2 4 τmax T c J τmax 48634 MPa Ans Problem 541 The motor A develops a power of 300 W and turns its connected pulley at 90 revmin Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is τallow 85 MPa Unit used rpm 2π 60 rad s Given rA 60mm ωA 90 rpm rB 150mm P 300W τallow 85MPa Solution ωB ωA rA rB ωB 377 rad s TA P ωA TA 31831 N m TB P ωB TB 79577 N m Allowable Shear Stress For shaft A τallow TA c J c dA 2 J π 32 dA 4 τallow 16TA π dA 3 dA 3 16TA π τallow dA 1240 mm Ans For shaft B τallow TB c J c dB 2 J π 32 dB 4 τallow 16TB π dB 3 dB 3 16TB π τallow dB 1683 mm Ans Problem 542 The motor delivers 400 kW to the steel shaft AB which is tubular and has an outer diameter of 50 mm and an inner diameter of 46 mm Determine the smallest angular velocity at which it can rotate if the allowable shear stress for material is τallow 175 MPa Unit used rpm 2π 60 rad s Given do 50mm di 46mm P 400kW τallow 175MPa Solution c do 2 J π 2 do 2 4 di 2 4 τallow T c J T J τallow c T 121813 N m ω P T ω 32837 rad s ω 3135714 rpm Ans Problem 543 The motor delivers 40 kW while turning at a constant rate of 1350 rpm at A Using the belt and pulley system this loading is delivered to the steel blower shaft BC Determine to the nearest multiples of 5mm the smallest diameter of this shaft if the allowable shear stress for steel if τallow 84 MPa Unit used rpm 2π 60 rad s Given rA 100mm rB 200mm P 40kW ω 1350 rpm τallow 84MPa Solution ω 14137 rad s TA P ω TA 282942 N m TA 2rA F F TB 2rB F F TB rB rA TA TB 56588 N m Max shear stress c d 2 J π 2 d 2 4 τallow TB c J π 2 d 2 4 TB d 2 τallow d 16TB π τallow 1 3 d 3249 mm Use d 35mm Ans Problem 544 The propellers of a ship are connected to a solid A36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm If the power output is 45 MW when the shaft rotates at 20 rads determine the maximum torsional stress in the shaft and its angle of twist Given do 340mm di 260mm L 60m P 4500kW G 75GPa ω 20 rad s Solution T P ω T 225kN m Maximum Shear Stress c do 2 J π 32 do 4 di 4 τmax T c J τmax 4431 MPa Ans Angle of Twist φ T L G J φ 02085 rad φ 11946 deg Ans Problem 545 A shaft is subjected to a torque T Compare the effectiveness of using the tube shown in the figure with that of a solid section of radius c To do this compute the percent increase in torsional stress and angle of twist per unit length for the tube versus the solid section Given ro c ri 05c Solution Maximum Shear Stress For solid shaft τsmax T c J c ro Js π 2 ro 4 τsmax 2T π ro 3 For the tube τtmax T c J c ro Jt π 2 ro 4 ri 4 Jt 15π 32 ro 4 τtmax 32T 15π ro 3 increase in shear stress τ τtmax τsmax τsmax 100 τ 32 15 2 2 100 τ 667 Ans Angle of Twist For solid shaft φs T L G Js For the tube φt T L G Jt φ 1 Js 1 Jt 1 Js 100 increase in angle of twist φ φt τsmax τsmax 100 φ π 2 15π 32 15π 32 100 φ 667 Ans Problem 546 The tubular drive shaft for the propeller of a hovercraft is 6 m long If the motor delivers 4 MW of power to the shaft when the propellers rotate at 25 rads determine the required inner diameter of the shaft if the outer diameter is 250 mmWhat is the angle of twist of the shaft when it is operating Tak τallow 90 MPa and G 75 GPa Given do 250mm L 6m P 4000kW τallow 90MPa G 75GPa ω 25 rad s Solution T P ω T 160kN m Maximum Shear Stress τallow T c J c do 2 J π 32 do 4 di 4 τallow 16T do π do 4 di 4 di 4 do 4 16T do π τallow di 2013 mm Ans Angle of Twist J π 32 do 4 di 4 φ T L G J φ 005760 rad φ 330 deg Ans Problem 547 The A36 steel axle is made from tubes AB and CD and a solid section BC It is supported on smooth bearings that allow it to rotate freely If the gears fixed to its ends are subjected to 85Nm torques determine the angle of twist of gear A relative to gear D The tubes have an outer diameter of 30 mm and an inner diameter of 20 mm The solid section has a diameter of 40 mm Given do 30mm di 20mm ds 40mm LAB 04m LBC 025m LCD 04m T 85N m G 75GPa Solution TAB T TBC T TCD T Angle of Twist JAB π 32 do 4 di 4 JBC π 32 ds 4 JCD JAB φ 2 TAB LAB G JAB TBC LBC G JBC φ 001534 rad φ 0879 deg Ans Problem 548 The A36 steel axle is made from tubes AB and CD and a solid section BC It is supported on smooth bearings that allow it to rotate freely If the gears fixed to its ends are subjected to 85Nm torques determine the angle of twist of the end B of the solid section relative to end C The tubes have an oute diameter of 30 mm and an inner diameter of 20 mmThe solid section has a diameter of 40 mm Given do 30mm di 20mm ds 40mm LAB 04m LBC 025m LCD 04m T 85N m G 75GPa Solution TAB T TBC T TCD T Angle of Twist JBC π 32 ds 4 φ TBC LBC G JBC φ 0001127 rad φ 00646 deg Ans Problem 549 The hydrofoil boat has an A36 steel propeller shaft that is 30 m long It is connected to an inline diesel engine that delivers a maximum power of 2000 kW and causes the shaft to rotate at 1700 rpm If the outer diameter of the shaft is 200 mm and the wall thickness is 10 mm determine the maximum shear stress developed in the shaft Also what is the wind up or angle of twist in the shaft at full power Unit used rpm 2π 60 rad s Given do 200mm t 10mm P 2000kW ω 1700rpm G 75GPa L 30m Solution ω 17802 rad s T P ω T 11234467 N m Max shear stress di do 2t c do 2 J π 2 do 2 4 di 2 4 τmax T c J τmax 20797 MPa Ans Angle of Twist φ T L G J φ 00832 rad φ 4766 deg Ans Problem 550 The splined ends and gears attached to the A36 steel shaft are subjected to the torques shown Determine the angle of twist of gear C with respect to gear D The shaft has a diameter of 40 mm Given LAC 03m LCD 04m LDE 05m do 40mm G 75GPa TA 300 N m TC 500N m TD 200N m TE 400 N m Solution TAC TA TCD TA TC TDE TA TC TD Angle of Twist J π 32 do 4 φCD TCD LCD G J φCD 0004244 rad φCD 0243 deg Ans Problem 551 The 20mmdiameter A36 steel shaft is subjected to the torques shown Determine the angle of twist of the end B Given LAD 02m LDC 06m LCB 08m TB 80N m TC 20 N m TD 30N m do 20mm G 75GPa Solution TCB TB TDC TB TC TAD TB TC TD Angle of Twist J π 32 do 4 φ TCB LCB G J TDC LDC G J TAD LAD G J φ 0100162 rad φ 5739 deg Ans Problem 552 The 8mmdiameter A36 bolt is screwed tightly into a block at A Determine the couple forces F that should be applied to the wrench so that the maximum shear stress in the bolt becomes 18 MPa Also compute the corresponding displacement of each force F needed to cause this stress Assume that the wrench is rigid Given do 8mm L 80mm a 150mm G 75GPa τallow 18MPa Solution Allowable Shear Stress c do 2 J π 32 do 4 τallow T c J T τallow J c T 18096 N m Equilibrium T F 2a 0 F T 2a F 603 N Ans Angle of Twist φ T L G J φ 000480 rad Displacement s a φ s 0720 mm Ans Problem 553 The turbine develops 150 kW of power which is transmitted to the gears such that C receives 70 and D receives 30 If the rotation of the 100mmdiameter A36 steel shaft is ω 800 revmin determine the absolute maximum shear stress in the shaft and the angle of twist of end E of the shaft relative to B The journal bearing at E allows the shaft to turn freely about its axis Unit used rpm 2π 60 rad s Given do 100mm ω 800 rpm LBC 3m LCD 4m LDE 2m P 150kW G 75GPa TC 07T TD 03T Solution ω 8378 rad s T P ω T 1790 kN m TC 07T TC 1253 kN m TD 03T TD 0537 kN m TBC T TCD 03T TDE 0 Maximum Shear Stress Maximum torque occurs in region BC c do 2 J π 32 do 4 τmax T c J τmax 9119 MPa Ans Angle of Twist φEB TBC LBC G J TCD LCD G J TDE LDE G J φEB 0010213 rad φEB 05852 deg Ans Problem 554 The turbine develops 150 kW of power which is transmitted to the gears such that both C and D receive an equal amount If the rotation of the 100mmdiameter A36 steel shaft is ω 500 revmin determine the absolute maximum shear stress in the shaft and the rotation of end B of the shaft relativ to E The journal bearing at C allows the shaft to turn freely about its axis Unit used rpm 2π 60 rad s Given do 100mm ω 500 rpm LBC 3m LCD 4m LDE 2m P 150kW G 75GPa TC 05T TD 05T Solution ω 5236 rad s T P ω T 2865 kN m TC 05T TC 1432 kN m TD 05T TD 1432 kN m TBC T TCD 05T TDE 0 Maximum Shear Stress Maximum torque occurs in region BC c do 2 J π 32 do 4 τmax T c J τmax 1459 MPa Ans Angle of Twist φBE TBC LBC G J TCD LCD G J TDE LDE G J φBE 0019454 rad φBE 11146 deg Ans Problem 555 The motor delivers 33 kW to the 304 stainless steel shaft while it rotates at 20 Hz The shaft is supported on smooth bearings at A and B which allow free rotation of the shaft The gears C and D fixed to the shaft remove 20 kW and 12 kW respectively Determine the diameter of the shaft to the nearest mm if the allowable shear stress is τallow 56 MPa and the allowable angle of twist of C with respect to D is 020 Given LAC 250mm LCD 200mm LDB 150mm P 32kW PC 20 kW PD 12 kW f 20Hz G 75GPa φallow 020deg τallow 56MPa Solution ω 2 π f ω 1256637 rad s TAC P ω TAC 254648 N m TCD P PC ω TCD 95493 N m Tmax max TAC TCD Tmax 2546479 N m Max shear stress Assume failure due to shear stress c d 2 J π 2 d 2 4 τallow Tmax c J π 2 d 2 4 Tmax d 2 τallow d 16Tmax π τallow 1 3 d 285041 mm Angle of Twist Assume failure due to angle of twist limitation occured between C and D J π 2 d 2 4 φallow TCD LCD G J φallow 0003491 rad π 2 d 2 4 TCD LCD G φallow d 2 2TCD LCD π G φallow 025 d 293601 mm Use d 30mm Ans Problem 556 The motor delivers 33 kW to the 304 stainless steel solid shaft while it rotates at 20 Hz The shaft has a diameter of 375 mm and is supported on smooth bearings at A and B which allow free rotation of the shaft The gears C and D fixed to the shaft remove 20 kW and 12 kW respectively Determine the absolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D Given LAC 250mm LCD 200mm LDB 150mm P 32kW PC 20 kW PD 12 kW f 20Hz G 75GPa d 375mm Solution ω 2 π f ω 12566 rad s TAC P ω TAC 254648 N m TCD P PC ω TCD 95493 N m Tmax max TAC TCD Tmax 25465 N m Max shear stress Maximum shear stress occured between A and C c d 2 J π 2 d 2 4 τmax Tmax c J τmax 2459 MPa Ans Angle of Twist φCD TCD LCD G J φCD 0001312 rad φCD 0075152 deg Ans Problem 557 The motor produces a torque of T 20 Nm on gear A If gear C is suddenly locked so it does not turn yet B can freely turn determine the angle of twist of F with respect to E and F with respect to D of the L2steel shaft which has an inner diameter of 30 mm and an outer diameter of 50 mm Also calculate the absolute maximum shear stress in the shaft The shaft is supported on journal bearings at G and H Given do 50mm di 30mm rA 30mm rF 100mm G 75GPa T 20N m LEF 06m Solution Equilibrium T F rA 0 F T rA F 66667 N T F rF 0 T F rF T 6667 N m Angle of Twist J π 32 do 4 di 4 Since shaft is held fixed at C the torque is only in region EF of the shaft φFE T LEF G J φFE 9986 10 4 rad Ans Since the torque in region ED is zero φFD φFE φFD 9986 10 4 rad Ans Maximum Shear Stress c do 2 τmax T c J τmax 3121 MPa Ans Problem 558 The two shafts are made of A36 steel Each has a diameter of 25 mm and they are supported by bearings at A B and C which allow free rotation If the support at D is fixed determine the angle of twist of end B when the torques are applied to the assembly as shown Given LDH 250mm LHE 750mm LAG 200mm LGF 250mm LFB 300mm d 25mm rE 150mm rF 100mm TH 120N m TG 60N m G 75GPa Solution Internal Torque AS shown on FBD At F TFTG P TG rF P 600 N At E TE P rE TE 90 N m Angle of Twist J π 2 d 2 4 φE TH LDH G J TE LDH LHE G J φE 0020861 rad φE rE φF rF φF rE rF φE φF 0031291 rad Since there is no torque applied between F and B φB φF φB 0031291 rad Ans φB 1793 deg Problem 559 The two shafts are made of A36 steel Each has a diameter of 25 mm and they are supported by bearings at A B and C which allow free rotation If the support at D is fixed determine the angle of twist of end A when the torques are applied to the assembly as shown Given LDH 250mm LHE 750mm LAG 200mm LGF 250mm LFB 300mm d 25mm rE 150mm rF 100mm TH 120N m TG 60N m G 75GPa Solution Internal Torque AS shown on FBD At F TFTG P TG rF P 600 N At E TE P rE TE 90 N m Angle of Twist J π 2 d 2 4 φE TH LDH G J TE LDH LHE G J φE 0020861 rad φE rE φF rF φF rE rF φE φF 0031291 rad Since there is no torque applied between A and G φAF TG LGF G J φAF 0005215 rad φA φAF φF φA 0036506 rad φA 2092 deg Ans Problem 560 Consider the general problem of a circular shaft made from m segments each having a radius of cm and shearing modulus Gm If there are n torques on the shaft as shown write a computer program that can be used to determine the angle of twist of its end A Show an application of the program using the values L1 05 m c1 002 m G1 30 GPa L2 15 m c2 005 m G2 15 GPa T1 450 Nm d1 025 m T2 600 Nm d2 08 m Problem 561 The 30mmdiameter shafts are made of L2 tool steel and are supported on journal bearings that allow the shaft to rotate freely If the motor at A develops a torque of T 45 Nm on the shaft AB while the turbine at E is fixed from turning determine the amount of rotation of gears B and C Given LAB 15m LDC 05m LCE 075m TA 45N m rB 50mm rC 75mm do 30mm G 75GPa Solution TAB TA Equilibrium TAB F rB 0 F TAB rB F 90000 N T F rC 0 T F rC T 6750 N m TCE T Angle of Twist J π 32 do 4 φB TAB LAB G J φC TCE LCE G J φB 0011318 rad φC 0008488 rad φB 0648 deg Ans φC 0486 deg Ans Problem 562 The 60mmdiameter solid shaft is made of A36 steel and is subjected to the distributed and concentrated torsional loadings shown Determine the angle of twist at the free end A of the shaft due to these loadings Given LAC 06m LCB 08m do 60mm TA 400N m TC 600 N m G 75GPa q 2000 N m m Solution Internal Torque As shown in the torque diagram TAC TA TCB TA TC q x Angle of Twist J π 32 do 4 φA TAC LAC G J 1 G J 0 LCB x TCB d φA TAC LAC G J 1 G J 0 LCB x TA TC q x d φA 0007545 rad φA 0432 deg Ans Problem 563 When drilling a well the deep end of the drill pipe is assumed to encounter a torsional resistance TA Furthermore soil friction along the sides of the pipe creates a linear distribution of torque per unit length varying from zero at the surface B to t0 at A Determine the necessary torque TB that must be supplied by the drive unit to turn the pipe Also what is the relative angle of twist of one end of the pipe with respect to the other end at the instant the pipe is about to turn The pipe has an outer radius ro and an inner radius ri The shear modulus is G Problem 564 The assembly is made of A36 steel and consists of a solid rod 15 mm in diameter connected to the inside of a tube using a rigid disk at B Determine the angle of twist at A The tube has an outer diameter of 30 mm and wall thickness of 3 mm Given do 30mm t 3mm dr 15mm LAB 03m LBC 03m G 75GPa TA 50N m TB 30N m Solution di do 2t TAB TA TBC TA TB Angle of Twist JAB π 32 dr 4 JBC π 32 do 4 di 4 φA TAB LAB G JAB TBC LBC G JBC φA 004706 rad φA 2696 deg Ans Problem 565 The device serves as a compact torsional spring It is made of A36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B The ring at A can also be assumed rigid and is fixed from rotating If a torque of T 025 Nm is applied to the shaft determine the angle of twist at the end C and the maximum shear stress in the tube and shaft Given LBC 600mm LBA 300mm rti 1875mm rto 25mm r 125mm TC 025kN m G 75GPa Solution Internal Torque AS shown on FBD TBC TC Inner shaft TBA TC Outer tube Max Shear Stress Inner shaft cBC r JBC π 2 r4 τBC TBC cBC JBC τBC 8149 MPa Ans cBA rto JBA π 2 rto 4 rti 4 Outer tube τBA TBC cBA JBA τBA 149 MPa Ans Angle of Twist φB TBA LBA G JBA φB 0002384 rad φCB TBC LBC G JBC φCB 0052152 rad φC φCB φB φC 0054536 rad φC 3125 deg Ans Problem 566 The device serves as a compact torsion spring It is made of A36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B The ring at A can also be assumed rigid and is fixed from rotating If the allowable shear stress for the material is τallow 84 MPa and the angle of twist at C is limited to φallow 3 determine the maximum torque T that can be applied at the end C Given LBC 600mm LBA 300mm rti 1875mm rto 25mm τallow 84MPa r 125mm φallow 3deg G 75GPa Solution Internal Torque AS shown on FBD Inner shaft TBCTC Outer tube TBATC Allowable Shear Stress Assume failure due to shear stress Inner shaft cBC r JBC π 2 r4 TBC τallow JBC cBC TBC 25771 N m cBA rto JBA π 2 rto 4 rti 4 Outer tube TBA τallow JBA cBA TBA 140934 N m Angle of Twist Assume failure due to angle of twist limitation maximum occurred at C φC φCB φB φCB TC LBC G JBC φB TC LBA G JBA Thus φallow TC LBC G JBC TC LBA G JBA TC G φallow LBC JBC LBA JBA TC 24002 N m Ans controls Problem 567 The shaft has a radius c and is subjected to a torque per unit length of t0 which is distributed uniformly over the shafts entire length L If it is fixed at its far end A determine the angle of twist φ o end B The shear modulus is G Solution Internal Torque As shown in the torque diagram T x to x Angle of Twist J π 2 c4 φB 1 G J 0 L x T x d φB to G J 0 L x x d φB to L2 2G J φB to L2 πG c4 φB to L2 πG c4 Ans Problem 568 The A36 bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t k x2 Nm m where x is in meters If a torque of T 50 Nm is applied to the bol head determine the constant k and the amount of twist in the 50mm length of the shank Assume the shank has a constant radius of 4 mm Given ro 4mm L 50mm TA 50N m G 75GPa t k x2 N m m Solution c ro Internal Torque As shown in the torque diagram t x k x2 Equilibrium TA 0 L x t x d 0 TA 0 L x k x2 d 0 TA k L3 3 0 k 3TA L3 k 1200 MPa Ans Hence T x TA x k x2 d T x TA k x3 3 Angle of Twist J π 2 c4 φ 1 G J 0 L x T x d φ 1 G J 0 L x TA k x3 3 d φ 006217 rad φ 3562 deg Ans Problem 569 Solve Prob 568 if the distributed torque is t k x23 Nm m Given ro 4mm L 50mm TA 50N m G 75GPa t k x 2 3 N m m Solution c ro Internal Torque As shown in the torque diagram Equilibrium TA 0 L x t x d 0 TA 0 L x k x 2 3 d 0 Let unit m 4 3 TA 3k L 5 3 5 0 k 5TA 3 L 5 3 unit k 001228 MPa Ans Hence T x TA x k x 2 3 d T x TA 3k x 5 3 5 Angle of Twist J π 2 c4 φ 1 G J 0 L x T x d φ 1 G J 0 L x TA 3k x 5 3 5 unit d φ 005181 rad φ 2968 deg Ans Problem 570 The contour of the surface of the shaft is defined by the equation y e ax where a is a constant If the shaft is subjected to a torque T at its ends determine the angle of twist of end A with respect to end B The shear modulus is G Problem 571 The A36 steel shaft has a diameter of 50 mm and is subjected to the distributed and concentrated loadings shown Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x Given do 50mm L 05m TC 250 N m G 75GPa q 200 N m m Solution Support Reaction RA TC q L RA 15000 N m Internal Torque As shown in the torque diagram T x RA q x The maximum torque occurs at C Tmax RA q L Tmax 25000 N m Maximum Shear Stress c 05 do J π 2 c4 τmax Tmax c J τmax 1019 MPa Ans Angle of Twist φx 1 G J 0 x x T x d φx 1 G J 0 x x RA q x d φx 1 G J RA x q x2 2 At C x L φC 1 G J RA L q L2 2 φC 000217 rad φC 0125 deg Ans For L x 2L Since Tx 0 then φx φC x1 0 001 L L x2 L 101 L 2L φ1 x1 1 G J RA x1 q x1 2 2 1 N m φ2 x2 φC 1 N m 0 02 04 06 08 0 0001 0002 Distance m Twist rad φ1 x1 φ2 x2 x1 x2 Problem 572 A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft If the ring is held fixed and a torque T is applied to the rigid shaft determine the angle of twist of the shaft The shear modulus of the rubber is G Hint As shown in the figure the deformation of the element at radius r can be determined from rdθ drγ Use this expression along with τ T2π r2h from Prob 528 to obtain the result Problem 573 The A36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft Given do 50mm LAC 04m LBC 08m G 75GPa TC 300N m Solution c 05 do J π 2 c4 Equilibrium Given TA TB TC 0 1 Compatibility φCA φCB TA LAC G J TB LBC G J 2 Solving Eqs 1 and 2 Guess TA 1N m TB 1N m TA TB Find TA TB TA TB 200 100 N m Maximum Shear Stress τACmax TA c J τACmax 815 MPa Ans τBCmax TB c J τBCmax 407 MPa Ans Problem 574 The bronze C86100 pipe has an outer diameter of 375 mm and a thickness of 3 mm The coupling on it at C is being tightened using a wrench If the torque developed at A is 16 Nm determine the magnitude F of the couple forces The pipe is fixed supported at end B Given LCB 200mm LCA 250mm Lw 300mm do 375mm t 3mm TA 16N m G 38GPa Solution ro 05do ri ro t Compatibility φCB φCA TB LCB G J TB LCA G J TB LCB TA LCA TB LCA LCB TA TB 2000 N m Equilibrium F Lw TB TA 0 F TB TA Lw F 12000 N Ans Problem 575 The bronze C86100 pipe has an outer diameter of 375 mm and a thickness of 3 mm The coupling on it at C is being tightened using a wrench If the applied force is F 100 N determine the maximum shear stress in the pipe Given LCB 200mm LCA 250mm Lw 300mm do 375mm t 3mm F 100N G 38GPa Solution ro 05do ri ro t Compatibility φCB φCA TB LCB G J TB LCA G J Given TB LCB TA LCA 1 Equilibrium F Lw TB TA 0 2 Initial guess TA 1N m TB 2N m Solving 1 and 2 TA TB Find TA TB TA TB 1333 1667 N m Max Shear Stress c ro J π 2 ro 4 ri 4 τmax TB c J τmax 321 MPa Ans Problem 576 The steel shaft is made from two segments AC has a diameter of 12 mm and CB has a diameter of 25 mm If it is fixed at its ends A and B and subjected to a torque of 750 Nm determine the maximum shear stress in the shaft Gst 75 GPa Given LAC 125mm LCD 200mm LDB 300mm dAC 12mm dCD 25mm dDB 25mm TD 750N m G 75GPa Solution JAC π 2 dAC 2 4 JCD π 2 dCD 2 4 JDB π 2 dDB 2 4 Compatibility φDA φDB Given TA LAC G JAC TA LCD G JCD TB LDB G JDB 1 Equilibrium TD TB TA 0 2 Initial guess TA 1N m TB 2N m Solving 1 and 2 TA TB Find TA TB TA TB 7882 67118 N m Max Shear Stress cAC dAC 2 τAC TA cAC JAC τAC 2323 MPa cDB dDB 2 τDB TB cDB JDB τDB 21877 MPa τmax max τAC τDB τmax 23230 MPa Ans Problem 577 The shaft is made of L2 tool steel has a diameter of 40 mm and is fixed at its ends A and B If it is subjected to the couple determine the maximum shear stress in regions AC and CB Given do 40mm LAC 04m LBC 06m G 75GPa PC 2kN rC 50mm Solution c 05 do J π 2 c4 Equilibrium Given TA TB PC 2rC 0 1 Compatibility φCA φCB TA LAC G J TB LBC G J 2 Solving Eqs 1 and 2 Guess TA 1N m TB 1N m TA TB Find TA TB TA TB 120 80 N m Maximum Shear Stress τACmax TA c J τACmax 955 MPa Ans τBCmax TB c J τBCmax 637 MPa Ans Problem 578 The composite shaft consists of a midsection that includes the 20mmdiameter solid shaft and a tube that is welded to the rigid flanges at A and B Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D The shaft is subjected to a torque of 800 Nm The material is A36 steel Given LCA 100mm LAB 150mm LBD 100mm ds 20mm dto 60mm t 5mm T 800N m G 75GPa Solution dti dto 2t Js π 2 ds 2 4 Jt π 2 dto 2 4 dti 2 4 Compatibility φs φt Ts LAB G Js Tt LAB G Jt Given Ts Js Tt Jt 1 Equilibrium T Ts Tt 0 2 Initial guess Ts 1N m Tt 2N m Solving 1 and 2 Ts Tt Find Ts Tt Ts Tt 1863 78137 N m Angle of Twist φCD Ts LCA G Js Tt LAB G Jt Ts LBD G Js φCD 0005535 rad φCD 0317 deg Ans Problem 579 The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core If it is fixed to a rigid support at A and a torque of T 50 Nm is applied to it at C determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel Take Gst 80 GPa Gbr 40 GPa Given LAB 15m LBC 1m rs 20mm rto 20mm rti 10mm T 50N m Gst 80GPa Gbr 40GPa Solution Js π 2 rs 4 Jt π 2 rto 4 rti 4 Jbr π 2 rti 4 Compatibility φst φbr Tst LBC Gst Jt Tbr LBC Gbr Jbr Given Tst Gst Jt Tbr Gbr Jbr 1 Equilibrium T Tst Tbr 0 2 Initial guess Tst 1N m Tbr 2N m Solving 1 and 2 Tst Tbr Find Tst Tbr Tst Tbr 4839 161 N m Angle of Twist φC T LAB Gst Js Tbr LBC Gbr Jbr φC 0006297 rad φC 0361 deg Ans Max Shear Stress cAB rs τAB T cAB Js τAB 398 MPa cBC rto τBC Tst cBC Jt τBC 411 MPa τst max τAB τBC τst 411 MPa Ans γst τst Gst γst 5134 10 6 rad Ans τbr Tbr rti Jbr τbr 103 MPa Ans γbr τbr Gbr γbr 2567 10 6 rad Ans Problem 580 The two 1mlong shafts are made of 2014T6 aluminum Each has a diameter of 30 mm and they are connected using the gears fixed to their ends Their other ends are attached to fixed supports at A and B They are also supported by bearings at C and D which allow free rotation of the shafts along their axes If a torque of 900 Nm is applied to the top gear as shown determine the maximum shear stress in each shaft Given L 1000mm d 30mm rA 80mm rB 40mm T 900N m G 27GPa Solution J π 2 d 2 4 Compatibility φE rA φF rB TA L G J rA TB L G J rB Given TA rA TB rB 1 Equilibrium T TA F rA 0 2 TB F rB 0 3 Initial guess TA 1N m TB 2N m F 1N Solving 1 2 and 3 TA TB F Find TA TB F TA TB 180 360 N m F 900000 N Max Shear Stress c d 2 τAC TA c J τAC 3395 MPa Ans τBD TB c J τBD 6791 MPa Ans Problem 581 The two shafts are made of A36 steel Each has a diameter of 25 mm and they are connected using the gears fixed to their ends Their other ends are attached to fixed supports at A and B They are also supported by journal bearings at C and D which allow free rotation of the shafts along their axes If a torque of 500 Nm is applied to the gear at E as shown determine the reactions at A and B Given LAE 15m LBF 075m do 25mm rE 100mm rF 50mm TE 500N m G 75GPa Solution J π 32 do 4 Compatibility φE rE φF rF TA LAE G J rE TB LBF G J rF Given TA LAE rE TB LBF rF 1 Equilibrium TA F rE TE 0 2 TB F rF 0 3 Initial guess TA 1N m TB 2N m F 1N Solving 1 2 and 3 TA TB F Find TA TB F TA TB 5556 22222 N m A F 444444 N Problem 582 Determine the rotation of the gear at E in Prob 581 Given LAE 15m LBF 075m do 25mm rE 100mm rF 50mm TE 500N m G 75GPa Solution J π 32 do 4 Compatibility φE rE φF rF TA LAE G J rE TB LBF G J rF Given TA LAE rE TB LBF rF 1 Equilibrium TA F rE TE 0 2 TB F rF 0 3 Initial guess TA 1N m TB 2N m F 1N Solving 1 2 and 3 TA TB F Find TA TB F TA TB 5556 22222 N m F 444444 N Angle of Twist φE TA LAE G J φE 002897 rad φE 1660 deg Ans Problem 583 The A36 steel shaft is made from two segments AC has a diameter of 10 mm and CB has a diameter of 20 mm If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 300 Nmm along segment CB determine the absolute maximum shear stress in the shaft Given LAC 01m LCB 04m dAC 10mm dCB 20mm G 75GPa q 300 N m m Solution JAC π 2 dAC 2 4 JCB π 2 dCB 2 4 Compatibility φCA φCB Given TA LAC G JAC TB LCB 05q LCB 2 G JCB 1 Equilibrium q LCB TA TB 0 2 Initial guess TA 1N m TB 2N m Solving 1 and 2 TA TB Find TA TB TA TB 12 108 N m Max Shear Stress cAC dAC 2 τAC TA cAC JAC τAC 6112 MPa Ans cCB dCB 2 τCB TB cCB JCB τCB 6875 MPa Ans τmax max τAC τCB τmax 6875 MPa Ans Problem 584 The tapered shaft is confined by the fixed supports at A and B If a torque T is applied at its midpoin determine the reactions at the supports Problem 585 A portion of the A36 steel shaft is subjected to a linearly distributed torsional loading If the shaft has the dimensions shown determine the reactions at the fixed supports A and C Segment AB has a diameter of 30 mm and segment BC has a diameter of 15 mm Given LAB 12m LBC 096m dAB 30mm dBC 15mm G 75GPa q 15 kN m m Solution JAB π 2 dAB 2 4 JBC π 2 dBC 2 4 qx 1 x Lab q TR qx x 05 q qx x TR 1 x Lab q x 05 x Lab q x TR 1 05x Lab q x β 0 LAB x 1 05x LAB x d β 048 m2 Compatibility φBA φBC Given TA LAB q β G JAB TC LBC G JBC 1 Equilibrium 05q LAB TA TC 0 2 Initial guess TA 1kN m TC 2kN m Solving 1 and 2 TA TC Find TA TC TA TC 87826 2174 N m Ans Problem 586 Determine the rotation of joint B and the absolute maximum shear stress in the shaft in Prob 585 Given LAB 12m LBC 096m dAB 30mm dBC 15mm G 75GPa q 15 kN m m Solution JAB π 2 dAB 2 4 JBC π 2 dBC 2 4 qx 1 x Lab q TR qx x 05 q qx x TR 1 x Lab q x 05 x Lab q x TR 1 05x Lab q x β 0 LAB x 1 05x LAB x d β 048 m2 Compatibility φBA φBC Given TA LAB q β G JAB TC LBC G JBC 1 Equilibrium 05q LAB TA TC 0 2 Initial guess TA 1kN m TC 2kN m Solving 1 and 2 TA TC Find TA TC TA TC 87826 2174 N m Angle of Twist φB TC LBC G JBC φB 0055987 rad φB 3208 deg Ans Max Shear Stress cAB dAB 2 τAB TA cAB JAB τAB 16566 MPa cBC dBC 2 τBC TC cBC JBC τBC 328 MPa τmax max τAB τBC τmax 16566 MPa Ans Problem 587 The shaft of radius c is subjected to a distributed torque t measured as torquelength of shaft Determine the reactions at the fixed supports A and B Problem 588 Compare the values of the maximum elastic shear stress and the angle of twist developed in 304 stainless steel shafts having circular and square cross sections Each shaft has the same crosssectiona area of 5600 mm2 length of 900 mm and is subjected to a torque of 500 Nm Given L 900mm A 5600mm2 G 75GPa T 500N m Solution Max Shear Stress For circular shaft A π r2 r A π c r J π 2 r4 τCmax T c J τCmax 423 MPa Ans For squarr shaft A a2 a A τSmax T 481 a3 τSmax 574 MPa Ans Angle of Twist For circular shaft φC T L G J φC 0001202 rad φC 00689 deg Ans For sqaure shaft φS 710 T L G a4 φS 0001358 rad φS 00778 deg Ans Note The sqaure shaft has a greater maximum shear stress and angle of twist Problem 589 The shaft is made of red brass C83400 and has an elliptical cross section If it is subjected to the torsional loading shown determine the maximum shear stress within regions AC and BC and the ang of twist φ of end B relative to end A Given LAC 2m LCB 15m a 50mm b 20mm TA 50N m TB 30N m G 37GPa TC 20N m Solution Max Shear Stress τBC 2TB π a b2 τBC 0955 MPa Ans τAC 2TA π a b2 τAC 1592 MPa Ans Angle of Twist φBA n a2 b2 Tn Ln π a3 b3 G φBA a2 b2 TB LCB TA LAC π a3 b3 G φBA 0003618 rad φBA 02073 deg Ans Problem 590 Solve Prob 589 for the maximum shear stress within regions AC and BC and the angle of twist φ of end B relative to C Given LAC 2m LCB 15m a 50mm b 20mm TA 50N m TB 30N m G 37GPa TC 20N m Solution Max Shear Stress τBC 2TB π a b2 τBC 0955 MPa Ans τAC 2TA π a b2 τAC 1592 MPa Ans Angle of Twist φBC n a2 b2 Tn Ln π a3 b3 G φBC a2 b2 TB LCB π a3 b3 G φBC 0001123 rad φBC 00643 deg Ans Problem 591 The steel shaft is 300 mm long and is screwed into the wall using a wrench Determine the largest couple forces F that can be applied to the shaft without causing the steel to yield τY 56 MPa Given L 300mm Lw 400mm a 25mm τY 56MPa Solution Max Shear Stress τmax 481T a3 T a3τY 481 T 18191 N m Equilibrium F Lw T 0 F T Lw F 45478 N Ans Problem 592 The steel shaft is 300 mm long and is screwed into the wall using a wrench Determine the maximum shear stress in the shaft and the amount of displacement that each couple force undergoes if the couple forces have a magnitude of F 150 N Gst 75 GPa Given L 300mm Lw 400mm a 25mm F 150N Gst 75GPa Solution Equilibrium F Lw T 0 T F Lw T 60N m Max Shear Stress τmax 481T a3 τmax 1847 MPa Ans Angle of Twist φ 710 T L Gst a4 φ 0004362 rad δF φ Lw 2 δF 0872 mm Ans Problem 593 The shaft is made of plastic and has an elliptical crosssection If it is subjected to the torsional loadin shown determine the shear stress at point A and show the shear stress on a volume element located at this point Also determine the angle of twist φ at the end B Gp 15 GPa Given LOC 2m LCB 15m a 50mm b 20mm TB 50N m TC 40N m G 15GPa Solution TOC TB TC Shear Stress τA 2TOC π a b2 τA 2865 MPa Ans Angle of Twist φB n a2 b2 Tn Ln π a3 b3 G φB a2 b2 TB LCB TOC LOC π a3 b3 G φB 0015693 rad φB 08991 deg Ans Problem 594 The square shaft is used at the end of a drive cable in order to register the rotation of the cable on a gauge If it has the dimensions shown and is subjected to a torque of 8 Nm determine the shear stres in the shaft at point A Sketch the shear stress on a volume element located at this point Given a 5mm T 8N m Solution Maximum Shear Stress τAmax 481T a3 τAmax 3078 MPa Ans Problem 595 The brass wire has a triangular cross section 2 mm on a side If the yield stress for brass is τY 205 MPa determine the maximum torque T to which it can be subjected so that the wire will not yield If this torque is applied to a segment 4 m long determine the greatest angle of twist of one end of the wire relative to the other end that will not cause permanent damage to the wire Gbr 37 GPa Given a 2mm L 4m G 37GPa τY 205MPa Solution Allowable Shear Stress τallow 20T a3 τallow τY T τY a3 20 T 00820 N m Ans Angle of Twist φ 46T L G a4 φ 2549 rad Ans Problem 596 It is intended to manufacture a circular bar to resist torque however the bar is made elliptical in the process of manufacturing with one dimension smaller than the other by a factor k as shown Determine the factor by which the maximum shear stress is increased Given a 05 do b 05 k do Solution Maximum Shear Stress For the circular shaft τcmax T c J c 05 do J π 32 do 4 τcmax 16T π do 3 For the elliptical shaft τemax 2T π a b2 τemax 2T π 05 do 05 k do 2 τtmax 16T π k2 do 3 Factor of increase in shear stress Fτ τemax τcmax Fτ 16T π k2 do 3 16T π do 3 Fτ 1 k2 Ans Problem 597 The 2014T6 aluminum strut is fixed between the two walls at A and B If it has a 50 mm by 50 mm square cross section and it is subjected to the torsional loading shown determine the reactions at the fixed supports Also what is the angle of twist at C Given LAC 600mm LCD 600mm a 50mm LDB 600mm TC 60N m TD 30N m G 27GPa Solution T TC TD T 90N m L LAC LCD LDB Compatibility φT φB 0 710 T LAC G a4 710 TD LCD G a4 710 TB L G a4 0 TB T LAC TD LCD L TB 40N m Ans Equilibrium TA TB T 0 TA T TB TA 50N m Ans Angle of Twist φC 710 TA LAC G a4 φC 0001262 rad φC 00723 deg Ans Problem 598 The 304 stainless steel tube has a thickness of 10 mm If the allowable shear stress is τallow 80 MPa determine the maximum torque T that it can transmit Also what is the angle of twist of one end of th tube with respect to the other if the tube is 4 m long Neglect the stress concentrations at the corners The mean dimensions are shown Given a 70mm b 30mm L 4m τallow 80 MPa t 10mm G 75GPa Solution Section properties S Σds Am a b Am 2100mm2 S 2a 2 b S 200mm Average shear stress τavg T 2t Am τavg τallow T 2 t Am τallow T 336 kN m Ans Angle of Twist φ T L 4Am 2 G s 1 t d φ T L 4Am 2 G S t φ 0203175 rad φ 11641 deg Ans Problem 599 The 304 stainless steel tube has a thickness of 10 mm If the applied torque is T 50 Nm determine the average shear stress in the tube Neglect the stress concentrations at the corners The mean dimensions are shown Given a 70mm b 30mm t 10mm T 50N m G 75GPa Solution Section properties Am a b Am 2100mm2 Average shear stress τavg T 2t Am τavg 119 MPa Ans Problem 5100 Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed 84 MPa when a torque of T 25 kNm is applied to the tube Neglect stress concentrations at the corners The mean dimensions of the tube are shown Given a 100mm b 50mm T 25kN m τallow 84MPa Solution Section properties Am a b Am 5000mm2 Average shear stress τavg T 2t Am t T 2 Am τallow t 298 mm Ans Problem 5101 Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to exceed 84 MPa Neglect stress concentrations at the corners The mean dimensions of the tube are shown and the tube has a thickness of 3 mm Given a 100mm b 50mm t 3mm τallow 84MPa Solution Section properties Am a b Am 5000mm2 Average shear stress τavg T 2t Am T 2 t Am τallow T 252 kN m Ans Problem 5102 A tube having the dimensions shown is subjected to a torque of T 50 Nm Neglecting the stress concentrations at its corners determine the average shear stress in the tube at points A and B Show the shear stress on volume elements located at these points Given ao 50mm bo 50mm ta 3mm tb 5mm T 50N m Solution Section properties a ao ta b bo tb Am a b Am 2115mm2 Average shear stress τAavg T 2ta Am τAavg 394 MPa Ans τBavg T 2tb Am τBavg 236 MPa Ans Problem 5103 The tube is made of plastic is 5 mm thick and has the mean dimensions shown Determine the average shear stress at points A and B if it is subjected to the torque of T 5 Nm Show the shear stress on volume elements located at these points Given a 80mm b 110mm t 5mm c1 40mm c2 30mm T 5N m Solution Section properties c c1 2 c2 2 c 50mm Am a b 1 2 a c2 Am 10000mm2 Average shear stress τAavg T 2t Am τAavg 005 MPa Ans τBavg T 2t Am τBavg 005 MPa Ans Problem 5104 The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of t 5 mm If the allowable shear stress is τallow 56 MPa and the tube is to resist a torque of T 375 Nm determine the necessary dimension b The mean area for the ellipse is Am π b 05b Given T 375N m t 5mm τallow 56MPa Solution Section properties Am π b 05 b Average shear stress τavg T 2t Am τavg T 2t π b 05 b b T t π τallow b 2065 mm Ans Problem 5105 The tube is made of plastic is 5 mm thick and has the mean dimensions shown Determine the average shear stress at points A and B if the tube is subjected to the torque of T 500 Nm Show the shear stress on volume elements located at these points Neglect stress concentrations at the corners Given a 40mm b 100mm t 5mm c1 20mm c2 30mm T 500N m Solution Section properties c c1 2 c2 2 c 3606 mm Am a b 2 1 2 a c2 Am 5200mm2 Average shear stress τAavg T 2t Am τAavg 962 MPa Ans τBavg T 2t Am τBavg 962 MPa Ans Problem 5106 The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of t 5 mm If the allowable shear stress is τallow 56 MPa determine the necessary dimension b needed to resist the torque shown The mean area for the ellipse is Am π b 05b Given T1 75N m τallow 56MPa T2 120 N m t 5mm T3 450N m Solution Section properties Am π b 05 b Internal torque Tmax T1 T2 T3 Tmax 405N m Average shear stress τavg T 2t Am τavg T 2t π b 05 b b Tmax t π τallow b 2146 mm Ans Problem 5107 The symmetric tube is made from a highstrength steel having the mean dimensions shown and a thickness of 5 mm If it is subjected to a torque of T 40 Nm determine the average shear stress developed at points A and B Indicate the shear stress on volume elements located at these points Given a 40mm b 60mm t 5mm T 40N m Solution Section properties Am a a 4 a b Am 11200mm2 Average shear stress τAavg T 2t Am τAavg 0357 MPa Ans τBavg T 2t Am τBavg 0357 MPa Ans Problem 5108 Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle By what percentage is the torsional strength reduced when the eccentricity e is onefourth of the difference in the radii Given e a b 4 Solution For the aligned tube Section properties t a b Am π a b 2 2 Average shear stress τAavg T 2t Am T τAavg 2t Am T τAavg 2 a b π a b 2 2 For the eccentric tube Section properties Am π a b 2 2 t a e 2 e 2 b t a e b t a a b 4 b t 3 4 a b Average shear stress τAavg T 2t Am T τAavg 2t Am T τAavg 2 3 4 a b π a b 2 2 Factor of increase in shear stress FT T T Fτ τAavg 2 3 4 a b π a b 2 2 τAavg 2 a b π a b 2 2 FT 3 4 reduction in strength T 1 FT 100 T 2500 Ans Problem 5109 For a given average shear stress determine the factor by which the torquecarrying capacity is increased if the halfcircular sections are reversed from the dashedline positions to the section shown The tube is 25 mm thick Given a 30mm b 45mm r 15mm t 25mm Solution Section properties am a t rm r 05t Am am b 2 π 2 rm 2 Am 64354 mm2 Am am b 2 π 2 rm 2 Am 183146 mm2 Average shear stress τavg T 2t Am T τavg 2t Am T τavg 2t Am Hence the factor of increase is α T T α Am Am α 285 Ans Problem 5110 For a given maximum shear stress determine the factor by which the torque carrying capacity is increased if the halfcircular section is reversed from the dashedline position to the section shown The tube is 25 mm thick Given a 30mm b 45mm r 15mm t 25mm Solution Section properties am a t bm b 05t rm r 05t Am am bm π 2 rm 2 Am 90615 mm2 Am am bm π 2 rm 2 Am 150010 mm2 Average shear stress τavg T 2t Am T τavg 2t Am T τavg 2t Am Hence the factor of increase is α T T α Am Am α 166 Ans Problem 5111 The steel used for the shaft has an allowable shear stress of τallow 8 MPa If the members are connected with a fillet weld of radius r 4 mm determine the maximum torque T that can be applied Given D 50mm d 20mm r 4mm τallow 8MPa Solution Stress Concentration Factor D d 250 r d 020 From Fig 536 K 125 Allowable Shear Stress c d 2 J π 32 d4 τallow K 05T c J T τallow J 05K c T 2011 N m Ans Problem 5112 The shaft is used to transmit 660 W while turning at 450 rpm Determine the maximum shear stress in the shaft The segments are connected together using a fillet weld having a radius of 1875 mm Unit used rpm 2π 60 rad s Given D 25mm d 125mm r 1875mm ω 450rpm P 660W Solution T P ω T 1401 N m Stress Concentration Factor D d 200 r d 015 From Fig 536 K 130 Max shear stress c d 2 J π 2 d 2 4 τmax K T c J τmax 4748 MPa Ans Problem 5113 The shaft is fixed to the wall at A and is subjected to the torques shown Determine the maximum shear stress in the shaft A fillet weld having a radius of 45 mm is used to connect the shafts at B Given D 60mm d 30mm r 45mm TC 250N m TD 300 N m TE 800N m Solution Internal Torque As shown in the torque diagram TCD TC TDB TC TD TBE TC TD TEA TC TD TE Maximum Shear Stress For segment CD c d 2 J π 32 d4 τCD TCD c J τCD 4716 MPa Ans Max For segment EA c D 2 J π 32 D4 τEA TEA c J τEA 1768 MPa Ans For the fillet Stress Concentration Factor D d 200 r d 015 From Fig 536 K 130 Max shear stress τmax K TDB c J τmax 1226 MPa Ans Problem 5114 The builtup shaft is to be designed to rotate at 720 rpm while transmitting 30 kW of power Is this possible The allowable shear stress is τallow 12 MPa Unit used rpm 2π 60 rad s Given d 60mm D 75mm ω 720rpm P 30kW τallow 12MPa Solution ω 7540 rad s T P ω T 39789 N m Maximum Shear Stress c d 2 J π 32 d4 τallow K T c J K τallow J T c K 128 Stress Concentration Factor D d 125 K 128 From Fig 536 r d 0133 r 0133d r 798 mm Check D d 2 750 mm r 798mm No It si impossible Problem 5115 The builtup shaft is designed to rotate at 540 rpm If the radius of the fillet weld connecting the shaft is r 720 mm and the allowable shear stress for the material is τallow 55 MPa determine the maximum power the shaft can transmit Unit used rpm 2π 60 rad s Given d 60mm D 75mm ω 540rpm r 720mm τallow 55MPa Solution Stress Concentration Factor D d 125 r d 012 From Fig 536 K 130 Maximum Shear Stress c d 2 J π 32 d4 τallow K T c J T τallow J K c T 17943 kN m Maximum Power ω 5655 rad s P T ω P 10147 kW Ans Problem 5116 The steel used for the shaft has an allowable shear stress of τallow 8 MPa If the members are connected together with a fillet weld of radius r 225 mm determine the maximum torque T that can be applied Given d 15mm D 30mm r 225mm τallow 8MPa Solution Stress Concentration Factor D d 200 r d 015 From Fig 536 K 130 Maximum Shear Stress c d 2 J π 32 d4 τallow K 05T c J T τallow J 05K c T 8156 N m Ans Problem 5117 A solid shaft is subjected to the torque T which causes the material to yield If the material is elasticplastic show that the torque can be expressed in terms of the angle of twist φ of the shaft as T 43 TY 1 φY 3 4φ3 where TY and φY are the torque and angle of twist when the material begins to yield Problem 5118 A solid shaft having a diameter of 50 mm is made of elasticplastic material having a yield stress of τY 112 MPa and shear modulus of G 84 GPa Determine the torque required to develop an elastic core in the shaft having a diameter of 25 mm Also what is the plastic torque Given d 50mm G 84GPa dY 25mm τY 112MPa Solution c d 2 ρY dY 2 Elasticplastic torque Use Eq 526 from the text T π τY 6 4c3 ρY 3 T 3551 kN m Ans Plastic torque Use Eq 527 from the text TP 2π τY 3 c3 TP 3665 kN m Ans Problem 5119 Determine the torque needed to twist a short 3mmdiameter steel wire through several revolutions if it is made from steel assumed to be elasticplastic and having a yield stress of τY 80 MPa Assume that the material becomes fully plastic Given d 3mm τY 80MPa Solution c d 2 Plastic torque Use Eq 527 from the text TP 2π τY 3 c3 TP 0565 N m Ans Problem 5120 A solid shaft has a diameter of 40 mm and length of 1 m It is made from an elasticplastic material having a yield stress of τY 100 MPa Determine the maximum elastic torque TY and the corresponding angle of twist What is the angle of twist if the torque is increased to T 12TY G 80 GPa Given d 40mm G 80GPa L 1m τY 100MPa Solution c d 2 J π 32 d4 Maximum Elastic Torque τY TY c J TY τY J c TY 1257 kN m Ans Angle of Twist γY τY G γY 000125 rad φ γY L c φ 00625 rad φ 3581 deg Ans Elasticplastic torque T 12TY Use Eq 526 from the text T π τY 6 4c3 ρY 3 12TY π τY 6 4c3 ρY 3 ρY 3 4c3 72 TY π τY ρY 1474 mm Angle of Twist φ γY L ρY φ 0084826 rad φ 486 deg Ans Problem 5121 The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diamet segment Determine the radius of the elastic core produced in the smaller diameter segment Neglect the stress concentration at the fillet Given do 60mm ds 55mm Solution co do 2 Jo π 32 do 4 cs ds 2 Js π 32 ds 4 Set τY 1MPa Maximum Elastic Torque For the larger diameter segment τY TY c J TY τY Jo co TY 4241 N m Elasticplastic torque For the smaller diameter segment TP 2π τY 3 cs 3 TP 4356 N m TY Applying Eq 526 from the text T π τY 6 4cs 3 ρY 3 TY π τY 6 4cs 3 ρY 3 ρY 3 4cs 3 6 TY π τY ρY 1298 mm Ans Problem 5122 A bar having a circular cross section of 75 mm diameter is subjected to a torque of 12 kNm If the material is elasticplastic with τY 120 MPa determine the radius of the elastic core Given d 75mm T 12kN m τY 120MPa Solution c d 2 Elasticplastic torque Use Eq 526 from the text T π τY 6 4c3 ρY 3 ρY 3 4c3 6T π τY ρY 2712 mm Ans Problem 5123 A tubular shaft has an inner diameter of 20 mm an outer diameter of 40 mm and a length of 1 m It i made of an elastic perfectly plastic material having a yield stress of τY 100 MPa Determine the maximum torque it can transmitWhat is the angle of twist of one end with respect to the other end if the shear strain on the inner surface of the tube is about to yield G 80 GPa Given do 40mm G 80GPa L 1m di 20mm τY 100MPa Solution ρY di 2 Plastic Torque co 05do ci 05di TP 2π ci co ρ τY ρ2 d TP 1466 kN m Ans Angle of Twist γY τY G γY 000125 rad φ γY L ρY φ 0125 rad φ 7162 deg Ans Problem 5124 The 2mlong tube is made from an elasticplastic material as shown Determine the applied torque T which subjects the material of the tubes outer edge to a shearing strain of γmax 0008 rad What would be the permanent angle of twist of the tube when the torque is removed Sketch the residual stress distribution of the tube Given ro 45mm ri 40mm L 2m γY 0003 τY 240MPa γmax 0008 Solution Determine if it is fully plastic φmax γmax L ro φmax 035556 rad However φmax γY L ρY ρY γY L φmax ρY 1688 mm ri 40mm Therefore the tube is filly plastic Also at r ri γr ri ro γmax γr 000711 γY 0003 γmax ro γr ri Again the tube is filly plastic Plastic Torque Tp 2π ri ro ρ τY ρ2 d Tp 1363 kN m Ans Angle of Twist When the torque is removed The equal but opposite torque TP is applied G τY γY J π 2 ro 4 ri 4 φ Tp L G J φ 014085 rad Permanent angle of twist φr φmax φ φr 021470 rad φr 1230 deg Ans Shear Stresses At r ro τpo Tp ro J τpo 2535 MPa At r ri τpi Tp ri J τpi 2254 MPa Problem 5125 The tube has a length of 2 m and is made of an elasticplastic material as shown Determine the torqu needed to just cause the material to become fully plastic What is the permanent angle of twist of the tube when this torque is removed Given ro 100mm ri 60mm L 2m γY 0007 τY 350MPa Solution At Just Fully Plastic ρY ri φp γY L ρY φp 023333 rad Plastic Torque Tp 2π ri ro ρ τY ρ2 d Tp 57470 kN m Ans Angle of Twist When the torque is removed The equal but opposite torque TP is applied G τY γY J π 2 ro 4 ri 4 φ Tp L G J φ 016814 rad Permanent angle of twist φr φp φ φr 006520 rad φr 374 deg Ans Problem 5126 The shaft is made from a strainhardening material having a τ γ diagram as shown Determine the torque T that must be applied to the shaft in order to create an elastic core in the shaft having a radius of ρc 125 mm Given r 15mm ρc 125mm γa 0005 τa 70MPa γb 0010 τb 105MPa Solution c r ρa ρc For linear strain variations against ρ γ ρ γa ρa γ γa ρa ρ Path 1 τ1 τa γa γ τ1 τa ρa ρ Path 2 τ2 τa γ γa τb τa γb γa τ2 τb τa γb γa γ γa τa τ2 τb τa γb γa ρ ρa 1 γa τa T 2π 0 c ρ τ ρ2 d T 2π 0 ρa ρ τ1 ρ2 d 2π ρa c ρ τ2 ρ2 d T 2π τa ρa 0 ρa ρ ρ3 d 2π γa τb τa γb γa ρa c ρ ρ ρa 1 ρ2 d 2π τa ρa c ρ ρ2 d T 43427 N m Ans Problem 5127 The 2mlong tube is made of an elastic perfectly plastic material as shown Determine the applied torque T that subjects the material at the tubes outer edge to a shear strain of γmax 0006 rad What would be the permanent angle of twist of the tube when this torque is removed Sketch the residual stress distribution in the tube Given ro 35mm ri 30mm L 2m γY 0003 τY 210MPa γmax 0006 Solution At Fully Plastic ρY ri φp γmax L ro φp 034286 rad Also at r ri γr ri ro γmax γr 000514 γmax ro γr ri γY 0003 Confirm that tube is filly plastic Plastic Torque Tp 2π ri ro ρ τY ρ2 d Tp 698 kN m Ans Angle of Twist When the torque is removed The equal but opposite torque TP is applied G τY γY J π 2 ro 4 ri 4 φ Tp L G J φ 018389 rad Permanent angle of twist φr φp φ φr 015897 rad φr 911 deg Ans Residual Shear Stresses At r ro τpo Tp ro J τpo 2253 MPa τro τY τpo τro 153 MPa At r ri τpi Tp ri J τpi 1931 MPa τri τY τpi τri 169 MPa Problem 5128 The shear stressstrain diagram for a solid 50mm diameter shaft can be approximated as shown in the figure Determine the torque required to cause a maximum shear stress in the shaft of 125 MPa If the shaft is 3 m long what is the corresponding angle of twist Given do 50mm L 3m γ1 00025 τ1 50MPa γmax 0010 τmax 125MPa Solution ro 05 do τρ Function At r ρ1 γmax ro γ1 ρ1 ρ1 γ1 γmax ro ρ1 625 mm For the region 0 r ρ1 k1 τ1 0 ρ1 0 k1 800 MPa mm k1 τ 0 ρ 0 τ k1 ρ For the region ρ1 r r0 k2 τmax τ1 ro ρ1 k2 400 MPa mm k2 τ τ1 ρ ρ1 τ τ1 k2 ρ ρ1 Plastic Torque Tp 2π 0 ρ1 ρ τ ρ2 d 2π ρ1 ro ρ τ ρ2 d Tp 2π 0 ρ1 ρ k1 ρ ρ2 d 2π ρ1 ro ρ τ1 k2 ρ ρ1 ρ2 d Tp 327 kN m Ans Angle of Twist φmax γmax L ro φmax 120000 rad φmax 6875 deg Ans Problem 5129 The shaft consists of two sections that are rigidly connected If the material is elastic perfectly plastic as shown determine the largest torque T that can be applied to the shaft Also draw the shearstress distribution over a radial line for each section Neglect the effect of stress concentration Given r1 20mm r2 25mm γY 0002 τY 70MPa Solution Plastic Torque For the smallerdiameter segment c r1 Tp 2π 0 c ρ τY ρ2 d Tp 1173 kN m Ans Max Shear Stress For the biggerdiameter segment c r2 J π 2 r2 4 τmax Tp c J τmax 4779 MPa Ans Problem 5130 The shaft is made of an elasticperfectly plastic material as shown Plot the shearstress distribution acting along a radial line if it is subjected to a torque of T 2 kNm What is the residual stress distribution in the shaft when the torque is removed Given ro 20mm T 2kN m γY 0001875 τY 150MPa Solution c ro J π 2 ro 4 Maximum Elastic Torque τY TY c J TY τY J c TY 1885 kN m T 2 kNm Plastic Torque TP 2π τY 3 c3 TP 2513 kN m T 2 kNm Therefore it is elasticplastic Elasticplastic Torque Applying Eq 526 from the text T π τY 6 4c3 ρY 3 ρY 3 4c3 6 T π τY ρY 1870 mm Residual Shear Stresses When the torque is removed The equal but opposite torque T is applied At r ro τpo T ro J τpo 15915 MPa τro τY τpo τro 915 MPa At r ρY τpi T ρY J τpi 14878 MPa τri τY τpi τri 122 MPa Problem 5131 A 40mmdiameter shaft is made from an elasticplastic material as shown Determine the radius of its elastic core if it is subjected to a torque of T 300 Nm If the shaft is 250 mm long determine the angle of twist Given d 40mm L 250mm T 300N m γY 0006 τY 21MPa Solution c d 2 Elasticplastic torque Use Eq 526 from the text T π τY 6 4c3 ρY 3 ρY 3 4c3 6T π τY ρY 1677 mm Ans Angle of Twist φ γY L ρY φ 0089445 rad φ 51248 deg Ans Problem 5132 A torque is applied to the shaft having a radius of 100 mm If the material obeys a shear stressstrain relation of τ 20γ 13 MPa determine the torque that must be applied to the shaft so that the maximum shear strain becomes 0005 rad Given ro 100mm γmax 0005 τ 20 3 γ unit Solution c ro τρ Function unit MPa γ ρ c γmax τ 20 3 γ unit τ 20 3 ρ c γmax unit Ultimate Torque T 2π 0 c ρ τ ρ2 d T 2π unit 0 c ρ 20 3 ρ c γmax ρ2 d T 6446 kN m Ans Problem 5133 The shaft is made of an elasticperfectly plastic material as shown Determine the torque that the shaf can transmit if the allowable angle of twist is 0375 rad Also determine the permanent angle of twist once the torque is removed The shaft is 2mlong Given ro 20mm φallow 0375rad γY 0001875 τY 150MPa L 2m Solution c ro Angle of Twist γmax φallow c L γmax 000375 rad γmax c γY ρY ρY γY γmax c ρY 1000 mm Elasticplastic Torque Applying Eq 526 from the text T π τY 6 4c3 ρY 3 T 2435 kN m Ans Permanent Angle of Twist When the torque is removed The equal but opposite torque T is applied G τY γY J π 2 ro 4 φ T L G J φ 024219 rad Permanent angle of twist φr φallow φ φr 013281 rad φr 761 deg Ans Problem 5134 Consider a thinwalled tube of mean radius r and thickness t Show that the maximum shear stress in the tube due to an applied torque T approaches the average shear stress computed from Eq 518 as rt approaches infinity Problem 5135 The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm When it is rotating at 60 rads it transmits 30 kW of power from the engine E to the generator G Determine the smallest thickness of the shaft if the allowable shear stress is τallow 150 MPa and the shaft is restricted not to twist more than 008 rad Given do 60mm L 3m ω 60 rad s P 30kW G 75GPa τallow 150MPa φallow 008rad Solution T P ω T 500N m Allowable Shear Stress Assume failure due to shear strss ro do 2 c ro J π 2 ro 4 ri 4 τallow T c J J T c τallow π 2 ro 4 ri 4 T ro τallow ri 4 ro 4 2T ro π τallow ri 29392 mm Angle of Twist Assume failure due to angle of twist limitation φ T L G J J T L G φallow π 2 ro 4 ri 4 T L G φallow ri 4 ro 4 2T L π G φallow ri 28403 mm Choose the smallest value of ri ri min ri ri ri 28403 mm t ro ri t 1597 mm Ans Problem 5136 The 304 stainless solid steel shaft is 3 m long and has a diameter of 50 mm It is required to transmit 40 kW of power from the engine E to the generator G Determine the smallest angular velocity the shaft can have if it is restricted not to twist more than 15 Given do 50mm L 3m P 40kW G 75GPa φallow 15deg Solution Angle of Twist ro do 2 c ro J π 2 ro 4 φ T L G J T G J L φallow T 40160 N m Angular Velocity T P ω ω P T ω 996 rad s Ans Problem 5137 The drilling pipe on an oil rig is made from steel pipe having an outside diameter of 112 mm and a thickness of 6 mm If the pipe is turning at 650 revmin while being powered by a 12kW motor determine the maximum shear stress in the pipe Unit used rpm 2π 60 rad s Given do 112mm t 6mm ω 650 rpm P 12kW τallow 70MPa Solution ω 6807 rad s T P ω T 17629 N m Max stress c do 2 di do 2t J π 2 do 2 4 di 2 4 τmax T c J τmax 175 MPa Ans Problem 5138 The tapered shaft is made from 2014T6 aluminum alloy and has a radius which can be described by the function r 002 1 x32 m where x is in meters Determine the angle of twist of its end A if it i subjected to a torque of 450 Nm Given T 450N m L 4m G 27GPa r 002 1 x3 unit Solution unit m Angle of Twist J π 2 r4 r 002 1 x3 unit φ 0 L x T G J d φ unit 0 L m x 2T G π 002 1 x3 unit 4 d φ 002771 rad φ 1588 deg Ans Problem 5139 The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at 1500 revmin Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowabl shear stress is τallow 56 MPa and the vibrations limit the angle of twist of the shaft to 005 rad The shaft is 06 m long and made of L2 tool steel Unit used rpm 2π 60 rad s Given L 600mm ω 1500 rpm P 660kW τallow 56MPa G 75GPa φallow 005rad Solution ω 15708 rad s T P ω T 4202 kN m Allowable shear stress Assume failure due to shear stress c d 2 J π 2 d 2 4 τallow T c J Thus d 2 2 π T τallow 1 3 d 7257 mm Angle of Twist Assume failure due to angle of twist limitation J π 2 d 2 4 φ T L G J Thus d 2 2 π T L G φallow 025 d 5115 mm Shear stress failure controls the design Hence Use d 75mm Ans Problem 5140 The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at 1500 revmin Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowabl shear stress is τallow 75 MPa and the vibrations limit the angle of twist of the shaft to 003 rad The shaft is 06 m long and made of L2 tool steel Unit used rpm 2π 60 rad s Given L 600mm ω 1500 rpm P 660kW τallow 75MPa G 75GPa φallow 003rad Solution ω 15708 rad s T P ω T 4202 kN m Allowable shear stress Assume failure due to shear stress c d 2 J π 2 d 2 4 τallow T c J Thus d 2 2 π T τallow 1 3 d 6583 mm Angle of Twist Assume failure due to angle of twist limitation J π 2 d 2 4 φ T L G J Thus d 2 2 π T L G φallow 025 d 5812 mm Shear stress failure controls the design Hence Use d 70mm Ans Problem 5141 The material of which each of three shafts is made has a yield stress of τY and a shear modulus of G Determine which shaft geometry will resist the largest torque without yielding What percentage of th torque can be carried by the other two shafts Assume that each shaft is made of the same amount of material and that it has the same crosssectional area A Given AΟ A Asq A A A Solution For circular shaft AΟ πc2 c A π J π 2 c4 J A2 2π τmax T c J τmax 2 π T A A T A A 2 π τY αΟ 1 2 π αΟ 02821 For square shaft Asq a2 a A τmax 481T a3 τmax 481T A A T A A 481 τY αsq 1 481 αsq 02079 For triangular shaft θ 60deg A a 2 a sin θ a 2 A 4 3 τmax 20T a3 τmax 5 4 27 T 2A A T 2A A 5 4 27 τY α 2 5 4 27 α 01755 The circular shaft will carry the largest torque Ans For square shaft sq αsq αΟ 100 sq 737 Ans For triangular shaft α αΟ 100 622 Ans Problem 5142 The A36 steel circular tube is subjected to a torque of 10 kNm Determine the shear stress at the mean radius ρ 60 mm and compute the angle of twist of the tube if it is 4 m long and fixed at its far end Solve the problem using Eqs 57 and 515 and by using Eqs 518 and 520 Given ρ 60mm t 5mm L 4m T 10kN m G 75GPa Solution Section properties ro ρ 05t ri ρ 05t Σds 2π ρ Σds 37699 mm Am π ρ2 Am 1130973 mm2 J π 2 ro 4 ri 4 J 679762110 mm4 Shear Stress Applying Eq 57 c ρ τ T c J τ 8827 MPa Ans Applying Eq 518 τavg T 2 t Am τavg 8842 MPa Ans Angle of Twist Applying Eq 520 φ T L J G φ 007846 rad φ 4495 deg Ans Applying Eq 520 φ T L 4Am 2 G s 1 t d φ T L 4Am 2 G Σds t φ 007860 rad φ 4503 deg Ans Problem 5143 The aluminum tube has a thickness of 5 mm and the outer crosssectional dimensions shown Determine the maximum average shear stress in the tube If the tube has a length of 5 m determine th angle of twist Gal 28 GPa Given ao 150mm bo 100mm L 4m t 5mm TA 280N m TB 135N m LAB 2m LBC 3m G 28GPa Solution Section properties S Σds a ao t b bo t Am a b Am 13775mm2 S 2a 2 b S 480mm Maximum Average shear stress TAB TA TBC TA TB Tmax max TAB TBC Tmax 280N m τavgmax Tmax 2t Am τavgmax 203 MPa Ans Angle of Twist φ T L 4Am 2 G s 1 t d φ TAB LAB TBC LBC 4Am 2 G S t φ 0004495 rad φ 0258 deg Ans Problem 61 Draw the shear and moment diagrams for the shaft The bearings at A and B exert only vertical reactions on the shaft Given a 250mm b 800mm F 24kN Solution Given Equilibrium ΣFy0 A B F 0 ΣΜA0 F a B b 0 Guess A 1kN B 1kN A B Find A B A B 3150 750 kN x1 0 001 a a x2 a 101 a a b V1 x1 F 1 kN V2 x2 F A 1 kN M1 x1 F x1 kN m M2 x2 F x2 A x2 a 1 kN m 0 05 1 40 20 0 20 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 05 1 5 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 62 The load binder is used to support a load If the force applied to the handle is 250 N determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC Given a 300mm b 75mm F 250N Solution Given Equilibrium ΣFy0 T1 T2 F 0 ΣΜB0 F a T2 b 0 Guess T1 1N T2 1N T1 T2 Find T1 T2 T1 T2 125 100 kN Ans x1 0 001 a a x2 a 101 a a b V1 x1 F 1 kN V2 x2 F T1 1 kN M1 x1 F x1 N m M2 x2 F x2 T1 x2 a 1 N m 0 02 1 0 1 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 02 100 50 0 Distane m Moment Nm M1 x1 M2 x2 x1 x2 Problem 63 Draw the shear and moment diagrams for the shaft The bearings at A and D exert only vertical reactions on the shaft The loading is applied to the pulleys at B and C and E Given a 350mm b 500mm c 375mm d 300mm B 400N C 550N E 175N Solution Equilibrium Given ΣFy0 A D B C E 0 ΣΜD0 A a b c B b c C c E d 0 Guess A 1N D 1N A D Find A D A D 41122 71378 N x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c x4 a b c 101 a b c a b c d V1 x1 A 1 N V2 x2 A B 1 N V3 x3 A B C 1 N V4 x4 A B C D 1 N M1 x1 A x1 N m M2 x2 A x2 B x2 a 1 N m M3 x3 A x3 B x3 a C x3 a b 1 N m M4 x4 A x4 B x4 a C x4 a b D x4 a b c 1 N m 0 02 04 06 08 1 12 14 1000 500 0 500 Distance m Shear N V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 02 04 06 08 1 12 14 100 0 100 200 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 64 Draw the shear and moment diagrams for the beam Given a 1m F 10kN Solution Equilibrium Given ΣFy0 A 4F B 0 ΣΜB0 A 5a F 4a 3a 2a a 0 Guess A 1kN B 1kN A B Find A B A B 20 20 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a x5 4a 101 4a 5a V1 x1 A 1 kN V2 x2 A F 1 kN V3 x3 A 2F 1 kN V4 x4 A 3F 1 kN V5 x5 A 4F 1 kN M1 x1 A x1 kN m M2 x2 A x2 F x2 a 1 kN m M3 x3 A x3 F x3 a F x3 2a 1 kN m M4 x4 A x4 F x4 a F x4 2a F x4 3a 1 kN m M5 x5 A x5 F x5 a F x5 2a F x5 3a F x5 4a 1 kN m 0 1 2 3 4 5 20 0 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 V5 x5 x1 x2 x3 x4 x5 0 1 2 3 4 5 0 10 20 30 Distance m Moment kNm M1 x1 M2 x2 M3 x3 M4 x4 M5 x5 x1 x2 x3 x4 x5 Problem 65 A reinforced concrete pier is used to support the stringers for a bridge deck Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown Assume the columns a A and B exert only vertical reactions on the pier Given a 1m b 15m F1 60kN F2 35kN Solution L 4a 2 b Equilibrium Given ΣFy0 A B 2F1 3F2 0 ΣΜB0 A 2a 2 b F1 L 2a F2 3 b 3a 0 Guess A 1kN B 1kN A B Find A B A B 1125 1125 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 2a b x4 2a b 101 2a b 2a 2 b x5 2a 2 b 101 2a 2 b 3a 2 b x6 3a 2 b 101 3a 2 b 4a 2 b V1 x1 F1 1 kN V2 x2 A F1 1 kN V3 x3 A F1 F2 1 kN V4 x4 A F1 2F2 1 kN V5 x5 A F1 3F2 1 kN V6 x6 A F1 3F2 B 1 kN M1 x1 F1 x1 kN m M2 x2 A x2 a F1 x2 1 kN m M3 x3 A x3 a F1 x3 F2 x3 2a 1 kN m M4 x4 A x4 a F1 x4 F2 2 x4 2 a b 1 kN m M5 x5 A x5 a F1 x5 F2 3 x5 2 a 3 b 1 kN m M6 x6 A x6 a F1 x6 F2 3 x6 2 a 3 b B x6 L a 1 kN 0 1 2 3 4 5 6 7 50 0 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 V5 x5 V6 x6 x1 x2 x3 x4 x5 x6 0 1 2 3 4 5 6 7 60 40 20 0 20 Distance m Moment kNm M1 x1 M2 x2 M3 x3 M4 x4 M5 x5 M6 x6 x1 x2 x3 x4 x5 x6 Problem 66 Draw the shear and moment diagrams for the shaft The bearings at A and B exert only vertical reactions on the shaft Also express the shear and moment in the shaft as a function of x within the region 125 mm x 725 mm Given a 125mm b 600mm c 75mm F1 08kN F2 15kN Solution L a b c Equilibrium Given ΣFy0 A F1 F2 B 0 ΣΜB0 A L F1 b c F2 c 0 Guess A 1kN B 1kN A B Find A B A B 08156 14844 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A 1 kN V2 x2 A F1 1 kN V3 x3 A F1 F2 1 kN M1 x1 A x1 N m M2 x2 A x2 F1 x2 a 1 N m M3 x3 A x3 F1 x3 a F2 x3 a b 1 N m 0 01 02 03 04 05 06 07 2 1 0 1 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 02 04 06 0 50 100 150 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 67 Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout the shaft as a function of x The bearings at A and B exert only vertical reactions on the shaft Given a 09m b 06m c 03m d 015m F1 4kN F2 25kN Solution Equilibrium Given ΣFy0 A F1 B F2 0 ΣΜB0 A a b F1 b F2 c d 0 Guess A 1kN B 1kN A B Find A B A B 085 565 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A 1 kN V2 x2 A F1 1 kN V3 x3 A F1 B 1 kN M1 x1 A x1 N m M2 x2 A x2 F1 x2 a 1 N m M3 x3 A x3 F1 x3 a B x3 a b 1 N m 0 05 1 15 4 2 0 2 4 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 05 1 15 1000 500 0 500 1000 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 68 Draw the shear and moment diagrams for the pipe The end screw is subjected to a horizontal force of 5 kN Hint The reactions at the pin C must be replaced by equivalent loadings at point B on the axis the pipe Given a 400mm h 80mm F 5kN Solution Given Equilibrium ΣFy0 A C 0 ΣΜC0 A a F h 0 Guess A 1N C 1N A C Find A C A C 100 100 kN Ans x1 0 001 a a V1 x1 A 1 kN M1 x1 A x1 N m 0 02 04 2 1 0 1 Distance m Shear kN V1 x1 x1 0 02 0 600 400 200 0 Distane m Moment Nm M1 x1 x1 Problem 69 Draw the shear and moment diagrams for the beam Hint The 100kN load must be replaced by equivalent loadings at point C on the axis of the beam Given a 1m b 1m c 1m d 025m F1 75kN F2 100kN Solution Equilibrium Given ΣFy0 A F1 B 0 ΣΜC0 A a b c F1 b c F2 d 0 Guess A 1kN B 1kN A B Find A B A B 5833 1667 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A 1 kN V2 x2 A F1 1 kN V3 x3 A F1 1 kN M1 x1 A x1 kN m M2 x2 A x2 F1 x2 a 1 kN m M3 x3 A x3 F1 x3 a F2 d 1 kN m 0 05 1 15 2 25 3 0 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 05 1 15 2 25 3 0 20 40 60 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 610 The engine crane is used to support the engine which has a weight of 6 kN Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown Given a 09m b 15m c 12m W 6kN Solution d a2 c2 v c d h a d Equilibrium Given ΣFy0 Ay B v W 0 ΣΜA0 B v a W a b 0 ΣFx0 Ax B h 0 Guess Ax 1kN Ay 1kN B 1kN Ax Ay B Find Ax Ay B Ax Ay B 12 10 20 kN x1 0 001 a a x2 a 101 a a b V1 x1 Ay 1 kN V2 x2 Ay B v 1 kN M1 x1 Ay x1 kN m M2 x2 Ay x2 B v x2 a 1 kN m 0 1 2 10 0 10 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 10 5 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 611 Draw the shear and moment diagrams for the compound beam It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force although it can support a moment and axial load Set a 1m P 1kN Solution Equilibrium Given ΣFy0 A P C P 0 ΣΜB0 P 2a C a 0 Guess A 1kN C 1kN A C Find A C A C 0 2 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a V1 x1 A 1 kN V2 x2 A P 1 kN V3 x3 A P 1 kN V4 x4 A P C 1 kN 0 05 1 15 2 25 3 35 4 2 1 0 1 2 Distance m Shear P kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 MA C 3a P 4a P a MA 100 kN m M1 x1 MA A x1 kN m M2 x2 MA A x2 P x2 a 1 kN m M3 x3 A P x3 2a 1 kN m M4 x4 A P x4 2a C x4 3a 1 kN m 0 1 2 3 4 1 0 1 Distance m Moment P kNm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 612 Draw the shear and moment diagrams for the compound beam which is pin connected at B Given a 1m b 15m c 1m d 1m F1 30kN F2 40kN Solution Equilibrium Given ΣFy0 F1 A F2 C 0 ΣΜB0 F1 a b A b 0 Guess A 1kN C 1kN A C Find A C A C 50 20 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c x4 a b c 101 a b c a b c d V1 x1 F1 1 kN V2 x2 F1 A 1 kN V3 x3 F1 A 1 kN V4 x4 F1 A F2 1 kN 0 1 2 3 4 40 20 0 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 M1 x1 F1 x1 kN m M2 x2 F1 x2 A x2 a 1 kN m M3 x3 F1 A x3 a b 1 kN m M4 x4 F1 A x4 a b F2 x4 a b c 1 kN m 0 1 2 3 4 30 20 10 0 10 20 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 613 Draw the shear and moment diagrams for the beam Set a 1m Mo 1kN m Solution Equilibrium Given ΣFy0 A B 0 ΣΜB0 A 3a 2Mo Mo 0 Guess A 1kN B 1kN A B Find A B A B 033 033 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a V1 x1 A 1 kN V2 x2 A 1 kN V3 x3 A 1 kN M1 x1 Mo A x1 kN m M2 x2 2 Mo A x2 1 kN m M3 x3 2 Mo A x3 Mo 1 kN m 0 1 2 3 05 0 05 Distance m Shear Moa kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 0 1 2 Distance m Moment Mo kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 614 Consider the general problem of a simply supported beam subjected to n concentrated loads Write a computer program that can be used to determine the internal shear and moment at any specified location x along the beam and plot the shear and moment diagrams for the beam Show an applicatio of the program using the values P1 25 kN d1 15 m P2 4 kN d2 45 m L1 3 m L 45 m Problem 615 The beam is subjected to the uniformly distributed moment m Momentlength Draw the shear and moment diagrams for the beam Set L 1m mo 1 kN m m Solution Given Equilibrium ΣFy0 A 0 ΣΜA0 MA mo L x1 0 001 L L V1 x1 A 1 kN M1 x1 MA A x1 mo x1 1 kN m 0 05 1 0 05 1 15 Distance m Shear kN V1 x1 x1 0 05 1 0 05 1 Distane m Moment mL kNm M1 x1 x1 Problem 616 Draw the shear and moment diagrams for the beam Given a 25m w 10 kN m b 25m Solution Equilibrium A w a w a ΣFy0 A 0kN ΣΜA0 MA w a 05a w b a 05b MA 6250 kN m x1 0 001 a a x2 a 101 a a b V1 x1 A w x1 1 kN V2 x2 A w a w x2 a 1 kN M1 x1 MA A x1 05w x1 2 1 kN m M2 x2 MA A x2 w a x2 05a 05w x2 a 2 1 kN m 0 2 4 20 0 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 2 4 0 50 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 617 The 75kg man sits in the center of the boat which has a uniform width and a weight per linear foot of 50 Nm Determine the maximum bending moment exerted on the boat Assume that the water exerts a uniform distributed load upward on the bottom of the boat Given a 25m Mw 75kg b 25m w 50 N m Solution W Mw g Equilibrium ΣFy0 q W w 2a 2a q 1971 N m x1 0 001 a a x2 a 101 a a b V1 x1 q w x1 1 N V2 x2 q w x2 W 1 N M1 x1 05 q w x1 2 1 N m M2 x2 05 q w x2 2 W x2 a 1 N m 0 2 4 500 0 500 Distance m Shear N V1 x1 V2 x2 x1 x2 0 2 4 0 200 400 Distane m Moment Nm M1 x1 M2 x2 x1 x2 Problem 618 Draw the shear and moment diagrams for the beam It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force although it can support a moment an axial load Set L 1m w 1 kN m A 0 Solution Given Equilibrium ΣFy0 A B w L 0 B w L ΣΜA0 MA A L w L L 2 0 MA w L2 2 x1 0 001 L L V1 x1 A w x1 1 kN M1 x1 MA A x1 w 2 x1 2 1 kN m 0 05 1 1 0 Distance m Shear wL kN V1 x1 x1 0 05 1 0 02 04 06 Distane m Moment wLL kNm M1 x1 x1 Problem 619 Draw the shear and moment diagrams for the beam Given a 15m w 30 kN m Mo 45kN m Solution Equilibrium Given ΣFy0 w a A B 0 ΣΜA0 w a 05a Mo B 2a 0 Guess A 1kN B 1kN A B Find A B A B 4125 375 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a V1 x1 w x1 kN V2 x2 w a A 1 kN V3 x3 w a A 1 kN M1 x1 05 w x1 2 kN m M2 x2 w a x2 05a A x2 a 1 kN m M3 x3 w a x3 05a A x3 a Mo 1 kN m 0 1 2 3 4 40 20 0 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 40 20 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 620 Draw the shear and moment diagrams for the beam and determine the shear and moment throughout the beam as functions of x Given a 24m b 12m P1 50kN w 30 kN m P2 40kN M2 60kN m Solution Equilibrium A w a P1 P2 A 162kN ΣFy0 MA w a 05a P1 a P2 a b M2 ΣΜA0 MA 41040 kN m x1 0 001 a a x2 a 101 a a b V1 x1 A w x1 1 kN V2 x2 A w a P1 1 kN M1 x1 MA A x1 05w x1 2 1 kN m M2 x2 MA A x2 w a x2 05 a P1 x2 a 1 kN m 0 1 2 3 0 50 100 150 200 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 400 200 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 621 Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x where 12 m x 3 m Given a 12m Mo 300N m b 18m w 25 kN m c 12m Solution Equilibrium Given ΣFy0 w b A B 0 ΣΜB0 Mo A b w b 05b Mo 0 Guess A 1kN B 1kN A B Find A B A B 225 225 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 0 kN V2 x2 A w x2 a 1 kN V3 x3 A w b B 1 kN M1 x1 Mo N m M2 x2 Mo A x2 a 05w x2 a 2 1 N m M3 x3 Mo A x3 a w b x3 a 05 b B x3 a b 1 N m 0 1 2 3 4 2 0 2 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 200 0 200 400 600 800 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 622 Draw the shear and moment diagrams for the compound beamThe three segments are connected by pins at B and E Given a 2m b 1m F 3kN LBE a 2 b w 08 kN m LAB a b Solution L 3a 4 b Equilibrium Consider segment AB ΣΜB0 A a b F b 0 A b a b F A 100 kN ΣFy0 A B F 0 B F A B 200 kN Consider segment BE By symmetry E B D C ΣFy0 2C 2B w a 2 b 0 C B w 2 a 2 b C 360 kN D C D 360 kN E B E 200 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a 2 x4 a 2 b 101 a 2 b 2a 2 b x5 2a 2 b 101 2a 2 b 2a 3 b x6 2a 3 b 101 2a 3 b 2a 4 b x7 2a 4 b 101 2a 4 b 3a 4 b V1 x1 A 1 kN V2 x2 A F 1 kN V3 x3 A F w x3 LAB 1 kN V4 x4 A F C w x4 LAB 1 kN V5 x5 A F C D w x5 LAB V6 x6 A F C D w LBE 1 kN V7 x7 A 2F C D w LBE 1 kN M1 x1 A x1 kN m M2 x2 A x2 F x2 a 1 kN m M3 x3 B x3 LAB 05w x3 LAB 2 1 kN m M4 x4 B x4 LAB 05w x4 LAB 2 C x4 LAB b 1 kN m M5 x5 B x5 LAB 05w x5 LAB 2 C x5 LAB b D x5 LAB b a M6 x6 E x6 LAB LBE 1 kN m M7 x7 E x7 LAB LBE F x7 LAB LBE b 1 kN m 0 2 4 6 8 10 4 2 0 2 4 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 V5 x5 V6 x6 V7 x7 x1 x2 x3 x4 x5 x6 x7 0 2 4 6 8 10 3 2 1 0 1 2 Distance m Moment kNm M1 x1 M2 x2 M3 x3 M4 x4 M5 x5 M6 x6 M7 x7 x1 x2 x3 x4 x5 x6 x7 Problem 623 Draw the shear and moment diagrams for the beam Given a 15m Mo 30kN m w 30 kN m Solution Equilibrium Given ΣFy0 w a A w a B 0 ΣΜB0 Mo w a 25a A 2a w a 05a 0 Guess A 1kN B 1kN A B Find A B A B 5750 3250 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2 a 3a V1 x1 w x1 kN V2 x2 w a A 1 kN V3 x3 w a A w x3 2a 1 kN M1 x1 Mo 05w x1 2 kN m M2 x2 Mo w a x2 05a A x2 a 1 kN m M3 x3 Mo w a x3 05a A x3 a 05w x3 2a 2 1 kN m 0 1 2 3 4 40 20 0 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 10 0 10 20 30 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 624 The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed loading on the beam over its 06m length Draw the shear and moment diagrams for the beam if it supports a uniform loading of 30 kNm Given a 03m c 06m b 24m w 30 kN m Solution Equilibrium Given ΣFy0 A w b qB c 0 ΣΜA0 w b a 05b qB c a b 05c 0 Guess A 1kN qB 1 kN m A qB Find A qB A 3600 kN qB 6000 kN m x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A kN V2 x2 A w x2 a 1 kN V3 x3 A w b qB x3 a b 1 kN M1 x1 A x1 kN m M2 x2 A x2 05w x2 a 2 1 kN m M3 x3 A x3 w b x3 a 05 b 05qB x3 a b 2 1 kN m 0 05 1 15 2 25 3 40 20 0 20 40 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 05 1 15 2 25 3 0 10 20 30 40 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 625 Draw the shear and moment diagrams for the beam The two segments are joined together at B Given a 09m P 40kN b 15m w 50 kN m c 24m Solution Equilibrium Given ΣFy0 A P w c C 0 ΣΜB0 w c 05c C c 0 Guess A 1kN C 1kN A C Find A C A C 100 60 kN MA P a C w c a b MA 180kN m x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A kN V2 x2 A P 1 kN V3 x3 A P w x3 a b 1 kN M1 x1 MA A x1 kN m M2 x2 MA A x2 P x2 a 1 kN m M3 x3 MA A x3 P x3 a 05w x3 a b 2 1 kN m 0 1 2 3 4 50 0 50 100 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 200 100 0 100 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 626 Consider the general problem of a cantilevered beam subjected to n concentrated loads and a constant distributed loading w Write a computer program that can be used to determine the internal shear and moment at any specified location x along the beam and plot the shear and moment diagrams for the beam Show an application of the program using the values P1 4 kN d1 2 m w 800 Nm a1 2 m a2 4 m L 4 m Problem 627 Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum Draw the shear and moment diagrams for this condition Problem 628 Draw the shear and moment diagrams for the rod Only vertical reactions occur at its ends A and B Given a 900mm A 360N B 720N w 24 kN m Solution ΣFy0 A 05w xo a xo 0 xo A a 05w xo 51962 mm ΣΜ Mmax A xo 05w xo a xo xo 3 Mmax 12471 N m x 0 001 a a V x A 05w x a x 1 N M x A x 05w x a x x 3 1 N m 0 02 04 06 08 500 0 Distance m Shear N V x x 0 02 04 06 08 0 50 100 Distance m Moment Nm M x x Problem 629 Draw the shear and moment diagrams for the beam Given Set L 1m wo 1 kN m a L 3 Solution Equilibrium Given ΣFy0 A 2 05wo a wo a B 0 ΣΜB0 A 3 a 05 wo a 2a a 3 wo a 15a 05 wo a 2a 3 0 Guess A 1kN B 1kN A B Find A B A B 033 033 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a V1 x1 A wo 2 x1 a x1 1 kN V2 x2 A 05 wo a wo x2 a 1 kN V3 x3 A 05 wo a wo a wo x3 2a 1 05 x3 2a a 1 kN 0 02 04 06 08 04 02 0 02 04 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 M1 x1 A x1 wo 2 x1 a x1 x1 3 1 N m M2 x2 A x2 wo a 2 x2 2a 3 05wo x2 a 2 1 N m M3 x3 wo 2 x3 2 a 2 1 x3 2 a a 1 3 M3 x3 A x3 wo a 2 x3 2 a 3 wo a x3 15 a M3 x3 1 N m 0 02 04 06 08 0 20 40 60 80 100 120 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 630 Draw the shear and moment diagrams for the beam Set L 1m wo 1 kN m Solution Equilibrium Given ΣFy0 A B 05 wo L 0 ΣΜB0 A 2 L 3 wo 2 L L 3 0 Guess A 1kN B 1kN A B Find A B A B 025 025 kN Let a L 3 x1 0 001 a a x2 a 101 a 3a V1 x1 wo 2 x1 L x1 1 kN V2 x2 A wo 2 x2 L x2 1 kN M1 x1 wo 2 x1 L x1 x1 3 1 kN m M2 x2 A x2 a wo 2 x2 L x2 x2 3 1 kN m 0 02 04 06 08 02 0 02 Distance m Shear Wo kN V1 x1 V2 x2 x1 x2 0 02 04 06 08 0 002 004 Distance m Moment WoLL kNm M1 x1 M2 x2 x1 x2 Problem 631 The Tbeam is subjected to the loading shown Draw the shear and moment diagrams Given a 2m P 10kN b 3m w 3 kN m c 3m Solution Equilibrium Given ΣFy0 P A w c B 0 ΣΜB0 P a b c A b c w c 05c 0 Guess A 1kN B 1kN A B Find A B A B 1558 342 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 P kN V2 x2 P A 1 kN V3 x3 P A w x3 a b 1 kN M1 x1 P x1 kN m M2 x2 P x2 A x2 a 1 kN m M3 x3 P x3 A x3 a 05 w x3 a b 2 1 kN m 0 1 2 3 4 5 6 7 8 10 5 0 5 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 2 4 6 8 25 20 15 10 5 0 5 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 632 The ski supports the 900N 90kg weight of the man If the snow loading on its bottom surface is trapezoidal as shown determine the intensity w and then draw the shear and moment diagrams for the ski Given a 05m P 900N Solution Equilibrium ΣFy0 P w 2 2a 4a 0 w P 3a w 600 N m x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a V1 x1 w 2 x1 a x1 1 N V2 x2 05 w a w x2 a 1 N V3 x3 05 w a w a P w x3 2a 1 N V4 x4 05 w a w a P wa w x4 3a 1 05 x4 3a a 1 N 0 05 1 15 2 400 200 0 200 400 Distance m Shear N V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 M1 x1 w 2 x1 a x1 x1 3 1 N m M2 x2 w a 2 x2 2a 3 w 2 x2 a 2 1 N m M3 x3 w a 2 x3 2a 3 w 2 x3 a 2 P x3 2a 1 N m M4 x4 w 2 x4 3 a 2 1 x4 3 a a 1 3 M4 x4 w a 2 x4 2 a 3 2w a x4 2 a P x4 2a M4 x4 1 N m 0 05 1 15 2 0 50 100 150 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 633 Draw the shear and moment diagrams for the beam Given L 9m wo 50 kN m a 05L Solution Equilibrium Given ΣFy0 A 2 05wo a B 0 ΣΜB0 A L 05 wo a a a 3 05 wo a 2a 3 0 Guess A 1kN B 1kN A B Find A B A B 11250 11250 kN x1 0 001 a a x2 a 101 a 2a V1 x1 A wo x1 1 05 x1 a 1 kN V2 x2 A 05 wo a wo 2 x2 a a x2 a 1 kN M1 x1 A x1 wo 2 x1 2 1 x1 a 1 3 1 kN m M2 x2 wo 2 x2 a a x2 a x2 a 3 M2 x2 A x2 wo a 2 x2 a 3 M2 x2 1 kN m 0 5 100 0 100 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 5 0 100 200 Distance m Moment Nm M1 x1 M2 x2 x1 x2 Problem 634 Draw the shear and moment diagrams for the wood beam and determine the shear and moment throughout the beam as functions of x Given a 1m P 1kN b 15m w 2 kN m c 1m Solution Equilibrium Given ΣFy0 A w b B 2P 0 ΣΜB0 P a b A b w b 05b P c 0 Guess A 1kN B 1kN A B Find A B A B 250 250 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 P kN V2 x2 P A w x2 a 1 kN V3 x3 P A w b B 1 kN M1 x1 P x1 kN m M2 x2 P x2 A x2 a 05w x2 a 2 1 kN m M3 x3 P x3 A x3 a w b x3 a 05 b B x3 a b 1 kN m 0 05 1 15 2 25 3 35 2 1 0 1 2 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 05 1 15 2 25 3 35 1 08 06 04 02 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 635 The smooth pin is supported by two leaves A and B and subjected to a compressive load of 04 kNm caused by bar C Determine the intensity of the distributed load w0 of the leaves on the pin and draw the shear and moment diagrams for the pin Given L 100mm w 04 kN m a 02L Solution Equilibrium ΣFy0 2 05wo a w 3a 0 wo 3w wo 120 kN m x1 0 001 a a x2 a 101 a 4a x3 4a 101 4a L V1 x1 wo 2 x1 a x1 1 N V2 x2 wo 2 a w x2 a 1 N V3 x3 wo 2 a w 3a wo x3 4a 1 05 x3 4a a 1 N 0 002 004 006 008 10 0 10 Distance m Shear N V1 x1 V2 x2 V3 x3 x1 x2 x3 M1 x1 wo 2 x1 a x1 x1 3 1 N m M2 x2 wo a 2 x2 2a 3 05w x2 a 2 1 N m M3 x3 wo 2 x3 4 a 2 1 x3 4 a a 1 3 M3 x3 wo a 2 x3 2 a 3 w 3a x3 25 a M3 x3 1 N m 0 002 004 006 008 0 01 02 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 636 Draw the shear and moment diagrams for the beam Given a 36m b 18m MA 225kN m w 45 kN m Solution Equilibrium Given ΣFy0 A B 05w b 0 ΣΜB0 MA A a 05w b b 3 0 Guess A 1kN B 1kN A B Find A B A B 738 4788 kN x1 0 001 a a x2 a 101 a a b V1 x1 A kN V2 x2 A B w x2 a 1 05 x2 a b 1 kN M1 x1 MA A x1 1 kN m M2 x2 MA A x2 B x2 a w 2 x2 a 2 1 x2 a b 1 3 1 kN m 0 1 2 3 4 5 20 0 20 40 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 4 5 30 20 10 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 637 The compound beam consists of two segments that are pinned together at B Draw the shear and moment diagrams if it supports the distributed loading shown Set L 1m a L 3 w 1 kN m Solution Consider segment AB ΣMB0 A 2a w 2 2a L 2 a 2a 3 0 A 2w L 27 A 00741 kN Consider whole beam ABC MC A L w L 2 L 3 MC 5 w L2 54 MC 009259 kN m x1 0 001 2a 2a V x A w 2 x L x 1 kN M x A x w 2 x L x x 3 1 kN m 0 05 1 04 02 0 02 Distance m Shear WL kN V x x 0 05 1 01 005 0 Distance m Moment wLL kNm M x x Problem 638 Draw the shear and moment diagrams for the beam Given L 3m wo 12 kN m w1 18 kN m Solution Equilibrium ΣFy0 B wo w1 L 2 ΣΜB0 MB wo L L 2 w1 wo L 2 L 3 B 4500 kN MB 6300 kN m w w1 wo x1 0 001 L L V x wo x w 2 x L x 1 kN M x wo x x 2 w 2 x L x x 3 1 kN m 0 1 2 3 40 20 0 Distance m Shear kN V x x 0 1 2 3 50 0 Distance m Moment kNm M x x Problem 639 Draw the shear and moment diagrams for the beam and determine the shear and moment as functions of x Given a 3m wo 200 N m w1 400 N m Solution L 2a w w1 wo Equilibrium Given ΣFy0 A wo w1 a 2 B 0 ΣΜB0 A 2 a wo a a 2 w a 2 a 3 0 Guess A 1N B 1N A B Find A B A B 200 700 N Shear and Moment Functions For 0 x 3m V A V 200 N Ans M A x M 200x N m Ans For 3m x 6m V A wo x a w 2 x a a x a V 500 100 3 x2 N Ans M A x wo x a x a 2 w 2 x a a x a x a 3 M 600 500x 100 9 x3 N m Ans x1 0 001 a a x2 a 101 a L V1 x1 A N V2 x2 A wo x2 a w 2 x2 a a x2 a 1 N M1 x1 A x1 1 N m M2 x2 A x2 wo x2 a x2 a 2 w 2 x2 a L x2 a x2 a 3 1 N m 0 2 4 6 500 0 500 Distance m Shear N V1 x1 V2 x2 x1 x2 0 2 4 6 0 200 400 600 Distance m Moment Nm M1 x1 M2 x2 x1 x2 Problem 640 Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum Draw the shear and moment diagrams for this condition Solution Equilibrium ΣFy0 A 2P B 0 ΣΜA0 P L 2 B a P L 0 B 3 L 2a P A 4a 3L 2a P Internal Moment For positive moment Mmax A L 2 For negative moment Mmin P L a When Mmax Mmin A L 2 P L a 4a 3L 2a P L 2 P L a 4a 3L L 4a L a a 3 2 L Ans Set L 1m P 1kN a 3 2 L A 4a 3L 2a P B 3 L 2a P a 05L b a a c L a x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L V1 x1 A 1 kN V2 x2 A P 1 kN V3 x3 A P B 1 kN M1 x1 A x1 kN m M2 x2 A x2 P x2 a 1 kN m M3 x3 A x3 P x3 a B x3 a b 1 kN m 0 02 04 06 08 1 05 0 05 1 Distance m Shear P kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 02 04 06 08 01 0 01 Distance m Moment PL Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 641 Draw the shear and moment diagrams for the beam Given a 2m w0 8 kN m Solution unit kN m3 w unit 2 x2 Wx 0 x w x d Wx unit 0 x x 2x2 d Wx unit 2 3 x3 Wa unit 2 3 a3 ΣFy0 A 0 a w x d 0 A unit 0 a x 2x2 d A 533 kN xc 0 a w x x d A xc unit A 0 a x 2 x3 d xc 1500 m MA A xc MA 800 kN m x 0 001 a a V x A unit 2 3 x3 1 kN M x MA A x unit 2 3 x3 x 1 xc a 1 kN m 0 05 1 15 2 0 2 4 6 Distance m Shear kN V x x 0 05 1 15 2 8 6 4 2 0 Distance m Moment kNm M x x Problem 642 The truck is to be used to transport the concrete column If the column has a uniform weight of w forcelength determine the equal placement a of the supports from the ends so that the absolute maximum bending moment in the column is as small as possible Also draw the shear and moment diagrams for the column Solution Support Reactions By symmetry ABR ΣFy0 2R wL 0 R 05w L Internal Moment For negative moment Mmin 05 w a2 For positive moment Mmax w L 2 L 4 R L 2 a at midspan Mmax w L 8 4a L For optimal minimum Mmax Mmin w L 8 4a L 1 2 w a2 4a L L 4 a2 Let α a L α2 α 025 0 α 1 2 1 12 4 025 α 02071 a 02071L Ans Set L 1m w 1 kN m a αL R 05w L b L 2a x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L V1 x1 w x1 kN V2 x2 R w x2 1 kN V3 x3 2R w x3 1 kN M1 x1 w 2 x1 2 1 kN m M2 x2 R x2 a w 2 x2 2 1 kN m M3 x3 R x3 a R x3 a b w 2 x3 2 1 kN m 0 02 04 06 08 04 02 0 02 04 Distance m Shear wL kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 02 04 06 08 004 002 0 002 004 Distance m Moment wLL kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 643 A member having the dimensions shown is to be used to resist an internal bending moment of M 2 kNm Determine the maximum stress in the member if the moment is applied a about the z axis b about the y axis Sketch the stress distribution for each case Given d 120mm b 60mm Mz 2kN m My 2kN m Solution Iz 1 12 b d3 Iy 1 12 d b3 Maximum Stress σ M c I a About the z axis ymax d 2 σmax Mz ymax Iz σmax 1389 MPa Ans b About the y axis zmax b 2 σmax My zmax Iy σmax 2778 MPa Ans Problem 644 The steel rod having a diameter of 20 mm is subjected to an internal moment of M 300 Nm Determine the stress created at points A and B Also sketch a threedimensional view of the stress distribution acting over the cross section Given d 20mm M 300N m θ 45deg Solution I π 4 d 2 4 σ M y I yA d 2 σA M yA I σA 38197 MPa Ans yB d 2 sin θ σB M yB I σB 27009 MPa Ans Problem 645 The beam is subjected to a moment M Determine the percentage of this moment that is resisted by th stresses acting on both the top and bottom boardsA and B of the beam Given bf 200mm tf 25mm tw 25mm dw 150mm Solution D dw 2tf I 1 12 bf D3 bf 2tw dw 3 Bending Stress σ M c I Set M 1kN m co 05D σo M co I σo 1097143 MPa ci 05dw σi M ci I σi 0822857 MPa Resultant Force and Moment For board A or B F 1 2 σo σi bf tf F 4800 kN Centroid of force σi bf tf tf 2 1 2 σo σi bf tf tf 3 F yc yc 1 F σi bf tf tf 2 1 2 σo σi bf tf tf 3 yc 11905 mm M F D 2yc M 08457 kN m Hence M M M 100 M 8457 Ans Problem 646 Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of σD 30 MPa Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam Given bf 200mm tf 25mm dw 150mm tw 25mm σD 30MPa Solution D dw 2tf I 1 12 bf D3 bf 2tw dw 3 I 9114583333 mm4 Bending Stress σ M c I cD 05dw σD M cD I M σD I cD M 3646 kN m Ans cmax 05D σmax M cmax I σmax 4000 MPa Ans Problem 647 The slab of marble which can be assumed a linear elastic brittle material has a specific weight of 24 kNm3 and a thickness of 20 mm Calculate the maximum bending stress in the slab if it is supported a on its side and b on its edges If the fracture stress is σ f 15 MPa explain the consequences of supporting the slab in each position Given t 20mm L 15m d 05m γ 24 kN m3 σf 15MPa Solution w γ d t w 024 kN m Mmax 1 8 w L2 Is 1 12 t d3 Ie 1 12 d t3 Maximum Stress σ M c I a Supported on its side c1 d 2 σmax Mmax c1 Is σmax 0081 MPa Ans b Supported on its edges c2 t 2 σmax Mmax c2 Ie σmax 2025 MPa Ans σf 15 MPa The marble slab will break if it is supported as in case b Problem 648 The slab of marble which can be assumed a linear elastic brittle material has a specific weight of 24 kNm3 If it is supported on its edges as shown in b determine the minimum thickness it should have without causing it to breakThe fracture stress is σ f 15 MPa Given L 15m d 05m γ 24 kN m3 σf 15MPa Solution w γ d t Mmax 1 8 w L2 Mmax 1 8 γ d t L2 Maximum Stress c t 2 Ie 1 12 d t3 σmax Mmax c Ie σmax Mmax 6 d t2 Thus σmax 1 8 γ d t L2 6 d t2 t 3 γ L2 4 σf t 27mm Ans Problem 649 A beam has the cross section shown If it is made of steel that has an allowable stress of σallow 170 MPa determine the largest internal moment the beam can resist if the moment is applied a about the z axis b about the y axis Given bf 120mm tf 5mm d 120mm tw 5mm σallow 170MPa Solution D d 2tf Iz 1 12 bf D3 bf tw d3 Iy 2 1 12 tf bf 3 1 12 d tw 3 Bending Stress σallow M c I a About the z axis cz D 2 Mz σallow Iz cz Mz 1415 kN m Ans b About the y axis cy bf 2 My σallow Iy cy My 408 kN m Ans Problem 650 Two considerations have been proposed for the design of a beam Determine which one will support a moment of with the least amount of M 150 kNm bending stress What is that stress By what percentage is it more effective Given bf 200mm dw 300mm tfa 15mm twa 30mm tfb 30mm twb 15mm M 150kN m Solution Section Property For section a Da dw 2tfa Ia 1 12 bf Da 3 bf twa dw 3 For section b Db dw 2tfb Ib 1 12 bf Db 3 bf twb dw 3 Maximum Bending Stress σ M c I For section a cmax 05Da σmax M cmax Ia σmax 11435 MPa For section b cmax 05Db σmax M cmax Ib σmax 7472 MPa Ans By comparison section b will have the least amount of bending stress eff σmax σmax σmax 100 eff 5303 Ans Problem 651 The aluminum machine part is subjected to a moment of Determine the bending stress M 75 Nm created at points B and C on the cross section Sketch the results on a volume element located at each of these points Given bf 80mm tf 10mm tw 10mm dw 40mm M 75N m Solution D dw tf y Σ yi Ai Σ Ai yc bf tf 05 tf 2 dw tw 05dw tf bf tf 2dw tw yc 1750 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 tw dw 3 dw tw yc 05dw tf 2 I If 2Iw Bending Stress σ M c I At B cB yc σB M cB I σB 3612 MPa Ans At C cC yc tf σC M cC I σC 1548 MPa Ans Problem 652 The aluminum machine part is subjected to a moment of M 75 Nm Determine the maximum tensil and compressive bending stresses in the part Given bf 80mm tf 10mm tw 10mm dw 40mm M 75N m Solution D dw tf y Σ yi Ai Σ Ai yc bf tf 05 tf 2 dw tw 05dw tf bf tf 2dw tw yc 1750 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 tw dw 3 dw tw yc 05dw tf 2 I If 2Iw Bending Stress σ M c I For compression cc yc σcmax M cc I σcmax 3612 MPa Ans For tension ct D yc σtmax M ct I σtmax 6709 MPa Ans Problem 653 A beam is constructed from four pieces of wood glued together as shown If the moment acting on the cross section is M 450 Nm determine the resultant force the bending stress produces on the top board A and on the side board B Given bf 240mm tf 15mm tw 20mm dw 200mm M 450N m Solution D dw 2tf Iy 1 12 D bf 3 dw bf 2tw 3 Bending Stress σ M c I co 05bf σo M co Iy σo 0410251 MPa ci 05bf tw σi M ci Iy σi 0341876 MPa Resultant Force For board A or B FA σo 2 bf 2 tf σo 2 bf 2 tf FA 0kN Ans FB 1 2 σo σi dw tw FB 1504 kN Ans Problem 654 The aluminum strut has a crosssectional area in the form of a cross If it is subjected to the moment M 8 kNm determine the bending stress acting at points A and B and show the results acting on volume elements located at these points Given bf 50mm tf 20mm tw 20mm dw 220mm M 8kN m Solution I 1 12 tw dw 3 2 1 12 bf tf 3 I 1781333333 mm4 Bending Stress σ M c I At A cA 05dw σA M cA I σA 49401 MPa Ans At B cB 05tf σB M cB I σB 4491 MPa Ans Problem 655 The aluminum strut has a crosssectional area in the form of a cross If it is subjected to the moment M 8 kNm determine the maximum bending stress in the beam and sketch a threedimensional view of the stress distribution acting over the entire crosssectional area Given bf 50mm tf 20mm tw 20mm dw 220mm M 8kN m Solution I 1 12 tw dw 3 2 1 12 bf tf 3 I 1781333333 mm4 Bending Stress σ M c I cmax 05dw σmax M cmax I σmax 49401 MPa Ans At B cB 05tf σB M cB I σB 4491 MPa Problem 656 The beam is made from three boards nailed together as shown If the moment acting on the cross section is M 15 kNm determine the maximum bending stress in the beam Sketch a threedimensional view of the stress distribution acting over the cross section Given bf 250mm bf 150mm tf 38mm tw 25mm d 300mm M 15kN m Solution D d 2tf y Σ yi Ai Σ Ai yc bf tf 05 tf d tw 05d tf bf tf D 05tf bf tf d tw bf tf yc 15971 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 tw d3 d tw yc 05d tf 2 If 1 12 bf tf 3 bf tf yc D 05tf 2 I If Iw If Bending Stress σ M c I At B cmax D yc σmax M cmax I σmax 0684 MPa Ans At A cA cmax tf σA M cA I σA 0564 MPa At C cC yc tf σC M cC I σC 0385 MPa At D cD yc σD M cD I σD 0505 MPa Problem 657 Determine the resultant force the bending stresses produce on the top board A of the beam if M 15 kNm Given bf 250mm bf 150mm tf 38mm tw 25mm d 300mm M 15kN m Solution D d 2tf y Σ yi Ai Σ Ai yc bf tf 05 tf d tw 05d tf bf tf D 05tf bf tf d tw bf tf yc 15971 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 tw d3 d tw yc 05d tf 2 If 1 12 bf tf 3 bf tf yc D 05tf 2 I If Iw If Bending Stress σ M c I At C cC yc tf σC M cC I σC 0385 MPa At D cD yc σD M cD I σD 0505 MPa The resultant Force For top board A F 05 σC σD bf tf F 423 kN Ans Problem 658 The control level is used on a riding lawn mower Determine the maximum bending stress in the lever at section aa if a force of 100 N is applied to the handle The lever is supported by a pin at A and a wire at B Section aa is square 6 mm by 6 mm Given L 50mm b 6mm d 6mm F 100N Solution I 1 12 b d3 M F L M 500 N m Bending Stress σ M c I c d 2 σmax M c I σmax 13889 MPa Ans Problem 659 Determine the largest bending stress developed in the member if it is subjected to an internal bending moment of M 40 kNm Given bf 100mm tf 10mm rf 30mm tw 10mm dw 180mm M 40kN m Solution D dw tf 2rf y Σ yi Ai Σ Ai yc bf tf 05 tf dw tw 05dw tf π rf 2 rf dw tf bf tf dw tw π rf 2 yc 14341 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 tw dw 3 dw tw yc 05dw tf 2 If π 4 rf 4 π rf 2 yc rf dw tf 2 I If Iw If Bending Stress σ M c I Maximum stress occurs at the bottom fibre cmax yc σmax M cmax I σmax 12851 MPa Ans Problem 660 The tapered casting supports the loading shown Determine the bending stress at points A and B The cross section at section aa is given in the figure Given La 250mm P 750N Lb 375mm Lc 125mm b 100mm t 25mm d 75mm Solution Equilibrium ΣΜC0 F1 2Lb Lc P Lc Lb P Lb 0 F1 P Lc Lb P Lb 2 Lb Lc F1 75000 N Section aa D d 2t M F1 La M 18750 N m I 1 12 b D3 b d3 Bending Stress σ M c I cA D 2 σA M cA I σA 0918 MPa Ans cB d 2 σB M cB I σB 0551 MPa Ans Problem 661 If the shaft in Prob 61 has a diameter of 100 mm determine the absolute maximum bending stress in the shaft Given a 250mm b 800mm F 24kN do 100mm Solution Equilibrium Given ΣFy0 A B F 0 ΣΜA0 F a B b 0 Guess A 1kN B 1kN A B Find A B A B 3150 750 kN x1 0 001 a a x2 a 101 a a b M1 x1 F x1 kN m M2 x2 F x2 A x2 a 1 kN m 0 05 1 5 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Max Moment unit kN m M1 a 600 Bending Stress M M1 a unit I π do 4 64 c do 2 σ M c I σmax M c I σmax 6112 MPa Ans Problem 662 If the shaft in Prob 63 has a diameter of 40 mm determine the absolute maximum bending stress in the shaft Given a 350mm b 500mm c 375mm d 300mm do 40mm B 400N C 550N E 175N Solution Equilibrium Given ΣFy0 A D B C E 0 ΣΜD0 A a b c B b c C c E d 0 Guess A 1N D 1N A D Find A D A D 41122 71378 N x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c x4 a b c 101 a b c a b c d M1 x1 A x1 N m M2 x2 A x2 B x2 a 1 N m M3 x3 A x3 B x3 a C x3 a b 1 N m M4 x4 A x4 B x4 a C x4 a b D x4 a b c 1 N m 0 02 04 06 08 1 12 14 100 0 100 200 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Max Moment unit N m M3 a b 14954 Bending Stress M M3 a b unit I π do 4 64 c do 2 σ M c I σmax M c I σmax 238 MPa Ans Problem 663 If the shaft in Prob 66 has a diameter of 50 mm determine the absolute maximum bending stress in the shaft Given a 125mm b 600mm c 75mm F1 08kN F2 15kN do 50mm Solution L a b c Equilibrium Given ΣFy0 A F1 F2 B 0 ΣΜB0 A L F1 b c F2 c 0 Guess A 1kN B 1kN A B Find A B A B 08156 14844 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 A x1 N m M2 x2 A x2 F1 x2 a 1 N m M3 x3 A x3 F1 x3 a F2 x3 a b 1 N m 0 02 04 06 0 50 100 150 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit N m M2 a b 11133 Bending Stress M M2 a b unit I π do 4 64 c do 2 σ M c I σmax M c I σmax 9072 MPa Ans Problem 664 If the shaft in Prob 68 has a diameter of 30 mm and thickness of 10 mm determine the absolute maximum bending stress in the shaft Given a 400mm h 80mm F 5kN do 30mm t 10mm Solution Given di do 2t Equilibrium ΣFy0 A C 0 ΣΜC0 A a F h 0 Guess A 1N C 1N A C Find A C A C 100 100 kN x1 0 001 a a M1 x1 A x1 N m 0 02 0 600 400 200 0 Distane m Moment Nm M1 x1 x1 Max Moment unit N m M1 a 40000 Bending Stress M M1 a unit I π 64 do 4 di 4 c do 2 σ M c I σmax M c I σmax 1528 MPa Ans Problem 665 If the beam ACB in Prob 69 has a square cross section 150 mm by 150 mm determine the absolute maximum bending stress in the beam Given a 1m b 1m c 1m d 025m ao 150mm F1 75kN F2 100kN Solution Equilibrium Given ΣFy0 A F1 B 0 ΣΜC0 A a b c F1 b c F2 d 0 Guess A 1kN B 1kN A B Find A B A B 5833 1667 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 A x1 kN m M2 x2 A x2 F1 x2 a 1 kN m M3 x3 A x3 F1 x3 a F2 d 1 kN m 0 05 1 15 2 25 3 0 50 100 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit kN m M1 a 5833 Bending Stress M M1 a unit I ao 4 12 c ao 2 σ M c I σmax M c I σmax 1037 MPa Ans Problem 666 If the crane boom ABC in Prob 610 has a rectangular cross section with a base of 60 mm determine its required height h to the nearest multiples of 5 mm if the allowable bending stress is σallow 170 MPa Given a 09m b 15m bo 60mm c 12m W 6kN σallow 170MPa Solution d a2 c2 v c d h a d Equilibrium Given ΣFy0 Ay B v W 0 ΣΜA0 B v a W a b 0 ΣFx0 Ax B h 0 Guess Ax 1kN Ay 1kN B 1kN Ax Ay B Find Ax Ay B Ax Ay B 12 10 20 kN x1 0 001 a a x2 a 101 a a b M1 x1 Ay x1 kN m M2 x2 Ay x2 B v x2 a 1 kN m 0 1 2 10 5 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Max Moment unit kN m M1 a 900 Bending Stress M M1 a unit I bo ho 3 12 c ho 2 σ M c I ho 6 M bo σallow ho 7276 mm Use ho 75mm Ans Problem 667 If the crane boom ABC in Prob 610 has a rectangular cross section with a base of 50 mm and a height of 75 mm determine the absolute maximum bending stress in the boom Given a 09m b 15m bo 50mm c 12m W 6kN ho 75mm Solution d a2 c2 v c d h a d Equilibrium Given ΣFy0 Ay B v W 0 ΣΜA0 B v a W a b 0 ΣFx0 Ax B h 0 Guess Ax 1kN Ay 1kN B 1kN Ax Ay B Find Ax Ay B Ax Ay B 12 10 20 kN x1 0 001 a a x2 a 101 a a b M1 x1 Ay x1 kN m M2 x2 Ay x2 B v x2 a 1 kN m 0 1 2 10 5 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Max Moment unit kN m M1 a 900 Bending Stress M M1 a unit I bo ho 3 12 c ho 2 σ M c I σmax M c I σmax 192MPa Ans Problem 668 Determine the absolute maximum bending stress in the beam in Prob 624 The cross section is rectangular with a base of 75 mm and height of 100 mm Given a 03m b 24m c 06m bo 75mm ho 100mm w 30 kN m Solution Equilibrium Given ΣFy0 A w b qB c 0 ΣΜA0 w b a 05b qB c a b 05c 0 Guess A 1kN qB 1 kN m A qB Find A qB A 3600 kN qB 6000 kN m x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 A x1 kN m M2 x2 A x2 05w x2 a 2 1 kN m M3 x3 A x3 w b x3 a 05 b 05qB x3 a b 2 1 kN m 0 1 2 3 0 20 40 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit kN m b 05 b M2 a b 3240 Bending Stress M M2 a b unit I bo ho 3 12 c ho 2 σ M c I σmax M c I σmax 2592 MPa Ans Problem 669 Determine the absolute maximum bending stress in the beam in Prob 625 Each segment has a rectangular cross section with a base of 100 mm and height of 200 mm Given a 09m b 15m c 24m bo 100mm ho 200mm P 40kN w 50 kN m Solution Equilibrium Given ΣFy0 A P w c C 0 ΣΜB0 w c 05c C c 0 Guess A 1kN C 1kN A C Find A C A C 100 60 kN MA P a C w c a b MA 180kN m x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 MA A x1 kN m M2 x2 MA A x2 P x2 a 1 kN m M3 x3 MA A x3 P x3 a 05w x3 a b 2 1 kN m 0 2 4 200 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit kN m c 05 c M3 a b c 3600 Bending Stress M M2 a b c unit I bo ho 3 12 co ho 2 σ M c I σmax M co I σmax 108MPa Ans Problem 670 Determine the absolute maximum bending stress in the 20mmdiameter pin in Prob 635 Given L 100mm w 04 kN m a 02L do 20mm Solution Equilibrium ΣFy0 2 05wo a w 3a 0 wo 3w wo 120 kN m x1 0 001 a a x2 a 101 a 4a x3 4a 101 4a L M1 x1 wo 2 x1 a x1 x1 3 1 N m M2 x2 wo a 2 x2 2a 3 05w x2 a 2 1 N m M3 x3 wo 2 x3 4 a 2 1 x3 4 a a 1 3 M3 x3 wo a 2 x3 2 a 3 w 3a x3 25 a M3 x3 1 N m 0 002 004 006 008 0 01 02 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit N m M2 5L 0260 Bending Stress M M2 5L unit I π 64 do 4 c do 2 σ M c I σmax M c I σmax 0331 MPa Ans Problem 671 The member has a cross section with the dimensions shown Determine the largest internal moment M that can be applied without exceeding allowable tensile and compressive stresses of σ t allow 150 MPa and σ c allow 100 MPa respectively Given bf 100mm tf 10mm rf 30mm tw 10mm dw 180mm σtallow 150MPa σcallow 100MPa Solution D dw tf 2rf y Σ yi Ai Σ Ai yc bf tf 05 tf dw tw 05dw tf π rf 2 rf dw tf bf tf dw tw π rf 2 yc 14341 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 tw dw 3 dw tw yc 05dw tf 2 If π 4 rf 4 π rf 2 yc rf dw tf 2 I If Iw If I 4463960823 mm4 Maximum Bending Stress σ M c I Assume failure due to tensile stress ctmax yc σtmax M ctmax I Mt σtallow I ctmax Mt 4669 kN m Assume failure due to compressive stress ccmax D yc σcmax M ccmax I Mc σcallow I ccmax Mc 4188 kN m Mallow min Mt Mc Mallow 4188 kN m Ans Problem 672 Determine the absolute maximum bending stress in the 30mmdiameter shaft which is subjected to th concentrated forces The sleeve bearings at A and B support only vertical forces Given a 08m b 12m c 06m F1 06kN F2 04kN do 30mm Solution L a b c Equilibrium Given ΣFy0 A F1 F2 B 0 ΣΜB0 A b F1 a b F2 c 0 Guess A 1kN B 1kN A B Find A B A B 08 02 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 F1 x1 N m M2 x2 F1 x2 A x2 a 1 N m M3 x3 F1 x3 A x3 a B x3 a b 1 N m 0 05 1 15 2 25 600 400 200 0 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit N m M1 a 480000 Bending Stress M M1 a unit I π 64 do 4 c do 2 σ M c I σmax M c I σmax 1811 MPa Ans Problem 673 Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces The sleeve bearings at A and B support only vertical forces and the allowable bending stress is σallow 160 MPa Given a 08m b 12m c 06m F1 06kN F2 04kN σallow 160MPa Solution L a b c Equilibrium Given ΣFy0 A F1 F2 B 0 ΣΜB0 A b F1 a b F2 c 0 Guess A 1kN B 1kN A B Find A B A B 08 02 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 F1 x1 N m M2 x2 F1 x2 A x2 a 1 N m M3 x3 F1 x3 A x3 a B x3 a b 1 N m 0 05 1 15 2 25 600 400 200 0 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit N m M1 a 480000 Bending Stress M M1 a unit I π 64 do 4 c do 2 σ M c I σallow M 32 π do 3 do 3 32 M π σallow do 3126 mm Ans Problem 674 Determine the absolute maximum bending stress in the 40mmdiameter shaft which is subjected to the concentrated forces The sleeve bearings at A and B support only vertical forces Given a 300mm b 450mm c 375mm do 40mm F1 2kN F2 15kN Solution Equilibrium Given ΣFy0 A F1 B F2 0 ΣΜB0 A a b F1 b F2 c 0 Guess A 1kN B 1kN A B Find A B A B 045 305 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A 1 kN V2 x2 A F1 1 kN V3 x3 A F1 B 1 kN M1 x1 A x1 N m M2 x2 A x2 F1 x2 a 1 N m M3 x3 A x3 F1 x3 a B x3 a b 1 N m 0 02 04 06 08 1 2 1 0 1 2 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 02 04 06 08 1 1000 500 0 500 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit N m M3 a b 56250 Bending Stress M M2 a b unit I π do 4 64 co do 2 σ M c I σmax M co I σmax 8952 MPa Ans Problem 675 Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces The sleeve bearings at A and B support only vertical forces and the allowable bending stress is σallow 150 MPa Given a 300mm b 450mm c 375mm σallow 150MPa F1 2kN F2 15kN Solution Equilibrium Given ΣFy0 A F1 B F2 0 ΣΜB0 A a b F1 b F2 c 0 Guess A 1kN B 1kN A B Find A B A B 045 305 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 A x1 N m M2 x2 A x2 F1 x2 a 1 N m M3 x3 A x3 F1 x3 a B x3 a b 1 N m 0 05 1 1000 500 0 500 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit N m M3 a b 56250 Bending Stress M M3 a b unit I π do 4 64 co do 2 σ M co I do 3 32M π σallow do 3368 mm Ans Problem 676 The bolster or main supporting girder of a truck body is subjected to the uniform distributed load Determine the bending stress at points A and B Given L1 24m L2 36m b 150mm tf 20mm d 300mm tw 12mm w 25 kN m Solution By symmetry F1 R F2 R Equilibrium ΣFy0 w L1 L2 2R 0 R 05 w L1 L2 MAB R L1 05w L1 2 Section properties D d 2 tf I 1 12 b D3 b tw d3 Bending Stress σ M c I cB d 2 σB MAB cB I σB 896 MPa Ans cA d 2 tf σA MAB cA I σA 7765 MPa Ans Problem 677 A portion of the femur can be modeled as a tube having an inner diameter of 95 mm and an outer diameter of 32 mm Determine the maximum elastic static force P that can be applied to its center without causing failure Assume the bone to be roller supported at its ends The σε diagram for the bone mass is shown and is the same in tension as in compression Given L1 100mm L2 100mm di 95mm do 32mm εe 002 mm mm σe 875MPa εr 006 mm mm σr 161MPa Solution By symmetry R 05P Mmax R L1 Mmax 05P L1 Section properties I π 64 do 4 di 4 Bending Stress σ M c I c do 2 M 2σ I do σmax σe Requires P 2σe I 05 L1 do P 5586 N Ans Problem 678 If the beam in Prob 620 has a rectangular cross section with a width of 200 mm and a height of 400 mm determine the absolute maximum bending stress in the beam Given a 24m b 12m P1 50kN w 30 kN m M2 60kN m P2 40kN bo 200mm do 400mm Solution Equilibrium A w a P1 P2 A 162kN ΣFy0 MA w a 05a P1 a P2 a b M2 ΣΜA0 MA 41040 kN m As indicated in the moment diagram the maximum moment is MA Section properties I 1 12 bo do 3 Bending Stress σ M c I co do 2 σmax MA co I σmax 7695 MPa Ans Problem 679 If the shaft has a diameter of 375 mm determine the absolute maximum bending stress in the shaft Given a 450mm b 600mm c 300mm do 375mm F1 1000N F2 750N Solution Equilibrium Given ΣFy0 A 2F1 B 2F2 0 ΣΜB0 2 F1 a b A b 2F2 c 0 Guess A 1kN B 1kN A B Find A B A B 275 075 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 2 F1 x1 kN m M2 x2 2 F1 x2 A x2 a 1 kN m M3 x3 2 F1 x3 A x3 a B x3 a b 1 kN m 0 02 04 06 08 1 12 1 05 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit kN m M1 a 0900 Bending Stress Mmax M1 a unit I π do 4 64 co do 2 σ M c I σmax Mmax co I σmax 17384 MPa Ans Problem 680 If the beam has a square cross section of 225 mm on each side determine the absolute maximum bending stress in the beam Given a 25m b 25m P 6kN w 15 kN m bo 225mm do 225mm Solution Equilibrium A w a P A 435 kN ΣFy0 MA w a 05a P a b ΣΜA0 MA 7688 kN m x1 0 001 a a x2 a 101 a a b M1 x1 MA A x1 05w x1 2 1 kN m M2 x2 MA A x2 w a x2 05 a 1 kN m 0 2 4 50 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 As indicated in the moment diagram the maximum moment is MA Section properties I 1 12 bo do 3 Bending Stress σ M c I co do 2 σmax MA co I σmax 4049 MPa Ans Problem 681 The beam is subjected to the load P at its center Determine the placement a of the supports so that th absolute maximum bending stress in the beam is as large as possible What is this stress Solution Equilibrium By symmetry ABR ΣFy0 2R P 0 R 05P Max Moment Mmax R 05L a Mmax 05P 05L a For the largest Mmax require a 0 Ans Mmax P L 4 Bending Stress σ M c I I b d3 12 c d 2 σmax Mmax c I σmax P L 2 3 b d2 Ans Problem 682 If the beam in Prob 623 has a cross section as shown determine the absolute maximum bending stress in the beam Given Mo 30kN m w 30 kN m a 15m b 100mm tf 12mm d 168mm tw 6mm Solution Equilibrium Given ΣFy0 w a A w a B 0 ΣΜB0 Mo w a 25a A 2a w a 05a 0 Guess A 1kN B 1kN A B Find A B A B 5750 3250 kN x1 0 001 a a x2 a 101 a 2a x3 2a 101 2 a 3a M1 x1 Mo 05w x1 2 kN m M2 x2 Mo w a x2 05a A x2 a 1 kN m M3 x3 Mo w a x3 05a A x3 a 05w x3 2a 2 1 kN m 0 1 2 3 4 0 20 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 As indicated in the moment diagram the maximum moment is Mo Section properties D d 2 tf I 1 12 b D3 b tw d3 Bending Stress σ M c I co D 2 σmax Mo co I σmax 13187 MPa Ans Problem 683 The pin is used to connect the three links together Due to wear the load is distributed over the top and bottom of the pin as shown on the freebody diagram If the diameter of the pin is 10 mm determine the maximum bending stress on the crosssectional area at the center section aa For the solution it is first necessary to determine the load intensities w1 and w2 Given P 2kN a 25mm b 375mm do 10mm Solution Maa P a 3 b 2 P b 2 Maa 166667 N m Bending Stress I π do 4 64 co do 2 σ M c I σmax Maa co I σmax 16977 MPa Ans Problem 684 A shaft is made of a polymer having an elliptical crosssection If it resists an internal moment of M 50 Nm determine the maximum bending stress developed in the material a using the flexure formul where Iz 14 π 008 m004 m3 b using integration Sketch a threedimensional view of the stress distribution acting over the crosssectional area Given a 80mm b 40mm Mz 50N m Solution a Using the flexure formula cmax b Iz π a b3 4 Iz 402123860 mm4 σmax Mz cmax Iz σmax 0497 MPa Ans b Using integration y2 b2 z2 a2 1 z a b b2 y2 Izo A A y2 d Izo b b y y2 2z d Izo b b y 2y2 a b b2 y2 d Izo 402163198 mm4 Bending Stress σmax Mz cmax Izo σmax 0497 MPa Ans Problem 685 Solve Prob 684 if the moment is M 50 Nm applied about the y axis instead of the z axis Here Iy 14 π 004 m008 m3 Given a 80mm b 40mm My 50N m Solution a Using the flexure formula cmax a Iy π b a3 4 Iy 1608495439 mm4 σmax My cmax Iy σmax 0249 MPa Ans b Using integration y2 b2 z2 a2 1 y b a a2 z2 Iyo A A z2 d Iyo b b z z2 2y d Iyo a a z 2z2 b a a2 z2 d Iyo 1608652794 mm4 Bending Stress σmax My cmax Iyo σmax 0249 MPa Ans Problem 686 The simply supported beam is made from four 16mmdiameter rods which are bundled as shown Determine the maximum bending stress in the beam due to the loading shown Given L1 05m L2 15m do 16mm P 400N Solution By symmetry F1 R F2 R Equilibrium ΣFy0 2 P 2 R 0 R P Mmax R L1 05L2 P 05L2 Section properties A π 4 do 2 I 4 π 64 do 4 A do 2 2 Bending Stress σ M c I cmax do σmax Mmax cmax I σmax 4974 MPa Ans Problem 687 Solve Prob 686 if the bundle is rotated 45 and set on the supports Given L1 05m L2 15m do 16mm P 400N Solution By symmetry F1 R F2 R Equilibrium ΣFy0 2 P 2 R 0 R P Mmax R L1 05L2 P 05L2 Section properties A π 4 do 2 d 05 do 2 do 2 I 2 π 64 do 4 2 π 64 do 4 A d2 Bending Stress σ M c I cmax d do 2 σmax Mmax cmax I σmax 6004 MPa Ans Problem 688 The steel beam has the crosssectional area shown Determine the largest intensity of distributed load w0 that it can support so that the maximum bending stress in the beam does not exceed σ max 150 MPa Given L1 4m L2 4m b 200mm t 8mm d 250mm σmax 150MPa Solution By symmetry F1 R F2 R Equilibrium ΣFy0 05 wo L1 L2 2R 0 R 025 wo L1 L2 Mmax R L1 05wo L1 L1 3 Mmax 025 wo L1 L2 L1 05wo L1 L1 3 Section properties D d 2 t I 1 12 b D3 b t d3 Bending Stress σ M c I cmax D 2 σmax 025 wo L1 L2 L1 05wo L1 L1 3 D 2I wo σmax 2I D 1 025 L1 L2 L1 05L1 L1 3 wo 1347 kN m Ans Problem 689 The steel beam has the crosssectional area shown If w0 10 kNm determine the maximum bending stress in the beam Given L1 4m L2 4m b 200mm t 8mm d 250mm wo 10 kN m Solution By symmetry F1 R F2 R Equilibrium ΣFy0 05 wo L1 L2 2R 0 R 025 wo L1 L2 Mmax R L1 05wo L1 L1 3 Section properties D d 2 t I 1 12 b D3 b t d3 Bending Stress σ M c I cmax D 2 σmax Mmax cmax I σmax 11138 MPa Ans Problem 690 The beam has a rectangular cross section as shown Determine the largest load P that can be supported on its overhanging ends so that the bending stress in the beam does not exceed σ max 10 MPa Given a 05m bo 50mm σallow 10MPa do 100mm Solution By symmetry A R B R Equilibrium ΣFy0 2 P 2 R 0 R P Mmax P 15a R 05a Mmax P a Section properties I 1 12 bo do 3 Bending Stress σ M c I cmax do 2 σ P a do 2I P σallow 2I a do P 167 kN Ans R P x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a M1 x1 P x1 kN m M2 x2 P x2 R x2 a 1 kN m M3 x3 P x3 R x3 a R x3 2a 1 kN m 0 05 1 15 1 05 0 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 691 The beam has the rectangular cross section shown If P 15 kN determine the maximum bending stress in the beam Sketch the stress distribution acting over the cross section Given a 05m bo 50mm P 15kN do 100mm Solution By symmetry A R B R Equilibrium ΣFy0 2 P 2 R 0 R P Mmax P 15a R 05a Mmax P a Mmax 075 kN m Section properties I 1 12 bo do 3 Bending Stress cmax do 2 σmax Mmax cmax I σmax 900 MPa Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a M1 x1 P x1 kN m M2 x2 P x2 R x2 a 1 kN m M3 x3 P x3 R x3 a R x3 2a 1 kN m 0 05 1 15 1 05 0 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 692 The beam is subjected to the loading shown If its crosssectional dimension a 180 mm determine the absolute maximum bending stress in the beam Given L1 2m L2 1m P 60kN a 180mm w 40 kN m Solution Equilibrium Given ΣFy0 A B P w L1 0 ΣΜA0 P L1 L2 B L1 05w L1 2 0 Guess A 1kN B 1kN A B Find A B A B 1000 13000 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 M1 x1 A x1 05w x1 2 1 kN m M2 x2 A x2 w L1 x2 05 L1 B x2 L1 1 kN m 0 1 2 3 60 40 20 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Max Moment unit kN m M1 L1 60000 Mmax M1 L1 unit Mmax 6000 kN m Section properties yc Σ yi Ai Σ Ai yc a a 3 a 6 a 2 2a 3 2a 3 a a 3 a 2 2a 3 yc 7500 mm Iw 1 12 a 2 2a 3 3 a 2 2a 3 yc 2a 3 2 If 1 12 a a 3 3 a a 3 yc a 6 2 I If Iw Bending Stress σ M c I cmax a yc σmax Mmax cmax I σmax 10511 MPa Ans Problem 693 The beam is subjected to the loading shown Determine its required crosssectional dimension a if the allowable bending stress for the material is σallo w 150 MPa Given L1 2m L2 1m P 60kN σallow 150MPa w 40 kN m Solution Equilibrium Given ΣFy0 A B P w L1 0 ΣΜA0 P L1 L2 B L1 05w L1 2 0 Guess A 1kN B 1kN A B Find A B A B 1000 13000 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 M1 x1 A x1 05w x1 2 1 kN m M2 x2 A x2 w L1 x2 05 L1 B x2 L1 1 kN m 0 1 2 3 60 40 20 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Max Moment unit kN m M1 L1 60000 Mmax M1 L1 unit Mmax 6000 kN m Section properties yc Σ yi Ai Σ Ai yc a a 3 a 6 a 2 2a 3 2a 3 a a 3 a 2 2a 3 yc 5a 12 Iw 1 12 a 2 2a 3 3 a 2 2a 3 yc 2a 3 2 Iw 43a4 1296 If 1 12 a a 3 3 a a 3 yc a 6 2 If 31a4 1296 I If Iw I 37a4 648 Bending Stress σ M c I cmax a yc cmax 7a 12 I Mmax cmax σallow a 3 648Mmax 7 12 37σallow a 15988 mm Ans Problem 694 The wing spar ABD of a light plane is made from 2014T6 aluminum and has a crosssectional area of 1000 mm2 a depth of 80 mm and a moment of inertia about its neutral axis of 1662 106 mm4 Determine the absolute maximum bending stress in the spar if the anticipated loading is to be as shown Assume A B and C are pins Connection is made along the central longitudinal axis of the spar Given a 1m b 2m A 1000m2 wo 15 kN m I 1662 106 mm4 do 80mm Solution L a b Equilibrium Given ΣFy0 A B 05wo L 0 ΣΜA0 B a 05wo L L 3 0 Guess A 1kN B 1kN A B Find A B A B 000 2250 kN x1 0 001 a a x2 a 101 a a b M1 x1 A x1 wo 2 x1 2 1 x1 L 1 3 1 kN m M2 x2 A x2 B x2 a wo 2 x2 2 1 x2 L 1 3 1 kN m 0 1 2 3 0 5 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Max Moment unit kN m M1 a 6667 Mmax M1 a unit Bending Stress σ M c I cmax do 2 σmax Mmax cmax I σmax 16045 MPa Ans σY 414 MPa Problem 695 The boat has a weight of 115 kN and a center of gravity at G If it rests on the trailer at the smooth contact A and can be considered pinned at B determine the absolute maximum bending stress developed in the main strut of the trailer Consider the strut to be a boxbeam having the dimensions shown and pinned at C Given a 09m b 18m c 12m d 03m bo 45mm do 75mm bi 38mm di 45mm W 115kN Solution Equilibrium for boat Given ΣFy0 A W B 0 ΣΜB0 A a b W b d 0 Guess A 1kN B 1kN A B Find A B A B 6389 5111 kN Equilibrium for assembly Given ΣFy0 D A B C 0 ΣΜC0 A a b c D b c B c 0 Guess C 1kN D 1kN C D Find C D C D 115 1035 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c M1 x1 A x1 kN m M2 x2 A x2 D x2 a 1 kN m M3 x3 A x3 D x3 a B x3 a b 1 kN m 0 1 2 3 5 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Max Moment unit kN m M1 a 5750 Mmax M1 a unit Section properties I 1 12 bo do 3 bi di 3 Bending Stress σ M c I cmax do 2 σmax Mmax cmax I σmax 1667 MPa Ans Problem 696 The beam supports the load of 25 kN Determine the absolute maximum bending stress in the beam if the sides of its triangular cross section are a 150 mm Given a 150mm L 06m P 25kN Solution Mmax P L Section Property I 1 36 a a sin 60deg 3 Maximum Bending Stress σ M c I cmax 2 3 a sin 60deg σmax Mmax cmax I σmax 1422 MPa Ans Problem 697 The beam supports the load of 25 kN Determine the required size a of the sides of its triangular cross section if the allowable bending stress is σ allow 126 MPa Given L 06m P 25kN σallow 126MPa Solution Mmax P L Section Property I 1 36 a a sin 60deg 3 Maximum Bending Stress σ M c I cmax 2 3 a sin 60deg σ Mmax cmax I I Mmax cmax σallow a 24 sin 60deg 2 Mmax σallow 1 3 a 1562 mm Ans Problem 698 The wood beam is subjected to the uniform load of w 3 kNm If the allowable bending stress for the material is σ allow 10 MPa determine the required dimension b of its cross section Assume the support at A is a pin and B is a roller Given L1 2m σallow 10MPa L2 1m w 3 kN m Solution Equilibrium Given ΣFy0 A B w L1 0 ΣΜA0 B L1 L2 05w L1 2 0 Guess A 1kN B 1kN A B Find A B A B 400 200 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 M1 x1 A x1 05w x1 2 1 kN m M2 x2 A x2 w L1 x2 05 L1 1 kN m 0 1 2 3 0 2 Distane m Moment kNm M1 x1 M2 x2 x1 x2 As indicated in the moment diagram the maximum moment occurs in L1 such that V10 V1 A w xc xc A w xc 1333 m Max Moment unit kN m M1 xc 2667 Mmax M1 xc unit Mmax 2667 kN m Section properties I 1 12 b 15 b 3 Bending Stress σ M co I co 15 b 2 I Mmax co σallow b 6 Mmax 225 σallow 1 3 b 893 mm Ans Problem 699 The wood beam has a rectangular cross section in the proportion shown Determine its required dimension b if the allowable bending stress is σ allow 10 MPa Given L1 2m σallow 10MPa L2 2m w 05 kN m Solution Equilibrium Given ΣFy0 A B w L1 0 ΣΜA0 B L1 L2 05w L1 2 0 Guess A 1kN B 1kN A B Find A B A B 075 025 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 M1 x1 A x1 05w x1 2 1 kN m M2 x2 A x2 w L1 x2 05 L1 1 kN m 0 2 0 05 1 Distane m Moment kNm M1 x1 M2 x2 x1 x2 As indicated in the moment diagram the maximum moment occurs in L1 such that V10 V1 A w xc xc A w xc 1500 m Max Moment unit kN m M1 xc 05625 Mmax M1 xc unit Mmax 05625 kN m Section properties I 1 12 b 15 b 3 Bending Stress σ M co I co 15 b 2 I Mmax co σallow b 3 6 Mmax 225 σallow b 531 mm Ans Problem 6100 A beam is made of a material that has a modulus of elasticity in compression different from that given for tension Determine the location c of the neutral axis and derive an expression for the maximum tensile stress in the beam having the dimensions shown if it is subjected to the bending moment M Problem 6101 The beam has a rectangular cross section and is subjected to a bending moment M If the material from which it is made has a different modulus of elasticity for tension and compression as shown determine the location c of the neutral axis and the maximum compressive stress in the beam Problem 6102 The box beam is subjected to a bending moment of M 25 kNm directed as shown Determine the maximum bending stress in the beam and the orientation of the neutral axis Given ay 3 az 4 ar 5 M 25kN m bo 150mm do 150mm bi 100mm di 100mm Solution Internal Moment Components My ay ar M Mz az ar M Section Property Iy 1 12 do bo 3 di bi 3 Iz 1 12 bo do 3 bi di 3 Maximum Bending Stress By inspection maximum bending stress occurs at B and D σ Mz y Iz My z Iy At B yB 05do zB 05bo σB Mz yB Iz My zB Iy σB 775 MPa C Ans At D yD 05 do zD 05 bo σD Mz yD Iz My zD Iy σD 775 MPa T Ans Orientation of Neutral Axis tan α Iz Iy tan θ θ atan ay az α atan Iz Iy tan θ α 3687 deg Ans y bo tan α y 11250 mm Problem 6103 Determine the maximum magnitude of the bending moment M so that the bending stress in the member does not exceed 100 MPa Given ay 3 az 4 ar 5 σallow 100MPa bo 150mm do 150mm bi 100mm di 100mm Solution Internal Moment Components My ay ar M Mz az ar M Section Property Iy 1 12 do bo 3 di bi 3 Iz 1 12 bo do 3 bi di 3 Maximum Bending Stress By inspection maximum bending stress occurs at B and D Apply the flexure formula for biaxial bending at either point B or D σ Mz y Iz My z Iy At B yB 05do zB 05bo σB Mz yB Iz My zB Iy σallow az ar M yB Iz ay ar M zB Iy M σallow az ar yB Iz ay ar zB Iy M 3224 kN m Ans Problem 6104 The beam has a rectangular cross section If it is subjected to a bending moment of M 3500 Nm directed as shown determine the maximum bending stress in the beam and the orientation of the neutral axis Given M 35kN m θ 30deg b 150mm d 300mm Solution θ 180deg θ θ 150deg Internal Moment Components My M sin θ Mz M cos θ Section Property Iy 1 12 d b3 Iz 1 12 b d3 Maximum Bending Stress σ Mz y Iz My z Iy At A yA 05d zA 05b σA Mz yA Iz My zA Iy σA 2903 MPa T Ans At B yB 05 d zB 05 b σB Mz yB Iz My zB Iy σB 2903 MPa C Ans At C yC 05d zC 05 b σC Mz yC Iz My zC Iy σC 0208 MPa C At D yD 05 d zD 05b σD Mz yD Iz My zD Iy σD 0208 MPa T Orientation of Neutral Axis tan α Iz Iy tan θ α atan Iz Iy tan θ α 6659 deg Ans z 05 b 05d tan α z 1005 mm Problem 6105 The Tbeam is subjected to a bending moment of M 15 kNm directed as shown Determine the maximum bending stress in the beam and the orientation of the neutral axis The location of the centroid C must be determined Given M 15kN m θ 60deg bf 300mm tf 50mm tw 50mm dw 200mm Solution θ 180deg θ θ 120deg Internal Moment Components My M sin θ Mz M cos θ Section Property Iy 1 12 tf bf 3 dw tw 3 Iy 11041666667 mm4 yc Σ yi Ai Σ Ai yc bf tf 05tf tw dw 05dw tf bf tf tw dw yc 7500 mm Iz 1 12 bf tf 3 bf tf 05tf yc 2 1 12 tw dw 3 tw dw 05dw tf yc 2 Iz 13020833333 mm4 Maximum Bending Stress σ Mz y Iz My z Iy At A yA yc zA 05bf σA Mz yA Iz My zA Iy σA 2197 MPa T Ans At B yB yc zB 05 bf σB Mz yB Iz My zB Iy σB 1333 MPa C At D yD tf dw yc zD 05 tw σD Mz yD Iz My zD Iy σD 1302 MPa C Orientation of Neutral Axis tan α Iz Iy tan θ α atan Iz Iy tan θ α 6391 deg Ans z 05bf yc tan α z 18672 mm Problem 6106 If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M 520 Nm and is directed as shown determine the bending stress at points A and B The location y of the centroid C of the struts crosssectional area must be determinedAlso specify the orientation o the neutral axis θ atan 5 12 Given M 520N m bf 400mm tf 20mm tw 20mm dw 180mm Solution D dw tf θ 180deg θ Internal Moment Components My M sin θ Mz M cos θ Section Property Iy 1 12 D bf 3 dw bf 2tw 3 Iy 36682666667 mm4 yc Σ yi Ai Σ Ai yc bf tf 05tf 2 tw dw 05dw tf bf tf 2 tw dw yc 5737 mm Iz 1 12 bf tf 3 bf tf 05tf yc 2 2 1 12 tw dw 3 tw dw 05dw tf yc 2 Iz 5760140351 mm4 Maximum Bending Stress σ Mz y Iz My z Iy At A yA yc D zA 05 bf σA Mz yA Iz My zA Iy σA 1298 MPa C Ans At B yB yc zB 05bf σB Mz yB Iz My zB Iy σB 0587 MPa T Orientation of Neutral Axis tan α Iz Iy tan θ α atan Iz Iy tan θ α 374 deg Ans Problem 6107 The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M 520 Nm and is directed as shown Determine the maximum bending stress in the strut The location y of the centroid C of the struts crosssectional area must be determinedAlso specify the orientation of the neutral axis θ atan 5 12 Given M 520N m bf 400mm tf 20mm tw 20mm dw 180mm Solution D dw tf θ 180deg θ Internal Moment Components My M sin θ Mz M cos θ Section Property Iy 1 12 D bf 3 dw bf 2tw 3 Iy 36682666667 mm4 yc Σ yi Ai Σ Ai yc bf tf 05tf 2 tw dw 05dw tf bf tf 2 tw dw yc 5737 mm Iz 1 12 bf tf 3 bf tf 05tf yc 2 2 1 12 tw dw 3 tw dw 05dw tf yc 2 Iz 5760140351 mm4 Maximum Bending Stress σ Mz y Iz My z Iy By inspection the maximum bending stress can occur at either point A or B At A yA yc D zA 05 bf σA Mz yA Iz My zA Iy σA 1298 MPa C Ans At B yB yc zB 05bf σB Mz yB Iz My zB Iy σB 0587 MPa T Orientation of Neutral Axis tan α Iz Iy tan θ α atan Iz Iy tan θ α 374 deg Ans Problem 6108 The 30mmdiameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shown It is supported on two journal bearings at A and B which offer no resistance to axial loading Furthermore the coupling to the motor at C can be assumed not to offer any support to the shaft Determine the maximum bending stress developed in the shaft Given a 1m Fy 150N Fz 400N do 30mm Solution Equilibrium In xy plane Given ΣFy0 Ay By 2Fy 0 ΣΜA0 2 Fy a By 2a 0 Guess Ay 1N By 1N Ay By Find Ay By Ay By 045 015 kN Equilibrium In xz plane by symmetry Az Bz Rz ΣFz0 2Rz 2Fz 0 Rz Fz Rz 400 N Internal Moment Components The shaft is subjected to two bending moment components My and Mz The moment disgram for each component is drawn x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a Mz1 x1 2 Fy x1 N m Mz2 x2 2 Fy x2 Ay x2 a 1 N m Mz3 x3 2 Fy x3 Ay x3 a 1 N m My1 x1 0 My2 x2 Rz x2 a N m My3 x3 Rz x3 a 2Fz x3 2a 1 N 0 1 2 3 400 200 0 Distane m Moment Nm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 0 1 2 0 500 Distane m Moment Nm My2 x2 My3 x3 x2 x3 Maximum Bending Stress unit N m Since all the axes through the circles center for circular shaft are principal axes then the resultant moment M My 2 Mz 205 can be used to determine the maximum bending stress The maximum bending stress moment occurs at E x2a Mmax Mz2 2a 2 My2 2a 2 unit Mmax 42720 N m Hence cmax do 2 I π 64 do 4 σmax Mmax cmax I σmax 1612 MPa Ans M1 x1 Mz1 x1 2 My1 x1 2 M2 x2 Mz2 x2 2 My2 x2 2 M3 x3 Mz3 x3 2 My3 x3 2 0 05 1 15 2 25 3 0 200 400 Distane m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 6109 The shaft is subjected to the vertical and horizontal loadings of two pulleys D and E as shown It is supported on two journal bearings at A and B which offer no resistance to axial loading Furthermorethe coupling to the motor at C can be assumed not to offer any support to the shaft Determine the required diameter d of the shaft if the allowable bending stress for the material is σ allow 180 MPa Given a 1m Fy 150N Fz 400N σallow 180MPa Solution Equilibrium In xy plane Given ΣFy0 Ay By 2Fy 0 ΣΜA0 2 Fy a By 2a 0 Guess Ay 1N By 1N Ay By Find Ay By Ay By 045 015 kN x1 0 001 a a x2 a 101 a 3a Mz1 x1 2 Fy x1 N m Mz2 x2 2 Fy x2 Ay x2 a 1 N m Equilibrium In xz plane by symmetry Az Bz Rz ΣFz0 2Rz 2Fz 0 Rz Fz Rz 400 N x1 a 101 a 2a x2 2a 101 2a 3a My1 x1 Rz x1 a N m My2 x2 Rz x2 a 2Fz x2 2a 1 N m Internal Moment Components The shaft is subjected to two bending moment components My and Mz The moment disgram for each component is drawn 0 1 2 3 400 200 0 Distane m Moment Nm Mz1 x1 Mz2 x2 x1 x2 0 1 2 0 500 Distane m Moment Nm My1 x1 My2 x2 x1 x2 Maximum Bending Stress unit N m Since all the axes through the circles center for circular shaft are principal axes then the resultant moment M My 2 Mz 205 can be used to determine the maximum bending stress The maximum bending stress moment occurs at E x2a Mmax Mz2 2a 2 My1 2a 2 unit Mmax 42720 N m Hence cmax do 2 I π 64 do 4 σallow Mmax cmax I π 64 do 4 Mmax σallow do 2 do 3 32Mmax π σallow do 2891 mm Ans Problem 6110 The board is used as a simply supported floor joist If a bending moment of M 12 kNm is applied 3 from the z axis determine the stress developed in the board at the corner A Compare this stress with that developed by the same moment applied along the z axis θ 0 What is the angle a for the neutral axis when θ 3 Comment Normally floor boards would be nailed to the top of the beam so that θ 0 nearly and the high stress due to misalignment would not occur Given M 12kN m θ 3deg b 50mm d 150mm Solution Internal Moment Components My M sin θ Mz M cos θ yc 05d Section Property Iy 1 12 d b3 Iz 1 12 b d3 Maximum Bending Stress σ Mz y Iz My z Iy At A yA yc zA 05 b σA Mz yA Iz My zA Iy σA 740 MPa T Ans Orientation of Neutral Axis tan α Iz Iy tan θ α atan Iz Iy tan θ α 2525 deg Ans When θ 0 My 0 Mz M σA Mz yA Iz My zA Iy σA 640 MPa T Ans Problem 6111 Consider the general case of a prismatic beam subjected to bendingmoment components My and Mz as shown when the x y z axes pass through the centroid of the cross section If the material is linearelastic the normal stress in the beam is a linear function of position such that σ a by cz Using the equilibrium conditions determine the constants a b and c and show that the normal stress can be determined from the equation σ Mz Iy My Iyz y My Iz Mz Iyz z Iy Iz Iyz 2 where the moments and products of inertia are defined in Appendix A A z A y A yσ dA zσ dA M σ dA M 0 Given Linear function σx a by z Solution Equilibrium Conditios A A A A A x z dA c y dA b dA a 0 czdA by a σ dA 0 0 A 2 A A y A y A x y z dA c yz dA b zdA a M czdA by za zσ dA M M A A 2 A z A z A x z yz dA c y dA b ydA a M czdA by ya yσ dA M M A A 0 zdA ydA The integrals are defined in Appendix A Note that c 0 b 0 dA a 0 A Thus y yz y c I b I a 0 M yz y z c I b I a 0 M 0 0 since A a Solving for a b and c Ans 2 yz z y y z yz y I I I I where D M I D M 1 b Ans I M I D M 1 c yz z z y Ans Bending Stress I z M I D M I y 1 M I D M 1 σ z y yz z yz y y z x QED In matrix form σ 1 D Mz My Iy Iyz Iyz Iz y z Problem 6112 The 65mmdiameter steel shaft is subjected to the two loads that act in the directions shown If the journal bearings at A and B do not exert an axial force on the shaft determine the absolute maximum bending stress developed in the shaft Given a 125m b 1m F 4kN L 2a b do 65mm θ 30deg Solution Equilibrium In xy plane by symmetry Ay By Ry ΣFy0 2Ry 2F cos θ 0 Ry F cos θ Ry 3464 kN Equilibrium In xz plane by antisymmetry Az Bz Rz ΣΜB0 Az L F sin θ b 0 Rz F sin θ b L Rz 0571 kN Internal Moment Components The shaft is subjected to two bending moment components My and Mz The moment disgram for each component is drawn x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L Mz1 x1 Ry x1 kN m Mz2 x2 Ry x2 F cos θ x2 a 1 kN m Mz3 x3 Ry x3 F cos θ x3 a F cos θ x3 a b 1 kN m My1 x1 Rz x1 kN m My2 x2 Rz x2 F sin θ x2 a 1 kN m My3 x3 Rz x3 F sin θ x3 a F sin θ x3 a b 1 kN m 0 2 0 5 Distane m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 0 2 1 0 1 Distane m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Maximum Bending Stress unit kN m Since all the axes through the circles center for circular shaft are principal axes then the resultant moment M My 2 Mz 205 can be used to determine the maximum bending stress The maximum bending stress moment occurs at xa Mmax Mz1 a 2 My1 a 2 unit Mmax 4389 kN m Hence cmax do 2 I π 64 do 4 σmax Mmax cmax I σmax 1628 MPa Ans M1 x1 Mz1 x1 2 My1 x1 2 M2 x2 Mz2 x2 2 My2 x2 2 M3 x3 Mz3 x3 2 My3 x3 2 0 1 2 3 0 5 Distane m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 6113 The steel shaft is subjected to the two loads that act in the directions shown If the journal bearings at A and B do not exert an axial force on the shaft determine the required diameter of the shaft if the allowable bending stress is σ allow 180 MPa Given a 125m b 1m F 4kN L 2a b θ 30deg σallow 180MPa Solution Equilibrium In xy plane by symmetry Ay By Ry ΣFy0 2Ry 2F cos θ 0 Ry F cos θ Ry 3464 kN Equilibrium In xz plane by antisymmetry Az Bz Rz ΣΜB0 Az L F sin θ b 0 Rz F sin θ b L Rz 0571 kN Internal Moment Components The shaft is subjected to two bending moment components My and Mz The moment disgram for each component is drawn x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L Mz1 x1 Ry x1 kN m Mz2 x2 Ry x2 F cos θ x2 a 1 kN m Mz3 x3 Ry x3 F cos θ x3 a F cos θ x3 a b 1 kN m My1 x1 Rz x1 kN m My2 x2 Rz x2 F sin θ x2 a 1 kN m My3 x3 Rz x3 F sin θ x3 a F sin θ x3 a b 1 kN m 0 2 0 5 Distane m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 0 2 1 0 1 Distane m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Maximum Bending Stress unit kN m Since all the axes through the circles center for circular shaft are principal axes then the resultant moment M My 2 Mz 205 can be used to determine the maximum bending stress The maximum bending stress moment occurs at xa Mmax Mz1 a 2 My1 a 2 unit Mmax 4389 kN m Hence cmax do 2 I π 64 do 4 σallow Mmax cmax I π 64 do 4 Mmax σallow do 2 do 3 32Mmax π σallow do 6286 mm Ans Problem 6114 Using the techniques outlined in Appendix A Example A4 or A5 the Z section has principal mome of inertia of Iy 0060103 m4 and Iz 0471103 m4 computed about the principal axes of inertia y and z respectively If the section is subjected to an internal moment of M 250 Nm directed horizontally as shown determine the stress produced at point A Solve the problem using Eq 617 Given M 250N m θ 329deg bf 300mm tf 50mm tw 50mm dw 150mm Iy 0060 10 3 m4 Iz 0471 10 3 m4 Solution Internal Moment Components My M cos θ Mz M sin θ Coordinates of Point A yA 05bf zA dw 05tf yA zA cos θ sin θ sin θ cos θ yA zA yA zA 22100 6546 mm Bending Stress σA Mz yA Iz My zA Iy σA 0293 MPa C Ans Problem 6115 Solve Prob 6114 using the equation developed in Prob 6111 Given M 250N m θ 329deg bf 300mm tf 50mm tw 50mm dw 150mm Iy 0060 10 3 m4 Iz 0471 10 3 m4 Solution Internal Moment Components My M Mz 0 Section Property czw 05dw 05tf cyw 05bf 05tw Iy 1 12 bf tf 3 2 1 12 tw dw 3 tw dw czw 2 Iy 18125 10 6 m4 Iz 1 12 tf bf 3 2 1 12 dw tw 3 dw tw cyw 2 Iz 35000 10 6 m4 Iyz tw dw czw cyw tw dw czw cyw Iyz 18750 10 6 m4 Coordinates of Point A yA 05bf zA dw 05tf yA 150mm zA 175 mm Bending Stress Using formula developed in Prob 6111 D Iy Iz Iyz 2 σA 1 D Mz My Iy Iyz Iyz Iz yA zA σA 0293 MPa C Ans Problem 6116 Using the techniques outlined in Appendix A Example A4 or A5 the Z section has principal mome of inertia of Iy 0060103 m4 and Iz 0471103 m4 computed about the principal axes of inertia y and z respectively If the section is subjected to an internal moment of M 250 Nm directed horizontally as shown determine the stress produced at point B Solve the problem using Eq 617 Given M 250N m θ 329deg bf 300mm tf 50mm tw 50mm dw 150mm Iy 0060 10 3 m4 Iz 0471 10 3 m4 Solution Internal Moment Components My M cos θ Mz M sin θ Coordinates of Point B yB 05 bf zB dw 05tf yB zB cos θ sin θ sin θ cos θ yB zB yB zB 22100 6546 mm Bending Stress σB Mz yB Iz My zB Iy σB 0293 MPa T Ans Problem 6117 For the section Iy 31711062 m4 Iz 114106 m4 Iyz 151106 m4 Using the techniques outlined in Appendix A the members crosssectional area has principal moments of inertia of Iy 290106 m4 and Iz 117106 m4 computed about the principal axes of inertia y and z respectively If the section is subjected to a moment of M 2500 Nm directed as shown determine the stress produced at point A using Eq 617 Given M 2500N m θ 1010deg h1 80mm h2 140mm b1 60mm b2 60mm Iy 290 10 6 m4 Iz 117 10 6 m4 Solution θ θ Internal Moment Components My M sin θ Mz M cos θ Coordinates of Point A yA h2 zA b2 yA zA cos θ sin θ sin θ cos θ yA zA yA zA 14835 3452 mm Bending Stress σA Mz yA Iz My zA Iy σA 2599 MPa T Ans Problem 6118 Solve Prob 6117 using the equation developed in Prob 6111 Given M 2500N m θ 1010deg h1 80mm h2 140mm b1 60mm b2 60mm Iy 317 10 6 m4 Iz 114 10 6 m4 Iyz 151 10 6 m4 Solution θ θ Internal Moment Components My 0 Mz M Coordinates of Point A yA h2 zA b2 yA zA cos θ sin θ sin θ cos θ yA zA yA zA 14835 3452 mm Bending Stress Using formula developed in Prob 6111 D Iy Iz Iyz 2 σA 1 D Mz My Iy Iyz Iyz Iz yA zA σA 2608 MPa T Ans Problem 6119 The composite beam is made of 6061T6 aluminum A and C83400 red brass B Determine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals What maximum moment will this beam support if the allowable bending stress for the aluminum is σ allow al 128 MPa and for the brass σ allow br 35 MPa Given b 150mm hal 50mm Eal 689GPa Ebr 101GPa σalallow 128MPa σbrallow 35MPa Solution Section Property n Eal Ebr n 0682178 Abr b h Aal n b hal yc Σyi Ai ΣA yc Aal 05 hal Abr hal 05 h Aal Abr Given yc hal hal Aal 05 hal b h hal 05 h Aal b h Guess h 10mm h Find h h 4130 mm Ans Abr b h Ibr 1 12 b h3 Abr 05h 2 Ial 1 12 n b hal 3 Aal yc 05hal 2 I Ibr Ial I 778510817 mm4 Allowable Bending Stress σ M y I Assume failure of red brass ybr h Mbr σbrallow I ybr Mbr 660 kN m Assume failure of aluminum yal hal Mal σalallow I n yal Mal 2922 kN m Mallow min Mbr Mal Mallow 660 kN m Ans Problem 6120 The composite beam is made of 6061T6 aluminum A and C83400 red brass B If the height h 4 mm determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is σ allow al 128 MPa and for the brass σ allow br 35 MPa Given b 150mm hal 50mm hbr 40mm Eal 689GPa Ebr 101GPa σalallow 128MPa σbrallow 35MPa Solution Section Property n Eal Ebr n 0682178 Abr b hbr Aal n b hal yc Σyi Ai ΣA yc Aal 05 hal Abr hal 05 hbr Aal Abr yc 49289 mm Ibr 1 12 b hbr 3 Abr hal 05hbr yc 2 Ial 1 12 n b hal 3 Aal yc 05hal 2 I Ibr Ial I 745798763 mm4 Allowable Bending Stress σ M y I Assume failure of red brass ybr hal hbr yc Mbr σbrallow I ybr Mbr 641 kN m Assume failure of aluminum yal yc Mal σalallow I n yal Mal 2839 kN m Mallow min Mbr Mal Mallow 641 kN m Ans Problem 6121 A wood beam is reinforced with steel straps at its top and bottom as shown Determine the maximum bending stress developed in the wood and steel if the beam is subjected to a bending moment of M kNm Sketch the stress distribution acting over the cross section Take Ew 11 GPa Est 200 GPa Given b 200mm hw 300mm tst 20mm Ew 11GPa Est 200GPa M 5kN m Solution Section Property n Est Ew n 18181818 Aw b hw Ast n b tst Ist 1 12 n b tst 3 Ast 05hw 05tst 2 I 1 12 b hw 3 2Ist I 417848484848 mm4 Maximum Bending Stress σ M y I For wood beam yw 05hw σwmax M yw I σwmax 0179 MPa Ans For steel straps yst 05hw tst σstmax n M yst I σstmax 3699 MPa Ans At yst 05hw σst n M yst I σst 3263 MPa Problem 6122 The Douglas Fir beam is reinforced with A36 steel straps at its center and sides Determine the maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz 10 kNm Sketch the stress distribution acting over the cross section Given bst 12mm d 150mm bw 50mm Mz 10kN m Ew 131GPa Est 200GPa Solution Section Property n Ew Est n 00655 Iz 1 12 3bst d3 n 2bw d3 Maximum Bending Stress σ Mz y Iz ymax 05d σst Mz ymax Iz σst 627 MPa Ans σw n σst σw 410 MPa Ans Problem 6123 The steel channel is used to reinforce the wood beam Determine the maximum stress in the steel and in the wood if the beam is subjected to a moment of M 12 kNm Est 200 GPa Ew 12GPa Given bst 399mm dst 100mm tst 12mm bw 375mm dw 88mm M 12kN m Ew 12GPa Est 200GPa Solution df dw n Ew Est n 006 Section Property As1 2tst df As2 bst tst Aw n bw dw yc Aw As1 05dw tst As2 05tst Aw As1 As2 yc 2904 mm Iw 1 12 n bw dw 3 Aw 05dw tst yc 2 Is1 1 12 2tst df 3 As1 05dw tst yc 2 Is2 1 12 bst tst 3 As2 05tst yc 2 I Iw Is1 Is2 Maximum Bending Stress σ M c I cmax dst yc σst M cmax I σst 1037 MPa Ans σw n σst σw 062 MPa Ans Problem 6124 The Douglas Fir beam is reinforced with A36 steel straps at its sides Determine the maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz 4 kNm Sketc the stress distribution acting over the cross section Given bst 15mm d 350mm bw 200mm Mz 4kN m Ew 131GPa Est 200GPa Solution Section Property n Ew Est n 00655 Iz 1 12 2bst d3 n bw d3 Maximum Bending Stress σ Mz y Iz ymax 05d σst Mz ymax Iz σst 4546 MPa Ans σw n σst σw 0298 MPa Ans Problem 6125 The composite beam is made of A36 steel A bonded to C83400 red brass B and has the cross section shown If it is subjected to a moment of M 65 kNm determine the maximum stress in the brass and steel Also what is the stress in each material at the seam where they are bonded together Given b 125mm hst 100mm hbr 100mm Est 200GPa Ebr 101GPa M 65kN m Solution Section Property n Ebr Est n 0505 Ast b hst Abr n b hbr yc Σyi Ai ΣA yc Abr 05 hbr Ast hbr 05 hst Abr Ast yc 116445 mm Ist 1 12 b hst 3 Ast hbr 05hst yc 2 Ibr 1 12 n b hbr 3 Abr yc 05hbr 2 I Ist Ibr I 5762060493 mm4 Maximum Bending Stress σ M y I For steel yst hbr hst yc σstmax M yst I σstmax 9426 MPa Ans For red brass ybr yc σbrmax n M ybr I σbrmax 6634 MPa Ans Bending Stress at the Seam yseam yc hbr σst M yseam I σst 1855 MPa Ans σbr n σst σbr 0937 MPa Ans Problem 6126 The composite beam is made of A36 steel A bonded to C83400 red brass B and has the cross section shown If the allowable bending stress for the steel is σ allowst 180 MPa and for the brass σ allowbr 60 MPa determine the maximum moment M that can be applied to the beam Given b 125mm hst 100mm hbr 100mm Est 200GPa Ebr 101GPa σstallow 180MPa σbrallow 60MPa Solution Section Property n Ebr Est n 0505 Ast b hst Abr n b hbr yc Σyi Ai ΣA yc Abr 05 hbr Ast hbr 05 hst Abr Ast yc 116445 mm Ist 1 12 b hst 3 Ast hbr 05hst yc 2 Ibr 1 12 n b hbr 3 Abr yc 05hbr 2 I Ist Ibr I 5762060493 mm4 Allowable Bending Stress σ M y I Assume failure of red brass ybr yc Mbr σbrallow I n ybr Mbr 5879 kN m Assume failure of steel yst hbr hst yc Mst σstallow I yst Mst 12413 kN m Mallow min Mbr Mst Mallow 5879 kN m Ans Problem 6127 The reinforced concrete beam is made using two steel reinforcing rods If the allowable tensile stress for the steel is σ st allow 280 MPa and the allowable compressive stress for the concrete is σ conc allow 21 MPa determine the maximum moment M that can be applied to the section Assume the concrete cannot support a tensile stress Est 200 GPa Econc 265 GPa Given bf 550mm df 100mm bw 150mm dw 450mm dr 25mm hr 50mm Econc 265GPa Est 200GPa σcallow 21MPa σstallow 280MPa Solution n Est Econc Section Property n 754717 Ast n 2 π 05dr 2 Given Ast dw hw hr bf df 05df hw bw hw 05hw 0 Guess hw 10mm hw Find hw hw 341 mm Ist Ast dw hw hr 2 If 1 12 bf df 3 bf df 05df hw 2 Iw 1 12 bw hw 3 bw hw 05hw 2 I Ist If Iw σ M y I Assume concrete fails ymax df hw Mconc σcallow I ymax Mconc 27783 kN m Assume steel fails yst dw hw hr Mst σstallow I n yst Mst 12798 kN m Mallow min Mconc Mst Mallow 12798 kN m Ans Problem 6128 Determine the maximum uniform distributed load w0 that can be supported by the reinforced concrete beam if the allowable tensile stress for the steel is σ st allow 200 MPa and the allowable compressive stress for the concrete is σ conc allow 20 MPa Assume the concrete cannot support a tensile stress Take Est 200 GPa Econc 25 GPa Given b 250mm d 500mm dr 16mm hr 50mm Econc 25GPa Est 200GPa σcallow 20MPa L 25m σstallow 200MPa Solution n Est Econc Section Property n 8 Ast n 2 π 05dr 2 Given Ast d h hr b h 05h 0 Guess h 10mm h Find h h 9551 mm I Ast d h hr 2 1 12 b h3 b h 05h 2 σ M y I Assume concrete fails ymax h Mconc σcallow I ymax Mconc 9985 kN m Assume steel fails yst d h hr Mst σstallow I n yst Mst 3363 kN m Thus steel fails first Maximum moment ocurs over the middle support Mmax wo L2 2 wo 2 Mst L2 wo 1076 kN m Ans Problem 6129 A bimetallic strip is made from pieces of 2014T6 aluminum and C83400 red brass having the cross section shown A temperature increase causes its neutral surface to be bent into a circular arc having a radius of 400 mm Determine the moment that must be acting on its cross section due to the thermal stress Take Eal 74 GPa Ebr 102 GPa Given bbr 6mm dbr 2mm bal 6mm dal 2mm Eal 74GPa ρ 400mm Ebr 102GPa Solution Transform the section to brass Section Property n Eal Ebr n 072549 Aal n bal dal Abr bbr dbr yc Aal 05dal Abr 05dbr dal Aal Abr yc 216 mm Ial 1 12 n bal dal 3 Aal 05dal yc 2 Ibr 1 12 bbr dbr 3 Abr 05dbr dal yc 2 I Ial Ibr Maximum Bending Stress 1 ρ M E I M Ebr I ρ M 691 N m Ans Problem 6130 The fork is used as part of a nosewheel assembly for an airplane If the maximum wheel reaction at the end of the fork is 45 kN determine the maximum bending stress in the curved portion of the fork at section aa There the crosssectional area is circular having a diameter of 50 mm Given rc 250mm a 150mm F 45kN θ 30deg do 50mm Solution Internal Moment ΣΜC0 M F a rc sin θ 0 M F a rc sin θ M 1125 N m Section Property ro 05do A π ro 2 A Ar r dA Σ I IAr 2 π rc rc 2 ro 2 R A IAr R π ro 2 2 π rc rc 2 ro 2 R 249373 mm Bending Stress σ M R r A r rc R rA rc ro σA M R rA A rA rc R σA 991 MPa rB rc ro σB M R rB A rB rc R σB 852 MPa σmax max σA σB σmax 991 MPa T Ans Given bf 75mm bf 150mm tf 10mm Problem 6131 Determine the greatest magnitude of the applied forces P if the allowable bending stress is σ allowc 50 MPa in compression and σ allowt 120 MPa in tension tw 10mm dw 150mm ri 250mm σcallow 50 MPa σtallow 120MPa Solution Internal Moment M Pdwtf kNm is positive since it tends to increase the beams radius of curvature Section Property re ri dw 2tf A bf tf dw tw bf tf A 3750mm2 r Σ ri Ai Σ Ai rc bf tf ri 05tf dw tw ri 05dw tf bf tf re 05tf A rc 319000 mm A Ar r dA Σ I IAr bf ln ri tf ri tw ln re tf ri tf bf ln re re tf IAr 12245 mm R A IAr R 306243 mm Normal Stress σ M R r A r rc R M σ A r rc R R r Assume tension failure r ri P dw tf σtallow A r rc R R r P σtallow A r rc R dw tf R r P 15948 kN Assume compression failure r re P dw tf σcallow A r rc R R r P σcallow A r rc R dw tf R r P 552 kN Pallow min P P Pallow 5520 kN Ans Given bf 75mm bf 150mm tf 10mm Problem 6132 If P 6 kN determine the maximum tensile and compressive bending stresses in the beam tw 10mm dw 150mm ri 250mm P 6kN Solution Internal Moment M P dw tf M 0960 kN m M is positive since it tends to increase the beams radius of curvature Section Property re ri dw 2tf A bf tf dw tw bf tf A 3750mm2 r Σ ri Ai Σ Ai rc bf tf ri 05tf dw tw ri 05dw tf bf tf re 05tf A rc 319000 mm A Ar r dA Σ I IAr bf ln ri tf ri tw ln re tf ri tf bf ln re re tf IAr 12245 mm R A IAr R 306243 mm Normal Stress σ M R r A r rc R Maximum tensile stress r ri σtmax M R r A r rc R σtmax 451 MPa T Ans Maximum compressive stress r re σcmax M R r A r rc R σcmax 544 MPa C Ans Problem 6133 The curved beam is subjected to a bending moment of M 900 Nm as shown Determine the stress at points A and B and show the stress on a volume element located at each of these points Given bf 100mm tf 20mm ri 400mm tw 15mm dw 150mm M 900 N m θ 30deg Solution Internal Moment M is negative since it tends to decrease the beams radius of curvature Section Property re ri dw tf A bf tf dw tw A 4250mm2 r Σ ri Ai Σ Ai rc dw tw ri 05dw bf tf re 05tf A rc 515000 mm A Ar r dA Σ I IAr tw ln re tf ri bf ln re re tf IAr 8349 mm R A IAr R 509067 mm Normal Stress σ M R r A r rc R At A rA re σA M R rA A rA rc R σA 382 MPa T Ans At B rB ri σB M R rB A rB rc R σB 973 MPa C Ans Problem 6134 The curved beam is subjected to a bending moment of M 900 Nm Determine the stress at point C Given bf 100mm tf 20mm ri 400mm tw 15mm dw 150mm M 900 N m θ 30deg Solution Internal Moment M is negative since it tends to decrease the beams radius of curvature Section Property re ri dw tf A bf tf dw tw A 4250mm2 r Σ ri Ai Σ Ai rc dw tw ri 05dw bf tf re 05tf A rc 515000 mm A Ar r dA Σ I IAr tw ln re tf ri bf ln re re tf IAr 8349 mm R A IAr R 509067 mm Normal Stress σ M R r A r rc R At C rC re tf σC M R rC A rC rc R σC 266 MPa T Ans Problem 6135 The curved bar used on a machine has a rectangular cross section If the bar is subjected to a couple as shown determine the maximum tensile and compressive stress acting at section aa Sketch the stress distribution on the section in three dimensions Given b 50mm h 75mm ri 1625mm P 250N θ 60deg a 150mm Solution Internal Moment M P a sin θ h cos θ M 41851 N m M is positive since it tends to increase the beams radius of curvature Section Property re ri h A b h A 3750mm2 rc ri 05 h rc 200mm A Ar r dA Σ I IAr b ln re ri IAr 18974 mm R A IAr R 197634 mm Normal Stress σ M R r A r rc R At A rA re σA M R rA A rA rc R σA 0792 MPa C Ans At B rB ri σB M R rB A rB rc R σB 1020 MPa T Ans Problem 6136 The circular spring clamp produces a compressive force of 3 N on the plates Determine the maximum bending stress produced in the spring at A The spring has a rectangular cross section as shown Given b 20mm h 10mm ri 200mm P 3N a 220mm Solution Internal Moment M P a M 0660 N m M is positive since it tends to increase the beams radius of curvature Section Property re ri h A b h A 200mm2 rc ri 05 h rc 205mm A Ar r dA Σ I IAr b ln re ri IAr 0976 mm R A IAr R 204959 mm Normal Stress σ M R r A r rc R Maximum tensile stress rA ri σtmax M R rA A rA rc R σtmax 201 MPa T Ans Maximum compressive stress rA re σcmax M R rA A rA rc R σcmax 195 MPa C Ans Problem 6137 Determine the maximum compressive force the spring clamp can exert on the plates if the allowable bending stress for the clamp is σ allow 4 MPa Given b 20mm h 10mm ri 200mm a 220mm σtallow 4MPa σcallow 4 MPa Solution Section Property re ri h A b h A 200mm2 rc ri 05 h rc 205mm A Ar r dA Σ I IAr b ln re ri IAr 0976 mm R A IAr R 204959 mm Internal Moment Mmax P a R M is positive since it tends to increase the beams radius of curvature Normal Stress σ M R r A r rc R M σ A r rc R R r Assume tension failure r ri P a R σtallow A r rc R R r P σtallow A r rc R a R R r P 3087 N Assume compression failure r re P a R σcallow A r rc R R r P σcallow A r rc R a R R r P 3189 N Pallow min P P Pallow 3087 N Ans Problem 6138 While in flight the curved rib on the jet plane is subjected to an anticipated moment of M 16 Nm a the section Determine the maximum bending stress in the rib at this section and sketch a two dimensional view of the stress distribution Given bf 30mm tf 5mm tw 5mm dw 20mm M 16N m ri 600mm Solution Internal Moment M is positive since it tends to increase the beams radius of curvature Section Property re ri dw 2tf A 2bf tf dw tw A 400mm2 r Σ ri Ai Σ Ai rc bf tf ri 05tf dw tw ri 05dw tf bf tf re 05tf A rc 615000 mm A Ar r dA Σ I IAr bf ln ri tf ri tw ln re tf ri tf bf ln re re tf IAr 0651 mm R A IAr R 614793 mm Normal Stress σ M R r A r rc R Maximum tensile stress r ri σtmax M R r A r rc R σtmax 477 MPa T Ans Maximum compressive stress r re σcmax M R r A r rc R σcmax 467 MPa C Ans Problem 6139 The steel rod has a circular cross section If it is gripped at its ends and a couple moment of M 15 Nm is developed at each grip determine the stress acting at points A and B and at the centroid C Given rci 50mm rce 75mm ro 12mm M 15N m Solution Internal Moment M 15Nm is positive since it tends to increase the beams radius of curvature Section Property A π ro 2 rc 05 rci rce A Ar r dA Σ I IAr 2 π rc rc 2 ro 2 R A IAr R 61919 mm Normal Stress σ M R r A r rc R rA rci σA M R rA A rA rc R σA 13594 MPa T Ans rB rce σB M R rB A rB rc R σB 09947 MPa C Ans rC rc σC M R rC A rC rc R σC 00531 MPa C Ans Problem 6140 The curved bar used on a machine has a rectangular cross section If the bar is subjected to a couple as shown determine the maximum tensile and compressive stresses acting at section aa Sketch the stress distribution on the section in three dimensions Given b 50mm h 75mm ri 100mm P 250N d 50mm d 150mm Solution Internal Moment M P d M 375 N m M is positive since it tends to increase the beams radius of curvature Section Property re ri h A b h A 3750mm2 rc ri 05 h rc 1375 mm A Ar r dA Σ I IAr b ln re ri IAr 27981 mm R A IAr R 134021 mm Normal Stress σ M R r A r rc R Maximum tensile stress r ri σtmax M R r A r rc R σtmax 0978 MPa T Ans Maximum compressive stress r re σcmax M R r A r rc R σcmax 0673 MPa C Ans Problem 6141 The member has an elliptical cross section If it is subjected to a moment of M 50 Nm determine the stress at points A and B Is the stress at point A which is located on the member near the wall the same as that at A Explain Given ri 100mm La 150mm Lb 75mm M 50N m Solution Internal Moment M is positive since it tends to increase the members radius of curvature Section Property ao 05La bo 05Lb re ri La A π ao bo A 883573 mm2 rc ri ao rc 175mm A Ar r dA Σ I IAr 2π bo ao rc rc 2 ao 2 R A IAr R π ao bo ao 2 π bo rc rc 2 ao 2 R 166557 mm Bending Stress σ M R r A r rc R rA ri σA M R rA A rA rc R σA 0446 MPa T Ans rB re σB M R rB A rB rc R σB 0224 MPa C Ans No The stress at point A is not the same as that at A because of localized stress concentration Ans Problem 6142 The member has an elliptical cross section If the allowable bending stress is σ allow 125 MPa determine the maximum moment M that can be applied to the member Given ri 100mm La 150mm Lb 75mm σtallow 125MPa σcallow 125 MPa Solution Internal Moment M is positive since it tends to increase the members radius of curvature Section Property ao 05La bo 05Lb re ri La A π ao bo A 883573 mm2 rc ri ao rc 175mm A Ar r dA Σ I IAr 2π bo ao rc rc 2 ao 2 R A IAr R π ao bo ao 2 π bo rc rc 2 ao 2 R 166557 mm Normal Stress σ M R r A r rc R M σ A r rc R R r Assume tension failure r ri M σtallow A r rc R R r M σtallow A r rc R R r M 1401 kN m Assume compression failure r re M σcallow A r rc R R r M σcallow A r rc R R r M 2794 kN m Mallow min M M Mallow 1401 kN m Ans Problem 6143 The bar has a thickness of 625 mm and is made of a material having an allowable bending stress of σ allow 126 MPa Determine the maximum moment M that can be applied Given t 625mm r 625mm w 100mm h 25mm σallow 126MPa Solution Section Property I 1 12 t h3 I 813802 mm4 Stress Concentration Factor w h 4 r h 025 From Fig 648 K 145 Maximum Moment σ K M c I c 05 h Mmax σallow I K c Mmax 5657 N m Ans Problem 6144 The bar has a thickness of 125 mm and is subjected to a moment of 90 Nm Determine the maximum bending stress in the bar Given t 125mm r 625mm w 100mm h 25mm M 90N m Solution Section Property I 1 12 t h3 I 1627604 mm4 Stress Concentration Factor w h 4 r h 025 From Fig 648 K 145 Maximum Bending Stress c 05 h σmax K M c I σmax 1002 MPa Ans Problem 6145 The bar is subjected to a moment of M 40 Nm Determine the smallest radius r of the fillets so that an allowable bending stress of σ allow 124 MPa is not exceeded Given w 80mm h 20mm t 7mm M 40N m σallow 124MPa Solution Section Property I 1 12 t h3 I 466667 mm4 Allowable Bending Stress σ K M c I c 05 h K σallow I M c K 145 Stress Concentration Factor From Fig 648 with K 145 w h 4 then r h 025 r 025 h r 500 mm Ans Problem 6146 The bar is subjected to a moment of M 175 N m If r 5 mm determine the maximum bending stress in the material Given w 80mm h 20mm t 7mm r 5mm M 175N m Solution Section Property I 1 12 t h3 I 466667 mm4 Stress Concentration Factor w h 4 r h 025 From Fig 648 K 145 Maximum Bending Stress c 05 h σmax K M c I σmax 544 MPa Ans Problem 6147 The bar is subjected to a moment of M 20 Nm Determine the maximum bending stress in the bar and sketch approximately how the stress varies over the critical section Given w 30mm h 10mm t 5mm r 15mm M 20N m Solution Section Property I 1 12 t h3 I 41667 mm4 Stress Concentration Factor w h 3 r h 015 From Fig 648 K 16 Maximum Bending Stress c 05 h σmax K M c I σmax 384MPa Ans Problem 6148 The allowable bending stress for the bar is σ allow 175 MPa Determine the maximum moment M that can be applied to the bar Given w 30mm h 10mm t 5mm r 15mm σallow 175MPa Solution Section Property I 1 12 t h3 I 41667 mm4 Stress Concentration Factor w h 3 r h 015 From Fig 648 K 16 Maximum Moment σ K M c I c 05 h M σallow I K c M 911 N m Ans Problem 6149 Determine the maximum bending stress developed in the bar if it is subjected to the couples shown The bar has a thickness of 6 mm Given t 6mm w 108mm h1 72mm h2 36mm r1 72mm r2 27mm M1 20N m M2 75N m Mw 125N m Solution Section Property For the larger section 1 For the smaller section 2 I1 1 12 t h1 3 I1 186624mm4 I2 1 12 t h2 3 I2 23328mm4 Stress Concentration Factor For the larger section 1 For the smaller section 2 w h1 15 r1 h1 01 w h2 3 r2 h2 075 From Fig 648 K1 1755 From Fig 648 K2 115 Maximum Moment σ K M c I For the larger section 1 For the smaller section 2 c1 05 h1 σ1 K1 M1 c1 I1 c2 05 h2 σ2 K2 M2 c2 I2 σ1 677 MPa σ2 666 MPa σmax max σ1 σ2 σmax 677 MPa Ans Problem 6150 Determine the length L of the center portion of the bar so that the maximum bending stress at A B and C is the sameThe bar has a thickness of 10 mm Given w 60mm t 10mm h 40mm r 7mm a 200mm P 350N Solution Section Property I 1 12 t h3 I 5333333 mm4 Support Reaction By symmetry A B R ΣFy0 2R P 0 Stress Concentration Factor R 05P R 175N w h 15 r h 0175 Internal Moment MA R a MA 3500 N m From Fig 648 K 15 MB MA Maximum Bending Stresses at A and B MC R a 05L c 05 h σAmax K MA c I σAmax 19688 MPa σBmax σAmax At Section CC σCmax σAmax Require I 1 12 t w3 I 180000mm4 Maximum Bending Stress c 05 w σCmax MC c I σAmax R a 05L c I L 2σAmax I R c 2a L 950mm Ans Problem 6151 If the radius of each notch on the plate is r 10 mm determine the largest moment M that can be applied The allowable bending stress for the material is σ allow 180 MPa Given w 165mm h 125mm t 20mm r 10mm σallow 180MPa Solution Section Property I 1 12 t h3 I 325520833 mm4 Stress Concentration Factor b 05 w h b r 2 r h 008 From Fig 650 K 21 Maximum Moment σ K M c I c 05 h M σallow I K c M 4464 kN m Ans Problem 6152 The stepped bar has a thickness of 15 mm Determine the maximum moment that can be applied to its ends if it is made of a material having an allowable bending stress of σ allow 200 MPa Given w 45mm t 15mm h1 30mm h2 10mm r1 3mm r2 6mm σallow 200MPa Solution Section Property For the larger section 1 For the smaller section 2 I1 1 12 t h1 3 I1 33750mm4 I2 1 12 t h2 3 I2 1250mm4 Stress Concentration Factor For the larger section 1 For the smaller section 2 w h1 15 r1 h1 01 h1 h2 3 r2 h2 06 From Fig 648 K1 175 From Fig 648 K2 12 Maximum Moment σ K M c I For the larger section 1 For the smaller section 2 c1 05 h1 M1 σallow I1 K1 c1 c2 05 h2 M2 σallow I2 K2 c2 M1 25714 N m M2 4167 N m Mallow min M1 M2 Mallow 4167 N m Ans Problem 6153 The bar has a thickness of 125 mm and is made of a material having an allowable bending stress of σ allow 140 MPa Determine the maximum moment M that can be applied Given t 125mm r 75mm w 150mm h 50mm σallow 140MPa Solution Section Property I 1 12 t h3 I 13020833 mm4 Stress Concentration Factor w h 3 r h 015 From Fig 648 K 16 Maximum Moment σ K M c I c 05 h M σallow I K c M 45573 N m Ans Problem 6154 The bar has a thickness of 125 mm and is subjected to a moment of 900 Nm Determine the maximum bending stress in the bar Given t 125mm r 75mm w 150mm h 50mm M 900N m Solution Section Property I 1 12 t h3 I 13020833 mm4 Stress Concentration Factor w h 3 r h 015 From Fig 648 K 16 Maximum Bending Stress c 05 h σmax K M c I σmax 2765 MPa Ans Problem 6155 The simply supported notched bar is subjected to two forces P Determine the largest magnitude of P that can be applied without causing the material to yield The material is A36 steel Each notch has a radius of r 3 mm Given t 12mm r 3mm w 42mm h 30mm a 500mm σY 250MPa Solution Section Property I 1 12 t h3 I 2700000 mm4 Support Reaction By symmstry R1R2R Stress Concentration Factor ΣFy0 2R 2P 0 R P b 05 w h Internal Moment At midspan b r 2 r h 01 MC R a From Fig 650 K 192 MC P a Maximum Moment σY K MC c I c 05 h MC σY I K c P a σY I K c P σY I a K c P 46875 N Ans Problem 6156 The simply supported notched bar is subjected to the two loads each having a magnitude of P 500 N Determine the maximum bending stress developed in the bar and sketch the bendingstress distribution acting over the cross section at the center of the bar Each notch has a radius of r 3 mm Given t 12mm r 3mm w 42mm h 30mm a 500mm P 500N Solution Section Property I 1 12 t h3 I 2700000 mm4 Support Reaction By symmstry R1R2R Stress Concentration Factor ΣFy0 2R 2P 0 R P b 05 w h Internal Moment At midspan b r 2 r h 01 MC R a From Fig 650 K 192 MC 250N m Maximum Bending Stress c 05 h σmax K MC c I σmax 2667 MPa Ans Problem 6157 A rectangular A36 steel bar has a width of 25 mm and height of 75 mm Determine the moment applied about the horizontal axis that will cause half the bar to yield Given b 25mm σY 250MPa d 75mm Solution de 05d dp 05d Elasticplastic Moment M σY b de 2 2de 3 σY b dp 2 de dp 2 M 952 kN m Ans Problem 6158 The box beam is made of an elastic perfectly plastic material for which σY 250 MPa Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released Given bo 200mm do 200mm bi 150mm di 150mm σY 250MPa Solution Section Property tb 05 do di td 05 bo bi I 1 12 bo do 3 bi di 3 I 9114583333 mm4 Plastic Moment Mp σY bo tb do tb σY 2td di 2 di 2 Mp 28906250 N m Modulus of Rupture The modulus of rupture σr can be determined using the flexure formula with the application of reverse plastic moment Mp c 05do σr Mp c I σr 31714 MPa Residul Bending Stress σt σr σY σt 6714 MPa Ans σb σr σY σb 6714 MPa Ans Problem 6159 The box beam is made of an elastic plastic material for which σY 250 MPa Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released Given bf 200mm dw 200mm tf 15mm tw 20mm σY 250MPa Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 8278333333 mm4 Plastic Moment Mp σY bf tf D tf σY tw dw 2 dw 2 Mp 21125 kN m Modulus of Rupture The modulus of rupture σr can be determined using the flexure formula with the application of reverse plastic moment Mp c 05D σr Mp c I σr 29346 MPa Residul Bending Stress σt σr σY σt 4346 MPa Ans σb σr σY σb 4346 MPa Ans Problem 6160 Determine the plastic section modulus and the shape factor of the beams cross section Set a mm Given bf 2a tf a dw 2a tw a Solution Section Property A bf tf dw tw yc Σ yi Ai Σ Ai yc bf tf 05tf dw tw 05dw tf A yc 125 a I 1 12 bf tf 3 bf tf 05tf yc 2 1 12 tw dw 3 tw dw 05dw tf yc 2 I 308 a4 Maximum Elastic Moment c dw tf yc c 175 a σY MY c I MY σY I c I c 17619 a3 MY 17619 a3 σY Plastic Moment 0 A σdA σY tw d σY tw dw d σY bf tf 0 d tw dw bf tf 2tw d 2a darm 05tf dw 05d darm 150 a Mp σY tw d darm Mp σY tw d darm tw d darm 300 a3 Mp 300 a3 σY Shape Factor k Mp MY k 300 a3 17619 a3 k 170 Ans Plastic Section Modulus z Mp σY z 300 a3 Ans Problem 6161 The beam is made of elastic perfectly plastic material Determine the maximum elastic moment and th plastic moment that can be applied to the cross section Take a 50 mmand σY 230 MPa Given a 50mm bf 2a tf a σY 230MPa dw 2a tw a Solution Section Property A bf tf dw tw yc Σ yi Ai Σ Ai yc bf tf 05tf dw tw 05dw tf A yc 625 mm I 1 12 bf tf 3 bf tf 05tf yc 2 1 12 tw dw 3 tw dw 05dw tf yc 2 I 1927083333 mm4 Maximum Elastic Moment c dw tf yc c 8750 mm σY MY c I MY σY I c MY 5065 kN m Ans Plastic Moment 0 A σdA σY tw d σY tw dw d σY bf tf 0 d tw dw bf tf 2tw d 100mm darm 05tf dw 05d darm 7500 mm Mp σY tw d darm Mp 8625 kN m Ans Problem 6162 The rod has a circular cross section If it is made of an elastic plastic material determine the shape factor and the plastic section modulus Z Set r mm Solution Section Property A πr2 I π 4 r4 Maximum Elastic Moment c r σY MY c I MY σY I c I c 07854 r3 MY 07854 r3 σY Plastic Moment darm 2 4r 3π Mp σY A 2 darm Mp σY A 2 darm A 2 darm 13333 r3 Mp 13333 r3 σY Shape Factor k Mp MY k 13333 r3 07854 r3 k 170 Ans Plastic Section Modulus z Mp σY z 1333 r3 Ans Problem 6163 The rod has a circular cross section If it is made of an elastic plastic material determine the maximum elastic moment and plastic moment that can be applied to the cross sectionTake r 75 mm σY 250 MPa Given r 75mm σY 250MPa Solution Section Property A πr2 I π 4 r4 Maximum Elastic Moment c r σY MY c I MY σY I c MY 8283 kN m Ans Plastic Moment darm 2 4r 3π Mp σY A 2 darm Mp 14063 kN m Ans Problem 6164 Determine the plastic section modulus and the shape factor of the cross section Set a mm Given bf 3a tf a dw 3a tw a Solution Section Property A bf tf dw tf tw A 5a2 I 1 12 bf tf 3 1 12 tw dw 3 tf 3 I 241667 a4 Maximum Elastic Moment c 05dw c 15 a σY MY c I MY σY I c I c 161111 a3 MY 161111 a3 σY Plastic Moment dwarm tf 05 dw tf dwarm 200 a dfarm 05tf dfarm 050 a Mp σY bf tf 2 dfarm σY tw dw tf 2 dwarm Mp σY bf tf 2 dfarm tw dw tf 2 dwarm bf tf 2 dfarm tw dw tf 2 dwarm 275 a3 Mp 275 a3 σY Shape Factor k Mp MY k 275 a3 161111 a3 k 171 Ans Plastic Section Modulus z Mp σY z 275 a3 Ans Problem 6165 The beam is made of elastic perfectly plastic material Determine the maximum elastic moment and th plastic moment that can be applied to the cross section Take a 50 mm and σY 250 MPa Given a 50mm σY 250MPa bf 3a tf a dw 3a tw a Solution Section Property A bf tf dw tf tw A 5a2 I 1 12 bf tf 3 1 12 tw dw 3 tf 3 I 241667 a4 Maximum Elastic Moment c 05dw c 15 a σY MY c I MY σY I c MY 5035 kN m Ans Plastic Moment dwarm tf 05 dw tf dwarm 200 a dfarm 05tf dfarm 050 a Mp σY bf tf 2 dfarm σY tw dw tf 2 dwarm Mp 8594 kN m Ans Problem 6166 The beam is made of an elastic perfectly plastic material Determine the plastic moment Mp that can be supported by a beam having the cross section shown σY 210 MPa Given ro 50mm tw 25mm σY 210MPa ri 25mm dw 250mm Solution Plastic Moment A1 π ro 2 ri 2 darm1 dw 2ro A2 tw 05dw darm2 05dw Mp σY A1 darm1 σY A2 darm2 Mp 515kN m Ans Problem 6167 Determine the plastic moment Mp that can be supported by a beam having the cross section shown σY 210 MPa Given ro 50mm tw 25mm σY 210MPa ri 25mm dw 250mm Solution A1 π ro 2 ri 2 A2 tw dw d A3 twd 0 A σ dA σY A1 σY A2 σY A3 0 A1 A2 A3 0 π ro 2 ri 2 tw dw d twd 0 d π ro 2 ri 2 tw dw 2tw d 24281 mm Plastic Moment A1 π ro 2 ri 2 darm1 ro dw d A2 tw dw d darm2 05 dw d A3 twd darm3 05d Mp σY A1 darm1 σY A2 darm2 σY A3 darm3 Mp 2256 kN m Ans Problem 6168 Determine the plastic section modulus and the shape factor for the member having the tubular cross section Set d mm Solution Section Property A π 4 2d 2 d2 A 235619 d2 I π 64 2d 4 d4 I 073631 d4 Maximum Elastic Moment c d σY MY c I MY σY I c I c 073631 d3 MY 073631 d3 σY Plastic Moment yc Σ yi Ai Σ Ai yc 05 π 4 2d 2 4d 3π 05 π 4 d2 1 2 4d 3π 05A yc 049515 d darm 2yc Mp σY A 2 darm Mp σY A 2 darm A 2 darm 116667 d3 Mp 116667 d3 σY Shape Factor k Mp MY k 116667 d3 073631 d3 k 158 Ans Plastic Section Modulus z Mp σY z 116667 d3 Ans Problem 6169 Determine the plastic section modulus and the shape factor for the member Solution Set b mm h mm Section Property A 1 2 b h I 1 36 b h3 Maximum Elastic Moment c 2 3 h σY MY c I MY σY I c I c 1 24 b h2 MY 1 24 b h2 σY Plastic Moment From the geometry b d h b A 1 2 b d Atrp 1 2 b b h d A 1 2 d h b d Atrp 1 2 b d h b h d 0 A σdA σY A σY Atrp 0 A Atrp 1 2 d h b d 1 2 b d h b h d d h 2 b b 2 Note The centroid of a trapezoidal area was used in the calculation hc h d 3 2 b b b b darm 1 3 d h d hc darm 1 3 d h d b 2 b 3 b b darm 1 3 h 2 h h 2 b 2 2 b 3 b 2 b darm 4 2 2 h 6 Mp σY A 2 darm Mp σY A 2 darm A 2 darm 2 2 6 b h2 Mp 2 2 6 b h2 σY Shape Factor k Mp MY k 2 2 6 b h2 1 24 b h2 k 234 Ans Plastic Section Modulus z Mp σY z 2 2 6 b h2 Ans Problem 6170 The member is made of elastic perfectly plastic material for which σY 230 MPa Determine the maximum elastic moment and the plastic moment that can be applied to the cross sectionTake b 50 mm and h 80 mm Given b 50mm h 80mm σY 230MPa Solution Section Property A 1 2 b h A 2000mm2 I 1 36 b h3 I 71111111 mm4 Maximum Elastic Moment c 2 3 h σY MY c I MY σY I c MY 307 kN m Ans Plastic Moment From the geometry b d h b A 1 2 b d Atrap 1 2 b b h d A 1 2 d h b d Atrap 1 2 b d h b h d 0 A σdA σY A σY Atrap 0 A Atrap 1 2 d h b d 1 2 b d h b h d d h 2 b b 2 Note The centroid of a trapezoidal area was used in the calculation hc h d 3 2 b b b b darm 1 3 d h d hc darm 1 3 d h d b 2 b 3 b b darm 1 3 h 2 h h 2 b 2 2 b 3 b 2 b darm 4 2 2 h 6 darm 3124 mm Mp σY A 2 darm Mp 719 kN m Ans Problem 6171 The wideflange member is made from an elasticplastic material Determine the shape factor and the plastic section modulus Z Set b mm h mm t mm Given bf b tf t D h tw t Solution dw D 2 tf dw h 2t Section Property A bf D bf tw dw A b h b t h 2t I 1 12 bf D3 1 12 bf tw dw 3 I 1 12 b h3 1 12 b t h 2t 3 Maximum Elastic Moment c 05D c 05h σY MY c I MY σY I c I c 1 6 b h2 1 6 h b t h 2t 3 MY 1 6 h b h3 b t h 2 t 3 σY Plastic Moment dwarm 05dw dwarm h 2t 2 dfarm D tf dfarm h t Mp σY bf tf dfarm σY tw dw 2 dwarm Mp σY b t h t t h 2t 2 h 2t 2 Mp b t h t t 4 h 2t 2 σY Shape Factor k Mp MY k b t h t t 4 h 2t 2 1 6 h b h3 b t h 2 t 3 k 3 h 2 4b t h t t h 2t 2 b h3 b t h 2 t 3 Ans Plastic Section Modulus z Mp σY z b t h t t 4 h 2t 2 Ans Problem 6172 The beam is made of an elasticplastic material for which σY 200 MPa If the largest moment in the beam occurs within the center section aa determine the magnitude of each force P that causes this moment to be a the largest elastic moment and b the largest plastic moment Given a 2m b 100mm h 200mm σY 200MPa Solution Section Property A b h A 20000mm2 I 1 12 b h3 I 6666666667 mm4 a Maximum Elastic Moment c 05 h MY P a σY MY c I σY P a c I P σY I a c P 6667 kN Ans b Plastic Moment darm h 2 Mp σY A 2 darm Mp 20000 kN m P Mp a P 10000 kN Ans Problem 6173 The beam is made of phenolic a structural plastic that has the stressstrain curve shown If a portion of the curve can be represented by the equation σ 5106ε 12 MPa determine the magnitude of w the distributed load that can be applied to the beam without causing the maximum strain in its fibers a the critical section to exceed εmax 0005 mmmm Given b 150mm h 150mm σ2 5 106 ε L 2m εmax 0005 mm mm Solution Stressstrain Relationship unit MPa When εmax 0005 σmax unit 5 106 εmax σmax 15811 MPa Resultant Internal Forces The resultant internal forces T and C can be evaluated from the volume of stress block which is a paraboloid T C T 2 3 σmax b h 2 T 118585 kN darm 2 3 5 h 2 darm 90mm Maximum Internal Moment Mmax T darm Mmax 10673 kN m By observation the maximum moment occurs over the middle support Mmax w L w Mmax L w 5336 m kN m Ans Problem 6174 The box beam is made from an elastic plastic material for which σY 175 MPa Determine the intensity of the distributed load w0 that will cause the moment to be a the largest elastic moment and b the largest plastic moment Given bo 200mm do 400mm bi 150mm di 300mm L 3m σY 175MPa Solution Support Reaction By symmstry R1R2R ΣFy0 2R 2 05 wo L 0 R 05 wo L Maximum Moment M R L L 3 05 wo L M wo L2 3 a Elastic Analysis σ M c I c 05do I 1 12 bo do 3 bi di 3 MY σY I c MY 63802 kN m wo 3 MY L2 wo 21267 kN m Ans b Plastic Analysis tw 05 bo bi tf 05 do di Af bo tf darm1 do tf Aw di tw darm2 05di Mp σY Af darm1 σY Aw darm2 Mp 8094 kN m wo 3 Mp L2 wo 26979 kN m Ans Problem 6175 The beam is made of a polyester that has the stressstrain curve shown If the curve can be represented by the equation σ 140 tan115ε MPa where tan115ε is in radians determine the magnitude of the force P that can be applied to the beam without causing the maximum strain in its fibers at the critical section to exceed εmax 0003 mmmm Given b 50mm h 100mm σ 140atan 15ε L 24m εmax 0003 mm mm Solution Support Reaction By symmstry R1R2R ΣFy0 2R 2P 0 R P Maximum Moment M R L M P L Stressstrain Relationship unit MPa The bending stress can bs expressed in terms of y using ε εmax 05h y σ 140atan 15 εmax 05h y unit When εmax 0003 ymax 05h σmax 140 atan 15 εmax 05h ymax unit σmax 630 MPa A yσdA M Resultant Internal Moment M 2 0 05h y y 140atan 15 εmax 05h y unit b d M 52479 N m P M L P 21866 N Ans Problem 6176 The stressstrain diagram for a titanium alloy can be approximated by the two straight lines If a strut made of this material is subjected to bending determine the moment resisted by the strut if the maximum stress reaches a value of a σA and b σB Given b 50mm d 75mm σA 980MPa σB 1260MPa εA 001 mm mm εB 004 mm mm Solution Maximum Elastic Moment Since the stress is linearly related to strain up to point A the flexure formula can be applied σ M c I c 05d I b d3 MY σA I c MY 55125 kN m Ans UItimate Moment yA εA εB 05d h 05d yA C1 T1 T1 σA σB 2 b h C2 T2 T2 σA 2 b yA Note The centroid of a trapezoidal area was used in the calculation of moment hc h 3 2σB σA σB σA darm1 2 yA hc darm2 2 3 2yA M T1 darm1 T2 darm2 M 7854 kN m Ans σ not zero Problem 6177 A beam is made from polypropylene plastic and has a stressstrain diagram that can be approximated by the curve shown If the beam is subjected to a maximum tensile and compressive strain of ε 002 mmmm determine the maximum moment M Given b 30mm h 100mm σ 10 4 ε MPa εmax 002 mm mm Solution Stressstrain Relationship unit MPa The bending stress can bs expressed in terms of y using ε εmax 05h y σ 10 4 εmax 05h y unit A yσdA M Resultant Internal Moment M 2 0 05h y y 10 4 εmax 05h y unit b d M 0251 kN m Ans Problem 6178 The bar is made of an aluminum alloy having a stressstrain diagram that can be approximated by the straight line segments shown Assuming that this diagram is the same for both tension and compression determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is εmax 003 Given b 75mm d 100mm σA 420MPa σB 560MPa εA 0006 mm mm εB 0025 mm mm σC 630MPa εC 005 mm mm εmax 003 mm mm Solution Maximum Stress σmax σB εmax εB σC σB εC εB σmax σC σB εC εB εmax εB σB σmax 574MPa Maximum Moment yA εA εmax 05d yB εB εmax 05d h1 05d yB h2 yB yA C1 T1 T1 σB σmax 2 b h1 C2 T2 T2 σA σB 2 b h2 C3 T3 T3 σA 2 b yA Note The centroid of a trapezoidal area was used in the calculation of moment hc1 h1 3 2σmax σB σmax σB hc2 h2 3 2σB σA σB σA darm1 2 yB hc1 darm2 2 yA hc2 darm3 2 3 2yA M T1 darm1 T2 darm2 T3 darm3 M 9648 kN m Ans Problem 6179 The bar is made of an aluminum alloy having a stressstrain diagram that can be approximated by the straight line segments shown Assuming that this diagram is the same for both tension and compression determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is εmax 005 Given b 75mm d 100mm σA 420MPa σB 560MPa εA 0006 mm mm εB 0025 mm mm σC 630MPa εC 005 mm mm Solution εmax εC Stressstrain Relationship σ1 ε σA εA σ1 σA εA ε σ2 σA ε εA σB σA εB εA σ2 σB σA εB εA ε εA σA σ3 σB ε εB σC σB εC εB σ3 σC σB εC εB ε εB σB yA εA εmax 05d yB εB εmax 05d Strain ε εmax 05d y σ1 σA εA εmax 05d y for 0 y yA σ2 σB σA εB εA εmax 05d y εA σA for yA y yB σ3 σC σB εC εB εmax 05d y εB σB for yB y 05d A yσdA M Resultant Moment M1 2 0 yA y y σA εA εmax 05d y b d M1 076 kN m M2 2 yA yB y y σB σA εB εA εmax 05d y εA σA b d M2 2228 kN m M3 2 yA yB y y σC σB εC εB εmax 05d y εB σB b d M3 238 kN m M M1 M2 M3 M 4684 kN m Ans Note The solution can also be obtained from stress blocks as in Prob 6178 Problem 6180 The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectly plastic in compression Determine the maximum bending moment M that can be supported by the bea so that the compressive material at the outer edge starts to yield Solution A1 a d C 1 2σY A1 A2 a h d T σY A2 0 A σ dA C T 0 1 2σY A1 σY A2 0 1 2A1 A2 0 1 2a d a h d 0 d 2 h 3 Plastic Moment darm 2 3d 1 2 h d darm 2 3 2 h 3 1 2 h 2 h 3 darm 11 h 18 Mp σY A2 darm Mp σY a h d 11 h 18 Mp 11 a h2 54 σY Ans Problem 6181 The plexiglass bar has a stressstrain curve that can be approximated by the straightline segments shown Determine the largest moment M that can be applied to the bar before it fails Given b 20mm h 20mm σt1 40MPa σt2 60MPa εt1 002 mm mm εt2 004 mm mm σc1 80 MPa σc2 100 MPa εc1 004 mm mm εc2 006 mm mm Solution Assume failure due to tension and εc εc1 A1 b h d C 1 2σc A1 A2 1 2 b d T1 1 2σt1 A2 A3 1 2 b d T2 1 2 σt1 σt2 A3 0 A σ dA C T1 T2 0 1 2σc b h d 1 2σt1 1 2 b d 1 2 σt1 σt2 1 2 b d 0 σc h d 1 σt1 05σt2 Try σc 74833MPa then d h σc σt1 05σt2 σc d 10334 mm σc σc1 Check From the strain diagram εc h d d εt2 εc 0037417 mm mm OK εc εc1 From the σε diagram σ εc εc1 σc1 σ 74833 MPa OK Close to assumed value Hence C 1 2σc b h d C 72336 kN T1 1 2σt1 1 2 b d T1 20667 kN T2 1 2 σt1 σt2 1 2 b d T2 51668 kN Ultimate Moment darm1 2 3 h d darm1 64442 mm darm2 2 3 d 2 darm2 34446 mm Note The centroid of a trapezoidal area was used in the calculation hc 05d 3 2σt1 σt2 σt1 σt2 hc 24112 mm darm3 d hc darm3 79225 mm Mult C darm1 T1 darm2 T2 darm3 Mult 9467 N m Ans Problem 6182 The beam is made from three boards nailed together as shown If the moment acting on the cross section is M 650 Nm determine the resultant force the bending stress produces on the top board Given bf 290mm tf 15mm tw 20mm dw 125mm M 650N m Solution D dw tf y Σ yi Ai Σ Ai yc bf tf 05 tf 2 dw tw 05dw tf bf tf 2dw tw yc 44933 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 2tw dw 3 2tw dw yc 05dw tf 2 I If Iw I 1799037489 mm4 Bending Stress σ M c I At B cB yc tf σB M cB I σB 10815 MPa At A cA yc σA M cA I σA 16235 MPa Resultant Force For the top board F 05 σA σB bf tf F 5883 kN Ans Problem 6183 The beam is made from three boards nailed together as shown Determine the maximum tensile and compressive stresses in the beam Given bf 290mm tf 15mm tw 20mm dw 125mm M 650N m Solution D dw tf y Σ yi Ai Σ Ai yc bf tf 05 tf 2 dw tw 05dw tf bf tf 2dw tw yc 44933 mm If 1 12 bf tf 3 bf tf yc 05tf 2 Iw 1 12 2tw dw 3 2tw dw yc 05dw tf 2 I If Iw I 1799037489 mm4 Maximum Bending Stress σ M c I For compression cc yc σcmax M cc I σcmax 1623 MPa Ans For tension ct D yc σtmax M ct I σtmax 3435 MPa Ans Problem 6184 Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x where 0 x 18 m Given a 24m b 12m P 40kN w 30 kN m M 75kN m Solution Equilibrium A w a P A 112kN 3054kNm 112kN 30x ΣFy0 MA w a 05a P a b M ΣΜA0 MA 30540 kN m x1 0 001 a a x2 a 101 a a b V1 x1 A w x1 1 kN V2 x2 A w a 1 kN Ans M1 x1 MA A x1 05w x1 2 1 kN m Ans M2 x2 MA A x2 w a x2 05 a M 1 kN m 0 1 2 3 0 50 100 150 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 400 200 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 6185 Draw the shear and moment diagrams for the beam Hint The 100kN load must be replaced by equivalent loadings at point C on the axis of the beam Given a 12m b 12m c 12m d 03m F1 75kN F2 100kN Solution Equilibrium Given ΣFy0 A F1 B 0 ΣΜC0 A a b c F1 b c F2 d 0 Guess A 1kN B 1kN A B Find A B A B 5833 1667 kN x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c V1 x1 A 1 kN V2 x2 A F1 1 kN V3 x3 A F1 1 kN M1 x1 A x1 kN m M2 x2 A x2 F1 x2 a 1 kN m M3 x3 A x3 F1 x3 a F2 d 1 kN m 0 1 2 3 0 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 0 50 100 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 6186 Determine the plastic section modulus and the shape factor for the wideflange beam Given bf 180mm tf 20mm dw 180mm tw 30mm Solution D dw 2 tf Section Property A bf D bf tw dw A 12600mm2 I 1 12 bf D3 1 12 bf tw dw 3 I 86820000mm4 Set σY MPa Maximum Elastic Moment c 05D σY MY c I MY I c σY MY σY 78927273 mm3 Plastic Moment dwarm 05dw dwarm 90mm dfarm D tf dfarm 200mm Mp bf tf dfarm tw dw 2 dwarm σY Mp σY 96300000 mm3 Plastic Section Modulus z Mp σY z 963 10 6 m3 Ans Shape Factor k Mp MY k 122 Ans Problem 6187 Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt gear and flywheel The bearings at A and B exert only vertical reactions on the shaft Given a 200mm b 400mm c 300mm d 200mm C 450N D 300 N E 150N Solution Equilibrium Given ΣFy0 A B C D E 0 ΣΜB0 A a b c C b c D c E d 0 Guess A 1N B 1N A B Find A B A B 21667 8333 N x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c x4 a b c 101 a b c a b c d V1 x1 A 1 N V2 x2 A C 1 N V3 x3 A C D 1 N V4 x4 A C D B 1 N M1 x1 A x1 N m M2 x2 A x2 C x2 a 1 N m M3 x3 A x3 C x3 a D x3 a b 1 N m M4 x4 A x4 C x4 a D x4 a b B x4 a b c 1 N m 0 02 04 06 08 1 400 200 0 200 400 Distance m Shear N V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 02 04 06 08 1 50 0 50 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 6188 The beam is constructed from four pieces of wood glued together as shown If the internal bending moment is M 120 kNm determine the maximum bending stress in the beam Sketch a threedimensional view of the stress distribution acting over the cross section Given bo 300mm do 300mm bi 250mm di 250mm M 120kN m Solution Section Property I 1 12 bo do 3 bi di 3 Maximum Bending Stress σ M c I cmax 05do σmax M cmax I σmax 5151 MPa Ans Problem 6189 The beam is constructed from four pieces of wood glued together as shown If the internal bending moment is M 120 kNm determine the resultant force the bending moment exerts on the top and bottom boards of the beam Given bo 300mm do 300mm bi 250mm di 250mm M 120kN m Solution Section Property t 05 do di I 1 12 bo do 3 bi di 3 Maximum Bending Stress σ M c I co 05do σo M co I σo 51505 MPa ci 05di σi M ci I σi 42921 MPa Resultant Force F 1 2 σo σi t bo F 35410 kN Ans Problem 6190 For the section Iy 317106 m4 Iz 114106 m4 Iyz 151106 m4 Using the techniques outline in Appendix A the members crosssectional area has principal moments of inertia of Iy 29106 m and Iz 117106 m4 computed about the principal axes of inertia y and z respectively If the sectio is subjected to a moment of M 2 kNm directed as shown determine the stress produced at point A a using Eq 611 and b using the equation developed in Prob 6111 Given M 2000N m θ 1010deg b1 80mm b2 140mm h1 60mm h2 60mm Iy 317 10 6 m4 Iz 114 10 6 m4 Iyz 151 10 6 m4 Iy 290 10 6 m4 Iz 117 10 6 m4 Solution θ θ Coordinates of Point A yA b2 zA h2 yA zA cos θ sin θ sin θ cos θ yA zA yA zA 14835 3452 mm a Using Eq 611 Internal Moment Components My M sin θ Mz M cos θ Bending Stress σA Mz yA Iz My zA Iy σA 2079 MPa C Ans b Using the equation developed in Prob 6111 Internal Moment Components My 0 Mz M Bending Stress Using formula developed in Prob 6111 D Iy Iz Iyz 2 σA 1 D Mz My Iy Iyz Iyz Iz yA zA σA 2086 MPa C Ans Problem 6191 The strut has a square cross section a by a and is subjected to the bending moment M applied at an angle θ as shown Determine the maximum bending stress in terms of a M and θ What angle θ wil give the largest bending stress in the strut Specify the orientation of the neutral axis for this case Solution Internal Moment Components My M sin θ Mz M cos θ Section Property Iy a4 12 Iz a4 12 Maximum Bending Stress By inspection maximum bending stress occurs at A and B At A yA 05a zA 05 a σ Mz y Iz My z Iy σ 12 M cos θ 05a a4 12 M sin θ 05 a a4 σ 6 M a3 cos θ sin θ Ans θ σ d d 6 M a3 sin θ cos θ θ σ d d 0 sin θ cos θ 0 tan θ 1 θ 45deg Ans Orientation of Neutral Axis θ θ tan α Iz Iy tan θ Iz Iy 1 α atan 1 tan θ α 4500 deg Ans Problem 71 If the beam is subjected to a shear of V 15 kN determine the webs shear stress at A and B Indicate the shearstress components on a volume element located at these points Set w 125 mm Show that the neutral axis is located at y 01747 m from the bottom and INA 02182103 m4 Given bf 200mm bf 125mm tf 30mm tw 25mm dw 250mm V 15kN Solution Section Property D dw 2tf A bf tf dw tw bf tf A 16000mm2 y Σ yi Ai Σ Ai yc bf tf 05tf dw tw 05dw tf bf tf D 05tf A yc 17469 mm If 1 12 bf tf 3 bf tf 05tf yc 2 Iw 1 12 tw dw 3 tw dw 05dw tf yc 2 If 1 12 bf tf 3 bf tf D 05tf yc 2 I If Iw If I 21818 10 6 m4 Q Σ yi A1 QA D 05tf yc bf tf QA 72187500 mm3 QB yc 05tf bf tf QB 59882812 mm3 Shear Stress τ V Q I t τA V QA I tw τA 199 MPa Ans τB V QB I tw τB 165 MPa Ans Problem 72 If the wideflange beam is subjected to a shear of V 30 kN determine the maximum shear stress in the beam Set w 200 mm Given bf 200mm tf 30mm tw 25mm dw 250mm V 30kN Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 26865 10 6 m4 Q Σ yi A1 Qmax bf tf D 2 tf 2 dw 2 tw dw 4 Qmax 103531250 mm3 Shear Stress τ V Q I t Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I tw τmax 462 MPa Ans Problem 73 If the wideflange beam is subjected to a shear of V 30 kN determine the shear force resisted by the web of the beam Set w 200 mm Given bf 200mm dw 250mm tf 30mm tw 25mm V 30kN Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 26865 10 6 m4 Af 05D y bf yf 05 05D y y yf 05 05D y Qf Af yf Qf 05 05D y bf 05D y Qf 05bf 025D2 y2 Shear Stress τ V Q I t τf V I bf Qf τf V I bf 05bf 025D2 y2 A 0 dA V τ Resultant Shear Force For the flange Vf V I bf 05dw 05D y 05 bf 025 D2 y2 bf d Vf 146 kN Vw V 2Vf Vw 2709 kN Ans Problem 74 If the wideflange beam is subjected to a shear of V 125 kN determine the maximum shear stress in the beam Given bf 200mm dw 250mm tf 25mm tw 25mm V 125kN Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 Qmax bf tf D 2 tf 2 dw 2 tw dw 4 Maximum Shear Stress τ V Q I t Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I tw τmax 1987 MPa Ans Problem 75 If the wideflange beam is subjected to a shear of V 125 kN determine the shear force resisted by the web of the beam Given bf 200mm dw 250mm tf 25mm tw 25mm V 125kN Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 A1 bf tf y1c 05 D tf A2 05dw y tw y2c 05 05dw y y y2c 05 05dw y Qw A1 y1c A2 y2c Qw 05bf tf D tf 05 05dw y tw 05dw y Qw 05bf tf D tf 05tw 025dw 2 y2 Shear Stress τ V Q I t τw V I tw Qw τw V I tw 05 bf tf D tf 05 tw 025 dw 2 y2 A 0 dA V τ Resultant Shear Force For the web Vw 05 dw 05dw y V I tw 05 bf tf D tf 05 tw 025 dw 2 y2 tw d Vw 11504 kN Ans Problem 76 The beam has a rectangular cross section and is made of wood having an allowable shear stress of τallow 112 MPa If it is subjected to a shear of V 20 kN determine the smallest dimension a of its bottom and 15a of its sides Given V 20kN τallow 112MPa Solution Section Property I 1 12 a 15a 3 t a Qmax 15a 2 a 15a 4 Allowablwe Shear Stress τ V Q I t I t V Qmax τallow 1 12 a 15a 3 a V τallow 15a 2 a 15a 4 a V τallow a 4226 mm Ans Problem 77 The beam has a rectangular cross section and is made of wood If it is subjected to a shear of V 20 kN and a 250 mm determine the maximum shear stress and plot the shearstress variation over the cross section Sketch the result in three dimensions Given a 250mm V 20kN Solution Section Property b a d 15a I 1 12 b d3 Qmax d 2 b d 4 Maximum Shear Stress τ V Q I b Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I b τmax 0320 MPa Ans Problem 78 Determine the maximum shear stress in the strut if it is subjected to a shear force of V 20 kN Given bf 120mm tf 12mm tw 80mm dw 60mm V 20kN Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 521 10 6 m4 Q Σ yi A1 Qmax bf tf D 2 tf 2 dw 2 tw dw 4 Qmax 8784000 mm3 Shear Stress τ V Q I t Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I tw τmax 422 MPa Ans Problem 79 Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is τallow 40 MPa Given bf 120mm tf 12mm tw 80mm dw 60mm τallow 40MPa Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 521 10 6 m4 Q Σ yi A1 Qmax bf tf D 2 tf 2 dw 2 tw dw 4 Qmax 8784000 mm3 Shear Stress τ V Q I t Maximum shear stress occurs at the point where the neutral axis passes through the section τallow V Qmax I tw V I tw τallow Qmax V 18969 kN Ans Problem 710 Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a shear force of V 15 kN Given bf 120mm tf 12mm tw 80mm dw 60mm V 15kN Solution Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 521 10 6 m4 Q Σ yi A1 QA bf tf D 2 tf 2 QA 5184000 mm3 Qmax bf tf D 2 tf 2 dw 2 tw dw 4 Qmax 8784000 mm3 Shear Stress τ V Q I t Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I tw τmax 316 MPa Ans τwA V QA I tw τwA 187 MPa Ans τfA V QA I bf τfA 124 MPa Ans Problem 711 If the pipe is subjected to a shear of V 75 kN determine the maximum shear stress in the pipe Given ro 60mm ri 50mm V 75kN Solution Section Property t ro ri I π 4 ro 4 ri 4 Qmax 4ro 3π π ro 2 2 4ri 3π π ri 2 2 Maximum Shear Stress τ V Q I b Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I 2t τmax 4317 MPa Ans Problem 712 The strut is subjected to a vertical shear of V 130 kN Plot the intensity of the shearstress distribution acting over the crosssectional area and compute the resultant shear force developed in the vertical segment AB Given bf 350mm tf 50mm tw 50mm dw 350mm V 130kN Solution Section Property a 05 bf tw I 1 12 bf tf 3 tw dw 3 tw tf 3 I 18177 10 6 m4 Q Σ yi A1 QC a tw a 2 tf 2 QC 750000mm3 QD a tw a 2 tf 2 tf 2 bf tf 4 QD 859375mm3 τ V Q I t Shear Stress τwC V QC I tw τfC V QC I bf τD V QD I bf τwC 1073 MPa τfC 153 MPa τD 176 MPa A 0 dA V τ Resultant Shear Force Aw 05dw y tw yw 05 05dw y y yw 05 05dw y Qw Aw yw Qw 05 05dw y tw 05dw y Q 05tw 025dw 2 y2 τw V I tw Qw τw V I tw 05tw 025dw 2 y2 VAB V I tw 05tf 05dw y 05 tw 025 dw 2 y2 tw d VAB 5029 kN Ans Problem 713 The steel rod has a radius of 30 mm If it is subjected to a shear of V 25 kN determine the maximum shear stress Given r 30mm V 25kN Solution Section Property I π 4 r4 Qmax 4r 3π π r2 2 Maximum Shear Stress τ V Q I b Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I 2r τmax 1179 MPa Ans Problem 714 If the Tbeam is subjected to a vertical shear of V 60 kN determine the maximum shear stress in the beam Also compute the shearstress jump at the flangeweb junction AB Sketch the variation of the shearstress intensity over the entire cross section Given bf 300mm dw 150mm tf 75mm tw 100mm V 60kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf dw tw bf tf dw tw yc 8250 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I I1 I2 Qmax tw D yc D yc 2 QAB bf tf yc 05tf Shear Stress τ V Q I b τmax V Qmax I tw τmax 3993 MPa Ans τfAB V QAB I bf τfAB 1327 MPa Ans τwAB V QAB I tw τwAB 3982 MPa Ans Problem 715 If the Tbeam is subjected to a vertical shear of V 60 kN determine the vertical shear force resisted by the flange Given bf 300mm dw 150mm tf 75mm tw 100mm V 60kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf dw tw bf tf dw tw yc 8250 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I I1 I2 Af yc y bf yfc 05 yc y y yfc 05 yc y Q Af yfc Q 05 yc y bf yc y Q 05bf yc 2 y2 Shear Stress τ V Q I b τf V I bf Q τf V I bf 05bf yc 2 y2 A 0 dA V τ Resultant Shear Force For the flange yo yc tf Vf yo yc y V I bf 05bf yc 2 y2 bf d Vf 1908 kN Ans Problem 716 The Tbeam is subjected to the loading shown Determine the maximum transverse shear stress in the beam at the critical section Given L1 2m L2 2m L3 3m bf 100mm dw 100mm tf 20mm tw 20mm P 20kN w 8 kN m Solution L L1 L2 L3 Support Reaction Equilibrium Given ΣFy0 A P w L3 B 0 ΣΜB0 AL P L2 L3 w L3 05L3 0 Guess A 1kN B 1kN A B Find A B A B 1943 2457 kN Section Property D dw tf yc 05tf bf tf 05dw tf dw tw bf tf dw tw yc 4000 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I I1 I2 I 533333333 mm4 Qmax tw D yc D yc 2 Qmax 64000mm3 Maximum Shear Stress τ V Q I b Vmax B τmax Vmax Qmax I tw τmax 1474 MPa Ans Shear Force Diagram x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 A kN V2 x2 A P 1 kN V3 x3 A P w x3 L1 L2 1 kN 0 2 4 6 20 0 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 717 Determine the largest end forces P that the member can support if the allowable shear stress is τallow 70 MPa The supports at A and B only exert vertical reactions on the beam Given L1 1m τallow 70MPa L2 2m w 3 kN m L3 1m do 100mm bo 160mm di 60mm bi 80mm Solution Section Property yc 05do bo do 05di bi di bo do bi di yc 5857 mm I1 1 12 bo do 3 bo do 05do yc 2 I2 1 12 bi di 3 bi di 05di yc 2 I I1 I2 Qmax Af yfc Af yc bo bi yc 05yc Qmax 05yc 2 bo bi Maximum Shear Stress τ V Q I b Vmax P τallow P I bo bi 05yc 2 bo bi τallow P I 05yc 2 P I 05yc 2 τallow P 37342 kN Ans Shear Force Diagram L L1 L2 L3 Equilibrium Given ΣFy0 A w L2 B 2P 0 ΣΜB0 P L1 L2 A L2 w L2 05L2 P L3 0 Guess A 1kN B 1kN A B Find A B A B 37642 37642 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 P kN V2 x2 P A w x2 L1 1 kN V3 x3 P A w L2 B 1 kN 0 1 2 3 4 0 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 718 If the force P 4 kN determine the maximum shear stress in the beam at the critical section The supports at A and B only exert vertical reactions on the beam Given L1 1m P 4kN L2 2m w 3 kN m L3 1m do 100mm bo 160mm di 60mm bi 80mm Solution Section Property yc 05do bo do 05di bi di bo do bi di yc 5857 mm I1 1 12 bo do 3 bo do 05do yc 2 I2 1 12 bi di 3 bi di 05di yc 2 I I1 I2 Qmax Af yfc Af yc bo bi yc 05yc Qmax 05yc 2 bo bi Maximum Shear Stress τ V Q I b Vmax P τmax P I bo bi 05yc 2 bo bi τmax P I 05yc 2 τmax 0750 MPa Ans Shear Force Diagram L L1 L2 L3 Equilibrium Given ΣFy0 A w L2 B 2P 0 ΣΜB0 P L1 L2 A L2 w L2 05L2 P L3 0 Guess A 1kN B 1kN A B Find A B A B 700 700 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 P kN V2 x2 P A w x2 L1 1 kN V3 x3 P A w L2 B 1 kN 0 1 2 3 4 10 0 10 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 719 Plot the shearstress distribution over the cross section of a rod that has a radius c By what factor is the maximum shear stress greater than the average shear stress acting over the cross section Problem 720 Develop an expression for the average vertical component of shear stress acting on the horizontal plane through the shaft located a distance y from the neutral axis Problem 721 Railroad ties must be designed to resist large shear loadings If the tie is subjected to the 150kN rail loadings and the gravel bed exerts a distributed reaction as shown determine the intensity w for equilibrium and find the maximum shear stress in the tie Given L1 045m P 150kN L2 090m w 3 kN m L3 045m d 150mm b 200mm Solution Equilibrium ΣFy0 05w L1 w L2 05w L3 2P 0 w 2P 05L1 L2 05L3 w 22222 kN m Section Property I 1 12 b d3 Qmax 05 b d 025 d Shear Force Diagram L L1 L2 L3 x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 05 x1 w x1 L1 1 kN V2 x2 05w L1 P w x2 L1 1 kN V3 x3 05 w L1 2 P w L2 w x3 L1 L2 1 x3 L1 L2 2L3 1 kN 0 05 1 15 100 0 100 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Maximum Shear Stress τ V Q I b Vmax V2 L1 kN τmax Vmax I b Qmax τmax 5MPa Ans Problem 722 The beam is subjected to a uniform load w Determine the placement a of the supports so that the shear stress in the beam is as small as possible What is this stress Set w kN m a m L 5m Given L1 a L3 a L2 L 2a Solution Equilibrium By equilibrium A B R ΣFy0 2R w L 0 R 05w L Shear Force Diagram x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 w x1 1 kN V2 x2 w x2 R 1 kN V3 x3 w x3 2 R 1 kN 0 1 2 3 4 5 0 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Require V1 a V2 a w a w a R 2w a 05w L a 025L Vmax w a Vmax 025w L Section Property I 1 12 b d3 Qmax b d 2 d 4 Maximum Shear Stress τ V Q I b τmax Vmax I b Qmax τmax 025w L 1 12 b d3 b b d 2 d 4 τmax 3w L 8 b d Ans Problem 723 The timber beam is to be notched at its ends as shown If it is to support the loading shown determine the smallest depth d of the beam at the notch if the allowable shear stress is τallow 3 MPa The beam has a width of 200 mm Given L1 12m P1 125kN L2 18m P2 250kN L3 18m P3 125kN L4 12m b 200mm τallow 3MPa Solution Equilibrium By symmetry R1R R2R ΣFy0 P1 P2 P3 2R 0 R 05 P1 P2 P3 R 2500 kN Section Property I 1 12 b d3 Qmax 05 b d 025 d Qmax 0125 b d2 Maximum Shear Stress τ V Q I b Vmax R τallow R 0125b d2 1 12 b d3 b d 12 0125 R τallow b d 625 mm Ans Problem 724 The beam is made from three boards glued together at the seams A and B If it is subjected to the loading shown determine the shear stress developed in the glued joints at section aa The supports at C and D exert only vertical reactions on the beam Given bf 150mm dw 200mm tf 40mm tw 50mm P 25kN Solution Equilibrium By symmetry RCR RDR ΣFy0 3P 2R 0 R 15P R 3750 kN Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 QA bf tf 05D 05tf QB QA Shear Stress τ V Q I b Vaa R P τA Vaa QA I tw τA 0747 MPa Ans τB Vaa QB I tw τB 0747 MPa Ans Problem 725 The beam is made from three boards glued together at the seams A and B If it is subjected to the loading shown determine the maximum shear stress developed in the glued joints The supports at C and D exert only vertical reactions on the beam Given bf 150mm dw 200mm tf 40mm tw 50mm P 25kN Solution Equilibrium By symmetry RCR RDR ΣFy0 3P 2R 0 R 15P R 3750 kN Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 QA bf tf 05D 05tf QB QA Shear Stress τ V Q I b Vmax R τA Vmax QA I tw τA 224 MPa Ans τB Vmax QB I tw τB 224 MPa Ans Problem 726 The beam is made from three boards glued together at the seams A and B If it is subjected to the loading shown determine the maximum vertical shear force resisted by the top flange of the beam The supports at C and D exert only vertical reactions on the beam Given bf 150mm dw 200mm tf 40mm tw 50mm P 25kN Solution Equilibrium By symmetry RCR RDR ΣFy0 3P 2R 0 R 15P Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 yc 05D Af yc y bf yfc 05 yc y y yfc 05 yc y Q Af yfc Q 05 yc y bf yc y Q 05bf yc 2 y2 Shear Stress in flange τ V Q I b V R τ R I bf Q τ R I 05 yc 2 y2 Resultant Shear Force For the flange yo yc tf Vf A A τ d Vf yo yc y R I 05 yc 2 y2 bf d Vf 236 kN Ans Problem 727 Determine the shear stress at points B and C located on the web of the fiberglass beam Given bf 100mm dw 150mm tf 18mm tw 12mm L1 2m L2 06m L3 2m wo 25 kN m w1 3 kN m Solution L L1 L2 L3 Equilibrium Given ΣFy0 A wo L1 05 w1 L3 D 0 ΣΜD0 A L wo L1 L 05L1 05 w1 L3 2L3 3 0 Guess A 1kN D 1kN A D Find A D A D 4783 3217 kN Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 QB bf tf 05D 05tf QC QB Shear Stress τ V Q I b VBC A wo 05L1 τB VBC QB I tw τB 0572 MPa Ans τC VBC QC I tw τC 0572 MPa Ans 100mm 84mm 18mm 18mm 12mm 75mm 75mm Problem 728 Determine the maximum shear stress acting in the fiberglass beam at the critical section Given bf 100mm dw 150mm tf 18mm tw 12mm L1 2m L2 06m L3 2m wo 25 kN m w1 3 kN m Solution L L1 L2 L3 Equilibrium Given ΣFy0 A wo L1 05 w1 L3 D 0 ΣΜD0 A L wo L1 L 05L1 05 w1 L3 2L3 3 0 Guess A 1kN D 1kN A D Find A D A D 4783 3217 kN Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 Qmax bf tf 05D 05tf 05dw tw 025dw Shear Stress τ V Q I b Vmax A τmax Vmax Qmax I tw τmax 1467 MPa Ans Shear Force Diagram L L1 L2 L3 x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 A wo x1 1 kN V2 x2 A wo L1 1 kN V3 x3 A wo L1 w1 x3 L1 L2 1 x3 L1 L2 2L3 1 kN 0 1 2 3 4 5 0 5 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 729 The beam is made from three plastic pieces glued together at the seams A and B If it is subjected to the loading shown determine the shear stress developed in the glued joints at the critical section The supports at C and D exert only vertical reactions on the beam Given bf 200mm dw 200mm tf 50mm tw 50mm wo 3 kN m L 25m Solution Equilibrium By symmetry RCR RDR ΣFy0 wo L 2R 0 R 05 wo L R 375 kN Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 QA bf tf 05D 05tf QB QA Shear Stress τ V Q I b Vmax R τA Vmax QA I tw τA 0225 MPa Ans τB Vmax QB I tw τB 0225 MPa Ans Problem 730 The beam is made from three plastic pieces glued together at the seams A and B If it is subjected to the loading shown determine the vertical shear force resisted by the top flange of the beam at the critical section The supports at C and D exert only vertical reactions on the beam Given bf 200mm dw 200mm tf 50mm tw 50mm L 25m wo 3 kN m Solution Equilibrium By symmetry RCR RDR ΣFy0 wo L 2R 0 R 05 wo L R 375 kN Section Property D dw 2tf I 1 12 bf D3 1 12 tw dw 3 yc 05D Af yc y bf yfc 05 yc y y yfc 05 yc y Q Af yfc Q 05 yc y bf yc y Q 05bf yc 2 y2 Shear Stress in flange τ V Q I b Vmax R τ R I bf Q τ R I 05 yc 2 y2 Resultant Shear Force For the flange yo yc tf Vf A A τ d Vf yo yc y R I 05 yc 2 y2 bf d Vf 030 kN Ans Problem 731 Determine the variation of the shear stress over the cross section of a hollow rivet What is the maximum shear stress in the rivet Also show that if then τ max 2 VA 0 1 r r Problem 732 The beam has a square cross section and is subjected to the shear force V Sketch the shearstress distribution over the cross section and specify the maximum shear stress Also from the neutral axis locate where a crack along the member will first start to appear due to shear Problem 733 Write a computer program that can be used to determine the maximum shear stress in the beam that has the cross section shown and is subjected to a specified constant distributed load w and concentrated force P Show an application of the program using the values L 4 m a 2 m P 15 kN d1 0 d2 2 m w 400 Nm t1 15 mm t2 20 mm b 50 mm and h 150 mm Problem 734 The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment Mp PL at the fixed support If the material is elasticplastic then at a distance x L the moment M P x creates a region of plastic yielding with an associated elastic core having a height 2y This situation has been described by Eq 630 and the moment M is distributed over the cross section as shown in Fig 654e Prove that the maximum shear stress developed in the beam is given by τ max 32 PAwhere A 2yb the crosssectional area of the elastic core Problem 735 The beam in Fig 654f is subjected to a fully plastic moment Mp Prove that the longitudinal and transverse shear stresses in the beam are zero Hint Consider an element of the beam as shown in Fig 74d Problem 736 The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 150 mm If each nail can support a 25kN shear force determine the maximum shear force V that can be applied to the beam Given b 150mm d1 50mm d2 50mm sn 150mm Fallow 25kN Solution Section Property D d1 d2 I 1 12 b D3 Q b d1 05d1 Shear Flow q V Q I There are two rows of nails Hence the allowable shear flow is qallow 2Fallow sn 2Fallow sn Vmax Q I Vmax 2 I Fallow Q sn Vmax 2222 kN Ans Problem 737 The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 150 mm If an internal shear force of V 3 kN is applied to the boards determine the shear force resisted by each nail Given b 150mm d1 50mm d2 50mm sn 150mm V 3kN Solution Section Property D d1 d2 I 1 12 b D3 Q b d1 05d1 Shear Flow q V Q I q 4500 kN m There are two rows of nails Hence the shear force resisted by each nail is F q 2 sn F 337 kN Ans Problem 738 A beam is constructed from five boards bolted together as shown Determine the maximum shear force developed in each bolt if the bolts are spaced s 250 mm apart and the applied shear is V 35 kN Given d1 250mm d2 350mm t 25mm a 100mm s 250mm V 35kN Solution a d2 d1 a a 200mm h a 05d1 h 325mm Section Property yc 2 d1 t 05d1 a 3 d2 t 05d2 2 d1 t 3 d2 t yc 22339 mm I1 1 12 2t d1 3 2t d1 05d1 a yc 2 I2 1 12 3 t d2 3 3 t d2 05d2 yc 2 I I1 I2 I 52359711022 mm4 Q d1 2t h yc Q 127016129 mm3 Shear Flow q V Q I q 8490 kN m There are four planes on the bolt Hencs the shear force resisted by each shear plane of the bolt is F q 4 s F 531 kN Ans Problem 739 A beam is constructed from five boards bolted together as shown Determine the maximum spacing s of the bolts if they can each resist a shear of 20 kN and the applied shear is V 45 kN Given d1 250mm d2 350mm t 25mm a 100mm V 45kN Fallow 20kN Solution a d2 d1 a a 200mm h a 05d1 h 325mm Section Property yc 2 d1 t 05d1 a 3 d2 t 05d2 2 d1 t 3 d2 t yc 22339 mm I1 1 12 2t d1 3 2t d1 05d1 a yc 2 I2 1 12 3 t d2 3 3 t d2 05d2 yc 2 I I1 I2 I 52359711022 mm4 Q d1 2t h yc Q 127016129 mm3 Shear Flow q V Q I q 10916 kN m Since there are four planes on the bolt the allowable shear flow is qallow 4Fallow s s 4Fallow q s 7329 mm Ans Problem 740 The beam is subjected to a shear of V 800 N Determine the average shear stress developed in the nails along the sides A and B if the nails are spaced s 100 mm apart Each nail has a diameter of 2 mm Given bf 250mm dw 150mm t 30mm a 100mm s 100mm V 800N do 2mm Solution h 05t h 15mm Section Property yc bf t 05t dw 2t 05dw bf t 2 dw t yc 4773 mm I1 1 12 bf t3 bf t 05t yc 2 I2 1 12 2 t dw 3 2 t dw 05dw yc 2 I I1 I2 I 3216477273 mm4 Q bf t yc h Q 24545455 mm3 Shear Flow q V Q I q 6105 kN m F q s F 06105 kN Since each side of the beam resists this shear force then Ao π 4 do 2 τavg F 2Ao τavg 9716 MPa Ans Problem 741 The beam is fabricated from two equivalent structural tees and two plates Each plate has a height of 150 mm and a thickness of 12 mm If a shear of V 250 kN is applied to the cross section determine the maximum spacing of the bolts Each bolt can resist a shear force of 75 kN Given bf 75mm dw 75mm t 12mm dp 150mm dg 50mm V 250kN Fallow 75kN Solution Section Property D 2dw 2t dg IT 1 12 bf D3 1 12 t dg 3 1 12 bf t D 2t 3 IP 1 12 2t dp 3 I IT IP Q bf t 05D 05t t dw 05dw 05dg Shear Flow q V Q I Since there are two shear planes on the bolt the allowable shear flow is q 2F sn 2Fallow sn V Q I sn 2 I Fallow V Q sn 1380 mm Ans Problem 742 The beam is fabricated from two equivalent structural tees and two plates Each plate has a height of 150 mm and a thickness of 12 mm If the bolts are spaced at s 200 mm determine the maximum shear force V that can be applied to the cross section Each bolt can resist a shear force of 75 kN Given bf 75mm dw 75mm t 12mm dp 150mm dg 50mm sn 200mm Fallow 75kN Solution Section Property D 2dw 2t dg IT 1 12 bf D3 1 12 t dg 3 1 12 bf t D 2t 3 IP 1 12 2t dp 3 I IT IP q V Q I Q bf t 05D 05t t dw 05dw 05dg Shear Flow Since there are two shear planes on the bolt the allowable shear flow is q 2F sn 2Fallow sn V Q I V 2 I Fallow sn Q V 1725 kN Ans Problem 743 The doubleweb girder is constructed from two plywood sheets that are secured to wood members at its top and bottom If each fastener can support 3 kN in single shear determine the required spacing s of the fasteners needed to support the loading P 15 kN Assume A is pinned and B is a roller Given bi 150mm di 250mm t 12mm do 350mm P 15kN Fallow 3kN Solution Equilibrium By symmetry AR BR ΣFy0 2R P 0 R 05P Section Property bo bi 2t d 05 do di I 1 12 bo do 3 1 12 bi di 3 Q bi d 05do 05d Shear Flow q V Q I Vmax R Since there are two shear planes on the bolt the allowable shear flow is q 2F sn 2Fallow sn Vmax Q I sn 2 I Fallow Vmax Q sn 3032 mm Ans Problem 744 The doubleweb girder is constructed from two plywood sheets that are secured to wood members at its top and bottom The allowable bending stress for the wood is σallow 56 MPa and the allowable shear stress is τallow 21 MPa If the fasteners are spaced s 150 mm and each fastener can support 3 kN in single shear determine the maximum load P that can be applied to the beam Given bi 150mm di 250mm t 12mm do 350mm sn 150mm σallow 56MPa Fallow 3kN τallow 21MPa Solution Equilibrium By symmetry AR BR ΣFy0 2R P 0 R 05P Section Property bo bi 2t d 05 do di I 1 12 bo do 3 1 12 bi di 3 Q bi d 05do 05d Shear Flow q V Q I Vmax R q 2F sn Since there are two shear planes on the bolt the allowable shear flow is 2Fallow sn Vmax Q I 2Fallow sn 05P Q I P 4 I Fallow sn Q P 3032 kN Ans Problem 745 The beam is made from three polystyrene strips that are glued together as shown If the glue has a shear strength of 80 kPa determine the maximum load P that can be applied without causing the glue to lose its bond Given bf 30mm tf 40mm tw 20mm dw 60mm τallow 0080MPa Solution Equilibrium By equilibrium A B R ΣFy0 2R P 2 1 4 P 0 R 075P Maximum Shear Vmax R Vmax 3 4P Section Property D dw 2tf I 1 12 bf D3 bf tw dw 3 I 668 10 6 m4 Q Σ yi A1 Q bf tf D 2 tf 2 Q 60000mm3 Shear Stress τ V Q I t τallow Vmax Q I tw Vmax τallow I tw Q 3 4P τallow I tw Q P 4τallow I tw 3Q P 0238 kN Ans Problem 746 The beam is made from four boards nailed together as shown If the nails can each support a shear force of 500 N determine their required spacings s and s if the beam is subjected to a shear of V 35 kN Given bf 250mm dw 250mm tf 25mm tw 40mm tb 25mm db 75mm V 35kN Fallow 05kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf dw tw 05db 2tb db bf tf dw tw 2 tb db yc 8594 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I3 1 12 2tb db 3 2tb db 05db yc 2 I I1 I2 I3 QC tb db yc 05db QD dw tw D yc 05dw Shear Flow q V Q I The allowable shear flow at points C and D are qC F sn qD F sn Fallow sn V QC I Fallow sn V QD I sn I Fallow V QC sn I Fallow V QD sn 2166 mm sn 307 mm Ans Problem 747 The beam is fabricated from two equivalent channels and two plates Each plate has a height of 150 mm and a thickness of 12 mm If a shear of V 250 kN is applied to the cross section determine the maximum spacing of the bolts Each bolt can resist a shear force of 75 kN Given bf 300mm dw 88mm t 12mm dp 150mm dg 50mm V 250kN Fallow 75kN Solution Section Property D 2dw 2t dg IU 1 12 bf D3 1 12 2t dg 3 1 12 bf 2t D 2t 3 IP 1 12 2t dp 3 I IU IP Q bf t 05D 05t 2t dw 05dw 05dg Shear Flow q V Q I Since there are two rows of bolts the allowable shear flow is q 2F sn 2Fallow sn V Q I sn 2 I Fallow V Q sn 1376 mm Ans Problem 748 A builtup timber beam is made from n boards each having a rectangular cross section Write a computer program that can be used to determine the maximum shear stress in the beam when it is subjected to any shear V Show an application of the program using a cross section that is in the form of a T and a box Problem 749 The timber Tbeam is subjected to a load consisting of n concentrated forces Pn If the allowable shear Vnail for each of the nails is known write a computer program that will specify the nail spacing between each load Show an application of the program using the values L 45 m a1 12 m P1 3 kN a2 24 m P2 75 kN b1 375 mm h1 250 mm b2 200 mm h2 25 mm and Vnail 1 kN Problem 750 The strut is constructed from three pieces of plastic that are glued together as shown If the allowable shear stress for the plastic is τallow 56 MPa and each glue joint can withstand 50 kNm determine the largest allowable distributed loading w that can be applied to the strut Given L1 1m τallow 56MPa L2 2m qallow 50 kN m L3 1m bf 74mm tf 25mm dw 75mm tw 12mm Solution Section Property D dw tf yc 05tf bf tf 05dw tf 2tw dw bf tf 2tw dw yc 3716 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 2tw dw 3 2tw dw 05dw tf yc 2 I I1 I2 Qmax 2tw D yc D yc 2 QA bf tf yc 05tf Allowable Shear Stress τ V Q I t Vmax w L1 τallow w L1 I 2tw 2tw D yc D yc 2 τallow w L1 2I D yc 2 w 2I L1 D yc 2 τallow w 913 kN m Shear Flow Assume the beam fails at the glue joint and the allowable shear flow is 2 qallow 2qallow V Q I 2qallow w L1 QA I w 2 I qallow L1 QA w 706 kN m Controls Ans Shear Force Diagram L L1 L2 L3 Equilibrium By symmetry R1 R R2 R ΣFy0 2R w L 0 R 05w L x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 w x1 kN V2 x2 w x2 R 1 kN V3 x3 w x3 2R 1 kN 0 1 2 3 4 10 0 10 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 751 The strut is constructed from three pieces of plastic that are glued together as shown If the distributed load w 3 kNm determine the shear flow that must be resisted by each glue joint Given L1 1m L2 2m w 3 kN m L3 1m bf 74mm tf 25mm dw 75mm tw 12mm Solution Section Property D dw tf yc 05tf bf tf 05dw tf 2tw dw bf tf 2tw dw yc 3716 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 2tw dw 3 2tw dw 05dw tf yc 2 I I1 I2 QA bf tf yc 05tf Shear Flow Since there are two glue joints hence 2q V Q I Vmax w L1 q Vmax QA 2I q 2124 kN m Ans Shear Force Diagram L L1 L2 L3 Equilibrium By symmetry R1 R R2 R ΣFy0 2R w L 0 R 05w L x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 w x1 kN V2 x2 w x2 R 1 kN V3 x3 w x3 2R 1 kN 0 1 2 3 4 10 0 10 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 752 The beam is subjected to the loading shown where P 7 kN Determine the average shear stress developed in the nails within region AB of the beam The nails are located on each side of the beam and are spaced 100 mm apart Each nail has a diameter of 5 mm Given bf 250mm dw 150mm t 30mm a 2m s 100mm do 5mm P 3kN P 7kN Solution Section Property I 1 12 bf 2 t dw 3 bf dw 2 t 3 I 72000000mm4 Q bf t 05dw 05t Q 450000mm3 Maximum Shear Vmax P P Vmax 10kN Shear Flow q Vmax Q I q 62500 kN m There are two rows of nails Hence the sher force resisted by each nail is F q 2 s F 3125 kN Ao π 4 do 2 τavg F Ao τavg 1592 MPa Ans Problem 753 The beam is constructed from four boards which are nailed together If the nails are on both sides of the beam and each can resist a shear of 3 kN determine the maximum load P that can be applied to the end of the beam Given bf 250mm dw 150mm t 30mm a 2m s 100mm P 3kN Fallow 3kN Solution Section Property I 1 12 bf 2 t dw 3 bf dw 2 t 3 I 72000000mm4 Q bf t 05dw 05t Q 450000mm3 Maximum Shear Vmax P P There are two rows of nails Hence the allowable sher flow is qallow 2Fallow s qallow 6000 kN m Shear Flow qallow Vmax Q I qallow P P Q I P qallow I Q P P 660 kN Ans Problem 754 The member consists of two plastic channel strips 12 mm thick bonded together at A and B If the glue can support an allowable shear stress of τallow 42 MPa determine the maximum intensity w0 of the triangular distributed loading that can be applied to the member based on the strength of the glue Given bo 150mm t 12mm do 150mm L 4m τallow 42MPa Solution Equilibrium By symmetry AR BR ΣFy0 2R 05wo L 0 R 025wo L Section Property bi bo 2t di do 2t I 1 12 bo do 3 1 12 bi di 3 Q bo t 05do 05t 2 t 05 di 05di Shear Flow q V Q I Vmax R Since there are two planes of glue the allowable shear flow is 2t τallow 2t τallow Vmax Q I 2 t τallow 025wo L Q I wo 8t I τallow L Q wo 973 kN m Ans Problem 755 The member consists of two plastic channel strips 12 mm thick glued together at A and B If the distributed load has a maximum intensity of w0 50 kNm determine the maximum shear stress resisted by the glue Given bo 150mm do 150mm L 4m t 12mm wo 50 kN m Solution Equilibrium By symmetry AR BR ΣFy0 2R 05wo L 0 R 025wo L Section Property bi bo 2t di do 2t I 1 12 bo do 3 1 12 bi di 3 Q bo t 05do 05t 2 t 05 di 05di Allowable Shear Stress τ V Q I b Vmax R τmax Vmax Q I 2t τmax 2158 MPa Ans Problem 756 A shear force of V 18 kN is applied to the symmetric box girder Determine the shear flow at A and B Given bf 125mm dw 300mm t 10mm dm 200mm dg 30mm V 18kN Solution Section Property I1 1 12 bf t3 bf t 05dw 05t 2 I2 1 12 bf t3 bf t 05t 05dm 2 I3 1 12 t dw 3 I 2I1 2I2 2I3 I 12516666667 mm4 QA bf t 05dw 05t QA 181250mm3 QB bf t 05t 05dm QB 131250mm3 Shear Flow qA 1 2 V QA I qA 1303 kN m Ans qB 1 2 V QB I qB 944 kN m Ans Problem 757 A shear force of V 18 kN is applied to the box girder Determine the shear flow at C Given bf 125mm dw 300mm t 10mm dm 200mm dg 30mm V 18kN Solution Section Property I1 1 12 bf t3 bf t 05dw 05t 2 I2 1 12 bf t3 bf t 05t 05dm 2 I3 1 12 t dw 3 I 2I1 2I2 2I3 I 12516666667 mm4 QC bf t 05dw 05t bf t 05t 05dm 2 05dw t 025dw QC 537500mm3 Shear Flow qC 1 2 V QC I qC 3865 kN m Ans Problem 758 The channel is subjected to a shear of V 75 kN Determine the shear flow developed at point A Given bf 400mm dw 200mm tf 30mm tw 30mm V 75kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf 2tw dw bf tf 2tw dw yc 7250 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 2tw dw 3 2tw dw 05dw tf yc 2 I I1 I2 I 120250000mm4 QA bf tf yc 05tf QA 690000mm3 Shear Flow qA 1 2 V QA I qA 2152 kN m Ans Problem 759 The channel is subjected to a shear of V 75 kN Determine the maximum shear flow in the channel Given bf 400mm dw 200mm tf 30mm tw 30mm V 75kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf 2tw dw bf tf 2tw dw yc 7250 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 2tw dw 3 2tw dw 05dw tf yc 2 I I1 I2 I 120250000mm4 Qmax tw D yc 1 2 D yc Qmax 37209375 mm3 Shear Flow qmax V Qmax I qmax 2321 kN m Ans Problem 760 The beam supports a vertical shear of V 35 kN Determine the resultant force developed in segment AB of the beam Given bf 125mm dw 250mm tf 12mm tw 12mm V 35kN Solution Section Property D dw 2tf I 1 12 2tf bf 3 1 12 dw tw 3 Q Af yfc Q tf 05bf y 05 05bf y y Q 05tf 025bf 2 y2 Shear Flow q V Q I q V tf 2I 025bf 2 y2 Resultant Shear Force For AB yo 05tw VAB A q y d VAB yo 05bf y V tf 2I 025bf 2 y2 d VAB 743 kN Ans Problem 761 The aluminum strut is 10 mm thick and has the cross section shown If it is subjected to a shear of V 150 N determine the shear flow at points A and B Given bf 60mm bf 80mm tf 10mm dw 40mm tw 10mm V 150N Solution Section Property D dw 2tf A bf tf 2dw tw bf tf A 2200mm2 y Σ yi Ai Σ Ai yc bf tf 05tf 2 dw tw 05dw tf bf tf D 05tf A yc 2773 mm If 1 12 bf tf 3 bf tf 05tf yc 2 Iw 1 12 tw dw 3 tw dw 05dw tf yc 2 If 1 12 bf tf 3 bf tf D 05tf yc 2 I If 2Iw If I 9819697 10 9 m4 QA 05bf tf yc 05tf QA 909091 mm3 QB bf tf D 05tf yc QB 1636364 mm3 Shear Flow qA V QA I qA 139 kN m Ans qB 1 2 V QB I qB 125 kN m Ans Problem 762 The aluminum strut is 10 mm thick and has the cross section shown If it is subjected to a shear of V 150 N determine the maximum shear flow in the strut Given bf 60mm bf 80mm tf 10mm dw 40mm tw 10mm V 150N Solution Section Property D dw 2tf A bf tf 2dw tw bf tf A 2200mm2 y Σ yi Ai Σ Ai yc bf tf 05tf 2 dw tw 05dw tf bf tf D 05tf A yc 2773 mm If 1 12 bf tf 3 bf tf 05tf yc 2 Iw 1 12 tw dw 3 tw dw 05dw tf yc 2 If 1 12 bf tf 3 bf tf D 05tf yc 2 I If 2Iw If I 9819697 10 9 m4 Qmax bf tf D 05tf yc 2 tw D yc tf 1 2 D yc tf Qmax 2132438 mm3 Shear Flow qmax 1 2 V Qmax I qmax 163 kN m Ans Problem 763 The angle is subjected to a shear of V 10 kN Sketch the distribution of shear flow along the leg AB Indicate numerical values at all peaks Given L 125mm t 6mm θ 45deg V 10kN Solution Section Property h L cos θ b t sin θ I 1 12 2b h3 Q A yc Q t 05h y sin θ 05 05h y y Q t 2sin θ 025h2 y2 Shear Flow q V Q I q V t 2I sin θ 025h2 y2 At y 0 q qmax qmax V t 2I sin θ 025h2 qmax 8485 kN m Ans Problem 764 The beam is subjected to a shear force of V 25 kN Determine the shear flow at points A and B Given bf 274mm tf 12mm bf 250mm tf 12mm dw 200mm tw 12mm dw 50mm V 25kN Solution Section Property D dw tf D D dw yc 05tf bf tf 05dw tf 2tw dw D 05tf bf tf bf tf 2tw dw bf tf yc 9247 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 2tw dw 3 2tw dw 05dw tf yc 2 I3 1 12 bf tf 3 bf tf D 05tf yc 2 I I1 I2 I3 QA bf tf yc 05tf QB bf tf D 05tf yc Shear Flow qA V QA 2I qA 6509 kN m Ans qB V QB 2I qB 4363 kN m Ans Problem 765 The beam is constructed from four plates and is subjected to a shear force of V 25 kN Determine the maximum shear flow in the cross section Given bf 274mm tf 12mm bf 250mm tf 12mm dw 200mm tw 12mm dw 50mm V 25kN Solution Section Property D dw tf D D dw yc 05tf bf tf 05dw tf 2tw dw D 05tf bf tf bf tf 2tw dw bf tf yc 9247 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 2tw dw 3 2tw dw 05dw tf yc 2 I3 1 12 bf tf 3 bf tf D 05tf yc 2 I I1 I2 I3 Qmax bf tf yc 05tf 2tw yc tf 2 yc tf Maximum Shear Flow qmax V Qmax 2I qmax 8288 kN m Ans Problem 766 A shear force of V 18 kN is applied to the box girder Determine the position d of the stiffener plates BE and FG so that the shear flow at A is twice as great as the shear flow at B Use the centerline dimensions for the calculation All plates are 10 mm thick Given t 10mm bf 135mm t V 18kN dw t 290mm Solution Section Property QA bf t 05dw 05t QA 181250mm3 QB bf t d Shear Flow qA 1 2 V QA I qB 1 2 V QB I Require qA 2qB 1 2 V QA I 2 1 2 V QB I QA 2QB QA 2 bf t d d QA 2bf t d 7250 mm Ans Problem 767 The pipe is subjected to a shear force of V 40 kN Determine the shear flow in the pipe at points A and B Given ri 150mm t 5mm V 40kN Solution Section Property ro ri t I π 4 ro 4 ri 4 Since a 0 then QA 0 QB 4ro 3π π ro 2 2 4ri 3π π ri 2 2 Shear Flow qA V QA I qA 000 kN m Ans qB V QB 2I qB 8348 kN m Ans Problem 768 Determine the location e of the shear center point O for the thinwalled member having the cross section shown where b2 b1 The member segments have the same thickness t Problem 769 Determine the location e of the shear center point O for the thinwalled member having the cross section shownThe member segments have the same thickness t Problem 770 Determine the location e of the shear center point O for the thinwalled member having the cross section shown The member segments have the same thickness t Problem 771 Determine the location e of the shear center point O for the thinwalled member having the cross section shown The member segments have the same thickness t Problem 772 Determine the location e of the shear center point O for the thinwalled member having the cross section shown The member segments have the same thickness t Problem 773 Determine the location e of the shear center point O for the thinwalled member having the cross section shown The member segments have the same thickness t Problem 774 Determine the location e of the shear center point O for the thinwalled member having the cross section shown The member segments have the same thickness t Given dθ 150mm dv 150mm θ 30deg Solution Set t 1mm Section Property hθ dθ sin θ bθ t sin θ Iθ 1 12 bθ hθ 3 dθ t dv hθ 2 2 Iv 1 12 t dv 3 I 2Iθ Iv yc 05dv hθ 05x sin θ Q A yc Q x t 05dv hθ 05x sin θ Shear Flow Resultant q V Q I V P q P x t I 05dv hθ 05x sin θ F1 P 0 dθ x x t I 05dv hθ 05x sin θ d Shear Center Summing moment about point A P e F1 dv cos θ e F1 P dv cos θ e 0 dθ x dv cos θ x t I 05dv hθ 05x sin θ d e 4330 mm Ans Problem 775 Determine the location e of the shear center point O for the thinwalled member having a slit along its side Given a 100mm b 100mm Solution Set t mm P kN Section Property h 2a I 1 12 2t h3 2 b t a2 I 33333 a3t Q1 y t 1 2 y Q1 t 2y2 Q2 a t 1 2 a x t a Q1 a t 2 a 2x Shear Flow Resultant q1 V Q1 I q1 P t y2 2I q2 V Q2 I q2 P a t a 2x 2I Fw 0 a q1 y d Fw 0 a y P t y2 2I d Fw 005 P Ff 0 b q2 x d Ff 0 b x P a t a 2x 2I d Ff 03 P Shear Center Summing moment about point A P e 2Fw b Ff h e 1 P 2Fw b Ff h e 70mm Ans Problem 776 Determine the location e of the shear center point O for the thinwalled member having a slit along its side Each element has aconstant thickness t Given a mm b a Solution Set t mm P kN Section Property h 2a I 1 12 2t h3 2 b t a2 I 33333 a3t Q1 y t 1 2 y Q1 t 2y2 Q2 a t 1 2 a x t a Q1 a t 2 a 2x Shear Flow Resultant q1 V Q1 I q1 P t y2 2I q2 V Q2 I q2 P a t a 2x 2I Fw 0 a q1 y d Fw 0 a y P t y2 2I d Fw 005 P Ff 0 b q2 x d Ff 0 b x P a t a 2x 2I d Ff 03 P Shear Center Summing moment about point A P e 2Fw b Ff h e 1 P 2Fw b Ff h e 07 a Ans Problem 777 Determine the location e of the shear center point O for the thinwalled member having the cross section shown Given a mm θ 60deg Solution Set t mm P kN Section Property b a sin θ b 086603 a t t cos θ t 2t h a 2 I 1 12 t a3 1 12 t a3 I 025 a3t Q1 y t 1 2 y Q1 t 2y2 Q2 h t 1 2 h h y t y 1 2 h y Q2 1 2 th2 t h2 y2 Shear Flow Resultant q1 V Q1 I q1 P t y2 2I q2 V Q2 I q2 P th2 t h2 y2 2I F 0 h q1 y d F 0 h y P t y2 2I d F 01667 P F 0 h q2 y d F 0 h y P th2 t h2 y2 2I d F 06667 P Shear Center Summing moment about point A P e 2 F b F 0 e 1 P 2F b e 11547 a Ans Problem 778 If the angle has a thickness of 3 mm a height h 100 mm and it is subjected to a shear of V 50 N determine the shear flow at point A and the maximum shear flow in the angle Given h 100mm t 3mm θ 45deg V 50N Solution Section Property t t sin θ t 42426 mm I 1 12 2t h3 Q A yc Q t 05h y 1 2 05h y y Q t 2 025h2 y2 Shear Flow q V Q I q V t 2I 025h2 y2 At A yA 05 h qA 0 Ans At y 0 q qmax qmax V t 2I 025h2 qmax 375 N m Ans Problem 779 The angle is subjected to a shear of V 10 kN Sketch the distribution of shear flow along the leg AB Indicate numerical values at all peaks The thickness is 6 mm and the legs AB are 125 mm Given L 125mm t 6mm θ 45deg V 10kN Solution Section Property h L cos θ t t sin θ t 84853 mm I 1 12 2t h3 Q A yc Q t 05h y 1 2 05h y y Q t 2 025h2 y2 Shear Flow q V Q I q V t 2I 025h2 y2 Ans At y 0 q qmax qmax V t 2I 025h2 qmax 8485 kN m Ans Problem 780 Determine the placement e for the force P so that the beam bends downward without twisting Take h 200 mm Given h1 100mm bf 300mm h2 200mm Solution Set t mm P kN Section Property I 1 12 bf t3 t h1 3 t h2 3 Qw2 t 05h2 y y 1 2 05h2 y Qw2 t 2 025h2 2 y2 Shear Flow Resultant qw2 V Qw2 I qw2 P t 025h2 2 y2 2I Fw2 05 h2 05h2 qw2 x d Fw2 05 h2 05h2 y P t 025h2 2 y2 2I d Fw2 08889 P Shear Center Summing moment about point A P e Fw2 bf t e 1 P Fw2 bf t e 2675 mm Ans Problem 781 A force P is applied to the web of the beam as shown If e 250 mm determine the height h of the right flange so that the beam will deflect downward without twisting The member segments have the same thickness t Given h1 100mm bf 300mm e 250mm Solution Set t mm P kN Shear Center Summing moment about point A P e Fw2 bf t Fw2 e bf t P Assume bft equal to bf Fw2 e bf P Section Property Assume bf t3 negligible I 1 12 t h1 3 t h2 3 Qw2 t 05h2 y y 1 2 05h2 y Qw2 t 2 025h2 2 y2 Shear Flow Resultant qw2 V Qw2 I qw2 P t 025h2 2 y2 2I Fw2 05 h2 05h2 qw2 x d e bf P 05 h2 05h2 y 6P t 025h2 2 y2 t h1 3 t h2 3 d Given e h1 3 h2 3 6bf 05 h2 05h2 y 025h2 2 y2 d Guess h2 10mm h2 Find h2 h2 1710 mm Ans Problem 782 Determine the location e of the shear center point O for the thinwalled member having the cross section shown Problem 783 Determine the location e of the shear center point O for the tube having a slit along its length Problem 784 The beam is fabricated from four boards nailed together as shown Determine the shear force each nail along the sides C and the top D must resist if the nails are uniformly spaced s 75 mm The beam is subjected to a shear of V 225 kN Given bf 250mm dw 300mm tf 25mm tw 25mm tb 25mm db 100mm V 225kN sn 75mm Solution Section Property D dw tf yc 05tf bf tf 05dw tf dw tw 05db 2tb db bf tf dw tw 2 tb db yc 8750 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I3 1 12 2tb db 3 2tb db 05db yc 2 I I1 I2 I3 QC tb db yc 05db QD dw tw D yc 05dw Shear Flow q V Q I The allowable shear flow at points C and D are qC FC sn qD FD sn FC sn V QC I FD sn V QD I FC V QC sn I FD V QD sn I FC 0987 kN FD 6906 kN Ans Problem 785 The beam is constructed from four boards glued together at their seams If the glue can withstand 15 kNm what is the maximum vertical shear V that the beam can support Given bf 100mm dw 249mm tf 12mm tw 12mm di 75mm qallow 15 kN m Solution Section Property I 2 12 tw dw 3 2 12 bf tf 3 2bf tf di tf 2 2 Q bf tf di tf 2 Shear Flow Since there are two glue joints hence 2q V Q I Vmax 2I qallow Q Vmax 2037 kN Ans Problem 786 Solve Prob 785 if the beam is rotated 90 from the position shown Given bf 100mm dw 249mm tf 12mm tw 12mm di 75mm qallow 15 kN m Solution Section Property I 2 12 tf bf 3 2 12 dw tw 3 2dw tw bf tw 2 2 Q dw tw bf tw 2 Shear Flow Since there are two glue joints hence 2q V Q I Vmax 2I qallow Q Vmax 3731 kN Ans Problem 787 The member is subjected to a shear force of V 2 kN Determine the shear flow at points A B and C The thickness of each thinwalled segment is 15 mm Given bf 200mm dw 300mm tf 15mm tw 15mm tb 15mm db 115mm V 2kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf dw tw 05db 2tb db bf tf dw tw 2 tb db yc 8798 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I3 1 12 2tb db 3 2tb db 05db yc 2 I I1 I2 I3 I 8693904538 mm4 QA 0 QB tb db yc 05db QB 5257705 mm3 QC QB tf 05bf 05tw yc 05tf QC 16424229 mm3 Shear Flow q V Q I qA 0 Ans qB V QB I qB 1210 kN m Ans qC V QC I qC 3778 kN m Ans Problem 788 The member is subjected to a shear force of V 2 kN Determine the maximum shear flow in the member All segments of the cross section are 15 mm thick Given bf 200mm dw 300mm tf 15mm tw 15mm tb 15mm db 115mm V 2kN Solution Section Property D dw tf yc 05tf bf tf 05dw tf dw tw 05db 2tb db bf tf dw tw 2 tb db yc 8798 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I3 1 12 2tb db 3 2tb db 05db yc 2 I I1 I2 I3 I 8693904538 mm4 Qmax tw D yc 1 2 D yc Qmax 38653747 mm3 Shear Flow q V Q I Maximum shear flow occurs at the point where the neutral axis passes through the section qmax V Qmax I qmax 8892 kN m Ans Problem 789 The beam is made from three thin plates welded together as shown If it is subjected to a shear of V 48 kN determine the shear flow at points A and B Also calculate the maximum shear stress in the beam Given bf 215mm tf 15mm tw 15mm dw 315mm V 48kN h 200mm Solution Section Property a 05 bf tw yc bf tw tf h 05tf dw tw 05dw bf tw tf dw tw yc 17692 mm If 1 12 bf tw tf 3 bf tw tf h 05 tf yc 2 Iw 1 12 tw dw 3 tw dw yc 05dw 2 I If Iw I 4371347 10 6 m4 QA a tw dw yc 05a QA 13212379 mm3 QB a tf h 05 tf yc QB 4587379 mm3 Qmax tw yc 1 2 yc Qmax 23474845 mm3 Shear Flow q V Q I qA V QA I qA 1451 kN m Ans qB V QB I qB 5037 kN m Ans Maximum Shear Stress τ V Q I b Maximum shear stress occurs at the point where the neutral axis passes through the section τmax V Qmax I tw τmax 1718 MPa Ans Problem 790 A steel plate having a thickness of 6 mm is formed into the thinwalled section shown If it is subjected to a shear force of V 125 kN determine the shear stress at points A and C Indicate the results on volume elements located at these points Given bf 100mm dw 50mm bf 25mm t 6mm V 125kN Solution Section Property D dw 2t mm mm yc 05t 2bf t 05dw t 2dw t D 05t bf t 2bf t 2dw t bf t yc 3660 mm I1 1 12 bf t3 bf t 05t yc 2 I2 1 12 t dw 3 dw t 05dw t yc 2 I3 1 12 bf t3 bf t D 05t yc 2 I 2I1 2I2 I3 QA bf t yc 05t QC bf t yc 05t dw t 05dw t yc 05bf t D 05t yc QC 000 mm3 since A 0 Shear Stress τ V Q I t τA V QA I t τA 1335 MPa Ans τC V QC I t τC 0MPa Ans Problem 791 A steel plate having a thickness of 6 mm is formed into the thinwalled section shown If it is subjected to a shear force of V 125 kN determine the shear stress at point B Given bf 100mm dw 50mm bf 25mm t 6mm V 125kN Solution Section Property D dw 2t yc 05t 2bf t 05dw t 2dw t D 05t bf t 2bf t 2dw t bf t yc 3660 mm I1 1 12 bf t3 bf t 05t yc 2 I2 1 12 t dw 3 dw t 05dw t yc 2 I3 1 12 bf t3 bf t D 05t yc 2 I 2I1 2I2 I3 QB bf t D 05t yc Shear Stress τ V Q I b τB V QB I 2t τB 1781 MPa Ans Problem 792 Determine the location e of the shear center point O for the thinwalled member having the cross section shown Problem 793 Sketch the intensity of the shearstress distribution acting over the beams crosssectional area and determine the resultant shear force acting on the segment AB The shear acting at the section is V 175 kN Show that INA 34082106 mm4 Given b1 200mm d1 200mm V 175kN b2 50mm d2 150mm Solution Section Property D d1 d2 yc 05d1 b1 d1 05d2 d1 b2 d2 b1 d1 b2 d2 yc 12763 mm yc D yc I1 1 12 b1 d1 3 b1 d1 05d1 yc 2 I2 1 12 b2 d2 3 b2 d2 D 05d2 yc 2 I I1 I2 I 34082 106 mm4 QED A1 yc y1 b1 y1c 05 yc y1 y1 y1c 05 yc y1 Q1 A1 y1c Q1 05 yc y1 b1 yc y1 Q1 05b1 yc 2 y1 2 A2 yc y2 b2 y2c 05 yc y2 y2 y2c 05 yc y2 Q2 A2 y2c Q2 05 yc y2 b2 yc y2 Q2 05b2 yc 2 y2 2 Shear Stress τ V Q I b τCB V I 05 yc 2 y1 2 τ1B V I 05yc 2 At B y1 0 τ1B 418 MPa At C y1 yc d1 τ1C V I 05 yc 2 y1 2 τ1C 284 MPa τAB V INA 05 yc 2 y2 2 At C y2 yc d2 τ2C V I 05 yc 2 y2 2 τ2C 1135 MPa Resultant Shear Force For segment AB yo yc d2 VAB A τAB A d VAB yo yc y V I 05 yc 2 y2 b2 d VAB 4978 kN Ans Problem 81 A spherical gas tank has an inner radius of r 15 m If it is subjected to an internal pressure of p 300 kPa determine its required thickness if the maximum normal stress is not to exceed 12 MPa Given r 15m p 03MPa σallow 12MPa Solution Normal Stress σallow p r 2 t t p r 2 σallow t 1875 mm Ans Problem 82 A pressurized spherical tank is to be made of 125mmthick steel If it is subjected to an internal pressure of p 14 MPa determine its outer radius if the maximum normal stress is not to exceed 105 MPa Given t 125mm p 14MPa σallow 105MPa Solution Normal Stress σ p r 2 t ri 2 t σallow p ri 18750 m ro ri t ro 18875 m Ans Problem 83 The thinwalled cylinder can be supported in one of two ways as shown Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 05 MPa The wall has a thickness of 6 mm and the inner diameter of the cylinder is 200 mm Given t 6mm p 05MPa ri 200mm Solution Case a Hoop Stress σ1 p ri t σ1 1667 MPa Ans Normal Stress σ2 0 Ans Case b Hoop Stress σ1 p ri t σ1 1667 MPa Ans Normal Stress σ2 p ri 2 t σ2 833 MPa Ans Problem 84 The tank of the air compressor is subjected to an internal pressure of 063 MPa If the internal diameter of the tank is 550 mm and the wall thickness is 6 mm determine the stress components acting at point A Draw a volume element of the material at this point and show the results on the element Given t 6mm p 063MPa di 550mm Solution ri 05di Hoop Stress α ri t α 4583 Since α 10 then thinwall analysis can be used σ1 p ri t σ1 2888 MPa Ans Longitudinal Stress σ2 p ri 2 t σ2 1444 MPa Ans Problem 85 The openended pipe has a wall thickness of 2 mm and an internal diameter of 40 mm Calculate the pressure that ice exerted on the interior wall of the pipe to cause it to burst in the manner shown The maximum stress that the material can support at freezing temperatures is σmax 360 MPa Show the stress acting on a small element of material just before the pipe fails Given t 2mm σallow 360MPa di 40mm Solution ri 05di Hoop Stress α ri t α 1000 Since α 10 then thinwall analysis can be used σ1 σallow σ1 p ri t p σallow t ri p 360 MPa Ans Longitudinal Stress Since the pipe is open at both neds then σ2 0 Ans Problem 86 The openended polyvinyl chloride pipe has an inner diameter of 100 mm and thickness of 5 mm If it carries flowing water at 042 MPa pressure determine the state of stress in the walls of the pipe Given t 5mm p 042MPa di 100mm Solution ri 05di Hoop Stress σ1 p ri t σ1 42 MPa Ans Normal Stress σ2 0 Ans There is no stress componenet in the longitudinal direction since pipe has open ends Problem 87 If the flow of water within the pipe in Prob 86 is stopped due to the closing of a valve determine the state of stress in the walls of the pipe Neglect the weight of the water Assume the supports only exert vertical forces on the pipe Given t 5mm p 042MPa di 100mm Solution ri 05di Hoop Stress σ1 p ri t σ1 42 MPa Ans Normal Stress σ2 p ri 2 t σ2 21 MPa Ans Problem 88 The A36steel band is 50 mm wide and is secured around the smooth rigid cylinder If the bolts are tightened so that the tension in them is 2 kN determine the normal stress in the band the pressure exerted on the cylinder and the distance half the band stretches Given t 3mm b 50mm r 200mm F 2kN E 200GPa Solution rb r 05t Lb π rb Tensile Stress in the Band σ1 F b t σ1 1333 MPa Ans Hoop Stress σ p r t p tσ1 rb p 0199 MPa Ans Stectch δ ε1 Lb ε1 σ1 E δ σ1 Lb E δ 00422 mm Ans Problem 89 The 304 stainless steel band initially fits snugly around the smooth rigid cylinder If the band is then subjected to a nonlinear temperature drop of T 12 sin2θ C where θ is in radians determine the circumferential stress in the band Unit used C deg Given t 04mm b 25mm r 250mm E 193GPa T 12 sin θ 2 α 17 10 6 1 C Solution Compatibility Since the band is fitted to a rigid cylinder which does not deform under load then δF δT 0 P 2π r A E 0 2π θ α T r d 0 2π r E P A 12 0 2π θ α sin θ 2 r d 0 However P A σc 2π r E σc 12 α r 0 2π θ sin θ 2 d σc 6 α E π 0 2π θ sin θ 2 d C σc 1969 MPa Ans Problem 810 The barrel is filled to the top with water Determine the distance s that the top hoop should be placed from the bottom hoop so that the tensile force in each hoop is the same Also what is the force in each hoop The barrel has an inner diameter of 12 m Neglect its wall thickness Assume that only the hoops resist the water pressure Note Water develops pressure in the barrel according to Pascals law p 001z MPa where z is the depth from the surface of the water in meter Given d 12m p 001z MPa h 24m h 06m Solution r 05d P 0 h z p 2r d P 2r 0 h 001z z d MPa m P 3456 kN Equilibrium for the Steel Hoop ΣFy0 P 4F 0 F 025P F 864 kN ΣΜBase0 P h 3 2F h 2F h s 0 s P h 6F h h s 400mm Ans Problem 811 A wood pipe having an inner diameter of 09 m is bound together using steel hoops having a crosssectional area of 125 mm2 If the allowable stress for the hoops is σallow 84 MPa determine their maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressure of 28 kPa Assume each hoop supports the pressure loading acting along the length s of the pipe Given d 09m p 28 10 3 MPa As 125mm2 σallow 84MPa Solution r 05d P p 2r s F σallow As Equilibrium for the Steel Hoop From the FBD ΣFy0 P 2F 0 p 2r s 2σallow As 0 s σallow As p r s 83333 mm Ans Problem 812 A boiler is constructed of 8mm thick steel plates that are fastened together at their ends using a butt joint consisting of two 8mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown If the steam pressure in the boiler is 135 MPa determine a the circumferential stress in the boilers plate apart from the seam b the circumferential stress in the outer cover plate along the rivet line aa and c the shear stress in the rivets Given to 8mm ri 750mm p 135MPa tc 8mm db 10mm s 50mm Solution a Hoop Stress σ1 p ri to σ1 1266 MPa Ans b Hoop Stress in cover plate along line aa Consider a width of s mm Fo in boiler plate Fo in cover plates σ1 s to σ1 s db 2 tc σ1 σ1 s to s db 2 tc σ1 791 MPa Ans c Shear Stress in Rivet rb 05db From the FBD ΣFy0 Fb σ1 s to 0 Fb σ1 s to τavg 1 2 Fb π rb 2 τavg 3223 MPa Ans Problem 813 The ring having the dimensions shown is placed over a flexible membrane which is pumped up with a pressure p Determine the change in the internal radius of the ring after this pressure is applied The modulus of elasticity for the ring is E Solution Equilibrium for the Ring ΣFy0 2P 2p ri w 0 P p ri w Hoop Stress and Strain for the Ring σ1 P A σ1 p ri w ro ri w σ1 p ri ro ri Using Hookes Law ε1 σ1 E ε1 p ri ro ri E 1 However ε1 2 π ri1 2 π ri 2 π r ε1 ri1 ri r ε1 δri r Then from Eq1 δri r p ri ro ri E δri p ri 2 ro ri E Ans Problem 814 A closedended pressure vessel is fabricated by crosswinding glass filaments over a mandrel so that the wall thickness t of the vessel is composed entirely of filament and an epoxy binder as shown in the figure Consider a segment of the vessel of width w and wrapped at an angle θ If the vessel is subjected to an internal pressure p show that the force in the segment is Fθ σ0 wt where σ0 is the stress in the filaments Also show that the stresses in the hoop and longitudinal directions are σh σ0 sin2 θ and σl σ0 cos2 θ respectively At what angle θ optimum winding angle would the filaments have to be wound so that the hoop and longitudinal stresses are equivalent Problem 815 The steel bracket is used to connect the ends of two cables If the allowable normal stress for the steel is σallow 168 MPa determine the largest tensile force P that can be applied to the cables The bracket has a thickness of 12 mm and a width of 18 mm Given b 18mm a 50mm t 12mm σallow 168MPa Solution Internal Force and Moment ao a 05 b N P M P ao Section Property A t b I 1 12 t b3 Alowable Normal Stress σ N A M c I The maximum normal stress occurs at the bottom of the steel bracket cmax 05 b σallow P A P ao cmax I P σallow 1 A ao cmax I P 1756 kN Ans Problem 816 The steel bracket is used to connect the ends of two cables If the applied force P 25 kN determine the maximum normal stress in the bracket The bracket has a thickness of 12 mm and a width of 18 mm Given b 18mm a 50mm t 12mm P 25kN Solution Internal Force and Moment ao a 05 b N P M P ao Section Property A t b I 1 12 t b3 Alowable Normal Stress σ N A M c I The maximum normal stress occurs at the bottom of the steel bracket cmax 05 b σmax N A M cmax I σmax 2392 MPa Ans Problem 817 The joint is subjected to a force of 125 kN as shown Sketch the normalstress distribution acting over section aa if the member has a rectangular cross section of width 12 mm and thickness 18 mm Given b 18mm ah 50mm av 32mm t 12mm P 125kN v 3 h 4 r 5 Solution Internal Force and Moment ΣFx0 P h r N 0 N P h r ΣFy0 V P v r 0 V P v r ΣΜA0 M P h r av P v r ah 0 M P v r ah P h r av Section Property A b t I 1 12 b t3 Normal Stress σ N A M c I ctop 05 t σtop N A M ctop I σtop 1736 MPa T Ans cbot 05 t σbot N A M cbot I σbot 810 MPa C Ans Location of zero stress σtop σbot yo t yo yo t σtop σbot σtop yo 818 mm Problem 818 The joint is subjected to a force of 125 kN as shown Determine the state of stress at points A and B and sketch the results on differential elements located at these points The member has a rectangular crosssectional area of width 12 mm and thickness 18 mm Given b 18mm ah 50mm av 32mm t 12mm P 125kN v 3 h 4 r 5 Solution Internal Force and Moment ΣFx0 P h r N 0 N P h r ΣFy0 V P v r 0 V P v r ΣΜA0 M P h r av P v r ah 0 M P v r ah P h r av Section Property A b t I 1 12 b t3 QA 05 t b 025t QB 0 since A 0 Normal Stress σ N A M c I cA 0 σA N A M cA I σA 463 MPa T Ans cB 05 t σB N A M cB I σB 810 MPa C Ans Shear Stress τ V Q I b τA V QA I b τA 521 MPa Ans τB V QB I b τB 0MPa Ans Problem 819 The coping saw has an adjustable blade that is tightened with a tension of 40 N Determine the state of stress in the frame at points A and B Given t 3mm a 100mm h 8mm P 40N Solution Internal Force and Moment At A NA P MA P a At B NB 0 MB P 05a Section Property A t h I 1 12 t h3 State of Stress σ N A M c I At A cA 05 h σA NA A MA cA I σA 1233 MPa T Ans At B cB 05 h σB NB A MB cB I σB 625 MPa T Ans Problem 820 Determine the maximum and minimum normal stress in the bracket at section a when the load is applied at x 0 Given a 30mm b 20mm xe 05a P 4kN Solution Internal Force and Moment N P M P xe M 0060 kN m Section Property A a b A 600mm2 I 1 12 b a3 I 45000mm4 Normal Stress σ N A M c I cmax 05a σt N A M cmax I σt 1333 MPa T cmin 05 a σc N A M cmin I σc 2667 MPa C σmax max σt σc σmax 2667 MPa Ans σmin min σt σc σmin 1333 MPa Ans Problem 821 Determine the maximum and minimum normal stress in the bracket at section a when the load is applied at x 50 mm Given a 30mm b 20mm P 4 kN xe 05a 50mm Solution Internal Force and Moment N P M P xe M 0140 kN m Section Property A a b A 600mm2 I 1 12 b a3 I 45000mm4 Normal Stress σ N A M c I cmax 05a σt N A M cmax I σt 4000 MPa T cmin 05 a σc N A M cmin I σc 5333 MPa C σmax max σt σc σmax 5333 MPa Ans σmin min σt σc σmin 4000 MPa Ans Problem 822 The vertical force P acts on the bottom of the plate having a negligible weight Determine the maximum distance d to the edge of the plate at which it can be applied so that it produces no compressive stresses on the plate at section aa The plate has a thickness of 10 mm and P acts along the centerline of this thickness Given t 10mm b 150mm Solution Internal Force and Moment N P xe d 05b M P xe Section Property A t b A 1500mm2 I 1 12 t b3 I 2812500mm4 Normal Stress Require σmin 0 I Mc A N σ cmin 05 b 0 P A P xe cmin I 0 P A P d 05b 05 b I d 2I A b b 2 d 100mm Ans Problem 823 The vertical force P 600 N acts on the bottom of the plate having a negligible weight The plate has a thickness of 10 mm and P acts along the centerline of this thickness such that d 100 mm Plot the distribution of normal stress acting along section aa Given t 10mm b 150mm d 100mm P 06kN Solution Internal Force and Moment N P xe d 05b M P xe M 0015 kN m Section Property A t b A 1500mm2 I 1 12 t b3 I 2812500mm4 Normal Stress σ N A M c I cA 05 b σA N A M cA I σA 0MPa C Ans cB 05 b σB N A M cB I σB 0800 MPa T Ans Problem 824 The gondola and passengers have a weight of 75 kN and center of gravity at G The suspender arm AE has a square crosssectional area of 38 mm by 38 mm and is pin connected at its ends A and E Determine the largest tensile stress developed in regions AB and DC of the arm Given b 38mm d 38mm ah 375mm L1 12m L2 165m W 75kN Solution Section Property A b d I 1 12 b d3 Segment AB NAB W MAB 0 Maximum Normal Stress σ N A M c I σAB NAB A σAB 519 MPa T Ans Segment DC NDC W MDC W ah Maximum Normal Stress σ N A M c I cmax 05d σDC NDC A MDC cmax I σDC 31273 MPa T Ans Problem 825 The stepped support is subjected to the bearing load of 50 kN Determine the maximum and minimum compressive stress in the material Given a 100mm b 100mm P 50kN xe 05a 30mm Solution Internal Force and Moment N P M P xe M 1000 kN m Section Property For the bottom portion of the stepped support A a b A 10000mm2 I 1 12 b a3 I 833333333 mm4 Normal Stress σ N A M c I cmax 05a σt N A M cmax I σt 1MPa T cmin 05 a σc N A M cmin I σc 1100 MPa C σcmax max 0 σc σcmax 11MPa Ans σcmin min 0 σc σcmin 0MPa Ans Problem 826 The bar has a diameter of 40 mm If it is subjected to a force of 800 N as shown determine the stress components that act at point A and show the results on a volume element located at this point Given do 40mm a 200mm P 08kN θ 30deg Solution Internal Force and Moment At section AB N P sin θ N 0400 kN V P cos θ V 0693 kN M V a M 01386 kN m Section Property ro 05do A π ro 2 A 125664 mm2 I π 4 ro 4 I 12566371 mm4 QA 4ro 3π A 2 QA 533333 mm3 Normal Stress σ N A M c I cA 0 σA N A M cA I σA 0318 MPa T Ans Shear Stress τ V Q I b τA V QA I do τA 0735 MPa Ans Problem 827 Solve Prob 826 for point B Given do 40mm a 200mm P 08kN θ 30deg Solution Internal Force and Moment At section AB N P sin θ N 0400 kN V P cos θ V 0693 kN M V a M 01386 kN m Section Property ro 05do A π ro 2 A 125664 mm2 I π 4 ro 4 I 12566371 mm4 QB 0 since A0 Normal Stress σ N A M c I cB ro σB N A M cB I σB 2173 MPa C Ans Shear Stress τ V Q I b τA V QB I do τA 0MPa Ans Problem 828 Since concrete can support little or no tension this problem can be avoided by using wires or rods to prestress the concrete once it is formed Consider the simply supported beam shown which has a rectangular cross section of 450 mm by 300 mm If concrete has a specific weight of 24 kNm3 determine the required tension in rod AB which runs through the beam so that no tensile stress is developed in the concrete at its center section aa Neglect the size of the rod and any deflection of the beam Given b 300mm d 450mm d 400mm γ 24 kN m3 L 24m Solution a d d w γ b d Support Reactions By symmetry RA R RA R 2R w L 0 R 05w L Internal Force and Moment ΣFx0 T N 0 N T ΣΜO0 M T 05d a R 05L 05w L 025L 0 M R 025 L T 05 d a Section Property A b d I 1 12 b d3 Normal Stress σa N A M c I Requires σa 0 0 T A M ca I ca 05d 0 T A R 025 L T 05 d a ca I T R 025 L 05 d a I A ca T 9331 kN Ans Problem 829 Solve Prob 828 if the rod has a diameter of 12 mm Use the transformed area method discussed in Sec 66 Est 200 GPa Ec 25 GPa Given b 300mm d 450mm d 400mm do 12mm Est 200GPa Ec 25GPa L 24m γ 24 kN m3 Solution a d d w γ b d Support Reactions By symmetry RA R RA R 2R w L 0 R 05w L Internal Force and Moment ΣFx0 T N 0 N T ΣΜO0 M T d yc a R 05L 05w L 025L 0 M R 025 L T d yc a Section Property n Est Ec Aconc n 1 π 4 do 2 A b d Aconc yc b d 05d Aconc d A I 1 12 b d3 b d 05d yc 2 Aconc d yc 2 Normal Stress σa N A M c I Requires σa 0 0 T A M ca I ca d yc 0 T A R 025 L T d yc a ca I T R 025 L d yc a I A ca T 9343 kN Ans Problem 830 The block is subjected to the two axial loads shown Determine the normal stress developed at points A and B Neglect the weight of the block Given b 50mm d 75mm P1 250N P2 500N Solution Internal Force and Moment ΣFx0 N P1 P2 0 N P1 P2 ΣΜz0 Mz P1 05d P2 05d 0 Mz 05 d P2 P1 ΣΜy0 My P1 05b P2 05b 0 My 05 b P2 P1 Section Property A b d Iz 1 12 b d3 Iy 1 12 d b3 Normal Stress σ N A Mz y Iz My z Iy At A yA 05d zA 05b σA N A Mz yA Iz My zA Iy σA 0200 MPa C Ans At B yB 05d zB 05 b σB N A Mz yB Iz My zB Iy σB 0600 MPa C Ans Solution T Given b 50mm d 75mm P1 250N P2 500N ΣΜz0 Internal Force and Moment ΣFx0 N P1 P2 0 N P1 P2 Ans Mz P1 05d P2 05d 0 Mz 05 d P2 P1 ΣΜy0 My P1 05b P2 05b 0 My 05 b P2 P1 Section Property A b d Iz 1 12 b d3 Iy 1 12 d b3 Normal Stress σ N A Mz y Iz My z Iy At A yA 05d zA 05b σA N A Mz yA Iz My zA Iy σA 0200 MPa C Ans At B yB 05d zB 05 b σB N A Mz yB Iz My zB Iy σB 0600 MPa C Ans At C yC 05 d zC 05 b σC N A Mz yC Iz My zC Iy σC 0200 MPa C Ans At D yD 05 d zD 05b σD N A Mz yD Iz My zD Iy σD 0200 MPa Problem 831 The block is subjected to the two axial loads shown Sketch the normal stress distribution acting over the cross section at section aa Neglect the weight of the block Problem 832 A bar having a square cross section of 30 mm by 30 mm is 2 m long and is held upward If it has a mass of 5 kgm determine the largest angle θ measured from the vertical at which it can be supported before it is subjected to a tensile stress near the grip Given b 30mm L 2m t 30mm mo 5 kg m Solution W mo g L W 00981 kN Internal Force and Moment ΣFy0 N W cos θ 0 N W cos θ ΣΜO0 M W sin θ 05 L 0 M 05W L sin θ Section Property A b t I 1 12 b t3 Normal Stress Require σmax 0 I Mc A N σ cmax 05 b 0 N A M cmax I 0 W cos θ b t 05W L sin θ 05 b 1 12 b t3 tan θ t2 3 L b θ atan t2 3 L b θ 000500 rad θ 0286 deg Ans Problem 833 Solve Prob 832 if the bar has a circular cross section of 30mm diameter Given do 30mm L 2m mo 5 kg m Solution W mo g L W 00981 kN Internal Force and Moment ΣFy0 N W cos θ 0 N W cos θ ΣΜO0 M W sin θ 05 L 0 M 05W L sin θ Section Property ro 05do A π ro 2 A 70686 mm2 I π 4 ro 4 I 3976078 mm4 Normal Stress Require σmax 0 I Mc A N σ cmax ro 0 N A M cmax I 0 W cos θ A 05W L sin θ ro I tan θ 2I A L ro θ atan 2I A L ro θ 000375 rad θ 0215 deg Ans Problem 834 The wideflange beam is subjected to the loading shown Determine the stress components at points A and B and show the results on a volume element at each of these points Use the shear formula to compute the shear stress Given b 100mm d 150mm t 12mm dB 50mm P1 25kN P2 125kN P3 15kN L1 05m L2 1m L3 15m Solution L 3 L1 L2 L3 Support Reactions Given ΣFy0 R1 R2 P1 P2 P3 0 ΣΜR20 R1 L P1 L L1 P2 L2 L3 P3 L3 0 Guess R1 1N R2 1N R1 R2 Find R1 R2 R1 R2 1563 1438 kN At Section AB M R1 2L1 V R1 P1 Section Property D d 2t A 2 b t d t I 1 12 b D3 1 12 QA 0 since A 0 Normal Stress σ M y I yA 05D σA M yA I σA 7098 MPa C Ans yB 05d dB σB M yB I σB 204 MPa T Ans Shear Stress τ V Q I t τA V QA I b τA 000 MPa Ans τB V QB I t τB 7265 MPa Ans b t d3 QB b t 05D 05t dB t 05D t 05dB Problem 835 The cantilevered beam is used to support the load of 8 kN Determine the state of stress at points A and B and sketch the results on differential elements located at each of these points Given bo 100mm d 100mm t 10mm L 3m dA 25mm dB 45mm P 8kN Solution Internal Force and Moment At Section AB V P V 8kN M P L M 24kN m Section Property bi bo 2t A 2 d t bi t A 2800mm2 I 1 12 bi t3 1 12 2t d3 I 167333333 mm4 QA dA t 05d 05dA QA 9375mm3 QB dB t 05d 05dB QB 12375mm3 Normal Stress σ M y I yA 05d dA σA M yA I σA 3586 MPa T Ans yB 05d dB σB M yB I σB 717 MPa T Ans Shear Stress τ V Q I t τA V QA I t τA 448 MPa Ans τB V QB I t τB 592 MPa Ans Problem 836 The cylinder of negligible weight rests on a smooth floor Determine the eccentric distance ey at which the load can be placed so that the normal stress at point A is zero Solution Internal Force and Moment V P M P ey Section Property A π r2 I π 4 r4 Normal Stress Require σmax 0 I Mc A N σ cA r 0 N A M cr I 0 P π r2 4P ey r π r4 ey r 4 Ans Problem 837 The beam supports the loading shown Determine the state of stress at points E and F at section aa and represent the results on a differential volume element located at each of these points Given bf 150mm dw 200mm tf 10mm tw 15mm a 1m b 2m yB 33m w 10 kN m yD 03m Solution L 2 a b Support Reactions Given Dx L yB yD Dy 1 ΣFy0 Cy Dy w 2a 0 2 ΣΜC0 w 2a a Dy L Dx yD 0 3 Solving Eqs1 2 and 3 Guess Cy 1N Dx 1N Dy 1N Cy Dx Dy Find Cy Dx Dy Cy Dx Dy 154545 60606 45455 kN Internal Force and Moment At Section aa N Dx N 60606 kN V w a Dy V 54545 kN M Dy a b Dx yD w a 05a M 104545 kN m Section Property do dw 2tf A 2 bf tf dw tw A 6000mm2 I 1 12 bf do 3 1 12 bf tw dw 3 I 43100000mm4 QE bf tf 05do 05tf 05dw tw 025dw QE 232500mm3 QF 0 since A 0 I Mc A N σ Normal Stress cE 0 σE N A M cE I σE 101 MPa C Ans cF 05 do σF N A M cF I σF 2769 MPa C Ans Shear Stress τ V Q I t τE V QE I tw τE 196 MPa Ans τF V QF I bf τF 0MPa Ans Problem 838 The metal link is subjected to the axial force of P 7 kN Its original cross section is to be altered by cutting a circular groove into one side Determine the distance a the groove can penetrate into the cross section so that the tensile stress does not exceed σallow 175 MPa Offer a better way to remove this depth of material from the cross section and calculate the tensile stress for this case Neglect the effects of stress concentration Given h 80mm t 25mm P 7kN σallow 175MPa Solution Internal Force and Moment At narrow section ΣFx0 P N 0 N P ΣΜbase0 M N h a 2 P h 2 0 M 05P a Section Property A h a t I t 12 h a 3 Normal Stress Require σmax σallow cmax 05 d a σmax N A M cmax I σallow P h a t 05P a 05 h a t 12 h a 3 Given σallow t h a 2 P h a 3P a 0 1 Solving Eq1 Guess a 1mm a Find a a 6194 mm Ans Better way To remove material equally from both sides such that M0 A h a t σmax N A 0 σmax 1550 MPa Ans Problem 839 Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN Indicate the result as a differential volume element Given bf 150mm dw 200mm tf 20mm tw 15mm a 2m b 075m c 1m hG 375mm rG 250mm P 4kN Solution L a b c Support Reactions Given ΣFx0 Cx P 0 Cx P ΣΜD0 P hg rG a Cy L 0 Cy hG rG L P Internal Force and Moment At Section AB N Cx N 4kN V Cy V 06667 kN M Cy c M 06667 kN m Section Property do dw 2tf A 2 bf tf dw tw A 9000mm2 I 1 12 bf do 3 1 12 bf tw dw 3 I 82800000mm4 QA bf tf 05do 05tf 05dw tw 025dw QA 405000mm3 I Mc A N σ Normal Stress cA 0 σA N A M cA I σA 0444 MPa T Ans Shear Stress τ V Q I t τA V QA I tw τA 0217 MPa Ans Problem 840 Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN Indicate the result as a differential volume element Given bf 150mm dw 200mm tf 20mm tw 15mm a 2m b 075m c 1m hG 375mm rG 250mm P 4kN Solution L a b c Support Reactions Given ΣFx0 Cx P 0 Cx P ΣΜD0 P hg rG a Cy L 0 Cy hG rG L P Internal Force and Moment At Section AB N Cx N 4kN V Cy V 06667 kN M Cy c M 06667 kN m Section Property do dw 2tf A 2 bf tf dw tw A 9000mm2 I 1 12 bf do 3 1 12 bf tw dw 3 I 82800000mm4 QB 0 since A 0 I Mc A N σ Normal Stress cB 05 do σB N A M cB I σB 0522 MPa C Ans Shear Stress τ V Q I t τB V QB I tw τB 0MPa Ans Problem 841 The bearing pin supports the load of 35 kN Determine the stress components in the support member at point A The support is 12 mm thick Given b 12mm d 18mm W 35kN L1 50mm L2 75mm a 32mm θ 30deg Solution Internal Force and Moment ΣFx0 N W cos θ 0 N W cos θ ΣFy0 V W sin θ 0 V W sin θ ΣΜ0 M W a L1 sin θ 0 M W a L1 sin θ Section Property A b d I 1 12 b d3 QA 0 since A 0 Normal Stress σ N A M y I At A yA 05 d σA N A M yA I σA 2378 MPa C Ans Shear Stress τ V Q I t τA V QA I b τA 0MPa Ans Problem 842 The bearing pin supports the load of 35 kN Determine the stress components in the support member at point B The support is 12 mm thick Given b 12mm d 18mm W 35kN L1 50mm L2 75mm a 32mm θ 30deg Solution Internal Force and Moment ΣFx0 N W cos θ 0 N W cos θ ΣFy0 V W sin θ 0 V W sin θ ΣΜ0 M W a L1 sin θ 0 M W a L1 sin θ Section Property A b d I 1 12 b d3 QB 0 since A 0 Normal Stress σ N A M y I At B yB 05d σB N A M yB I σB 5184 MPa T Ans Shear Stress τ V Q I t τB V QB I b τB 0MPa Ans p 8 kN m2 Given b 36m h 18m ro 75mm ri 68mm W Nx 0 W 75kN d 09m Solution P p b h Internal Force and Moment ΣFx0 Ans Nx W ΣFy0 Vy P 0 Vy P ΣFz0 Vz 0 ΣΜx0 Mx P 05b 0 Mx 05 P b ΣΜy0 My W 05b 0 My 05 W b ΣΜz0 Mz P 05h d 0 Mz P 05h d Nx 75 kN Vy 5184 kN Vz 0kN Mx 93312 kN m My 135 kN m Mz 93312 kN m Section Property A π ro 2 ri 2 Iy π 4 ro 4 ri 4 Iz Iy QCy 4ro 3π π 2 ro 2 4ri 3π π 2 ri 2 QDz QCy QDy 0 QCz 0 J π 2 ro 4 ri 4 Normal Stress σ N A Mz y Iz My z Iy At C yC 0 zC ri σC N A Mz yC Iz My zC Iy σC 1139 MPa T Ans At D yD ro zD 0 σD N A Mz yD Iz My zD Iy σD 8685 MPa T Problem 843 The uniform sign has a weight of 75 kN and is supported by the pipe AB which has an inner radius of 68 mm and an outer radius of 75 mm If the face of the sign is subjected to a uniform wind pressure of p 8 kNm2 determine the state of stress at points C and D Show the results on a differential volume element located at each of these points Neglect the thickness of the sign and assume that it is supported along the outside edge of the pipe Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τt T ρ J At C t 2 ro ri ρ ri τvy Vy QCy Iz t τt Mx ρ J τCxy τvy τt τCxy 3608 MPa Ans τCxz 0 Ans τCyz 0 Ans At D t 2 ro ri ρ ro τvz Vz QDz Iy t τt Mx ρ J τDxz τvz τt τDxz 4343 MPa Ans τDxy 0 Ans τDyz 0 Ans Problem 844 Solve Prob 843 for points E and F Given b 36m h 18m ro 75mm ri 68mm p 8 kN m2 W 75kN d 09m Solution P p b h Internal Force and Moment ΣFx0 W Nx 0 Nx W ΣFy0 Vy P 0 Vy P ΣFz0 Vz 0 ΣΜx0 Mx P 05b 0 Mx 05 P b ΣΜy0 My W 05b 0 My 05 W b ΣΜz0 Mz P 05h d 0 Mz P 05h d Nx 75 kN Vy 5184 kN Vz 0kN Mx 93312 kN m My 135 kN m Mz 93312 kN m Section Property A π ro 2 ri 2 Iy π 4 ro 4 ri 4 Iz Iy QFy 4ro 3π π 2 ro 2 4ri 3π π 2 ri 2 QEz QFy QEy 0 QFz 0 J π 2 ro 4 ri 4 Normal Stress σ N A Mz y Iz My z Iy At F yF 0 zF ro σF N A Mz yF Iz My zF Iy σF 1257 MPa C Ans At E yE ro zE 0 σE N A Mz yE Iz My zE Iy σE 8685 MPa C Ans Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τt T ρ J At F t 2 ro ri ρ ro τvy Vy QFy Iz t τt Mx ρ J τFxy τvy τt τFxy 4672 MPa Ans τFxz 0 Ans τFyz 0 Ans At E t 2 ro ri ρ ro τvz Vz QEz Iy t τt Mx ρ J τExz τvz τt τExz 4343 MPa Ans τExy 0 Ans τEyz 0 Ans Problem 845 The bar has a diameter of 40 mm If it is subjected to the two force components at its end as shown determine the state of stress at point A and show the results on a differential volume element located at this point Given a 100mm b 150mm do 40mm Py 03kN Pz 05 kN Solution Internal Force and Moment ΣFx0 Nx 0 ΣFy0 Vy Py 0 Vy Py ΣFz0 Vz Pz 0 Vz Pz ΣΜx0 Mx 0 ΣΜy0 My Pz b 0 My Pz b ΣΜz0 Mz Py b 0 Mz Py b Nx 0kN Vy 03 kN Vz 05 kN Mx 0kN m My 0075 kN m Mz 0045 kN m Section Property ro 05do A π ro 2 J π 2 ro 4 Iy π 64 do 4 Iz Iy QAz 0 QAy 4ro 3π A 2 Normal Stress σ Nx A Mz y Iz My z Iy At A yA 0 zA ro σA Nx A Mz yA Iz My zA Iy σA 119 MPa T Ans Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τt T ρ J At A ty do ρ ro τvy Vy QAy Iz ty τt Mx ρ J τvy 0318 MPa τt 0MPa τAxy τvy τt τAxy 0318 MPa Ans τAxz 0 Ans τAyz 0 Ans Problem 846 Solve Prob 845 for point B Given a 100mm b 150mm do 40mm Py 03kN Pz 05 kN Solution Internal Force and Moment ΣFx0 Nx 0 ΣFy0 Vy Py 0 Vy Py ΣFz0 Vz Pz 0 Vz Pz ΣΜx0 Mx 0 ΣΜy0 My Pz b 0 My Pz b ΣΜz0 Mz Py b 0 Mz Py b Nx 0kN Vy 03 kN Vz 05 kN Mx 0kN m My 0075 kN m Mz 0045 kN m Section Property ro 05do A π ro 2 J π 2 ro 4 Iy π 64 do 4 Iz Iy QBy 0 QBz 4ro 3π A 2 Normal Stress σ Nx A Mz y Iz My z Iy At B yB ro zB 0 σB Nx A Mz yB Iz My zB Iy σB 716 MPa C Ans Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τt T ρ J At B tz do ρ ro τvz Vz QBz Iy tz τt Mx ρ J τvz 0531 MPa τt 0MPa τBxz τvz τt τBxz 0531 MPa Ans τBxy 0 Ans τByz 0 Ans Problem 847 The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mg and center of mass at G If the pipe has an outer diameter of 70 mm and a wall thickness of 10 mmdetermine the state of stress acting at point C Show the results on a differential volume element located at this point Neglect the weight of the pipe Given do 70mm t 10mm hA 75mm L 3m Mo 3000kg θ 45deg Solution W Mo g W 2942 kN Equilibrium ΣFy0 P W 0 P W Support Reactions By symmetry Ay By R P 2R 0 R 05 P R 1471 kN Also Ax Bx and tan θ Ay Ax Ax R tan θ Ax 1471 kN Internal Force and Moment At Section CD N Ax N 1471 kN V R 05W V 0kN M V 05 L Ax hA M 1103 kN m Section Property di do 2t A π 4 do 2 di 2 A 188496 mm2 I π 64 do 4 di 4 I 87179196 mm4 I Mc A N σ Normal Stress cC 05do σC N A M cC I σC 5210 MPa C Ans Shear Stress τ V Q I t τC 0 since V 0 Ans Problem 848 The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mg and center of mass at G If the pipe has an outer diameter of 70 mm and a wall thickness of 10 mmdetermine the state of stress acting at point D Show the results on a differential volume element located at this point Neglect the weight of the pipe Given do 70mm t 10mm hA 75mm L 3m Mo 3000kg θ 45deg Solution W Mo g W 2942 kN Equilibrium ΣFy0 P W 0 P W Support Reactions By symmetry Ay By R P 2R 0 R 05 P R 1471 kN Also Ax Bx and tan θ Ay Ax Ax R tan θ Ax 1471 kN Internal Force and Moment At Section CD N Ax N 1471 kN V R 05W V 0kN M V 05 L Ax hA M 1103 kN m Section Property di do 2t A π 4 do 2 di 2 A 188496 mm2 I π 64 do 4 di 4 I 87179196 mm4 I Mc A N σ Normal Stress cD 0 σD N A M cD I σD 780 MPa C Ans Shear Stress τ V Q I t τD 0 since V 0 Ans Problem 849 The sign is subjected to the uniform wind loading Determine the stress components at points A and B on the 100mmdiameter supporting post Show the results on a volume element located at each of these points Given do 100mm a 3m po 1500Pa bo 2m ho 1m Solution Px po bo ho Px 300 kN Internal Force and Moment ΣFx0 Vx Px 0 Vx Px ΣFy0 Vy 0 ΣFz0 Nz 0 ΣΜx0 Mx 0 ΣΜy0 My Px a 05ho 0 My Px a 05ho ΣΜz0 Mz Px 05bo Mz Px 05bo Vx 3kN Vy 0kN Nz 0kN Mx 0kN m My 105 kN m Mz 3 kN m Section Property ro 05 do A π ro 2 A 785398 mm2 I π 4 ro 4 I 490873852 mm4 J π 2 ro 4 J 981747704 mm4 QA 0 since A 0 QB 4ro 3π A 2 At A xA ro At B xB 0 ρA ro ρB ro Normal Stress I Mc A N σ σA Nz A My xA I σB Nz A My xB I σA 1070 MPa T Ans σB 00 MPa Ans Shear Stress τ T ρ J V Q I t τA Mz ρA J since QA 0 τB Mz ρB J Vx QB I do τA 1528 MPa Ans τB 1477 MPa Ans Problem 850 The sign is subjected to the uniform wind loading Determine the stress components at points C and D on the 100mmdiameter supporting post Show the results on a volume element located at each of these points Given do 100mm a 3m po 1500Pa bo 2m ho 1m Solution Px po bo ho Px 300 kN Internal Force and Moment ΣFx0 Vx Px 0 Vx Px ΣFy0 Vy 0 ΣFz0 Nz 0 ΣΜx0 Mx 0 ΣΜy0 My Px a 05ho 0 My Px a 05ho ΣΜz0 Mz Px 05bo Mz Px 05bo Vx 3kN Vy 0kN Nz 0kN Mx 0kN m My 105 kN m Mz 3 kN m Section Property ro 05 do A π ro 2 A 785398 mm2 I π 4 ro 4 I 490873852 mm4 J π 2 ro 4 J 981747704 mm4 QC 0 since A 0 QD 4ro 3π A 2 At C xC ro At D xD 0 ρC ro ρD ro Normal Stress I Mc A N σ σC Nz A My xC I σD Nz A My xD I σC 1070 MPa C Ans σD 0MPa Ans Shear Stress τ T ρ J V Q I t τC Mz ρC J since QC 0 τD Mz ρD J Vx QD I do τC 1528 MPa Ans τD 1579 MPa Ans Problem 851 The 18mmdiameter shaft is subjected to the loading shown Determine the stress components at point A Sketch the results on a volume element located at this point The journal bearing at C can exert only force components Cy and Cz on the shaft and the thrust bearing at D can exert force components Dx Dy and Dz on the shaft Given do 18mm L 500mm ax 50mm ay 200mm P 600N a 250mm Solution ro 05do Support Reactions at C Tx 0 My 0 Mz 0 Cx 0 Cy 0 Cz P Internal Force and Moment at A N 0 Vy 0 Vz Cz Tx 0 My Cz a Mz 0 Section Property Iy π 4 ro 4 Iz Iy A π ro 2 QA 0 J π 2 ro 4 Normal Stress σ N A Mz y Iz My z Iy At A yA 0 zA ro σA N A Mz yA Iz My zA Iy σA 2620 MPa C Ans Shear Stress τ V Q I b At A b 0 τA Vz QA Iy b τA 000 Ans Problem 852 Solve Prob 851 for the stress components at point B Given do 18mm L 500mm ax 50mm ay 200mm P 600N a 250mm Solution ro 05do Support Reactions at C Tx 0 My 0 Mz 0 Cx 0 Cy 0 Cz P Internal Force and Moment at B N 0 Vy 0 Vz Cz Tx 0 My Cz a Mz 0 Section Property Iy π 4 ro 4 Iz Iy A π ro 2 J π 2 ro 4 QB 4ro 3π π 2 ro 2 Normal Stress σ N A Mz y Iz My z Iy At B yB ro zB 0 σA N A Mz yB Iz My zB Iy σA 00 MPa Ans Shear Stress τ V Q I b At B b 2 ro τB Vz QB Iy b τB 314 MPa Ans Problem 853 The solid rod is subjected to the loading shown Determine the state of stress developed in the material at point A and show the results on a differential volume element at this point Given ro 30mm a 150mm Px 10 kN Py 1kN Pz 15kN Tx 02kN m Solution Internal Force and Moment At Section A ΣFx0 Nx Px 0 Nx Px ΣFy0 Vy 0 Nx 10kN ΣFz0 Vz 0 Vy 0kN ΣΜx0 Mx Tx 0 Mx Tx Vz 0kN ΣΜy0 My 0 Mx 02 kN m ΣΜz0 Mz Px ro 0 Mz Px ro My 0kN m Section Property ρ ro Mz 03 kN m A π ro 2 A 282743 mm2 I π 4 ro 4 I 63617251 mm4 J π 2 ro 4 J 127234502 mm4 QAz 4ro 3π A 2 QAy 0 since A 0 y y z z x I z M I M y A N σ Normal Stress yA ro zA 0 σA Nx A Mz yA I My zA I σA 177 MPa T Ans Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τvz 0 since Vz 0 τt T ρ J τt Mx ρ J τt 472 MPa τxz τvz τt τxz 4716 MPa Ans τxy 0 Ans τyz 0 Ans Problem 854 The solid rod is subjected to the loading shown Determine the state of stress at point B and show the results on a differential volume element at this point Given ro 30mm a 150mm Px 10 kN Py 10kN Pz 15kN Tx 02kN m Solution Internal Force and Moment At Section B ΣFx0 Nx Px 0 Nx Px ΣFy0 Vy Py 0 Vy Py ΣFz0 Vz 0 Nx 10kN ΣΜx0 Mx Py ro Tx 0 Mx Py ro Tx Vy 10 kN ΣΜy0 My 0 Vz 0kN ΣΜz0 Mz Px ro Py a 0 Mz Px ro Py a Mx 01 kN m My 0kN m Section Property ρ ro Mz 18 kN m A π ro 2 A 282743 mm2 I π 4 ro 4 I 63617251 mm4 J π 2 ro 4 J 127234502 mm4 QBz 4ro 3π A 2 QBy 0 since A 0 y y z z x I z M I M y A N σ Normal Stress yB ro zB 0 σB Nx A Mz yB I My zB I σB 813 MPa C Ans Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τvz 0 since Vz 0 τt T ρ J τt Mx ρ J τt 236 MPa τxz τvz τt τxz 2358 MPa Ans τxy 0 Ans τyz 0 Ans Problem 855 The solid rod is subjected to the loading shown Determine the state of stress at point C and show the results on a differential volume element at this point Given ro 30mm a 150mm Px 10 kN Py 10kN Pz 15kN Tx 02kN m Solution Internal Force and Moment At Section A ΣFx0 Nx Px 0 Nx Px ΣFy0 Vy Py 0 Vy Py ΣFz0 Vz Pz 0 Vz Pz Nx 10kN ΣΜx0 Mx Py ro Pz ro Tx 0 Mx Py ro Pz ro Tx Vy 10 kN ΣΜy0 My Pz a 0 My Pz a Vz 15 kN ΣΜz0 Mz Px ro Py 3a 0 Mz Px ro 3Py a Mx 035 kN m Section Property ρ ro My 225 kN m Mz 48 kN m A π ro 2 A 282743 mm2 I π 4 ro 4 I 63617251 mm4 J π 2 ro 4 J 127234502 mm4 QCy 4ro 3π A 2 QCz 0 since A 0 y y z z x I z M I M y A N σ Normal Stress yC 0 zC ro σC Nx A Mz yC I My zC I σC 1026 MPa C Ans Shear Stress The transverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula for τv and the torsional formula for τt respectively τv V Q I t τvy Vy QCy I 2ro τvy 472 MPa τt T ρ J τt Mx ρ J τt 825 MPa τxy τvy τt τxy 354 MPa Ans τxz 0 Ans τyz 0 Ans Problem 856 The 25mmdiameter rod is subjected to the loads shown Determine the state of stress at point A and show the results on a differential element located at this point Given ax 200 mm az 75mm do 25mm Px 375 N Py 400 N Pz 500N Solution ro 05do Internal Force and Moment ΣFx0 Nx Px 0 Nx Px ΣFy0 Vy Py 0 Vy Py ΣFz0 Vz Pz 0 Vz Pz ΣΜx0 Tx Py az 0 Tx Py az ΣΜy0 My Pz ax Px az 0 My Pz ax Px az ΣΜz0 Mz Py ax 0 Mz Py ax Section Property A π ro 2 Iy π 4 ro 4 Iz Iy QAy 0 QAz 4ro 3π π 2 ro 2 J π 2 ro 4 Normal Stress σ N A Mz y Iz My z Iy At A yA ro zA 0 σA Nx A Mz yA Iz My zA Iy σA 529 MPa T Ans Shear Stress τv V Q I b τt T ρ J At A bz 2 ro ρ ro τvz Vz QAz Iy bz τt Tx ρ J τAxz τvz τt τAxz 1114 MPa Ans τAxy 0 Ans Problem 857 The 25mmdiameter rod is subjected to the loads shown Determine the state of stress at point B and show the results on a differential element located at this point Given ax 200 mm az 75mm do 25mm Px 375 N Py 400 N Pz 500N Solution ro 05do Internal Force and Moment ΣFx0 Nx Px 0 Nx Px ΣFy0 Vy Py 0 Vy Py ΣFz0 Vz Pz 0 Vz Pz ΣΜx0 Tx Py az 0 Tx Py az ΣΜy0 My Pz ax Px az 0 My Pz ax Px az ΣΜz0 Mz Py ax 0 Mz Py ax Section Property A π ro 2 Iy π 4 ro 4 Iz Iy QBz 0 QBy 4ro 3π π 2 ro 2 J π 2 ro 4 Normal Stress σ N A Mz y Iz My z Iy At B yB 0 zB ro σB Nx A Mz yB Iz My zB Iy σB 461 MPa C Ans Shear Stress τv V Q I b τt T ρ J At B by 2 ro ρ ro τvy Vy QBy Iz by τt Tx ρ J τBxy τvy τt τBxy 1086 MPa Ans τBxz 0 Ans Problem 858 The crane boom is subjected to the load of 25 kN Determine the state of stress at points A and B Show the results on a differential volume element located at each of these points Given b 75mm d 76mm t 12mm ah 15m av 24m P 25kN v 4 h 3 r 5 Solution Internal Force and Moment ΣFx0 V P h r 0 V P h r ΣFy0 N P v r 0 N P v r ΣΜO0 M P h r av P v r ah 0 M P v r ah P h r av Section Property D d 2t A b D t d I 1 12 b D3 1 12 b t d3 QA 0 QB 0 since A 0 Normal Stress σ N A M c I cA 05D σA N A M cA I σA 8334 MPa T Ans cB 05 D σB N A M cB I σB 8395 MPa C Ans Shear Stress τ V Q I b τA V QA I b τA 0MPa Ans τB V QB I b τB 0MPa Ans Problem 859 The masonry pier is subjected to the 800kN load Determine the equation of the line y f x along which the load can be placed without causing a tensile stress in the pier Neglect the weight of the pier Given a 15m b 225m Pz 800 kN xA a yA b Solution Section Property A 2a 2 b A 135 m2 Ix 1 12 2a 2 b 3 Ix 2278125 m4 Iy 1 12 2b 2 a 3 Iy 10125 m4 Force and Moment Mx Pz y My Pz x Normal Stress Require σA 0 σA Pz A Mx yA Ix My xA Iy 0 Pz A Pz y yA Ix Pz x xA Iy 0 1 A yA Ix y xA Iy x 0 1 4a b 3 4a b2 y 3 4b a2 x y b 3 b a x y 075 15 x Ans Problem 860 The masonry pier is subjected to the 800kN load If x 025 m and y 05 m determine the normal stress at each corner A B C D not shown and plot the stress distribution over the cross section Neglect the weight of the pier Unit Used kPa 103Pa Given a 15m b 225m Pz 800 kN xA a xB a xC a xD a yA b yB b yC b yD b x 025m y 05m Solution Section Property A 2a 2 b A 135 m2 Ix 1 12 2a 2 b 3 Ix 2278125 m4 Iy 1 12 2b 2 a 3 Iy 10125 m4 Force and Moment Mx Pz y Mx 400 kN m My Pz x My 200kN m Normal Stress σ Pz A Mx y Ix My x Iy σA Pz A Mx yA Ix My xA Iy σA 9877 kPa T Ans σB Pz A Mx yB Ix My xB Iy σB 4938 kPa C Ans σC Pz A Mx yC Ix My xC Iy σC 1284 kPa C Ans σD Pz A Mx yD Ix My xD Iy σD 691 kPa C Ans Problem 861 The symmetrically loaded spreader bar is used to lift the 10kN 1tonne tank Determine the state of stress at points A and B and indicate the results on a differential volume elements Given b 25mm d 50mm a 045m L 12m W 10kN θ 30deg Solution Support Reactions ΣFy0 W 2F cos θ 0 F W 2 cos θ Internal Force and Moment ΣFx0 F sin θ N 0 N F sin θ ΣFy0 V F cos θ 0 V F cos θ ΣΜB0 M F cos θ a 0 M F cos θ a Section Property A b d I 1 12 b d3 QB 05 d b 025d QA 0 since A 0 Normal Stress σ N A M c I cA 05d σA N A M cA I σA 21831 MPa T Ans cB 0 σB N A M cB I σB 231 MPa T Ans Shear Stress τ V Q I b τA V QA I b τA 0MPa Ans τB V QB I b τB 600 MPa Ans Problem 862 A post having the dimensions shown is subjected to the bearing load P Specify the region to which this load can be applied without causing tensile stress to be developed at points A B C and D Problem 863 The man has a mass of 100 kg and center of mass at G If he holds himself in the position shown determine the maximum tensile and compressive stress developed in the curved bar at section aa He is supported uniformly by two bars each having a diameter of 25 mm Assume the floor is smooth Given Ri 150mm do 25mm mo 100kg e 03m a 035m b 1m Solution Equilibrium For the man ΣΜtoe0 mo g b 2P a b 0 P mo g b 2 a b P 03632 kN Section Property ro 05do rc Ri ro A π ro 2 A 49087 mm2 A Ar r dA Σ I IAr 2π rc rc 2 ro 2 IAr 30252 mm R A IAr R 162259 mm Internal Force and Moment As shown on BFBD The internal moment must be computed about the neutral axis M is negative since it tends to decrease the bars radius of curvature N P N 03632 kN M P R e M 016790 kN m Maximum Normal Stress For tensile stress σt N A M R r2 A r1 rc R r2 rc ro σt N A M R r2 A r2 rc R σt 1027 MPa T Ans For compressive stress σc N A M R r1 A r1 rc R r1 rc ro σc N A M R r1 A r1 rc R σc 1169 MPa C Ans Problem 864 The block is subjected to the three axial loads shown Determine the normal stress developed at points A and B Neglect the weight of the block Given b 100mm b 50mm d 75mm d 125mm P1 500N P2 1250N P3 250N Solution B b 2b D d 2d Internal Force and Moment ΣFz0 N P1 P2 P3 0 N P1 P2 P3 ΣΜx0 Mx P1 05d P2 05d P3 05D 0 Mx P1 05 d P2 05 d P3 05 D ΣΜy0 My P1 05B P2 05B P3 05 b 0 My P1 05 B P2 05 B P3 05 b Section Property A B D 4b d Ix 1 12 b D3 2 12 b d3 Iy 1 12 d B3 2 12 d b3 Normal Stress σ N A Mx y Ix My x Iy At A xA 05B yA 05 d σA N A Mx yA Ix My xA Iy σA 01703 MPa C Ans At B xB 05b yB 05 D σB N A Mx yB Ix My xB Iy σB 00977 MPa C Ans Problem 865 If P 15 kN plot the distribution of stress acting over the cross section aa of the offset link Given ho 50mm P 15kN to 10mm a 30mm Solution Section Property A ho to A 500mm2 I 1 12 to ho 3 I 10416667 mm4 Moment M P a 05ho M 0825 kN m Normal Stress I Mc A N σ σA P A M 05ho I σA 228MPa T Ans σB P A M 05ho I σB 168 MPa C Ans y σA ho y σB y ho σA σA σB y 2879 mm Problem 866 Determine the magnitude of the load P that will cause a maximum normal stress σmax 200 MPa of in the link at section aa Given ho 50mm a 30mm to 10mm σallow 200MPa Solution Section Property A ho to A 500mm2 I 1 12 to ho 3 I 10416667 mm4 Moment M P a 05ho Normal Stress The maximum normal stress occurs at A I Mc A N σ σA P A M 05ho I σallow P A P a 05ho 05ho I P σallow A I I A a 05ho 05ho P 1316 kN Ans Problem 867 Air pressure in the cylinder is increased by exerting forces P 2 kN on the two pistons each having a radius of 45 mm If the cylinder has a wall thickness of 2 mm determine the state of stress in the wall of the cylinder Given t 2mm P 2kN ri 45mm Solution A πri 2 p P A p 03144 MPa Hoop Stress α ri t α 2250 Since α 10 then thinwall analysis can be used σ1 p ri t σ1 707 MPa Ans Longitudinal Stress σ2 0 Ans The pressure p is supported by the surface of the pistons in the longitudinal direction Problem 868 Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress component in the cylinder does not exceed 3 MPa Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm Given t 2mm σallow 3MPa ri 45mm Solution A πri 2 p P A Hoop Stress α ri t α 2250 Since α 10 then thinwall analysis can be used σ1 p ri t σallow P ri A t P σallow ri A t P 0848 kN Ans Problem 869 The screw of the clamp exerts a compressive force of 25 kN on the wood blocks Determine the maximum normal stress developed along section aa The cross section there is rectangular 18 mm by 12 mm Given b 12mm d 18mm a 100mm P 25kN Solution Internal Force and Moment N P M P a Section Property A b d I 1 12 b d3 Normal Stress σ N A M c I cmax 05d σmax N A M cmax I σmax 3974 MPa T Ans Problem 870 The wall hanger has a thickness of 6 mm and is used to support the vertical reactions of the beam that is loaded as shown If the load is transferred uniformly to each strap of the hanger determine the state of stress at points C and D of the strap at B Assume the vertical reaction F at this end acts in the center and on the edge of the bracket as shown Given t 6mm d 50mm P 50kN L1 12m L2 18m w 30 kN m Solution L L1 L2 Support Reactions Given ΣFy0 FA P w L2 FB 0 ΣΜB0 FA L P L 05L1 w L2 05L2 0 Guess FA 1kN FB 1kN FA FB Find FA FB FA FB 5620 4780 kN Section Property A 2t d I 2 12 t d3 At Section CD P FB M P 05d V 0 Stresses σ P A M y I τ V Q I b At C yC 0 σC P A M yC I σC 7967 MPa T Ans τC 0 Ans At D yD 05 d σD P A M yD I σD 15933 MPa C Ans τD 0 Ans Problem 871 The support is subjected to the compressive load P Determine the absolute maximum and minimum normal stress acting in the material Problem 872 The support has a circular cross section with a radius that increases linearly with depth If it is subjected to the compressive load P determine the maximum and minimum normal stress acting in the material Problem 873 The cap on the cylindrical tank is bolted to the tank along the flanges The tank has an inner diameter of 15 m and a wall thickness of 18 mm If the largest normal stress is not to exceed 150 MPa determine the maximum pressure the tank can sustain Also compute the number of bolts required to attach the cap to the tank if each bolt has a diameter of 20 mm The allowable stress for the bolts is σallowb 180 MPa Given t 18mm di 15m σallow 150MPa db 20mm σballow 180MPa Solution ri 05di Hoop Stress α ri t α 4167 Since α 10 then thinwall analysis can be used σ1 p ri t σallow p ri t p σallow t ri p 36 MPa Ans Force Equilibrium for the Cap A πri 2 ΣFy0 p A Fb 0 Fb p A Fb 636173 kN Allowable Normal Stress for Bolts Ab π 4 db 2 σballow Fb n Ab n Fb σballow Ab n 1125 Use n 113 Ans Problem 874 The cap on the cylindrical tank is bolted to the tank along the flanges The tank has an inner diameter of 15 m and a wall thickness of 18 mm If the pressure in the tank is p 120 MPa determine the force in the 16 bolts that are used to attach the cap to the tank Also specify the state of stress in the wall of the tank Given t 18mm di 15m p 120MPa n 16 Solution ri 05di Hoop Stress α ri t α 4167 Since α 10 then thinwall analysis can be used σ1 p ri t σ1 50MPa Ans Longitudinal Stress σ2 p ri 2 t σ2 25MPa Ans Force Equilibrium for the Cap A πri 2 ΣFy0 p A 16 Fb 0 Fb p A 16 Fb 1325 kN Ans Problem 875 The crowbar is used to pull out the nail at A If a force of 40 N is required determine the stress components in the bar at points D and E Show the results on a differential volume element located at each of these points The bar has a circular cross section with a diameter of 12 mm No slipping occurs at B Given do 12mm dA 60mm hA 75mm a 125mm dP 300mm F 40N hP 300mm Solution rp dP 2 hP 2 Support Reactions ΣΜB0 F hA P rp 0 P F hA rp Internal Force and Moment ΣFx0 N 0 ΣFy0 V P 0 V P ΣΜO0 M P a 0 M P a Section Property ro 05do A π ro 2 I π 4 ro 4 QE 4ro 3π 05π ro 2 QD 0 since A 0 Normal Stress σ N A M c I cD ro σD N A M cD I σD 521 MPa T Ans cE 0 σE N A M cE I σE 000 MPa Ans Shear Stress τ V Q I b bE 2 ro τE V QE I bE τE 00834 MPa Ans τD 0 Ans Problem 876 The screw of the clamp exerts a compressive force of 25 kN on the wood blocks Sketch the stress distribution along section aa of the clamp The cross section there is rectangular 18 mm by 12 mm Given b 12mm d 18mm a 100mm P 25kN Solution Internal Force and Moment N P M P a Section Property A b d I 1 12 b d3 Normal Stress σ N A M c I cmax 05d σmax N A M cmax I σmax 3974 MPa T Ans cmin 05d σmin N A M cmin I σmin 3742 MPa C Ans y d y σmin σmax y d σmin σmin σmax y 873 mm Problem 877 The clamp is made from members AB and AC which are pin connected at A If the compressive force at C and B is 180 N determine the state of stress at point F and indicate the results on a differential volume element The screw DE is subjected only to a tensile force along its axis Given h 15mm t 15mm P 180N a 30mm b 40mm Solution Support Reactions ΣΜO0 P b a FDE a 0 FDE b a a P FDE 0420 kN Internal Force and Moment ΣFy0 N 0 ΣFx0 V FDE P 0 V P FDE ΣΜO0 M P b 05a FDE 05a 0 M FDE 05a P b 05a M 360 N m Section Property A h t A 225mm2 I 1 12 t h3 I 421875 mm4 QF 0 since A 0 Normal Stress σ N A M c I cF 05 h σF N A M cF I σF 640 MPa C Ans Shear Stress τ V Q I b τF V QF I t τF 0MPa Ans Problem 878 The eye is subjected to the force of 250 N Determine the maximum tensile and compressive stresses at section aa The cross section is circular and has a diameter of 6 mm Use the curvedbeam formula to compute the bending stress Given Ri 30mm do 6mm P 0250kN Solution ro 05do rc Ri ro Section Property A π ro 2 A 2827 mm2 A Ar r dA Σ I IAr 2π rc rc 2 ro 2 IAr 08586 mm R A IAr R 32932 mm Internal Force and Moment As shown on BFBD The internal moment must be computed about the neutral axis M is positive since it tends to increase the beams radius of curvature N P M P R Maximum Normal Stress For tensile stress σt N A M R r1 A r1 rc R r1 rc ro σt N A M R r1 A r1 rc R σt 4253 MPa T Ans For compressive stress σc N A M R r2 A r1 rc R r2 rc ro σc N A M R r2 A r2 rc R σc 3544 MPa C Ans Problem 879 Solve Prob 878 if the cross section is square having dimensions of 6 mm by 6 mm Given Ri 30mm do 6mm P 0250kN Solution ro 05do rc Ri ro Section Property A do 2 A 3600 mm2 A Ar r dA Σ I IAr do ln rc ro rc ro IAr 10939 mm R A IAr R 3291 mm Internal Force and Moment As shown on BFBD The internal moment must be computed about the neutral axis M is positive since it tends to increase the beams radius of curvature N P M P R Maximum Normal Stress For tensile stress σt N A M R r1 A r1 rc R r1 rc ro σt N A M R r1 A r1 rc R σt 2502 MPa T Ans For compressive stress σc N A M R r2 A r1 rc R r2 rc ro σc N A M R r2 A r2 rc R σc 2084 MPa C Ans Problem 91 Prove that the sum of the normal stresses σx σy σx σy is constant See Figs 92a and 92b Solution Stress Transformation Equations Applying Eqs 91 and 93 of the text σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ 1 σy σx σy 2 σx σy 2 cos 2θ τxy sin 2θ 2 1 2 LHS σx σy RHS σx σy 2 σx σy 2 RHS σx σy Hence σx σy σx σy QED Problem 92 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 91 Given σx 3MPa σy 5MPa τxy 8 MPa φ 40deg Solution Set A m2 θ 180deg φ Force Equilibrium For the sectioned element Ax A cos φ Ay A sin φ Fxx σx Ax Fxy τxy Ax Fyy σy Ay Fyx τxy Ay Given ΣFx0 Fx Fxy sin θ Fxx cos θ Fyx cos θ Fyy sin θ 0 ΣFy0 Fy Fxy cos θ Fxx sin θ Fyx sin θ Fyy cos θ 0 Guess Fx 1kN Fy 1kN Fx Fy Find Fx Fy Fx Fy 405211 40438 kN Normal and Shear Stress σ 0 A F A lim σx Fx A σx 4052 MPa Ans τxy Fy A τxy 0404 MPa Ans The negative signs indicate that the sense of σx and τxy are opposite to that shown in FBD Problem 93 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 91 Given σx 0200 MPa σy 0350MPa φ 50deg τxy 0MPa Solution Set A m2 θ 180deg φ Force Equilibrium For the sectioned element Ax A cos φ Ay A sin φ Fxx σx Ax Fxy τxy Ax Fxy 000 Fyy σy Ay Fyx τxy Ay Fyx 000 Given ΣFx0 Fx Fxy sin θ Fxx cos θ Fyx cos θ Fyy sin θ 0 ΣFy0 Fy Fxy cos θ Fxx sin θ Fyx sin φ Fyy cos θ 0 Guess Fx 1kN Fy 1kN Fx Fy Find Fx Fy Fx Fy 12275 27082 kN Normal and Shear Stress σ 0 A F A lim σx Fx A σx 0123 MPa Ans τxy Fy A τxy 0271 MPa Ans Problem 94 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 91 Given σx 0650 MPa σy 0400MPa φ 60deg τxy 0MPa Solution Set A m2 θ 90deg φ Force Equilibrium For the sectioned element Ax A sin φ Ay A cos φ Fxx σx Ax Fxy τxy Ax Fxy 000 Fyy σy Ay Fyx τxy Ay Fyx 000 Given ΣFx0 Fx Fxy sin θ Fxx cos θ Fyx cos θ Fyy sin θ 0 ΣFy0 Fy Fxy cos θ Fxx sin θ Fyx sin θ Fyy cos θ 0 Guess Fx 1kN Fy 1kN Fx Fy Find Fx Fy Fx Fy 38750 45466 kN Normal and Shear Stress σ 0 A F A lim σx Fx A σx 0387 MPa Ans τxy Fy A τxy 0455 MPa Ans The negative signs indicate that the sense of σx is opposite to that shown in FBD Problem 95 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 91 Given σx 60 MPa σy 50 MPa φ 30deg τxy 28MPa Solution Set A m2 θ 180deg φ Force Equilibrium For the sectioned element Ax A cos φ Ay A sin φ Fxx σx Ax Fxy τxy Ax Fxy 2424871 kN Fyy σy Ay Fyx τxy Ay Fyx 1400000 kN Given ΣFx0 Fx Fxy sin θ Fxx cos θ Fyx cos θ Fyy sin θ 0 ΣFy0 Fy Fxy cos θ Fxx sin θ Fyx sin θ Fyy cos θ 0 Guess Fx 1kN Fy 1kN Fx Fy Find Fx Fy Fx Fy 3325129 1833013 kN Normal and Shear Stress σ 0 A F A lim σx Fx A σx 33251 MPa Ans τxy Fy A τxy 18330 MPa Ans The negative signs indicate that the sense of σx is opposite to that shown in FBD Problem 96 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 91 Given σx 90MPa σy 50MPa τxy 35 MPa φ 60deg Solution Set A m2 θ 90deg φ Force Equilibrium For the sectioned element Ax A sin φ Ay A cos φ Fxx σx Ax Fxy τxy Ax Fyy σy Ay Fyx τxy Ay Given ΣFx0 Fx Fxy sin θ Fxx cos θ Fyx cos θ Fyy sin θ 0 ΣFy0 Fy Fxy cos θ Fxx sin θ Fyx sin θ Fyy cos θ 0 Guess Fx 1kN Fy 1kN Fx Fy Find Fx Fy Fx Fy 4968911 3482051 kN Normal and Shear Stress σ 0 A F A lim σx Fx A σx 4969 MPa Ans τxy Fy A τxy 3482 MPa Ans The negative signs indicate that the sense of τxy isopposite to that shown in FBD Problem 97 Solve Prob 92 using the stresstransformation equations developed in Sec 92 Given σx 5MPa σy 3MPa τxy 8MPa φ 40deg Solution θ 90deg φ Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 405 MPa Ans The negative signs indicate that the sense of σx is a compressive stress Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 0404 MPa Ans The negative signs indicate that the sense of τxy is in the y direction Problem 98 Solve Prob 94 using the stresstransformation equations developed in Sec 92 Given σx 0650 MPa σy 0400MPa φ 60deg τxy 0MPa Solution θ 90deg φ Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 0387 MPa Ans The negative signs indicate that the sense of σx is a compressive stress Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 0455 MPa Ans Problem 99 Solve Prob 96 using the stresstransformation equations developed in Sec 92 Show the result on a sketch Given σx 90MPa σy 50MPa τxy 35 MPa φ 60deg Solution θ 90deg φ Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 4969 MPa Ans Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 3482 MPa Ans The negative signs indicate that the sense of τxy is in the y direction Problem 910 Determine the equivalent state of stress on an element if the element is oriented 30 counterclockwise from the element shown Use the stresstransformation equations Unit Used kPa 1000Pa Given σx 0kPa σy 300 kPa θ 30deg τxy 950kPa Solution Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 7477 kPa Ans σy σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σy 10477 kPa Ans Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 3451 kPa Ans Problem 911 Determine the equivalent state of stress on an element if the element is oriented 60 clockwise from the element shown Given σx 0300MPa σy 0MPa θ 60 deg τxy 0120MPa Solution Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 00289 MPa Ans σy σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σy 0329 MPa Ans Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 00699 MPa Ans Problem 912 Solve Prob 96 using the stresstransformation equations Given σx 90MPa σy 50MPa φ 60deg τxy 35 MPa Solution θ 90deg φ Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 4969 MPa Ans Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 3482 MPa Ans Problem 913 The state of stress at a point is shown on the element Determine a the principal stresses and b the maximum inplane shear stress and average normal stress at the point Specify the orientation of the element in each case Given σx 45MPa σy 60 MPa τxy 30MPa Solution a Principal Stress σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 5297 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 6797 MPa Ans Orientation of Principal Stress tan 2θp 2τxy σx σy θp 1 2 atan 2τxy σx σy θp θp 90deg θp 1487 deg θp 7513 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σx σx σy 2 σx σy 2 cos 2θp τxy sin 2θp σx 5297 MPa Therefore θp1 θp θp1 1487 deg Ans θp2 θp θp2 7513 deg Ans b τmax σx σy 2 2 τxy 2 τmax 6047 MPa Ans σavg σx σy 2 σavg 750 MPa Ans Orientation of Maximum Inplane Shear Stress tan 2θs σx σy 2τxy θs 1 2 atan σx σy 2τxy θs θs 90deg θs 3013 deg θs 5987 deg By observation in order to preserve equilibrium along AB τmax has to act in the direction shown in the figure Problem 914 The state of stress at a point is shown on the element Determine a the principal stresses and b the maximum inplane shear stress and average normal stress at the point Specify the orientation of the element in each case Given σx 180MPa σy 0MPa τxy 150 MPa Solution a Principal Stress σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 26493 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 8493 MPa Ans Orientation of Principal Stress tan 2θp 2τxy σx σy θp 1 2 atan 2τxy σx σy θp θp 90deg θp 2952 deg θp 6048 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σx σx σy 2 σx σy 2 cos 2θp τxy sin 2θp σx 26493 MPa Therefore θp1 θp θp1 2952 deg Ans θp2 θp θp2 6048 deg Ans b τmax σx σy 2 2 τxy 2 τmax 17493 MPa Ans σavg σx σy 2 σavg 9000 MPa Ans Orientation of Maximum Inplane Shear Stress tan 2θs σx σy 2τxy θs 1 2 atan σx σy 2τxy θs θs 90deg θs 1548 deg θs 7452 deg By observation in order to preserve equilibrium along AB τmax has to act in the direction shown in the figure Problem 915 The state of stress at a point is shown on the element Determine a the principal stresses and b the maximum inplane shear stress and average normal stress at the point Specify the orientation of the element in each case Given σx 30 MPa σy 0MPa τxy 12 MPa Solution a Principal Stress σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 421 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 3421 MPa Ans Orientation of Principal Stress tan 2θp 2τxy σx σy θp 1 2 atan 2τxy σx σy θp θp 90deg θp 1933 deg θp 7067 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σx σx σy 2 σx σy 2 cos 2θp τxy sin 2θp σx 3421 MPa Therefore θp1 θp θp1 7067 deg Ans θp2 θp θp2 1933 deg Ans b τmax σx σy 2 2 τxy 2 τmax 1921 MPa Ans σavg σx σy 2 σavg 1500 MPa Ans Orientation of Maximum Inplane Shear Stress tan 2θs σx σy 2τxy θs 1 2 atan σx σy 2τxy θs θs 90deg θs 2567 deg θs 6433 deg By observation in order to preserve equilibrium along AB τmax has to act in the direction shown in the figure Problem 916 The state of stress at a point is shown on the element Determine a the principal stresses and b the maximum inplane shear stress and average normal stress at the point Specify the orientation of the element in each case Given σx 200 MPa σy 250MPa τxy 175MPa Solution a Principal Stress σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 31004 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 26004 MPa Ans Orientation of Principal Stress tan 2θp 2τxy σx σy θp 1 2 atan 2τxy σx σy θp θp 90deg θp 1894 deg θp 7106 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σx σx σy 2 σx σy 2 cos 2θp τxy sin 2θp σx 26004 MPa Therefore θp1 θp θp1 7106 deg Ans θp2 θp θp2 1894 deg Ans b τmax σx σy 2 2 τxy 2 τmax 28504 MPa Ans σavg σx σy 2 σavg 2500 MPa Ans Orientation of Maximum Inplane Shear Stress tan 2θs σx σy 2τxy θs 1 2 atan σx σy 2τxy θs θs 90deg θs 2606 deg θs 6394 deg By observation in order to preserve equilibrium along AB τmax has to act in the direction shown in the figure Problem 917 A point on a thin plate is subjected to the two successive states of stress shown Determine the resultant state of stress represented on the element oriented as shown on the right Given a σxa 200 MPa τxya 0MPa σya 350 MPa θa 30 deg a σxb 0 τxyb 58MPa σyb 0 θb 25deg Solution Stress Transformation Equations Applying Eqs 91 92 and 93 of the text For element a σxa σxa σya 2 σxa σya 2 cos 2θa τxya sin 2θa σxa 23750 MPa σya σxa σya 2 σxa σya 2 cos 2θa τxya sin 2θa σya 31250 MPa τxya σxa σya 2 sin 2θa τxya cos 2θa τxya 6495 MPa For element b σxb σxb σyb 2 σxb σyb 2 cos 2θb τxyb sin 2θb σxb 4443 MPa σyb σxb σyb 2 σxb σyb 2 cos 2θb τxyb sin 2θb σyb 4443 MPa τxyb σxb σyb 2 sin 2θb τxyb cos 2θb τxyb 3728 MPa Combining the stress componenets of two elements yields σx σxa σxb σx 1931 MPa Ans σy σya σyb σy 3569 MPa Ans τxy τxya τxyb τxy 1022 MPa Ans Problem 918 The steel bar has a thickness of 12 mm and is subjected to the edge loading shown Determine the principal stresses developed in the bar Given d 50mm t 12mm q 4 kN m L 500mm Solution Normal and Shear Stress In accordance with the established sign convention σx 0MPa σy 0MPa τxy q t τxy 0333 MPa Inplane Principal Stress Apply Eq 95 σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 0333 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 0333 MPa Ans Problem 919 The steel plate has a thickness of 10 mm and is subjected to the edge loading shown Determine the maximum inplane shear stress and the average normal stress developed in the steel Given ax 300mm ay 100mm t 10mm qx 30 kN m qy 40 kN m Solution Normal and Shear Stress In accordance with the established sign convention σx qx t σx 300 MPa σy qy t σy 400 MPa τxy 0 Maximum Inplane Shear Stress Apply Eq 97 τmax σx σy 2 2 τxy 2 τmax 0500 MPa Ans Average Normal Stress Apply Eq 98 σavg σx σy 2 σavg 350 MPa Ans Problem 920 The stress acting on two planes at a point is indicated Determine the shear stress on plane aa and the principal stresses at the point Given σa 80MPa σb 60MPa θ 45deg β 60deg Solution σx σb sin β τxy σb cos β Given σa σx σy 2 σx σy 2 cos 2θ τxy sin 2θ τa σx σy 2 sin 2θ τxy cos 2θ Guess σy 1MPa τa 1MPa σy τa Find σy τa σy τa 4804 196 MPa Ans Principal Stress σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 8006 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 1994 MPa Ans Problem 921 The stress acting on two planes at a point is indicated Determine the normal stress σb and the principal stresses at the point Given σa 4MPa τxy 2 MPa φbb 45deg β 60deg Solution Stress Transformation Equations Applying Eqs 93 and 91 with θ φbb 90deg σy σa sin β τxy σa cos β σx σb Given σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ τxy σx σy 2 sin 2θ τxy cos 2θ Guess σx 1MPa σx 1MPa σx σx Find σx σx σx σx 7464 7464 MPa σb σx σb 7464 MPa Ans Inplane Principal Stress Applying Eq 95 σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 829 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 264 MPa Ans Problem 922 The clamp bears down on the smooth surface at E by tightening the bolt If the tensile force in the bolt is 40 kN determine the principal stresses at points A and B and show the results on elements located at each of these points The crosssectional area at A and B is shown in the adjacent figure Given h 50mm t 30mm P 40kN a 100mm b 200mm Solution L 3a b Support Reactions ΣΜO0 E L P a b 0 E P a b L E 24kN Internal Force and Moment At Section AB ΣFx0 E V 0 V E ΣΜO0 M E a 0 M E a Section Property A h t A 1500mm2 I 1 12 t h3 I 312500mm4 QB 05 h t 025 h QB 9375mm3 QA 0 since A 0 Normal Stress σ M c I cA 05 h σA M cA I σA 19200 MPa cB 0 σB M cB I σB 0MPa Shear Stress τ V Q I t τA V QA I t τA 000 MPa τB V QB I t τB 2400 MPa Inplane Principal Stress At A σxA σA σyA 0 τxy τA Since no shear stress acts upon the element σA1 σyA σA1 0MPa Ans σA2 σxA σA2 192 MPa Ans At B σxB σB σyB 0 τxy τB 2 2 12 2 2 xy y x y x τ σ σ σ σ σ σB1 τxy σB1 24MPa Ans σB2 τxy σB2 24 MPa Ans Orientation of Principal Plane Applying Eq 94 for point B tan 2θp 2τB σxB σyB θp 1 2 90deg θp θp 90deg θp 45deg θp 45 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σxB σxB σyB 2 σxB σyB 2 cos 2θp τB sin 2θp σxB 2400 MPa Therefore θp1 θp θp1 4500 deg Ans θp2 θp θp2 4500 deg Ans Problem 923 Solve Prob 922 for points C and D Given h 50mm t 30mm P 40kN a 100mm b 200mm hD 10mm Solution L 3a b Support Reactions ΣΜE0 P 2a R L 0 R 2P a L R 16kN Internal Force and Moment At Section CD ΣFx0 R V 0 V R ΣΜO0 M R b 0 M R b Section Property A h t A 1500mm2 I 1 12 t h3 I 312500mm4 QD hD t 05 h 05 hD QD 6000mm3 QC 0 since A 0 Normal Stress σ M c I cC 05h σC M cC I σC 25600 MPa cD 05 h hD σD M cD I σD 1536 MPa Shear Stress τ V Q I t τC V QC I t τC 000 MPa τD V QD I t τD 1024 MPa Inplane Principal Stress At C σxC 0 σyC σC τxy τC Since no shear stress acts upon the element σC1 σyC σC1 256MPa Ans σC2 σxC σC2 0MPa Ans At D σxD 0 σyD σD τxy τD σD1 σxD σyD 2 σxD σyD 2 2 τD 2 σD1 0680 MPa Ans σD2 σxD σyD 2 σxD σyD 2 2 τD 2 σD2 15428 MPa Ans Orientation of Principal Plane Applying Eq 94 for point D tan 2θp 2τD σxD σyD θp 1 2 atan 2τD σxD σyD θp θp 90deg θp 3797 deg θp 86203 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σxD σxD σyD 2 σxD σyD 2 cos 2θp τD sin 2θp σxD 0680 MPa Therefore θp1 θp θp1 380 deg Ans θp2 θp θp2 8620 deg Ans Problem 924 The grains of wood in the board make an angle of 20 with the horizontal as shown Determine the normal and shear stress that act perpendicular to the grains if the board is subjected to an axial load of 250 N Unit Used kPa 1000Pa Given P 250N φ 20deg h 60mm t 25mm Solution σx P h t σx 01667 MPa σy 0 τxy 0 θ 90deg φ σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 1950 kPa Ans τxy σx σy 2 sin 2θ τxy cos 2θ τxy 5357 kPa Ans Problem 925 The wooden block will fail if the shear stress acting along the grain is 385 MPa If the normal stress σx 28 MPa determine the necessary compressive stress σy that will cause failure Given σx 28MPa τxy 0MPa θgrain 58deg τxy 385MPa Solution θ θgrain 90deg Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy σx σy 2 sin 2θ σy 2τxy sin 2θ σx σy 5767 MPa Ans Problem 926 The Tbeam is subjected to the distributed loading that is applied along its centerline Determine the principal stresses at points A and B and show the results on elements located at each of these points Given bf 150mm tf 20mm dw 150mm tw 20mm a 2m b 1m hB 50mm w 12 kN m Solution Internal Force and Moment At Section AB ΣFy0 V w a 0 V w a ΣΜA0 M w a 05a b 0 M w a 05a b Section Property D dw tf yc 05tf bf tf 05dw tf dw tw bf tf dw tw yc 5250 mm I1 1 12 bf tf 3 bf tf 05tf yc 2 I2 1 12 tw dw 3 dw tw 05dw tf yc 2 I I1 I2 I 1656250000 mm4 QA 0 since A 0 QB hB tw D yc 05 hB QB 92500mm3 Normal Stress σ M c I cA yc σA M cA I σA 15215 MPa cB D yc hB σB M cB I σB 19562 MPa Shear Stress τ V Q I t τA V QA I bf τA 000 MPa τB V QB I tw τB 6702 MPa Inplane Principal Stress At A σxA σA σyA 0 τxy τA Since no shear stress acts upon the element σA1 σxA σA1 15215 MPa Ans σA2 σyA σA2 0MPa Ans At B σxB σB σyB 0 τxy τB σB1 σxB σyB 2 σxB σyB 2 2 τB 2 σB1 0229 MPa Ans σB2 σxB σyB 2 σxB σyB 2 2 τB 2 σB2 195852 MPa Ans Orientation of Principal Plane Applying Eq 94 for point B tan 2θp 2τB σxB σyB θp 1 2 atan 2 τB σxB σyB θp θp 90deg θp 196 deg θp 8804 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σxB σxB σyB 2 σxB σyB 2 cos 2θp τB sin 2θp σxB 19494 MPa Therefore θp1 θp θp1 8804 deg Ans θp2 θp θp2 196 deg Ans Problem 927 The bent rod has a diameter of 15 mm and is subjected to the force of 600 N Determine the principal stresses and the maximum inplane shear stress that are developed at point A and point B Show the results on properly oriented elements located at these points Given do 15mm a 50mm P 06kN Solution Internal Force and Moment At Section AB ΣFx0 N P 0 N P ΣΜO0 M P a 0 M P a Section Property A π do 2 4 A 17671 mm2 I π do 4 64 I 248505 mm4 I Mc A N σ12 Normal Stress cA 05do σA N A M cA I σA 8715 MPa cB 05do σB N A M cB I σB 9394 MPa Inplane Principal Stress At A σxA σA σyA 0 τxy 0 Since no shear stress acts upon the element σA1 σyA σA1 0MPa Ans σA2 σxA σA2 8715 MPa Ans At B σxB σB σyB 0 τxy 0 Since no shear stress acts upon the element σB1 σxB σB1 9394 MPa Ans σB2 σyB σB2 0MPa Ans Maximum Inplane Shear Stress Applying Eq 97 τAmax σxA σyA 2 2 τxy 2 τAmax 436 MPa Ans τBmax σxB σyB 2 2 τxy 2 τBmax 470 MPa Ans Orientation of Plane for Maximum Inplane Shear Stress Applying Eq 96 tan2θ SA θsA 45deg Ans tan 2θsA σxA σyA 2τxy Ans θsA θsA 90deg θsA 45 deg Ans tan2θ SB tan 2θsB σxB σyB 2τxy θsB 45 deg Ans Ans θsB θsB 90deg θsB 45deg Ans By observation in order to preserve equilibrium along AB τmax has to act in the direction shown in the figure Average Normal Stress Applying Eq 98 σavgA σxA σyA 2 σavgA 4357 MPa σavgB σxB σyB 2 σavgB 4697 MPa Problem 928 The simply supported beam is subjected to the traction stress τ0 on its top surface Determine the principal stresses at points A and B Problem 929 The beam has a rectangular cross section and is subjected to the loadings shown Determine the principal stresses and the maximum inplane shear stress that are developed at point A and point B These points are just to the left of the 10kN load Show the results on properly oriented elements located at these points Given b 150mm d 375mm F 10kN P 5kN L 12m Solution Support Reactions By symmetry R1R R2 R ΣFy0 2R F 0 R 05F ΣFx0 H1 P 0 H1 P Internal Force and Moment At Section AB ΣFx0 H1 N 0 N H1 ΣFy0 R V 0 V R ΣΜO0 M R 05L 0 M 05R L Section Property A b d I 1 12 b d3 QA 0 QB 0 since A 0 Normal Stress σ N A M c I cA 05 d σA N A M cA I σA 0942 MPa cB 05d σB N A M cB I σB 0764 MPa Shear Stress Since QA QB 0 τA 0 τB 0 Inplane Principal Stress At A σxA σA σyA 0 τxy 0 Since no shear stress acts upon the element σA1 σyA σA1 0MPa Ans σA2 σxA σA2 0942 MPa Ans At B σxB σB σyB 0 τxy 0 Since no shear stress acts upon the element σB1 σxB σB1 0764 MPa Ans σB2 σyB σB2 0MPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmaxA σxA σyA 2 2 τxy 2 τmaxA 0471 MPa Ans τmaxB σxB σyB 2 2 τxy 2 τmaxB 0382 MPa Ans Orientation of Plane for Maximum Inplane Shear Stress Applying Eq 96 tan2θ SA θsA 45deg Ans tan 2θsA σxA σyA 2τxy Ans θsA θsA 90deg θsA 45 deg Ans tan2θ SB tan 2θsB σxB σyB 2τxy θsB 45 deg Ans Ans θsB θsB 90deg θsB 45deg Ans By observation in order to preserve equilibrium along AB τmax has to act in the direction shown in the figure Average Normal Stress Applying Eq 98 σavgA σxA σyA 2 σavgA 0471 MPa Ans σavgB σxB σyB 2 σavgB 0382 MPa Ans Problem 930 The wideflange beam is subjected to the loading shown Determine the principal stress in the beam at point A and at point B These points are located at the top and bottom of the web respectively Although it is not very accurate use the shear formula to compute the shear stress Given bf 200mm tf 10mm tw 10mm dw 200mm P 25kN θ 30deg a 3m w 8 kN m Solution Internal Force and Moment At Section AB ΣFx0 P cos θ N 0 N P cos θ ΣFy0 V P sin θ w a 0 V P sin θ w a ΣΜO0 M P sin θ a w a 05a 0 M P a sin θ 05w a2 Section Property D dw 2tf A bf D bf tw dw A 6000mm2 I 1 12 bf D3 bf tw dw 3 I 5080 10 6 m4 QA bf tf D 2 tf 2 QA 210000mm3 QB QA Normal Stress σ N A M c I cA 05dw σA N A M cA I σA 148293 MPa cB 05 dw σB N A M cB I σB 141077 MPa Shear Stress τ V Q I t τA V QA I tw τA 1509 MPa τB V QB I tw τB 1509 MPa Inplane Principal Stress At A σxA σA σyA 0 τxy τA σA1 σxA σyA 2 σxA σyA 2 2 τA 2 σA1 1498 MPa Ans σA2 σxA σyA 2 σxA σyA 2 2 τA 2 σA2 152 MPa Ans At B σxB σB σyB 0 τxy τB σB1 σxB σyB 2 σxB σyB 2 2 τB 2 σB1 160 MPa Ans σB2 σxB σyB 2 σxB σyB 2 2 τB 2 σB2 1427 MPa Ans Problem 931 The shaft has a diameter d and is subjected to the loadings shown Determine the principal stresses and the maximum inplane shear stress that is developed anywhere on the surface of the shaft Problem 932 A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown Determine the shear stress acting along the seam which is at 30 from the vertical when the tube is subjected to an axial force of 10 N The paper is 1 mm thick and the tube has an outer diameter of 30 mm Unit used kPa 1000Pa Given do 30mm t 1mm θ 30deg P 10N Solution Section Property di do 2t A π 4 do 2 di 2 A 9111 mm2 Normal Stress σx P A σx 10976 kPa σy 0 τxy 0 Shear stress along the seam τxy σx σy 2 sin 2θ τxy cos 2θ τxy 4753 kPa Ans Problem 933 Solve Prob 932 for the normal stress acting perpendicular to the seam Unit used kPa 1000Pa Given do 30mm t 1mm θ 30deg P 10N Solution Section Property di do 2t A π 4 do 2 di 2 A 9111 mm2 Normal Stress σx P A σx 10976 kPa σy 0 τxy 0 Normal stress perpendicular to the seam σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 8232 kPa Ans Problem 934 The shaft has a diameter d and is subjected to the loadings shown Determine the principal stresses and the maximum inplane shear stress that is developed at point A The bearings only support vertical reactions Problem 935 The drill pipe has an outer diameter of 75 mm a wall thickness of 6 mm and a weight of 08 kNm If it is subjected to a torque and axial load as shown determine a the principal stresses and b the maximum inplane shear stress at a point on its surface at section a Given do 75mm t 6mm L 6m P 75kN Mx 12kN m w 08 kN m Solution Internal Force and Moment At section a ΣFx0 N P w L 0 N P w L ΣΜx0 T Mx 0 T Mx Section Property di do 2t A π 4 do 2 di 2 J π 32 do 4 di 4 Normal Stress σ N A σ 9457 MPa Shear Stress c 05do τ T c J τ 28850 MPa a Inplane Principal Stresses σx 0 σy σ τxy τ for any point on the shafts surface Applying Eq 95 σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 2451 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 3396 MPa Ans b Maximum Inplane Shear Stress Applying Eq 97 τmax σx σy 2 2 τxy 2 τmax 2924 MPa Ans Problem 936 The internal loadings at a section of the beam are shown Determine the principal stresses at point A Also compute the maximum inplane shear stress at this point Given bf 200mm tf 50mm tw 50mm dw 200mm Px 500 kN My 30 kN m Py 800 kN Mz 40kN m Solution Section Property D dw 2tf A bf D bf tw dw A 30000mm2 Iz 1 12 bf D3 bf tw dw 3 Iz 35000 10 6 m4 Iy 1 12 2tf bf 3 dw tw 3 Iy 6875 10 6 m4 QA 0 since A 0 Normal Stress yA 05D zA 05bf σA Px A Mz yA Iz My zA Iy σA 77446 MPa Shear Stress Since QA 0 τA 0 Inplane Principal Stress σx σA σy 0 τxy 0 Since no shear stress acts upon the element σ1 σy σ1 0MPa Ans σ2 σx σ2 7745 MPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmax σx σy 2 2 τxy 2 τmax 3872 MPa Ans Problem 937 Solve Prob 936 for point B Given bf 200mm tf 50mm tw 50mm dw 200mm Px 500 kN My 30 kN m Py 800 kN Mz 40kN m Solution Section Property D dw 2tf A bf D bf tw dw A 30000mm2 Iz 1 12 bf D3 bf tw dw 3 Iz 35000 10 6 m4 Iy 1 12 2tf bf 3 dw tw 3 Iy 6875 10 6 m4 QB 0 since A 0 Normal Stress yB 05 D zB 05 bf σB Px A Mz yB Iz My zB Iy σB 44113 MPa Shear Stress Since QB 0 τB 0 Inplane Principal Stress σx σB σy 0 τxy 0 Since no shear stress acts upon the element σ1 σx σ1 44113 MPa Ans σ2 σy σ2 000 MPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmax σx σy 2 2 τxy 2 τmax 2206 MPa Ans Problem 938 Solve Prob 936 for point C located in the center on the bottom of the web Given bf 200mm tf 50mm tw 50mm dw 200mm Px 500 kN My 30 kN m Py 800 kN Mz 40kN m Solution Section Property D dw 2tf A bf D bf tw dw A 30000mm2 Iz 1 12 bf D3 bf tw dw 3 Iz 35000 10 6 m4 Iy 1 12 2tf bf 3 dw tw 3 Iy 6875 10 6 m4 QC bf tf D 2 tf 2 QC 1250000mm3 Normal Stress yC 05 dw zC 0 σC Px A Mz yC Iz My zC Iy σC 5238 MPa Shear Stress τ V Q I t τC Py QC Iz tw τC 5714 MPa Inplane Principal Stress σx σC σy 0 τxy τC σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 5458 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 5982 MPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmax σx σy 2 2 τxy 2 τmax 5720 MPa Ans Problem 939 The wideflange beam is subjected to the 50kN force Determine the principal stresses in the beam at point A located on the web at the bottom of the upper flange Although it is not very accurate use the shear formula to calculate the shear stress Given bf 200mm tf 12mm tw 10mm dw 250mm P 50kN a 3m Solution Internal Force and Moment At Section AB ΣFy0 V P 0 V P ΣΜO0 M P a 0 M P a Section Property D dw 2tf A bf D bf tw dw A 7300mm2 I 1 12 bf D3 bf tw dw 3 I 9545 10 6 m4 QA bf tf D 2 tf 2 QA 314400mm3 Normal Stress σ M c I cA 05dw σA M cA I σA 196435 MPa Shear Stress τ V Q I t τA V QA I tw τA 1647 MPa Inplane Principal Stress σx σA σy 0 τxy τA σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 19781 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 137 MPa Ans Problem 940 Solve Prob 939 for point B located on the web at the top of the bottom flange Given bf 200mm tf 12mm tw 10mm dw 250mm P 50kN a 3m Solution Internal Force and Moment At Section AB ΣFy0 V P 0 V P ΣΜO0 M P a 0 M P a Section Property D dw 2tf A bf D bf tw dw A 7300mm2 I 1 12 bf D3 bf tw dw 3 I 9545 10 6 m4 QB bf tf D 2 tf 2 QB 314400mm3 Normal Stress σ M c I cB 05 dw σB M cB I σB 196435 MPa Shear Stress τ V Q I t τB V QB I tw τB 1647 MPa Inplane Principal Stress σx σB σy 0 τxy τB σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 137 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 19781 MPa Ans Problem 941 The bolt is fixed to its support at C If a force of 90 N is applied to the wrench to tighten it determine the principal stresses developed in the bolt shank at point A Represent the results on an element located at this point The shank has a diameter of 6 mm Given do 6mm a 150mm L 50mm P 90N Solution Internal Force and Moment At section AB Mx P L Ty P a Section Property I π 64 do 4 J π 32 do 4 cA 05do Normal Stress σA Mx cA I σA 21221 MPa Shear Stress τA Ty cA J τA 31831 MPa Inplane Principal Stresses Applying Eq 95 σx σA σy 0 τxy τA σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 44163 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 22942 MPa Ans Orientation of Principal Stress tan 2θp 2τxy σx σy θp 1 2 atan 2τxy σx σy θp θp 90deg θp 35783 deg θp 54217 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σx σx σy 2 σx σy 2 cos 2θp τxy sin 2θp σx 44163 MPa Therefore θp1 θp θp1 3578 deg Ans θp2 θp θp2 5422 deg Ans Problem 942 Solve Prob 941 for point B Given do 6mm a 150mm L 50mm P 90N Solution Internal Force and Moment At section AB Mx P L Ty P a Vz P Section Property ro 05do I π 64 do 4 J π 32 do 4 QB 4ro 3π π ro 2 2 Normal Stress cBσ 0 σB Mx cBσ I σB 0MPa Shear Stress bB do cBτ ro τB Vz QB I bB Ty cBτ J τB 31407 MPa Inplane Principal Stresses Applying Eq 95 σx σB σy 0 τxy τB σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 31407 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 31407 MPa Ans Orientation of Principal Stress tan2θ P tan 2θp 2τxy σx σy θp1 45deg Ans θp2 θp1 90deg θp2 45 deg Ans Problem 943 The beam has a rectangular cross section and is subjected to the loadings shown Determine the principal stresses that are developed at point A and point B which are located just to the left of the 20kN load Show the results on elements located at these points Given b 100mm d 200mm F 20kN P 10kN L 4m Solution Support Reactions By symmetry R1R R2 R ΣFy0 2R F 0 R 05F ΣFx0 H1 P 0 H1 P Internal Force and Moment At Section AB ΣFx0 H1 N 0 N H1 ΣFy0 R V 0 V R ΣΜO0 M R 05L 0 M 05R L Section Property A b d I 1 12 b d3 QA 0 since A 0 QB b 05d 025d Normal Stress σ N A M c I cA 05 d σA N A M cA I σA 305 MPa cB 0 σB N A M cB I σB 05 MPa Shear Stress Since QA 0 τA 0 τB V QB I b τB 075 MPa Inplane Principal Stress Applying Eq 95 At A σxA σA σyA 0 τxy 0 Since no shear stress acts upon the element σA1 σyA σA1 0MPa Ans σA2 σxA σA2 3050 MPa Ans At B σxB σB σyB 0 τxy τB σB1 σxB σyB 2 σxB σyB 2 2 τB 2 σB1 0541 MPa Ans σB2 σxB σyB 2 σxB σyB 2 2 τB 2 σB2 1041 MPa Ans Orientation of Principal Plane Applying Eq 94 for point B tan 2θp 2τB σxB σyB θp 1 2 atan 2τB σxB σyB θp 35783 deg θp θp 90deg θp 54217 deg Use Eq 91 to determine the principal plane of σ1 and σ2 σxB σxB σyB 2 σxB σyB 2 cos 2θp τB sin 2θp σxB 104 MPa Therefore θp1 θp θp1 5422 deg Ans θp2 θp θp2 3578 deg Ans Problem 944 The solid propeller shaft on a ship extends outward from the hull During operation it turns at ω 15 rads when the engine develops 900 kW of power This causes a thrust of F 123 MN on the shaft If the shaft has an outer diameter of 250 mm determine the principal stresses at any point located on the surface of the shaft Given do 250mm L 075m F 1230kN P 900kW ω 15 rad s Solution Internal Force and Moment As shown on FBD To P ω To 6000 kN m N F Section Property A π do 2 4 A 4908739 mm2 J π do 4 32 J 38349519697 mm4 Normal Stress σa N A σa 2506 MPa Shear Stress cmax 05 do τo To cmax J τo 1956 MPa Inplane Principal Stresses Applying Eq 95 σx σa σy 0 τxy τo σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 1070 MPa Ans σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 3575 MPa Ans Problem 945 The solid propeller shaft on a ship extends outward from the hull During operation it turns at ω 15 rads when the engine develops 900 kW of power This causes a thrust of F 123 MN on the shaft If the shaft has a diameter of 250 mm determine the maximum inplane shear stress at any point located on the surface of the shaft Given do 250mm L 075m F 1230kN P 900kW ω 15 rad s Solution Internal Force and Moment As shown on FBD To P ω To 6000 kN m N F Section Property A π do 2 4 A 4908739 mm2 J π do 4 32 J 38349519697 mm4 Normal Stress σa N A σa 2506 MPa Shear Stress cmax 05 do τo To cmax J τo 1956 MPa Maximum Inplane Shear Stress Applying Eq 97 σx σa σy 0 τxy τo τmax σx σy 2 2 τxy 2 τmax 232 MPa Ans L 250mm Given do 75mm di 68mm At section AB P 100N a 300mm Solution Internal Force and Moment Ans Tx P a My P L Vz P Section Property ro 05do ri 05di I π 64 do 4 di 4 J π 32 do 4 di 4 QAz 4ro 3π π ro 2 2 4ri 3π π ri 2 2 Normal Stress cAσ 0 σA My cAσ I σA 0MPa Shear Stress bA do di cAτ ro τA Vz QAz I bA Tx cAτ J τA 0863 MPa Inplane Principal Stresses Applying Eq 95 σx σA σz 0 τxz τA σ1 σx σz 2 σx σz 2 2 τxz 2 σ1 0863 MPa Ans σ2 σx σz 2 σx σz 2 2 τxz 2 σ2 0863 MPa Problem 946 The steel pipe has an inner diameter of 68mm and an outer diameter of 75 mm If it is fixed at C and subjected to the horizontal 100N force acting on the handle of the pipe wrench at its end determine the principal stresses in the pipe at point A which is located on the surface of the pipe L 250mm Given do 75mm di 68mm At section AB P 100N a 300mm Solution Internal Force and Moment Ans Tx P a My P L Vz P Section Property ro 05do ri 05di I π 64 do 4 di 4 J π 32 do 4 di 4 QBz 0 Since A0 Normal Stress cBσ ro σB My cBσ I σB 1862 MPa Shear Stress cBτ ro τB Tx cBτ J τB 1117 MPa Inplane Principal Stresses Applying Eq 95 σx σB σz 0 τxz τB σ1 σx σz 2 σx σz 2 2 τxz 2 σ1 2385 MPa Ans σ2 σx σz 2 σx σz 2 2 τxz 2 σ2 0523 MPa Problem 947 Solve Prob 946 for point B which is located on the surface of the pipe Problem 948 The cantilevered beam is subjected to the load at its end Determine the principal stresses in the beam at points A and B Given b 120mm h 150mm P 15kN L 12m yA 45mm zA 60mm yθ 4 5 zθ 3 5 yB 75mm zB 20 mm Solution Internal Force and Moment At Section AB ΣFz0 Vz P zθ 0 Vz P zθ ΣFy0 Vy P yθ 0 Vy P yθ Mz P yθ L My P zθ L Section Property Iz 1 12 b h3 Iz 3375 10 6 m4 Iy 1 12 h b3 Iy 2160 10 6 m4 QAy b 05h yA yA 05 05h yA QAy 216000mm3 QBz h 05b zB zB 05 05b zB QBz 240000mm3 QAz 0 since A 0 QBy 0 since A 0 Normal Stress σA Mz yA Iz My zA Iy σA 108 MPa σB Mz yB Iz My zB Iy σB 420 MPa Shear Stress τ V Q I t τA Vy QAy Iz b τA 0640 MPa τB Vz QBz Iy h τB 0667 MPa Inplane Principal Stress Applying Eq 95 At A σxA σA σyA 0 τxy 0 σA1 σxA σyA 2 σxA σyA 2 2 τA 2 σA1 00378 MPa Ans σA2 σxA σyA 2 σxA σyA 2 2 τA 2 σA2 1084 MPa Ans At B σxB σB σzB 0 τxz τB σB1 σxB σzB 2 σxB σzB 2 2 τB 2 σB1 4201 MPa Ans σB2 σxB σzB 2 σxB σzB 2 2 τB 2 σB2 00106 MPa Ans Problem 949 The box beam is subjected to the loading shown Determine the principal stresses in the beam at points A and B Given bo 200mm bi 150mm do 200mm di 150mm L1 09m L2 15m P1 4kN P2 6kN Solution L L1 2L2 Support Reactions Given ΣFy0 R1 R2 P1 P2 0 ΣΜR20 Ans Guess R1 1kN R2 1kN R1 R2 Find R1 R2 R1 R2 82 18 kN Internal Force and Moment At section AB N 0 V P1 R1 M P1 L1 05L2 R1 05L2 Section Property I 1 12 bo do 3 bi di 3 QA 0 QB 0 Since A0 For point A τA 0 cA 05 do σA M cA I σA 0494 MPa σ1 σA σ1 0494 MPa Ans σ2 0 σ2 0000 MPa Ans For point B τB 0 cB 05 di σB M cB I σB 037 MPa σ1 0 σ1 0000 MPa Ans σ2 σB σ2 0370 MPa P1 L R1 L L1 P2 L2 0 Problem 950 A bar has a circular cross section with a diameter of 25 mm It is subjected to a torque and a bending moment At the point of maximum bending stress the principal stresses are 140 MPa and 70 MPa Determine the torque and the bending moment Given do 6mm σ1 140MPa σ2 70 MPa Solution Inplane Principal Stresses Applying Eq 95 Given σy 0 σ1 σx σy 2 σx σy 2 2 τxy 2 1 σ2 σx σy 2 σx σy 2 2 τxy 2 2 Solving Eqs 1 and 2 Guess σx 2MPa τxy 1MPa σx τxy Find σx τxy σx τxy 7000 9899 MPa Section Property ro 05do I π 64 do 4 J π 32 do 4 Normal Stress σ M c I c ro M σx I c M 1484 N m Ans Shear Stress τ T c J c ro T τxy J c T 4199 N m Ans Problem 951 The internal loadings at a section of the beam consist of an axial force of 500 N a shear force of 800 N and two moment components of 30 Nm and 40 Nm Determine the principal stresses at point A Also calculate the maximum inplane shear stress at this point Unit Used kPa 1000Pa Given b 100mm h 200mm Px 05kN yA 100mm zA 0mm Py 08kN My 004 kN m Mz 003kN m Solution Section Property A b h A 20000mm2 Iz 1 12 b h3 Iz 6667 10 6 m4 Iy 1 12 h b3 Iy 1667 10 6 m4 QAy 0 since A 0 Normal Stress σA Px A Mz yA Iz My zA Iy σA 200 kPa Shear Stress Since QA 0 τA 0 Inplane Principal Stress σx σA σy 0 τxy 0 Since no shear stress acts upon the element σ1 σy σ1 0kPa Ans σ2 σx σ2 20 kPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmax σx σy 2 2 τxy 2 τmax 10kPa Ans Problem 952 The internal loadings at a section of the beam consist of an axial force of 500 N a shear force of 800 N and two moment components of 30 Nm and 40 Nm Determine the principal stresses at point B Also calculate the maximum inplane shear stress at this point Unit Used kPa 1000Pa Given b 100mm h 200mm Px 05kN yB 0mm zB 50 mm Py 08kN My 004 kN m Mz 003kN m Solution Section Property A b h A 20000mm2 Iz 1 12 b h3 Iz 6667 10 6 m4 Iy 1 12 h b3 Iy 1667 10 6 m4 QBy b 05h 025h Normal Stress σB Px A Mz yB Iz My zB Iy σB 1450 kPa Shear Stress τB Py QBy Iz b τB 6000 kPa Inplane Principal Stress σxB σB σyB 0 τxy τB σB1 σxB σyB 2 σxB σyB 2 2 τB 2 σB1 1666 kPa Ans σB2 σxB σyB 2 σxB σyB 2 2 τB 2 σB2 2161 kPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmax σxB σyB 2 2 τxy 2 τmax 9411 kPa Ans Problem 953 The internal loadings at a section of the beam consist of an axial force of 500 N a shear force of 800 N and two moment components of 30 Nm and 40 Nm Determine the principal stresses at point C Also calculate the maximum inplane shear stress at this point Unit Used kPa 1000Pa Given b 100mm h 200mm Px 05kN yC 50 mm zC 0mm Py 08kN My 004 kN m Mz 003kN m Solution Section Property A b h A 20000mm2 Iz 1 12 b h3 Iz 6667 10 6 m4 Iy 1 12 h b3 Iy 1667 10 6 m4 QCy b 05h yC yC 05 05h yC QCy 375000mm3 Normal Stress σC Px A Mz yC Iz My zC Iy σC 475 kPa Shear Stress τC Py QCy Iz b τC 4500 kPa Inplane Principal Stress σxC σC σyC 0 τxy τC σC1 σxC σyC 2 σxC σyC 2 2 τC 2 σC1 7463 kPa Ans σC2 σxC σyC 2 σxC σyC 2 2 τC 2 σC2 2713 kPa Ans Maximum Inplane Shear Stress Applying Eq 97 τmax σxC σyC 2 2 τxy 2 τmax 5088 kPa Ans Problem 954 The beam has a rectangular cross section and is subjected to the loads shown Write a computer program that can be used to determine the principal stresses at points A B C and D Show an application of the program using the values h 300 mm b 200 mm Nx 2 kN Vy 15 kN Vz 0 My 0 and Mz 225 kNm Problem 955 The member has a rectangular cross section and is subjected to the loading shown Write a computer program that can be used to determine the principal stresses at points A B and C Show an application of the program using the values b 150 mm h 200 mm P 15 kN x 75 mm z 50 mm Vx 300 N and Vz 600 N Problem 956 Solve Prob 94 using Mohrs circle Given σx 0650 MPa σy 0400MPa φ 60deg τxy 0MPa Solution θ 90deg φ θ 3000 deg Center σc σx σy 2 σc 0125 MPa Radius R σx σc R 0525 MPa Coordinates A σx 0 B σy 0 C σc 0 Stresses σx σc R cos 2θ σx 0387 MPa Ans τxy R sin 2θ τxy 0455 MPa Ans Problem 957 Solve Prob 92 using Mohrs circle Given σx 5MPa σy 3MPa τxy 8MPa φ 40deg Solution Center σc σx σy 2 σc 4MPa Radius R σx σc 2 τxy 2 R 8062 MPa Angles θ 90deg φ θ 130deg φ atan τxy σx σc φ 82875 deg α 180deg 2θ φ α 2875 deg Stresses σx σc R cos α σx 4052 MPa Ans τxy R sin α τxy 0404 MPa Ans Problem 958 Solve Prob 93 using Mohrs circle Given σx 0350MPa σy 0200 MPa φ 50deg τxy 0MPa Solution θ 90deg φ θ 140deg Center σc σx σy 2 σc 0075 MPa Radius R σx σc 2 τxy 2 R 0275 MPa Coordinates A σx 0 C σc 0 Angles α 360deg 2θ α 80deg Stresses represented by coordinates of point P σx σc R cos α σx 0123 MPa Ans τxy R sin α τxy 0271 MPa Ans Problem 959 Solve Prob 910 using Mohrs circle Given σx 0MPa σy 0300 MPa θ 30deg τxy 0950MPa Solution Center σc σx σy 2 σc 015 MPa Radius R σx σc 2 τxy 2 R 0962 MPa Angles φ atan τxy σx σc φ 81027 deg α 2θ φ α 21027 deg Stresses σx σc R cos α σx 0748 MPa Ans σy σc R cos α σy 1048 MPa Ans τxy R sin α τxy 0345 MPa Ans Problem 960 Solve Prob 96 using Mohrs circle Given σx 90MPa σy 50MPa τxy 35 MPa φ 60deg Solution Center σc σx σy 2 σc 70MPa Radius R σx σc 2 τxy 2 R 40311 MPa Angles θ 90deg φ φ atan τxy σx σc φ 60255 deg α 180deg 2θ φ α 59745 deg Stresses σx σc R cos α σx 49689 MPa Ans τxy R sin α τxy 34821 MPa Ans Problem 961 Solve Prob 911 using Mohrs circle Given σx 0300MPa σy 0MPa θ 60 deg τxy 0120MPa Solution Center σc σx σy 2 σc 015 MPa Radius R σx σc 2 τxy 2 R 01921 MPa Coordinates A σx τxy C σc 0 Angles φ atan τxy σx σc φ 38660 deg α 180deg 2θ φ α 21340 deg Stresses represented by coordinates of points P and Q σx σc R cos α σx 00289 MPa Ans σy σc R cos α σy 0329 MPa Ans τxy R sin α τxy 00699 MPa Ans Problem 962 Solve Prob 913 using Mohrs circle Given σx 45MPa σy 60 MPa τxy 30MPa Solution Center σc σx σy 2 σc 75 MPa Radius R σx σc 2 τxy 2 R 60467 MPa Coordinates A σx τxy C σc 0 Stresses σ1 σc R σ1 5297 MPa Ans σ2 σc R σ2 6797 MPa Ans τmax R τmax 6047 MPa Ans σavg σc σavg 75 MPa Ans Angles θp1 1 2 atan τxy σx σc θp1 14872 deg Counterclockwise Ans 2θs1 90deg 2θp1 θs1 45deg θp1 θs1 30128 deg Clockwise Ans Problem 963 Solve Prob 914 using Mohrs circle Given σx 180MPa σy 0MPa τxy 150 MPa Solution Center σc σx σy 2 σc 90MPa Radius R σx σc 2 τxy 2 R 174929 MPa Coordinates A σx τxy B σy τxy C σc 0 Stresses σ1 σc R σ1 26493 MPa Ans σ2 σc R σ2 8493 MPa Ans τmax R τmax 17493 MPa Ans σavg σc σavg 90MPa Ans Angles θp 1 2 atan τxy σx σc θp 29518 deg Clockwise Ans 2θs 90deg 2θp θs 45deg θp θs 15482 deg Counterclockwise Ans Problem 964 Solve Prob 916 using Mohrs circle Given σx 200 MPa σy 250MPa τxy 175MPa Solution Center σc σx σy 2 σc 25MPa Radius R σx σc 2 τxy 2 R 285044 MPa Coordinates A σx τxy B σy τxy C σc 0 Stresses σ1 σc R σ1 31004 MPa Ans σ2 σc R σ2 26004 MPa Ans τmax R τmax 28504 MPa Ans σavg σc σavg 25MPa Ans Angles θp 1 2 atan τxy σx σc θp 18937 deg Clockwise Ans 2θs 90deg 2θp θs 45deg θp θs 26063 deg Counterclockwise Ans Problem 965 Solve Prob 915 using Mohrs circle Given σx 30 MPa σy 0MPa τxy 12 MPa Solution Center σc σx σy 2 σc 15 MPa Radius R σx σc 2 τxy 2 R 19209 MPa Coordinates A σx τxy C σc 0 Stresses σ1 σc R σ1 421 MPa Ans σ2 σc R σ2 3421 MPa Ans τmax R τmax 1921 MPa Ans σavg σc σavg 15 MPa Ans Angles θp2 1 2 atan τxy σx σc θp2 19330 deg Countrclockwise Ans 2θs2 2θp2 90deg θs2 θp2 45deg θs2 64330 deg Countrclockwise Ans Problem 966 Determine the equivalent state of stress if an element is oriented 20 clockwise from the element shown Show the result on the element Given σx 3MPa σy 2 MPa θ 20 deg τxy 4 MPa Solution Center σc σx σy 2 σc 05 MPa Radius R σx σc 2 τxy 2 R 4717 MPa Coordinates A σx τxy C σc 0 Angles φ atan τxy σx σc φ 57995 deg α φ 2θ α 17995 deg Stresses represented by coordinates of points P and Q σx σc R cos α σx 4986 MPa Ans σy σc R cos α σy 3986 MPa Ans τxy R sin α τxy 1457 MPa Ans Problem 967 Determine the equivalent state of stress if an element is oriented 60 counterclockwise from the element shown Given σx 0750MPa σy 0800 MPa θ 60deg τxy 0450MPa Solution Center σc σx σy 2 σc 0025 MPa Radius R σx σc 2 τxy 2 R 08962 MPa Coordinates A σx τxy B σy τxy C σc 0 Angles φ atan τxy σx σc φ 30141 deg α 2θ φ α 89859 deg Stresses σx σc R cos α σx 00228 MPa Ans σy σc R cos α σy 00272 MPa Ans τxy R sin α τxy 0896 MPa Ans Problem 968 Determine the equivalent state of stress if an element is oriented 30 clockwise from the element shown Given σx 350MPa σy 230MPa θ 30deg τxy 480 MPa Solution Center σc σx σy 2 σc 290MPa Radius R σx σc 2 τxy 2 R 4837355 MPa Coordinates A σx τxy B σy τxy C σc 0 Angles φ atan τxy σx σc φ 82875 deg α 2 θ φ α 22875 deg Stresses σx σc R cos α σx 7356922 MPa Ans σy σc R cos α σy 1556922 MPa Ans τxy R sin α τxy 188038 MPa Ans Problem 969 Determine the equivalent state of stress if an element is oriented 30 clockwise from the element shown Show the result on the element Given σx 7MPa σy 8MPa θ 30 deg τxy 15MPa Solution Center σc σx σy 2 σc 75 MPa Radius R σx σc 2 τxy 2 R 150083 MPa Coordinates A σx τxy C σc 0 Angles φ atan τxy σx σc φ 88091 deg α 180deg 2 θ φ α 31909 deg Stresses represented by coordinates of points P and Q σx σc R cos α σx 5240 MPa Ans σy σc R cos α σy 20240 MPa Ans τxy R sin α τxy 7933 MPa Ans Problem 970 Determine a the principal stress and b the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case Given σx 350MPa σy 200 MPa τxy 500MPa Solution Center σc σx σy 2 σc 75MPa Radius R σx σc 2 τxy 2 R 570636 MPa Coordinates A σx τxy C σc 0 a Principal Stresses σ1 σc R σ1 64564 MPa Ans σ2 σc R σ2 49564 MPa Ans Angles θp1 1 2 atan τxy σx σc θp1 30595 deg Counterclockwise Ans b Maximum Inplane Shear Stress represented by coordinates of point E τmax R τmax 57064 MPa Ans σavg σc σavg 75MPa Ans 2θs 90deg 2 θp1 θs 45deg θp1 θs 14405 deg Clockwise Ans Problem 971 Determine a the principal stress and b the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case Given σx 10MPa σy 80MPa τxy 60 MPa Solution Center σc σx σy 2 σc 45MPa Radius R σx σc 2 τxy 2 R 69462 MPa Coordinates A σx τxy C σc 0 a Principal Stresses σ1 σc R σ1 11446 MPa Ans σ2 σc R σ2 2446 MPa Ans Angles θp2 1 2 atan τxy σx σc θp1 90deg θp2 θp1 60128 deg Clockwise Ans b Maximum Inplane Shear Stress represented by coordinates of point E τmax R τmax 6946 MPa Ans σavg σc σavg 45MPa Ans 2θs2 90deg 2 θp2 θs2 45deg θp2 θs2 15128 deg Clockwise Ans Problem 972 Determine a the principal stress and b the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case Given σx 0MPa σy 50MPa τxy 30 MPa Solution Center σc σx σy 2 σc 25MPa Radius R σx σc 2 τxy 2 R 39051 MPa Coordinates A σx τxy B σy τxy C σc 0 a Principal Stresses σ1 σc R σ1 6405 MPa Ans σ2 σc R σ2 1405 MPa Ans Angles θp 1 2 atan τxy σx σc θp 25097 deg Clockwise b Maximum Inplane Shear Stress represented by coordinates of point E τmax R τmax 3905 MPa Ans σavg σc σavg 25MPa Ans 2θs2 90deg 2 θp θs2 45deg θp θs2 19903 deg Counterclockwise Problem 973 Determine a the principal stress and b the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case Given σx 12 MPa σy 8 MPa τxy 4MPa Solution Center σc σx σy 2 σc 10 MPa Radius R σx σc 2 τxy 2 R 4472 MPa Coordinates A σx τxy B σy τxy C σc 0 a Principal Stresses σ1 σc R σ1 553 MPa Ans σ2 σc R σ2 1447 MPa Ans Angles θp 1 2 atan τxy σx σc θp 31717 deg Clockwise b Maximum Inplane Shear Stress represented by coordinates of point E τmax R τmax 447 MPa Ans σavg σc σavg 10 MPa Ans 2θs2 90deg 2 θp θs2 45deg θp θs2 13283 deg Counterclockwise Problem 974 Determine a the principal stress and b the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case Given σx 45MPa σy 30MPa τxy 50 MPa Solution Center σc σx σy 2 σc 375 MPa Radius R σx σc 2 τxy 2 R 50559 MPa Coordinates A σx τxy B σy τxy C σc 0 a Principal Stresses σ1 σc R σ1 8806 MPa Ans σ2 σc R σ2 1306 MPa Ans Angles θp 1 2 atan τxy σx σc θp 40735 deg Clockwise b Maximum Inplane Shear Stress represented by coordinates of point E τmax R τmax 5056 MPa Ans σavg σc σavg 375 MPa Ans 2θs2 90deg 2 θp θs2 45deg θp θs2 4265 deg Counterclockwise Problem 975 The square steel plate has a thickness of 12 mm and is subjected to the edge loading shown Determine the principal stresses developed in the steel Given σx 0MPa σy 0MPa qxy 32 kN m L 100mm t 12mm Solution τxy qxy t τxy 0267 MPa Center σc σx σy 2 σc 0MPa Radius R σx σc 2 τxy 2 R 0267 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 0267 MPa Ans σ2 σc R σ2 0267 MPa Ans σ1 11560 MPa Given σx 105MPa σy 0MPa τxy 35 MPa Solution Center σc σx σy 2 σc 525 MPa Radius R σx σc 2 τxy 2 R 63097 MPa Coordinates A σx τxy C σc 0 a InplanePrincipal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R Ans Ans σ2 σc R σ2 1060 MPa Ans Angles θp 1 2 atan τxy σx σc θp 16845 deg Clockwise b Maximum Inplane Shear Stresses Represented by the coordinates of point E on the circle τmax R τmax 6310 MPa Ans 2θs 90deg 2 θp θs 45deg θp θs 28155 deg Counterclockwise Problem 976 Determine a the principal stress and b the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case Problem 977 Draw Mohrs circle that describes each of the following states of stress Given σx 30 MPa σy 30MPa τxy 0MPa Solution Cases a and b are the same Center σc σx σy 2 σc 0MPa Radius R σx σc 2 τxy 2 R 30MPa Coordinates A σx τxy C σc 0 Problem 978 Draw Mohrs circle that describes each of the following states of stress a Given σx 08MPa τxy 0MPa σy 06 MPa Solution Center σc σx σy 2 σc 01 MPa Radius R σx σc 2 τxy 2 R 07 MPa Coordinates A σx τxy C σc 0 b Given σx 0MPa τxy 0MPa σy 2 MPa Solution Center σc σx σy 2 σc 1 MPa Radius R σx σc 2 τxy 2 R 1MPa Coordinates A σx τxy C σc 0 c Given σx 0MPa τxy 20MPa σy 0MPa Solution Center σc σx σy 2 σc 0MPa Radius R σx σc 2 τxy 2 R 20MPa Coordinates A σx τxy C σc 0 Problem 979 A point on a thin plate is subjected to two successive states of stress as shown Determine the resulting state of stress with reference to an element oriented as shown on the bottom Unit used kPa 1000Pa Given σxa 50kPa σya 0 τxya 0 σyb 18 kPa σxb 0 τxyb 45 kPa θa 30deg βb 50deg Solution For element a Center σca σxa σya 2 σca 25kPa Radius Ra σxa σca 2 τxya 2 Ra 25kPa Coordinates A σxa τxya B σxa τxya C σca 0 σxa σca Ra cos 2θa σya σca Ra cos 2θa τxya Ra sin 2θa For element b θb 90deg βb Center σcb σxb σyb 2 σcb 9 kPa Radius Rb σxb σcb 2 τxyb 2 Rb 45891 kPa Coordinates A σxb τxyb B σxa τxyb C σcb 0 Angles φ atan τxyb σxb σcb φ 7869 deg αb 2θb φ αb 1310 deg σxb σcb Rb cos αb σyb σcb Rb cos αb τxyb Rb sin αb Resultants σx σxa σxb σx 7438 kPa Ans σy σya σyb σy 4238 kPa Ans τxy τxya τxyb τxy 2270 kPa Ans Problem 980 Mohrs circle for the state of stress in Fig 915a is shown in Fig 915b Show that finding the coordinates of point Pσx τxy on the circle gives the same value as the stresstransformation Eqs 91 and 92 Problem 981 The cantilevered rectangular bar is subjected to the force of 25 kN Determine the principal stresses at point A Given b 80mm d 160mm cA 40mm P 25kN L 04m bA 40 mm rv 3 5 rh 4 5 Solution Internal Force and Moment At Section AB ΣFx0 P rh N 0 N P rh ΣFy0 P rv V 0 V P rv ΣΜO0 M P rv L 0 M P rv L Section Property dA 05d cA A b d I 1 12 b d3 QA b dA 05dA cA Normal Stress σA N A M cA I σA 10352 MPa Shear Stress τA V QA I b τA 1318 MPa Construction of Mohrs Circle σx σA σy 0MPa τxy τA Center σc σx σy 2 σc 5176 MPa Radius R σx σc 2 τxy 2 R 5341 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 1052 MPa Ans σ2 σc R σ2 0165 MPa Ans Problem 982 Solve Prob 981 for the principal stresses at point B Given b 80mm d 160mm cB 25 mm P 25kN L 04m bB 25mm rv 3 5 rh 4 5 Solution Internal Force and Moment At Section AB ΣFx0 P rh N 0 N P rh ΣFy0 P rv V 0 V P rv ΣΜO0 M P rv L 0 M P rv L Section Property dB 05d cB A b d I 1 12 b d3 QB b dB 05dB cB Normal Stress σB N A M cB I σB 3931 MPa Shear Stress τB V QB I b τB 1586 MPa Construction of Mohrs Circle σx σB σy 0MPa τxy τB Center σc σx σy 2 σc 1965 MPa Radius R σx σc 2 τxy 2 R 2526 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 0560 MPa Ans σ2 σc R σ2 4491 MPa Ans Problem 983 The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at B If a man having a weight of 1500 N 150 kg stands in the center of the tread determine the principal stresses developed in the supporting truck on the cross section at point C The stairs move at constant velocity Given b 12mm d 50mm cC 0mm W 15kN hC 150mm θ 30deg vA 450mm hA 375mm hB 150mm Solution Support Reactions ΣΜA0 W hA By hB 0 By W hA hB Bx By tan θ Internal Force and Moment At Section C ΣFx0 Bx V 0 V Bx ΣFy0 By N 0 N By ΣΜO0 Bx hC M 0 M Bx hC Section Property dC 05d A b d I 1 12 b d3 QC b dC 05dC Normal Stress σC N A M cC I σC 6250 MPa Shear Stress τC V QC I b τC 5413 MPa Construction of Mohrs Circle σx 0MPa σy σC τxy τC Center σc σx σy 2 σc 3125 MPa Radius R σx σc 2 τxy 2 R 625 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 3125 MPa Ans σ2 σc R σ2 9375 MPa Ans Problem 984 The pedal crank for a bicycle has the cross section shown If it is fixed to the gear at B and does not rotate while subjected to a force of 400 N determine the principal stresses in the material on the cross section at point C Given b 75mm d 20mm cC 5mm P 04kN L 100mm Solution Internal Force and Moment At Section C ΣFy0 V P 0 V P ΣΜO0 M P L 0 M P L Section Property dC 05d cC A b d I 1 12 b d3 QC b dC cC 05dC Normal Stress σC M cC I σC 40000 MPa Shear Stress τC V QC I b τC 3000 MPa Construction of Mohrs Circle σx σC σy 0MPa τxy τC Center σc σx σy 2 σc 20MPa Radius R σx σc 2 τxy 2 R 20224 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 40224 MPa Ans σ2 σc R σ2 0224 MPa Ans Problem 985 The frame supports the distributed loading of 200 Nm Determine the normal and shear stresses at point D that act perpendicular and parallel respectively to the grains The grains at this point make an angle of 30 with the horizontal as shown Unit Used kPa 1000Pa Given b 100mm a 1m w 02 kN m h 200mm L 25m hD 75mm θ 60deg Solution Support Reactions Due to symmetry BCR ΣFy0 2R w L 0 R 05w L Internal Force and Moment At Section D ΣFy0 V R w a 0 V R w a ΣΜO0 M R a wa 05a 0 M R a w a 05a Section Property cD 05h hD A b h I 1 12 b h3 QD b hD cD 05hD Normal Stress σD M cD I σD 5625 kPa Shear Stress τD V QD I b τD 3516 kPa Construction of Mohrs Circle σx σD σy 0MPa τxy τD Center σc σx σy 2 σc 28125 kPa Radius R σx σc 2 τxy 2 R 28344 kPa Coordinates A σx τxy C σc 0 Angles φ atan τxy σx σc φ 7125 deg α 180deg 2 θ φ α 52875 deg Stresses on the Rotated Element represented by coordinates of point P σx σc R cos α σx 1102 kPa Ans τxy R sin α τxy 2260 kPa Ans Problem 986 The frame supports the distributed loading of 200 Nm Determine the normal and shear stresses at point E that act perpendicular and parallel respectively to the grains The grains at this point make an angle of 60 with the horizontal as shown Unit Used kPa 1000Pa Given b 50mm w 02 kN m h 100mm L 25m θ 60deg Solution Support Reactions Due to symmetry BCR ΣFy0 2R w L 0 R 05w L Internal Force At Section E ΣFy0 N R 0 N R Section Property A b h A 5000mm2 Normal Stress σE N A σE 50 kPa Shear Stress τD 0 Construction of Mohrs Circle σx σE σy 0MPa τxy τD Center σc σx σy 2 σc 25 kPa Radius R σx σc 2 τxy 2 R 25kPa Coordinates A σx τxy C σc 0 Angles φ atan τxy σx σc φ 0000 deg α 180deg 2 θ φ α 60000 deg Stresses on the Rotated Element represented by coordinates of point P σx σc R cos α σx 1250 kPa Ans τxy R sin α τxy 2165 kPa Ans Problem 987 The bent rod has a diameter of 15 mm and is subjected to the force of 600 N Determine the principal stresses and the maximum inplane shear stress that are developed at point A and point B Show the results on elements located at these points Given do 15mm P 06kN a 50mm L 01m Solution Internal Force At Section AB N P N 0600 kN M P a M 0030 kN m Section Property A π 4 do 2 A 17671 mm2 I π 64 do 4 I 0 m2 mm2 QA 0 since A 0 QB 0 since A 0 Normal Stress cA 05do σA N A M cA I σA 8715 MPa cB 05do σB N A M cB I σB 9394 MPa Shear Stress τA 0 since QA 0 τB 0 since QB 0 For A Construction of Mohrs Circle σx σA σy 0MPa τxy τA Center σc σx σy 2 σc 43573 MPa Radius R σx σc 2 τxy 2 R 43573 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and A represent σ1 and σ2 respectively σ1 σc R σ1 0MPa Ans σ2 σc R σ2 8715 MPa Ans Maximum Inplane Shear Stresses Represented by the coordinates of point E on the circle τmax R τmax 4357 MPa Ans 2θs 90deg θs 45deg Counterclockwise Ans For B Construction of Mohrs Circle σx σB σy 0MPa τxy τB Center σc σx σy 2 σc 46968 MPa Radius R σx σc 2 τxy 2 R 46968 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points A and B represent σ1 and σ2 respectively σ1 σc R σ1 9394 MPa Ans σ2 σc R σ2 0MPa Ans Maximum Inplane Shear Stresses Represented by the coordinates of point E on the circle τmax R τmax 4697 MPa Ans 2θs 90deg θs 45deg Clockwise Ans Problem 988 Draw the three Mohrs circles that describe each of the following states of stress a σmax 6MPa σint 0 σmin 0 b σmax 50MPa σint 0 σmin 40 MPa c Unit used kPa 1000Pa σmax 600kPa σint 200kPa σmin 100kPa d σmax 0 σint 7 MPa σmin 9 MPa e σmax 30 MPa σint 30 MPa σmin 30 MPa Problem 989 Draw the three Mohrs circles that describe each of the following states of stress a σ1 15MPa σ2 0 σ3 15 MPa τmax 15MPa b σ1 65MPa σ2 65 MPa σ3 65 MPa τmax 65MPa Problem 990 The stress at a point is shown on the element Determine the principal stresses and the absolute maximum shear stress Given σx 100 MPa σy 90MPa σz 80 MPa τxy 0MPa τyz 40MPa τxz 0MPa Solution Construction of Mohrs Circle in yz Plane Center σc σy σz 2 σc 5MPa Radius R σy σc 2 τyz 2 R 93941 MPa Coordinates A σy τyz C σc 0 Inplane Principal Stresses The coordinates of points A and B represent σ1 and σ2 respectively σ1 σc R σ1 98941 MPa σ2 σc R σ2 88941 MPa Construction of Three Circles From the results obtained above σmax σ1 σint σ2 σmin σx σmax 9894 MPa Ans σint 8894 MPa Ans σmin 10000 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τabsmax σmax σmin 2 τabsmax 9947 MPa Ans Problem 991 The stress at a point is shown on the element Determine the principal stresses and the absolute maximum shear stress Given σx 0MPa σy 7MPa σz 0MPa τxy 0MPa τyz 50MPa τxz 5MPa Solution Construction of Mohrs Circle in xz Plane Center σc σx σz 2 σc 0MPa Radius R σx σc 2 τxz 2 R 5MPa Coordinates A σy τxz C σc 0 Inplane Principal Stresses The coordinates of points A and B represent σ1 and σ2 respectively σ1 σc R σ1 5MPa σ2 σc R σ2 5 MPa Construction of Three Circles From the results obtained above σmax σy σint σ1 σmin σ2 σmax 7MPa Ans σint 5MPa Ans σmin 5 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 6MPa Ans Problem 992 The stress at a point is shown on the element Determine the principal stresses and the absolute maximum shear stress Given σx 150MPa σy 0MPa σz 90MPa τxy 0MPa τyz 80 MPa τxz 0MPa Solution Construction of Mohrs Circle in yz Plane Center σc σy σz 2 σc 45MPa Radius R σz σc 2 τyz 2 R 91788 MPa Coordinates A σy τyz B σz τyz C σc 0 Inplane Principal Stresses The coordinates of points A and B represent σ1 and σ2 respectively σ1 σc R σ1 136788 MPa σ2 σc R σ2 46788 MPa Construction of Three Circles From the results obtained above σmax σx σint σ1 σmin σ2 σmax 15000 MPa Ans σint 13679 MPa Ans σmin 4679 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 9839 MPa Ans Problem 993 The principal stresses acting at a point in a body are shown Draw the three Mohrs circles that describe this state of stress and find the maximum inplane shear stresses and associated average normal stresses for the xy yz and xz planes For each case show the results on the element oriented in the appropriate direction Given σx 40MPa σy 40 MPa σz 40 MPa τxy 0MPa τyz 0MPa τxz 0MPa Solution Construction of Three Circles σmax σx σint σy σmin σz σmax 4000 MPa σint 4000 MPa σmin 4000 MPa For xy Plane σavg σx σy 2 σavg 0MPa Ans τmax σx σy 2 τmax 40MPa Ans For yz Plane σavg σy σz 2 σavg 40 MPa Ans τmax σy σz 2 τmax 0MPa Ans For xy Plane σavg σx σz 2 σavg 0MPa Ans τmax σx σz 2 τmax 40MPa Ans Problem 994 Consider the general case of plane stress as shown Write a computer program that will show a plot of the three Mohrs circles for the element and will also calculate the maximum inplane shear stress and the absolute maximum shear stress Problem 995 The solid shaft is subjected to a torque bending moment and shear force as shown Determine the principal stresses acting at points A and B and the absolute maximum shear stress Given ro 25mm L 450mm P 08kN To 0045kN m Mo 03kN m Solution ρ ro Internal Force and Moment At Section AB Vy P Tx To Mz Mo P L Section Property J π 2 ro 4 A π ro 2 Iz π 4 ro 4 QA 0 Since A0 QB 4 ro 3 π 05A Normal Stress cA ro cB 0 σA Mz cA Iz σA 4889 MPa σB Mz cB Iz σB 0MPa Shear Stress bB 2 ro τA Tx ρ J τA 1833 MPa τB Vy QB Iz bB Tx ρ J τB 1290 MPa For Point A Construction of Mohrs Circle σx σA σz 0MPa τxz τA Center σc σx σz 2 σc 2445 MPa Radius R σz σc 2 τxz 2 R 3056 MPa Coordinates A σz τxz C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 5500 MPa σ2 σc R σ2 0611 MPa Three Mohrs Circlesσy 0 From the results obtained above σmax σ1 σint σy σmin σ2 σmax 550 MPa Ans σint 000 MPa Ans σmin 0611 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 306 MPa Ans For Point B Construction of Mohrs Circle σx σB σz 0MPa τxz τB Center σc σx σz 2 σc 0MPa Radius R σz σc 2 τxz 2 R 1290 MPa Coordinates A σz τxz C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 1290 MPa σ2 σc R σ2 1290 MPa Three Mohrs Circlesσy 0 From the results obtained above σmax σ1 σint σy σmin σ2 σmax 129 MPa Ans σint 000 MPa Ans σmin 129 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 129 MPa Ans Problem 996 The bolt is fixed to its support at C If a force of 90 kN is applied to the wrench to tighten it determine the principal stresses and the absolute maximum shear stress developed in the bolt shank at point A Represent the results on an element located at this point The shank has a diameter of 6 mm Given do 6mm a 50mm P 90N L 150mm Solution ρ 05do Internal Force and Moment At Section AB Vy P Mz P a Tx P L Section Property A π 4 do 2 Iz π 64 do 4 J π 32 do 4 QA 0 Since A0 Normal Stress cA 05do σA Mz cA Iz σA 21221 MPa Shear Stress τA Tx ρ J τA 31831 MPa Construction of Mohrs Circle in xz Plane σx σA σz 0MPa τxz τA Center σc σx σz 2 σc 1061 MPa Radius R σz σc 2 τxz 2 R 33553 MPa Coordinates A σz τxz C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 441631 MPa σ2 σc R σ2 229425 MPa Angles θp2 1 2 atan τxz σz σc θp2 3578 deg 2θp1 180deg 2 θp2 θp1 90deg θp2 θp1 5422 deg Clockwise Construction of Three Circles σy 0 From the results obtained above σmax σ1 σint σy σmin σ2 σmax 44163 MPa Ans σint 000 MPa Ans σmin 22942 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 33553 MPa Ans And the orientation is θs 1 2 atan σz σc τxz θs 922 deg Problem 997 Solve Prob 996 for point B Given do 6mm a 50mm P 90N L 150mm Solution ρ 05do Internal Force and Moment At Section AB Vy P Mz P a Tx P L Section Property A π 4 do 2 Iz π 64 do 4 J π 32 do 4 QB 4ρ 3π A 2 Normal Stress cB 0 σB Mz cB Iz σB 0MPa Shear Stress bB do τB Vy QB Iz bB Tx ρ J τB 31407 MPa Construction of Mohrs Circle in xz Plane σx σB σz 0MPa τxz τB Center σc σx σz 2 σc 0MPa Radius R σz σc 2 τxz 2 R 31407 MPa Coordinates A σz τxz C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 31407 MPa σ2 σc R σ2 31407 MPa Angles θp1 1 2 90deg θp1 4500 deg Clockwise Construction of Three Circles σy 0 From the results obtained above σmax σ1 σint σy σmin σ2 σmax 31407 MPa Ans σint 000 MPa Ans σmin 31407 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 31407 MPa Ans Problem 998 The stress at a point is shown on the element Determine the principal stresses and the absolute maximum shear stress Given σx 90 MPa σy 70MPa σz 120 MPa τxy 30MPa τyz 0MPa τxz 0MPa Solution Construction of Mohrs Circle in xy Plane Center σc σx σy 2 σc 10 MPa Radius R σx σc 2 τxy 2 R 8544 MPa Coordinates A σy τxy B σx τxy C σc 0 Inplane Principal Stresses The coordinates of points representing σ1 and σ2 respectively are σ1 σc R σ1 75440 MPa σ2 σc R σ2 95440 MPa Construction of Three Circles From the results obtained above σmax σ1 σint σ2 σmin σz σmax 7544 MPa Ans σint 9544 MPa Ans σmin 12000 MPa Ans Absolute Maximum Shear Stress From the three Mohrs circles τmax σmax σmin 2 τmax 9772 MPa Ans Problem 999 The cylindrical pressure vessel has an inner radius of 125 m and a wall thickness of 15 mm It is made from steel plates that are welded along a 45 seam with the horizontal Determine the normal and shear stress components along this seam if the vessel is subjected to an internal pressure of 3 MPa Given t 15mm ri 1250mm p 3MPa θ 45deg Solution Hoop Stress α ri t α 8333 Since α 10 then thinwall analysis can be used σ1 p ri t σ1 250MPa Longitudinal Stress σ2 p ri 2 t σ2 125MPa Construction of Mohrs Circle σx σ2 σy σ1 τxy 0 Center σc σx σy 2 σc 1875 MPa Radius R σy σc 2 τxy 2 R 625 MPa Coordinates A σx τxz C σc 0 Stresses represented by coordinates of point P on the circle σx σc σx 1875 MPa Ans τxy R τxy 625 MPa Ans Problem 9100 Determine the equivalent state of stress if an element is oriented 40 clockwise from the element shown Use Mohrs circle Given σx 6MPa σy 10 MPa θ 40deg τxy 0MPa Solution Center σc σx σy 2 σc 2 MPa Radius R σx σc R 8MPa Coordinates A σx 0 B σy 0 C σc 0 Stresses σx σc R cos 2θ σx 0611 MPa Ans τxy R sin 2θ τxy 7878 MPa Ans σy σc R cos 2θ σy 3389 MPa Ans Problem 9101 The internal loadings at a cross section through the 150mmdiameter drive shaft of a turbine consist of an axial force of 125 kN a bending moment of 12 kNm and a torsional moment of 225 kNm Determine the principal stresses at point A Also compute the maximum inplane shear stress at this point Given do 150mm N 125 kN Mz 12kN m Tx 225kN m Solution ρ 05do Section Property A π 4 do 2 Iz π 64 do 4 J π 32 do 4 Normal Stress cA ρ σA N A Mz cA Iz σA 433 MPa Shear Stress τA Tx ρ J τA 34 MPa Construction of Mohrs Circle in xy Plane σx σA σy 0MPa τxy τA Center σc σx σy 2 σc 216 MPa Radius R σx σc 2 τxy 2 R 403 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 1862 MPa σ2 σc R σ2 6191 MPa Maximum Inplane Shear Stress Represented by the coordinates of point E on the circle τmax R τmax 403 MPa Ans Problem 9102 The internal loadings at a cross section through the 150mmdiameter drive shaft of a turbine consist of an axial force of 125 kN a bending moment of 12 kNm and a torsional moment of 225 kNm Determine the principal stresses at point B Also compute the maximum inplane shear stress at this point Given do 150mm N 125 kN Mz 12kN m Tx 225kN m Solution ρ 05do Section Property A π 4 do 2 Iz π 64 do 4 J π 32 do 4 Normal Stress cB 0 σB N A Mz cB Iz σB 0707 MPa Shear Stress τB Tx ρ J τB 3395 MPa Construction of Mohrs Circle in xy Plane σx σB σy 0MPa τxy τB Center σc σx σy 2 σc 0354 MPa Radius R σx σc 2 τxy 2 R 3414 MPa Coordinates A σx τxy C σc 0 Inplane Principal Stresses The coordinates of points B and D represent σ1 and σ2 respectively σ1 σc R σ1 3060 MPa σ2 σc R σ2 3767 MPa Maximum Inplane Shear Stress Represented by the coordinates of point E on the circle τmax R τmax 341 MPa Ans Problem 9103 Determine the equivalent state of stress on an element if it is oriented 30 clockwise from the element shown Use the stresstransformation equations Given σx 0MPa σy 0300 MPa θ 30 deg τxy 0950MPa Solution Normal Stress σx σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σx 0898 MPa Ans σy σx σy 2 σx σy 2 cos 2θ τxy sin 2θ σy 0598 MPa Ans Shear Stress τxy σx σy 2 sin 2θ τxy cos 2θ τxy 0605 MPa Ans Problem 9104 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Ans Given σx 50 MPa σy 100 MPa φ 30deg τxy 28MPa Solution Construction of Mohrs Circle θ 90deg φ Center σc σx σy 2 σc 75 MPa Radius R σy σc 2 τxy 2 R 3754 MPa Coordinates A σx τxz C σc 0 Angles φ atan τxy σx σc φ 48240 deg α 360deg 2θ φ α 71760 deg Stresses represented by coordinates of point P on the circle σx σc R cos α σx 6325 MPa Ans τxy R sin α τxy 3565 MPa Problem 101 Prove that the sum of the normal strains in perpendicular directions is constant Solution Stress Transformation Equations Applying Eqs 105 and 107 of the text εx εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ 1 εy εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ 2 1 2 LHS εx εy RHS εx εy 2 εx εy 2 RHS εx εy Hence εx εy εx εy QED Problem 102 The state of strain at the point on the leaf of the caster assembly has components of εx 400106 εy 860106 and γxy 375106 Use the straintransformation equations to determine the equivalent inplane strains on an element oriented at an angle of θ 30 counterclockwise from the original position Sketch the deformed element due to these strains within the xy plane Given εx 400 10 6 γxy 375 10 6 εy 860 10 6 θ 30deg Solution Stress Transformation Equations εx εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εx 7738 10 6 Ans εy εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εy 38262 10 6 Ans γxy 2 εx εy 2 sin 2θ γxy 2 cos 2θ γxy 1279 10 3 Ans Problem 103 The state of strain at the point on the pin leaf has components of εx 200106 εy 180106 and γxy 300106 Use the straintransformation equations to determine the equivalent inplane strains on an element oriented at an angle of θ 60 counterclockwise from the original position Sketch the deformed element due to these strains within the xy plane Given εx 200 10 6 γxy 300 10 6 εy 180 10 6 θ 60deg Solution Stress Transformation Equations εx εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εx 5510 10 6 Ans εy εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εy 32490 10 6 Ans γxy 2 εx εy 2 sin 2θ γxy 2 cos 2θ γxy 132679 10 6 Ans Problem 104 Solve Prob 103 for an element oriented θ 30 clockwise Given εx 200 10 6 γxy 300 10 6 εy 180 10 6 θ 30 deg Solution Stress Transformation Equations εx εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εx 32490 10 6 Ans εy εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εy 5510 10 6 Ans γxy 2 εx εy 2 sin 2θ γxy 2 cos 2θ γxy 132679 10 6 Ans Problem 105 Due to the load P the state of strain at the point on the bracket has components of εx 500106 εy 350106 and γxy 430106 Use the straintransformation equations to determine the equivalent inplane strains on an element oriented at an angle of θ 30 clockwise from the original position Sketch the deformed element due to these strains within the xy plane Given εx 500 10 6 γxy 430 10 6 εy 350 10 6 θ 30 deg Solution Stress Transformation Equations εx εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εx 64870 10 6 Ans εy εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ εy 20130 10 6 Ans γxy 2 εx εy 2 sin 2θ γxy 2 cos 2θ γxy 85096 10 6 Ans Problem 106 The state of strain at the point on a wrench has components εx 120106 εy 180106 γxy 150106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 120 10 6 εy 180 10 6 γxy 150 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 13771 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 19771 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 13283 deg θp 76717 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 13771 10 6 Therefore θp1 θp θp1 1328 deg Ans θp2 θp θp2 7672 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 33541 10 6 Ans εavg εx εy 2 εavg 30 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 31717 deg θs 58283 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 33541 10 6 Therefore θs1 θs θs1 3172 deg Ans θs2 θs θs2 5828 deg Ans Problem 107 The state of strain at the point on the gear tooth has components εx 850106 εy 480106 γxy 650106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 850 10 6 εy 480 10 6 γxy 650 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 1039 10 3 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 29103 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 30175 deg θp 120175 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 1039 10 3 Therefore θp1 θp θp1 3018 deg Ans θp2 θp θp2 12018 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 74793 10 6 Ans εavg εx εy 2 εavg 665 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 14825 deg θs 75175 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 74793 10 6 Therefore θs1 θs θs1 1482 deg Ans θs2 θs θs2 7518 deg Ans Problem 108 The state of strain at the point on the gear tooth has the components εx 520106 εy 760106 γxy 750106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 520 10 6 εy 760 10 6 γxy 750 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 621772 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 86177 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 15184 deg θp 74816 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 621772 10 6 Therefore θp1 θp θp1 1518 deg Ans θp2 θp θp2 7482 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 148 10 3 Ans εavg εx εy 2 εavg 120 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 29816 deg θs 60184 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 148 10 3 Therefore θs1 θs θs1 2982 deg Ans θs2 θs θs2 6018 deg Ans Problem 109 The state of strain at the point on the spanner wrench has components εx 260106 εy 320106 γxy 180106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 260 10 6 εy 320 10 6 γxy 180 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 38487 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 19513 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 35783 deg θp 54217 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 195132 10 6 Therefore θp1 θp θp1 5422 deg Ans θp2 θp θp2 3578 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 18974 10 6 Ans εavg εx εy 2 εavg 290 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 9217 deg θs 80783 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 18974 10 6 Therefore θs1 θs θs1 922 deg Ans θs2 θs θs2 8078 deg Ans Problem 1010 The state of strain at the point on the arm has components εx 250106 εy 450106 γxy 825106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 250 10 6 εy 450 10 6 γxy 825 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 44098 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 64098 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 24843 deg θp 65157 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 440977 10 6 Therefore θp1 θp θp1 2484 deg Ans θp2 θp θp2 6516 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 1082 10 3 Ans εavg εx εy 2 εavg 100 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 20157 deg θs 69843 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 1082 10 3 Therefore θs1 θs θs1 2016 deg Ans θs2 θs θs2 6984 deg Ans Problem 1011 The state of strain at the point on the fan blade has components εx 250106 εy 450106 γxy 825106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 250 10 6 εy 450 10 6 γxy 825 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 44098 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 64098 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 24843 deg θp 65157 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 440977 10 6 Therefore θp1 θp θp1 2484 deg Ans θp2 θp θp2 6516 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 1082 10 3 Ans εavg εx εy 2 εavg 100 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 20157 deg θs 69843 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 1082 10 3 Therefore θs1 θs θs1 2016 deg Ans θs2 θs θs2 6984 deg Ans Problem 1012 A strain gauge is mounted on the 25mmdiameter A36 steel shaft in the manner shown When the shaft isrotating with an angular velocity of ω 1760 revmin using a slip ring the reading on the strain gauge is ε 800106 Determine the power output of the motor Assume the shaft is only subjected to a torque rpm 2π rad 60sec Unit used Given εx 800 10 6 ω 1760rpm εy 0 εx 0 θ 60deg G 75GPa do 25mm Solution Section Property ρ 05do J π 32 do 4 Stress Transformation Equations εx εx εy 2 εx εy 2 cos 2θ γxy 2 sin 2θ γxy 2 sin 2θ εx εx εy 2 εx εy 2 cos 2θ γxy 1848 10 3 Shear Stress τ T ρ J τ G γxy T G J γxy ρ T 04251 kN m P T ω P 7835 kW Ans Problem 1013 The state of strain at the point on the support has components εx 350106 εy 400106 γxy 675106 Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane Given εx 350 10 6 εy 400 10 6 γxy 675 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 71342 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 3658 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 42882 deg θp 47118 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 36575 10 6 Therefore θp1 θp θp1 4712 deg Ans θp2 θp θp2 4288 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 676849 10 6 Ans εavg εx εy 2 εavg 375 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 2118 deg θs 87882 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 676849 10 6 Therefore θs1 θs θs1 212 deg Ans θs2 θs θs2 8788 deg Ans Problem 1014 Consider the general case of plane strain where εx εy γxy and are known Write a computer program that can be used to determine the normal and shear strain εx and γxy on the plane of an element oriented θ from the horizontal Also compute the principal strains and the elements orientation and the maximum inplane shear strain the average normal strain and the elements orientation Problem 1015 Solve Prob 102 using Mohrs circle Given εx 400 10 6 γxy 375 10 6 εy 860 10 6 θ 30deg Solution Construction of Mohrs Circle Center εc εx εy 2 εc 230 10 6 Radius R εy εc 2 γxy 2 2 R 65731 10 6 Coordinates A εx 05 γxy C εc 0 Angles φ atan 05 γxy εy εc φ 16574 deg α 2θ φ α 76574 deg Strain on the inclined Element represented by coordinates of points P and Q εx εc R cos α εx 7738 10 6 Ans εy εc R cos α εy 38262 10 6 Ans γxy 2R sin α γxy 1279 10 3 Ans Problem 1016 Solve Prob 104 using Mohrs circle Given εx 200 10 6 γxy 300 10 6 εy 180 10 6 θ 30 deg Solution Construction of Mohrs Circle Center εc εx εy 2 εc 190 10 6 Radius R εy εc 2 γxy 2 2 R 15033 10 6 Coordinates A εx 05 γxy C εc 0 Angles φ atan 05 γxy εy εc φ 86186 deg α 2θ φ α 26186 deg Strain on the inclined Element represented by coordinates of points P and Q εx εc R cos α εx 3249 10 6 Ans εy εc R cos α εy 551 10 6 Ans γxy 2 R sin α γxy 1327 10 6 Ans Problem 1017 Solve Prob 103 using Mohrs circle Given εx 200 10 6 γxy 300 10 6 εy 180 10 6 θ 60deg Solution Construction of Mohrs Circle Center εc εx εy 2 εc 190 10 6 Radius R εy εc 2 γxy 2 2 R 15033 10 6 Coordinates A εx 05 γxy C εc 0 Angles φ atan 05 γxy εy εc φ 86186 deg α 2θ φ α 206186 deg Strain on the inclined Element represented by coordinates of points P and Q εx εc R cos α εx 551 10 6 Ans εy εc R cos α εy 3249 10 6 Ans γxy 2 R sin α γxy 1327 10 6 Ans Problem 1018 Solve Prob 105 using Mohrs circle Given εx 500 10 6 γxy 430 10 6 εy 350 10 6 θ 30 deg Solution Construction of Mohrs Circle Center εc εx εy 2 εc 425 10 6 Radius R εy εc 2 γxy 2 2 R 22771 10 6 Coordinates A εx 05 γxy C εc 0 Angles φ atan 05 γxy εy εc φ 70769 deg α 2θ φ α 10769 deg Strain on the inclined Element represented by coordinates of points P and Q εx εc R cos α εx 6487 10 6 Ans εy εc R cos α εy 2013 10 6 Ans γxy 2 R sin α γxy 851 10 6 Ans Problem 1019 Solve Prob 106 using Mohrs circle Given εx 120 10 6 εy 180 10 6 γxy 150 10 6 Solution Construction of Mohrs Circle Center εc εx εy 2 εc 30 10 6 Radius R εy εc 2 γxy 2 2 R 16771 10 6 Coordinates A εx 05 γxy C εc 0 Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 13771 10 6 Ans ε2 εc R ε2 19771 10 6 Ans Orientation of Principal Strain tan 2θp1 05γxy εx εc θp1 1 2 atan 05γxy εx εc θp1 1328 deg Ans θp2 θp1 90deg θp2 7672 deg Ans Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 33541 10 6 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 31717 deg Clockwise Ans Problem 1020 Solve Prob 108 using Mohrs circle Given εx 520 10 6 εy 760 10 6 γxy 750 10 6 Solution Construction of Mohrs Circle Center εc εx εy 2 εc 120 10 6 Radius R εy εc 2 γxy 2 2 R 74177 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 62177 10 6 Ans ε2 εc R ε2 86177 10 6 Ans Orientation of Principal Strain tan 2θp1 05γxy εx εc θp1 1 2 atan 05γxy εx εc θp1 1518 deg Ans θp2 θp1 90deg θp2 10518 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 1484 10 3 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 29816 deg Clockwise Ans Problem 1021 Solve Prob 107 using Mohrs circle Given εx 850 10 6 εy 480 10 6 γxy 650 10 6 Solution Construction of Mohrs Circle Center εc εx εy 2 εc 665 10 6 Radius R εy εc 2 γxy 2 2 R 37397 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 1039 10 3 Ans ε2 εc R ε2 29103 10 6 Ans Orientation of Principal Strain tan 2θp1 05γxy εx εc θp1 1 2 atan 05γxy εx εc θp1 3018 deg Ans θp2 θp1 90deg θp2 5982 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 74793 10 6 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 14825 deg Clockwise Ans Problem 1022 Solve Prob 109 using Mohrs circle Given εx 260 10 6 εy 320 10 6 γxy 180 10 6 Solution Construction of Mohrs Circle Center εc εx εy 2 εc 290 10 6 Radius R εy εc 2 γxy 2 2 R 9487 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 38487 10 6 Ans ε2 εc R ε2 19513 10 6 Ans Orientation of Principal Strain tan 2θp2 05 γxy εx εc θp2 1 2 atan 05 γxy εx εc θp2 3578 deg Ans θp1 90deg θp2 θp1 5422 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 18974 10 6 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 9217 deg Counterclockwise Ans Problem 1023 The strain at point A on the bracket has components εx 300106 εy 550106 γxy 650106 εz 0 Determine a the principal strains at A b the maximum shear strain in the xy plane and c the absolute maximum shear strain Given εx 300 10 6 εy 550 10 6 γxy 650 10 6 εz 0 Solution Construction of Mohrs Circle for xy plane Center εc εx εy 2 εc 425 10 6 Radius R εy εc 2 γxy 2 2 R 34821 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains ε1 εc R ε1 77321 10 6 Ans ε2 εc R ε2 7679 10 6 Ans b Maximum Inplane Shear Strain γmax 2R γmax 69642 10 6 Ans c Absolute Maximum Shear Strain Construction of Three Circles From the results obtained above εmax ε1 εint ε2 εmin 0 Absolute Maximum Shear Strain From the three Mohrs circles γabsmax εmax εmin γabsmax 77321 10 6 Ans Problem 1024 The strain at a point has components of εx 480106 εy 650106 γxy 780106 εz 0 Determine a the principal strains b the maximum shear strain in the xy plane and c the absolute maximum shear strain Given εx 480 10 6 εy 650 10 6 γxy 780 10 6 εz 0 Solution Construction of Mohrs Circle for xy plane Center εc εx εy 2 εc 85 10 6 Radius R εy εc 2 γxy 2 2 R 68653 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 77153 10 6 Ans ε2 εc R ε2 60153 10 6 Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2R γmax 1373 10 3 Ans c Absolute Maximum Shear Strain Construction of Three Circles From the results obtained above εmax ε1 εint 0 εmin ε2 Absolute Maximum Shear Strain From the three Mohrs circles γabsmax εmax εmin γabsmax 1373 10 3 Ans Problem 1025 The strain at a point on a pressurevessel wall has components of εx 350106 εy 460106 γxy 560106 εz 0 Determine a the principal strains at the point b the maximum shear strain in the xy plane and c the absolute maximum shear strain Given εx 350 10 6 εy 460 10 6 γxy 560 10 6 εz 0 Solution Construction of Mohrs Circle for xy plane Center εc εx εy 2 εc 55 10 6 Radius R εy εc 2 γxy 2 2 R 49237 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 43737 10 6 Ans ε2 εc R ε2 54737 10 6 Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 9847 10 6 Ans c Absolute Maximum Shear Strain Construction of Three Circles From the results obtained above εmax ε1 εint 0 εmin ε2 Absolute Maximum Shear Strain From the three Mohrs circles γabsmax εmax εmin γabsmax 9847 10 6 Ans Problem 1026 The strain at point A on the leg of the angle has components of εx 140106 εy 180106 γxy 125106 εz 0 Determine a the principal strains at A b the maximum shear strain in the xy plane and c the absolute maximum shear strain Given εx 140 10 6 εy 180 10 6 γxy 125 10 6 εz 0 Solution Construction of Mohrs Circle for xy plane Center εc εx εy 2 εc 20 10 6 Radius R εy εc 2 γxy 2 2 R 17177 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains ε1 εc R ε1 19177 10 6 Ans ε2 εc R ε2 15177 10 6 Ans b Maximum Inplane Shear Strain γmax 2 R γmax 3435 10 6 Ans c Absolute Maximum Shear Strain Construction of Three Circles From the results obtained above εmax ε1 εint 0 εmin ε2 Absolute Maximum Shear Strain From the three Mohrs circles γabsmax εmax εmin γabsmax 3435 10 6 Ans Problem 1027 The steel bar is subjected to the tensile load of 25 kN If it is 12 mm thick determine the absolute maximum shear strain E 200 GPa ν 03 Given d 50mm t 12mm L 375mm N 25kN E 200GPa ν 03 Solution Stress σx N d t σy 0 σz 0 Strain εx σx E εx 20833 10 5 εy ν εx εy 625 10 6 εz ν εx εz 625 10 6 γxy 0 Construction of Mohrs Circle in xy Plane Center εc εx εy 2 εc 72917 10 6 Radius R εx εc 2 05γxy 2 R 13542 10 5 Coordinates A εx 0 C εc 0 Inplane Principal Strains ε1 εc R ε1 20833 10 5 ε2 εc R ε2 625 10 6 Similarly from Mohrs Circle in xz Plane ε3 εc R ε3 625 10 6 Absolute Maximum Inplane Shear Strain γabsmax ε1 ε2 γabsmax 27083 10 5 Ans Given εa 475 10 6 εb 250 10 6 εc 360 10 6 Problem 1028 The 45 strain rosette is mounted on the surface of an aluminum plate The following readings are obtained for each gauge εa 475106 εb 250106 and εc 360106 Determine the inplane principal strains θa 0deg θb 45 deg θc 90 deg Solution Strain Rosettes 450 Applying Eq 1016 Given εa εx cos θa 2 εy sin θa 2 γxy sin θa cos θa 1 εb εx cos θb 2 εy sin θb 2 γxy sin θb cos θb 2 εc εx cos θc 2 εy sin θc 2 γxy sin θc cos θc 3 Solving Eqs1 2 and 3 Guess εx 10 6 εy 10 6 γxy 10 6 εx εy γxy Find εx εy γxy εx εy γxy 475 10 6 360 10 6 385 10 6 Construction of Mohrs Circle Center εc εx εy 2 εc 575 10 6 Radius R εy εc 2 γxy 2 2 R 45974 10 6 Coordinates A εx 05 γxy C εc 0 Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 51724 10 6 Ans ε2 εc R ε2 40224 10 6 Ans Problem 1029 The 60 strain rosette is mounted on the surface of the bracket The following readings are obtained for each gauge εa 780106 εb 400106 and εc 500106Determine a the principal strains and b the maximum inplane shear strain and associated average normal strain In each case show the deformed element due to these strains Given εa 780 10 6 θa 0deg εb 400 10 6 θb 60deg εc 500 10 6 θc 120deg Solution Strain Rosettes 600 Applying Eq 1016 Given εa εx cos θa 2 εy sin θa 2 γxy sin θa cos θa 1 εb εx cos θb 2 εy sin θb 2 γxy sin θb cos θb 2 εc εx cos θc 2 εy sin θc 2 γxy sin θc cos θc 3 Solving Eqs1 2 and 3 Guess εx 10 6 εy 10 6 γxy 10 6 εx εy γxy Find εx εy γxy εx εy γxy 780 10 6 860 10 6 11547 10 6 Construction of Mohrs Circle Center εc εx εy 2 εc 40 10 6 Radius R εy εc 2 γxy 2 2 R 82203 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 86203 10 6 Ans ε2 εc R ε2 78203 10 6 Ans Orientation of Principal Strain tan 2θp2 05γxy εx εc θp2 1 2 atan 05γxy εx εc θp2 201 deg Ans θp1 90deg θp2 θp1 8799 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 1644 10 3 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 42986 deg Clockwise Ans Problem 1030 The 45 strain rosette is mounted near the tooth of the wrench The following readings are obtained for each gauge εa 800106 εb 520106 and εc 450106 Determine a the inplane principal strains and b the maximum inplane shear strain and associated average normal strain In each case show the deformed element due to these strains Given εa 800 10 6 θa 135 deg εb 520 10 6 θb 90 deg εc 450 10 6 θc 45 deg Solution Strain Rosettes 450 Applying Eq 1016 Given εa εx cos θa 2 εy sin θa 2 γxy sin θa cos θa 1 εb εx cos θb 2 εy sin θb 2 γxy sin θb cos θb 2 εc εx cos θc 2 εy sin θc 2 γxy sin θc cos θc 3 Solving Eqs1 2 and 3 Guess εx 10 6 εy 10 6 γxy 10 6 εx εy γxy Find εx εy γxy εx εy γxy 170 10 6 520 10 6 125 10 3 Construction of Mohrs Circle Center εc εx εy 2 εc 175 10 6 Radius R εy εc 2 γxy 2 2 R 7139 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 88890 10 6 Ans ε2 εc R ε2 53890 10 6 Ans Orientation of Principal Strain tan 2θp2 05 γxy εx εc θp2 1 2 atan 05 γxy εx εc θp2 3055 deg Ans θp1 90deg θp2 θp1 5945 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2R γmax 1428 10 3 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 14449 deg Counterclockwise Ans Problem 1031 The 60 strain rosette is mounted on a beam The following readings are obtained from each gauge εa 150106 εb 330106 and εc 400106 Determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case show the deformed element due to these strains Given εa 150 10 6 θa 30 deg εb 330 10 6 θb 30deg εc 400 10 6 θc 90deg Solution Strain Rosettes 600 Applying Eq 1016 Given εa εx cos θa 2 εy sin θa 2 γxy sin θa cos θa 1 εb εx cos θb 2 εy sin θb 2 γxy sin θb cos θb 2 εc εx cos θc 2 εy sin θc 2 γxy sin θc cos θc 3 Solving Eqs1 2 and 3 Guess εx 10 6 εy 10 6 γxy 10 6 εx εy γxy Find εx εy γxy εx εy γxy 25333 10 6 400 10 6 55426 10 6 Construction of Mohrs Circle Center εc εx εy 2 εc 7333 10 6 Radius R εy εc 2 γxy 2 2 R 42838 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 50172 10 6 Ans ε2 εc R ε2 35505 10 6 Ans Orientation of Principal Strain tan 2θp2 05γxy εx εc θp2 1 2 atan 05γxy εx εc θp2 2015 deg Ans θp1 90deg θp2 θp1 6985 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 856764 10 6 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 24845 deg Clockwise Ans Problem 1032 The 45 strain rosette is mounted on a steel shaft The following readings are obtained from each gauge εa 800106 εb 520106 and εc 450106 Determine the inplane principal strains and their orientation Given εa 800 10 6 εb 520 10 6 εc 450 10 6 θa 45 deg θb 0deg θc 45deg Solution Strain Rosettes 450 Applying Eq 1016 Given εa εx cos θa 2 εy sin θa 2 γxy sin θa cos θa 1 εb εx cos θb 2 εy sin θb 2 γxy sin θb cos θb 2 εc εx cos θc 2 εy sin θc 2 γxy sin θc cos θc 3 Solving Eqs1 2 and 3 Guess εx 10 6 εy 10 6 γxy 10 6 εx εy γxy Find εx εy γxy εx εy γxy 520 10 6 170 10 6 125 10 3 Construction of Mohrs Circle Center εc εx εy 2 εc 175 10 6 Radius R εy εc 2 γxy 2 2 R 7139 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains ε1 εc R ε1 88890 10 6 Ans ε2 εc R ε2 53890 10 6 Ans Orientation of Principal Strain tan 2θp 05γxy εx εc θp 1 2 atan 05γxy εx εc θp 3055 deg Ans Problem 1033 Consider the general orientation of three strain gauges at a point as shown Write a computer program that can be used to determine the principal inplane strains and the maximum inplane shear strain at the point Show an application of the program using the values θa 40 εa 160106 θb 125 εb 100106 θc 220 εc 80106 Problem 1034 For the case of plane stress show that Hookes law can be written as 1 2 y x x E νε ε ν σ 1 2 x y y E νε ε ν σ Problem 1035 Use Hookes law Eq 1018 to develop the straintransformation equations Eqs 105 and 106 from the stresstransformation equations Eqs 91 and 92 Problem 1036 Abar of copper alloy is loaded in a tension machine and it is determined that εx 940106 and σx 100 MPa σy 0 σz 0 Determine the modulus of elasticity Ecu and the dilatation ecu of the copper νcu 035 Given σx 100MPa σy 0 σz 0 εx 940 10 6 νcu 035 Solution εx 1 E σx ν σy σz Ecu σx νcu σy σz εx Ecu 10638 GPa Ans εcu 1 2νcu Ecu σx σy σz εcu 2820 10 4 Ans Problem 1037 The principal plane stresses and associated strains in a plane at a point are σ1 250 MPa σ2 112 MPa ε1 102103 ε2 0180103 Determine the modulus of elasticity and Poissons ratio Given σ1 250MPa σ2 112MPa ε1 102 10 3 ε2 0180 10 3 Solution σ3 0MPa Given ε1 1 E σ1 ν σ2 σ3 1 ε2 1 E σ2 ν σ1 σ3 2 Solving 1 and 2 Guess E 1GPa ν 10 6 E ν Find E ν E 21277 GPa Ans ν 0295 Ans Problem 1038 Determine the bulk modulus for hard rubber if Er 5 GPa νr 043 Given Er 5GPa νr 043 Solution Bulk Modulus Applying Eq 1025 κ Er 3 1 2νr κ 1190 GPa Ans Problem 1039 The principal strains at a point on the aluminum fuselage of a jet aircraft are ε1 780106 and ε2 400106 Determine the associated principal stresses at the point in the same plane Eal 70 GPa νal 033 Hint See Prob 1034 Given ε1 780 10 6 E 70GPa ε2 400 10 6 ν 033 Solution Plane stress σ3 0MPa Use the formula developed in Prob 1034 σ1 E 1 ν2 ε1 ν ε2 σ1 7164 MPa Ans σ2 E 1 ν2 ε2 ν ε1 σ2 5164 MPa Ans Problem 1040 The rod is made of aluminum 2014T6 If it is subjected to the tensile load of 700 N and has a diameter of 20 mm determine the absolute maximum shear strain in the rod at a point on its surface Given Px 700N do 20mm σy 0 σz 0 E 731GPa ν 035 Solution σx 4Px π do 2 σx 2228 MPa Normal Strains Apply the general Hookes Law εx 1 E σx ν σy σz εx 30481 10 6 εy 1 E σy ν σx σz εy 10668 10 6 εz 1 E σz ν σx σy εz 10668 10 6 Principal Strains From the results obtained above εmax εx εmax 30481 10 6 εmin εz εmin 10668 10 6 Absolute Maximum Shear Strain From the three Mohrs circles γabsmax εmax εmin γabsmax 41149 10 6 Ans Problem 1041 The rod is made of aluminum 2014T6 If it is subjected to the tensile load of 700 N and has a diameter of 20 mm determine the principal strains at a point on the surface of the rod Given Px 700N do 20mm σy 0 σz 0 E 731GPa ν 035 Solution σx 4Px π do 2 σx 2228 MPa Normal Strains Apply the general Hookes Law εx 1 E σx ν σy σz εx 30481 10 6 εy 1 E σy ν σx σz εy 10668 10 6 εz 1 E σz ν σx σy εz 10668 10 6 Principal Strains From the results obtained above εmax εx εmax 30481 10 6 Ans εint εy εint 10668 10 6 Ans εmin εz εmin 10668 10 6 Ans Problem 1042 A rod has a radius of 10 mm If it is subjected to an axial load of 15 N such that the axial strain in the rod is εx 275106 determine the modulus of elasticity E and the change in its diameter ν 023 Given Px 15N σy 0 ro 10mm σz 0 ν 023 εx 275 10 6 Solution Normal Stresses σx Px π ro 2 σx 00477 MPa Normal Strains Appling the generalized Hookes Law εx 1 E σx ν σy σz E 1 εx σx ν σy σz E 1736 GPa Ans εy ν εx εy 63250 10 9 εz ν εx εz 63250 10 9 Thus d εy 2 ro d 1265 10 6 mm Ans Problem 1043 The principal strains at a point on the aluminum surface of a tank are ε1 630106 and ε2 350106 If this is a case of plane stress determine the associated principal stresses at the point in the same plane Eal 70 GPa νal 033 Hint See Prob 1034 Given ε1 630 10 6 Eal 70GPa ε2 350 10 6 νal 033 Solution Plane stress σ3 0MPa Use the formula developed in Prob 1034 σ1 Eal 1 νal 2 ε1 νal ε2 σ1 5856 MPa Ans σ2 Eal 1 νal 2 ε2 νal ε1 σ2 4383 MPa Ans Problem 1044 A uniform edge load of 100 kNm and 70 kNm is applied to the polystyrene specimen If the specimen is originally square and has dimensions of a 50 mm b 50 mm and a thickness of t 6 mm determine its new dimensions a b and t after the load is applied Ep 4 GPa νp 025 Given a 50mm b 50mm t 6mm qa 100 kN m qb 70 kN m E 4GPa ν 025 Solution Plane stress σz 0MPa σx qa t σx 1667 MPa σy qb t σy 1167 MPa Normal Strains Apply the general Hookes Law εx 1 E σx ν σy σz εx 3438 10 3 εy 1 E σy ν σx σz εy 1875 10 3 εz 1 E σz ν σx σy εz 1771 10 3 The new dimensions for the new specimen are a a 1 εy a 5009 mm Ans b b 1 εx b 5017 mm Ans t t 1 εz t 599 mm Ans Problem 1045 The principal stresses at a point are shown If the material is graphite for which Eg 56 GPa νg 023 determine the principal strains Given σx 70MPa σy 105 MPa σz 182 MPa E 56GPa ν 023 Solution Normal Strains Apply the general Hookes Law εx 1 E σx ν σy σz εy 1 E σy ν σx σz εy 14150 10 3 εz 1 E σz ν σx σy εz 31063 10 3 Principal Strains From the results obtained above εmax εx εmax 00243 Ans εint εy εint 00142 Ans εmin εz εmin 00311 Ans εx 24287 10 3 Problem 1046 The shaft has a radius of 15 mm and is made of L2 tool steel Determine the strains in the x and y directions if a torque T 2 kNm is applied to the shaft Given ro 15mm T 2kN m G 75GPa θ 45deg Solution Section Property J π ro 4 2 Shear Stress τ T ro J Shear Stressstrain Relationship Applying Hookes Law γxy τ G γxy 503 10 3 Strain Rosettes For pure shear εx 0 εy 0 Applying Eq 1015 θx θ θy θ 90deg εx εx cos θx 2 εy sin θx 2 γxy sin θx cos θx εx 252 10 3 Ans εy εx cos θy 2 εy sin θy 2 γxy sin θy cos θy εy 252 10 3 Ans Problem 1047 The cross section of the rectangular beam is subjected to the bending moment M Determine an expression for the increase in length of lines AB and CD The material has a modulus of elasticity E and Poissons ratio is ν Problem 1048 The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm A strain gauge having a length of 20mm is attached to it and it is observed to increase in length by 0012 mm when the vessel is pressurized Determine the pressure causing this deformation and find the maximum inplane shear stress and the absolute maximum shear stress at a point on the outer surface of the vessel The material is steel for which Est 200 GPa and νst 03 Given t 10mm di 2m L 20mm E 200GPa ν 03 L 0012mm Solution Normal Strains Appling the generalized Hookes Law with εmax L L εmax 600 10 6 εint εmax εint 600 10 6 εmax 1 E σmax ν σint σmin σmax E εmax ν σint σmin 50 p E εmax ν 50 p 0 p E εmax 50 1 ν p 343 MPa Ans Normal Stresses ri 05di α ri t α 100 Since α 10 then thinwall analysis can be used This is a plane stress problem where σmin 0 since there is no load acting on the outer surface of the wall σmax p ri 2 t σmax 17143 MPa σint σmax σint 17143 MPa Maximum Inplane Shear Stress Sperical Surface Mohrs circle is simply a dot As the result the state of stress is the same consisting of two normal stresses with zero shear stress regardless of the orientation of the element τmax 0 Ans Absolute Maximum Shear Stress τabsmax σmax σmin 2 τabsmax 8571 MPa Ans Problem 1049 A rod has a radius of 10 mm If it is subjected to an axial load of 15 N such that the axial strain in the rod is εx 275106 determine the modulus of elasticity E and the change in its diameter ν 023 Given Px 15N σy 0 ro 10mm σz 0 ν 023 εx 275 10 6 Solution Normal Stresses σx Px π ro 2 σx 00477 MPa Normal Strains Appling the generalized Hookes Law εx 1 E σx ν σy σz E 1 εx σx ν σy σz E 1736 GPa Ans εy ν εx εy 63250 10 9 εz ν εx εz 63250 10 9 Thus d εy 2 ro d 1265 10 6 mm Ans Problem 1050 A single strain gauge placed in the vertical plane on the outer surface and at an angle of 60 to the axis of the pipe gives a reading at point A of εA 250106 Determine the vertical force P if the pipe has an outer diameter of 25 mm and an inner diameter of 15 mmThe pipe is made of C86100 bronze Given do 25mm di 15mm G 38GPa a 200mm L 150mm θ 60deg εA 250 10 6 Solution ρo 05do ρi 05di Strain Rosettes For pure shear εx 0 εy 0 Applying Eq 1015 εA εx εy γxy sin θ cos θ γxy εA εx εy sin θ cos θ Shear Stressstrain Relationship Applying Hookes Law τA G γxy Internal Force and Moment At Section A Vy P Mz P a Tx P L Section Property A π 4 do 2 di 2 Iz π 64 do 4 di 4 J π 32 do 4 di 4 QA 4ρo 3π π ρo 2 2 4ρi 3π π ρi 2 2 Normal Stress cA 0 σA Mz cA Iz σA 0MPa Shear Stress in xy plane bA do di τA Vy QA Iz bA Tx ρo J G γxy P QA Iz bA P L ρo J P G γxy QA Iz bA L ρo J P 0438 kN Ans a 200mm Given do 25mm di 15mm G 38GPa ρi 05di L 150mm θ 60deg εA 250 10 6 Solution ρo 05do Ans Strain Rosettes For pure shear εx 0 εy 0 Applying Eq 1015 εA εx εy γxy sin θ cos θ γxy εA εx εy sin θ cos θ γxy 57735 10 6 Construction of Mohrs Circle in xy Plane Center εc εx εy 2 εc 0 Radius R εx εc 2 05γxy 2 R 288675 10 6 Coordinates A εx 05γxy C εc 0 Inplane Principal Strains The coordinates of points B and D represent ε1 and ε2 respectively ε1 εc R ε1 288675 10 6 ε2 εc R ε2 288675 10 6 Principal Stresses Since σxσyσz0 then from the generalized Hookes Law εz 0 From the results obtained above we have εmax ε1 εmax 2887 10 6 Ans εint εz εint 0 Ans εmin ε2 εmin 2887 10 6 Problem 1051 A single strain gauge placed in the vertical plane on the outer surface and at an angle of 60 to the axis of the pipe gives a reading at point A of εA 250106 Determine the principal strains in the pipe at point A The pipe has an outer diameter of 25 mm and an inner diameter of 15 mm and is made of C86100 bronze Problem 1052 A material is subjected to principal stresses σx and σy Determine the orientation θ of a strain gauge placed at the point so that its reading of normal strain responds only to σy and not σx The material constants are E and ν Problem 1053 The principal stresses at a point are shown in the figure If the material is aluminum for which Eal 70 GPa and νal 033 determine the principal strains Given σx 70MPa σy 105 MPa σz 182 MPa E 70GPa ν 033 Solution Apply the general Hookes Law εx 1 E σx ν σy σz εx 2353 10 3 Ans εy 1 E σy ν σx σz εy 9720 10 4 Ans εz 1 E σz ν σx σy εz 2435 10 3 Ans Ans Problem 1054 A thinwalled cylindrical pressure vessel has an inner radius r thickness t and length L If it is subjected to an internal pressure p show that the increase in its inner radius is dr rε1 pr21 ½ ν Et and the increase in its length is L pLr ½ ν Et Using these results show that the change in internal volume becomes dV π r21 ε12 1 ε2L π r2L Since ε1and ε2 are small quantities show further that the change in volume per unit volume called volumetric strain can be written as dV V pr 25 2ν Et Problem 1055 The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the bending stress that would occur if flat ends were used The bending stresses at the seam where the caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder respectively This requires the radial expansion to be the same for both the hemispheres and cylinder Show that this ratio is tc th 2 ν 1 ν Assume that the vessel is made of the same material and both the cylinder and hemispheres have the same inner radius If the cylinder is to have a thickness of 12 mm what is the required thickness of the hemispheres Take ν 03 Given tc 12mm ν 03 Solution For cylindrical vessel ο1 p r tc ο2 p r 2tc ο3 0 ε1 1 E σ1 ν σ2 σ3 ε1 1 E p r tc ν p r 2tc ε1 p r E tc 1 ν 2 dr ε1 r dr p r2 E tc 1 ν 2 1 For hemispherical end cap ο1 p r 2th ο2 p r 2th ο3 0 ε1 1 E σ1 ν σ2 σ3 ε1 1 E p r 2th ν p r 2th ε1 p r 2E th 1 ν dr ε1 r dr p r2 2E th 1 ν 2 Equate Eqs 1 and 2 p r2 E tc 1 ν 2 p r2 2E th 1 ν 1 tc 1 ν 2 1 2th 1 ν tc th 2 ν 1 ν QED Hence th 1 ν 2 ν tc th 494 mm Ans Problem 1056 TheA36 steel pipe is subjected to the axial loading of 60 kN Determine the change in volume of the material after the load is applied Given do 40mm di 30mm L 05m P 60kN E 200GPa ν 032 Solution Section Property A π 4 do 2 di 2 Normal Stress The pipe is subjected to uniaxial load Therefore σx P A σx 10913 MPa σy 0 σz 0 Dilation Apply Eq 1023 δV V 1 2ν E σx σy σz δV 1 2ν E σx σy σz A L δV 5400 mm3 Ans Given Unit used C deg Given H 150mm H 03mm E 689GPa ν 035 T 110C α 24 10 6 1 C Solution Normal Strains Since the aluminum is confined at the sides by a rigid container and allowed to expand in the zdirection εx 0 εy 0 εz H H Applying the generalized Hookes Law with the additional thernal strain εT α T Ans E E 10 3 Scale down to avoid floatingpoint error during calculation εx 1 E σx ν σy σz εT 1 εy 1 E σy ν σx σz εT 2 εz 1 E σz ν σy σx εT 3 Solving 1 2 and 3 Guess σx 1MPa σy 2MPa σz 3MPa σx σy σz Find σx σy σz σx σy σz 103 σx σy σz Scale back up σx σy σz 4872 4872 3852 MPa Problem 1057 The smooth rigidbody cavity is filled with liquid 6061T6 aluminum When cooled it is 03 mm from the top of the cavity If the top of the cavity is covered and the temperature is increased by 110C determine the stress components σx σy and σz in the aluminum Hint Use Eqs 1018 with an additional strain term of αT Eq 44 Problem 1058 The smooth rigidbody cavity is filled with liquid 6061T6 aluminum When cooled it is 03 mm from the top of the cavity If the top of the cavity is not covered and the temperature is increased by 110C determine the strain components εx εy and εz in the aluminum Hint Use Eqs 1018 with an additional strain term of αT Eq 44 Unit used C deg Given H 150mm H 03mm E 689GPa ν 035 T 110C α 24 10 6 1 C Solution Normal Strains Since the aluminum is confined at the sides by a rigid container then εx 0 εy 0 Ans and since it is not restrained in zdirection σz 0 Applying the generalized Hookes Law with the additional thernal strain εT α T Given εx 1 E σx ν σy σz εT 1 εy 1 E σy ν σx σz εT 2 Solving 1 and 2 Guess σx 1MPa σy 2MPa σx σy Find σx σy σx σy 2798 2798 MPa εz 1 E σz ν σy σx εT εz 548 10 3 Ans Problem 1059 The thinwalled cylindrical pressure vessel of inner radius r and thickness t is subjected to an internal pressure p If the material constants are E and ν determine the strains in the circumferential and longitudinal directions Using these results compute the increase in both the diameter and the length of a steel pressure vessel filled with air and having an internal gauge pressure of 15 MPa The vessel is 3 m long and has an inner radius of 05 m and a wall thick of 10 mm Est 200 GPa νst 03 Given t 10mm r 05m L 3m E 200GPa ν 03 p 15MPa Solution Normal Stresses α r t α 50 Since α 10 then thinwall analysis can be used σ1 p r t σ2 p r 2 t σ3 0 Normal Strains Appling the generalized Hookes Law εcir 1 E σ1 ν σ2 σ3 εcir 1 E p r t ν p r 2 t 0 εcir p r 2E t 2 ν εcir 31875 10 3 Ans εlong 1 E σ2 ν σ1 σ3 εlong 1 E p r 2t ν p r t 0 εlong p r 2E t 1 2ν εlong 750 10 6 Ans Deformations d εcir 2r d 319 mm Ans L εlong L L 225 mm Ans Problem 1060 Estimate the increase in volume of the tank in Prob 1059 Suggestion Use the results of Prob 1054 as a check Given t 10mm r 05m L 3m E 200GPa ν 03 p 15MPa Solution Section Property V π r2 L Normal Stresses α r t α 50 Since α 10 then thinwall analysis can be used σ1 p r t σ2 p r 2 t σ3 0 Normal Strains Appling the generalized Hookes Law εcir 1 E σ1 ν σ2 σ3 εcir 31875 10 3 εlong 1 E σ2 ν σ1 σ3 εlong 750 10 6 Deformations r εcir r r 159 mm L εlong L L 225 mm V π r r 2 L L πr2 L V 00168 m3 Ans Or appling the result of Prob 1054 V V p r E t 25 2ν V p r E t 25 2ν π r2 L V 00168 m3 Ans Problem 1061 A soft material is placed within the confines of a rigid cylinder which rests on a rigid support Assuming that εx 0 and εy 0 determine the factor by which the modulus of elasticity will be increased when a load is applied if ν 03 for the material Given ν 03 Solution Normal Strains Since the material is confined in a rigid cylinder εx 0 εy 0 Appling the generalized Hookes Law εx 1 E σx ν σy σz σx ν σy σz 1 εy 1 E σy ν σx σz σy ν σx σz 2 Solving Eqs 1 and 2 σx ν 1 ν σz σy ν 1 ν σz Thus εz 1 E σz ν σx σy εz 1 E σz ν ν 1 ν σz σz εz σz E 1 2ν2 1 ν εz σz E 1 ν 1 2ν 1 ν Hence when the material is not being confined and undergoes the same normal strain of εz then the required modulus of elasticity is E σz εz E 1 ν 1 ν 1 2ν E The increased factor is κ E E κ 1 ν 1 ν 1 2ν κ 135 Ans Problem 1062 A thinwalled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p Show that the increase in the volume within the vessel is V 2pπ r4Et1ν Use a smallstrain analysis Problem 1063 A material is subjected to plane stress Express the distortionenergy theory of failure in terms of σx σy and τxy Problem 1064 A material is subjected to plane stress Express the maximumshearstress theory of failure in terms of σx σy and τxy Assume that the principal stresses are of different algebraic signs Problem 1065 The components of plane stress at a critical point on an A36 structural steel shell are shown Determine if failure yielding has occurred on the basis of the maximumshearstress theory Given σx 75 MPa σy 125MPa τxy 80 MPa Solution Inplane Principal Stress Applying Eq 95 σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 15306 MPa σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 10306 MPa Maximum Shear Stress Theory σ1 and σ2 have opposite signs Therefore fσ σ1 σ2 fσ 25612 MPa σY 250 MPa Ans Based on the result obtained above the material yields according to the maximum shear stress theory Ans Problem 1066 The components of plane stress at a critical point on an A36 structural steel shell are shown Determine if failure yielding has occurred on the basis of the maximumdistortionenergy theory Given σx 75 MPa σy 125MPa τxy 80 MPa Solution Inplane Principal Stress Applying Eq 95 σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 15306 MPa σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 10306 MPa Maximum Distortion Energy Theory σ1 2 σ1 σ2 σ2 2 σY 2 fσ σ1 2 σ1 σ2 σ2 2 fσ 49825MPa2 σY 2 62500 MPa2 Ans Based on the result obtained above the material does not yield according to the maximum distortion energy theory Ans Problem 1067 The yield stress for a zirconiummagnesium alloy is σY 107 MPa If a machine part is made of this material and a critical point in the material is subjected to inplane principal stresses σ1 and σ2 05σ1 determine the magnitude of σ1 that will cause yielding according to the maximumshearstress theory Given σY 107MPa σ2 05 σ1 Solution σ1 σ2 σY σ1 05 σ1 σY 15σ1 σY σ1 σY 15 σ1 7133 MPa Ans Problem 1068 Solve Prob 1067 using the maximumdistortionenergy theory Given σY 107MPa σ2 05 σ1 Solution σ1 2 σ1 σ2 σ2 2 σY 2 σ1 2 σ1 05 σ1 05 σ1 2 σY 2 175σ1 2 σY 2 σ1 σY 2 175 σ1 8088 MPa Ans Problem 1069 If a shaft is made of a material of which σY 350 MPa determine the maximum torsional shear stress required to cause yielding using the maximumdistortionenergy theory Given σY 350MPa Solution σ1 τ σ2 τ σ1 2 σ1 σ2 σ2 2 σY 2 τ2 τ τ τ 2 σY 2 3τ2 σY 2 τ σY 2 3 τ 2021 MPa Ans Problem 1070 Solve Prob 1069 using the maximumshearstress theory Both principal stresses have opposite signs Given σY 350MPa Solution σ1 τ σ2 τ σ1 σ2 σY τ τ σY 2τ σY τ σY 2 τ 1750 MPa Ans Problem 1071 The yield stress for a plastic material is σY 110 MPa If this material is subjected to plane stress and elastic failure occurs when one principal stress is 120 MPa what is the smallest magnitude of the other principal stress Use the maximumdistortionenergy theory Given σY 110MPa σ1 120MPa Solution Given σ1 2 σ1 σ2 σ2 2 σY 2 1 Solving Eq 1 Guess σ2 1MPa σ2 Find σ2 σ2 2394 MPa Ans Problem 1072 Solve Prob 1071 using the maximumshearstress theory Both principal stresses have the same sign Given σY 110MPa σ1 120MPa Solution Since σ1 σY 110 MPa the material will fail for any σ2 Ans Problem 1073 The plate is made of Tobin bronze which yields at σY 175 MPa Using the maximumshearstress theory determine the maximum tensile stress σx that can be applied to the plate if a tensile stress σy 075σx is also applied Given σY 175MPa σy 075σx Solution σ1 σx σ2 075σx σ1 and σ1 have the same signs so σ2 σY 075σx σY 075σx σY σx σY 075 σx 2333 MPa Or σ1 σY σx σY σx σY σx 1750 MPa Controls Ans Problem 1074 The plate is made of Tobin bronze which yields at σY 175 MPa Using the maximumdistortion energy theory determine the maximum tensile stress σx that can be applied to the plate if a tensile stress σy 075σx is also applied Given σY 175MPa σy 075σx Solution σ1 σx σ2 075σx σ1 2 σ1 σ2 σ2 2 σY 2 σx 2 σx 075σx 075σx 2 σY 2 08125σx 2 σY 2 σ1 σY 2 08175 σ1 19355 MPa Ans Problem 1075 An aluminum alloy 6061T6 is to be used for a solid drive shaft such that it transmits 33 kW at 2400 revmin Using a factor of safety of 2 with respect to yielding determine the smallestdiameter shaft that can be selected based on the maximumshearstress theory Unit Used rpm 2π 60 rad s Given P 33kW ω 2400rpm σY 255MPa Fsafety 2 Solution Torsion T P ω T 01313 kN m Section Property J π 2 ρ4 Shear Stress τ T ρ J τ 2T π ρ3 Principal Stresses σ1 τ σ2 τ Maximum shear stress theory Both principal stresses have opposite sign hence σ1 σ2 σY Fsafety τ τ σY Fsafety 4T π ρ3 σY Fsafety ρ 3 4T π Fsafety σY do 2ρ do 2189 mm Ans Problem 1076 Solve Prob 1075 using the maximumdistortionenergy theory Unit Used rpm 2π 60 rad s Given P 33kW ω 2400rpm σY 255MPa Fsafety 2 Solution Torsion T P ω T 01313 kN m Section Property J π 2 ρ4 Shear Stress τ T ρ J τ 2T π ρ3 Principal Stresses σ1 τ σ2 τ Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σY Fsafety 2 τ2 τ τ τ 2 σY Fsafety 2 3 τ σY Fsafety 3 2T π ρ3 σY Fsafety ρ 3 2 3 T π Fsafety σY do 2ρ do 2087 mm Ans Problem 1077 An aluminum alloy is to be used for a drive shaft such that it transmits 20 kW at 1500 revmin Using a factor of safety of 25 with respect to yielding determine the smallestdiameter shaft that can be selected based on the maximumdistortionenergy theory σY 25 MPa Unit Used rpm 2π 60 rad s Given P 30kW ω 1500rpm σY 25MPa Fsafety 25 Solution Torsion T P ω T 01910 kN m Section Property J π 2 ρ4 Shear Stress τ T ρ J τ 2T π ρ3 Principal Stresses σ1 τ σ2 τ Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σY Fsafety 2 τ2 τ τ τ 2 σY Fsafety 2 3 τ σY Fsafety 3 2T π ρ3 σY Fsafety ρ 3 2 3 T π Fsafety σY do 2ρ do 5523 mm Ans Problem 1078 A bar with a square crosssectional area is made of a material having a yield stress of σY 840 MPa If the bar is subjected to a bending moment of 10 kNm determine the required size of the bar according to the maximumdistortionenergy theory Use a factor of safety of 15 with respect to yielding Given M 10kN m σY 840MPa Fsafety 15 Solution Section Property I a4 12 Normal Stresses σy 0 c a 2 σx Mc I σx 6M a3 Inplane Principal Stresses Since no shaer stress acts on the element σ1 σx σ2 σy Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σY Fsafety 2 σx 2 σx 0 02 σY Fsafety 2 σx σY Fsafety 6M a3 σY Fsafety a 3 6M Fsafety σY a 4750 mm Ans Problem 1079 Solve Prob 1078 using the maximumshearstress theory Given M 10kN m σY 840MPa Fsafety 15 Solution Section Property I a4 12 Normal Stresses σy 0 c a 2 σx Mc I σx 6M a3 Inplane Principal Stresses Since no shaer stress acts on the element σ1 σx σ2 σy Maximum shear stress theory σ2 0 σ2 σY Fsafety OK σ1 σY Fsafety 6M a3 σY Fsafety a 3 6M Fsafety σY a 4750 mm Ans Problem 1080 The principal plane stresses acting on a differential element are shown If the material is machine steel having a yield stress of σY 700 MPa determine the factor of safety with respect to yielding using the maximumdistortionenergy theory Given σx 480 MPa σy 475 MPa τxy 0 θ 30deg σY 700MPa Solution Principal Stresses σ1 σy σ2 σx Maximum Distortion Eenergy Theory σ1 2 σ1 σ2 σ2 2 σallow 2 σallow σ1 2 σ1 σ2 σ2 2 σallow 4775 MPa Fsafety σY σallow Fsafety 147 Ans Problem 1081 The principal plane stresses acting on a differential element are shown If the material is machine steel having a yield stress of σY 700 MPa determine the factor of safety with respect to yielding if the maximumshearstress theory is considered Given σx 80MPa σy 50 MPa τxy 0 θ 0deg σY 700MPa Solution Principal Stresses σmax σx σmin σy Maximum Shear Stress Theory τabsmax σmax σmin 2 τabsmax 65MPa τmax σY 2 τmax 350MPa Fsafety τmax τabsmax Fsafety 538 Ans Problem 1082 The state of stress acting at a critical point on a machine element is shown in the figure Determine the smallest yield stress for a steel that might be selected for the part based on the maximumshearstress theory Given σx 56MPa σy 70 MPa τxy 28MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 6194 MPa σ2 7594 MPa Maximum shear stress theory Both principal stresses have opposite sign hence σ1 σ2 σY σY σ1 σ2 σY 1379 MPa Ans Problem 1083 The yield stress for a uranium alloy is σY 160 MPa If a machine part is made of this material and a critical point in the material is subjected to plane stress such that the principal stresses are σ1 and σ2 025σ1 determine the magnitude of σ1 that will cause yielding according to the maximum distortionenergy theory Given σ2 025σ1 σY 160MPa Solution Maximum Distortion Eenergy Theory σ1 2 σ1 σ2 σ2 2 σY 2 σ1 2 σ1 025σ1 025σ1 2 σY 2 08125 σ1 2 σY 2 Principal Stress σ1 σY 08125 σ1 1775 MPa Ans Problem 1084 Solve Prob 1083 using the maximumshearstress theory Given σ2 025σ1 σY 160MPa Solution Principal Stresses This is a plane stress case σmax σ1 σint 025σ1 σmin 0 Maximum Shear Stress Theory τallow σY 2 τallow 80MPa τabsmax σmax σmin 2 τabsmax σ1 2 τabsmax τallow σ1 2 80 MPa σ1 160MPa Ans Problem 1085 An aluminum alloy is to be used for a solid drive shaft such that it transmits 25 kW at 1200 revmin Using a factor of safety of 25 with respect to yielding determine the smallestdiameter shaft that can be selected based on the maximumshearstress theory σY 70 MPa Unit Used rpm 2π 60 rad s Given P 25kW ω 1200rpm σY 70MPa Fsafety 25 Solution Torsion T P ω T 01989 kN m Section Property J π 2 ρ4 Shear Stress τ T ρ J τ 2T π ρ3 Principal Stresses σ1 τ σ2 τ Maximum shear stress theory Both principal stresses have opposite sign hence σ1 σ2 σY Fsafety τ τ σY Fsafety 4T π ρ3 σY Fsafety ρ 3 4T π Fsafety σY do 2ρ do 4167 mm Ans Problem 1086 The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure Determine the smallest yield stress for a steel that can be selected for the member based on the maximumshearstress theory Given σx 560MPa σy 0MPa τxy 175MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 61019 MPa σ2 5019 MPa Maximum shear stress theory Both principal stresses have opposite sign hence σ1 σ2 σY σY σ1 σ2 σY 6604 MPa Ans Problem 1087 Solve Prob 1086 using the maximumdistortionenergy theory Given σx 560MPa σy 0MPa τxy 175MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 61019 MPa σ2 5019 MPa Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σY 2 σY σ1 2 σ1 σ2 σ2 2 σY 6368 MPa Ans Problem 1088 If a machine part is made of titanium Ti6A14V and a critical point in the material is subjected to plane stress such that the principal stresses are σ1 and σ2 05σ1 determine the magnitude of σ1 in MPa that will cause yielding according to a the maximumshearstress theory and b the maximumdistortionenergy theory Given σ2 05σ1 σY 924MPa Solution a Maximum shear stress theory Both principal stresses have the same signs so σ1 σY Controls σ2 σY 05σ1 σY σ1 2σY Hence σ1 σY σ1 9240 MPa Ans b Maximum Distortion Eenergy Theory σ1 2 σ1 σ2 σ2 2 σY 2 σ1 2 σ1 05σ1 05σ1 2 σY 2 075 σ1 2 σY 2 Principal Stress σ1 σY 075 σ1 10669 MPa Ans Problem 1089 Derive an expression for an equivalent torque Te that if applied alone to a solid bar with a circular cross section would cause the same energy of distortion as the combination of an applied bending moment M and torque T Problem 1090 An aluminum alloy 6061T6 is to be used for a drive shaft such that it transmits 40 kW at 1800 revmin Using a factor of safety of FS 2 with respect to yielding determine the smallestdiameter shaft that can be selected based on the maximumdistortionenergy theory Unit Used rpm 2π 60 rad s Given P 40kW ω 1800rpm σY 255MPa Fsafety 2 Solution Torsion T P ω T 02122 kN m Section Property J π 2 ρ4 Shear Stress τ T ρ J τ 2T π ρ3 Principal Stresses σ1 τ σ2 τ Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σY Fsafety 2 τ2 τ τ τ 2 σY Fsafety 2 3τ σY Fsafety 2 3 T π ρ3 σY Fsafety ρ 3 2 3 T π Fsafety σY do 2ρ do 2449 mm Ans Problem 1091 Derive an expression for an equivalent bending moment Me that if applied alone to a solid bar with a circular cross section would cause the same energy of distortion as the combination of an applied bending moment M and torque T Problem 1092 The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 345 kNm a bending moment of 225 kNm and an axial thrust of 125 kN If the yield points for tension and shear are σY 700 MPa and τY 350 MPa respectively determine the required diameter of the shaft using the maximumshearstress theory Given Nx 125 kN M 225 kN m T 345kN m σY 700MPa τY 350MPa Solution Section Property A π 2 ρ2 I π 4 ρ4 J π 2 ρ4 Normal Stress σy 0 σx Nx A M ρ I σx 2Nx π ρ2 4M π ρ3 Shear Stress τxy T ρ J τxy 2T π ρ3 Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 Maximum shear stress theory Assume σ1 and σ2 have opposite sign hence σ1 σ2 σY 2 σx σy 2 2 τxy 2 σY σx σy 2 2 τxy 2 σY 2 4 1 4 2Nx π ρ2 4M π ρ3 2 2T π ρ3 2 σY 2 4 1 Given 2ρ Nx 4M 2 4T 2 π ρ3 σY 2 Solving Eq 1 Guess ρ 10mm ρ Find ρ ρ 1968 mm τxy 2T π ρ3 Check signs σx 2Nx π ρ2 4M π ρ3 σ1 σx 2 σx 2 2 τxy 2 σ2 σx 2 σx 2 2 τxy 2 σ1 15165 MPa σ2 54835 MPa σ1 and σ2 are of opposite sign OK Therefore do 2ρ do 3935 mm Ans Problem 1093 The element is subjected to the stresses shown If σY 350 ksi determine the factor of safety for this loading based on a the maximumshearstress theory and b the maximumdistortionenergy theory Given σx 84MPa σy 56 MPa τxy 49MPa σY 350MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 9945 MPa σ2 7145 MPa a Maximum shear stress theory Both principal stresses have opposite sign hence σallow σ1 σ2 σallow 1709 MPa Factor of safety is Fsafety σY σallow Fsafety 205 Ans b Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σallow 2 σallow σ1 2 σ1 σ2 σ2 2 σallow 1487 MPa Factor of safety is Fsafety σY σallow Fsafety 235 Ans Problem 1094 The state of stress acting at a critical point on a wrench is shown in the figure Determine the smallest yield stress for steel that might be selected for the part based on the maximumdistortionenergy theory Given σx 175MPa σy 0MPa τxy 70MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 19955 MPa σ2 2455 MPa Maximum distortion energy theory σ1 2 σ1 σ2 σ2 2 σY 2 σY σ1 2 σ1 σ2 σ2 2 σY 2129 MPa Ans Problem 1095 The state of stress acting at a critical point on a wrench is shown in the figure Determine the smallest yield stress for steel that might be selected for the part based on the maximumshearstress theory Given σx 175MPa σy 0MPa τxy 70MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 19955 MPa σ2 2455 MPa Maximum shear stress theory Both principal stresses have opposite sign hence σY σ1 σ2 σY 2241 MPa Ans Problem 1096 The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 Nm and an axial compressive force of 2 kN Determine if it fails according to the maximumnormalstress theory The ultimate stress of the concrete is σult 28 MPa Given do 50mm σult 28MPa P 2kN T 05kN m Solution Section Property A π 4 do 2 J π 32 do 4 Normal Stress σ P A σ 1019 MPa Shear Stress ρ 05do τ T ρ J τ 2037 MPa Inplane Principal Stresses σx 0 σy σ τxy τ for any point on the shafts surface Applying Eq 95 σ1 σx σy 2 σx σy 2 2 τxy 2 σ1 1987 MPa σult 28 MPa σ2 σx σy 2 σx σy 2 2 τxy 2 σ2 2089 MPa σult 28 MPa Failure Criteria σ1 σult OK σ2 σult OK Based on the result obtained above the material does not fail according to the maximum normalstress theory Ans Problem 1097 If a solid shaft having a diameter d is subjected to a torque T and moment M show that by the maximumnormalstress theory the maximum allowable principal stress is 16 2 2 3 T M M d allow π σ Problem 1098 The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are σ1 prt σ2 pr2t and σ3 0 If the yield stress is σy determine the maximum value of p based on a the maximumshearstress theory and b the maximumdistortionenergy theory Given σ1 p r t σ2 p r 2t σ3 0 Solution a Maximum Shear Stress Theory σ2 05σ1 Both principal stresses have the same signs so σ1 σY Controls σ2 σY 05σ1 σY σ1 2σY Hence p r t σY p t r σY Ans b Maximum Distortion Eenergy Theory σ1 2 σ1 σ2 σ2 2 σY 2 σ1 2 σ1 05σ1 05σ1 2 σY 2 075 σ1 2 σY 2 Principal Stress σ1 σY 075 Hence p r t σY 075 p 2t 3 r σY Ans Problem 1099 A thinwalled spherical pressure vessel has an inner radius r thickness t and is subjected to an internal pressure p If the material constants are E and v determine the strain in the circumferential direction in terms of the stated parameters Solution Normal Stresses This is a plane stress problem where σmin 0 since there is no load acting on the outer surface of the wall σ1 p r 2 t σ2 σ1 Normal Strains Appling the generalized Hookes Law ε1 1 E σ1 ν σ2 σ3 ε1 1 E σ1 ν σ2 ε1 σ1 E 1 ν ε2 1 E σ2 ν σ1 σ3 ε2 1 E σ2 ν σ1 ε2 σ1 E 1 ν Hence εcir σ1 E 1 ν εcir p r 2E t 1 ν Ans Problem 10100 The strain at point A on the shell has components εx 250106 εy 400106 γxy 275106 εz 0 Determine a the principal strains at A b the maximum shear strain in the xy plane and c the absolute maximum shear strain Given εx 250 10 6 εy 400 10 6 γxy 275 10 6 εz 0 Solution Construction of Mohrs Circle Center εc εx εy 2 εc 325 10 6 Radius R εy εc 2 γxy 2 2 R 15662 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains ε1 εc R ε1 48162 10 6 Ans ε2 εc R ε2 16838 10 6 Ans b Maximum Inplane Shear Strain γmax 2R γmax 31325 10 6 Ans c Absolute Maximum Shear Strain From the results obtained above εmax ε1 εmax 481625 10 6 εmin εz εmin 0 γabsmax εmax εmin γabsmax 48162 10 6 Ans Problem 10101 A differential element is subjected to plane strain that has the following components εx 950106 εy 420106 γxy 325106 Use the straintransformation equations and determine a the principal strains and b the maximum inplane shear strain and the associated average strain In each case specify the orientation of the element and show how the strains deform the element Given εx 950 10 6 εy 420 10 6 γxy 325 10 6 Solution a Inplane Principal Strains Applying Eq 109 ε1 εx εy 2 εx εy 2 2 γxy 2 2 ε1 99586 10 6 Ans ε2 εx εy 2 εx εy 2 2 γxy 2 2 ε2 37414 10 6 Ans Orientation of Principal Strain tan 2θp γxy εx εy θp 1 2 atan γxy εx εy θp θp 90deg θp 15758 deg θp 74242 deg Use Eq 105 to determine the direction of ε1 and ε2 εx εx εy 2 εx εy 2 cos 2θp γxy 2 sin 2θp εx 995856 10 6 Therefore θp1 θp θp1 1576 deg Ans θp2 θp θp2 7424 deg Ans b Maximum Inplane Shear Strain Applying Eq 1011 γmax 2 εx εy 2 2 γxy 2 2 γmax 621711 10 6 Ans εavg εx εy 2 εavg 685 10 6 Ans Orientation of Principal Strain tan 2θs εx εy γxy θs 1 2 atan εx εy γxy θs θs 90deg θs 29242 deg θs 119242 deg Use Eq 106 to determine the sign of γmax γxy 2 εx εy 2 sin 2θs γxy 2 cos 2θs γxy 621711 10 6 Therefore θs1 θs θs1 2924 deg Ans θs2 θs θs2 11924 deg Ans Problem 10102 The components of plane stress at a critical point on a thin steel shell are shown Determine if failure yielding has occurred on the basis of the maximumdistortionenergy theory The yield stress for the steel is σY 650 MPa Given σx 55 MPa σy 340MPa τxy 65MPa σY 650MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 35042 MPa σ2 6542 MPa Maximum Distortion Energy Theory σ1 2 σ1 σ2 σ2 2 σY 2 fσ σ1 2 σ1 σ2 σ2 2 fσ 150000MPa2 σY 2 422500 MPa2 Ans Based on the result obtained above the material does not yield according to the maximum distortion energy theory Ans Problem 10103 Solve Prob 10102 using the maximumshearstress theory Given σx 55 MPa σy 340MPa τxy 65MPa σY 650MPa Solution Principal Stresses σ1 σx σy 2 σx σy 2 2 τxy 2 σ2 σx σy 2 σx σy 2 2 τxy 2 σ1 35042 MPa σ2 6542 MPa Maximum Shear Stress Theory Both principal stresses have opposite sign hence σ1 σ2 σY fσ σ1 σ2 fσ 41584 MPa σY 650 MPa Ans Based on the result obtained above the material does not yield according to the maximum shear stress theory Ans Problem 10104 The 60 strain rosette is mounted on a beam The following readings are obtained for each gauge εa 600106 εb 700106 and εc 350106 Determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case show the deformed element due to these strains Given εa 600 10 6 θa 150deg εb 700 10 6 θb 150 deg εc 350 10 6 θc 90 deg Solution Strain Rosettes 600 Applying Eq 1016 Given εa εx cos θa 2 εy sin θa 2 γxy sin θa cos θa 1 εb εx cos θb 2 εy sin θb 2 γxy sin θb cos θb 2 εc εx cos θc 2 εy sin θc 2 γxy sin θc cos θc 3 Solving Eqs1 2 and 3 Guess εx 10 6 εy 10 6 γxy 10 6 εx εy γxy Find εx εy γxy εx εy γxy 183333 10 6 350 10 6 1501 10 3 Construction of Mohrs Circle Center εc εx εy 2 εc 8333 10 6 Radius R εy εc 2 γxy 2 2 R 79652 10 6 Coordinates A εx 05 γxy C εc 0 a Inplane Principal Strains represented by coordinates of points B and D ε1 εc R ε1 87985 10 6 Ans ε2 εc R ε2 71319 10 6 Ans Orientation of Principal Strain tan 2θp2 05γxy εx εc θp2 1 2 atan 05γxy εx εc θp2 3522 deg Ans θp1 90deg θp2 θp1 5478 deg Ans b Maximum Inplane Shear Strain represented by coordinates of point E γmax 2 R γmax 1593 10 3 Ans Orientation of Maximum Inplane Shear Strain tan 2θs εx εc 05γxy θs 1 2 atan εx εc 05γxy θs 978 deg Clockwise Ans Problem 10105 The aluminum beam has the rectangular cross section shown If it is subjected to a bending moment of M 75 kNm determine the increase in the 50mm dimension at the top of the beam and the decrease in this dimension at the bottom Eal 70 GPa νal 03 Given M 75kN m b 50mm h 75mm E 70GPa ν 03 Solution Section Property I b h3 12 Normal Stresses c h 2 σz M c I Lateral Strain and deformation εx ν σz E εx 68571 10 6 At the top b εx b b 003429 mm At the bottom b εx b b 003429 mm The negative sign indicates shortening Problem 1101 The simply supported beam is made of timber that has an allowable bending stress of σallow 65 MPa and an allowable shear stress of τallow 500 kPa Determine its dimensions if it is to be rectangular and have a heighttowidth ratio of 125 Given σallow 65MPa La 2m Lb 4m τallow 05MPa wo 8 kN m h 125 b Solution L 2La Lb Support Reactions By symmetry RLRRR ΣFy0 2R wo L 0 R 05wo L R 32kN Maximum Moment and Shear Vmax wo La Vmax 16kN Mmax wo La 05 La Mmax 16kN m Section Property I b h3 12 Sx I 05h Sx b h2 6 Sx b 125 b 2 6 Sx 25 b3 96 Bending Stress Sreqd Mmax σallow 25 b3 96 Mmax σallow b 3 96Mmax 25σallow b 2114 mm Ans h 125 b h 2643 mm Ans Shear Check I b h3 12 Qmax 05b h 025 h τmax Vmax Qmax I b τmax 0429 MPa τallow 05 MPa OK x1 0 001 La La x2 La 101 La La Lb x3 La Lb 101 La Lb L V1 x1 wo x1 kN V2 x2 wo x2 R 1 kN V3 x3 wo x3 2R 1 kN M1 x1 05 wo x1 2 kN m M2 x2 05 wo x2 2 R x2 La 1 kN m M3 x3 05 wo x3 2 R x3 La R x3 La Lb 1 kN m 0 2 4 6 8 20 10 0 10 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 2 4 6 8 20 10 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1102 The joists of a floor in a warehouse are to be selected using square timber beams made of oak If each beam is to be designed to carry 15 kNm over a simply supported span of 75 m determine the dimension a of its square cross section to the nearest multiples of 5mm The allowable bending stress is σallow 32 MPa and the allowable shear stress is τallow 0875 MPa Given σallow 32MPa L 75m τallow 0875MPa w 15 kN m Solution Support Reactions By symmetry RLRRR ΣFy0 2R w L 0 R 05w L Maximum Moment and Shear Vmax R Vmax 563 kN Mmax R 05L w 05L 025L Mmax 1055 kN m Section Property I a4 12 Qmax 05a a 025 a Bending Stress cmax 05a a 3 6Mmax σallow σmax M cmax I σallow 12Mmax 05a a4 a 12552 mm Use 130mm Ans Shear Stress I a4 12 Qmax 05a a 025 a τmax Vmax Qmax I a τmax 0536 MPa τallow 0875 MPa OK x 0 001 L L V x R w x 1 kN M x R x w x 05x 1 kN m 0 2 4 6 Distance m Shear kN V x x 0 2 4 6 5 10 Distance m Moment kNm M x x Problem 1103 The timber beam is to be loaded as shown If the ends support only vertical forces determine the greatest magnitude of P that can be applied σallow 25 MPa τallow 700 kPa Given σallow 25MPa a 4m τallow 07MPa bf 150mm df 30mm tw 40mm dw 120mm Solution L 2a Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df tw dw 05dw df bf df tw dw yc 5371 mm I 1 12 bf df 3 bf df 05df yc 2 1 12 tw dw 3 tw dw 05dw df yc 2 I 1916201613 mm4 Qmax h yc tw 05 h yc Qmax 18543652 mm3 Support Reactions By symmetry ABR ΣFy0 2R P 0 R 05P Maximum Load Assume failure due to bending moment Mmax R a Mmax 05P a cmax h yc σallow Mmax cmax I σallow 05P a h yc I P 2I σallow a h yc P 249 kN Ans Check Shear R 05P Vmax R Mmax R a τmax Vmax Qmax I tw τmax 0301 MPa τallow 07 MPa OK Problem 1104 Select the lightestweight steel wideflange beam from Appendix B that will safely support the machine loading shown The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 98 MPa Given σallow 168MPa a 06m τallow 98MPa P 25kN Solution L 5a Support Reactions By symmetry RLRRR ΣFy0 2R 4P 0 R 2P Maximum Moment and Shear Vmax R Vmax 50kN Mmax R 2a P a Mmax 45kN m Bending Stress Sreqd Mmax σallow Sreqd 26786 103 mm3 Select W 310x24 Sx 281 103 mm3 d 305mm tw 559mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 2933 MPa τallow 98 MPa OK Hence Use W 310x24 Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a x5 4a 101 4a 5a V1 x1 R 1 kN V2 x2 R P 1 kN V3 x3 R 2P 1 kN V4 x4 R 3P 1 kN V5 x5 R 4P 1 kN M1 x1 R x1 kN m M2 x2 R x2 P x2 a 1 kN m M3 x3 R x3 P x3 a P x3 2a 1 kN m M4 x4 R x4 P x4 a P x4 2a P x4 3a 1 kN m M5 x5 R x5 P x5 a P x5 2a P x5 3a P x5 4a 1 kN m 0 05 1 15 2 25 3 50 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 V5 x5 x1 x2 x3 x4 x5 0 05 1 15 2 25 3 20 40 Distance m Moment kNm M1 x1 M2 x2 M3 x3 M4 x4 M5 x5 x1 x2 x3 x4 x5 Problem 1105 The simply supported beam is made of timber that has an allowable bending stress of σallow 7 MPa and an allowable shear stress of τallow 05 MPa Determine its dimensions if it is to be rectangular and have a heighttowidth ratio of 125 Given σallow 7MPa L 2m τallow 05MPa wo 75 kN m h 125 b Solution Support Reactions By symmetry RLRRR ΣFy0 2R 05wo 2L 0 R 05 wo L Maximum Moment and Shear Vmax R Vmax 75kN Mmax R L 05wo L L 3 Mmax 100kN m Section Property I b h3 12 Sx I 05h Sx b h2 6 Sx b 125 b 2 6 Sx 25 b3 96 Bending Stress Sreqd Mmax σallow 25 b3 96 Mmax σallow b 3 96Mmax 25σallow b 3800 mm Shear Stress Provide a shear stress check h 125 b τmax 15Vmax b h τmax 062 MPa τallow 05 MPa NG Shear controls τallow 15Vmax b h τallow 15Vmax b 125b b 15Vmax 125τallow b 4243 mm Ans x1 0 001 L L x2 L 101 L 2L V1 x1 R wo 2 x1 L x1 1 kN V2 x2 R 05 wo L wo x2 L 1 05 x2 L L 1 kN M1 x1 R x1 wo 2 x1 L x1 x1 3 1 kN m M2 x2 R x2 wo L 2 x2 2 L 3 wo 2 x2 L 2 1 x2 L L 1 3 1 kN m 0 1 2 3 4 0 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 4 0 50 100 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1106 The wooden beam has a rectangular cross section and is used to support a load of 6 kN If the allowable bending stress is σallow 14 MPa and the allowable shear stress is τallow 5 MPa determine the height h of the cross section to the nearest multiples of 5mm if it is to be rectangular and have a width of b 75 mm Assume the supports at A and B only exert vertical reactions on the beam Given σallow 14MPa L1 12m τallow 5MPa L2 18m P 6kN b 75mm Solution L L1 L2 Support Reactions ΣFy0 A B P 0 1 ΣΜB0 A L P L2 0 2 Solving Eqs 1 and 2 A P L2 L B P L1 L Maximum Moment and Shear Vmax max A B Vmax 360 kN Mmax A L1 Mmax 432 kN m Section Property I b h3 12 Qmax 05h b 025 h Bending Stress cmax 05h h 6Mmax b σallow σmax M cmax I σallow 12Mmax 05h b h3 h 15712 mm Use 160mm Ans Shear Stress I b h3 12 Qmax 05h b 025 h τmax Vmax Qmax I b τmax 0458 MPa τallow 5 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 A 1 kN V2 x2 A P 1 kN M1 x1 A x1 kN m M2 x2 A x2 P x2 L1 1 kN m 0 1 2 3 5 0 5 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 0 5 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1107 Solve Prob 116 if the cross section has an unknown width but is to be square ie h b Given σallow 14MPa L1 12m τallow 5MPa L2 18m P 6kN b h Solution L L1 L2 Support Reactions ΣFy0 A B P 0 1 ΣΜB0 A L P L2 0 2 Solving Eqs 1 and 2 A P L2 L B P L1 L Maximum Moment and Shear Vmax max A B Vmax 360 kN Mmax A L1 Mmax 432 kN m Section Property I b h3 12 Qmax 05h b 025 h Bending Stress cmax 05h b h h 3 6Mmax σallow σmax M cmax I σallow 12Mmax 05h h h3 h 12279 mm Use 125mm Ans Shear Stress b h I b h3 12 Qmax 05h b 025 h τmax Vmax Qmax I b τmax 0358 MPa τallow 5 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 A 1 kN V2 x2 A P 1 kN M1 x1 A x1 kN m M2 x2 A x2 P x2 L1 1 kN m 0 1 2 3 5 0 5 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 0 5 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1108 The simply supported beam is composed of two W310 X 33 sections built up as shown Determine the maximum uniform loading w the beam will support if the allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 100 MPa Given σallow 160MPa L 8m τallow 100MPa Use W 310x33 Solution Support Reactions By symmetry RLRRR ΣFy0 2R w L 0 R 05w L Maximum Moment and Shear Vmax R Mmax R 05L w 05L 025L Mmax 05w L 05L w 05L 025L Mmax 0125 w L2 Section Property One W 310x33 tw 660mm Ix 650 106 mm4 A 4180mm2 d 313mm Two W 310x33 Ic 2 Ix A 05d 2 Sc Ic d Maximum Loading Assume moment controls cmax d σmax Mmax cmax Ic σallow 0125 w L2 d 2 Ix A 05d 2 w 16σallow L2 d Ix A 05d 2 w 2139 kN m Ans Check Shear Neglect area of flanges Aw 2d tw R 05w L Vmax R τmax Vmax Aw τmax 2071 MPa τallow 100 MPa OK x 0 001 L L V x R w x 1 kN M x R x w x 05x 1 kN m 0 5 100 0 100 Distance m Shear kN V x x 0 5 0 100 200 Distance m Moment kNm M x x Problem 1109 The simply supported beam is composed of two W310 X 33 sections built up as shown Determine if the beam will safely support a loading of w 30 kNm The allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 100 MPa Given σallow 160MPa L 8m Use W 310x33 τallow 100MPa w 30 kN m Solution Support Reactions By symmetry RLRRR ΣFy0 2R w L 0 R 05w L Maximum Moment and Shear Vmax R Mmax R 05L w 05L 025L Mmax 05w L 05L w 05L 025L Mmax 0125 w L2 Section Property One W 310x33 tw 660mm Ix 650 106 mm4 A 4180mm2 d 313mm Two W 310x33 Ic 2 Ix A 05d 2 Sc Ic d Bending Stress cmax d σmax Mmax cmax Ic σmax 2244 MPa σallow 160 MPa Not OK The beam fails due to bending stress criteria Ans Check Shear Neglect area of flanges Aw 2d tw τmax Vmax Aw τmax 2904 MPa τallow 100 MPa OK x 0 001 L L V x R w x 1 kN M x R x w x 05x 1 kN m 0 5 0 Distance m Shear kN V x x 0 5 0 200 Distance m Moment kNm M x x Problem 1110 Select the lightestweight steel wideflange beam from Appendix B that will safely support the loading shown where w 100 kNm and P 25 kN The allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 100 MPa Given σallow 160MPa L1 24m τallow 100MPa L2 18m P 25kN w 100 kN m Solution L L1 L2 Support Reactions Given ΣFy0 A B P w L1 0 1 ΣΜB0 A L1 P L2 05w L1 2 0 2 Solving Eqs 1 and 2 Guess A 1kN B 1kN A B Find A B A B 10125 16375 kN Maximum Moment and Shear Vmax P B Vmax 13875 kN When V0 xo L1 A A B xo A L1 A B xo 0917 m Mmax A xo 05w xo 2 Mmax 508 kN m Bending Stress Assume bending controls the design Sreqd Mmax σallow Sreqd 3175101 mm3 Select W 310x33 Sx 415 103 mm3 d 313mm tw 660mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 6717 MPa τallow 100 MPa OK Hence Use W 310x33 Ans x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 A w x1 1 kN V2 x2 A w L1 B 1 kN M1 x1 A x1 05w x1 2 1 kN m M2 x2 A x2 w L1 x2 05 L1 B x2 L1 1 kN m 0 2 4 100 0 100 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 2 4 50 0 50 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1111 Select the lightestweight steel wideflange beam having the shortest height from Appendix B that will safely support the loading shown where w 0 and P 50 kN The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa Given σallow 168MPa L1 24m τallow 100MPa L2 18m P 50kN w 0 kN m Solution L L1 L2 Support Reactions Given ΣFy0 A B P 0 1 ΣΜB0 A L1 P L2 0 2 Solving Eqs 1 and 2 Guess A 1kN B 1kN A B Find A B A B 3750 8750 kN Maximum Moment and Shear Vmax P Vmax 50kN Mmax P L2 Mmax 90kN m Bending Stress Assume bending controls the design Sreqd Mmax σallow Sreqd 5357143 mm3 Three choices of wide flange section having the weight of 39 kgm can be made W 310x39 W 360x39 and W 410x39 However the shortest is the W 310x39 Select W 310x39 Sx 547 103 mm3 d 310mm tw 584mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 2762 MPa τallow 100 MPa OK Hence Use W 310x39 Ans x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 A 1 kN V2 x2 A B 1 kN M1 x1 A x1 1 kN m M2 x2 A x2 B x2 L1 1 kN m 0 2 4 50 0 50 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 2 4 100 50 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1112 Determine the minimum width of the beam to the nearest multiples of 5mm that will safely support the loading of P 40 kN The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 105 MPa Given σallow 168MPa L1 2m τallow 105MPa L2 2m P 40kN h 150mm Solution L L1 L2 Support Reactions Given ΣFy0 A B P 0 1 ΣΜB0 A L1 P L 0 2 Solving Eqs 1 and 2 Guess A 1kN B 1kN A B Find A B A B 80 40 kN Maximum Moment and Shear Vmax P Vmax 40kN Mmax P L1 Mmax 80kN m Section Property I b h3 12 Sx I 05h Sx b h2 6 Qmax 05h b 025 h Bending Stress Assume bending controls the design Sreqd Mmax σallow b h2 6 Mmax σallow b 6Mmax h2 σallow b 12698 mm Use 130mm Ans Check Shear I b h3 12 Qmax 05h b 025 h τmax Vmax Qmax I b τmax 3150 MPa τallow 105 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 P 1 kN V2 x2 P A 1 kN M1 x1 P x1 1 kN m M2 x2 P x2 A x2 L1 1 kN m 0 2 4 50 0 50 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 2 4 100 50 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1113 Select the lightestweight steel wideflange beam from Appendix B that will safely support the loading shown The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa Given σallow 168MPa L1 2m τallow 100MPa L2 3m P 75kN wo 75 kN m Solution L L1 L2 Support Reactions Given ΣFy0 A B P 05wo L2 0 1 ΣΜA0 P L1 05wo L2 L2 3 B L2 0 2 Solving Eqs 1 and 2 Guess A 1kN B 2kN A B Find A B A B 20000 1250 kN Maximum Moment and Shear Vmax P A Vmax 125kN Mmax P L1 Mmax 150 kN m Bending Stress Assume bending controls the design Sreqd Mmax σallow Sreqd 8928571 mm3 Select W 410x53 Sx 923 103 mm3 d 403mm tw 749mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 4141 MPa τallow 100 MPa OK Hence Use W 410x53 Ans x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 P kN V2 x2 P A wo x2 L1 1 05 x2 L1 L2 1 kN M1 x1 P x1 1 kN m M2 x2 P x2 A x2 L1 wo 2 x2 L1 2 1 x2 L1 L2 1 3 1 kN m 0 1 2 3 4 5 100 0 100 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 4 5 150 100 50 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1114 Select the lightestweight steel structural wideflange beam with the shortest depth from Appendix B that will safely support the loading shown The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa Given σallow 168MPa L 2m τallow 100MPa wo 120 kN m Solution Support Reactions Given ΣFy0 R 05wo L 0 R 05wo L ΣΜA0 M 05wo L L 3 0 M wo L2 6 Maximum Moment and Shear Vmax R Vmax 120kN Mmax M Mmax 80 kN m Bending Stress Assume bending controls the design Sreqd Mmax σallow Sreqd 4761905 mm3 Select W 310x39 Sx 547 103 mm3 d 310mm tw 584mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 6628 MPa τallow 100 MPa OK Hence Use W 310x39 Ans Problem 1115 Select the shortest and lightestweight steel wideflange beam from Appendix B that will safely support the loading shown The allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 84 MPa Given σallow 160MPa a 12m τallow 84MPa P1 20kN P2 50kN P3 30kN Solution L 4a Support Reactions Given ΣFy0 A B P1 P2 P3 0 1 ΣΜB0 A L P1 3a P2 2 a P3 a 0 2 Solving Eqs 1 and 2 Guess A 1kN B 2kN A B Find A B A B 4750 5250 kN Maximum Moment and Shear Vmax max A B Vmax 525 kN Mmax A 2a P1 a Mmax 90kN m Bending Stress Assume bending controls the design Sreqd Mmax σallow Sreqd 562500mm3 Select W 250x58 Sx 693 103 mm3 d 252mm tw 800mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 2604 MPa τallow 84 MPa OK Hence Use W 250x58 Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a 4a V1 x1 A 1 kN V2 x2 A P1 1 kN V3 x3 A P1 P2 1 kN V4 x4 A P1 P2 P3 1 kN M1 x1 A x1 kN m M2 x2 A x2 P1 x2 a 1 kN m M3 x3 A x3 P1 x3 a P2 x3 2a 1 kN m M4 x4 A x4 P1 x4 a P2 x4 2a P3 x4 3a 1 kN m 0 1 2 3 4 50 0 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 1 2 3 4 0 50 100 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 1116 The beam is made of a ceramic material having an allowable bending stress of σallow 5 MPa and an allowable shear stress of τallow 28 MPa Determine the width b of the beam if the height h 2b Given σallow 5MPa L1 50mm τallow 28MPa Lo 150mm P1 75N w 12 kN m P2 50N h 2b Solution L 2L1 Lo Support Reactions Given ΣFy0 A B P1 P2 w Lo 0 1 ΣΜB0 A Lo P1 L1 Lo w Lo 05Lo P2 L1 0 2 Solving Eqs 1 and 2 Guess A 1N B 1N A B Find A B A B 17333 13167 N Maximum Moment and Shear Vmax P1 A Vmax 9833 N Mmax P1 L1 Mmax 375 N m Section Property I b h3 12 h 2b Sx I 05h Sx b h2 6 Sx 4b3 6 Bending Stress Assume bending controls the design Sreqd Mmax σallow 4b3 6 Mmax σallow b 3 6Mmax 4σallow b 1040 mm Ans Check Shear h 2 b τmax 15Vmax h b τmax 0682 MPa τallow 28 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L1 Lo x3 L1 Lo 101 L1 Lo L V1 x1 P1 N V2 x2 P1 A w x2 L1 1 N V3 x3 P1 A w Lo B 1 N M1 x1 P1 x1 N m M2 x2 P1 x2 A x2 L1 05w x2 L1 2 1 N m M3 x3 P1 x3 A x3 L1 w Lo x3 L1 05 Lo B x3 L1 Lo 1 N m 0 005 01 015 02 100 0 100 Distance m Shear N V1 x1 V2 x2 V3 x3 x1 x2 x3 0 005 01 015 02 4 2 0 2 Distance m Moment Nm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1117 The steel cantilevered Tbeam is made from two plates welded together as shown Determine the maximum loads P that can be safely supported on the beam if the allowable bending stress is σallow 170 MPa and the allowable shear stress is τallow 95 MPa Given σallow 170MPa a 2m τallow 95MPa bf 150mm df 15mm tw 15mm dw 150mm Solution L 2a Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df tw dw 05dw df bf df tw dw yc 4875 mm I 1 12 bf df 3 bf df 05df yc 2 1 12 tw dw 3 tw dw 05dw df yc 2 I 1191796875 mm4 Qmax h yc tw 05 h yc Qmax 10135547 mm3 Support Reactions ΣFy0 R P P 0 R 2P ΣΜA0 MA P a P 2a 0 MA 3P a Maximum Load Assume failure due to bending moment Mmax MA cmax h yc σallow Mmax cmax I σallow 3P a h yc I P I σallow 3a h yc P 290 kN Ans Check Shear R 2P Vmax R MA 3P a τmax Vmax Qmax I tw τmax 3294 MPa τallow 95 MPa OK x1 0 001 a a x2 a 101 a L V1 x1 R 1 kN V2 x2 R P 1 kN M1 x1 MA R x1 1 kN m M2 x2 MA R x2 P x2 a 1 kN m 0 2 4 0 2 4 6 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 2 4 20 10 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1118 Draw the shear and moment diagrams for the W310 X 21 beam and check if the beam will safely support the loading The allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 84 MPa Given σallow 160MPa L1 1m τallow 84MPa L2 4m Mo 75kN m w 25 kN m Solution L L1 L2 Support Reactions Given ΣFy0 A B w L2 0 1 ΣΜA0 Mo 05w L2 2 B L2 0 2 Solving Eqs 1 and 2 Guess A 1kN B 2kN A B Find A B A B 6875 3125 kN Maximum Moment and Shear Vmax max A B Vmax 6875 kN Mmax Mo Mmax 75kN m Bending Stress Assume bending controls the design Use W 310x21 Sx 244 103 mm3 d 303mm tw 508mm Sreqd Mmax σallow Sreqd 468750mm3 Sx 244 103 mm3 No Good Shear Stress Provide a shear stress check τmax Vmax tw d τmax 4466 MPa τallow 84 MPa OK Hence the wide flange section W 310x21 fails due to the bending stress and will not safely support the loading Ans x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 0 kN V2 x2 A w x2 L1 1 kN M1 x1 Mo 1 kN m M2 x2 Mo A x2 L1 w 2 x2 L1 2 1 kN m 0 1 2 3 4 5 0 50 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 4 5 100 50 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1119 Select the lightestweight steel wideflange beam from Appendix B that will safely support the loading shown The allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 84 MPa Given σallow 160MPa L1 1m τallow 84MPa L2 4m Mo 75kN m w 25 kN m Solution L L1 L2 Support Reactions Given ΣFy0 A B w L2 0 1 ΣΜA0 Mo 05w L2 2 B L2 0 2 Solving Eqs 1 and 2 Guess A 1kN B 2kN A B Find A B A B 6875 3125 kN Maximum Moment and Shear Vmax max A B Vmax 6875 kN Mmax Mo Mmax 75kN m Bending Stress Assume bending controls the design Sreqd Mmax σallow Sreqd 468750mm3 Select W 360x33 Sx 475 103 mm3 d 349mm tw 584mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 3373 MPa τallow 84 MPa OK Hence Use W 360x33 Ans x1 0 001 L1 L1 x2 L1 101 L1 L V1 x1 0 kN V2 x2 A w x2 L1 1 kN M1 x1 Mo 1 kN m M2 x2 Mo A x2 L1 w 2 x2 L1 2 1 kN m 0 1 2 3 4 5 0 50 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 4 5 100 50 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1120 The compound beam is made from two sections which are pinned together at B Use Appendix B and select the light wideflange beam that would be safe for each section if the allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa The beam supports a pipe loading of 6 kN and 9 kN as shown Given σallow 168MPa L1 36m τallow 100MPa L2 24m P1 6kN L3 3m P2 9kN Solution L L1 L2 L3 Support Reactions Given For segment BC ΣFy0 B C P2 0 1 ΣΜB0 C L2 L3 P2 L2 0 2 Solving Eqs 1 and 2 Guess C 1kN B 2kN C B Find C B C B 4 5 kN For segment AB A P1 B A 11kN MA P1 05L1 B L1 MA 2880 kN m Maximum Moment and Shear For segment AB Vmax max A B Vmax 11kN Mmax MA Mmax 288 kN m For segment BC Vmax max B C Vmax 5kN Mmax C L3 Mmax 12kN m Bending Stress Assume bending controls the design For segment AB Sreqd Mmax σallow Sreqd 1714286 mm3 Select W 250x18 Sx 179 103 mm3 d 251mm tw 483mm For segment BC Sreqd Mmax σallow Sreqd 714286 mm3 Select W 150x14 Sx 912 103 mm3 d 150mm tw 432mm Shear Stress Provide a shear stress check For segment AB τmax Vmax tw d τmax 907 MPa τallow 100 MPa OK Hence Use W 250x18 Ans For segment BC τmax Vmax tw d τmax 772 MPa τallow 100 MPa OK Hence Use W 150x14 Ans Set a 05L1 b 05L1 c L2 d L3 x1 0 001 a a x2 a 101 a a b x3 a b 101 a b a b c x4 a b c 101 a b c a b c d V1 x1 A 1 kN V2 x2 A P1 1 kN V3 x3 A P1 1 kN V4 x4 A P1 P2 1 kN M1 x1 MA A x1 kN m M2 x2 MA A x2 P1 x2 a 1 kN m M3 x3 B x3 a b 1 kN m M4 x4 B x4 a b P2 x4 a b c 1 kN m 0 2 4 6 8 10 0 10 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 2 4 6 8 20 0 20 Distance m Moment kNm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 1121 The steel beam has an allowable bending stress σallow 140 MPa and an allowable shear stress of τallow 90 MPa Determine the maximum load that can safely be supported Given σallow 140MPa a 2m τallow 90MPa bf 120mm df 20mm tw 20mm dw 150mm Solution L 2a Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df tw dw 05dw df bf df tw dw yc 5722 mm I 1 12 bf df 3 bf df 05df yc 2 1 12 tw dw 3 tw dw 05dw df yc 2 I 1533833333 mm4 Qmax h yc tw 05 h yc Qmax 12718827 mm3 Support Reactions By symmetry RR P ΣFy0 Rc P P 0 Rc 2P Maximum Load Assume failure due to bending moment Mmax P a cmax h yc σallow Mmax cmax I σallow P a h yc I P I σallow a h yc P 952 kN Ans Check Shear Vmax P Rc 2P τmax Vmax Qmax I tw τmax 3947 MPa τallow 90 MPa OK x1 0 001 a a x2 a 101 a L V1 x1 P 1 kN V2 x2 P Rc 1 kN M1 x1 P x1 1 kN m M2 x2 P x2 Rc x2 a 1 kN m 0 2 4 10 0 10 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 2 4 20 10 0 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1122 The timber beam has a rectangular cross section If the width of the beam is 150 mm determine its height h so that it simultaneously reaches its allowable bending stress of σallow 10 MPa and an allowable shear stress of τallow 035 MPa Also what is the maximum load P that the beam can then support Given σallow 10MPa L 15m τallow 035MPa b 150mm Solution Support Reactions By symmetry ABR ΣFy0 2R P 0 R 05P Section Property I b h3 12 Sx I 05h Sx b h2 6 Qmax 05h b 025 h Qmax 0125b h2 Maximum Moment and Shear Vmax R Vmax 05P Mmax R L Mmax 05P L If shear conrols τallow Vmax Qmax I b I b Qmax Vmax τallow b h3 b 12 0125b h2 05P τallow b h 3 P 4τallow 1 If bending conrols Sreqd Mmax σallow b h2 6 05P L σallow 2 Solving Eqs 1 and 2 h 4τallow L σallow h 210mm Ans From Eq 1 P 4 3 b h τallow P 1470 kN Ans R 05P R 735 kN x1 0 001 L L x2 L 101 L 2L V1 x1 R 1 kN V2 x2 R P 1 kN M1 x1 R x1 1 kN m M2 x2 R x2 P x2 L 1 kN m 0 1 2 3 0 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 0 10 20 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1123 The beam is to be used to support the machine which has a weight of 80 kN and a center of gravity at G If the maximum bending stress is not to exceed σallow 160 MPa determine the required width b of the flanges The supports at B and C are smooth Given W 80kN σallow 160MPa L1 18m L2 09m L3 15m L4 18m dw 175mm df 12mm tw 12mm Solution L L1 L2 L3 L4 Support Reactions Given For machine BC ΣFy0 B C W 0 1 ΣΜC0 B L2 L3 W L3 0 2 Solving Eqs 1 and 2 Guess B 1kN C 2kN B C Find B C B C 50 30 kN For beam AD ΣFy0 A D W 0 3 ΣΜD0 A L W L3 L4 0 4 Solving Eq 4 A W L3 L4 L A 44kN From Eq 3 D W A D 36kN Section Property h 2df dw I 1 12 b h3 b tw dw 3 I 1 12 b h3 dw 3 tw dw 3 Bending Stress Mmax A L1 cmax 05h σallow Mmax cmax I I Mmax cmax σallow 1 12 b h3 dw 3 tw dw 3 A L1 05h σallow b 12A L1 05h σallow tw dw 3 h3 dw 3 b 2089 mm Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 L3 x3 L1 L2 L3 101 L1 L2 L3 L V1 x1 A kN V2 x2 A B 1 kN V3 x3 A B C 1 kN M1 x1 A x1 kN m M2 x2 A x2 B x2 L1 1 kN m M3 x3 A x3 B x3 L1 C x3 L1 L2 L3 1 kN m 0 1 2 3 4 5 6 50 0 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 5 6 0 50 100 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1124 The beam has a flange width b 200 mm If the maximum bending stress is not to exceed σallow 160 MPa determine the greatest weight of the machine that the beam can support The center of gravity for the machine is at G and the supports at B and C are smooth Given L1 18m L2 09m L3 15m L4 18m dw 175mm df 12mm tw 12mm b 200mm σallow 160MPa Solution L L1 L2 L3 L4 Support Reactions For beam AD ΣFy0 A D W 0 1 ΣΜD0 A L W L3 L4 0 2 Solving Eq 2 A W L3 L4 L From Eq 1 D W A Section Property h 2df dw I 1 12 b h3 b tw dw 3 I 4737977500 mm4 Bending Stress Mmax A L1 cmax 05h σallow Mmax cmax I Mmax I σallow cmax W I σallow cmax L L1 L3 L4 W L3 L4 L L1 I σallow cmax W 7696 kN Ans Evaluate Support Reactions For beam AD For machine BC Given From Eqs 1 and 2 ΣFy0 B C W 0 3 A W L3 L4 L ΣΜC0 B L2 L3 W L3 0 4 D W A Solving Eqs 3 and 4 Guess B 1kN C 2kN A 4233 kN B C Find B C B C 481 2886 kN D 3463 kN x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 L3 x3 L1 L2 L3 101 L1 L2 L3 L V1 x1 A kN V2 x2 A B 1 kN V3 x3 A B C 1 kN M1 x1 A x1 kN m M2 x2 A x2 B x2 L1 1 kN m M3 x3 A x3 B x3 L1 C x3 L1 L2 L3 1 kN m 0 1 2 3 4 5 6 50 0 50 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 5 6 0 50 100 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1125 The box beam has an allowable bending stress of σallow 10 MPa and an allowable shear stress of τallow 775 kPa Determine the maximum intensity w of the distributed loading that it can safely support Also determine the maximum safe nail spacing for each third of the length of the beam Each nail can resist a shear force of 200 N Given Vallow 02kN L 6m bf 210mm df 30mm tw 30mm dw 190mm σallow 10MPa τallow 0775MPa Solution Section Property h 2df dw I 1 12 bf h3 1 12 bf 2tw dw 3 I 187700000mm4 QA 05 h 05df bf 2tw df QA 495000mm3 Qmax 1 8 bf h2 1 8 bf 2tw dw 2 Qmax 963750mm3 Support Reactions By symmetry RLRRR ΣFy0 2R wo L 0 R 05wo L Maximum Moment and Shear Vmax R Vmax 05 wo L Mmax R 05L wo 05L 025 L Mmax 0125 wo L2 Maximum Load Assume failure due to bending cmax 05 h σallow Mmax cmax I σallow 0125 wo L2 05 h I wo 16 I σallow h L2 wo 3337 kN m Assume failure due to shear τallow Vmax Qmax I 2tw τallow 05 wo L Qmax I 2tw wo 4 I τallow tw L Qmax wo 3019 kN m w min wo wo w 3019 kN m Ans Shear Flow Since there are two rows of nails the allowable shear flow is q 2Vs Region AB 0 x 2m and CD 4m x 6m Vmax 05 w L q1 Vmax QA I q1 2388 kN m s1 2Vallow q1 s1 167 mm Ans Region BC 2m x 4m Vmax Vmax w L 3 q2 Vmax QA I q2 796 kN m s2 2Vallow q2 s2 502 mm Ans R 05w L x 0 001 L L V x R w x 1 kN M x R x w x 05x 1 kN m 0 2 4 6 10 0 10 Distance m Shear kN V x x 0 2 4 6 0 10 20 Distance m Moment kNm M x x Problem 1126 The beam is constructed from three boards as shown If each nail can support a shear force of 250 N determine the maximum spacing of the nails s s and s for regions AB BC and CD respectively Given Vallow 250N a 15m bf 200mm df 25mm tw 25mm dw 150mm P1 4kN P2 6kN Solution L 3a Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df 2 tw dw 05dw df bf df 2 tw dw yc 6500 mm I 1 12 bf df 3 bf df 05df yc 2 2 1 12 tw dw 3 tw dw 05dw df yc 2 Q yc 05df bf df I 3729166667 mm4 Q 26250000 mm3 Support Reactions Given ΣFy0 B D P1 P2 0 1 ΣΜD0 P1 3a B 2a P2 a 0 2 Solving Eqs 1 and 2 Guess B 1kN D 2kN B D Find B D B D 9 1 kN Region AB VAB P1 q VAB Q I q 2816 kN m s Vallow 05q s 1776 mm Ans Region BC VBC P1 B q VBC Q I q 3520 kN m s Vallow 05q s 1421 mm Ans Region CD VCD P1 B P2 q VCD Q I q 704 kN m s Vallow 05q s 7103 mm Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a L V1 x1 P1 kN V2 x2 P1 B 1 kN V3 x3 P1 B P2 1 kN 0 1 2 3 4 10 5 0 5 10 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 1127 The beam is constructed from two boards as shown If each nail can support a shear force of 1 kN determine the maximum spacing of the nails s s and s to the nearest multiples of 5 mm for regions ABBC and CD respectively Given Vallow 1kN a 15m bf 200mm df 25mm tw 25mm dw 150mm P1 25kN P2 75kN Solution L 3a Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df tw dw 05dw df bf df tw dw yc 5000 mm I 1 12 bf df 3 bf df 05df yc 2 1 12 tw dw 3 tw dw 05dw df yc 2 Q yc 05df bf df I 2369791667 mm4 Q 18750000 mm3 Support Reactions Given ΣFy0 B D P1 P2 0 1 ΣΜD0 P1 3a B 2a P2 a 0 2 Solving Eqs 1 and 2 Guess B 1kN D 2kN B D Find B D B D 75 25 kN Region AB VAB P1 q VAB Q I q 1978 kN m s Vallow q s 5056 mm Use s 55mm Ans Region BC VBC P1 B q VBC Q I q 3956 kN m s Vallow q s 2528 mm Use s 30mm Ans Region CD VCD P1 B P2 q VCD Q I q 1978 kN m s Vallow q s 5056 mm Use s 55mm Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a L V1 x1 P1 kN V2 x2 P1 B 1 kN V3 x3 P1 B P2 1 kN 0 1 2 3 4 10 5 0 5 10 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 Problem 1128 Draw the shear and moment diagrams for the shaft and determine its required diameter to the nearest multiples of 5mm if σallow 50 kN and τallow 20 kN The bearings at A and D exert only vertical reactions on the shaft The loading is applied to the pulleys at B C and E Take P 550 N Given σallow 50MPa L1 350mm τallow 20MPa L2 500mm PB 400N L3 375mm PC 550N L4 300mm PE 175N Solution Lo L1 L2 L3 L Lo L4 Support Reactions Given ΣFy0 A D PB PC PE 0 1 ΣΜD0 A Lo PB L2 L3 PC L3 PE L4 0 2 Solving Eqs 1 and 2 Guess A 1N D 1N A D Find A D A D 41122 71378 N Maximum Moment and Shear Vmax A PB PC Vmax 53878 N Mmax A L1 L2 PB L2 Mmax 14954 N m Section Property I π do 4 64 Sx I 05 do Sx π do 3 32 Qmax 4 05 do 3π 1 2 π do 2 4 Qmax do 3 12 Bending Stress Assume bending controls the design Sreqd Mmax σallow π do 3 32 Mmax σallow do 3 32Mmax π σallow do 3123 mm Use 35mm Ans Check Shear I π do 4 64 Qmax do 3 12 τmax Vmax Qmax I do τmax 0938 MPa τallow 20 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L1 L2 L3 x4 L1 L2 L3 101 L1 L2 L3 L V1 x1 A 1 N V2 x2 A PB 1 kN V3 x3 A PB PC 1 N V4 x4 A PB PC D 1 N M1 x1 A x1 N m M2 x2 A x2 PB x2 L1 1 N m M3 x3 A x3 PB x3 L1 PC x3 L1 L2 1 N m M4 x4 A x4 PB x4 L1 PC x4 L1 L2 D L4 x4 L 1 N m 0 02 04 06 08 1 12 14 500 0 500 Distance m Shear N V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 02 04 06 08 1 12 14 100 0 100 200 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 1129 Draw the shear and moment diagrams for the shaft and determine its required diameter to the nearest multiples of 5mm if σallow 50 kN and τallow 20 kN The bearings at A and D exert only vertical reactions on the shaft The loading is applied to the pulleys at B C and E Take P 400 N Given σallow 50MPa L1 350mm τallow 20MPa L2 500mm PB 400N L3 375mm PC 400N L4 300mm PE 175N Solution Lo L1 L2 L3 L Lo L4 Support Reactions Given ΣFy0 A D PB PC PE 0 1 ΣΜD0 A Lo PB L2 L3 PC L3 PE L4 0 2 Solving Eqs 1 and 2 Guess A 1N D 1N A D Find A D A D 36531 60969 N Maximum Moment and Shear Vmax A PB PC Vmax 43469 N Mmax A L1 Mmax 12786 N m Section Property I π do 4 64 Sx I 05 do Sx π do 3 32 Qmax 4 05 do 3π 1 2 π do 2 4 Qmax do 3 12 Bending Stress Assume bending controls the design Sreqd Mmax σallow π do 3 32 Mmax σallow do 3 32Mmax π σallow do 2964 mm Use 30mm Ans Check Shear I π do 4 64 Qmax do 3 12 τmax Vmax Qmax I do τmax 0840 MPa τallow 20 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L1 L2 L3 x4 L1 L2 L3 101 L1 L2 L3 L V1 x1 A 1 N V2 x2 A PB 1 kN V3 x3 A PB PC 1 N V4 x4 A PB PC D 1 N M1 x1 A x1 N m M2 x2 A x2 PB x2 L1 1 N m M3 x3 A x3 PB x3 L1 PC x3 L1 L2 1 N m M4 x4 A x4 PB x4 L1 PC x4 L1 L2 D L4 x4 L 1 N m 0 02 04 06 08 1 12 14 500 0 500 Distance m Shear N V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 02 04 06 08 1 12 14 100 0 100 200 Distance m Moment Nm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 1130 The overhang beam is constructed using two 50mm by 100mm pieces of wood braced as shown If the allowable bending stress is σallow 42 MPa determine the largest load P that can be applied Also determine the associated maximum spacing of nails s along the beam section AC if each nail can resist a shear force of 4 kN Assume the beam is pinconnected at A B and D Neglect the axial force developed in the beam along DA Given Vallow 4kN L 09m b 100mm ho 50mm σallow 42MPa Solution h 2ho Maximum Moment and Shear Vmax P Mmax P L Section Property I b h3 12 Sx I 05h Qmax 05h b 025 h Qmax 0125bh2 Maximum Load Sreqd Mmax σallow Mmax Sx σallow P L Sx σallow P Sx σallow L P 7778 N Ans Nail Spacing Vmax P Qmax 0125bh2 q Vmax Qmax I q 117 kN m smax Vallow q smax 3429 mm Ans Problem 1131 The tapered beam supports a concentrated force P at its center If it is made from a plate that has a constant width b determine the absolute maximum bending stress in the beam Problem 1132 Determine the variation of the radius r of the cantilevered beam that supports the uniform distributed load so that it has a constant maximum bending stress σmax throughout its length Problem 1133 Determine the variation in the depth d of a cantilevered beam that supports a concentrated force P at its end so that it has a constant maximum bending stress σallow throughout its length The beam has a constant width b0 Problem 1134 The beam is made into the shape of a frustum and has a diameter of 12 mm at A and a diameter of 300 mm at B If it supports a force of 750 N at A determine the absolute maximum bending stress in the beam and specify its location x Given P 750N L 900mm do 150mm d1 300mm Solution Section Property δr d1 do 2 r δr x δr L r δr L x L I π r4 4 S I r S π r3 4 S π δr 3 L x 3 4 L3 Bendiug Stress M P x σ M S σ P x 4 L3 π δr 3 L x 3 1 In order to have the absolute maximum bending stress x d σ d 0 Differentiate Eq 1 4P L3 π δr 3 x x L x 3 d d 0 L x 3 x d x d x x L x 3 d d 0 L x 3 x 3 L x 2 0 L x 2 L 2x 0 x L 2 x 450mm Ans Substituting into Eq 1 σmax P x 4 L3 π δr 3 L x 3 σmax 03018 MPa Ans Problem 1135 The beam has a width w and a depth that varies as shown If it supports a concentrated force P at its end determine the absolute maximum bending stress in the beam and specify its location x Problem 1136 The tapered beam supports a uniform distributed load w If it is made from a plate and has a constant width b determine the absolute maximum bending stress in the beam Problem 1137 The tapered simply supported beam supports the concentrated force P at its center Determine the absolute maximum bending stress in the beam Problem 1138 t0ta0t0nt0The bearings at A and D exert only y and z components of force on the shaft If τallow 60 MPa determine to the nearest millimeter the smallestdiameter shaft that will support the loading Use the maximumshearstress theory of failure Given a 300mm r 50mm τallow 60MPa PB 5kN PC 5kN Solution L 3a Support Reactions In xz plane ΣFz0 Az DZ PB 0 1 ΣΜD0 Az 3a PB 2a 0 2 Solving Eqs 1 and 2 Az 2 3PB Az 33333 kN Dz PB Az Dz 16667 kN In xy plane ΣFy0 Ay Dy PC 0 3 ΣΜD0 PC a Ay 3a 0 4 Solving Eqs 3 and 4 Ay 1 3PC Ay 16667 kN Dy PC Ay Dy 33333 kN Torsion occurs in segment BC TBC PB r TBC 0250 kN m Critical Section Located just to the left of gear C and just to the right of gear B where My Dz a Mz Dy a M My 2 Mz 2 M 1118 kN m T TBC T 0250 kN m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 2299 mm do 2c do 4599 mm Use do 46mm Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a My1 x1 Az x1 kN m My2 x2 Az x2 PB x2 a 1 kN m My3 x3 Az x3 PB x3 a 1 kN m 0 02 04 06 08 0 05 1 Distance m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Mz1 x1 Ay x1 kN m Mz2 x2 Ay x2 kN m Mz3 x3 Ay x3 PC x3 2a 1 kN m 0 02 04 06 08 1 05 0 Distance m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 Mx1 x1 0 Mx2 x2 r PB kN m Mx3 x3 0 0 02 04 06 08 05 0 Distance m Moment kNm Mx1 x1 Mx2 x2 Mx3 x3 x1 x2 x3 Problem 1139 Solve Prob 1138 using the maximumdistortionenergy theory of failure with σallow 180 MPa Given a 300mm r 50mm σallow 180MPa PB 5kN PC 5kN Solution L 3a Support Reactions In xz plane ΣFz0 Az DZ PB 0 1 ΣΜD0 Az 3a PB 2a 0 2 Solving Eqs 1 and 2 Az 2 3PB Az 33333 kN Dz PB Az Dz 16667 kN In xy plane ΣFy0 Ay Dy PC 0 3 ΣΜD0 PC a Ay 3a 0 4 Solving Eqs 3 and 4 Ay 1 3PC Ay 16667 kN Dy PC Ay Dy 33333 kN Torsion occurs in segment BC TBC PB r TBC 0250 kN m Critical Section Located just to the left of gear C and just to the right of gear B where My Dz a Mz Dy a M My 2 Mz 2 M 1118 kN m T TBC T 0250 kN m Maximum Distortion Energy Theory Applying Eq 95 σ1 05 σx σy 05 σx σy 2 τxy 2 σ2 05 σx σy 05 σx σy 2 τxy 2 where σy 0 σx M c I M c π 4 c4 4M c π c4 τxy T c J T c π 2 c4 2T c π c4 Let a 05σx and b 05σx 2 τxy 2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σx 2 3 05σx 2 τxy 2 2 σallow 2 σx 2 3τxy 2 σallow 2 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 2005 mm do 2c do 4009 mm Use do 41mm Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a My1 x1 Az x1 kN m My2 x2 Az x2 PB x2 a 1 kN m My3 x3 Az x3 PB x3 a 1 kN m 0 02 04 06 08 0 05 1 Distance m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Mz1 x1 Ay x1 kN m Mz2 x2 Ay x2 kN m Mz3 x3 Ay x3 PC x3 2a 1 kN m 0 02 04 06 08 1 05 0 Distance m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 Mx1 x1 0 Mx2 x2 r PB kN m Mx3 x3 0 0 02 04 06 08 05 0 Distance m Moment kNm Mx1 x1 Mx2 x2 Mx3 x3 x1 x2 x3 Problem 1140 The bearings at A and D exert only y and z components of force on the shaft If τallow 60 MPa determine to the nearest millimeter the smallestdiameter shaft that will support the loading Use the maximumshearstress theory of failure Given L1 200mm L2 400mm L3 350mm τallow 60MPa rB 50mm rC 75mm PB 3kN PC 2kN Solution L L1 L2 L3 Support Reactions In xz plane ΣFz0 Az DZ PC 0 1 ΣΜD0 Az L PC L3 0 2 Solving Eqs 1 and 2 Az L3 L PC Az 07368 kN Dz PB Az Dz 22632 kN In xy plane ΣFy0 Ay Dy PB 0 3 ΣΜD0 PB L2 L3 Ay L 0 4 Solving Eqs 3 and 4 Ay L2 L3 L PB Ay 23684 kN Dy PB Ay Dy 06316 kN Torsion occurs in segment BC TBC PB rB TBC 0150 kN m Critical Section Located just to right of gear B where My Az L1 Mz Ay L1 M My 2 Mz 2 M 0496 kN m T TBC T 0150 kN m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 1765 mm do 2c do 3530 mm Use do 36mm Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L My1 x1 Az x1 kN m My2 x2 Az x2 kN m My3 x3 Az x3 PC x3 L1 L2 1 kN m 0 02 04 06 08 0 02 04 Distance m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Mz1 x1 Ay x1 kN m Mz2 x2 Ay x2 PB x2 L1 1 kN m Mz3 x3 Ay x3 PB x3 L1 1 kN m 0 02 04 06 08 0 02 04 Distance m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 Mx1 x1 0 Mx2 x2 rB PB kN m Mx3 x3 0 0 02 04 06 08 0 01 Distance m Moment kNm Mx1 x1 Mx2 x2 Mx3 x3 x1 x2 x3 Problem 1141 The bearings at A and D exert only y and z components of force on the shaft If τallow 60 MPa determine to the nearest millimeter the smallestdiameter shaft that will support the loading Use the maximumdistortionenergy theory of failure σallow 130 MPa Given L1 200mm L2 400mm L3 350mm σallow 130MPa rB 50mm rC 75mm PB 3kN PC 2kN Solution L L1 L2 L3 Support Reactions In xz plane ΣFz0 Az DZ PC 0 1 ΣΜD0 Az L PC L3 0 2 Solving Eqs 1 and 2 Az L3 L PC Az 07368 kN Dz PB Az Dz 22632 kN In xy plane ΣFy0 Ay Dy PB 0 3 ΣΜD0 PB L2 L3 Ay L 0 4 Solving Eqs 3 and 4 Ay L2 L3 L PB Ay 23684 kN Dy PB Ay Dy 06316 kN Torsion occurs in segment BC TBC PB rB TBC 0150 kN m Critical Section Located just to right of gear B where My Az L1 Mz Ay L1 M My 2 Mz 2 M 0496 kN m T TBC T 0150 kN m Maximum Distortion Energy Theory Applying Eq 95 σ1 05 σx σy 05 σx σy 2 τxy 2 σ2 05 σx σy 05 σx σy 2 τxy 2 where σy 0 σx M c I M c π 4 c4 4M c π c4 τxy T c J T c π 2 c4 2T c π c4 Let a 05σx and b 05σx 2 τxy 2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σx 2 3 05σx 2 τxy 2 2 σallow 2 σx 2 3τxy 2 σallow 2 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 1713 mm do 2c do 3425 mm Use do 35mm Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L My1 x1 Az x1 kN m My2 x2 Az x2 kN m My3 x3 Az x3 PC x3 L1 L2 1 kN m 0 02 04 06 08 0 02 04 Distance m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Mz1 x1 Ay x1 kN m Mz2 x2 Ay x2 PB x2 L1 1 kN m Mz3 x3 Ay x3 PB x3 L1 1 kN m 0 02 04 06 08 0 02 04 Distance m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 Mx1 x1 0 Mx2 x2 rB PB kN m Mx3 x3 0 0 02 04 06 08 0 01 Distance m Moment kNm Mx1 x1 Mx2 x2 Mx3 x3 x1 x2 x3 L2 15m Given P1 1250N P2 750N r 150mm L1 03m L3 06m τallow 84MPa Solution L L1 L2 L3 Support Reactions Ps P1 P2 Ans In yz plane ΣFz0 Az BZ P1 P2 0 1 ΣΜB0 Az L P1 P2 L2 L3 0 2 Solving Eqs 1 and 2 Az 1 L P1 P2 L2 L3 Az 1750 N Bz P1 P2 Az Bz 250 N In xy plane ΣFx0 Ax Bx P1 P2 0 3 ΣΜB0 Ax L P1 P2 L3 0 4 Solving Eqs 3 and 4 Ax 1 L P1 P2 L3 Ax 500 N Bx P1 P2 Ax Bx 1500 N Torsion occurs in segment DC T P1 P2 r T 75N m Critical Section Located just to the left of point C Mx Bx L3 Mz Bz L3 M Mx 2 Mz 2 M 91241 N m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 1907 mm do 2c do 3815 mm Use do 39mm Problem 1142 The pulleys attached to the shaft are loaded as shown If the bearings at A and B exert only horizontal and vertical forces on the shaft determine the required diameter of the shaft to the nearest mm using the maximumshearstress theory of failure τallow 84 MPa y1 0 001 L1 L1 y2 L1 101 L1 L1 L2 y3 L1 L2 101 L1 L2 L Mz1 y1 Az y1 1 N m Mz2 y2 Az y2 Ps y2 L1 1 N m Mz3 y3 Az y3 Ps y3 L1 1 N m 0 1 2 0 500 Distance m Mz Nm Mz1 y1 Mz2 y2 Mz3 y3 y1 y2 y3 Mx1 y1 Ax y1 N m Mx2 y2 Ax y2 N m Mx3 y3 Ax y3 Ps y3 L1 L2 1 N m 0 1 2 0 500 1000 Distance m Mx Nm Mx1 y1 Mx2 y2 Mx3 y3 y1 y2 y3 My1 y1 0 My2 y2 T N m My3 y3 0 0 1 2 0 50 100 Distance m My Nm My1 y1 My2 y2 My3 y3 y1 y2 y3 Problem 1143 The pulleys attached to the shaft are loaded as shown If the bearings at A and B exert only horizontal and vertical forces on the shaft determine the required diameter of the shaft to the nearest mm Use the maximumditortionenergy theory of failure σallow 140 MPa Given P1 1250N P2 750N r 150mm L1 03m L2 15m L3 06m σallow 140MPa Solution L L1 L2 L3 Support Reactions Ps P1 P2 In yz plane ΣFz0 Az BZ P1 P2 0 1 ΣΜB0 Az L P1 P2 L2 L3 0 2 Solving Eqs 1 and 2 Az 1 L P1 P2 L2 L3 Az 1750 N Bz P1 P2 Az Bz 250 N In xy plane ΣFx0 Ax Bx P1 P2 0 3 ΣΜB0 Ax L P1 P2 L3 0 4 Solving Eqs 3 and 4 Ax 1 L P1 P2 L3 Ax 500 N Bx P1 P2 Ax Bx 1500 N Torsion occurs in segment DC T P1 P2 r T 75N m Critical Section Located just to the left of point C Mx Bx L3 Mz Bz L3 M Mx 2 Mz 2 M 91241 N m Maximum Distortion Energy Theory Both states of stress will yield the same result σ1 05σ 05σ 2 τ2 σ2 05σ 05σ 2 τ2 Let a 05σ and b 05σ 2 τ2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σ 2 3 05σ 2 τ2 2 σallow 2 σ2 3τ2 σallow 2 5 σ M c I M c π 4 c4 4M c π c4 τ T c J T c π 2 c4 2T c π c4 From Eq 5 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 2026 mm do 2c do 4052 mm Use do 41mm Ans y1 0 001 L1 L1 y2 L1 101 L1 L1 L2 y3 L1 L2 101 L1 L2 L Mz1 y1 Az y1 1 N m Mz2 y2 Az y2 Ps y2 L1 1 N m Mz3 y3 Az y3 Ps y3 L1 1 N m 0 1 2 0 500 Distance m Mz Nm Mz1 y1 Mz2 y2 Mz3 y3 y1 y2 y3 Mx1 y1 Ax y1 N m Mx2 y2 Ax y2 N m Mx3 y3 Ax y3 Ps y3 L1 L2 1 N m 0 1 2 0 500 1000 Distance m Mx Nm Mx1 y1 Mx2 y2 Mx3 y3 y1 y2 y3 My1 y1 0 My2 y2 T N m My3 y3 0 0 1 2 0 50 100 Distance m My Nm My1 y1 My2 y2 My3 y3 y1 y2 y3 Problem 1144 The shaft is supported on journal bearings that do not offer resistance to axial load If the allowable normal stress for the shaft is σallow 80 MPa determine to the nearest millimeter the smallest diameter of the shaft that will support the loading Use the maximumdistortionenergy theory of failure Given La 250mm Lb 500mm σallow 80MPa rD 150mm rC 100mm θ 30deg PD 020kN PC 035kN PD 010kN PC 015kN Solution L La Lb La Support Reactions In yz plane ΣFz0 Az BZ PC sin θ PD sin θ 0 1 ΣΜA0 Bz L PC sin θ L La PD sin θ La 0 2 Solving Eqs 1 and 2 Bz PC L La L PD La L sin θ Bz 015625 kN Az PC PD sin θ Bz Az 011875 kN In xy plane ΣFx0 Ax Bx PC cos θ PD cos θ 0 3 ΣΜA0 PC cos θ L La PD cos θ La Bx L 0 4 Solving Eqs 3 and 4 Bx PC L La L PD La L cos θ Bx 018403 kN Ax PC PD cos θ Bx Ax 005413 kN Torsion occurs in segment CD TCD PC rC TCD 0015 kN m Critical Section Located just to the left of gear C where Mx Bz La Mz Bx La M Mx 2 Mz 2 M 0060354 kN m T TCD T 0015 kN m Maximum Distortion Energy Theory Applying Eq 95 σ1 05 σx σy 05 σx σy 2 τxy 2 σ2 05 σx σy 05 σx σy 2 τxy 2 where σy 0 σx M c I M c π 4 c4 4M c π c4 τxy T c J T c π 2 c4 2T c π c4 Let a 05σx and b 05σx 2 τxy 2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σx 2 3 05σx 2 τxy 2 2 σallow 2 σx 2 3τxy 2 σallow 2 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 994 mm do 2c do 1988 mm Use do 20mm Ans y1 0 001 La La y2 La 101 La La Lb y3 La Lb 101 La Lb L Mx1 y1 Az y1 kN m Mx2 y2 Az y2 PD sin θ y2 La 1 kN m Mx3 y3 Az y3 PD sin θ y3 La PC sin θ y3 La Lb 1 kN m 0 02 04 06 08 0 002 004 Distance m Moment kNm Mx1 y1 Mx2 y2 Mx3 y3 y1 y2 y3 Mz1 y1 Ax y1 kN m Mz2 y2 Ax y2 PD cos θ y2 La 1 kN m Mz3 y3 Ax y3 PD cos θ y3 La PC cos θ y3 La Lb 1 kN m 0 02 04 06 08 0 005 Distance m Moment kNm Mz1 y1 Mz2 y2 Mz3 y3 y1 y2 y3 My1 y1 0 My2 y2 T kN m My3 y3 0 0 02 04 06 08 0 002 Distance m Moment kNm My1 y1 My2 y2 My3 y3 y1 y2 y3 Problem 1145 The shaft is supported on journal bearings that do not offer resistance to axial load If the allowable shear stress for the shaft is τallow 35 MPa determine to the nearest millimeter the smallest diameter of the shaft that will support the loading Use the maximumshearstress theory of failure Given La 250mm Lb 500mm τallow 35MPa rD 150mm rC 100mm θ 30deg PD 020kN PC 035kN PD 010kN PC 015kN Solution L La Lb La Support Reactions In yz plane ΣFz0 Az BZ PC sin θ PD sin θ 0 1 ΣΜA0 Bz L PC sin θ L La PD sin θ La 0 2 Solving Eqs 1 and 2 Bz PC L La L PD La L sin θ Bz 015625 kN Az PC PD sin θ Bz Az 011875 kN In xy plane ΣFx0 Ax Bx PC cos θ PD cos θ 0 3 ΣΜA0 PC cos θ L La PD cos θ La Bx L 0 4 Solving Eqs 3 and 4 Bx PC L La L PD La L cos θ Bx 018403 kN Ax PC PD cos θ Bx Ax 005413 kN Torsion occurs in segment CD TCD PC rC TCD 0015 kN m Critical Section Located just to the left of gear C where Mx Bz La Mz Bx La M Mx 2 Mz 2 M 0060354 kN m T TCD T 0015 kN m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 1042 mm do 2c do 2084 mm Use do 21mm Ans y1 0 001 La La y2 La 101 La La Lb y3 La Lb 101 La Lb L Mx1 y1 Az y1 kN m Mx2 y2 Az y2 PD sin θ y2 La 1 kN m Mx3 y3 Az y3 PD sin θ y3 La PC sin θ y3 La Lb 1 kN m 0 02 04 06 08 0 002 004 Distance m Moment kNm Mx1 y1 Mx2 y2 Mx3 y3 y1 y2 y3 Mz1 y1 Ax y1 kN m Mz2 y2 Ax y2 PD cos θ y2 La 1 kN m Mz3 y3 Ax y3 PD cos θ y3 La PC cos θ y3 La Lb 1 kN m 0 02 04 06 08 0 005 Distance m Moment kNm Mz1 y1 Mz2 y2 Mz3 y3 y1 y2 y3 My1 y1 0 My2 y2 T kN m My3 y3 0 0 02 04 06 08 0 002 Distance m Moment kNm My1 y1 My2 y2 My3 y3 y1 y2 y3 Problem 1146 The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft If the allowable normal stress for the shaft is σallow 105 MPa determine to the nearest mm the smallest diameter of the shaft that will support the gear loading Use the maximumdistortionenergy theory of failure Given La 200mm Lb 100mm PC 1kN r 100mm PD 025kN PE 125kN σallow 105MPa Solution L 3La Lb Support Reactions In yz plane ΣFz0 Az BZ PD 0 1 ΣΜB0 Az 2La Lb PD La Lb 0 2 Solving Eqs 1 and 2 Az La Lb 2La Lb PD Az 150 N Bz PD Az Bz 100 N In xy plane ΣFx0 Ax Bx PC PE 0 3 ΣΜB0 PC L Ax 2La Lb PE Lb 0 4 Solving Eqs 3 and 4 Ax 1 2La Lb PE Lb PC L Ax 1150 N Bx PE PC Ax Bx 1400 N Torsion occurs in segment DC TDC PC r TDC 100N m in segment DE TDE PE r TDE 125N m Critical Section Located at support A within segment DC Mx PC La Mz 0 M Mx 2 Mz 2 M 200N m T TDC T 100N m Maximum Distortion Energy Theory Applying Eq 95 σ1 05 σx σy 05 σx σy 2 τxy 2 σ2 05 σx σy 05 σx σy 2 τxy 2 where σy 0 σx M c I M c π 4 c4 4M c π c4 τxy T c J T c π 2 c4 2T c π c4 Let a 05σx and b 05σx 2 τxy 2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σx 2 3 05σx 2 τxy 2 2 σallow 2 σx 2 3τxy 2 σallow 2 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 1383 mm do 2c do 2765 mm Use do 28mm Ans y1 0 001 La La y2 La 101 La 2La y3 2La 101 2La 3La y4 3La 101 3La L Mz1 y1 0 y1 N m Mz2 y2 Az y2 La 1 N m Mz3 y3 Az y3 La PD y3 2La 1 N m Mz4 y4 Az y4 La PD y4 2La 1 N m 0 02 04 06 0 20 Distance m Moment Nm Mz1 y1 Mz2 y2 Mz3 y3 Mz4 y4 y1 y2 y3 y4 Mx1 y1 PC y1 N m Mx2 y2 PC y2 Ax y2 La 1 N m Mx3 y3 PC y3 Ax y3 La 1 N m Mx4 y4 PC y4 Ax y4 La PE y4 3La 1 N m 0 02 04 06 0 100 200 Distance m Moment Nm Mx1 y1 Mx2 y2 Mx3 y3 Mx4 y4 y1 y2 y3 y4 My1 y1 r PC N m My2 y2 r PC N m My3 y3 r PC r PD 1 N m My4 y4 r PC r PD r PE 1 N m 0 02 04 06 150 100 50 0 Distance m Moment Nm My1 y1 My2 y2 My3 y3 My4 y4 y1 y2 y3 y4 Problem 1147 The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft If the allowable normal stress for the shaft is σallow 105 MPa determine to the nearest mm the smallest diameter of the shaft that will support the gear loading Use the maximumshearstress theory of failure with τallow 42 MPa Given La 200mm Lb 100mm PC 1kN r 100mm PD 025kN PE 125kN τallow 42MPa Solution L 3La Lb Support Reactions In yz plane ΣFz0 Az BZ PD 0 1 ΣΜB0 Az 2La Lb PD La Lb 0 2 Solving Eqs 1 and 2 Az La Lb 2La Lb PD Az 150 N Bz PD Az Bz 100 N In xy plane ΣFx0 Ax Bx PC PE 0 3 ΣΜB0 PC L Ax 2La Lb PE Lb 0 4 Solving Eqs 3 and 4 Ax 1 2La Lb PE Lb PC L Ax 1150 N Bx PE PC Ax Bx 1400 N Torsion occurs in segment DC TDC PC r TDC 100N m in segment DE TDE PE r TDE 125N m Critical Section Located at support A within segment DC Mx PC La Mz 0 M Mx 2 Mz 2 M 20000 N m T TDC T 10000 N m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 150 mm do 2c do 300 mm Use do 31mm Ans y1 0 001 La La y2 La 101 La 2La y3 2La 101 2La 3La y4 3La 101 3La L Mz1 y1 0 y1 N m Mz2 y2 Az y2 La 1 N m Mz3 y3 Az y3 La PD y3 2La 1 N m Mz4 y4 Az y4 La PD y4 2La 1 N m 0 02 04 06 0 20 Distance m Moment Nm Mz1 y1 Mz2 y2 Mz3 y3 Mz4 y4 y1 y2 y3 y4 Mx1 y1 PC y1 N m Mx2 y2 PC y2 Ax y2 La 1 N m Mx3 y3 PC y3 Ax y3 La 1 N m Mx4 y4 PC y4 Ax y4 La PE y4 3La 1 N m 0 02 04 06 0 100 200 Distance m Moment Nm Mx1 y1 Mx2 y2 Mx3 y3 Mx4 y4 y1 y2 y3 y4 My1 y1 r PC N m My2 y2 r PC N m My3 y3 r PC r PD 1 N m My4 y4 r PC r PD r PE 1 N m 0 02 04 06 150 100 50 0 Distance m Moment Nm My1 y1 My2 y2 My3 y3 My4 y4 y1 y2 y3 y4 Problem 1148 The end gear connected to the shaft is subjected to the loading shown If the bearings at A and B exert only y and z components of force on the shaft determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading Use the maximumshearstress theory of failure with τallow 60 MPa Given L1 150mm L2 250mm L3 100mm τallow 60MPa rD 100mm rC 75mm rC 50mm PD 15kN Solution L L1 L2 L3 Equilibrium Torque at C ΣΜx0 PD rD FC rC 0 FC rD rC PD FC 200 kN Thus TC FC rC TC 0100 kN m Ans Support Reactions In xz plane ΣFz0 Az Bz PD 0 1 ΣΜB0 Az L L1 PD L 0 2 Solving Eqs 1 and 2 Az L L L1 PD Az 21429 kN Bz PD Az Bz 06429 kN In xy plane ΣFy0 Ay By FC 0 3 ΣΜB0 Ay L L1 FC L3 0 4 Solving Eqs 3 and 4 Ay L3 L L1 FC Ay 05714 kN By FC Ay By 14286 kN Torsion occurs in segment DC TDC PD rD TDC 0150 kN m Critical Section Located just to right of gear A where My PD L1 Mz 0 M My 2 Mz 2 M 0225 kN m T TDC T 0150 kN m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 1421 mm do 2c do 2842 mm Use do 29mm Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L My1 x1 PD x1 kN m My2 x2 PD x2 Az x2 L1 1 kN m My3 x3 PD x3 Az x3 L1 1 kN m 0 01 02 03 04 0 02 Distance m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Mz1 x1 0 Mz2 x2 Ay x2 L1 1 kN m Mz3 x3 Ay x3 L1 FC x3 L1 L2 1 kN m 0 01 02 03 04 0 01 02 Distance m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 Mx1 x1 T kN m Mx2 x2 T kN m Mx3 x3 0 0 01 02 03 04 0 02 Distance m Moment kNm Mx1 x1 Mx2 x2 Mx3 x3 x1 x2 x3 Problem 1149 The end gear connected to the shaft is subjected to the loading shown If the bearings at A and B exert only y and z components of force on the shaft determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading Use the maximumdistortionenergy theory of failure with σallow 80 MPa Given L1 150mm L2 250mm L3 100mm σallow 80MPa rD 100mm rC 75mm rC 50mm PD 15kN Solution L L1 L2 L3 Equilibrium Torque at C ΣΜx0 PD rD FC rC 0 FC rD rC PD FC 200 kN Thus TC FC rC TC 0100 kN m Ans Support Reactions In xz plane ΣFz0 Az Bz PD 0 1 ΣΜB0 Az L L1 PD L 0 2 Solving Eqs 1 and 2 Az L L L1 PD Az 21429 kN Bz PD Az Bz 06429 kN In xy plane ΣFy0 Ay By FC 0 3 ΣΜB0 Ay L L1 FC L3 0 4 Solving Eqs 3 and 4 Ay L3 L L1 FC Ay 05714 kN By FC Ay By 14286 kN Torsion occurs in segment DC TDC PD rD TDC 0150 kN m Critical Section Located just to right of gear A where My PD L1 Mz 0 M My 2 Mz 2 M 0225 kN m T TDC T 0150 kN m Maximum Distortion Energy Theory Applying Eq 95 σ1 05 σx σy 05 σx σy 2 τxy 2 σ2 05 σx σy 05 σx σy 2 τxy 2 where σy 0 σx M c I M c π 4 c4 4M c π c4 τxy T c J T c π 2 c4 2T c π c4 Let a 05σx and b 05σx 2 τxy 2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σx 2 3 05σx 2 τxy 2 2 σallow 2 σx 2 3τxy 2 σallow 2 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 1605 mm do 2c do 3210 mm Use do 33mm Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L My1 x1 PD x1 kN m My2 x2 PD x2 Az x2 L1 1 kN m My3 x3 PD x3 Az x3 L1 1 kN m 0 01 02 03 04 0 02 Distance m Moment kNm My1 x1 My2 x2 My3 x3 x1 x2 x3 Mz1 x1 0 Mz2 x2 Ay x2 L1 1 kN m Mz3 x3 Ay x3 L1 FC x3 L1 L2 1 kN m 0 01 02 03 04 0 01 02 Distance m Moment kNm Mz1 x1 Mz2 x2 Mz3 x3 x1 x2 x3 Mx1 x1 T kN m Mx2 x2 T kN m Mx3 x3 0 0 01 02 03 04 0 02 Distance m Moment kNm Mx1 x1 Mx2 x2 Mx3 x3 x1 x2 x3 Problem 1150 Draw the shear and moment diagrams for the shaft and then determine its required diameter to the nearest millimeter if σallow 140 MPa and τallow 80 MPa The bearings at A and B exert only vertical reactions on the shaft Given L1 125mm L2 600mm L3 75mm P1 08kN P2 15kN τallow 80MPa σallow 140MPa Solution L L1 L2 L3 Support Reactions ΣFz0 A B P1 P2 0 1 ΣΜB0 A L P1 L L1 P2 L3 0 2 Solving Eqs 1 and 2 A 1 L P1 L L1 P2 L3 A 0815625 kN B P1 P2 A B 1484375 kN Maximum Moment and Shear Vmax B Vmax 1484375 kN Mmax B L3 Mmax 0111328 kN m Section Property I π do 4 64 S I 05do S π do 3 32 Bending Stress Sreqd Mmax σallow π do 3 32 Mmax σallow do 3 32Mmax π σallow do 201 mm Use do 21mm Ans Shear Check Ao π do 2 4 I π do 4 64 Qmax 2 do 3π Ao 2 τmax Vmax Qmax I do τmax 5714 MPa τallow 80 MPa OK x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 A kN V2 x2 A P1 1 kN V3 x3 A P1 P2 1 kN M1 x1 A x1 kN m M2 x2 A x2 P1 x2 L1 1 kN m M3 x3 A x3 P1 x3 L1 P2 x3 L1 L2 1 kN m 0 02 04 06 2 1 0 1 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 02 04 06 0 005 01 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1151 The cantilevered beam has a circular cross section If it supports a force P at its end determine its radius y as a function of x so that it is subjected to a constant maximum bending stress σallow throughout its length Problem 1152 The simply supported beam is made of timber that has an allowable bending stress of σallow 8 MPa and an allowable shear stress of τallow 750 kPa Determine its dimensions if it is to be rectangular and have a heighttowidth ratio of hb 125 Given σallow 8MPa a 3m τallow 075MPa wo 03 kN m h 125 b Solution L 2a Support Reactions ΣFy0 A B 05 wo a 0 1 ΣΜA0 05 wo a 2a 3 B L 0 2 Solving Eqs 1 and 2 A wo a 3 B wo a 6 Maximum Moment and Shear Mmax occurs at x where Vx0 V A x wo 2 x a 0 A x wo 2 x a x 2a wo wo a 3 x 2 3 a Vmax max A B Vmax 0300 kN Mmax A x 05 x wo x a x 3 Mmax 04899 kN m Section Property I b h3 12 Sx I 05h Sx b h2 6 Sx b 125 b 2 6 Sx 25 b3 96 Bending Stress Sreqd Mmax σallow 25 b3 96 Mmax σallow b 3 96Mmax 25σallow b 617 mm Ans h 125 b h 772 mm Ans Shear Check I b h3 12 Qmax 05b h 025 h τmax Vmax Qmax I b τmax 0094 MPa τallow 075 MPa OK x1 0 001 a a x2 a 101 a L V1 x1 A wo 2 x1 a x1 1 kN V2 x2 A 05 wo a 1 kN M1 x1 A x1 wo 2 x1 a x1 x1 3 1 kN m M2 x2 A x2 wo a 2 x2 2 a 3 1 kN m 0 1 2 3 4 5 6 05 0 05 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 1 2 3 4 5 6 0 05 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1153 The beam is made in the shape of a frustum that has a diameter of 03 m at A and a diameter of 06 m at B If it supports a couple moment of 12 kNm at its end determine the absolute maximum bending stress in the beam and specify its location x Given M 12kN m L 18m do 03m d1 06m Solution Section Property δr d1 do 2 r δr x δr L r δr L x L I π r4 4 S I r S π r3 4 S π δr 3 L x 3 4 L3 Bendiug Stress σ M S σ M 4 L3 π δr 3 L x 3 1 Since σ is a decreasing function the maximum bending stress occurs at x 0 Ans Substituting x0 into Eq 1 σmax M 4 L3 π δr 3 L x 3 σmax 4527 MPa Ans Problem 1154 Select the lightestweight steel wideflange overhanging beam from Appendix B that will safely support the loading Assume the support at A is a pin and the support at B is a roller The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa Given σallow 168MPa P 10kN τallow 100MPa L1 24m L2 06m L3 12m Solution L L1 L2 L3 Support Reactions ΣFy0 A B 2P 0 1 ΣΜB0 A L1 P L2 P L2 L3 0 2 Solving Eqs 1 and 2 A 2L2 L3 L1 P A 10 kN B 2P A B 30kN Maximum Moment and Shear Vmax A B Vmax 20kN Mmax A L1 Mmax 24 kN m Bending Stress Sreqd Mmax σallow Sreqd 14285714 mm3 Select W 250x18 Sx 179 103 mm3 d 251mm tw 483mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 165 MPa τallow 100 MPa OK Hence Use W 250x18 Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L V1 x1 A kN V2 x2 A B 1 kN V3 x3 A B P 1 kN M1 x1 A x1 kN m M2 x2 A x2 B x2 L1 1 kN m M3 x3 A x3 B x3 L1 P x3 L1 L2 1 kN m 0 1 2 3 4 0 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 1 2 3 4 20 0 Distance m Moment kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1155 The bearings at A and B exert only x and z components of force on the steel shaft Determine the shafts diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of τallow 80 MPa Use the maximumshearstress theory of failure Given PC 75kN PD 5kN rC 50mm L1 150mm L2 350mm L3 250mm τallow 80MPa rD 75mm Solution L L1 L2 L3 Support Reactions In yz plane ΣFz0 Az BZ PC 0 1 ΣΜB0 Az L PC L3 0 2 Solving Eqs 1 and 2 Az L3 L PC Az 25 kN Bz PC Az Bz 5kN In xy plane ΣFx0 Ax Bx PD 0 3 ΣΜB0 Ax L PD L L1 0 4 Solving Eqs 3 and 4 Ax L L1 L PD Ax 4kN Bx PD Ax Bx 1kN Torsion occurs in segment DC T PC rC T 0375 kN m Critical Section Located just to the left of gear C Mx Bx L3 Mz Bz L3 M Mx 2 Mz 2 M 127475 kN m Maximum Shear Stress Theory c 3 2 π τallow M2 T2 c 2195 mm do 2c do 4390 mm Use do 44mm Ans y1 0 001 L1 L1 y2 L1 101 L1 L1 L2 y3 L1 L2 101 L1 L2 L Mz1 y1 Az y1 kN m Mz2 y2 Az y2 kN m Mz3 y3 Az y3 PC y3 L1 L2 1 kN m 0 05 0 1 Distance m Mz kNm Mz1 y1 Mz2 y2 Mz3 y3 y1 y2 y3 Mx1 y1 Ax y1 kN m Mx2 y2 Ax y2 PD y2 L1 1 kN m Mx3 y3 Ax y3 PD y3 L1 1 kN m 0 05 0 05 Distance m Mx kNm Mx1 y1 Mx2 y2 Mx3 y3 y1 y2 y3 My1 y1 0 My2 y2 T kN m My3 y3 0 0 05 0 05 Distance m My kNm My1 y1 My2 y2 My3 y3 y1 y2 y3 Problem 1156 The bearings at A and B exert only x and z components of force on the steel shaft Determine the shafts diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of τallow 80 MPa Use the maximumdistortionenergy theory of failure with σallow 200 MPa Given PC 75kN PD 5kN rC 50mm L1 150mm L2 350mm L3 250mm σallow 200MPa rD 75mm Solution L L1 L2 L3 Support Reactions In yz plane ΣFz0 Az BZ PC 0 1 ΣΜB0 Az L PC L3 0 2 Solving Eqs 1 and 2 Az L3 L PC Az 25 kN Bz PC Az Bz 5kN In xy plane ΣFx0 Ax Bx PD 0 3 ΣΜB0 Ax L PD L L1 0 4 Solving Eqs 3 and 4 Ax L L1 L PD Ax 4kN Bx PD Ax Bx 1kN Torsion occurs in segment DC T PC rC T 0375 kN m Critical Section Located just to the left of gear C Mx Bx L3 Mz Bz L3 M Mx 2 Mz 2 M 127475 kN m Maximum Distortion Energy Theory Applying Eq 95 σ1 05 σx σy 05 σx σy 2 τxy 2 σ2 05 σx σy 05 σx σy 2 τxy 2 where σy 0 σx M c I M c π 4 c4 4M c π c4 τxy T c J T c π 2 c4 2T c π c4 Let a 05σx and b 05σx 2 τxy 2 Then σ1 2 a b 2 σ2 2 a b 2 σ1 σ2 a b a b a2 b2 σ1 2 σ1 σ2 σ2 2 a b 2 a2 b2 a b 2 a2 3b2 Hence σ1 2 σ1 σ2 σ2 2 σallow 2 05σx 2 3 05σx 2 τxy 2 2 σallow 2 σx 2 3τxy 2 σallow 2 4M c π c4 2 3 2T c π c4 2 σallow 2 c 6 16M2 12T2 π2 σallow 2 c 2031 mm do 2c do 4061 mm Use do 41mm Ans y1 0 001 L1 L1 y2 L1 101 L1 L1 L2 y3 L1 L2 101 L1 L2 L Mz1 y1 Az y1 kN m Mz2 y2 Az y2 kN m Mz3 y3 Az y3 PC y3 L1 L2 1 kN m 0 05 0 1 Distance m Mz kNm Mz1 y1 Mz2 y2 Mz3 y3 y1 y2 y3 Mx1 y1 Ax y1 kN m Mx2 y2 Ax y2 PD y2 L1 1 kN m Mx3 y3 Ax y3 PD y3 L1 1 kN m 0 05 0 05 Distance m Mx kNm Mx1 y1 Mx2 y2 Mx3 y3 y1 y2 y3 My1 y1 0 My2 y2 T kN m My3 y3 0 0 05 0 05 Distance m My kNm My1 y1 My2 y2 My3 y3 y1 y2 y3 Problem 1157 Select the lightestweight steel wideflange beam from Appendix B that will safely support the loading shown The allowable bending stress is σallow 160 MPa and the allowable shear stress is τallow 84 MPa Given σallow 160MPa L1 3m τallow 84MPa L2 15m P1 40kN P2 50kN Solution L 2 L1 L2 Support Reactions By symmetry RARBR ΣFy0 2R 2P1 P2 0 R P1 05P2 Maximum Moment and Shear Vmax R Vmax 65kN Mmax R L1 L2 P1 L2 Mmax 2325 kN m Bending Stress Sreqd Mmax σallow Sreqd 145312500 mm3 Select W 460x74 Sx 1460 103 mm3 d 457mm tw 902mm Shear Stress Provide a shear stress check τmax Vmax tw d τmax 1577 MPa τallow 98 MPa OK Hence Use W 460x74 Ans x1 0 001 L1 L1 x2 L1 101 L1 L1 L2 x3 L1 L2 101 L1 L2 L1 2L2 x4 L1 2L2 101 L1 2L2 L V1 x1 R 1 kN V2 x2 R P1 1 kN V3 x3 R P1 P2 1 kN V4 x4 R P1 P2 P1 1 kN M1 x1 R x1 kN m M2 x2 R x2 P1 x2 L1 1 kN m M3 x3 R x3 P1 x3 L1 P2 x3 L1 L2 1 kN m M4 x4 R x4 P1 x4 L1 P2 x4 L1 L2 P1 L1 x4 L 1 kN m 0 2 4 6 8 100 0 100 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 2 4 6 8 0 100 200 300 Distance m Moment kNm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 1201 An L2 steel strap having a thickness of 3 mm and a width of 50 mm is bent into a circular arc of radius 15 m Determine the maximum bending stress in the strap Given b 50mm t 3mm ρ 15m E 200GPa Solution Section Property c t 2 I b t3 12 Moment Curvature Relationship 1 ρ M E I M E I ρ Bendiug Stress σ M c I σ E I ρ c I σ E c ρ σ 20MPa Ans Problem 1202 A picture is taken of a man performing a pole vault and the minimum radius of curvature of the pole is estimated by measurement to be 45 m If the pole is 40 mm in diameter and it is made of a glass reinforced plastic for which Eg 131 GPa determine the maximum bending stress in the pole Given do 40mm ρ 45m E 131GPa Solution Section Property c do 2 I π do 4 64 Moment Curvature Relationship 1 ρ M E I M E I ρ Bendiug Stress σ M c I σ E I ρ c I σ E c ρ σ 5822 MPa Ans Problem 1203 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for Specify the slope at A and the beams maximum deflection EI is constant 2 0 x L Problem 1204 Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates Specify the beams maximum deflection EI is constant Problem 1205 Determine the equations of the elastic curve using the x1 and x2 coordinates EI is constant Problem 1206 Determine the equations of the elastic curve for the beam using the x1 and x3 coordinates Specify the beams maximum deflection EI is constant Problem 1207 Determine the equations of the elastic curve for the shaft using the x1 and x2 coordinates Specify the slope at A and the displacement at the center of the shaft EI is constant Problem 1208 Determine the equations of the elastic curve for the shaft using the x1 and x3 coordinates Specify the slope at A and the deflection at the center of the shaft EI is constant Problem 1209 The beam is made of two rods and is subjected to the concentrated load P Determine the maximum deflection of the beam if the moments of inertia of the rods are IAB and IBC and the modulus of elasticity is E Problem 1210 The beam is made of two rods and is subjected to the concentrated load P Determine the slope at C The moments of inertia of the rods are IAB and IBC and the modulus of elasticity is E Problem 1211 The bar is supported by a roller constraint at B which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown determine the slope at A and the deflection at C EI is constant Problem 1212 Determine the deflection at B of the bar in Prob 1211 Problem 1213 The fence board weaves between the three smooth fixed posts If the posts remain along the same line determine the maximum bending stress in the board The board has a width of 150 mm and a thickness of 12 mm E 12 GPa Assume the displacement of each end of the board relative to its center is 75 mm Given b 150mm t 12mm L 24m E 12GPa δ 75mm Solution Support Reactions By symmetry RARCR ΣFy0 2R P 0 R 05P Moment Function M x R x M x 05 P x Mmax R 05L Mmax 025P L Section Property c t 2 I b t3 12 Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 P x 2 E I dv dx P x2 4 C1 1 E I v P x3 12 C1 x C2 2 Boundary Conditions Due to symmetry dvdx0 at xL2 Also v0 at x0 From Eq 1 0 P 4 L 2 2 C1 C1 P L2 16 From Eq 2 0 0 0 C2 C2 0 The Elastic Curve Substitute the values of C1 and C2 into Eq 2 E I v P x3 12 P L2 16 x v P x 48E I 4x2 3L2 3 Require at xL2 vδ From Eq 3 δ P 48E I L 2 4 L 2 2 3L2 δ P L3 48 E I P 48 E I L3 δ Bending Stress Mmax 025P L cmax t 2 σmax Mmax cmax I σmax 1125 MPa Ans Problem 1214 Determine the equation of the elastic curve for the beam using the x coordinate Specify the slope at A and the maximum deflection EI is constant Problem 1215 Determine the deflection at the center of the beam and the slope at B EI is constant Problem 1216 A torque wrench is used to tighten the nut on a bolt If the dial indicates that a torque of 90 Nm is applied when the bolt is fully tightened determine the force P acting at the handle and the distance s the needle moves along the scale Assume only the portion AB of the beam distorts The cross section is square having dimensions of 12 mm by 12 mm E 200 GPa Given b 12mm h 12mm L 045m E 200GPa δ 75mm R 03m Tz 90N m Solution Equations of Equilibrium ΣFy0 Ay P 0 1 ΣΜB0 Tz P L 0 2 Solving Eqs 1 and 2 P Tz L Ay P Ay 200N P 200 N Ans Moment Function M x Ay x Tz Section Property I b h3 12 Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 Ay x Tz E I dv dx Ay x2 2 Tz x C1 1 E I v Ay x3 6 Tz x2 2 C1 x C2 2 Boundary Conditions Due to symmetry dvdx0 at x0 and v0 at x0 From Eq 1 0 0 0 C1 C1 0 From Eq 2 0 0 0 0 C2 C2 0 The Elastic Curve Substitute the values of C1 and C2 into Eq 2 E I v Ay x3 6 Tz x2 2 v x2 6E I Ayx 3Tz 3 At xR vs From Eq 3 s R2 6E I AyR 3Tz s 911 mm Ans Problem 1217 The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft and at B by a thrust bearing that exerts horizontal and vertical reactions on the shaft Draw the bendingmoment diagram for the shaft and then from this diagram sketch the deflection or elastic curve for the shafts centerline Determine the equations of the elastic curve using the coordinates x1 and x2 EI is constant Given h 60mm L1 150mm P 5kN L2 400mm Solution L L1 L2 Support Reactions ΣFy0 A B 0 1 ΣΜB0 P h A L2 0 2 Solving Eqs 1 and 2 A P h L2 A 075 kN B A B 075 kN Moment Function M1 x1 P h M2 x2 B x2 Section Property EI kN m2 Slope and Elastic Curve EI d2 v1 dx1 2 M1 x1 EI d2 v2 dx2 2 M2 x2 EI d2v1 dx1 2 P h EI d2v2 dx2 2 B x2 EI dv1 dx1 P h x1 C1 1 EI dv2 dx2 B 2 x2 2 C3 3 EI v1 P h x1 2 2 C1 x1 C2 2 EI v2 B x2 3 6 C3 x2 C4 4 Boundary Conditions v10 at x1015m From Eq 2 0 P h 015m 2 2 C1 015m C2 5 v20 at x20 From Eq 4 0 0 0 C4 C4 0 Ans v20 at x204m From Eq 4 0 B 04m 3 6 C3 04m C4 C3 B 6 04m 2 C3 002 kN m2 Ans Continuity Condition dv1dx1 dv2dx2 at A x1015m and x204m From Eqs 1 and 3 P h 015m C1 B 2 04m 2 C3 C1 B 2 04m 2 C3 P h 015m C1 0005 kN m2 Ans From Eq 5 C2 P h 015m 2 2 C1 015m C2 000263 kN m3 Ans The Elastic Curve Substitute the values of C1 and C2 into Eq 2 and C3 and C4 into Eq 4 v1 1 EI P h 2 x1 2 C1 x1 C2 Ans v2 1 EI B 6 x2 3 C3 x2 C4 Ans BMD x1 0 001 L1 L1 x2 L1 101 L1 L M1 x1 P h kN m M2 x2 P h A x2 L1 1 kN m 0 02 04 0 02 04 Distane m Moment kNm M1 x1 M2 x2 x1 x2 Problem 1218 Determine the equations of the elastic curve using the coordinates x1 and x2 and specify the slope and deflection at C EI is constant Problem 1219 Determine the equations of the elastic curve using the coordinates x1 and x2 and specify the slope at A EI is constant Problem 1220 Determine the equations of the elastic curve using the coordinates x1 and x2 and specify the slope and deflection at B EI is constant Problem 1221 Determine the equations of the elastic curve using the coordinates x1 and x3 and specify the slope and deflection at B EI is constant Problem 1222 Determine the maximum slope and maximum deflection of the simplysupported beam which is subjected to the couple moment M0 EI is constant Problem 1223 The two wooden meter sticks are separated at their centers by a smooth rigid cylinder having a diameter of 50 mm Determine the force F that must be applied at each end in order to just make their ends touch Each stick has a width of 20 mm and a thickness of 5 mm Ew 11 GPa Given do 50mm L 05m b 20mm t 5mm E 11GPa Solution Section Property I b t3 12 Moment Function M x F x Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 F x E I dv dx F x2 2 C1 1 E I v F x3 6 C1 x C2 2 Boundary Conditions dvdx0 at xL From Eq 1 0 F L2 2 C1 C1 F L2 2 v0 at xL From Eq 2 0 F L3 6 C1 L C2 C2 F L3 6 C1 L C2 F L3 3 Require v 05 do at x0 From Eq 2 05 do E I 0 0 F L3 3 F 15do E I L3 F 1375 N Ans Problem 1224 The pipe can be assumed roller supported at its ends and by a rigid saddle C at its center The saddle rests on a cable that is connected to the supports Determine the force that should be developed in the cable if the saddle keeps the pipe from sagging or deflecting at its center The pipe and fluid within it have a combined weight of 2 kNm EI is constant Given e 03m L 375m w 2kN m Solution Moment Function M x P x 1 2 w x2 Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 P x w 2 x2 E I dv dx P x2 2 w 6 x3 C1 1 E I v P x3 6 w 24 x4 C1 x C2 2 Boundary Conditions v0 at x0 and at xL From Eq 2 0 0 0 0 C2 C2 0 0 P L3 6 w 24 L4 C1 L 3 Also dvdx0 at xL From Eq 1 0 P L2 2 w 6 L3 C1 4 Solving Eqs 3 and 4 for P P L2 6 w 24 L3 C1 P L2 2 w 6 L3 C1 P L2 3 w 8 L3 P 3w L 8 P 2813 kN Equations of Equilibrium ΣFy0 2P F w 2L 0 F 2w L 2P F 9375 N At C ΣFy0 2Tcable e e2 L2 F 0 Tcable e2 L2 2e F Tcable 5878 kN Ans Problem 1225 Determine the equations of the elastic curve using the coordinates x1 and x2 and specify the slope at C and displacement at B EI is constant Problem 1226 Determine the equations of the elastic curve using the coordinates x1 and x3 and specify the slope at B and deflection at C EI is constant Problem 1227 Determine the elastic curve for the simply supported beam using the x coordinate Also determine the slope at A and the maximum deflection of the beam EI is constant 2 0 x L Problem 1228 Determine the elastic curve for the cantilevered beam using the x coordinate Also determine the maximum slope and maximum deflection EI is constant Problem 1229 The beam is made of a material having a specific weight γ Determine the displacement and slope at its end A due to its weight The modulus of elasticity for the material is E Problem 1230 The beam is made of a material having a specific weight γ Determine the displacement and slope at its end A due to its weight The modulus of elasticity for the material is E Problem 1231 The leaf spring assembly is designed so that it is subjected to the same maximum stress throughout its length If the plates of each leaf have a thickness t and can slide freely between each other show that the spring must be in the form of a circular arc in order that the entire spring becomes flat when a large enough load P is applied What is the maximum normal stress in the spring Consider the spring to be made by cutting the n strips from the diamondshaped plate of thickness t and width b The modulus of elasticity for the material is E Hint Show that the radius of curvature of the spring is constant Problem 1232 The beam has a constant width b and is tapered as shown If it supports a load P at its end determine the deflection at BThe load P is applied a short distance s from the tapered end B where s L EI is constant Problem 1233 A thin flexible 6mlong rod having a weight of 10 Nm rests on the smooth surface If a force of 15 N is applied at its end to lift it determine the suspended length x and the maximum moment developed in the rod Given P 15N w 10N m Lmax 6m Solution Since the horizontal section has no curvature the moment in the rod is zero Hence R acts at the end of the suspended portion and this portion acts like a simply supported beam Thus Equations of Equilibrium ΣFy0 P R w x 0 1 ΣΜ00 P x w 2 x2 0 2 From Eqs 2 x 2P w x 3m Ans Maximum moment occurs at center Mmax P x 2 w 2 x 2 2 Mmax 1125 N m Ans Problem 1234 The shaft supports the two pulley loads shown Determine the equation of the elastic curve The bearings at A and B exert only vertical reactions on the shaft EI is constant Use Macaulay Function Ψ z Φ z z Given a m P kN EI kN m2 EIo 1 Solution Support Reactions ΣFy0 A B P 2P 0 1 ΣΜB0 A 2a P a 2P a 0 2 Solving Eqs 1 and 2 A P 2 B 7P 2 Moment Function M x A Ψ x 0 P Ψ x a B Ψ x 2a Slope and Elastic Curve EI d2 v dx2 M x EI d2v dx2 A Ψ x 0 P Ψ x a B Ψ x 2a EI dv dx A 2 Ψ x 0 2 P 2 Ψ x a 2 B 2 Ψ x 2a 2 C1 EI v A 6 Ψ x 0 3 P 6 Ψ x a 3 B 6 Ψ x 2a 3 C1 x C2 EI v A 6 x3 P 6 Ψ x a 3 B 6 Ψ x 2a 3 C1 x C2 1 Boundary Conditions v0 at x0 From Eq 1 0 0 0 0 0 C2 C2 0 Also v0 at x2a From Eq 1 0 A 6 2a 3 P 6 Ψ 2a a 3 0 C1 2a C1 A 12a 2a 3 P 12a a3 C1 5 12 P a2 Equation of Elastic Curve v 1 EI A 6 x3 P 6 Ψ x a 3 B 6 Ψ x 2a 3 C1 x m v x 1 EIo P 12 x3 P 6 Ψ x a 3 7P 12 Ψ x 2a 3 5 12 P a2 x m Ans Problem 1235 Determine the equation of the elastic curve Specify the slopes at A and B EI is constant Use Macaulay Function Ψ z Φ z z Given a m w kN m Solution L 2a Support Reactions ΣFy0 A B w a 0 1 ΣΜA0 wa 05a B 2 a 0 2 Solving Eqs 1 and 2 B 1 4 wa A w a B B 025 w a A 075 w a Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x 0 M x 05 w x2 05w Ψ x a 2 A x Slope and Elastic Curve EI kN m2 EIo 1 EI d2 v dx2 M x EI d2v dx2 05 w x2 05w Ψ x a 2 A x EI dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 x2 C1 3 EI v 1 24 w x4 1 24w Ψ x a 4 A 6 x3 C1 x C2 4 Boundary Conditions v0 at x0 and x2a From Eq 4 0 0 0 0 0 C2 C2 0 0 1 24 w 2a 4 1 24 w a4 A 6 2a 3 C1 2a C1 1 3w a3 1 48w a3 2A 3 a2 C1 3 16 w a3 Equations of Elastic Curve and Slope v x 1 EIo 1 24 w x4 1 24w Ψ x a 4 3w a 24 x3 3 16 w a3 x Ans θ x 1 EIo 1 6 w x3 1 6w Ψ x a 3 A 2 x2 C1 Slope at A Substitute x0 into Eq3 θ 0 3 16 w a3 EIo Ans Slope at B Substitute xL into Eq3 θ L 7 48 w a3 EIo Ans Problem 1236 The beam is subjected to the load shown Determine the equation of the elastic curve EI is constant Use Macaulay Function Ψ z Φ z z Given a 15m P 20kN w 6 kN m b 3m Solution L 2a b Support Reactions ΣFy0 A B P w a 0 1 ΣΜA0 wa 05a P a b B b 0 2 Solving Eqs 1 and 2 B 1 b wa 05a P a b A P w a B B 2775 kN A 125 kN Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x a B Ψ x a b M x 05 w x2 05w Ψ x a 2 A Ψ x a B Ψ x a b Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 05 w x2 05w Ψ x a 2 A Ψ x a B Ψ x a b E I dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 Ψ x a 2 B 2 Ψ x a b 2 C1 3 E I v 1 24 w x4 1 24w Ψ x a 4 A 6 Ψ x a 3 B 6 Ψ x a b 3 C1 x C2 4 Boundary Conditions v0 at xa and xab From Eq 4 0 1 24 w a4 0 0 0 C1 a C2 5 0 1 24 w a b 4 1 24w b4 A 6 b3 0 C1 a b C2 6 65 C1 1 b 1 24w a b 4 1 24w b4 A 6 b3 1 24 w a4 C1 25125 kN m2 From Eq 5 C2 1 24 w a4 C1 a C2 36422 kN m3 Equation of Elastic Curve ao w 24 a1 A 6 a2 B 6 a3 C1 a4 C2 ao 025 kN m a1 02083 kN a2 4625 kN a3 2513 kN m2 a4 3642 kN m3 Ans v 1 E I ao x4 ao Ψ x a 4 a1 Ψ x a 3 a2 Ψ x a b 3 a3 x a4 Ans Problem 1237 The shaft supports the two pulley loads shown Determine the equation of the elastic curve The bearings at A and B exert only vertical reactions on the shaft EI is constant Use Macaulay Function Ψ z Φ z z Given a 05m P1 200N P2 300N Solution Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 A 2a P1 a P2 a 0 2 Solving Eqs 1 and 2 A P1 P2 2 B P1 3P2 2 A 50 N B 550 N Moment Function M x A Ψ x 0 P1 Ψ x a B Ψ x 2a Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 A Ψ x 0 P1 Ψ x a B Ψ x 2a E I dv dx A 2 Ψ x 0 2 P1 2 Ψ x a 2 B 2 Ψ x 2a 2 C1 E I v A 6 Ψ x 0 3 P1 6 Ψ x a 3 B 6 Ψ x 2a 3 C1 x C2 1 Boundary Conditions v0 at x0 From Eq 1 0 0 0 0 0 C2 C2 0 Also v0 at x2a From Eq 1 0 A 6 Ψ 2a 0 3 P1 6 Ψ 2a a 3 0 C1 2a C1 A 12a 2a 3 P1 12a a3 C1 1250 N m2 Equation of Elastic Curve ao A 6 a1 P1 6 a2 B 6 a3 C1 ao 833 N a1 3333 N a2 9167 N a3 1250 N m2 Ans v 1 E I ao Ψ x 0 3 a1 Ψ x a 3 a2 Ψ x 2a 3 a3 x Ans Problem 1238 The beam is subjected to the load shown Determine the equation of the elastic curve EI is constant Use Macaulay Function Ψ z Φ z z Given a 4m P 50kN w 3 kN m b 3m Solution L a 2 b Support Reactions ΣFy0 A B P w a 0 1 ΣΜA0 wa 05a P a b B L 0 2 Solving Eqs 1 and 2 B 1 L wa 05a P a b A P w a B B 374 kN A 246 kN Moment Function Ans M x A x 05w x2 05w Ψ x a 2 P Ψ x a b Slope and Elastic Curve EI kN m2 EIo 1 EI d2 v dx2 M x EI d2v dx2 A x 05w x2 05w Ψ x a 2 P Ψ x a b EI dv dx A 2 x2 w 6 x3 w 6 Ψ x a 3 P 2 Ψ x a b 2 C1 3 EI v A 6 x3 w 24 x4 w 24 Ψ x a 4 P 6 Ψ x a b 3 C1 x C2 4 Boundary Conditions v0 at x0 From Eq 4 0 0 0 0 0 0 C2 C2 0 v0 at xL From Eq 4 0 A 6 L3 w 24 L4 w 24 L a 4 P 6 L a b 3 C1 L C1 A L2 6 w 24 L3 w 24L L a 4 P 6L L a b 3 C1 27870 kN m2 Equation of Elastic Curve ao A 6 a1 w 24 a2 P 6 a3 C1 a3 C2 ao 410 kN a1 0125 kN m a2 833 kN a3 000 N m2 Ans v 1 EI ao x3 a1 x4 a1 Ψ x a 4 a2 Ψ x a b 3 a3 x a4 M x A Ψ x 0 05w Ψ x 0 2 05 w Ψ x a 2 P Ψ x a b Problem 1239 The beam is subjected to the load shown Determine the displacement at x 7 m and the slope at A EI is constant Use Macaulay Function Ψ z Φ z z Given a 4m P 50kN w 3 kN m b 3m Solution L a 2 b Support Reactions ΣFy0 A B P w a 0 1 ΣΜA0 wa 05a P a b B L 0 2 Solving Eqs 1 and 2 B 1 L wa 05a P a b A P w a B B 374 kN A 246 kN Moment Function Ans M x A x 05w x2 05w Ψ x a 2 P Ψ x a b Slope and Elastic Curve EI kN m2 EIo 1 EI d2 v dx2 M x EI d2v dx2 A x 05w x2 05w Ψ x a 2 P Ψ x a b EI dv dx A 2 x2 w 6 x3 w 6 Ψ x a 3 P 2 Ψ x a b 2 C1 3 EI v A 6 x3 w 24 x4 w 24 Ψ x a 4 P 6 Ψ x a b 3 C1 x C2 4 Boundary Conditions v0 at x0 From Eq 4 0 0 0 0 0 0 C2 C2 0 v0 at xL From Eq 4 0 A 6 L3 w 24 L4 w 24 L a 4 P 6 L a b 3 C1 L C1 A L2 6 w 24 L3 w 24L L a 4 P 6L L a b 3 C1 27870 kN m2 Equations of Elastic Curve and Slope v x 1 EI A 6 x3 w 24 x4 w 24 Ψ x a 4 P 6 Ψ x a b 3 C1 x C2 Displacement at x7m Substitute x7m into Eq4 v 7m 83460 m EIo Ans θ x 1 EI A 2 x2 w 6 x3 w 6 Ψ x a 3 P 2 Ψ x a b 2 C1 Slope at A Substitute x0 into Eq3 θ 0 27870 1 EIo M x A Ψ x 0 05w Ψ x 0 2 05 w Ψ x a 2 P Ψ x a b Problem 1240 The beam is subjected to the load shown Determine the equation of the elastic curve EI is constant Use Macaulay Function Ψ z Φ z z Given a 24m P1 10kN MB 6kN m P2 20kN Solution Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 A 3a P1 2a P2 a MB 0 2 Solving Eqs 1 and 2 A 2P1 P2 3 MB 3a B P1 P2 A A 125 kN B 175 kN Moment Function M x A Ψ x 0 P1 Ψ x a P2 Ψ x 2a Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 A Ψ x 0 P1 Ψ x a P2 Ψ x 2a E I dv dx A 2 Ψ x 0 2 P1 2 Ψ x a 2 P2 2 Ψ x 2a 2 C1 E I v A 6 Ψ x 0 3 P1 6 Ψ x a 3 P2 6 Ψ x 2a 3 C1 x C2 1 Boundary Conditions v0 at x0 From Eq 1 0 0 0 0 C2 C2 0 Also v0 at x3a From Eq 1 0 A 6 Ψ 3a 0 3 P1 6 Ψ 3a a 3 P2 6 Ψ 3a 2a 3 C1 3a C1 A 18a 3a 3 P1 18a 2a 3 P2 18a a3 C1 7600 kN m2 Equation of Elastic Curve ao A 6 a1 P1 6 a2 P2 6 a3 C1 ao 208 kN a1 167 kN a2 333 kN a3 7600 kN m2 Ans v 1 E I ao Ψ x 0 3 a1 Ψ x a 3 a2 Ψ x 2a 3 a3 x Ans w 6 kN m Use Macaulay Function Ψ z Φ z z Given a 15m b 3m P 20kN Problem 1241 The beam is subjected to the load shown Determine the equation of the elastic curve EI is constant Solution L 2a b Support Reactions ΣFy0 A B P w a 0 1 ΣΜA0 wa 05a P a b B b 0 2 Solving Eqs 1 and 2 B 1 b wa 05a P a b A P w a B B 2775 kN A 125 kN Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x a B Ψ x a b M x 05 w x2 05w Ψ x a 2 A Ψ x a B Ψ x a b Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 05 w x2 05w Ψ x a 2 A Ψ x a B Ψ x a b E I dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 Ψ x a 2 B 2 Ψ x a b 2 C1 3 E I v 1 24 w x4 1 24w Ψ x a 4 A 6 Ψ x a 3 B 6 Ψ x a b 3 C1 x C2 4 Boundary Conditions v0 at xa and xab From Eq 4 0 1 24 w a4 0 0 0 C1 a C2 5 0 1 24 w a b 4 1 24w b4 A 6 b3 0 C1 a b C2 6 65 C1 1 b 1 24w a b 4 1 24w b4 A 6 b3 1 24 w a4 C1 25125 kN m2 From Eq 5 C2 1 24 w a4 C1 a C2 36422 kN m3 Equation of Elastic Curve ao w 24 a1 A 6 a2 B 6 a3 C1 a4 C2 ao 025 kN m a1 02083 kN a2 4625 kN a3 2513 kN m2 a4 3642 kN m3 Ans v 1 E I ao x4 ao Ψ x a 4 a1 Ψ x a 3 a2 Ψ x a b 3 a3 x a4 Ans w 3 kN m Problem 1242 The beam is subjected to the load shown Determine the equation of the slope and elastic curve EI is constant Use Macaulay Function Ψ z Φ z z Given a 5m b 3m Mo 15kN m Solution L a b Support Reactions ΣFy0 A B w a 0 1 ΣΜA0 wa 05a B a Mo 0 2 Solving Eqs 1 and 2 B 1 a wa 05a Mo A w a B B 105 kN A 45 kN Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x 0 B Ψ x a M x 05 w x2 05w Ψ x a 2 A x B Ψ x a Slope and Elastic Curve E I d2 v dx2 M x E I d2v dx2 05 w x2 05w Ψ x a 2 A x B Ψ x a E I dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 x2 B 2 Ψ x a 2 C1 3 E I v 1 24 w x4 1 24w Ψ x a 4 A 6 x3 B 6 Ψ x a 3 C1 x C2 4 Boundary Conditions v0 at x0 and xa From Eq 4 0 0 0 0 0 0 C2 C2 0 0 1 24 w a4 0 A 6 a3 0 C1 a C1 1 24w a3 A 6 a2 C1 3125 kN m2 Equation of Elastic Curve and slope ao w 24 a1 A 6 a2 B 6 a3 C1 ao 0125 kN m a1 075 kN a2 175 kN a3 3125 kN m2 Ans v 1 E I ao x4 ao Ψ x a 4 a1 x3 a2 Ψ x a 3 a3 x Ans θ 1 E I 4 ao x3 4 ao Ψ x a 3 3a1 x2 3a2 Ψ x a 2 a3 Ans Problem 1243 Determine the equation of the elastic curve Specify the slope at A and the displacement at C EI is constant Use Macaulay Function Ψ z Φ z z Given a m w kN m Solution L 2a Support Reactions ΣFy0 A B w a 0 1 ΣΜA0 wa 05a B 2 a 0 2 Solving Eqs 1 and 2 B 1 4 wa A w a B B 025 w a A 075 w a Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x 0 M x 05 w x2 05w Ψ x a 2 A x Slope and Elastic Curve EI kN m2 EIo 1 EI d2 v dx2 M x EI d2v dx2 05 w x2 05w Ψ x a 2 A x EI dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 x2 C1 3 EI v 1 24 w x4 1 24w Ψ x a 4 A 6 x3 C1 x C2 4 Boundary Conditions v0 at x0 and x2a From Eq 4 0 0 0 0 0 C2 C2 0 0 1 24 w 2a 4 1 24 w a4 A 6 2a 3 C1 2a C1 1 3w a3 1 48w a3 2A 3 a2 C1 3 16 w a3 Equations of Elastic Curve and Slope v x 1 EIo 1 24 w x4 1 24w Ψ x a 4 3w a 24 x3 3 16 w a3 x Displacement at C Substitute xa into Eq4 v a 5 48 w a4 EIo Ans θ x 1 EIo 1 6 w x3 1 6w Ψ x a 3 A 2 x2 C1 m Slope at A Substitute x0 into Eq3 θ 0 3 16 m w a3 EIo Ans Problem 1244 Determine the equation of the elastic curve Specify the slope at A and B EI is constant Use Macaulay Function Ψ z Φ z z Given a m w kN m Solution L 2a Support Reactions ΣFy0 A B w a 0 1 ΣΜA0 wa 05a B 2 a 0 2 Solving Eqs 1 and 2 B 1 4 wa A w a B B 025 w a A 075 w a Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x 0 M x 05 w x2 05w Ψ x a 2 A x Slope and Elastic Curve EI kN m2 EIo 1 EI d2 v dx2 M x EI d2v dx2 05 w x2 05w Ψ x a 2 A x EI dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 x2 C1 3 EI v 1 24 w x4 1 24w Ψ x a 4 A 6 x3 C1 x C2 4 Boundary Conditions v0 at x0 and x2a From Eq 4 0 0 0 0 0 C2 C2 0 0 1 24 w 2a 4 1 24 w a4 A 6 2a 3 C1 2a C1 1 3w a3 1 48w a3 2A 3 a2 C1 3 16 w a3 Equations of Elastic Curve and Slope v x 1 EIo 1 24 w x4 1 24w Ψ x a 4 3w a 24 x3 3 16 w a3 x Ans θ x 1 EIo 1 6 w x3 1 6w Ψ x a 3 A 2 x2 C1 Slope at A Substitute x0 into Eq3 θ 0 3 16 w a3 EIo Ans Slope at B Substitute xL into Eq3 θ L 7 48 w a3 EIo Ans Problem 1245 The beam is subjected to the load shown Determine the equation of the elastic curve EI is constant Use Macaulay Function Ψ z Φ z z Given a 15m P 20kN Solution L 4a Support Reactions ΣFy0 A B 2P 0 1 ΣΜB0 A 2a P 3a P a 0 2 Solving Eqs 1 and 2 A P B P Moment Function M x P Ψ x 0 A Ψ x a B Ψ x 3a Slope and Elastic Curve EI kN m2 EI d2 v dx2 M x EI d2v dx2 P Ψ x 0 A Ψ x a B Ψ x 3a EI dv dx P 2 Ψ x 0 2 A 2 Ψ x a 2 B 2 Ψ x 3a 2 C1 1 EI v P 6 Ψ x 0 3 A 6 Ψ x a 3 B 6 Ψ x 3a 3 C1 x C2 EI v P 6 x3 A 6 Ψ x a 3 B 6 Ψ x 3a 3 C1 x C2 2 Boundary Conditions Due to symmetry dvdx0 at x2a From Eq 1 0 P 2 2a 2 A 2 Ψ 2a a 2 0 C1 C1 2 P A 2 a2 C1 6750 kN m2 Also v0 at xa From Eq 2 0 P 6 a3 0 0 C1 a C2 C2 P 6 a3 C1 a C2 90 kN m3 Equation of Elastic Curve ao P 6 a1 A 6 a2 C1 a3 C2 ao 333333 N a1 33333 kN a2 675 kN m2 a3 90 kN m3 Ans v 1 E I ao x3 a1 Ψ x a 3 a1 Ψ x 3a 3 a2 x a3 Ans w 2 kN m Problem 1246 The beam is subjected to the load shown Determine the equation of the slope and elastic curve EI is constant Use Macaulay Function Ψ z Φ z z Given a 5m b 3m P 8kN Solution L a b Support Reactions ΣFy0 A B w a P 0 1 ΣΜA0 wa 05a B a P L 0 2 Solving Eqs 1 and 2 B 1 a wa 05a P L B 178 kN A w a P B A 02 kN Moment Function M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x 0 B Ψ x a M x 05 w x2 05w Ψ x a 2 A x B Ψ x a Slope and Elastic Curve EI d2 v dx2 M x EI d2v dx2 05 w x2 05w Ψ x a 2 A x B Ψ x a EI dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 x2 B 2 Ψ x a 2 C1 3 EI v 1 24 w x4 1 24w Ψ x a 4 A 6 x3 B 6 Ψ x a 3 C1 x C2 4 Boundary Conditions v0 at x0 and xa From Eq 4 0 0 0 0 0 0 C2 C2 0 0 1 24 w a4 0 A 6 a3 0 C1 a C1 1 24w a3 A 6 a2 C1 9583 kN m2 Equation of Elastic Curve and slope ao w 24 a1 A 6 a2 B 6 a3 C1 ao 00833 kN m a1 00333 kN a2 29667 kN a3 9583 kN m2 Ans v 1 EI ao x4 ao Ψ x a 4 a1 x3 a2 Ψ x a 3 a3 x Ans θ 1 EI 4 ao x3 4 ao Ψ x a 3 3a1 x2 3a2 Ψ x a 2 a3 Ans EI d2v dx2 05 w x2 05w Ψ x a 2 A x B Ψ x a w 2 kN m Problem 1247 The beam is subjected to the load shown Determine the slope at A and the displacement at C EI is constant Use Macaulay Function Ψ z Φ z z Given a 5m b 3m P 8kN Solution L a b Support Reactions ΣFy0 A B w a P 0 1 ΣΜA0 wa 05a B a P L 0 2 Solving Eqs 1 and 2 B 1 a wa 05a P L A w a P B B 178 kN A 02 kN Moment Function M x 05 w x2 05w Ψ x a 2 A x B Ψ x a Slope and Elastic Curve EI kN m2 EIo 1 EI d2 v dx2 M x EI dv dx 1 6 w x3 1 6w Ψ x a 3 A 2 x2 B 2 Ψ x a 2 C1 3 EI v 1 24 w x4 1 24w Ψ x a 4 A 6 x3 B 6 Ψ x a 3 C1 x C2 4 Boundary Conditions v0 at x0 and xa From Eq 4 0 0 0 0 0 0 C2 C2 0 0 1 24 w a4 0 A 6 a3 0 C1 a C1 1 24w a3 A 6 a2 C1 9583 kN m2 Equation of Elastic Curve and slope v x 1 EI 1 24 w x4 1 24w Ψ x a 4 A 6 x3 B 6 Ψ x a 3 C1 x Displacement at C Substitute xL into Eq4 v L 16075 m EIo Ans Ans θ x 1 EI 1 6 w x3 1 6w Ψ x a 3 A 2 x2 B 2 Ψ x a 2 C1 Slope at A Substitute x0 into Eq3 θ 0 958 1 EIo Ans M x 05 w Ψ x 0 2 05 w Ψ x a 2 A Ψ x 0 B Ψ x a Problem 1248 The beam is subjected to the load shown Determine the equation of the elastic curve Use Macaulay Function Ψ x Φ x x Given L1 2m L2 3m wo 150 kN m Solution L L1 L2 Support Reactions ΣFy0 A B wo L1 wo 2 L2 0 1 ΣΜA0 wo 2 L1 2 wo 2 L2 L2 3 B L2 0 2 Solving Eqs 1 and 2 B 3 L1 2 L2 2 6L2 wo A wo L1 05L2 B B 25 kN A 550kN Moment Function M x wo 2 Ψ x 0 2 wo 6L2 Ψ x L1 3 A Ψ x L1 Slope and Elastic Curve EI d2 v dx2 M x EI kN m2 EIo 1 EI d2v dx2 wo 2 Ψ x 0 2 wo 6L2 Ψ x L1 3 A Ψ x L1 EI dv dx wo 6 Ψ x 0 3 wo 24 L2 Ψ x L1 4 A 2 Ψ x L1 2 C1 3 EI v wo 24 Ψ x 0 4 wo 120 L2 Ψ x L1 5 A 6 Ψ x L1 3 C1 x C2 4 Boundary Conditions v0 at xL1 Given 5 From Eq 4 0 wo 24 L1 4 0 0 C1 L1 C2 Also v0 at xL From Eq 4 0 wo 24 L4 wo 120 L2 L L1 5 A 6 L L1 3 C1 L C2 6 Solving Eqs 5 and 6 Guess C1 1kN m2 C2 1kN m3 C1 C2 Find C1 C2 C1 410kN m2 C2 720 kN m3 Equation of Elastic Curve ao wo 24 a1 wo 120 L2 a2 A 6 a3 C1 a4 C2 ao 625 1 m kN a1 042 1 m2 kN a2 9167 kN Ans a3 41000 kN m2 a4 72000 kN m3 Ans v 1 E I ao Ψ x 0 4 a1 Ψ x L1 5 a2 Ψ x L1 3 a3 x a4 Ans wo 150 kN m A B wo L1 wo 2 L2 0 Problem 1249 Determine the displacement at C and the slope at A of the beam Use Macaulay Function Ψ x Φ x x Given L1 2m L2 3m Solution L L1 L2 Support Reactions ΣFy0 1 ΣΜA0 wo 2 L1 2 wo 2 L2 L2 3 B L2 0 2 Solving Eqs 1 and 2 B 3 L1 2 L2 2 6L2 wo A wo L1 05L2 B B 25 kN A 550kN Moment Function M x wo 2 Ψ x 0 2 wo 6L2 Ψ x L1 3 A Ψ x L1 Slope and Elastic Curve EI d2 v dx2 M x EI kN m2 EIo 1 EI d2v dx2 wo 2 Ψ x 0 2 wo 6L2 Ψ x L1 3 A Ψ x L1 EI dv dx wo 6 Ψ x 0 3 wo 24 L2 Ψ x L1 4 A 2 Ψ x L1 2 C1 3 EI v wo 24 Ψ x 0 4 wo 120 L2 Ψ x L1 5 A 6 Ψ x L1 3 C1 x C2 4 Boundary Conditions v0 at xL1 Given 5 From Eq 4 0 wo 24 L1 4 0 0 C1 L1 C2 Also v0 at xL From Eq 4 0 wo 24 L4 wo 120 L2 L L1 5 A 6 L L1 3 C1 L C2 6 Solving Eqs 5 and 6 Guess C1 1kN m2 C2 1kN m3 C1 C2 Find C1 C2 C1 410kN m2 C2 720 kN m3 Equation of Elastic Curve and Slope v x 1 EI wo 24 Ψ x 0 4 wo 120 L2 Ψ x L1 5 A 6 Ψ x L1 3 C1 x C2 7 Displacement at C Substitute x0 into Eq7 v 0m 720 m EIo Ans θ x 1 EI wo 6 Ψ x 0 3 wo 24 L2 Ψ x L1 4 A 2 Ψ x L1 2 C1 Slope at A Substitute xL1 into Eq3 θ L1 210 1 EIo Ans A B w L 0 w kN m Use Macaulay Function Ψ x Φ x x Given L m EIo 1 Problem 1250 Determine the equation of the elastic curve Specify the slope at A EI is constant Solution Support Reactions ΣFy0 1 ΣΜA0 05L wL B L 0 2 Solving Eqs 1 and 2 B 05 wL A w L B A 15 w L Moment Function M x 05w Ψ x 0 2 05 w Ψ x L 2 A Ψ x L Slope and Elastic Curve Ans EI kN m2 EI d2v dx2 05w Ψ x 0 2 05 w Ψ x L 2 A Ψ x L EI dv dx w 6 Ψ x 0 3 w 6 Ψ x L 3 A 2 Ψ x L 2 C1 3 EI v w 24 Ψ x 0 4 w 24 Ψ x L 4 A 6 Ψ x L 3 C1 x C2 4 Boundary Conditions Given v0 at xL From Eq 4 0 w 24 L4 0 0 C1 L C2 5 v0 at x2L From Eq 4 0 w 24 2L 4 w 24 L4 A 6 L3 C1 2L C2 6 Solving Eqs 5 and 6 Guess C1 1kN m2 C2 1kN m3 C1 C2 Find C1 C2 C1 1 3 w L3 C2 7 24 w L4 Equation of Elastic Curve and Slope v x 1 EI w 24 x4 w 24 Ψ x L 4 3w L 12 Ψ x L 3 1 3 w L3 x 7 24 w L4 Ans θ x 1 EI w 6 Ψ x 0 3 w 6 Ψ x L 3 9w L 12 Ψ x L 2 C1 Slope at A Substitute xL into Eq3 θ L 1 6 w L3 EI EI d2 v dx2 M x w kN m Use Macaulay Function Ψ x Φ x x Given L m Problem 1251 Determine the equation of the elastic curve Specify the deflection at C EI is constant Solution Support Reactions ΣFy0 A B w L 0 1 ΣΜA0 05L wL B L 0 2 Solving Eqs 1 and 2 B 05 wL A w L B A 15 w L Moment Function M x 05w Ψ x 0 2 05 w Ψ x L 2 A Ψ x L Slope and Elastic Curve Ans EI kN m2 EIo 1 EI d2v dx2 05w Ψ x 0 2 05 w Ψ x L 2 A Ψ x L EI dv dx w 6 Ψ x 0 3 w 6 Ψ x L 3 A 2 Ψ x L 2 C1 3 EI v w 24 Ψ x 0 4 w 24 Ψ x L 4 A 6 Ψ x L 3 C1 x C2 4 Boundary Conditions Given v0 at xL From Eq 4 0 w 24 L4 0 0 C1 L C2 5 v0 at x2L From Eq 4 0 w 24 2L 4 w 24 L4 A 6 L3 C1 2L C2 6 Solving Eqs 5 and 6 Guess C1 1kN m2 C2 1kN m3 C1 C2 Find C1 C2 C1 1 3 w L3 C2 7 24 w L4 Equation of Elastic Curve v x 1 EI w 24 x4 w 24 Ψ x L 4 3w L 12 Ψ x L 3 1 3 w L3 x 7 24 w L4 Ans v 0 7 24 w L4 EI EI d2 v dx2 M x w kN m Problem 1252 Determine the equation of the elastic curve Specify the slope at B EI is constant Use Macaulay Function Ψ x Φ x x Given L m Solution Support Reactions ΣFy0 A B w L 0 1 ΣΜA0 05L wL B L 0 2 Solving Eqs 1 and 2 B 05 wL A w L B A 15 w L Moment Function M x 05w Ψ x 0 2 05 w Ψ x L 2 A Ψ x L Slope and Elastic Curve Ans EI kN m2 EIo 1 EI d2v dx2 05w Ψ x 0 2 05 w Ψ x L 2 A Ψ x L EI dv dx w 6 Ψ x 0 3 w 6 Ψ x L 3 A 2 Ψ x L 2 C1 3 EI v w 24 Ψ x 0 4 w 24 Ψ x L 4 A 6 Ψ x L 3 C1 x C2 4 Boundary Conditions Given v0 at xL From Eq 4 0 w 24 L4 0 0 C1 L C2 5 v0 at x2L From Eq 4 0 w 24 2L 4 w 24 L4 A 6 L3 C1 2L C2 6 Solving Eqs 5 and 6 Guess C1 1kN m2 C2 1kN m3 C1 C2 Find C1 C2 C1 1 3 w L3 C2 7 24 w L4 Equation of Elastic Curve and Slope v x 1 EI w 24 x4 w 24 Ψ x L 4 3w L 12 Ψ x L 3 1 3 w L3 x 7 24 w L4 Ans θ x 1 EI w 6 Ψ x 0 3 w 6 Ψ x L 3 9w L 12 Ψ x L 2 C1 Slope at A Substitute x2L into Eq3 θ 2L 1 12 w L3 EI EI d2 v dx2 M x Problem 1253 The shaft is made of steel and has a diameter of 15 mm Determine its maximum deflection The bearings at A and B exert only vertical reactions on the shaft Est 200 GPa Use Macaulay Function Ψ x Φ x x Given a 200mm P1 250N b 300mm P2 80N do 15mm E 200GPa Solution L 2a b Section Property I πdo 4 64 Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 A L P1 a b P2 b 0 2 Solving Eqs 1 and 2 A P1 a b P2 a L A 201429 N B P1 P2 A B 128571 N Moment Function M x A Ψ x 0 P1 Ψ x a P2 Ψ x a b Slope and Elastic Curve E I d2v dx2 A Ψ x 0 P1 Ψ x a P2 Ψ x a b E I dv dx A 2 Ψ x 0 2 P1 2 Ψ x a 2 P2 2 Ψ x a b 2 C1 1 E I v A 6 Ψ x 0 3 P1 6 Ψ x a 3 P2 6 Ψ x a b 3 C1 x C2 2 Boundary Conditions v0 at x0 From Eq 2 0 0 0 0 C2 C2 0 v0 at xL From Eq 2 0 A 6 L3 P1 6 L a 3 P2 6 a3 C1 L C1 A 6 L2 P1 6L L a 3 P2 6L a3 C1 8857 N m2 Equation of Elastic Curve v x 1 E I A 6 x3 P1 6 Ψ x a 3 P2 6 Ψ x a b 3 C1 x 3 Maximum Deflection Assume vmax occurs at a x ab Given E I d2 v dx2 M x dvdx0 at x From Eq 1 0 A 2 x2 P1 2 x a 2 0 C1 4 Solving Eq3 Guess x 300 mm x Find x x 33005 mm For vmax substitute x into Eq3 v x 364 mm Ans Note The negative sign indicates downward displacement Problem 1254 Determine the slope and deflection at C EI is constant Given L1 8m L2 4m P 75kN Solution Support Reactions L L1 L2 ΣFy0 A B P 0 1 ΣΜB0 A L1 P L2 0 2 Solving Eqs 1 and 2 A P L2 L1 B P L1 L2 L1 A 375 kN B 1125 kN MEI Diagram Set EI kN m2 EIo 1 x1 0 001 L1 L1 x2 L1 101 L1 L M1 x1 A x1 1 EI M2 x2 A x2 B x2 L1 1 EI 0 5 10 200 0 Distance m MEI 1m M1 x1 M2 x2 x1 x2 MomentArea Theorems Slopes tBA 1 2 M1 L1 L1 L1 3 θA tBA L1 θA 400 1 EIo θCA 1 2 M1 L1 L1 1 2 M1 L1 L2 θC θCA θA θC 1400 1 EIo Ans Deflections tCA 1 2 M1 L1 L1 L1 3 L2 1 2 M1 L1 L2 2L2 3 C tCA L L1 tBA C 4800 m EIo Ans Problem 1255 Determine the slope and deflection at B EI is constant Problem 1256 If the bearings exert only vertical reactions on the shaft determine the slope at the bearings and the maximum deflection of the shaft EI is constant Problem 1257 Determine the slope at B and the deflection at C EI is constant Problem 1258 Determine the slope at C and the deflection at B EI is constant Problem 1259 The 60kg gymnast stands on the center of the simply supported balance beam If the beam is made of wood and has the cross section shown determine the maximum deflectionThe supports at A and B are assumed to be rigid Ew 12 GPa Given mo 60 kg L 27m bt 125mm bb 75mm h 150mm E 12GPa Solution W mo g Support Reactions By symmetry ABR ΣFy0 2R W 0 27 m 54 m m 79434 R 05W R 29420 N Section Property b bt bb yc Σ yi Ai Σ Ai yc bb h 05h 05b h h 3 bb h 05b h yc 6875 mm I 1 12 bb h3 bb h 05h yc 2 1 36 b h3 05 b h h 3 yc 2 I 2753906250 mm4 MEI Diagram x1 0 001 L L x2 L 101 L 2L M1 x1 R x1 1 E I M2 x2 R x2 W x2 L 1 E I 0 2 4 0 0002 Distance m MEI M1 x1 M2 x2 x1 x2 MomentArea Theorems Deflections tAC 1 2 M1 L L 2L 3 C tAC C 584 mm Due to symmetry the slope at midspan point C is zero Hence max C max 584 mm Ans Problem 1260 The shaft is supported by a journal bearing at A which exerts only vertical reactions on the shaft and by a thrust bearing at B which exerts both horizontal and vertical reactions on the shaft Determine the slope of the shaft at the bearings EI is constant Given L 03m a 01m P 400N Solution Support Reactions L 2L Due to antisymmetry B A ΣΜB0 A 2L P 2a 0 A a L P B A A 6667 N B 6667 N MEI Diagram Set EI kN m2 EIo 1 x1 0 001 L L x2 L 101 L 2L M1 x1 A x1 1 EI M2 x2 A x2 P 2a 1 EI 0 05 1 0 Distance m MEI 1m M1 x1 M2 x2 x1 x2 MomentArea Theorems Slopes tBA 1 2 M1 L L 2L 3 1 2 M1 L L L 3 L θA tBA L θA 0008 1 EIo Ans Due to antisymmetry or in a similar manner θB θA θB 0008 1 EIo Ans Problem 1261 The beam is subjected to the loading shown Determine the slope at A and the displacement at C Assume the support at A is a pin and B is a roller EI is constant Problem 1262 The rod is constructed from two shafts for which the moment of inetia of AB is I and BC is 2I Determine the maximum slope and deflection of the rod due to the loading The modulus of elasticity is E Problem 1263 Determine the deflection and slope at C EI is constant Problem 1264 If the bearings at A and B exert only vertical reactions on the shaft determine the slope at A EI is constant Problem 1265 If the bearings at A and B exert only vertical reactions on the shaft determine the slope at C EI is constant Problem 1266 Determine the deflection at C and the slope of the beam at A B and C EI is constant Given L1 6m L2 3m Mo 8kN m Solution L L1 L2 Support Reactions ΣFy0 A B 0 1 ΣΜB0 A L1 Mo 0 2 Solving Eqs 1 and 2 A Mo L1 B A A 1333 kN B 1333 kN MEI Diagram Set EI kN m2 EIo 1 x1 0 001 L1 L1 x2 L1 101 L1 L M1 x1 A x1 1 EI M2 x2 A x2 B x2 L1 1 EI 0 5 10 5 0 Distance m MEI 1m M1 x1 M2 x2 x1 x2 MomentArea Theorems Deflections tBA 1 2 M1 L1 L1 L1 3 tCA 1 2 M1 L1 L1 L1 3 L2 M1 L1 L2 L2 2 C tCA L L1 tBA C 84 m EIo Ans Slopes θA tBA L1 θA 8 1 EIo Ans θBA 1 2 M1 L1 L1 θB θBA θA θB 16 1 EIo Ans θCA 1 2 M1 L1 L1 M1 L1 L2 θC θCA θA θC 40 1 EIo Ans Problem 1267 The bar is supported by the roller constraint at C which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown determine the slope and displacement at A EI is constant Problem 1268 The acrobat has a weight of 750 N 75 kg and suspends himself uniformly from the center of the high bar Determine the maximum bending stress in the pipe bar and its maximum deflection The pipe is made of L2 steel and has an outer diameter of 25 mm and a wall thickness of 3 mm Given a 09m do 25mm W 750N b 045m t 3mm E 200GPa Solution L 2a b Support Reactions By symmetry ABR ΣFy0 2R W 0 R 05W R 37500 N Section Property ro 05do ri ro t I π 4 ro 4 ri 4 Maximum Bendiug Stress Mmax R a σmax Mmax ro I σmax 33017 MPa Ans σY 703 MPa OK MEI Diagram x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L M1 x1 R x1 1 E I M2 x2 R x2 W 2 x2 a 1 E I M3 x3 R x3 W x3 L 2 1 E I 0 1 2 0 01 02 Distance m MEI 1m M1 x1 M2 x2 M3 x3 x1 x2 x3 MomentArea Theorems Deflections tAC 1 2 M1 a a 2a 3 M1 a b 2 b 4 a C tAC C 6574 mm Due to symmetry the slope at midspan point C is zero Hence max C max 6574 mm Ans Problem 1269 Determine the value of a so that the displacement at C is equal to zero EI is constant Problem 1270 The beam is made of a ceramic material In order to obtain its modulus of elasticity it is subjected to the elastic loading shown If the moment of inertia is I and the beam has a measured maximum deflection determine E The supports at A and D exert only vertical reactions on the beam Problem 1271 Determine the maximum deflection of the shaft EI is constant The bearings exert only vertical reactions on the shaft Problem 1272 The beam is subjected to the load P as shown Determine the magnitude of force F that must be applied at the end of the overhang C so that the deflection at C is zero EI is constant Problem 1273 At what distance a should the bearing supports at A and B be placed so that the deflection at the center of the shaft is equal to the deflection at its ends The bearings exert only vertical reactions on the shaft EI is constant 0 1 2 0 001 Distance m MEI 1m M1 x1 M2 x2 M3 x3 x1 x2 x3 Given a 05m b 08m c 12m do 50mm P1 600N P2 1200N E 200GPa Solution L a b c Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 A L P1 b c P2 c 0 2 Solving Eqs 1 and 2 A P1 b c P2 c L B P1 P2 A A 1056 kN B 0744 kN Section Property I π do 4 64 MEI Diagram x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L M1 x1 A x1 1 E I M2 x2 A x2 P1 x2 a 1 E I M3 x3 A x3 P1 x3 a P2 x3 a b 1 E I Problem 1274 Determine the slope of the 50mmdiameter A36 steel shaft at the bearings at A and B The bearings exert only vertical reactions on the shaft MomentArea Theorems Slopes tBA 1 2 M3 L c c 2c 3 1 2 M2 a b M1 a b b 3 c M1 a b b 2 c 1 2 M1 a a L 2a 3 θA tBA L θA 001046 rad Ans θB tAB L θB 000968 rad Ans tAB 1 2 M3 L c c L 2c 3 1 2 M2 a b M1 a b 2 b 3 a M1 a b b 2 a 1 2 M1 a a 2a 3 0 1 2 0 001 Distance m MEI 1m M1 x1 M2 x2 M3 x3 x1 x2 x3 Given a 05m b 08m c 12m do 50mm P1 600N P2 1200N E 200GPa Solution L a b c Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 A L P1 b c P2 c 0 2 Solving Eqs 1 and 2 A P1 b c P2 c L B P1 P2 A A 1056 kN B 0744 kN Section Property I π do 4 64 MEI Diagram x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L M1 x1 A x1 1 E I M2 x2 A x2 P1 x2 a 1 E I M3 x3 A x3 P1 x3 a P2 x3 a b 1 E I Problem 1275 Determine the maximum deflection of the 50mm diameter A36 steel shaft It is supported by bearings at its ends A and B which only exert vertical reactions on the shaft MomentArea Theorems Deflections tBA 1 2 M3 L c c 2c 3 1 2 M2 a b M1 a b b 3 c M1 a b b 2 c 1 2 M1 a a L 2a 3 θA tBA L θA 001046 rad Ans Maximum Deflection Assume vmax occurs at point E a x ab Slopes θEA 1 2 M1 a a M1 a x 1 2 M2 a b M1 a x b x θEA 1 2 A a E I a A a E I x 1 2E I A a b P1 b A a x b x θEA 1 2E I A a2 2A a x A P1 x2 θE θA θEA θE θA 1 2E I A a2 2A a x A P1 x2 At point E where θE0 Given 0 θA 2E I A a2 2A a x A P1 x2 3 Solving Eq3 Guess x 07 m x Find x x 73327 mm 08m OK Deflection tAE 1 2 M1 a a 2a 3 M1 a x x 2 a 1 2 M2 a b M1 a x b x 2 x 3 a max tAE max 8161 mm Ans 0 05 1 01 005 Distance m MEI 1m M1 x1 M2 x2 M3 x3 x1 x2 x3 a 02m b 03m c 05m do 20mm Problem 1276 Determine the slope of the 20mmdiameter A36 steel shaft at the bearings at A and B The bearings exert only vertical forces on the shaft P1 800N P2 350N E 200GPa Solution L a b c Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 P1 L A b c P2 c 0 2 Solving Eqs 1 and 2 A P1 L P2 c b c B P1 P2 A A 121875 N B 6875 N Section Property I π do 4 64 MEI Diagram x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L M1 x1 P1 x1 1 E I M2 x2 P1 x2 A x2 a 1 E I M3 x3 P1 x3 A x3 a P2 x3 a b 1 E I Given MomentArea Theorems Slopes tBA 1 2 M3 a b c 2c 3 M3 a b b b 2 c 1 2 M2 a M3 a b b 2 b 3 c θA tBA b c θA 001811 rad Ans tAB 1 2 M3 a b c c 3 b M3 a b b b 2 1 2 M2 a M3 a b b b 3 θB tAB b c θB 000592 rad Ans 0 05 1 01 005 Distance m MEI 1m M1 x1 M2 x2 M3 x3 x1 x2 x3 Given a 02m b 03m c 05m do 20mm P1 800N P2 350N E 200GPa Solution L a b c Support Reactions ΣFy0 A B P1 P2 0 1 ΣΜB0 P1 L A b c P2 c 0 2 Solving Eqs 1 and 2 A P1 L P2 c b c B P1 P2 A A 121875 N B 6875 N Section Property I π do 4 64 MEI Diagram x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L M1 x1 P1 x1 1 E I M2 x2 P1 x2 A x2 a 1 E I M3 x3 P1 x3 A x3 a P2 x3 a b 1 E I Problem 1277 Determine the displacement of the 20mmdiameter A36 steel shaft at D The bearings at A and B exert only vertical reactions on the shaft MomentArea Theorems Deflection tDB 1 2 M3 a b c L 2c 3 M3 a b b b 2 a 1 2 M2 a M3 a b b b 3 a 1 2 M1 a a 2a 3 tAB 1 2 M3 a b c c 3 b M3 a b b b 2 1 2 M2 a M3 a b b b 3 D tDB L b c tAB D 498 mm Ans Problem 1278 The beam is subjected to the loading shown Determine the slope at B and deflection at C EI is constant Problem 1279 Determine the slope at B and the displacement at C The bearings at A and B exert only vertical reactions on the shaft EI is constant Problem 1280 Determine the displacement at D and the slope at C The bearings at A and B exert only vertical reactions on the shaft EI is constant Fh 1kN max C Given a 50mm do 32mm Fv 45kN Support Reactions b 650mm h 450mm E 200GPa Solution L a b max 178 mm ΣFy0 Fv A B 0 1 ΣΜA0 Fv a Fh h B b 0 2 Solving Eqs 1 and 2 B Fv a Fh h b A Fv B Ans Section Property I π 64 do 4 MEI Diagram x1 0 001 a a x2 a 101 a L M1 x1 Fh h Fv x1 1 E I 0 02 04 06 0 005 Distance m MEI 1m M1 x1 M2 x2 x1 x2 MomentArea Theorems Deflections tAB 1 2 M1 a b b 3 θB tAB b θB M1 a b 6 The maximum displacement occurs at point C where θC 0 Let x L x θCB 1 2 M1 a x b x θC θB θCB 0 M1 a b 6 1 2 M1 a x2 b x b 1 3 x 37528 mm b 650 mm OK C tBC tBC θBC 2x 3 θBC θCB C 1 2 M1 a x b x 2x 3 Problem 1281 The two force components act on the tire of the automobile as shown The tire is fixed to the axle which is supported by bearings at A and B Determine the maximum deflection of the axle Assume that the bearings resist only vertical loads The thrust on the axle is resisted at C The axle has a diameter of 32 mm and is made of A36 steel Neglect the effect of axial load on deflection M2 x2 Fh h Fv x2 A x2 a 1 E I EI kN m2 Given a 03m P 20kN b 09m Solution L 2a b Support Reactions By symmetry ABR ΣFy0 2R 2P 0 R P Maximum Bendiug Stress Mmax R a MEI Diagram Set Ans EIo 1 x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L M1 x1 R x1 1 EI M2 x2 R x2 P x2 a 1 EI M3 x3 R x3 P x3 a P x3 a b 1 EI 0 05 1 15 0 5 10 Distance m MEI 1m M1 x1 M2 x2 M3 x3 x1 x2 x3 MomentArea Theorems Deflections tAC 1 2 M1 a a 2a 3 M1 a b 2 b 4 a C tAC C 15975 m EIo Due to symmetry the slope at midspan point C is zero Hence max C max 15975 m EIo Problem 1282 Determine the displacement at D and the slope at C The bearings at A and B exert only vertical reactions on the shaft EI is constant Problem 1283 Beams made of fiberreinforced plastic may one day replace many of those made of A36 steel since they are onefourth the weight of steel and are corrosion resistant Using the table in Appendix B with σallow 160 MPa and τallow 84 MPa select the lightestweight steel wideflange beam that will safely support the 25kN load then compute its maximum deflection What would be the maximum deflection of this beam if it were made of a fiberreinforced plastic with Ep 126 GPa and had the same moment of inertia as the steel beam Given σallow 160MPa P 25kN L 3m τallow 84MPa E 126GPa Solution Support Reactions By symmetry ABR ΣFy0 2R P 0 R 05P R 1250 kN Maximum Moment and Shear Vmax R Vmax 125 kN Mmax R L Mmax 375 kN m Bending Stress Sreqd Mmax σallow Sreqd 23437500 mm3 Select W 310x21 Sx 244 103 mm3 d 303mm tw 508mm I Sx 05d Shear Stress Provide a shear stress check τmax Vmax tw d τmax 812 MPa τallow 84 MPa OK Hence Use W 310x21 Ans MEI Diagram x1 0 001 L L x2 L 101 L 2L M1 x1 R x1 1 E I M2 x2 R x2 P x2 L 1 E I 0 2 4 6 0 0005 Distance m MEI M1 x1 M2 x2 x1 x2 MomentArea Theorems Deflections tAC 1 2 M1 L L 2L 3 C tAC C 2415 mm Due to symmetry the slope at midspan point C is zero Hence max C max 2415 mm Ans Problem 1284 The simply supported shaft has a moment of inertia of 2I for region BC and a moment of inertia I for regions AB and CD Determine the maximum deflection of the shaft due to the load P The modulus of elasticity is E Problem 1285 The A36 steel shaft is used to support a rotor that exerts a uniform load of 5 kNm within the region CD of the shaft Determine the slope of the shaft at the bearings A and B The bearings exert only vertical reactions on the shaft Problem 1286 The beam is subjected to the loading shown Determine the slope at B and deflection at C EI is constant Problem 1287 Determine the slope of the shaft at A and the displacement at D The bearings at A and B exert only vertical reactions on the shaft EI is constant Problem 1288 Determine the slope at B and the displacement at C The member is an A36 steel structural tee for which I 30106 mm4 Given P 25kN L 1m w 25 kN m E 200GPa I 30 106 mm4 Solution Support Reactions By symmetry ABR ΣFy0 2R P 0 2R w 2L 0 R 05P R w L R 1250 kN R 2500 kN MEI Diagram x1 0 001 L L x2 L 101 L 2L For point load P M1 x1 R x1 1 E I M2 x2 R x2 P x2 L 1 E I For uniform load w M1 x1 R x1 05w x1 2 1 E I M2 x2 R x2 05 w x2 2 1 E I 0 1 2 0 0002 Distance m MEI M1 x1 M2 x2 x1 x2 0 1 2 0 0002 Distance m MEI M1 x1 M2 x2 x1 x2 MomentArea Theorems Slope Due to symmetry the slope at midspan point C is zero Hence θB θBC θBC 1 2 M1 L L 2 3 M1 L L θB θBC θB 000243 rad Ans Deflection tAC 1 2 M1 L L 2L 3 2 3 M1 L L 5L 8 C tAC max C max 156 mm Ans Problem 1289 The W200 X 71 cantilevered beam is made of A36 steel and is subjected to the loading shown Determine the displacement at its end A Given P 6kN Mo 3kN m E 200GPa a 24m Solution L 2a Use W 200x71 I 766 106 mm4 Elsastic Curve The elastic curves for the concentrated load and couple moment are drawn separately as shown Method of Superposition Using the table in Appendix C the required slope and displacement are A1 P L3 3E I C2 Mo a2 2E I θC2 Mo a E I A2 C2 θC2 a Hence the displacement at A is A A1 A2 A 1613 mm Ans Problem 1290 The W200 X 71 cantilevered beam is made of A36 steel and is subjected to the loading shown Determine the displacement at C and the slope at A Given P 6kN Mo 3kN m E 200GPa a 24m Solution L 2a Use W 200x71 I 766 106 mm4 Elsastic Curve The elastic curves for the concentrated load and couple moment are drawn separately as shown Method of Superposition Using the table in Appendix C the required slope and displacement are C1 P a2 6E I 3L a C2 Mo a2 2E I θA1 P L2 2E I θC2 Mo a E I θA2 θC2 Hence the slope at A is θA θA1 θA2 θA 000498 rad Ans Hence the displacement at C is C C1 C2 C 508 mm Ans Problem 1291 The W360 X 64 simply supported beam is made of A36 steel and is subjected to the loading shown Determine the deflection at its center C Given a 3m Mo 60kN m E 200GPa w 30 kN m Solution L 2a Use W 360x64 I 179 106 mm4 Elsastic Curve The elastic curves for the uniform load and couple moment are drawn separately as shown Method of Superposition Using the table in Appendix C the required displacements are C1 5w L4 768E I C2 Mo 6 E I a L a2 3 L a 2 L2 Hence the displacement at C is C C1 C2 C 1084 mm Ans Problem 1292 The W360 X 64 simply supported beam is made of A36 steel and is subjected to the loading shown Determine the slope at A and B Given a 3m Mo 60kN m E 200GPa w 30 kN m Solution L 2a Use W 360x64 I 179 106 mm4 Elsastic Curve The elastic curves for the uniform load and couple moment are drawn separately as shown Method of Superposition Using the table in Appendix C the required slopes are θA1 7w L3 384E I θA2 Mo L 6 E I Hence the slope at A is θA θA1 θA2 θA 000498 rad θA 0285 deg Ans θB1 3w L3 128E I θB2 Mo L 3 E I Hence the slope at B is θB θB1 θB2 θB 000759 rad θB 0435 deg Ans Problem 1293 Determine the moment M0 in terms of the load P and dimension a so that the deflection at the center of the beam is zero EI is constant Problem 1294 The beam supports the loading shown Code restrictions due to a plaster ceiling require the maximum deflection not to exceed 1360 of the span length Select the lightestweight A36 steel wideflange beam from Appendix B that will satisfy this requirement and safely support the load The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa Assume A is a roller and B is a pin Given σallow 168MPa a 36m τallow 100MPa w 60 kN m E 200GPa Solution L 2a ΣFy0 A B w a 0 1 ΣΜB0 A L 05 w a2 0 2 Solving Eqs 1 and 2 A wa2 2L B w a A A 54kN B 162kN Maximum Moment and Shear Vmax max A B Vmax 162kN Mmax occurs at x where Vx 0 V x A w Φ x a 0 A w Φ x a x A w a Mmax A x 05w x a 2 Mmax 2187 kN m Strength criterion Sreqd Mmax σallow Sreqd 130178571 mm3 Select W 410x74 Sx 1330 103 mm3 d 413mm tw 965mm I Sx 05d Shear Stress Provide a shear stress check τmax Vmax tw d τmax 4065 MPa τallow 100 MPa OK Deflection criterion allow L 360 allow 20mm Using the table in Appendix C max 0006563 w L4 E I max 1927 mm allow 20mm OK Hence Use W 410x74 Ans x1 0 001 a a x2 a 101 a L V1 x1 A 1 kN V2 x2 A w x2 a 1 kN M1 x1 A x1 1 kN m M2 x2 A x2 05 w x2 a 2 1 kN m 0 5 200 0 Distance m Shear kN V1 x1 V2 x2 x1 x2 0 5 0 200 Distance m M kNm M1 x1 M2 x2 x1 x2 Problem 1295 The simply supported beam carries a uniform load of 29 kNm Code restrictions due to a plaster ceiling require the maximum deflection not to exceed 1360 of the span length Select the lightest weight A36 steel wideflange beam from Appendix B that will satisfy this requirement and safely support the load The allowable bending stress is σallow 168 MPa and the allowable shear stress is τallow 100 MPa Assume A is a pin and B a roller support Given σallow 168MPa a 12m b 24m τallow 100MPa P 40kN E 200GPa w 29 kN m Solution L 2a b Support Reactions By symmetry ABR ΣFy0 2R 2P w L 0 R 05 2P w L R 10960 kN Maximum Moment and Shear Vmax R Vmax 1096 kN Mmax occurs at midspan Mmax R 05L P 05L a 05w 05L 2 Mmax 13152 kN m Strength criterion Sreqd Mmax σallow Sreqd 78285714 mm3 Select W 360x51 Sx 794 103 mm3 d 355mm tw 724mm I Sx 05d Shear Stress Provide a shear stress check τmax Vmax tw d τmax 4264 MPa τallow 100 MPa OK Deflection criterion allow L 360 allow 1333 mm Using the table in Appendix C max 5w L4 384E I 2 P a b 6E I L L2 a2 05L 2 max 1161 mm allow 20mm OK Hence Use W 360x51 Ans x1 0 001 a a x2 a 101 a a b x3 a b 101 a b L V1 x1 R w x1 1 kN V2 x2 R w x2 P 1 kN V3 x3 R w x3 2P 1 kN M1 x1 R x1 05 w x1 2 1 kN m M2 x2 R x2 05 w x2 2 P x2 a 1 kN m M3 x3 R x3 05 w x3 2 P x3 a P x3 a b 1 kN m 0 2 4 100 0 100 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 2 4 0 100 Distance m M kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 1296 The W250 X 45 cantilevered beam is made of A36 steel and is subjected to unsymmetrical bending caused by the applied moment Determine the deflection of the centroid at its end A due to the loading Hint Resolve the moment into components and use superposition Given L 45m Mo 25kN m E 200GPa θ 30deg Solution Use W 250x45 Ix 711 106 mm4 Iy 703 106 mm4 Displacements Using the table in Appendix C the required displacements are xmax Mo sin θ L2 2E Iy ymax Mo cos θ L2 2E Ix Hence the displacement at A is A xmax ymax A 1054 mm Ans Problem 1297 The assembly consists of a cantilevered beam CB and a simply supported beam AB If each beam is made of A36 steel and has a moment of inertia about its principal axis of Ix 46106 mm4 determine the displacement at the center D of beam BA Given P 75kN I 46 106 mm4 E 200GPa LDA 24m LBC 48m LBA 48m Solution Consider beam AB Support Reactions By symmetry ABR ΣFy0 2R P 0 R 05P Method of Superposition Using the table in Appendix C the required slope and displacement are Consider beam BC B R LBC 3 3E I Consider beam AB D1 P LBA 3 48E I D2 LDA LBA B Hence the displacement at D is D D1 D2 D 9391 mm Ans Problem 1298 The rod is pinned at its end A and attached to a torsional spring having a stiffness k which measures the torque per radian of rotation of the spring If a force P is always applied perpendicular to the end of the rod determine the displacement of the force EI is constant Problem 1299 The relay switch consists of a thin metal strip or armature AB that is made of red brass C83400 and is attracted to the solenoid S by a magnetic field Determine the smallest force F required to attract the armature at C in order that contact is made at the free end B Also what should the distance a be for this to occur The armature is fixed at A and has a moment of inertia of I 0181012 m4 Given B 2mm L 50mm E 101GPa I 018 10 12 m4 Solution LAC L LCB L Elsastic Curve The elastic curves for the concentrated load and couple moment are drawn separately as shown Method of Superposition Using the table in Appendix C the required slope and displacement are θC F LAC 2 2E I C F LAC 3 3E I 1 Hence the displacement at B is B C θC LCB B F LAC 3 3E I F LAC 2 2E I LCB F 6B E I 2 LAC 3 3 LCB LAC 2 F 0349 N Ans From Eq1 C F LAC 3 3E I C 0800 mm Ans Problem 12100 Determine the vertical deflection and slope at the end A of the bracket Assume that the bracket is fixed supported at its base and neglect the axial deformation of segment AB EI is constant Given h 75mm L 100mm P 400N w 4 kN m Solution Set EI 1kNm2 Elsastic Curve The elastic curves for the concentrated load uniform distributed load and couple moment are drawn separately as shown MB 05w L2 Method of Superposition Using the table in Appendix C the required slope and displacement are θA1 w L3 6EI θB2 MB h EI θA2 θB2 θB3 P h2 2EI θA3 θB3 A1 w L4 8EI A2 θB2 L A3 θB3 L Hence the slope at A is θA θA1 θA2 θB3 θA 000329 rad θA 000329 EI Ans The displacement at A is A A1 A2 A3 A 0313 mm A 0313 EI Ans Problem 12101 The W610 X 155 A36 steel beam is used to support the uniform distributed load and a concentrated force which is applied at its end If the force acts at an angle with the vertical as shown determine the horizontal and vertical displacement at point A Given L 3m P 25kN E 200GPa cx 3 5 cy 4 5 w 30 kN m Solution Use W 610x155 Ix 1290 106 mm4 Iy 108 106 mm4 Elsastic Curve The elastic curves for the concentrated load and uniform load are drawn separately as shown Py P cy Py 20kN Px P cx Px 15kN Method of Superposition Using the table in Appendix C the required displacement are A1 w L4 8E Ix A2 Py L3 3E Ix The vertical displacement at A is Ay A1 A2 Ay 1875 mm Ans The horizontal displacement at A is Ax Px L3 3E Iy Ax 6250 mm Ans Problem 12102 The framework consists of two A36 steel cantilevered beams CD and BA and a simply supported beam CB If each beam is made of steel and has a moment of inertia about its principal axis of Ix 46106 mm4 determine the deflection at the center G of beam CB Given P 75kN I 46 106 mm4 E 200GPa LDC 48m LBC 48m LAB 48m Solution Consider beam BC Support Reactions By symmetry BCR ΣFy0 2R P 0 R 05P Method of Superposition Using the table in Appendix C the required slope and displacement are Consider beam DC C R LDC 3 3E I Due to symmetry B C Consider beam BC G P LBC 3 48E I Hence the displacement at G is G C G G 16904 mm Ans Problem 12103 Determine the reactions at the supports A and B then draw the moment diagram EI is constant Problem 12104 Determine the reactions at the supports A and B then draw the shear and moment diagrams EI is constant Neglect the effect of axial load Problem 12105 Determine the reactions at the supports A B and C then draw the shear and moment diagrams EI is constant Problem 12106 Determine the reactions at the supports then draw the shear and moment diagram EI is constant Problem 12107 Determine the moment reactions at the supports A and B EI is constant Problem 12108 Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment EI is constant Problem 12109 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant Problem 12110 The beam has a constant E1I1 and is supported by the fixed wall at B and the rod AC If the rod has a crosssectional area A2 and the material has a modulus of elasticity E2 determine the force in the rod Problem 12111 Determine the moment reactions at the supports A and B and then draw the shear and moment diagrams Solve by expressing the internal moment in the beam in terms of Ay and MA EI is constant Problem 12112 Determine the moment reactions at the supports A and B EI is constant Problem 12113 Determine the moment reactions at the supports A and B then draw the shear and moment diagrams EI is constant Problem 12114 Determine the moment reactions at the supports A and B then draw the shear and moment diagrams EI is constant Problem 12115 Determine the reactions at the supports EI is constant b 18m Given P 25kN a 12m Solution L a b Due to symmetry the slope at support B is zero Hence θB 0 and tanθB 0 MomentArea Theorems Ans Set EI 1kNm2 A 1kN x1 0 001 a a x2 a 101 a L For point load P M1 x1 0 1 EI M2 x2 P x2 a 1 EI For reaction A M1 x1 A x1 1 EI M2 x2 A x2 1 EI 0 1 2 3 50 0 Distance m MEI M1 x1 M2 x2 x1 x2 0 1 2 3 0 5 Distance m MEI M1 x1 M2 x2 x1 x2 Deflection tAB1 1 2 M2 L b a 2 b 3 tAB2 1 2 M2 L L 2L 3 tAB tAB1 tAB2 Ay A Compatibility tAB 0 0 tAB1 tAB2 Ay A Ay A tAB1 tAB2 Ay 1080 kN Ans Support Reactions Due to symmetry Cy Ay Cy 1080 kN Ans ΣFy0 Ay By Cy 2P 0 By 2P Ay Cy By 2840 kN Problem 12116 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant x1 0 001 a a x3 L 101 L L b x2 a 101 a L x4 L b 101 L b 2L V1 x1 Ay 1 kN V3 x3 Ay P By 1 kN V2 x2 Ay P 1 kN V4 x3 Ay P By P 1 kN M1 x1 Ay x1 1 kN m M2 x2 Ay x2 P x2 a 1 kN m M3 x3 Ay x3 P x3 a By x3 L 1 kN m M4 x4 Ay x4 P x4 a By x4 L P x4 L b 1 kN m 0 1 2 3 4 5 6 20 0 20 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 1 2 3 4 5 6 20 0 20 Distance m M kNm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 12117 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant Support B is a thrust bearing Problem 12118 Determine the reactions at the supports EI is constant Problem 12119 Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment EI is constant Problem 12120 Determine the moment reactions at the supports A and B then draw the shear and moment diagrams EI is constant Problem 12121 Determine the reactions at the supports A and B EI is constant Problem 12122 Determine the reactions at the bearing supports A B and C of the shaft then draw the shear and moment diagrams EI is constant Each bearing exerts only vertical reactions on the shaft Given a 1m P 04kN b 2m Solution L 4a Set EI kNm2 Method of Superposition Using the table in Appendix C the required displacements are B1 P a 6 EI b L L2 a2 b2 B2 B L3 48EI The compatibity condition at B requires 2B1 B2 0 B2 2 B1 B L3 48E I 2B1 B 48 EI L3 2B1 B 0550 kN Ans Support Reactions Given ΣFy0 A B C 2P 0 1 ΣΜC0 A L P 3a B b P a 0 2 Solving Eqs 1 and 2 Guess A 1kN C 1kN A C Find A C A C 0125 0125 kN Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a 3a x4 3a 101 3a L V1 x1 A 1 kN V2 x2 A P 1 kN V3 x3 A P B 1 kN V4 x4 A P B P 1 kN M1 x1 A x1 1 kN m M2 x2 A x2 P x2 a 1 kN m M3 x3 A x3 P x3 a B x3 2a 1 kN m M4 x4 A x4 P x4 a B x4 2a P x4 3a 1 kN m 0 2 4 0 Distance m Shear kN V1 x1 V2 x2 V3 x3 V4 x4 x1 x2 x3 x4 0 2 4 01 0 01 Distance m M kNm M1 x1 M2 x2 M3 x3 M4 x4 x1 x2 x3 x4 Problem 12123 The A36 steel beam and rod are used to support the load of 40 kN If it is required that the allowable normal stress for the steel is σallow 125 MPa and the maximum deflection not exceed 125 mm determine the smallest diameter rod that should be used The beam is rectangular having a height of 125 mm and a thickness of 75 mm Given σallow 125MPa Lb 12m Lr 15m δallow 125mm b 75mm h 125mm E 200GPa P 40kN Solution Compatibility at B δr δb Ib b h3 12 Fr Lr E Ar P Fr Lb 3 3E Ib Strength criterion Assume rod reaches its maximum stress σallow Fr Ar σallow Lr E P Fr Lb 3 3E Ib Fr P 3σallow Lr Ib Lb 3 Fr 3603 kN Check Bending Stress c 05 h Mmax P Fr Lb σmax Mmax c Ib σmax 2441 MPa σallow 125 MPa OK Check Deflection δb P Fr Lb 3 3E Ib δb 094 mm δallow 125mm OK Hence from σallow Fr Ar Ar π do 2 4 σallow 4Fr π do 2 do 4Fr π σallow do 1916 mm Ans Problem 12124 Determine the reactions at the supports A B and C then draw the shear and moment diagrams EI is constant Given a 2m P 60kN w 50 kN m b 4m Solution L 2a b Set EI 1kNm2 Method of Superposition Using the table in Appendix C the required displacements are B1 5w L4 768EI B2 P a 6 EI b L L2 a2 b2 B3 B L3 48EI The compatibity condition at C requires B1 B2 B3 0 B3 B1 B2 B L3 48E I B1 B2 B 48 EI L3 B1 B2 B 16625 kN Ans Support Reactions Given ΣFy0 A B C P w b 0 1 ΣΜC0 A L P a b B b 05 w b2 0 2 Solving Eqs 1 and 2 Guess A 1kN C 1kN A C Find A C A C 1188 8188 kN Ans x1 0 001 a a x2 a 101 a 2a x3 2a 101 2a L V1 x1 A 1 kN V2 x2 A P 1 kN V3 x3 A P B w x3 2a 1 kN M1 x1 A x1 1 kN m M2 x2 A x2 P x2 a 1 kN m M3 x3 A x3 P x3 a B x3 2a 05 w x3 2a 2 1 kN m 0 5 200 0 200 Distance m Shear kN V1 x1 V2 x2 V3 x3 x1 x2 x3 0 5 100 0 100 Distance m M kNm M1 x1 M2 x2 M3 x3 x1 x2 x3 Problem 12125 Determine the reactions at support C EI is constant for both beams Problem 12126 Determine the reactions at A and B Assume the support at A only exerts a moment on the beam EI is constant Problem 12127 Determine the reactions at the supports A and B EI is constant Problem 12128 Each of the two members is made from 6061T6 aluminum and has a square cross section 25 mm X 25 mm They are pin connected at their ends and a jack is placed between them and opened until the force it exerts on each member is 25 kN Determine the greatest force P that can be applied to the center of the top member without causing either of the two members to yield For the analysis neglect the axial force in each member Assume the jack is rigid Given b 25mm h 25mm Ro 25kN L 4m E 689GPa σallow 255MPa Solution Compatibility at B δE δF P R Ro L3 48E I R Ro L3 48E I P R Ro R Ro R 05P Ro Criterion Maximum moment occursat center of each member Top member Mmax 1 2 P R Ro L 2 Mmax P L 8 Bottom member Mmax 1 2 R Ro L 2 Mmax P L 8 Both members will yieldat the same time I b h3 12 c 05 h σallow Mmax c I σallow P L c 8I P 8I σallow L c P 133 kN Ans Problem 12129 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant Problem 12130 The beam is supported by a pin at A a spring having a stiffness k at B and a roller at C Determine the force the spring exerts on the beam EI is constant Problem 12131 The beam AB has a moment of inertia I 200106 mm4 and rests on the smooth supports at its ends A 18mmdiameter rod CD is welded to the center of the beam and to the fixed support at D If the temperature of the rod is decreased by 80C determine the force developed in the rod The beam and rod are both made of A36 steel Unit used C deg Given rc 18mm Ib 200 106 mm4 Lc 125m Lb 3m T 80C α 12 10 6 1 C E 200GPa Solution Section Property Ac π 4 rc 2 Method of Superposition Using the table in Appendix C the required displacements are vC Fc Lb 3 48E Ib Using the axial force formula δF Fc Lc AcE The thermal contraction is δT α T Lc The compatibity condition at C requires vC δF δT vC δF δT Fc Lb 3 48E Ib Fc Lc AcE δT Fc δT Lb 3 48E Ib Lc AcE Fc 3107 kN Ans Problem 12132 Determine the deflection at the end B of the clamped A36 steel strip The spring has a stiffness of k 2 Nmm The strip is 5 mm wide and 10 mm high Also draw the shear and moment diagrams for the strip P 50N Given L 200mm h 10mm k 2 N mm t 5mm E 200GPa Solution Section Property I t h3 12 Method of Superposition Using the table in Appendix C the required displacements are B1 P L3 3 E I B2 F L3 3 E I F k B The compatibity condition at B requires B B1 B2 B P L3 3 E I k B L3 3 E I B P L3 3 E I k L3 B 150 mm Ans Support Reactions B k B P A B MA B L B 46992 N A 46992 N MA 940 N m x1 0 001 L L V1 x1 A N M1 x1 MA A x1 1 N m 0 01 0 50 Distance m Shear N V1 x1 x1 0 01 10 0 Distance m M Nm M1 x1 x1 Problem 12133 The beam is made from a soft elastic material having a constant EI If it is originally a distance from the surface of its end support determine the distance a at which it rests on this support when it is subjected to the uniform load w0 which is great enough to cause this to happen Problem 12134 The box frame is subjected to a uniform distributed loading w along each of its sides Determine the moment developed in each corner Neglect the deflection due to axial load EI is constant Problem 12135 Determine the equation of the elastic curve for the beam Specify the slope and displacement at A EI is constant Problem 12136 The wooden beam is subjected to the loading shown Assume the support at A is a pin and B is a roller Determine the slope at A and the displacement at C Use the momentarea theorems EI is constant Problem 12137 Determine the maximum deflection between the supports A and B EI is constant Use the method of integration Problem 12138 If the bearings at A and B exert only vertical reactions on the shaft determine the slope at B and the deflection at C EI is constant Use the momentarea theorems Problem 12139 The W200 X 36 simply supported beam is subjected to the loading shown Using the method of superposition determine the deflection at its center C The beam is made of A36 steel Given a 24m Mo 75kN m E 200GPa w 100 kN m Solution L 2a Use W 200x36 I 344 106 mm4 Elsastic Curve The elastic curves for the uniform load and couple moment are drawn separately as shown Method of Superposition Using the table in Appendix C the required displacement are C1 5w L4 768E I C2 Mo 6E I a L a2 3a L 2L2 Hence the displacement at C is C C1 C2 C 5180 mm Ans Problem 12140 The shaft is supported by a journal bearing at A which exerts only vertical reactions on the shaft and by a thrust bearing at B which exerts both horizontal and vertical reactions on the shaft Draw the bendingmoment diagram for the shaft and then from this diagram sketch the deflection or elastic curve for the shafts centerline Determine the equations of the elastic curve using the coordinates x1 and x2 EI is constant Given L 03m a 01m P 400N Solution Support Reactions Lo 2L Due to antisymmetry B A ΣΜB0 A 2L P 2a 0 A a L P B A A 13333 N B 13333 N Moment Function M1 x1 A x1 M2 x2 B x2 Slope and Elastic Curve E I d2 v1 dx1 2 M x1 E I d2 v2 dx2 2 M x2 E I d2v1 dx1 2 A x1 E I d2v2 dx2 2 B x2 E I dv1 dx1 A x1 2 2 C1 1 E I dv2 dx2 B x2 2 2 C3 3 E I v1 A x1 3 6 C1 x1 C2 2 E I v2 B x2 3 6 C3 x2 C4 4 Boundary Conditions v10 at x10 and v20 at x20 From Eq 2 0 0 0 C2 From Eq 4 0 0 0 C4 C2 0 C4 0 Continuity Condition 1 dv1 dx1 dv2 dx2 at x1L and at x2L From Eqs 1 and 3 A L2 2 C1 B L2 2 C3 C1 C3 5 Continuity Condition 2 v1 v2 at x1L and at x2L From Eqs 2 and 4 A L3 6 C1 L B L3 6 C3 L C1 B A L2 6 C3 6 Solving Eqs 5 and 6 C1 B A L2 12 C1 200 N m2 Ans C3 C1 C3 200 N m2 Ans Hence Co A 6 Co 2222 N Ans v1 1 EI Co x1 3 C1 x1 Ans v2 1 EI Co x2 3 C3 x2 Ans MEI diagram Set EI 1kN m2 x1 0 001 L L x2 L 101 L 2L M1 x1 A x1 1 EI M2 x2 A x2 P 2a 1 EI 0 02 04 0 Distance m MEI 1m M1 x1 M2 x2 x1 x2 Problem 12141 The rim on the flywheel has a thickness t width b and specific weight γ If the flywheel is rotating at a constant rate of ω determine the maximum moment developed in the rim Assume that the spokes do not deform Hint Due to symmetry of the loading the slope of the rim at each spoke is zero Consider the radius to be sufficiently large so that the segment AB can be considered as a straight beam fixed at both ends and loaded with a uniform centrifugal force per unit length Show that this force is w btγω2rg Problem 12142 Determine the moment reactions at the supports A and B Use the method of integration EI is constant Problem 12143 Using the method of superposition determine the magnitude of M0 in terms of the distributed load w and dimension a so that the deflection at the center of the beam is zero EI is constant Problem 131 Determine the critical buckling load for the column The material can be assumed rigid Solution Equilibrium The disturbing force F can be deternined by summing moments about point A ΣΜA0 P L θ F L 2 0 F 2P θ Spring Formula The restoring spring force Fs can be deternined using the spring formula Fs kx Fs k L 2 θ k L θ 2 Critical Buckling Load For the mechanism to be on the verge of buckling the disturbing force F must be equal to the spring force Fs Thus 2Pcr θ k L θ 2 Pcr k L 4 Ans Problem 132 The column consists of a rigid member that is pinned at its bottom and attached to a spring at its top If the spring is unstretched when the column is in the vertical position determine the critical load that can be placed on the column Solution Equilibrium ΣΜA0 P L sin θ k L sin θ L cos θ 0 P k L cos θ Since θ is small cos θ 1 Pcr k L Ans Problem 133 An A36 steel column has a length of 4 m and is pinned at both ends If the cross sectional area has the dimensions shown determine the critical load Given L 4m E 200 GPa σY 250MPa b 25mm a 10mm Solution Section Property bo 2 b a A 2 bo a a2 Ix a bo 3 12 2 b a3 12 Iy Ix Buckling Load Applying Eulers formula I min Ix Iy For pinned ends K 10 Pcr π2 E I K L 2 Pcr 227 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 2066 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 134 Solve Prob 133 if the column is fixed at its bottom and pinned at its top Given L 4m E 200 GPa σY 250MPa b 25mm a 10mm Solution Section Property bo 2 b a A 2 bo a a2 Ix a bo 3 12 2 b a3 12 Iy Ix Buckling Load Applying Eulers formula I min Ix Iy For fixedpinned ends K 07 Pcr π2 E I K L 2 Pcr 464 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 4215 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 135 A square bar is made from PVC plastic that has a modulus of elasticity of E 9 GPa and a yield strain of εY 0001 mmmm Determine its smallest crosssectional dimensions a so it does not fail from elastic buckling It is pinned at its ends and has a length of 1250 mm Given L 1250m E 9 GPa εY 0001 mm mm Solution Stressstrain Relationship σY E εY σY 900 MPa Section Property A a2 I a4 12 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 σY A π2 E I K L 2 I σY A K L 2 π2 E For pinned ends K 10 Thus a4 12 σY a2 10L 2 π2 E a 12σY L2 π2 E a 4359 mm Ans Problem 136 The rod is made from an A36 steel rod Determine the smallest diameter of the rod to the nearest mm that will support the load of P 25 kN without buckling The ends are rollersupported Given L 500mm E 200 GPa Pcr 25kN Solution Section Property I π do 4 64 A π do 2 4 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 I Pcr K L 2 π2 E For roller ends K 10 Thus π do 4 64 Pcr 10 L 2 π2 E do 4 64Pcr L2 π3 E do 1594 mm Use do 16mm Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 4Pcr π do 2 σcr 12434 MPa σY 250 MPa OK Problem 137 The rod is made from a 25mmdiameter steel rod Determine the critical buckling load if the ends are roller supported Est 200 GPa σY 350 MPa Given L 500mm E 200 GPa do 25mm σY 350MPa Solution Section Property I π do 4 64 A π do 2 4 Buckling Load Applying Eulers formula For roller ends K 10 Pcr π2 E I K L 2 Pcr 1514 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 30843 MPa σY 350 MPa OK Problem 138 An A36 steel column has a length of 5 m and is fixed at both ends If the crosssectional area has the dimensions shown determine the critical load Given E 200 GPa L 5m bo 100mm bi 80mm σY 250MPa ho 50mm hi 30mm Solution Section Property A bo ho bi hi Ix bo ho 3 12 bi hi 3 12 Iy ho bo 3 12 hi bi 3 12 Buckling Load Applying Eulers formula I min Ix Iy For fixed ends K 05 Pcr π2 E I K L 2 Pcr 2721 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 10467 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 139 An A36 steel column has a length of 45 m and is pinned at both ends If the crosssectional area has the dimensions shown determine the critical load Given E 200 GPa L 45m bf 200mm df 12mm σY 250MPa tw 12mm dw 150mm Solution D dw 2df Section Property A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 Iy 2df bf 3 12 dw tw 3 12 Buckling Load Applying Eulers formula I min Ix Iy For pinned ends K 10 Pcr π2 E I K L 2 Pcr 15617 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 23663 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 1310 The W250 X 67 is made of A36 steel and is used as a column that has a length of 455 m If its ends are assumed pin supported and it is subjected to an axial load of 500 kN determine the factor of safety with respect to buckling Given E 200 GPa L 455m σY 250MPa P 500kN Solution Use W 250x67 Ix 104 106 mm4 A 8560mm2 Iy 222 106 mm4 Buckling Load Applying Eulers formula I min Ix Iy For pinned ends K 10 Pcr π2 E I K L 2 Pcr 21167 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 24728 MPa σY 250 MPa OK Factor of safety Fsafety Pcr P Fsafety 423 Ans Problem 1311 The W250 X 67 is made of A36 steel and is used as a column that has a length of 455 m If the ends of the column are fixed supported can the column support the critical load without yielding Given E 200 GPa L 455m σY 250MPa P 500kN Solution Use W 250x67 Ix 104 106 mm4 A 8560mm2 Iy 222 106 mm4 Buckling Load Applying Eulers formula I min Ix Iy For fixed ends K 05 Pcr π2 E I K L 2 Pcr 84668 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 98911 MPa σY 250 MPa No Ans The column will yield before the axial force achieves the critical load Pcr and the Eulers formula is not valid Problem 1312 Determine the maximum force P that can be applied to the handle so that the A36 steel control rod AB does not buckle The rod has a diameter of 30 mm It is pin connected at its ends Given L 09m E 200 GPa a 06m do 30mm σY 250MPa b 09m Solution Section Property I π do 4 64 A π do 2 4 Buckling Load For rod AB For pinned ends K 10 Pcr π2 E I K L 2 Pcr 9689 kN FAB Pcr Equations of Equilibrium ΣΜh0 FAB a P b 0 P a b FAB P 646 kN Ans Critical Stress check σcr Pcr A σcr 13708 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 1313 The two steel channels are to be laced together to form a 9mlong bridge column assumed to be pin connected at its ends Each channel has a crosssectional area of A 1950 mm2 and moments of inertia Ix 2160106 mm4 Iy 015106 mm4 The centroid C of its area is located in the figure Determine the proper distance d between the centroids of the channels so that buckling occurs about the xx and y y axes due to the same load What is the value of this critical load Neglect the effect of the lacing Est 200 GPa σY 350 MPa Given E 200 GPa L 9m σY 350MPa A 1950mm2 Ix 2160 106 mm4 c1 65mm Iy 015 106 mm4 c2 30mm Solution Section Property Ix 2Ix A 2A Iy 2Iy 2A 05d 2 In order for the column to buckle about xx and yy axes at the same time Iy must be equal to Ix Iy Ix 2Iy 2A 05d 2 2Ix d 2 Ix Iy A d 20976 mm Ans 2c2 60 mm OK Iy 2Iy 2A 05d 2 Iy 4320 106 mm4 Buckling Load Applying Eulers formula I min Ix Iy For pinned ends K 10 Pcr π2 E I K L 2 Pcr 10528 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 26994 MPa σY 350 MPa OK Therefore Eulers formula is valid Problem 1314 The W200 X 100 is used as a structural A36 steel column that can be assumed fixed at its base and pinned at its top Determine the largest axial force P that can be applied without causing it to buckle Given E 200 GPa L 75m σY 250MPa Solution Use W 200x100 Ix 133 106 mm4 A 12700mm2 Iy 366 106 mm4 Buckling Load Applying Eulers formula I min Ix Iy For fixedpinned ends K 07 Pcr π2 E I K L 2 Pcr 26212 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 20639 MPa σY 250 MPa OK Problem 1315 Solve Prob 1314 if the column is assumed fixed at its bottom and free at its top Given E 200 GPa L 75m σY 250MPa Solution Use W 200x100 Ix 133 106 mm4 A 12700mm2 Iy 366 106 mm4 Buckling Load Applying Eulers formula I min Ix Iy For fixedfree ends K 20 Pcr π2 E I K L 2 Pcr 3211 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 2528 MPa σY 250 MPa OK Problem 1316 A steel column has a length of 9 m and is fixed at both ends If the crosssectional area has the dimensions shown determine the critical load Est 200 GPa σY 250 MPa Given E 200 GPa L 9m bf 200mm df 10mm σY 250MPa tw 10mm dw 150mm Solution D dw 2df Section Property A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 Iy 2df bf 3 12 dw tw 3 12 Buckling Load Applying Eulers formula I min Ix Iy For fixed ends K 05 Pcr π2 E I K L 2 Pcr 13009 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 23653 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 1317 Solve Prob 1316 if the column is pinned at its top and bottom Given E 200 GPa L 9m bf 200mm df 10mm σY 250MPa tw 10mm dw 150mm Solution D dw 2df Section Property A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 Iy 2df bf 3 12 dw tw 3 12 Buckling Load Applying Eulers formula I min Ix Iy For pinned ends K 10 Pcr π2 E I K L 2 Pcr 3252 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 5913 MPa σY 250 MPa OK Therefore Eulers formula is valid Problem 1318 The 36m A36 steel pipe column has an outer diameter of 75 mm and a thickness of 6 mm Determine the critical load if the ends are assumed to be pin connected Given E 200 GPa L 36m σY 250MPa do 75mm t 6mm Solution Section Property di do 2t A π 4 do 2 di 2 I π 64 do 4 di 4 Buckling Load Applying Eulers formula For pinned ends K 10 Pcr π2 E I K L 2 Pcr 1188 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 9133 MPa σY 250 MPa OK Problem 1319 The 36m A36 steel column has an outer diameter of 75 mm and a thickness of 6 mm Determine the critical load if the bottom is fixed and the top is pinned Given E 200 GPa L 36m σY 250MPa do 75mm t 6mm Solution Section Property di do 2t A π 4 do 2 di 2 I π 64 do 4 di 4 Buckling Load Applying Eulers formula For fixedpinned ends K 07 Pcr π2 E I K L 2 Pcr 2424 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 18638 MPa σY 250 MPa OK Problem 1320 The 3m wooden rectangular column has the dimensions shown Determine the critical load if the ends are assumed to be pin connected Ew 12 GPa σY 35 MPa Given E 12 GPa L 3m σY 35MPa b 50mm h 100mm Solution Section Property A b h Ix b h3 12 Iy h b3 12 Buckling Load I min Ix Iy Applying Eulers formula For pinned ends K 10 Pcr π2 E I K L 2 Pcr 137 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 274 MPa σY 35 MPa OK Problem 1321 The 3m column has the dimensions shown Determine the critical load if the bottom is fixed and the top is pinned Ew 12 GPa σY 35 MPa Given E 12 GPa L 3m σY 35MPa b 50mm h 100mm Solution Section Property A b h Ix b h3 12 Iy h b3 12 Buckling Load I min Ix Iy Applying Eulers formula For fixedpinned ends K 07 Pcr π2 E I K L 2 Pcr 28kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 560 MPa σY 35 MPa OK Problem 1322 The members of the truss are assumed to be pin connected If member BD is an A36 steel rod of radius 50 mm determine the maximum load P that can be supported by the truss without causing the member to buckle Given a 4m E 200 GPa h 3m σY 250MPa ro 50mm Solution Section Property I π ro 4 4 A π ro 2 Buckling Load LBD a Applying Eulers formula For pinned ends K 10 Pcr π2 E I K LBD 2 Pcr 6056 kN FBD Pcr Equations of Equilibrium ΣΜC0 R a FBD h 0 R h a FBD Support Reactions By symmetry ABR ΣFy0 2R 2P 0 P R P 45419 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 7711 MPa σY 250 MPa OK Problem 1323 Solve Prob 1322 in the case of member AB which has a radius of 50 mm Given a 4m E 200 GPa h 3m σY 250MPa ro 50mm Solution Section Property I π ro 4 4 A π ro 2 Buckling Load LAB a2 h2 Applying Eulers formula For pinned ends K 10 Pcr π2 E I K LAB 2 Pcr 3876 kN FAB Pcr At Joint A Equations of Equilibrium ΣFy0 R FAB h LAB 0 R FAB h LAB Support Reactions By symmetry ABR ΣFy0 2R 2P 0 P R P 23255 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 4935 MPa σY 250 MPa OK Problem 1324 The truss is made from A36 steel bars each of which has a circular cross section with a diameter of 40 mm Determine the maximum force P that can be applied without causing any of the members to buckle The members are pin connected at their ends Given a 12m E 200 GPa do 40mm h 09m σY 250MPa Solution Section Property I π do 4 64 A π 4 do 2 Buckling Load Member AB and BC are in compression LAB 2a LBC a2 h2 LAC LBC Applying Eulers formula Assume failure of rod AB Assume failure of rod BC For pinned ends K 10 For pinned ends K 10 Pcr π2 E I K LAB 2 Pcr π2 E I K LBC 2 Pcr 431 kN Pcr 1102 kN FAB Pcr FBC Pcr At Joint A Equations of Equilibrium ΣFy0 FAC h LAC P 0 FAC P LAC h ΣFx0 FAC a LAC FAB 0 FAB FAC a LAC FAB P a h Pcr P a h P Pcr h a P 3230 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3427 MPa σY 250 MPa OK Check Rod BC Support Reactions ΣΜD0 P 2a Bh h 0 Bh P 2a h At Joint B Equations of Equilibrium ΣFx0 FAB FBC a LAC Bh 0 P a h Pcr a LAC P 2a h 0 P Pcr h LAC P 6615 kN Not Control h 09m Given a 12m E 200 GPa Equations of Equilibrium P 50kN σY 250MPa Solution At Joint A OK ΣFy0 FAC h LAC P 0 FAC P LAC h ΣFx0 FAC a LAC FAB 0 FAB FAC a LAC FAB P a h Buckling Load LAB 2a Applying Eulers formula For pinned ends K 10 Pcr π2 E I K LAB 2 I 10 2a 2 π2 E Pcr Pcr FAB I 4a2 π2 E P a h Section Property I π do 4 64 A π 4 do 2 π do 4 64 4a2 π2 E P a h do 4 256a2 π3 E P a h do 4462 mm Use do 45mm Ans Check Critical Stress Eulers formula is only valid if σcr σY A π 4 do 2 σcr Pcr A σcr 4192 MPa σY 250 MPa Problem 1325 The truss is made from A36 steel bars each of which has a circular cross section If the applied load P 50 kN determine the diameter of member AB to the nearest multiples of 5mm that will prevent this member from buckling The members are pin supported at their ends Problem 1326 An L2 tool steel link in a forging machine is pin connected to the forks at its ends as shown Determine the maximum load P it can carry without buckling Use a factor of safety with respect to buckling of FS 175 Note from the figure on the left that the ends are pinned for buckling whereas from the figure on the right the ends are fixed Given E 200 GPa L 06m Fsafety 175 σY 703MPa b 36mm t 12mm Solution Section Property Ix t b3 12 Ix 4665600 mm4 Iy b t3 12 Iy 518400 mm4 A b t A 43200 mm2 Critical Buckling Load With respect to the xx axis for pinned ends K 10 Applying Eulers formula Pcr π2 E Ix K L 2 Pcr 2558 kN With respect to the yy axis for fixed ends K 05 Applying Eulers formula Pcr π2 E Iy K L 2 Pcr 1137 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 26319 MPa σY 703 MPa OK Factor of safety Fsafety Pcr P P Pcr Fsafety P 6497 kN Ans σY 250MPa Given h 36m θAB 45deg E 200 GPa P 30kN θBC 30deg 1 Solution At Joint B Equations of Equilibrium Given ΣFy0 FAB cos θAB FBC cos θBC P 0 OK ΣFx0 FAB sin θAB FBC sin θBC 0 2 Solving Eqs 1 and 2 Guess FAB 1kN FBC 1kN FAB FBC Find FAB FBC FAB FBC 1553 2196 kN Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 I 10 L 2 π2 E Pcr π do 4 64 L2 π2 E Pcr do 4 64L2 π3 E Pcr For Member AB LAB h cos θAB Pcr FAB L LAB dAB 4 64L2 π3 E Pcr dAB 4515 mm Use dAB 50mm Ans Check Critical Stress Eulers formula is only valid if σcr σY A π 4 dAB 2 σcr Pcr A σcr 791 MPa σY 250 MPa OK For Member BC LBC h cos θBC Pcr FBC L LBC dBC 4 64L2 π3 E Pcr dBC 4449 mm Use dBC 45mm Ans Check Critical Stress Eulers formula is only valid if σcr σY A π 4 dBC 2 σcr Pcr A σcr 1381 MPa σY 250 MPa Problem 1327 The linkage is made using two A36 steel rods each having a circular cross section Determine the diameter of each rod to the nearest multiples of 5mm that will support a load of P 30 kN Assume that the rods are pin connected at their ends Use a factor of safety with respect to buckling of 18 Problem 1328 The linkage is made using two A36 steel rods each having a circular cross section If each rod has a diameter of 20 mm determine the largest load it can support without causing any rod to buckle Assume that the rods are pinconnected at their ends Given h 36m θAB 45deg E 200 GPa do 20mm θBC 30deg σY 250MPa Solution Set P 1 At Joint B Equations of Equilibrium Given ΣFy0 FAB cos θAB FBC cos θBC P 0 1 ΣFx0 FAB sin θAB FBC sin θBC 0 2 Solving Eqs 1 and 2 Guess FAB 1 FBC 1 FAB FBC Find FAB FBC FAB FBC 05176 07321 FAB FBC FAB P FBC P kN Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 Pcr π2 E I L2 For Member AB LAB h cos θAB Pcr FAB L LAB FAB P π2 E I LAB 2 P π2 E I FAB LAB 2 P 116 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY Pcr FAB P σcr Pcr A σcr 190 MPa σY 250 MPa OK Check Member BC LBC h cos θBC FBC FBC P FBC 0846 kN Pcr π2 E I LBC 2 Pcr 08972 kN FBC 0846 kN OK Problem 1329 The A36 steel pipe has an outer diameter of 50 mm and a thickness of 12 mm If it is held in place by a guywire determine the largest vertical force P that can be applied without causing the pipe to buckle Assume that the ends of the pipe are pin connected Given L 42m θBC 30deg E 200 GPa do 50mm t 12mm σY 250MPa Solution Set P 1 At Joint B Equations of Equilibrium Given ΣFy0 FBC sin θBC P 0 1 ΣFx0 FBC cos θBC FAB 0 2 Solving Eqs 1 and 2 Guess FAB 1 FBC 1 FAB FBC Find FAB FBC FAB FBC 17321 20000 FAB FBC FAB P FBC P kN Section Property di do 2t A π 4 do 2 di 2 I π do 4 di 4 64 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 Pcr π2 E I L2 Pcr FAB FAB P π2 E I L2 P π2 E I FAB L2 P 1837 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY Pcr FAB P σcr Pcr A σcr 2221 MPa σY 250 MPa OK Problem 1330 The A36 steel pipe has an outer diameter of 55 mm If it is held in place by a guywire determine its required inner diameter to the nearest multiples of 5mm so that it can support a maximum vertical load of P 20 kN without causing the pipe to buckle Assume the ends of the pipe are pin connected Given L 42m θBC 30deg E 200 GPa do 55mm P 20kN σY 250MPa Solution At Joint B Equations of Equilibrium Given ΣFy0 FBC sin θBC P 0 1 ΣFx0 FBC cos θBC FAB 0 2 Solving Eqs 1 and 2 Guess FAB 1kN FBC 1kN FAB FBC Find FAB FBC FAB FBC 3464 4000 kN Section Property di do 2t A π 4 do 2 di 2 I π do 4 di 4 64 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 Pcr FAB FAB π2 E I L2 I FAB L2 π2 E π do 4 di 4 64 FAB L2 π2 E di 4 do 4 64 FAB L2 π3 E di 4107 mm Use di 40mm Ans Check Critical Stress Eulers formula is only valid if σcr σY A π 4 do 2 di 2 I π do 4 di 4 64 Pcr π2 E I K L 2 σcr Pcr A σcr 3235 MPa σY 250 MPa OK Problem 1331 The linkage is made using two A36 steel rods each having a circular cross section Determine the diameter of each rod to the nearest multiples of 5mm that will support the 45kN load Assume that the rods are pin connected at their ends Use a factor of safety with respect to buckling of FS 18 Given a 48m h 36m E 200 GPa b 27m P 45kN σY 250MPa Fsafety 18 Solution LAB a2 h2 LBC b2 h2 At Joint B Equations of Equilibrium Given ΣFy0 FAB h LAB FBC h LBC P 0 1 ΣFx0 FAB a LAB FBC b LBC 0 2 Solving Eqs 1 and 2 Guess FAB 1kN FBC 1kN FAB FBC Find FAB FBC FAB FBC 27 36 kN Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 I 10 L 2 π2 E Pcr π do 4 64 L2 π2 E Pcr do 4 64L2 π3 E Pcr For Member AB Pcr Fsafety FAB L LAB dAB 4 64L2 π3 E Pcr dAB 3666 mm Use dAB 40mm Ans Check Critical Stress Eulers formula is only valid if σcr σY A π 4 dAB 2 σcr Pcr A σcr 387 MPa σY 250 MPa OK For Member BC Pcr Fsafety FBC L LBC dBC 4 64L2 π3 E Pcr dBC 3411 mm Use dBC 40mm Ans Check Critical Stress Eulers formula is only valid if σcr σY A π 4 dBC 2 σcr Pcr A σcr 516 MPa σY 250 MPa OK Problem 1332 The linkage is made using two A36 steel rods each having a circular cross section If each rod has a diameter of 20 mm determine the largest load it can support without causing any rod to buckle Assume that the rods are pin connected at their ends Given a 48m h 36m E 200 GPa b 27m do 20mm σY 250MPa Fsafety 18 Solution LAB a2 h2 LBC b2 h2 Set P 1 At Joint B Equations of Equilibrium Given ΣFy0 FAB h LAB FBC h LBC P 0 1 ΣFx0 FAB a LAB FBC b LBC 0 2 Solving Eqs 1 and 2 Guess FAB 1 FBC 1 FAB FBC Find FAB FBC FAB FBC 06 08 FAB FBC FAB P FBC P kN Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 Pcr π2 E I L2 For Member AB Pcr FAB L LAB FAB P π2 E I LAB 2 P π2 E I FAB LAB 2 P 0718 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY Pcr FAB P σcr Pcr A σcr 137 MPa σY 250 MPa OK Check Member BC FBC FBC P FBC 0574 kN Pcr π2 E I LBC 2 Pcr 07656 kN FBC 0574 kN OK ah ah LAC Problem 1333 The steel bar AB of the frame is assumed to be pinconnected at its ends for yy axis buckling If P 18 kN determine the factor of safety with respect to buckling about the yy axis due to the applied loading Est 200 GPa σY 360 MPa Given h 100mm av 4m E 200 GPa b 50mm ah 3m σY 360MPa P 18kN LAB 6m LAC av 2 ah 2 Solution av av LAC At Joint B Equations of Equilibrium ΣFy0 FAC av FAB 0 1 ΣFx0 FAC ah P 0 2 Solving Eqs 1 and 2 FAC P ah FAB P av ah FAC 30kN FAB 24kN Section Property Iy h b3 12 A h b Buckling Load Applying Eulers formula For pinned ends K 10 Pcr π2 E Iy K LAB 2 Pcr 5712 kN Fsafety Pcr FAB Fsafety 238 Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 1142 MPa σY 360 MPa OK Problem 1334 Determine the maximum load P the frame can support without buckling member AB Assume that AB is made of steel and is pinned at its ends for yy axis buckling and fixed at its ends for xx axis buckling Est 200 GPa σY 360 MPa Given h 100mm av 4m E 200 GPa b 50mm ah 3m σY 360MPa P 18kN LAB 6m LAC av 2 ah 2 Solution av av LAC ah ah LAC At Joint B Equations of Equilibrium ΣFy0 FAC av FAB 0 1 ΣFx0 FAC ah P 0 2 Solving Eqs 1 and 2 FAC P ah FAB P av ah 3 FAC 30kN FAB 24kN Section Property A h b Ix b h3 12 Iy h b3 12 Buckling Load Applying Eulers formula About xx axix About yy axix For fixed ends For pinned ends Pcr π2 E Iy Ky LAB 2 Pcr 91385 kN Pcr 5712 kN Pcr min Pcr Pcr Pcr 5712 kN Substituting FABPcr into Eq3 P Pcr ah av P 4284 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 1142 MPa σY 360 MPa OK Pcr π2 E Ix Kx LAB 2 Kx 05 Ky 10 Problem 1335 Determine the maximum force P that can be applied to the handle so that the A36 steel control rod BC does not buckle The rod has a diameter of 25 mm Given a 350mm θ 45deg E 200 GPa b 250mm do 25mm σY 250MPa c 800mm P kN Solution Equations of Equilibrium For the handle ΣΜA0 P a FBC b cos θ 0 FBC P a b cos θ FBC 19799 P Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 Pcr FBC L c FBC π2 E I K c 2 P a b cos θ π2 E I K c 2 P π2 E I b cos θ a K c 2 P 29870 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY Pcr P a b cos θ σcr Pcr A σcr 12048 MPa σY 250 MPa OK Problem 1336 Determine the maximum allowable load P that can be applied to member BC without causing member AB to buckle Assume that AB is made of steel and is pinned at its ends for xx axis buckling and fixed at its ends for yy axis buckling Use a factor of safety with respect to buckling of FS 3 Est 200 GPa σY 360 MPa Given h 30mm L 2m E 200 GPa b 20mm Fsafety 3 σY 360MPa Solution P kN Support Reactions For member BC ΣFy0 FBA FC P 0 By symmetry FBA FC Hence FBA 05P 1 FC 05P Section Property A h b Ix b h3 12 Iy h b3 12 Buckling Load Applying Eulers formula About xx axix About yy axix For pinned ends Kx 10 For fixed ends Ky 05 Pcr π2 E Ix Kx L 2 Pcr π2 E Iy Ky L 2 Pcr 2221 kN Pcr 3948 kN Pcr min Pcr Pcr Pcr 22207 kN FBA Pcr Fsafety FBA 7402 kN From Eq1 P 2FBA P 1480 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3701 MPa σY 360 MPa OK Problem 1337 Determine the maximum allowable load P that can be applied to member BC without causing member AB to buckle Assume that AB is made of steel and is pinned at its ends for xx axis buckling and fixed at its ends for yy axis buckling Use a factor of safety with respect to buckling of FS 3 Est 200 GPa σY 360 MPa Given h 30mm L 2m E 200 GPa b 20mm Fsafety 3 σY 360MPa Solution P kN Support Reactions For member BC ΣFy0 FBA FC P 0 By symmetry FBA FC Hence FBA 05P 1 FC 05P Section Property A h b Ix b h3 12 Iy h b3 12 Buckling Load Applying Eulers formula About xx axix About yy axix For pinned ends Kx 10 For fixed ends Ky 05 Pcr π2 E Ix Kx L 2 Pcr π2 E Iy Ky L 2 Pcr 2221 kN Pcr 3948 kN Pcr min Pcr Pcr Pcr 22207 kN FBA Pcr Fsafety FBA 7402 kN From Eq1 P 2FBA P 1480 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3701 MPa σY 360 MPa OK Problem 1338 Determine if the frame can support a load of P 20 kN if the factor of safety with respect to buckling of member AB is FS 3 Assume that AB is made of steel and is pinned at its ends for xx axis buckling and fixed at its ends for yy axis buckling Est 200 GPa σY 360 MPa Given h 30mm L 2m E 200 GPa b 20mm Fsafety 3 σY 360MPa P 20kN Solution Support Reactions For member BC ΣFy0 FBA FC P 0 By symmetry FBA FC Hence FBA 05P 1 FC 05P Section Property A h b Ix b h3 12 Iy h b3 12 Buckling Load Applying Eulers formula About xx axix About yy axix For pinned ends Kx 10 For fixed ends Ky 05 Pcr π2 E Ix Kx L 2 Pcr π2 E Iy Ky L 2 Pcr 2221 kN Pcr 3948 kN Pcr min Pcr Pcr Pcr 22207 kN Factor of Safety FS Pcr FBA FS 2221 Fsafety Hence the frame cannot support the load with the required factor of safety Ans Problem 1339 The members of the truss are assumed to be pin connected If member GF is an A36 steel rod having a diameter of 50 mm determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle Given a 4m E 200 GPa h 3m σY 250MPa do 50mm Solution Section Property I π do 4 64 A π 4 do 2 Buckling Load LGF a Applying Eulers formula For pinned ends K 10 Pcr π2 E I K LGF 2 Pcr 378 kN FGF Pcr Equations of Equilibrium Use method of sections ΣΜB0 R a FGF h 0 R h a FGF Support Reactions By symmetry ABR ΣFy0 2R 2P 0 P R P 2839 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 1928 MPa σY 250 MPa OK Problem 1340 The members of the truss are assumed to be pin connected If member AG is an A36 steel rod having a diameter of 50 mm determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle Given a 4m E 200 GPa h 3m σY 250MPa do 50mm Solution Section Property I π do 4 64 A π 4 do 2 Buckling Load LAG a2 h2 Applying Eulers formula For pinned ends K 10 Pcr π2 E I K LAG 2 Pcr 242 kN FAG Pcr Equations of Equilibrium Use method of joints at Joint A ΣFy0 R FAG h LAG 0 R h LAG FAG Support Reactions By symmetry ABR ΣFy0 2R 2P 0 P R P 1453 kN Ans Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 1234 MPa σY 250 MPa OK w kN m Problem 1341 Determine the maximum distributed loading that can be applied to the wideflange beam so that the brace CD does not buckle The brace is an A36 steel rod having a diameter of 50 mm Given b 2m do 50mm E 200 GPa Lc 4m σY 250MPa Solution Lb 2 b Equations of Equilibrium For the beam ΣΜA0 w Lb 05Lb FCD b 0 FCD w Lb 2 2 b FCD 400 w m Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 Pcr FCD L Lc FCD π2 E I K Lc 2 w Lb 2 2 b π2 E I K Lc 2 w 2π2 E I b Lb 2 K Lc 2 w 9462 kN m Ans Check Critical Stress Eulers formula is only valid if σcr σY Pcr π2 E I K Lc 2 σcr Pcr A σcr 1928 MPa σY 250 MPa OK Problem 1342 The 50mm diameter C86100 bronze rod is fixed supported at A and has a gap of 2 mm from the wall at B Determine the increase in temperature T that will cause the rod to buckle Assume that the contact at B acts as a pin Unit used C deg Given L 1m do 50mm go 2mm α 17 10 6 1 C E 103GPa σY 345MPa Solution Section Property I π do 4 64 A π 4 do 2 Compatibity Condition This requires go δT δF go α T L F L A E F A E α T go L Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For fixedpinned ends K 07 Pcr F A E α T go L π2 E I K L 2 T 1 α π2 I A K L 2 go L T 30278 C Ans Check Critical Stress Eulers formula is only valid if σcr σY Pcr π2 E I K L 2 σcr Pcr A σcr 32416 MPa σY 345 MPa OK Problem 1343 Consider an ideal column as in Fig 1312c having both ends fixed Show that the critical load on the column is given by Pcr 4π2EIL2 Hint Due to the vertical deflection of the top of the column a constant moment M will be developed at the supports Show that d2vdx2 PEI v MEI The solution is of the form MP P EI x C P EI x C v cos sin 2 1 Problem 1344 Consider an ideal column as in Fig 1312d having one end fixed and the other pinnedShow that the critical load on the column is given by Pcr 2019 EIL2 Hint Due to the vertical deflection at the top of the column a constant moment M will be developed at the fixed support and horizontal reactive forces R will be developed at both supports Show that d2vdx2 PEI v REI Lx The solution is of the form After application of the boundary conditions show that Solve by trial and error for the smallest root x L R P P EI x C P EI x C v cos sin 2 1 P EI L P EI L tan However u 0 at x L Then 0 EIP sinPEI L cosPEI L tanPEI PEI Q E D By trial and error and choosing the smallest root we have PEI L 449341 Then Pext 2019EI L² Q E D Problem 1345 The column is supported at B by a support that does not permit rotation but allows vertical deflection Determine the critical load Pcr EI is constant Problem 1346 The ideal column is subjected to the force F at its midpoint and the axial load P Determine the maximum moment in the column at midspan EI is constant Hint Establish the differential equation for deflection Eq 131 The general solution is v A sin kx B cos kx c2xk2 where c2 F2EI k2 PEI Elastic Curve u F2P secPL2EI sinPL2EI Fx F2P secPL2EI sinPL2EI x x However u umax at x L2 Then umax F2PEIP secPL2EI sinPL2EI L2 F2P secPL2EI sinPL2EI L2 L2 Ans Maximum Moment The maximum moment occurs at x L2 From Eq 1 Mmax FL2 L2 P umax Ans Problem 1347 The ideal column has a weight w forcelength and rests in the horizontal position when it is subjected to the axial load P Determine the maximum moment in the column at midspan EI is constant Hint Establish the differential equation for deflection Eq 131 with the origin at the mid span The general solution is v A sin kx B cos kx w2P x2 wL2P x wEIP2 where k2 PEI Elastic Curve u w2PPL2EI EIP cosPL2EI x²2 L2 EIP However u umax at x L2 Then umax w2P tanPL2EI sinPL2EI L2 EIP cosPL2EI L²8 EIP Ans Maximum Moment The maximum moment occurs at x L2 From Eq 1 Mmax wL²8 P umax Ans Problem 1348 Determine the load P required to cause the A36 steel W200 X 22 column to fail either by buckling or by yielding The column is fixed at its base and free at its top Given E 200 GPa L 24m σY 250MPa ey 25mm Solution Use W 200x22 Ix 200 106 mm4 A 2860mm2 Iy 142 106 mm4 d 206mm rx 836mm Buckling about yy axis For fixedfree ends K 20 Pcr π2 E Iy K L 2 Pcr 12166 kN P Pcr P 12166 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 4254 MPa σY 250 MPa OK Check yielding about xx axis Applying the secant formula c 05d σmax P A 1 ey c rx 2 sec K L 2rx P E A σmax 5969 MPa σY 250 MPa OK Problem 1349 The wood column is assumed fixed at its base and pinned at its top Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield Ew 12 GPa σY 56 MPa Given E 12 GPa L 36m b 38mm σY 56MPa ey 44mm h 88mm Solution Section Property Ix b h3 12 Iy h b3 12 A b h rx Ix A Buckling about yy axis For fixedpinned ends K 07 Pcr π2 E Iy K L 2 Pcr 75 kN P Pcr P 750 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 224 MPa σY 56 MPa OK Check yielding about xx axis Applying the secant formula c 05h σmax P A 1 ey c rx 2 sec K L 2rx P E A σmax 1089 MPa σY 56 MPa OK Problem 1350 The wood column is fixed at its base and fixed at its top Determine the maximum eccentric load P that can be applied at its top without causing the column to buckle or yield Ew 12 GPa σY 56 MPa Given E 12 GPa L 36m b 38mm σY 56MPa ey 44mm h 88mm Solution Section Property Ix b h3 12 Iy h b3 12 A b h rx Ix A Buckling about yy axis For fixedfixed ends K 05 Pcr π2 E Iy K L 2 Pcr 1471 kN P Pcr P 1471 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 440 MPa σY 56 MPa OK Check yielding about xx axis Applying the secant formula c 05h σmax P A 1 ey c rx 2 sec K L 2rx P E A σmax 2135 MPa σY 56 MPa OK Problem 1351 The wood column has a square cross section with dimensions 100 mm by 100 mm It is fixed at its base and free at its top Determine the load P that can be applied to the edge of the column without causing the column to fail either by buckling or by yielding Ew 12 GPa σY 55 MPa Given E 12 GPa L 2m σY 55MPa e 120mm do 100mm Solution Section Property I do 4 12 A do 2 r I A Buckling For fixedfree ends K 20 Pcr π2 E I K L 2 Pcr 6169 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 617 MPa σY 55 MPa OK Check yielding Applying the secant formula c 05do Given σY Pmax A 1 e c r2 sec K L 2r Pmax E A 1 Solving Eq 1 by trial and error Guess Pmax 1kN Pmax Find Pmax Pmax 3137 kN Controls Hence Pallow min Pcr Pmax Pallow 3137 kN Ans Problem 1352 The W200 X 71 structural A36 steel column is fixed at its bottom and free at its top If it is subjected to the eccentric load of 375 kN determine the factor of safety with respect to either the initiation of buckling or yielding Given E 200 GPa L 36m P 375kN σY 250MPa ey 200mm Solution Use W 200x71 Ix 766 106 mm4 A 9100mm2 Iy 254 106 mm4 d 216mm rx 917mm Buckling about yy axis For fixedfree ends K 20 Pcr π2 E Iy K L 2 Pcr 96716 kN Pallow Pcr Pallow 96716 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 10628 MPa σY 250 MPa OK Check yielding about xx axis Applying the secant formula c 05d Given σY Pmax A 1 ey c rx 2 sec K L 2rx Pmax E A 1 Solving Eq 1 by trial and error Guess Pmax 1kN Pmax Find Pmax Pmax 53159 kN Controls Pallow 96716 kN Factor of Safety Fsafety Pmax P Fsafety 142 Ans Problem 1353 The W200 X 71 structural A36 steel column is fixed at its bottom and pinned at its top If it is subjected to the eccentric load of 375 kN determine if the column fails by yielding The column is braced so that it does not buckle about the yy axis Given E 200 GPa L 36m P 375kN σY 250MPa ey 200mm Solution Use W 200x71 Ix 766 106 mm4 A 9100mm2 Iy 254 106 mm4 d 216mm rx 917mm For fixedpined ends K 07 Yielding about xx axis Applying the secant formula c 05d Given σmax P A 1 ey c rx 2 sec K L 2rx P E A σmax 14916 MPa σY 250 MPa OK Hence the column does not fail by yielding Ans Problem 1354 The brass rod is fixed at one end and free at the other end If the eccentric load P 200 kN is applied determine the greatest allowable length L of the rod so that it does not buckle or yield Ebr 101 GPa σY 69 MPa Given E 101 GPa do 100mm P 200kN σY 69MPa e 10mm L m Solution Section Property I πdo 4 64 A πdo 2 4 r I A Buckling For fixedfree ends K 20 Pcr P Pcr π2 E I K L 2 Lcr π2 E I P K2 Lcr 2473 m Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 2546 MPa σY 69 MPa OK Check yielding Applying the secant formula c 05do Given σY P A 1 e c r2 sec K Lmax 2r P E A 1 Solving Eq 1 by trial and error Guess Lmax 1m Lmax Find Lmax Lmax 1706 m Controls Hence Lallow min Lcr Lmax Lallow 1706 m Ans Problem 1355 The brass rod is fixed at one end and free at the other end If the length of the rod is L 2 m determine the greatest allowable load P that can be applied so that the rod does not buckle or yield Also determine the largest sidesway deflection of the rod due to the loading Ebr 101 GPa σY 69 MPa Given E 101 GPa do 100mm L 2m σY 69MPa e 10mm Solution Section Property I πdo 4 64 A πdo 2 4 r I A Buckling For fixedfree ends K 20 Pcr π2 E I K L 2 Pcr 30582 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3894 MPa σY 69 MPa OK Check yielding Applying the secant formula c 05do Given σY Pmax A 1 e c r2 sec K L 2r Pmax E A 1 Solving Eq 1 by trial and error Guess Pmax 1kN Pmax Find Pmax Pmax 17370 kN Controls Hence Pallow min Pcr Pmax Pallow 17370 kN Ans Maximum Deflection max e sec K L 2 Pallow E I 1 max 1650 mm Ans Problem 1356 A W310 X 39 structural A36 steel column is fixed connected at its ends and has a length L 69 m Determine the maximum eccentric load P that can be applied so the column does not buckle or yeild Compare this value with an axial critical load P applied through the centroid of the column Given E 200 GPa L 69m σY 250MPa ey 150mm Solution Use W 310x39 Ix 848 106 mm4 A 4930mm2 Iy 723 106 mm4 d 310mm rx 131mm Buckling about yy axis For fixedends K 05 Pcr π2 E Iy K L 2 Pcr 119903 kN P Pcr P 119903 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 24321 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 50974 kN Controls P 1199 kN Hence Pmax min P P Pmax 50974 kN Ans Problem 1357 A W360 X 45 structural A36 steel column is fixed connected at its ends and has a length L 6 m Determine the maximum eccentric load P that can be applied so the column does not buckle or yeild Compare this value with an axial critical load P applied through the centroid of the column Given E 200 GPa L 69m σY 250MPa ey 150mm Solution Use W 360x45 Ix 121 106 mm4 A 5710mm2 Iy 816 106 mm4 d 352mm rx 146mm Buckling about yy axis For fixed ends K 05 Pcr π2 E Iy K L 2 Pcr 135326 kN P Pcr P 135326 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 23700 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 62410 kN Controls P 1353 kN Hence Pmax min P P Pmax 62410 kN Ans Problem 1358 Solve Prob 1357 if the column is fixed at its bottom and free at its top Given E 200 GPa L 69m σY 250MPa ey 150mm Solution Use W 360x45 Ix 121 106 mm4 A 5710mm2 Iy 816 106 mm4 d 352mm rx 146mm Buckling about yy axis For fixedfree ends K 20 Pcr π2 E Iy K L 2 Pcr 8458 kN P Pcr P 8458 kN Controls Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 1481 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 45781 kN P 85 kN Hence Pmax min P P Pmax 8458 kN Ans Problem 1359 The wood column is fixed at its base and can be assumed pin connected at its top Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield Ew 12 GPa σY 56 MPa Given E 12 GPa L 3m b 100mm σY 56MPa ey 125mm h 250mm Solution Section Property Ix b h3 12 Iy h b3 12 A b h rx Ix A Buckling about yy axis For fixedpinned ends K 07 Pcr π2 E Iy K L 2 Pcr 5595 kN P Pcr P 55950 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 2238 MPa σY 56 MPa OK Check yielding about xx axis Applying the secant formula c 05h Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 32008 kN Controls P 5595 kN Hence Pmax min P P Pmax 32008 kN Ans Problem 1360 The wood column is fixed at its base and can be assumed fixed connected at its top Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield Ew 12 GPa σY 56 MPa Given E 12 GPa L 3m b 100mm σY 56MPa ey 125mm h 250mm Solution Section Property Ix b h3 12 Iy h b3 12 A b h rx Ix A Buckling about yy axis For fixedfixed ends K 05 Pcr π2 E Iy K L 2 Pcr 109662 kN P Pcr P 109662 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 4386 MPa σY 56 MPa OK Check yielding about xx axis Applying the secant formula c 05h Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 33413 kN Controls P 10966 kN Hence Pmax min P P Pmax 33413 kN Ans Problem 1361 The aluminum column has the cross section shown If it is fixed at the bottom and free at the top determine the maximum force P that can be applied at A without causing it to buckle or yield Use a factor of safety of 3 with respect to buckling and yielding Eal 70 GPa σY 95 MPa Given bf 160mm dw 150mm L 5m tf 10mm tw 10mm e 5mm E 70 GPa σY 95MPa Fsafety 3 Solution Section Property D dw tf xc 05tf bf tf 05dw tf dw tw bf tf dw tw xc 4371 mm Iy 1 12 bf tf 3 bf tf 05tf xc 2 1 12 tw dw 3 dw tw 05dw tf xc 2 Iy 778067204 mm4 Ix 1 12 tf bf 3 1 12 dw tw 3 Ix 342583333 mm4 A D bf dw bf tw A 3100mm2 ry Iy A Buckling about xx axis For fixedfree ends K 20 Pcr π2 E Ix K L 2 Pcr 2367 kN P Pcr P 2367 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 763 MPa σY 95 MPa OK Yielding about yy axis Applying the secant formula c xc ey xc e Given σY P A 1 ey c ry 2 sec K L 2ry P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 4561 kN Hence Pmax min P P Pmax 2367 kN Pallow Pmax Fsafety Pallow 789 kN Ans Problem 1362 A W250 X 22 structural A36 steel member is used as a fixedconnected column Determine the maximum eccentric load P that can be applied so the column does not buckle or yield Compare this value with an axial critical load P applied through the centroid of the column Given E 200 GPa L 75m σY 250MPa ey 200mm Solution Use W 250x22 Ix 288 106 mm4 A 2850mm2 Iy 122 106 mm4 d 254mm rx 101mm Buckling about yy axis For fixed ends K 05 Pcr π2 E Iy K L 2 Pcr 17125 kN P Pcr P 17125 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 6009 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 19549 kN P 171 kN Hence Pmax min P P Pmax 17125 kN Ans Problem 1363 Solve Prob 1362 if the column is pinconnected at its ends Given E 200 GPa L 75m σY 250MPa ey 200mm Solution Use W 250x22 Ix 288 106 mm4 A 2850mm2 Iy 122 106 mm4 d 254mm rx 101mm Buckling about yy axis For pinned ends K 10 Pcr π2 E Iy K L 2 Pcr 4281 kN P Pcr P 4281 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 1502 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 17294 kN P 43 kN Hence Pmax min P P Pmax 4281 kN Ans Problem 1364 Solve Prob 1362 if the column is fixed at its bottom and pinned at its top Given E 200 GPa L 75m σY 250MPa ey 200mm Solution Use W 250x22 Ix 288 106 mm4 A 2850mm2 Iy 122 106 mm4 d 254mm rx 101mm Buckling about yy axis For fixedpinned ends K 07 Pcr π2 E Iy K L 2 Pcr 8737 kN P Pcr P 8737 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3066 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 18774 kN P 874 kN Hence Pmax min P P Pmax 8737 kN Ans Problem 1365 Determine the load P required to cause the steel W310 X 74structural A36 steel column to fail either by buckling or by yielding The column is fixed at its bottom and the cables at its top act as a pin to hold it Given E 200 GPa L 75m σY 250MPa ey 50mm Solution Use W 310x74 Ix 165 106 mm4 A 9480mm2 Iy 234 106 mm4 d 310mm rx 132mm Buckling about yy axis For fixedpinned ends K 07 Pcr π2 E Iy K L 2 Pcr 167582 kN P Pcr P 167582 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 17677 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 155114 kN Controls P 1676 kN Hence Pmax min P P Pmax 155114 kN Ans Problem 1366 Solve Prob 1365 if the column is an A36 steel W310 X 24 section Given E 200 GPa L 75m σY 250MPa ey 50mm Solution Use W 310x24 Ix 428 106 mm4 A 3040mm2 Iy 116 106 mm4 d 305mm rx 119mm Buckling about yy axis For fixedpinned ends K 07 Pcr π2 E Iy K L 2 Pcr 8307 kN P Pcr P 8307 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 2733 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 45916 kN P 83 kN Hence Pmax min P P Pmax 8307 kN Ans Problem 1367 The W360 X 79 structural A36 steel column is fixed at its base and free at its top If P 375 kN determine the sidesway deflection at its top and the maximum stress in the column Given E 200 GPa L 55m P 375kN σY 250MPa ey 250mm Solution Use W 360x79 Ix 227 106 mm4 A 10100mm2 Iy 242 106 mm4 d 354mm rx 150mm Buckling about yy axis For fixedfree ends K 20 Pcr π2 E Iy K L 2 Pcr 39478 kN P 375 kN OK Hence the column does not buckle about the yy axis Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3909 MPa σY 250 MPa OK Yielding about xx axis Applying the secant formula c 05d σmax P A 1 ey c rx 2 sec K L 2rx P E A σmax 12032 MPa Ans Since σmax σY 250 MPa the column does not yield Maximum Displacement vmax ey sec K L 2 P E Ix 1 vmax 3485 mm Ans Problem 1368 The W360 X 79 steel column is fixed at its base and free at its top Determine the maximum eccentric load P that it can support without causing it to either buckle or yield Est 200 GPa σY 350 MPa Given E 200 GPa L 55m σY 350MPa ey 250mm Solution Use W 360x79 Ix 227 106 mm4 A 10100mm2 Iy 242 106 mm4 d 354mm rx 150mm Buckling about yy axis For fixedfree ends K 20 Pcr π2 E Iy K L 2 Pcr 39478 kN P Pcr P 39478 kN Controls Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 3909 MPa σY 350 MPa OK Yielding about xx axis Applying the secant formula c 05d Given σY P A 1 ey c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess P 1kN P Find P P 93305 kN P 395 kN Hence Pmax min P P Pmax 39478 kN Ans Problem 1369 The aluminum rod is fixed at its base and free at its top If the eccentric load P 200 kN is applied determine the greatest allowable length L of the rod so that it does not buckle or yield Eal 72 GPa σY 410 MPa Given E 72 GPa do 200mm P 200kN σY 410MPa e 5mm L m Solution Section Property I πdo 4 64 A πdo 2 4 r I A Buckling For fixedfree ends K 20 Pcr P Pcr π2 E I K L 2 Lcr π2 E I P K2 Lcr 8352 m Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 637 MPa σY 410 MPa OK Check yielding Applying the secant formula c 05do Given σY P A 1 e c r2 sec K Lmax 2r P E A 1 Solving Eq 1 by trial and error Guess Lmax 1m Lmax Find Lmax Lmax 8336 m Controls Hence Lallow min Lcr Lmax Lallow 8336 m Ans Problem 1370 The aluminum rod is fixed at its base and free at its top If the length of the rod is L 2 m determine the greatest allowable load P that can be applied so that the rod does not buckle or yield Also determine the largest sidesway deflection of the rod due to the loading Eal 72 GPa σY 410 MPa Given E 72 GPa do 200mm L 2m σY 410MPa e 5mm Solution Section Property I πdo 4 64 A πdo 2 4 r I A Buckling For fixedfree ends K 20 Pcr π2 E I K L 2 Pcr 348821 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 11103 MPa σY 410 MPa OK Check yielding Applying the secant formula c 05do Given σY Pmax A 1 e c r2 sec K L 2r Pmax E A 1 Solving Eq 1 by trial and error Guess Pmax 1kN Pmax Find Pmax Pmax 32005 kN Controls Hence Pallow min Pcr Pmax Pallow 32005 kN Ans Maximum Deflection max e sec K L 2 Pallow E I 1 max 7061 mm Ans Problem 1371 The steel column supports the two eccentric loadings If it is assumed to be pinned at its top fixed at the bottom and fully braced against buckling about the yy axis determine the maximum deflection of the column and the maximum stress in the column Est 200 GPa σY 360 MPa Given E 200 GPa bf 100mm df 10mm e1 120mm P1 130kN σY 360MPa tw 10mm dw 100mm e2 80mm P2 50kN L 6m Solution Section Property D dw 2df A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 rx Ix A Buckling For fixedpinned ends K 07 Pcr π2 E Ix K L 2 Pcr 77211 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 25737 MPa σY 360 MPa OK Yielding about xx axis Eccentricity of the two applied loasds is e e1 P1 e2 P2 P1 P2 e 6444 mm Applying the secant formula c 05D σmax P1 P2 A 1 e c rx 2 sec K L 2rx P1 P2 E A σmax 1990 MPa Ans Since σmax σY 360MPa the column does not yield Maximum Deflection max e sec K L 2 P1 P2 E Ix 1 max 2433 mm Ans Problem 1372 The steel column supports the two eccentric loadings If it is assumed to be fixed at its top and bottom and braced against buckling about the yy axis determine the maximum deflection of the column and the maximum stress in the column Est 200 GPa σY 360 MPa Given E 200 GPa bf 100mm df 10mm e1 120mm P1 130kN σY 360MPa tw 10mm dw 100mm e2 80mm P2 50kN L 6m Solution Section Property D dw 2df A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 rx Ix A Buckling For fixed ends K 05 Pcr π2 E Ix K L 2 Pcr 151334 kN Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 50445 MPa σY 360 MPa OK Yielding about xx axis Eccentricity of the two applied loasds is e e1 P1 e2 P2 P1 P2 e 6444 mm Applying the secant formula c 05D σmax P1 P2 A 1 e c rx 2 sec K L 2rx P1 P2 E A σmax 1777 MPa Ans Since σmax σY 360MPa the column does not yield Maximum Deflection max e sec K L 2 P1 P2 E Ix 1 max 1077 mm Ans Problem 1373 A column of intermediate length buckles when the compressive strength is 280 MPa If the slenderness ratio is 60 determine the tangent modulus Given σcr 280MPa λ 60 Solution σcr π2 Et λ2 Et σcr λ2 π2 Et 10213 GPa Ans Problem 1374 Construct the buckling curve PA versus Lr for a column that has a bilinear stressstrain curve in compression as shown Given σ1 91MPa ε1 0002 σ2 126MPa ε2 0005 Solution Let λ L r σ P A From the graph E1 σ1 ε1 E1 4550 GPa E2 σ2 σ1 ε2 ε1 E2 1167 GPa σ λ relationship σcr π2 Et λ2 For Et E1 σ σcr σ π2 E1 λ2 When σ σ1 λ1 π E1 σ1 λ1 7025 For Et E2 σ σcr σ π2 E2 λ2 When σ σ1 λ1 π E2 σ1 λ1 3557 Buckling Curve x1 10 101 10 λ1 x2 λ1 101 λ1 λ1 x3 λ1 101 λ1 130 σ1 x1 π2 E2 x1 2 1 MPa σ2 x2 π2 E2 λ1 2 1 MPa σ3 x3 π2 E1 x3 2 1 MPa 50 100 0 200 400 Slenderless Ratio Stress MPa σ1 x1 σ2 x2 σ3 x3 x1 x2 x3 Problem 1375 Construct the buckling curve PA versus Lr for a column that has a bilinear stressstrain curve in compression as shown Given σ1 175MPa ε1 0001 σ2 350MPa ε2 0005 Solution Let λ L r σ P A From the graph E1 σ1 ε1 E1 17500 GPa E2 σ2 σ1 ε2 ε1 E2 4375 GPa σ λ relationship σcr π2 Et λ2 For Et E1 σ σcr σ π2 E1 λ2 When σ σ1 λ1 π E1 σ1 λ1 9935 For Et E2 σ σcr σ π2 E2 λ2 When σ σ1 λ1 π E2 σ1 λ1 4967 Buckling Curve x1 10 101 10 λ1 x2 λ1 101 λ1 λ1 x3 λ1 101 λ1 200 σ1 x1 π2 E2 x1 2 1 MPa σ2 x2 π2 E2 λ1 2 1 MPa σ3 x3 π2 E1 x3 2 1 MPa 50 100 150 200 0 200 400 Slenderless Ratio Stress MPa σ1 x1 σ2 x2 σ3 x3 x1 x2 x3 Problem 1376 Construct the buckling curve PA versus Lr for a column that has a bilinear stressstrain curve in compression as shown The column is pinned at its ends Given σ1 140MPa ε1 0001 σ2 260MPa ε2 0004 Solution Let λ L r σ P A From the graph E1 σ1 ε1 E1 14000 GPa E2 σ2 σ1 ε2 ε1 E2 4000 GPa σ λ relationship σcr π2 Et λ2 For Et E1 σ σcr σ π2 E1 λ2 When σ σ1 λ1 π E1 σ1 λ1 9935 For Et E2 σ σcr σ π2 E2 λ2 When σ σ1 λ1 π E2 σ1 λ1 5310 Buckling Curve x1 10 101 10 λ1 x2 λ1 101 λ1 λ1 x3 λ1 101 λ1 160 σ1 x1 π2 E2 x1 2 1 MPa σ2 x2 π2 E2 λ1 2 1 MPa σ3 x3 π2 E1 x3 2 1 MPa 50 100 150 0 200 400 Slenderless Ratio Stress MPa σ1 x1 σ2 x2 σ3 x3 x1 x2 x3 Problem 1377 Determine the largest length of a structural A36 steel rod if it is fixed supported and subjected to an axial load of 100 kN The rod has a diameter of 50 mm Use the AISC equations Given E 200 GPa P 100kN σY 250MPa do 50mm Solution Section Property A π do 2 4 I π do 4 64 r I A AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Assume a long column using Eq 1321 σ P A σ 5093 MPa σallow 12π2 E 23λ2 λ 12π2 E 23σ λ 14220 λc Assumption is correct Slenderness Ratio For fixed ends K 05 λ K L r L λ r K L 3555 m Ans Problem 1378 Determine the largest length of a W250 X 18 structural A36 steel section if it is fixed supported and is subjected to an axial load of 140 kN Use the AISC equations Given E 200 GPa P 140kN σY 250MPa Solution Section Property W 250x18 A 2280mm2 ry 201mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Assume a long column using Eq 1321 σ P A σ 6140 MPa σallow 12π2 E 23λ2 λ 12π2 E 23σ λ 12951 λc Assumption is correct Slenderness Ratio For fixed ends K 05 λ K L r L λ ry K L 5206 m Ans Problem 1379 Determine the largest length of a W310 X 67 structural A36 steel column if it is pin supported and subjected to an axial load of 1000 kN Use the AISC equations Given E 200 GPa P 1000kN σY 250MPa Solution Section Property W 310x67 A 8530mm2 ry 493mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Assume a long column using Eq 1321 σ P A σ 11723 MPa σallow 12π2 E 23λ2 λ 12π2 E 23σ λ 9373 λc Assumption is not correct Thus it is an intermediate column Using Eq 1323 Given σ 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 1 Solving Eq 1 by trial and error Guess λ 1 λ Find λ λ 6535 For pinned ends K 10 λ K L r L λ ry K L 3222 m Ans Problem 1380 Determine the largest length of a W200 X 46 structural A36 steel section if it is pin supported and is subjected to an axial load of 380 kN Use the AISC equations Given E 200 GPa P 380kN σY 250MPa Solution Section Property W 200x46 A 5890mm2 ry 510mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Assume a long column using Eq 1321 σ P A σ 6452 MPa σallow 12π2 E 23λ2 λ 12π2 E 23σ λ 12634 λc Assumption is correct Slenderness Ratio For pinned ends K 10 λ K L r L λ ry K L 6444 m Ans Problem 1381 Using the AISC equations check if a W150 X 14 structural A36 steel column that is 3 m long can support an axial load of 200 kN The ends are fixed Given E 200 GPa P 200kN σY 250MPa L 3m Solution Section Property W 150x14 A 1730mm2 ry 230mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 For fixed ends K 05 λ K L ry λ 65217 λc Thus it is an intermediate column Using Eq 1323 σallow 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 σallow 1173 MPa Pallow A σallow Pallow 2030 kN P 200 kN OK Thus the column is adequate Ans Problem 1382 Using the AISC equations select from Appendix B the lightestweight structural steel column that is 42 m long and supports an axial load of 200 kN The ends are pinned Take σY 350 MPa Given E 200 GPa P 200kN σY 350MPa L 42m Solution Section Property Try W 150x24 A 2860mm2 ry 368mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 10621 For pinned ends K 10 λ K L ry λ 114130 λc Thus it is an long column Using Eq 1321 σallow 12π2 E 23λ2 σallow 791 MPa Pallow A σallow Pallow 2261 kN P 200 kN OK Use W 150x24 Ans Problem 1383 Using the AISC equations select from Appendix B the lightestweight structural A36 steel column that is 36 m long and supports an axial load of 200 kN The ends are fixed Take σY 350 MPa Given E 200 GPa P 200kN σY 350MPa L 36m Solution Section Property Try W 150x14 A 1730mm2 ry 230mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 10621 For fixed ends K 05 λ K L ry λ 78261 λc Thus it is an intermediate column Using Eq 1323 σallow 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 σallow 1347 MPa Pallow A σallow Pallow 2330 kN P 200 kN OK Use W 150x14 Ans Problem 1384 Using the AISC equations select from Appendix B the lightestweight structural A36 steel column that is 42 m long and supports an axial load of 200 kN The ends are fixed Given E 200 GPa P 200kN σY 250MPa L 42m Solution Section Property Try W 150x18 A 2290mm2 ry 235mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 For fixed ends K 05 λ K L ry λ 89362 λc Thus it is an intermediate column Using Eq 1323 σallow 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 σallow 989 MPa Pallow A σallow Pallow 2265 kN P 200 kN OK Use W 150x18 Ans Problem 1385 Using the AISC equations select from Appendix B the lightestweight structural A36 steel column that is 9 m long and supports an axial load of 1000 kN The ends are fixed Given E 200 GPa P 1000kN σY 250MPa L 9m Solution Section Property Try W 250x80 A 10200mm2 ry 650mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 For fixed ends K 05 λ K L ry λ 69231 λc Thus it is an intermediate column Using Eq 1323 σallow 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 σallow 1145 MPa Pallow A σallow Pallow 11677 kN P 1000 kN OK Use W 250x80 Ans Problem 1386 Determine the largest length of a W200 X 46 structural A36 steel section if it is pin supported and is subjected to an axial load of 90 kN Use the AISC equations Given E 200 GPa P 90kN σY 250MPa Solution Section Property W 200x46 A 5890mm2 ry 510mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Assume a long column using Eq 1321 σ P A σ 1528 MPa σallow 12π2 E 23λ2 λ 12π2 E 23σ λ 25961 λc Assumption is correct Slenderness Ratio λmax 200 For pinned ends K 10 λmax K L r L λmax ry K L 1020 m Ans Problem 1387 Determine the largest length of a W150 X 22 structural A36 steel section if it is pin supported and subjected to an axial load of 350 kN Use the AISC equations Given E 200 GPa P 350kN σY 250MPa Solution Section Property W 150x22 A 2860mm2 ry 368mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Assume a long column using Eq 1321 σ P A σ 12238 MPa σallow 12π2 E 23λ2 λ 12π2 E 23σ λ 9174 λc Assumption is not correct Thus it is an intermediate column Using Eq 1323 Given σ 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 1 Solving Eq 1 by trial and error Guess λ 1 λ Find λ λ 5778 For pinned ends K 10 λ K L r L λ ry K L 2126 m Ans Problem 1388 The bar is made of a 2014T6 aluminum alloy Determine its thickness b if its width is 5b Assume that it is pin connected at its ends Given E 731 GPa P 3kN t b σY 414MPa L 24m h 5 b Solution Section Property A t h A b 5 b A 5 b2 Iy h t3 12 Iy 5 b b3 12 Iy 5 b4 12 ry Iy A ry 5 b4 12 5 b2 ry b 12 Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λ K L ry λ 12 K L b σ P A σ P 5 b2 Assume a long column using Eq 1326 σallow 378125MPa λ2 λ2 378125MPa σ 12 K L b 2 378125MPa 5 b2 P b 4 12 K2 L2 P 378125MPa 5 b 1820 mm Hence ry b 12 λ K L ry λ 45685 55 Assumption is correct Thus b 1820 mm Ans Problem 1389 The bar is made of a 2014T6 aluminum alloy Determine its thickness b if its width is 5b Assume that it isfixed connected at its ends Given E 731 GPa P 3kN t b σY 414MPa L 24m h 5 b Solution Section Property A t h A b 5 b A 5 b2 Iy h t3 12 Iy 5 b b3 12 Iy 5 b4 12 ry Iy A ry 5 b4 12 5 b2 ry b 12 Slenderness Ratio For fixed ends K 05 Aluminum Association Formula λ K L ry λ 12 K L b σ P A σ P 5 b2 Assume a long column using Eq 1326 σallow 378125MPa λ2 λ2 378125MPa σ 12 K L b 2 378125MPa 5 b2 P b 4 12 K2 L2 P 378125MPa 5 b 1287 mm Hence ry b 12 λ K L ry λ 32304 55 Assumption is correct Thus b 1287 mm Ans Problem 1390 The 50mmdiameter rod is used to support an axial load of 40 kN Determine its greatest allowable length L if it is made of 2014T6 aluminum Assume that the ends are pin connected Given E 731 GPa do 50mm σY 414MPa P 40kN Solution Section Property A π do 2 4 A 196350 mm2 I π do 4 64 I 30679616 mm4 ry I A ry 1250 mm Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λ K L ry σ P A Assume a long column using Eq 1326 σallow 378125MPa λ2 λ 378125MPa σ λ 13624 55 Assumption is correct Hence λ K L ry L λ ry K L 1703 m Ans Problem 1391 The 50mmdiameter rod is used to support an axial load of 40 kN Determine its greatest allowable length L if it is made of 2014T6 aluminum Assume that the ends are fixed connected Given E 731 GPa do 50mm σY 414MPa P 40kN Solution Section Property A π do 2 4 A 196350 mm2 I π do 4 64 I 30679616 mm4 ry I A ry 1250 mm Slenderness Ratio For fixed ends K 05 Aluminum Association Formula λ K L ry σ P A Assume a long column using Eq 1326 σallow 378125MPa λ2 λ 378125MPa σ λ 13624 55 Assumption is correct Hence λ K L ry L λ ry K L 3406 m Ans Problem 1392 The tube is 6 mm thick is made of a 2014T6 aluminum alloy and is fixed at its bottom and pinned at its top Determine the largest axial load that it can support Given E 731 GPa ao 150mm t 6mm σY 414MPa L 3m Solution Section Property ai ao 2t A ao 2 ai 2 A 3456mm2 I ao 4 ai 4 12 I 11964672mm4 r I A r 5884 mm Slenderness Ratio For fixedpinned ends K 07 Aluminum Association Formula λ K L r λ 3569 Since 12 λ 55 it is an intermediate column Applying Eq 1325 σallow 2145 1628λ MPa σallow 1564 MPa Hence the allowable load is Pallow σallow A Pallow 5405 kN Ans Problem 1393 The tube is 6 mm thick is made of a 2014T6 aluminum alloy and is fixed connected at its ends Determine the largest axial load that it can support Given E 731 GPa ao 150mm t 6mm σY 414MPa L 3m Solution Section Property ai ao 2t A ao 2 ai 2 A 345600 mm2 I ao 4 ai 4 12 I 1196467200 mm4 r I A r 5884 mm Slenderness Ratio For fixed ends K 05 Aluminum Association Formula λ K L r λ 2549 Since 12 λ 55 it is an intermediate column Applying Eq 1325 σallow 2145 1628λ MPa σallow 17300 MPa Hence the allowable load is Pallow σallow A Pallow 5979 kN Ans Problem 1394 The tube is 6 mm thick is made of a 2014T6 aluminum alloy and is pin connected at its ends Determine the largest axial load that it can support Given E 731 GPa ao 150mm t 6mm σY 414MPa L 3m Solution Section Property ai ao 2t A ao 2 ai 2 A 345600 mm2 I ao 4 ai 4 12 I 1196467200 mm4 r I A r 5884 mm Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λ K L r λ 5099 Since 12 λ 55 it is an intermediate column Applying Eq 1325 σallow 2145 1628λ MPa σallow 13149 MPa Hence the allowable load is Pallow σallow A Pallow 4544 kN Ans Problem 1395 The bar is made of aluminum alloy 2014T6 Determine its thickness b if its width is 15b Assume that it is pin connected at its ends Given E 731 GPa P 4kN t b σY 414MPa L 15m h 15 b Solution Section Property A t h A b 15 b A 15 b2 Iy h t3 12 Iy 15 b b3 12 Iy b4 8 ry Iy A ry b4 8 15 b2 ry b 12 Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λ K L ry λ 12 K L b σ P A σ P 15 b2 Assume a long column using Eq 1326 σallow 378125MPa λ2 λ2 378125MPa σ 12 K L b 2 378125MPa 15 b2 P b 4 12 K2 L2 P 378125MPa 15 b 2089 mm Hence ry b 12 λ K L ry λ 24875 55 Assumption is correct Thus b 2089 mm Ans Problem 1396 Using the AISC equations check if a column having the cross section shown can support an axial force of 1500 kN The column has a length of 4 m is made from A36 steel and its ends are pinned Given bf 300mm df 20mm P 1500kN tw 10mm dw 310mm L 4m E 200 GPa σY 250MPa Solution Section Property D dw 2df A bf D dw bf tw A 15100mm2 Iy 2df bf 3 12 dw tw 3 12 Iy 9002583333 mm4 ry Iy A ry 7721 mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 For pinnd ends K 10 λ K L ry λ 51804 λc Thus it is an intermediate column Using Eq 1323 σallow 1 λ2 2λc 2 σY 5 3 3λ 8λc λ3 8λc 3 σallow 1262 MPa Pallow A σallow Pallow 19058 kN P 1500 kN OK Thus the column is adequate Ans Problem 1397 A 15mlong rod is used in a machine to transmit an axial compressive load of 15 kN Determine its diameter if it is pin connected at its ends and is made of a 2014T6 aluminum alloy Given E 731 GPa L 15m σY 414MPa P 15kN Solution Section Property A π do 2 4 I π do 4 64 ry I A ry π do 4 64 4 π do 2 ry do 4 Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λ K L ry λ 4 K L do σ P A σ 4P π do 2 Assume a long column using Eq 1326 σallow 378125MPa λ2 λ2 378125MPa σ 4 K L do 2 378125MPa π do 2 4P do 4 64 K2 L2 P 378125MPa π do 3672 mm Hence ry do 4 λ K L ry λ 16339 55 Assumption is correct Thus do 3672 mm Ans Problem 1398 Solve Prob 1397 if the rod is fixed connected at ends Given E 731 GPa L 15m σY 414MPa P 15kN Solution Section Property A π do 2 4 I π do 4 64 ry I A ry π do 4 64 4 π do 2 ry do 4 Slenderness Ratio For fixed ends K 05 Aluminum Association Formula λ K L ry λ 4 K L do σ P A σ 4P π do 2 Assume a long column using Eq 1326 σallow 378125MPa λ2 λ2 378125MPa σ 4 K L do 2 378125MPa π do 2 4P do 4 64 K2 L2 P 378125MPa π do 2597 mm Hence ry do 4 λ K L ry λ 11554 55 Assumption is correct Thus do 2597 mm Ans Problem 1399 The timber column has a square cross section and is assumed to be pin connected at its top and bottom If it supports an axial load of 250 kN determine its side dimensions a to the nearest multiples of 10 mm Use the NFPA formulas Given L 42m P 250kN Solution Section Property A a2 Slenderness Ratio For pinned ends K 10 NFPA Timber Column Formula λ K L a σ P A σ P a2 Assume a long column using Eq 1329 σallow 3718MPa λ2 λ2 3718MPa σ K L a 2 3718MPa a2 P a 4 K2 L2 P 3718MPa a 18558 mm Thus λ K L a λ 2263 26 Assumption NG Assume an intermediate column using Eq 1328 σallow 825 1 1 3 λ 260 2 MPa λ2 2028 1 σ 825MPa K L a 2 2028 1 P 825MPa a2 a K2 L2 2028 P 825MPa a 19749 mm Thus λ K L a λ 2127 Since 11 λ 26 assumption of an intermediate column is correct Use a 200mm Ans Problem 13100 Solve Prob 1399 if the column is assumed to be fixed connected at its top and bottom Given L 42m P 250kN Solution Section Property A a2 Slenderness Ratio For fixed ends K 05 NFPA Timber Column Formula λ K L a σ P A σ P a2 Assume a long column using Eq 1329 σallow 3718MPa λ2 λ2 3718MPa σ K L a 2 3718MPa a2 P a 4 K2 L2 P 3718MPa a 13123 mm Thus λ K L a λ 1600 26 Assumption NG Assume an intermediate column using Eq 1328 σallow 825 1 1 3 λ 260 2 MPa λ2 2028 1 σ 825MPa K L a 2 2028 1 P 825MPa a2 a K2 L2 2028 P 825MPa a 18022 mm Thus λ K L a λ 1165 Since 11 λ 26 assumption of an intermediate column is correct Use a 190mm Ans Problem 13101 The wood column is used to support an axial load of P 150 kN If it is fixed at the bottom and free at the top determine the minimum width of the column based on the NFPA formulas Given L 24m b 150mm P 150kN Solution Section Property A b d Slenderness Ratio For fixedfree ends K 20 NFPA Timber Column Formula σ P A σ P b d σallow 3718MPa λ2 Assume a long column using Eq 1329 λx K L d if d b λy K L b if d b λx 2 3718MPa σ λy 2 3718MPa σ K L d 2 3718MPa b d P K L b 2 3718MPa b d P d K2 L2 P 3718MPa b3 d 3 K2 L2 P 3718MPa b d 18368 mm d 27542 mm b 150mm Assumption NG b 150mm Assumption OK check λ K L b λ 3200 Since 26 λ 50 assumption of a long column is correct Hence dmin max d d dmin 27542 mm Ans Problem 13102 The timber column has a length of 6 m and is pin connected at its ends Use the NFPA formulas to determine the largest axial force P that it can support Given L 6m a 200mm Solution Section Property A a2 A 40000mm2 Slenderness Ratio For pinned ends K 10 NFPA Timber Column Formula λ K L a λ 3000 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λ2 σallow 413 MPa Pmax σallow A Pmax 1652 kN Ans Problem 13103 The timber column has a length of 6 m and is fixed connected at its ends Use the NFPA formulas to determine the largest axial force P that it can support Given L 6m a 200mm Solution Section Property A a2 A 40000mm2 Slenderness Ratio For fixed ends K 05 NFPA Timber Column Formula λ K L a λ 1500 Since 11 λ 26 it is an intermediate column Applying Eq 1328 σallow 825 1 1 3 λ 260 2 MPa σallow 733 MPa Pmax σallow A Pmax 2934 kN Ans Problem 13104 The column is made of wood It is fixed at its bottom and free at its top Use the NFPA formulas to determine its greatest allowable length if it supports an axial load of P 10 kN Given b 100mm d 50mm P 10kN Solution Section Property A b d Slenderness Ratio For fixedfree ends K 20 NFPA Timber Column Formula λ K L d σ P A σ P b d Assume a long column using Eq 1329 σallow 3718MPa λ2 λ2 3718MPa σ K L d 2 3718MPa b d P L 3718MPa b d3 P K2 L 1078 m Ans Check λ K L d λ 4312 Since 26 λ 50 assumption of a long column is correct Problem 13105 The column is made of wood It is fixed at its bottom and free at its top Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L 12 m Given b 100mm d 50mm L 12m Solution Section Property A b d A 5000mm2 Slenderness Ratio For fixedfree ends K 20 NFPA Timber Column Formula λ K L d λ 4800 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λ2 σallow 161 MPa Pmax σallow A Pmax 807 kN Ans Problem 13106 The W360 X 79 structural A36 steel column supports an axial load of 400 kN in addition to an eccentric load P Determine the maximum allowable value of P based on the AISC equations of Sec 136 and Eq 1330 Assume the column is fixed at its base and at its top it is free to sway in the xz plane while it is pinned in the yz plane Given E 200 GPa L 36m Po 400kN σY 250MPa ey 250mm Solution Use W 360x79 Ix 227 106 mm4 A 10100mm2 ry 489mm Iy 242 106 mm4 d 354mm rx 150mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedpinned ends Kx 07 λy Ky L ry λy 14724 λx Kx L rx λx 1680 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 4750 MPa Maximum Stress Bending is about xx axis Mx P ey σmax σallow c 05d P Po P Applying Eq 1330 σmax P A M c I σallow Po P A P ey 05d Ix P σallow A Po 1 ey d 2rx 2 P 2690 kN Ans Problem 13107 The W310 X 67 structural A36 steel column supports an axial load of 400 kN in addition to an eccentric load of P 300 kN Determine if the column fails based on the AISC equations of Sec 136 and Eq 1330 Assume that the column is fixed at its base and at its top it is free to sway in the xz plane while it is pinned in the yz plane Given E 200 GPa L 36m Po 400kN σY 250MPa ey 250mm P 300kN Solution Use W 310x67 Ix 145 106 mm4 A 8530mm2 ry 493mm Iy 207 106 mm4 d 306mm rx 130mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedpinned ends Kx 07 λy Ky L ry λy 14604 λx Kx L rx λx 1938 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 4828 MPa Maximum Stress Bending is about xx axis Mx P ey c 05d P Po P Applying Eq 1330 σmax P A Mx c Ix σmax 16120 MPa σallow Since σmax σallow the column is not adequate Ans Problem 13108 The W200 X 22 structural A36 steel column is fixed at its top and bottom If it supports end moments of M 75 kNm determine the axial force P that can be applied Bending is about the xx axis Use the AISC equations of Sec 136 and Eq 1330 Given E 200 GPa L 48m Mx 75kN m σY 250MPa ey 250mm Solution Use W200x22 Ix 200 106 mm4 A 2860mm2 ry 223mm Iy 142 106 mm4 d 206mm rx 836mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixed ends Ky 05 For fixed ends Kx 05 λy Ky L ry λy 10762 λx Kx L rx λx 2871 Buckling about yy axis because λy λx Since λy λc it is an intermediate column Applying Eq 1323 σallow 1 λy 2 2λc 2 σY 5 3 3λy 8λc λy 3 8λc 3 σallow 8292 MPa Maximum Stress Bending is about xx axis σmax σallow c 05d Applying Eq 1330 σmax P A M c I σallow P A Mx 05d Ix P σallow A Mx 05d A Ix P 12667 kN Ans Problem 13109 The W200 X 22 structural A36 steel column is fixed at its top and bottom If it supports end moments of M 32 kNm determine the axial force P that can be applied Bending is about the xx axis Use the interaction formula with σballow 168 MPa Given E 200 GPa L 48m Mx 32kN m σY 250MPa ey 250mm σballow 168MPa Solution Use W200x22 Ix 200 106 mm4 A 2860mm2 ry 223mm Iy 142 106 mm4 d 206mm rx 836mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixed ends Ky 05 For fixed ends Kx 05 λy Ky L ry λy 10762 λx Kx L rx λx 2871 Buckling about yy axis because λy λx Since λy λc it is an intermediate column Applying Eq 1323 σaallow 1 λy 2 2λc 2 σY 5 3 3λy 8λc λy 3 8λc 3 σaallow 8292 MPa Bending Stress Bending is about xx axis c 05d σb Mx c Ix σb 16480 MPa Axial Stress σa P A Applying the Interaction Formula σa σaallow σb σballow 1 P A σaallow σb σballow 1 P A σaallow 1 σb σballow P 452 kN Ans Note σa P A ra σa σaallow ra 002 015 Therefore the method is allowed Problem 13110 The W310 X 33 structural A36 steel column is fixed at its bottom and free at its top Determine the greatest eccentric load P that can be applied using Eq 1330 and the AISC equations of Sec 136 Given E 200 GPa L 18m σY 250MPa ex 300mm Solution Use W 310x33 Ix 650 106 mm4 A 4180mm2 ry 214mm Iy 192 106 mm4 bf 1020mm rx 125mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedfree ends Kx 20 λy Ky L ry λy 16822 λx Kx L rx λx 2880 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 3639 MPa Maximum Stress Bending is about yy axis My P ex σmax σallow c 05bf Applying Eq 1330 σmax P A My c Iy σallow P A P ex 05bf Iy P σallow A 1 ex bf 2ry 2 P 442 kN Ans Problem 13111 The W250 X 22 structural A36 steel column is fixed at its bottom and free at its top Determine the greatest eccentric load P that can be applied using Eq 1330 and the AISC equations of Sec 136 Given E 200 GPa L 18m σY 250MPa ex 300mm Solution Use W 250x22 Ix 288 106 mm4 A 2850mm2 ry 207mm Iy 122 106 mm4 bf 1020mm rx 101mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedfree ends Kx 20 λy Ky L ry λy 17391 λx Kx L rx λx 3564 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 3405 MPa Maximum Stress Bending is about yy axis My P ex σmax σallow c 05bf Applying Eq 1330 σmax P A My c Iy σallow P A P ex 05bf Iy P σallow A 1 ex bf 2ry 2 P 264 kN Ans Problem 13112 The W250 X 22 structural A36 steel column is fixed at its bottom and free at its top If it is subjected to a load of P 10 kN determine if it is safe based on the AISC equations of Sec 136 and Eq 1330 Given E 200 GPa L 18m P 10kN σY 250MPa ex 300mm Solution Use W 250x22 Ix 288 106 mm4 A 2850mm2 ry 207mm Iy 122 106 mm4 bf 1020mm rx 101mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedfree ends Kx 20 λy Ky L ry λy 17391 λx Kx L rx λx 3564 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 3405 MPa Maximum Stress Bending is about yy axis My P ex My 300 kN m σmax σallow c 05bf c 51mm Applying Eq 1330 σmax P A My c Iy σmax 12892 MPa σallow Since σmax σallow the column is not adequate Ans Problem 13113 The W310 X 33 structural A36 steel column is fixed at its bottom and free at its top If it is subjected to a load of P 20 kN determine if it is safe based on the AISC equations of Sec 136 and Eq 1330 Given E 200 GPa L 18m P 20kN σY 250MPa ex 300mm Solution Use W 310x33 Ix 650 106 mm4 A 4180mm2 ry 214mm Iy 192 106 mm4 bf 1020mm rx 125mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedfree ends Kx 20 λy Ky L ry λy 16822 λx Kx L rx λx 2880 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 3639 MPa Maximum Stress Bending is about yy axis My P ex My 600 kN m σmax σallow c 05bf c 51mm Applying Eq 1330 σmax P A My c Iy σmax 16416 MPa σallow Since σmax σallow the column is not adequate Ans Problem 13114 A 6mlong column is made of aluminum alloy 2014T6 If it is pinned at its top and bottom and a compressive load P is applied at point A determine the maximum allowable magnitude of P using the equations of Sec 136 and Eq 1330 Given E 731 GPa bf 200mm df 10mm ey 105mm σY 414MPa dw 200mm tw 10mm L 6m Solution Section Property D dw 2df A bf D bf tw dw A 6000mm2 Ix bf D3 12 bf tw dw 3 12 Ix 5080 106 mm4 Iy dw tw 3 12 2df bf 3 12 Iy 1335 106 mm4 ry Iy A ry 4717 mm rx Ix A rx 9201 mm Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λy K L ry λy 12720 Since λ 55 it is a long column Applying Eq 1326 σallow 378125MPa λy 2 σallow 2337 MPa Maximum Stress Bending is about xx axis Mx P ey σmax σallow c 05D Applying Eq 1330 σmax P A Mx c Ix σallow P A P ey 05D Ix P σallow A 1 ey D 2rx 2 P 5931 kN Ans Problem 13115 A 6mlong column is made of aluminum alloy 2014T6 If it is pinned at its top and bottom and a compressive load P is applied at point A determine the maximum allowable magnitude of P using the equations of Sec 136 and the interaction formula with σballow 140 MPa Given E 731 GPa bf 200mm df 10mm L 6m ey 105mm σY 414MPa dw 200mm tw 10mm σballow 140MPa Solution Section Property D dw 2df A bf D bf tw dw A 6000mm2 Ix bf D3 12 bf tw dw 3 12 Ix 5080 106 mm4 Iy dw tw 3 12 2df bf 3 12 Iy 1335 106 mm4 ry Iy A ry 4717 mm rx Ix A rx 9201 mm Slenderness Ratio For pinned ends K 10 λy K L ry λy 12720 Aluminum Association Formula Since λ 55 it is a long column Applying Eq 1326 σaallow 378125MPa λy 2 σaallow 2337 MPa Bending Stress Bending is about xx axis Mx P ey c 05D σb Mx c Ix σb P ey 05D Ix Axial Stress σa P A σa σaallow σb σballow 1 Applying the Interaction Formula P A σaallow P ey 05D Ix σballow 1 P 1 1 A σaallow ey D 2Ix σballow P 11421 kN Ans Note σa P A ra σa σaallow ra 081 015 Therefore the method is not allowed Problem 13116 Check if the wood column is adequate for supporting the eccentric load of P 3 kN applied at its top It is fixed at its base and free at its top Use the NFPA equations of Sec 136 and Eq 1330 Given L 12m b 50mm h 100mm P 3kN ey 75mm Solution Section Property A b h A 5000mm2 Ix b h3 12 Ix 417 106 mm4 Slenderness Ratio For fixedfree ends K 20 NFPA Timber Column Formula λ K L b λ 4800 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λ2 σallow 161 MPa Maximum Stress Bending is about xx axis Mx P ey σmax σallow c 05h Applying Eq 1330 σmax P A Mx c Ix σallow Pallow A Pallow ey 05h Ix Pallow σallow A 1 ey h A 2Ix Pallow 147 kN Since Pallow P 3 kN the column is not adequate Ans Problem 13117 Determine the maximum allowable eccentric load P that can be applied to the wood column The column is fixed at its base and free at its top Use the NFPA equations of Sec 136 and Eq 1330 Given L 12m b 50mm h 100mm ey 75mm Solution Section Property A b h A 5000mm2 Ix b h3 12 Ix 417 106 mm4 Slenderness Ratio For fixedfree ends K 20 NFPA Timber Column Formula λ K L b λ 4800 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λ2 σallow 161 MPa Maximum Stress Bending is about xx axis Mx P ey σmax σallow c 05h Applying Eq 1330 σmax P A Mx c Ix σallow P A P ey 05h Ix P σallow A 1 ey h A 2Ix P 147 kN Ans Problem 13118 The W360 X 57 structural A36 steel column is fixed at its bottom and free at its top Determine the greatest eccentric load P that can be applied using Eq 1330 and the AISC equations of Sec 136 Given E 200 GPa L 3m Po 200kN σY 250MPa ex 400mm Solution Use W 360x57 Ix 160 106 mm4 A 7200mm2 ry 393mm Iy 111 106 mm4 bf 1720mm rx 149mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedfree ends Kx 20 λy Ky L ry λy 15267 λx Kx L rx λx 4027 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 4418 MPa Maximum Stress Bending is about yy axis My P ex σmax σallow c 05bf P Po P Applying Eq 1330 σmax P A My c Iy σallow Po P A P ex 05bf Iy P σallow A Po 1 ex bf 2ry 2 P 508 kN Ans Problem 13119 The W250 X 67 structural A36 steel column is fixed at its bottom and free at its top If it is subjected to a load of P 10 kN determine if it is safe based on the AISC equations of Sec 136 and Eq 1330 Given E 200 GPa L 3m Po 200kN σY 250MPa ex 400mm P 10kN Solution Use W 250x67 Ix 104 106 mm4 A 8560mm2 ry 509mm Iy 222 106 mm4 bf 2040mm rx 110mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 12566 Slenderness Ratios For fixedfree ends Ky 20 For fixedfree ends Kx 20 λy Ky L ry λy 11788 λx Kx L rx λx 5455 Buckling about yy axis because λy λx Since λy λc it is an intermediate column Applying Eq 1323 σallow 1 λy 2 2λc 2 σY 5 3 3λy 8λc λy 3 8λc 3 σallow 7310 MPa Maximum Stress Bending is about yy axis My P ex My 4kN m c 05bf P Po P Applying Eq 1330 σmax P A My c Iy σmax Po P A My 05bf Iy σmax 4291 MPa σallow Since σmax σallow the column is safe Ans Problem 13120 The 3mlong bar is made of aluminum alloy 2014T6 If it is fixed at its bottom and pinned at the top determine the maximum allowable eccentric load P that can be applied using the fomulas in Sec 136 and Eq 1330 Given E 731 GPa b 100mm h 152mm σY 414MPa L 3m ey 38mm Solution Section Property A b h A 15200mm2 Ix b h3 12 Ix 2927 106 mm4 Iy h b3 12 Iy 1267 106 mm4 ry Iy A ry 2887 mm rx Ix A rx 4388 mm Slenderness Ratio For fixedpinned ends K 07 Aluminum Association Formula λy K L ry λy 7275 Since λ 55 it is a long column Applying Eq 1326 σallow 378125MPa λy 2 σallow 7145 MPa Maximum Stress Bending is about xx axis Mx P ey σmax σallow c 05h Applying Eq 1330 σmax P A Mx c Ix σallow P A P ey 05h Ix P σallow A 1 ey h 2rx 2 P 43443 kN Ans Problem 13121 The 3mlong bar is made of aluminum alloy 2014T6 If it is fixed at its bottom and pinned at the top determine the maximum allowable eccentric load P that can be applied using the equations of Sec 136 and the interaction formula with σballow 126 MPa Given E 731 GPa b 100mm h 152mm σballow 126MPa σY 414MPa L 3m ey 38mm Solution Section Property A b h A 15200mm2 Ix b h3 12 Ix 2927 106 mm4 Iy h b3 12 Iy 1267 106 mm4 ry Iy A ry 2887 mm rx Ix A rx 4388 mm Slenderness Ratio For fixedpinned ends K 07 Aluminum Association Formula λy K L ry λy 7275 Since λ 55 it is a long column Applying Eq 1326 σaallow 378125MPa λy 2 σaallow 7145 MPa Bending Stress Bending is about xx axis Mx P ey c 05h σb Mx c Ix σb P ey 05h Ix Axial Stress σa P A σa σaallow σb σballow 1 Applying the Interaction Formula P A σaallow P ey 05h Ix σballow 1 P 1 1 A σaallow ey h 2Ix σballow P 58687 kN Ans Note σa P A ra σa σaallow ra 054 015 Therefore the method is not allowed Problem 13122 The 250mmdiameter utility pole supports the transformer that has a weight of 3 kN and center of gravity at G If the pole is fixed to the ground and free at its top determine if it is adequate according to the NFPA equations of Sec 136 and Eq 1330 Given do 250mm L 54m e 375mm P 3kN Solution Section Property A π do 2 4 A 4908739 mm2 I π do 4 64 I 191748 106 mm4 Slenderness Ratio For fixedfree ends K 20 NFPA Timber Column Formula λ K L do λ 4320 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λ2 σallow 199 MPa Maximum Stress Bending moment is M P e M 1125 kN m c 05do Applying Eq 1330 σmax P A M c I σmax 0795 MPa Ans σallow OK Problem 13123 Using the NFPA equations of Sec 136 and Eq 1330 determine the maximum allowable eccentric load P that can be applied to the wood column Assume that the column is pinned at both its top and bottom Given L 36m b 75mm h 150mm ex 18mm Solution Section Property A b h A 11250mm2 Iy h b3 12 Iy 527 106 mm4 Slenderness Ratio For pinned ends K 10 NFPA Timber Column Formula λy K L b λy 4800 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λy 2 σallow 161 MPa Maximum Stress Bending is about yy axis My P ex σmax σallow c 05b Applying Eq 1330 σmax P A My c Iy σallow P A P ex 05b Iy P σallow A 1 ex b A 2Iy P 744 kN Ans Problem 13124 Using the NFPA equations of Sec 136 and Eq 1330 determine the maximum allowable eccentric load P that can be applied to the wood column Assume that the column is pinned at the top and fixed at the bottom Given L 36m b 75mm h 150mm ex 18mm Solution Section Property A b h A 11250mm2 Iy h b3 12 Iy 527 106 mm4 Slenderness Ratio For fixedpinned ends K 07 NFPA Timber Column Formula λy K L b λy 3360 Since 26 λ 50 it is a long column Applying Eq 1329 σallow 3718MPa λy 2 σallow 329 MPa Maximum Stress Bending is about yy axis My P ex σmax σallow c 05b Applying Eq 1330 σmax P A My c Iy σallow P A P ex 05b Iy P σallow A 1 ex b A 2Iy P 1518 kN Ans Problem 13125 The wood column is 4 m long and is required to support the axial load of 25 kN If the cross section is square determine the dimension a of each of its sides using a factor of safety against buckling of FS 25 The column is assumed to be pinned at its top and bottom Use the Euler equation Ew 11 GPa and σY 10 MPa Given E 11 GPa L 4m Fsafety 25 σY 10MPa P 25kN Solution Section Property A a2 I a4 12 Buckling Load Applying Eulers formula For pinned ends K 10 Pcr Fsafety P Pcr π2 E I K L 2 Fsafety P π2 E a4 12 K L 2 a 4 12 K L 2 Fsafety P π2 E a 1025 mm Ans Critical Stress Eulers formula is only valid if σcr σY A a2 σcr Pcr A σcr 594 MPa σY 10 MPa OK Problem 13126 The wood column is 4 m long and is required to support the axial load of 25 kN If the cross section is square determine the dimension a of each of its sides using a factor of safety against buckling of FS 15 The column is assumed to be fixed at its top and bottom Use the Euler equation Ew 11 GPa and σY 10 MPa Given E 11 GPa L 4m Fsafety 15 σY 10MPa P 25kN Solution Section Property A a2 I a4 12 Buckling Load Applying Eulers formula For fixed ends K 05 Pcr Fsafety P Pcr π2 E I K L 2 Fsafety P π2 E a4 12 K L 2 a 4 12 K L 2 Fsafety P π2 E a 638 mm Ans Critical Stress Eulers formula is only valid if σcr σY A a2 σcr Pcr A σcr 921 MPa σY 10 MPa OK Problem 13127 The member has a symmetric cross section If it is pin connected at its ends determine the largest force it can support It is made of 2014T6 aluminum alloy Given E 731 GPa t 12mm L 15m σY 414MPa a 50mm Solution Section Property b 2a t A b2 4a2 A 2544mm2 I t b3 12 2a t3 12 I 14193 106 mm4 r I A r 2362 mm Slenderness Ratio For pinned ends K 10 Aluminum Association Formula λ K L r λ 6351 Since λ 55 it is a long column Applying Eq 1326 σallow 378125MPa λ2 σallow 9376 MPa Pallow σallow A Pallow 2385 kN Ans Problem 13128 A steel column has a length of 5 m and is free at one end and fixed at the other end If the cross sectional area has the dimensions shown determine the critical load Est 200 GPa and σY 360 MPa Given E 200 GPa bo 80mm do 60mm σY 360MPa bi 60mm di 50mm L 5m Solution Section Property yc 05do bo do 05di bi di bo do bi di yc 3833 mm Ix 1 12 bo do 3 bo do 05do yc 2 1 12 bi di 3 bi di 05di yc 2 Iy 1 12 do bo 3 1 12 di bi 3 Buckling Load Applying Eulers formula I min Ix Iy For fixedfree ends K 20 Pcr π2 E I K L 2 Pcr 1214 kN Ans Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 001 m2 A MPa σY 360 MPa OK Therefore Eulers formula is valid Problem 13129 The distributed loading is supported by two pinconnected columns each having a solid circular cross section If AB is made of aluminum and CD of steel determine the required diameter of each column so that both will be on the verge of buckling at the same time Est 200 GPa Eal 70 GPa σYst 250 MPa and σYal 100 MPa Given a 075m b 3m wo 18 kN m Eal 70 GPa σYal 100MPa Est 200 GPa σYst 250MPa L 5m Solution L 2a b Support Reactions Given ΣFy0 FAB FCD 05wo L 0 1 ΣΜE0 FAB b a FCD a 1 2 wo L 2L 3 0 2 Solving Eqs 1 and 2 Guess FAB 1kN FCD 1kN FAB FCD Find FAB FCD FAB FCD 30375 10125 kN Section Property I π do 4 64 A π 4 do 2 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 For pinned ends K 10 I 10 L 2 π2 E Pcr π do 4 64 L2 π2 E Pcr do 4 64L2 π3 E Pcr For Member AB Pcr FAB dal 4 64L2 π3 Eal Pcr dal 6879 mm Ans For Member CD Pcr FCD dst 4 64L2 π3 Est Pcr dst 4020 mm Ans Check Critical Stress Eulers formula is only valid if σcr σY Aal π 4 dal 2 σcr Pcr Aal σcr 817 MPa σYal 100 MPa OK Ast π 4 dst 2 σcr Pcr Ast σcr 798 MPa σY 250 MPa OK Problem 13130 The steel pipe is fixed supported at its ends If it is 4 m long and has an outer diameter of 50 mm determine its required thickness so that it can support an axial load of P 100 kN without buckling Est 200 GPa and σY 250 MPa Given L 4m E 200 GPa Pcr 100kN do 50mm σY 250MPa Solution Section Property I π do 4 di 4 64 A π do 2 di 2 4 Buckling Load Applying Eulers formula Pcr π2 E I K L 2 I Pcr K L 2 π2 E For fixed ends K 05 Thus π do 4 di 4 64 Pcr K L 2 π2 E di 4 do 4 64Pcr K L 2 π3 E di 3817 mm t 05 do di t 592 mm Ans Critical Stress Eulers formula is only valid if σcr σY A π do 2 di 2 4 σcr Pcr A σcr 12203 MPa σY 250 MPa OK Problem 13131 The steel column is assumed to be pin connected at its top and bottom and fully braced against buckling about the yy axis If it is subjected to an axial load of 200 kN determine the maximum moment M that can be applied to its ends without causing it to yield Est 200 GPa and σY 250 MPa Given E 200 GPa bf 150mm df 15mm P 200kN σY 250MPa tw 15mm dw 120mm L 8m Solution Section Property D dw 2df A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 rx Ix A Buckling For pinned ends K 10 Pcr π2 E Ix K L 2 Pcr 70159 kN P 200 kN OK Yielding about xx axis Applying the secant formula c 05D Given σY P A 1 emax c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess emax 1mm emax Find emax emax 22124 mm Maximum Moment Mmax emax P Mmax 4425 kN m Ans Problem 13132 The steel column is assumed to be fixed connected at its top and bottom and braced against buckling about the yy axis If it is subjected to an axial load of 200 kN determine the maximum moment M that can be applied to its ends without causing it to yield Est 200 GPa and σY 250 MPa Given E 200 GPa bf 150mm df 15mm P 200kN σY 250MPa tw 15mm dw 120mm L 8m Solution Section Property D dw 2df A bf D dw bf tw Ix bf D3 12 bf tw dw 3 12 rx Ix A Buckling For fixed ends K 05 Pcr π2 E Ix K L 2 Pcr 280636 kN P 200 kN OK Yielding about xx axis Applying the secant formula c 05D Given σY P A 1 emax c rx 2 sec K L 2rx P E A 1 Solving Eq 1 by trial and error Guess emax 1mm emax Find emax emax 30231 mm Maximum Moment Mmax emax P Mmax 6046 kN m Ans Problem 13133 The W250 X 67 steel column supports an axial load of 300 kN in addition to an eccentric load P Determine the maximum allowable value of P based on the AISC equations of Sec 136 and Eq 1330 Assume that in the xz plane Kx 10 and in the yz plane Ky 20 Est 200 GPa σY 350 MPa Given E 200 GPa L 3m Kx 10 Po 300kN σY 350MPa ey 200mm Ky 20 Solution Use W 250x67 Ix 104 106 mm4 A 8560mm2 ry 509mm Iy 222 106 mm4 d 257mm rx 110mm AISC Column Formula λ K L r From Eq 1322 λc 2π2 E σY λc 10621 Slenderness Ratios For yz plane Ky 20 For xz plane Kx 10 λy Ky L ry λy 11788 λx Kx L rx λx 2727 Buckling about yy axis because λy λx Since λc λy 200 it is a long column Applying Eq 1321 σallow 12π2 E 23λy 2 σallow 7412 MPa Maximum Stress Bending is about xx axis Mx P ey σmax σallow c 05d P Po P Applying Eq 1330 σmax P A M c I σallow Po P A P ey 05d Ix P σallow A Po 1 ey d 2rx 2 P 1071 kN Ans Problem 13134 The steel bar AB has a rectangular cross section If it is assumed to be pin connected at its ends determine if member AB will buckle if the distributed load w 2 kNm Use a factor of safety with respect to buckling of FS 15 Est 200 GPa and σY 360 MPa Given h 20mm E 200 GPa Lb 5m b 30mm σY 360MPa w 2 kN m L 3m Fsafety 15 Solution Support Reactions For member BC ΣFy0 FBA FC w Lb 0 By symmetry FBA FC Hence FBA 05w Lb 1 FC 05w Lb Section Property A h b Iy b h3 12 Ix h b3 12 Buckling Load Applying Eulers formula About xx axix About yy axix For pinned ends Kx 10 For pinned ends Ky 10 Pcr π2 E Ix Kx L 2 Pcr π2 E Iy Ky L 2 Pcr 987 kN Pcr 439 kN Pcr min Pcr Pcr Pcr 4386 kN Check Critical Stress Eulers formula is only valid if σcr σY σcr Pcr A σcr 731 MPa σYal 100 MPa OK Factor of Safety Pmax FBA Fsafety Pmax 75 kN Pcr 4386kN Hence memeber BA will buckle Ans Problem 1401 A material is subjected to a general state of plane stress Express the strain energy density in terms of the elastic constants E G and ν and the stress components σx σy and τxy Problem 1402 The strainenergy density must be the same whether the state of stress is represented by σx σy and τxy or by the principal stresses σ1 and σ2 This being the case equate the strainenergy expressions for each of these two cases and show that G E21 ν Problem 1403 Determine the strain energy in the rod assembly Portion AB is steel BC is brass and CD is aluminum Est 200 GPa Ebr 101 GPa and Eal 731 GPa Given Lst 03m dst 15mm Lbr 04m dbr 20mm Lal 02m dal 25mm Est 200GPa PA 3kN Ebr 101GPa PB 4kN Eal 731GPa PC 10 kN Solution Section Properties Ast π dst 2 4 Ast 17671 mm2 Abr π dbr 2 4 Abr 31416 mm2 Aal π dal 2 4 Aal 49087 mm2 Internal Axial Forces Pst PA Pbr PA PB Pal PA PB PC N L2EA U 2 i Axial Strain Energy Ui Pst 2 Lst 2 Est Ast Pbr 2 Lbr 2 Ebr Abr Pal 2 Lal 2 Eal Aal Ui 0372 J Ans Problem 1404 Determine the torsional strain energy in the A36 steel shaft The shaft has a radius of 40 mm Given LAB 06m TA 8kN m r 40mm LBC 04m TB 6 kN m G 75GPa LCD 05m TC 12 kN m Solution Section Properties J π r4 2 J 402123860 mm4 Internal Torsional Moment TAB TA TBC TA TB TCD TA TB TC T L2GJ U 2 i Axial Strain Energy Ui TAB 2 LAB 2 G J TBC 2 LBC 2 G J TCD 2 LCD 2 G J Ui 1492 J Ans Problem 1405 Determine the torsional strain energy in the A36 steel shaft The shaft has a radius of 30 mm Given L 05m r 30mm G 75GPa TA 0 TB 3kN m TC 4 kN m Solution Section Properties J π r4 2 J 127234502 mm4 Internal Torsional Moment TAB TA TBC TA TB TCD TA TB TC T L2GJ U 2 i Axial Strain Energy Ui TAB 2 L 2 G J TBC 2 L 2 G J TCD 2 L 2 G J Ui 2620 J Ans Problem 1406 Determine the bending strain energy in the beam EI is constant Solution Support Reactions By symmetry A B R ΣFy0 2R P 0 R 05P Internal Moment Function M R x L 0 2 i M 2EI dx U Bending Strain Energy Applying Eq 1417 Ui 1 2E I 0 L x M2 d 2 2E I 0 05L x M2 d 2 2E I 0 05L x R x 2 d Ui 1 4E I 0 05L x P x 2 d P2 4E I 0 05L x x2 d Ui P2 L3 96E I Ans P 30kN Given a 4m b 2m I 225 106 mm4 E 200GPa Solution L a b Use W 250x18 Ans Support Reactions ΣFy0 A B P 0 1 ΣΜB0 A a P b 0 2 Solving Eqs 1 and 2 A P b a B P L a A 15 kN B 45kN Internal Moment Function M1 A x1 M2 A a x2 M4 P x4 M3 P b x3 L 0 2 i M 2EI dx U Bending Strain Energy a Using coocrdinates x1 and x4 and applying Eq 1417 Ui 1 2E I 0 a x1 M1 2 d 0 b x4 M4 2 d Ui 1 2E I 0 a x1 A x1 2 d 0 b x4 P x4 2 d Ui 800J Ans b Using coocrdinates x2 and x3 and applying Eq 1417 Ui 1 2E I 0 a x M2 2 d 0 b x3 M3 2 d Ui 1 2E I 0 a x2 A a x2 2 d 0 b x3 P b x3 2 d Ui 800J Problem 1407 Determine the bending strain energy in the A36 structural steel W250 X 18 beam Obtain the answer using the coordinates a x1 and x4 and b x2 and x3 Problem 1408 Determine the total axial and bending strain energy in the A36 steel beam A 2300 mm2 I 95106 mm4 Given L 10m A 2300mm2 P 15kN I 95 106 mm4 E 200GPa w 15 kN m Solution Support Reactions ΣFy0 Av Bv w L 0 1 ΣΜB0 Av L w L 05L 0 2 Solving Eqs 1 and 2 Av 05w L Av 75 kN Bv 05w L Bv 75 kN ΣFx0 Ah P 0 Ah P Ah 15 kN Internal Moment and Force Function M Av x w x 05x Av x 05w x2 N Ah L 0 2 i N 2EA dx U Axial Strain Energy Uai 1 2E A 0 L x N2 d Uai 1 2E A 0 L x Ah 2 d Uai 24457 J L 0 2 i M 2EI dx U Bending Strain Energy Ubi 1 2E I 0 L x M2 d Ubi 1 2E I 0 L x Av x 05w x2 2 d Ubi 4934211 J Total Strain Energy Ui Uai Ubi Ui 4959 J Ans Problem 1409 Determine the total axial and bending strain energy in the A36 structural steel W200 X 86 beam Given a 3m P1 25kN θ 30deg P2 15kN E 200GPa Solution L 2a Use W 200x86 I 947 106 mm4 A 11000mm2 Support Reactions ΣFy0 Av Bv P1 P2 sin θ 0 1 ΣΜB0 Av 2a P1 a 0 2 Solving Eqs 1 and 2 Av 05P1 Av 125 kN Bv 05P1 05P2 Bv 20kN ΣFx0 Ah P2 cos θ 0 Ah P2 cos θ Ah 1299 kN Internal Moment and Force Function M1 Av x1 M2 Bv P2 sin θ x2 M2 Av x2 N1 Ah N2 Ah L 0 2 i N 2EA dx U Axial Strain Energy Uai 1 2E A 0 a x1 N1 2 d 0 a x2 N2 2 d Uai 1 2E A 0 a x1 Ah 2 d 0 a x2 Ah 2 d Uai 023 J L 0 2 i M 2EI dx U Bending Strain Energy Ubi 1 2E I 0 a x1 M1 2 d 0 a x2 M2 2 d Ubi 1 2E I 0 a x1 Av x1 2 d 0 a x2 Av x2 2 d Ubi 7425 J Total Strain Energy Ui Uai Ubi Ui 7448 J Ans Problem 1410 Determine the bending strain energy in the beam EI is constant Solution Support Reactions By symmetry A B R ΣMB0 R L Mo Mo 0 R 0 Internal Moment Function M Mo L 0 2 i M 2EI dx U Bending Strain Energy Applying Eq 1417 Ui 1 2E I 0 L x M2 d 1 2E I 0 L x Mo 2 d Ui Mo 2 4E I 0 L 1 x d Ui Mo 2 L 2E I Ans Problem 1411 Determine the bending strain energy in the A36 steel beam due to the loading shown Obtain the answer using the coordinates a x1 and x4 and b x2 and x3 I 21106 mm4 Unit used kJ 1000J Given a 3m b 15m w 120 kN m E 200GPa Solution L a b Section Property I 21 106 mm4 Support Reactions ΣFy0 A B w b 0 1 ΣΜB0 A a w b 05b 0 2 Solving Eqs 1 and 2 A w b2 2a B w b b 2a 2a A 45 kN B 225kN Internal Moment Function M1 A x1 M2 A a x2 M4 05 w x4 2 M3 05 w b x3 2 L 0 2 i M 2EI dx U Bending Strain Energy a Using coocrdinates x1 and x4 and applying Eq 1417 Ui 1 2E I 0 a x1 M1 2 d 0 b x4 M4 2 d Ui 1 2E I 0 a x1 A x1 2 d 0 b x4 05 w x4 2 2 d Ui 2821 kJ Ans b Using coocrdinates x2 and x3 and applying Eq 1417 Ui 1 2E I 0 a x2 M2 2 d 0 b x3 M3 2 d Ui 1 2E I 0 a x2 A a x2 2 d 0 b x3 05 w b x3 2 2 d Ui 2821 kJ Ans Problem 1412 Determine the bending strain energy in the cantilevered beam due to a uniform load w Solve the problem two ways a Apply Eq 1417 b The load w dx acting on a segment dx of the beam is displaced a distance y where y w x4 4L3x 3L424EI the equation of the elastic curve Hence the internal strain energy in the differential segment dx of the beam is equal to the external work ie dUi ½ w dxy Integrate this equation to obtain the total strain energy in the beam EI is constant Problem 1413 Determine the bending strain energy in the simply supported beam due to a uniform load w Solve the problem two ways a Apply Eq 1417 b The load w dx acting on a segment dx of the beam is displaced a distance y where y w x4 2Lx3 L3x24EI the equation of the elastic curve Hence the internal strain energy in the differential segment dx of the beam is equal to the external work ie dUi ½ w dxy Integrate this equation to obtain the total strain energy in the beam EI is constant Problem 1414 Determine the shear strain energy in the beam The beam has a rectangular cross section of area A and the shear modulus is G Problem 1415 The concrete column contains six 25mmdiameter steel reinforcing rods If the column supports a load of 1500 kN determine the strain energy in the column Est 200 GPA Ec 25 GPa Given L 15m rc 300mm dst 25mm Est 200GPa Econc 25GPa P 1500kN Solution Section Properties Ast 6 π dst 2 4 Ast 294524 mm2 Aconc π rc 2 Ast Aconc 27979810 mm2 Given Equilibrium ΣFy0 Pconc Pst P 0 1 Compatibility conc st Pconc L Aconc Econc Pst L Ast Est 2 Solving Eqs 1 and 2 Guess Pconc 1kN Pst 1kN Pconc Pst Find Pconc Pst Pconc Pst 138350 11650 kN N L2EA U 2 i Axial Strain Energy Ui Pconc 2 L 2 Econc Aconc Pst 2 L 2 Est Ast Ui 22251 J Ans Problem 1416 Determine the bending strain energy in the beam and the axial strain energy in each of the two rods The beam is made of 2014T6 aluminum and has a square cross section 50 mm by 50 mm The rods are made of A36 steel and have a circular cross section with a 20mm diameter Given Lst 2m Est 200GPa dst 20mm Lal 4m Eal 731GPa dal 50mm a 1m b 2m P 8kN Solution Section Properties Ast π dst 2 4 Ial dal 4 12 Support Reactions By symmetry A B Fst ΣFy0 2Fst 2P 0 Fst P Fst 8kN Internal Moment and Force Function Nst Fst M1 Fst x1 M2 Fst a x2 P x2 P a Axial Strain Energy Uai 1 2Est Ast 0 Lst x Nst 2 d Uai 1 2Est Ast 0 Lst x Fst 2 d Uai 1019 J Ans L 0 2 i M 2EI dx U Bending Strain Energy Using coocrdinates x1 and x2 and applying Eq 1417 Ubi 1 2Eal Ial 2 0 a x1 M1 2 d 0 b x2 M2 2 d Ubi 1 2Eal Ial 2 0 a x1 Fst x1 2 d 0 b x2 P a 2 d Ubi 22413 J Ans L 0 2 i N 2EA dx U Problem 1417 Determine the bending strain energy in the beam due to the distributed load EI is constant Solution Internal Moment Function M x 2 wo x L x 3 wo x3 6L L 0 2 i M 2EI dx U Bending Strain Energy Applying Eq 1417 Ui 1 2E I 0 L x M2 d 1 2E I 0 L x wo x3 6L 2 d Ui wo 2 72E I L2 0 L x x6 d Ui wo 2 L5 504E I Ans Problem 1418 The beam shown is tapered along its width If a force P is applied to its end determine the strain energy in the beam and compare this result with that of a beam that has a constant rectangular cross section of width b and height h Problem 1419 The bolt has a diameter of 10 mm and the link AB has a rectangular cross section that is 12 mm wide by 7 mm thick Determine the strain energy in the link due to bending and in the bolt due to axial force The bolt is tightened so that it has a tension of 500 N Both members are made of A36 steel Neglect the hole in the link Given Lo 60mm do 10mm d 12mm a 50mm b 30mm t 7mm E 200GPa P 05kN Solution Section Properties Ao π do 2 4 I d t3 12 Support Reactions Ans ΣFx0 P A B 0 1 ΣΜB0 A a b P b 0 2 Solving Eqs 1 and 2 A P b a b B P A A 01875 kN B 03125 kN Internal Moment and Force Function No P M1 A x1 M2 B x2 Axial Strain Energy Uai 1 2E Ao 0 Lo x No 2 d Uai 1 2E Ao 0 Lo x P2 d Uai 477 10 4 J Ans L 0 2 i M 2EI dx U Bending Strain Energy Using coocrdinates x1 and x2 and applying Eq 1417 Ubi 1 2 E I 0 a x1 M1 2 d 0 b x2 M2 2 d Ubi 1 2 E I 0 a x1 A x1 2 d 0 b x2 B x2 2 d Ubi 00171 J L 0 2 i N 2EA dx U Problem 1420 The steel beam is supported on two springs each having a stiffness of k 8 MNm Determine the strain energy in each of the springs and the bending strain energy in the beam Est 200 GPa I 5106 mm4 Given L 4m E 200GPa I 5 106 mm4 a 1m k 8000 kN m w 2 kN m b 2m Solution Support Reactions By symmetry A B R ΣFy0 2R w L 0 R 05w L R 4kN Internal Moment Function M1 w 2 x1 2 M2 w 2 a x2 2 R x2 2 U 2 i kδ Spring Strain Energy The spring deforms δ R k δ 0500 mm Uspi 1 2k δ2 Uspi 1000 J Ans L 0 2 i M 2EI dx U Bending Strain Energy Using coocrdinates x1 and x2 and applying Eq 1417 Ubi 1 2E I 2 0 a x1 M1 2 d 0 b x2 M2 2 d Ubi 1 2E I 2 0 a x1 w 2 x1 2 2 d 0 b x2 w 2 a x2 2 R x2 2 d Ubi 0400 J Ans Problem 1421 Determine the bending strain energy in the 50mmdiameter A36 steel rod due to the loading shown Given a 06m b 06m do 50mm E 200GPa P 400N Solution Section Properties I π do 4 64 Internal Moment Function M1 P x1 M2 P a L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 2 0 a x1 M1 2 d 0 b x2 M2 2 d Ui 1 2E I 2 0 a x1 P x1 2 d 0 b x2 P a 2 d Ui 0469 J Ans Problem 1422 The A36 steel bar consists of two segments one of circular cross section of radius r and one of square cross section If it is subjected to the axial loading of P determine the dimensions a of the square segment so that the strain energy within the square segment is the same as in the circular segment Problem 1423 Consider the thinwalled tube of Fig 530 Use the formula for shear stress τavg T2tAm Eq 518 and the general equation of shear strain energy Eq 1411 to show that the twist of the tube is given by Eq 520 Hint Equate the work done by the torque T to the strain energy in the tube determined from integrating the strain energy for a differential element Fig 144 over the volume of material Problem 1424 Determine the horizontal displacement of joint C AE is constant Problem 1425 Determine the horizontal displacement of joint A Each bar is made of A36 steel and has a cross sectional area of 950 mm2 Given a 09m b 12m A 950mm2 E 200GPa P 10kN Solution c a2 b2 Member Forces Applying the method of joints to Joint A Given ΣFy0 FAB FAD a c 0 1 ΣFx0 FAD b c P 0 2 Solving Eqs 1 and 2 Guess FAB 1kN FAD 1kN C FAB FAD Find FAB FAD FAB FAD 750 1250 kN T Applying the method of joints to Joint D Given ΣFy0 FCD FBD a c FAD a c 0 3 ΣFx0 FBD b c FAD b c 0 4 Solving Eqs 3 and 4 Guess FCD 1kN FBD 1kN C FBD FCD Find FBD FCD FBD FCD 1250 1500 kN T N L2EA U 2 i Axial Strain Energy Ui 1 2 E A FAB 2 2a FAD 2 c FBD 2 c FCD 2 b Ui 2211 J External Work The external work done by the force P is Ue 1 2 P Ah Conservation of Energy Ui Ue Ui 1 2 P Ah Ah 2 Ui P Ah 0442 mm Ans Problem 1426 Determine the vertical displacement of joint D AE is constant Given L m a 06L b 08L c L EA kN P kN Solution Member Forces By inspection of Joint D FAD 0 FCD P T Applying the method of joints to Joint C Given ΣFy0 FCA b c P 0 1 ΣFx0 FCB FCA a c 0 2 Solving Eqs 1 and 2 Guess FCA 1kN FCB 1kN C FCA FCB Find FCA FCB FCA FCB 125 075 P T Applying the method of joints to Joint A ΣFy0 FAB FCA b c 0 FAB FCA b c FAB 100 P T Axial Strain Energy Ui 1 2 EA FCD 2 b FCA 2 c FCB 2 a FAB 2 b Ui 1750 P2 L EA External Work The external work done by the force P is Ue 1 2 P Dv Conservation of Energy Ui Ue Ui 1 2 P Dv Dv 2 Ui P Dv 350 P L EA Ans Problem 1427 Determine the slope at the end B of the A36 steel beam I 80106 mm4 Given L 8m I 80 106 mm4 Ans Mo 6kN m Solution Support Reactions ΣFy0 A B 0 1 ΣΜB0 A L Mo 0 2 Solving Eqs 1 and 2 A Mo L B A A 075 kN B 075 kN Internal Moment Function M A x L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 0 L x A x 2 d Ui 3000 J External Work The external work done by the moment Mo is Ue 1 2 Mo θB Conservation of Energy Ui Ue Ui 1 2 Mo θB θB 2 Ui Mo θB 00010 rad E 200GPa Problem 1428 Determine the displacement of point B on the A36 steel beam I 977106 mm4 Given a 45m b 3m I 977 106 mm4 E 200GPa P 40kN Solution Internal Moment Function M P x L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 0 a x P x 2 d Ui 1243603 J External Work The external work done by the force P is Ue 1 2 P B Conservation of Energy Ui Ue Ui 1 2 P B B 2 Ui P B 6218 mm Ans Problem 1429 The cantilevered beam has a rectangular crosssectional area A a moment of inertia I and a modulus of elasticity E If a load P acts at point B as shown determine the displacement at B in the direction of P accounting for bending axial force and shear Problem 1430 Use the method of work and energy and determine the slope of the beam at point B EI is constant Problem 1431 Determine the slope at point A of the beam EI is constant Problem 1432 Determine the displacement of point B on the 2014T6 aluminum beam Given a 2m bf 150mm dw 150mm b 6m df 25mm tw 25mm P 20kN E 731GPa Solution L a b Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df tw dw 05dw df bf df tw dw yc 5625 mm I 1 12 bf df 3 bf df 05df yc 2 1 12 tw dw 3 tw dw 05dw df yc 2 I 2158203125 mm4 Support Reactions ΣFy0 A C P 0 1 ΣΜC0 A L P b 0 2 Solving Eqs 1 and 2 A P b L A 15kN C P A C 5kN Internal Moment Function M1 A x1 M2 C x2 L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 0 a x1 M1 2 d 0 b x2 M2 2 d Ui 1 2E I 0 a x1 A x1 2 d 0 b x2 C x2 2 d Ui 76063 J External Work The external work done by the force P is Ue 1 2 P B Conservation of Energy Ui Ue Ui 1 2 P B B 2 Ui P B 7606 mm Ans Problem 1433 The A36 steel bars are pin connected at C If they each have a diameter of 50 mm determine the displacement at E Given a 36m b 3m c 24m E 200GPa P 10kN do 50mm Solution Section Properties I π do 4 64 Support Reactions By symmetry ABR and D0 ΣFy0 2R P 0 R 05P R 5kN Internal Moment Function M1 R x1 M2 R x2 L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 0 05a x1 M1 2 d 0 05a x2 M2 2 d Ui 1 2E I 0 05a x1 R x1 2 d 0 05a x2 R x2 2 d Ui 792057 J External Work The external work done by the force P is Ue 1 2 P E Conservation of Energy Ui Ue Ui 1 2 P E E 2 Ui P E 15841 mm Ans Problem 1434 Determine the deflection of the beam at its center caused by shear The shear modulus is G Problem 1435 The A36 steel bars are pin connected at B and CIf they each have a diameter of 30 mm determine the slope at E Given a 3m b 2m do 30mm E 200GPa Mo 03kN m Solution Section Properties I π do 4 64 Equations of Equilibrium For member BC ΣFy0 B C 0 1 ΣΜB0 Ans 2 Solving Eqs 1 and 2 B Mo 2 b C B B 75 N C 75 N Internal Moment Function M1 B x1 M2 B x2 L 0 2 i M 2EI dx U Bending Strain Energy Using coocrdinates x1 and x2 and applying Eq 1417 Ubi 1 2E I 2 0 a x1 M1 2 d 2 0 b x2 M2 2 d Ubi 1 2E I 2 0 a x1 B x1 2 d 2 0 b x2 B x2 2 d Ubi 8252 J External Work The external work done by the moment Mo is Ue 1 2 Mo θE Conservation of Energy Ui Ue Ubi 1 2 Mo θE θE 2 Ubi Mo θE 005502 rad θE 3152 deg B 2 b Mo 0 Problem 1436 The A36 steel bars are pin connected at B If each has a square cross section determine the vertical displacement at B Given a 24m b 12m c 3m E 200GPa P 4kN do 75mm Solution Section Properties I do 4 12 Support Reactions A 0 ΣFy0 C D P 0 1 ΣΜC0 P b D c 0 2 Solving Eqs 1 and 2 D P b c D 16 kN C P D C 56 kN Internal Moment Function M1 P x1 M2 D x2 L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 0 b x1 M1 2 d 0 c x2 M2 2 d Ui 1 2E I 0 b x1 P x1 2 d 0 c x2 D x2 2 d Ui 30583 J External Work The external work done by the force P is Ue 1 2 P B Conservation of Energy Ui Ue Ui 1 2 P B B 2 Ui P B 1529 mm Ans Problem 1437 The rod has a circular cross section with a moment of inertia I If a vertical force P is applied at A determine the vertical displacement at this point Only consider the strain energy due to bending The modulus of elasticity is E Problem 1438 The load P causes the open coils of the spring to make an angle θ with the horizontal when the spring is stretched Show that for this position this causes a torque T PR cos θ and a bending moment M PR sin θ at the cross section Use these results to determine the maximum normal stress in the material Problem 1439 The coiled spring has n coils and is made from a material having a shear modulus G Determine the stretch of the spring when it is subjected to the load P Assume that the coils are close to each other so that θ 0 and the deflection is caused entirely by the torsional stress in the coil Problem 1440 A bar is 4 m long and has a diameter of 30 mm If it is to be used to absorb energy in tension from an impact loading determine the total amount of elastic energy that it can absorb if a it is made of steel for which Est 200 GPa σY 800 MPa and b it is made from an aluminum alloy for which Eal 70 GPa σY 405 MPa Unit used kJ 1000J Given L 4m E1 200GPa E2 70GPa do 30mm σY1 800MPa σY2 405MPa Solution Geometric Property V π do 2 4 L V 282743339 mm3 a Strain Energy εY1 σY1 E1 ur1 1 2 σY1 εY1 ur1 1 2E1 σY1 2 ur1 1600 106 J m3 Ui1 ur1 V Ui1 4524 kJ Ans b Strain Energy εY2 σY2 E2 ur2 1 2 σY2 εY2 ur1 1 2E2 σY2 2 ur2 1172 106 J m3 Ui2 ur2 V Ui2 3313 kJ Ans Problem 1441 Determine the diameter of a brass bar that is 24 m long if it is to be used to absorb 1200 J of energy in tension from an impact loading Take σY 70 MPa E 100GPa Given L 24m Ui 1200J E 100GPa σY 70MPa Solution Geometric Property V π do 2 4 L Strain Energy εY σY E ur 1 2 σY εY ur 2450000 J m3 Conservation of Energy Ui Ur Ui ur V Ui ur π do 2 4 L do 4 Ui π L ur do 16120 mm Ans Problem 1442 The collar has a weight of 250 N 25 kg and falls down the titanium bar If the bar has a diameter of 12 mm determine the maximum stress developed in the bar if the weight is a dropped from a height of h 300 mm b released from a height h 0 and c placed slowly on the flange at A Eti 120 GPa σY 420 MPa Given do 12mm L 12m W 250N E 120GPa σY 420MPa Solution Section Property A π do 2 4 Static Displacement st W L E A st 002210 mm a Drop Height h 300mm Impact factor η 1 1 2 h st η 16576 Pmax W η σmax Pmax A σmax 3664 MPa Ans b Drop Height h 0mm Impact factor η 1 1 2 h st η 200 Pmax W η σmax Pmax A σmax 4421 MPa Ans c No impact η 1 Pmax W η σmax Pmax A σmax 2210 MPa Ans Problem 1443 The collar has a weight of 250 N 25 kg and falls down the titanium bar If the bar has a diameter of 12 mm determine the largest height h at which the weight can be released and not permanently damage the bar after striking the flange at A Eti 120 GPa σY 420 MPa Given do 12mm L 12m W 250N E 120GPa σY 420MPa Solution Section Property A π do 2 4 Static Displacement st W L E A st 002210 mm Maximum Drop Height Impact factor η 1 1 2 h st σmax W A 1 1 2 h st σmax σY h st 2 σY A W 1 2 1 h 3948 mm Ans Problem 1444 The mass of 50 Mg is held just over the top of the steel post having a length of L 2 m and a cross sectional area of 001 m2 If the mass is released determine the maximum stress developed in the bar and its maximum deflection Est 200 GPa σY 600 MPa Given L 2m A 001m2 mo 50 103 kg E 200GPa σY 600MPa Solution Weight W mo g Static Displacement st W L E A st 04903 mm Drop Height h 0 Impact factor η 1 1 2 h st η 200 Pmax W η σmax Pmax A σmax 981 MPa Ans σY 600 MPa OK max st η max 0981 mm Ans Problem 1445 Determine the speed v of the 50Mg mass when it is just over the top of the steel post if after impact the maximum stress developed in the post is 550 MPa The post has a length of L 1 m and a crosssectional area of 001 m2 Est 200 GPa σY 600 MPa Given L 1m A 001m2 mo 50 103 kg E 200GPa σY 600MPa σmax 550MPa Solution Weight W mo g Equivalent Spring Stiffness k E A L k 200 106 kN m Maximum Displacement Pmax A σmax max Pmax k max 275 mm Conservation of Energy Ui Ur Given 1 2 mo v2 W max 1 2 k max 2 1 Solving Eq 1 Guess v 1 m s v Find v v 0499 m s Ans Problem 1446 The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm Determine the maximum axial stress developed in the bar if the 5kg collar is dropped from a height of h 100 mm Eal 70 GPa σY 410 MPa Given d1 5mm d2 10mm E 70GPa L1 02m L2 03m σY 410MPa mo 5kg h 01m Solution Weight W mo g Section Property A1 π d1 2 4 A2 π d2 2 4 Static Displacement st mo g L1 E A1 mo g L2 E A2 st 000981 mm Impact factor η 1 1 2 h st η 14378 Pmax W η σmax Pmax A1 σmax 3591 MPa Ans σY 410 MPa OK Problem 1447 The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm Determine the maximum height h from which the 5kg collar should be dropped so that it produces a maximum axial stress in the bar of σmax 300 MPa Eal 70 GPa σY 410 MPa Given d1 5mm d2 10mm E 70GPa L1 02m L2 03m σY 410MPa mo 5kg σmax 300MPa Solution Weight W mo g Section Property A1 π d1 2 4 A2 π d2 2 4 Static Displacement st W L1 E A1 W L2 E A2 st 000981 mm Maximum Drop Height Impact factor η 1 1 2 h st σmax W A1 1 1 2 h st h st 2 σmax A1 W 1 2 1 h 696 mm Ans Problem 1448 A steel cable having a diameter of 16 mm wraps over a drum and is used to lower an elevator having a weight of 4 kN 400kg The elevator is 45 m below the drum and is descending at the constant rate of 06 ms when the drum suddenly stops Determine the maximum stress developed in the cable when this occurs Est 200 GPa σY 350 MPa Given do 16mm E 200GPa L 45m W 4kN σY 350MPa v 06 m s Solution Section Property A π do 2 4 Eq Spring Constant k E A L k 89361 kN m Conservation of Energy Ui Ur Given 1 2 W g v2 W max 1 2 k max 2 1 Solving Eq 1 Guess max 10mm max Find max max 1805 mm Maximum Stress Pmax k max σmax Pmax A σmax 802 MPa Ans σY 350 MPa OK Problem 1449 Solve Prob 1448 if the elevator is descending at the constant rate of 09 ms Given do 16mm E 200GPa L 45m W 4kN σY 350MPa v 09 m s Solution Section Property A π do 2 4 Eq Spring Constant k E A L k 89361 kN m Conservation of Energy Ui Ur Given 1 2 W g v2 W max 1 2 k max 2 1 Solving Eq 1 Guess max 10mm max Find max max 2422 mm Maximum Stress Pmax k max σmax Pmax A σmax 1076 MPa Ans σY 350 MPa OK Problem 1450 The 250N 25kg weight is falling at 09 ms at the instant it is 06 m above the spring and post assembly Determine the maximum stress in the post if the spring has a stiffness of k 40 MNm The post has a diameter of 75 mm and a modulus of elasticity of E 48 GPa Assume the material will not yield Given do 75mm h 06m W 250N L 06m E 48GPa v 09 m s ksp 40 103 kN m Solution Section Property A π do 2 4 Eq Spring Constants kp E A L kp 35342917 kN m Equilibrium requires Fsp Fp Hence ksp sp kp p sp kp ksp p 1 Conservation of Energy Ui Ur 1 2 W g v2 W h max 1 2 ksp sp 2 1 2 kp p 2 2 However max p sp 3 Substituting Eqs1 and 3 into 2 Given 1 2 W g v2 W h p kp ksp p 1 2 ksp kp ksp p 2 1 2 kp p 2 4 Solving Eq 4 Guess p 10mm p Find p p 03044 mm Maximum Stress Pmax kp p σmax Pmax A σmax 244 MPa Ans Problem 1451 The A36 steel bolt is required to absorb the energy of a 2kg mass that falls h 30 mm If the bolt has a diameter of 4 mm determine its required length L so the stress in the bolt does not exceed 150 MPa Given do 4mm h 30mm mo 2 kg E 200GPa σallow 150MPa Solution Weight W mo g Section Property A π do 2 4 Static Displacement st W L E A Allowable Length Impact factor η 1 1 2 h st σallow W A η Given σallow W A 1 1 2 h E A W L 1 Solving Eq 1 Guess L 10mm L Find L L 8501 mm Ans Problem 1452 The A36 steel bolt is required to absorb the energy of a 2kg mass that falls h 30 mm If the bolt has a diameter of 4 mm and a length of L 200 mm determine if the stress in the bolt will exceed 175 MPa Given do 4mm L 200mm h 30mm mo 2 kg E 200GPa σallow 175MPa Solution W mo g Section Property A π do 2 4 Static Displacement st W L E A st 000156 mm Maximum Stress Impact factor η 1 1 2 h st η 19707 σmax W A η σmax 3076 MPa σallow 175 MPa Exceeded Ans Problem 1453 The A36 steel bolt is required to absorb the energy of a 2kg mass that falls along the 4mmdiameter bolt shank that is 150 mm long Determine the maximum height h of release so the stress in the bolt does no exceed 150 MPa Given do 4mm L 150mm mo 2 kg E 200GPa σallow 150MPa Solution Weight W mo g Section Property A π do 2 4 Static Displacement st W L E A Allowable Length Impact factor η 1 1 2 h st σmax W A 1 1 2 h st h st 2 σallow A W 1 2 1 h 5293 mm Ans Problem 1454 The collar has a mass of 5 kg and falls down the titanium Ti6A14V bar If the bar has a diameter of 20 mm determine the maximum stress developed in the bar if the weight is a dropped from a height of h 1 m b released from a height h 0 and c placed slowly on the flange at A Given do 20mm L 15m mo 5kg E 120GPa σY 924MPa Solution Weight W mo g Section Property A π do 2 4 Static Displacement st W L E A st 000195 mm a Drop Height h 1m Impact factor η 1 1 2 h st η 101349 Pmax W η σmax Pmax A σmax 1582 MPa Ans b Drop Height h 0mm Impact factor η 1 1 2 h st η 200 Pmax W η σmax Pmax A σmax 0312 MPa Ans c No impact η 1 Pmax W η σmax Pmax A σmax 0156 MPa Ans Note since all of the σmax σY 924 MPa the above analysis is valid Problem 1455 The collar has a mass of 5 kg and falls down the titanium Ti6A14V bar If the bar has a diameter of 20 mm determine if the weight can be released from rest at any point along the bar and not permanently damage the bar after striking the flange at A Given do 20mm L 15m mo 5kg E 120GPa σY 924MPa Solution Weight W mo g Section Property A π do 2 4 Static Displacement st W L E A st 000195 mm Drop Height hmax 15m Impact factor ηmax 1 1 2 hmax st ηmax 124104 Pmax W ηmax σmax Pmax A σmax 1937 MPa Note since σmax σY 924 MPa the weight can be released from rest at any point along the bar without causing permanently damage to the bar Ans Problem 1456 A cylinder having the dimensions shown is made from magnesium Am 1004T61 If it is struck by a rigid block having a weight of 4 kN and traveling at 06 ms determine the maximum stress in the cylinder Neglect the mass of the cylinder Given do 150mm E 447GPa W 4kN σY 152MPa L 05m v 06 m s Solution Section Property A π do 2 4 Eq Spring Constant k E A L k 157982841 kN m Conservation of Energy Ui Ur 1 2 W g v2 1 2 k max 2 max W g k v max 03049 mm Maximum Stress Pmax k max σmax Pmax A σmax 273 MPa Ans σY 152 MPa OK Problem 1457 The wideflange beam has a length of 2L a depth 2c and a constant EI Determine the maximum height h at which a weight W can be dropped on its end without exceeding a maximum elastic stress σmax in the beam Problem 1458 The sack of cement has a weight of 450 N 45 kg If it is dropped from rest at a height of h 12 m onto the center of the W250 X 58 structural steel A36 beam determine the maximum bending stress developed in the beam due to the impact Also what is the impact factor Given L 72m h 12m W 450N E 200GPa σY 250MPa Solution Use W 250x58 I 873 106 mm4 d 252mm Static Displacement st W L3 48E I st 020041 mm From Appendix C Maximum Stress Impact factor η 1 1 2 h st η 11044 Ans Maximum moment occurs at midspan where Mmax W L 4 Mmax 0810 kN m c 05d σmax Mmax c I η σmax 1291 MPa Ans Since σmax σY 250 MPa the above analysis is valid Problem 1459 The sack of cement has a weight of 450 N 45 kg Determine the maximum height h from which it can be dropped from rest onto the center of the W250 X 58 structural steel A36 beam so that the maximum bending stress due to impact does not exceed 210 MPa Given L 72m W 450N E 200GPa σallow 210MPa Solution Use W 250x58 I 873 106 mm4 d 252mm Static Displacement st W L3 48E I st 020041 mm From Appendix C Maximum Stress Impact factor η 1 1 2 h st Maximum moment occurs at midspan where Mmax W L 4 c 05d σmax Mmax c I η σmax σallow σallow W L d 8I 1 1 2 h st h st 2 8 I σallow W L d 1 2 1 h 3197 m Ans Problem 1460 A 200N 20kg weight is dropped from a height of h 06 m onto the center of the cantilevered A36 steel beam If the beam is a W250 X 22 determine the maximum bending stress developed in the beam Given L 15m h 06m W 200N E 200GPa σY 250MPa Solution Use W 250x22 I 288 106 mm4 d 254mm Static Displacement st W L3 3E I st 003906 mm From Appendix C Maximum Stress Impact factor η 1 1 2 h st η 17627 Maximum moment occurs at support A where Mmax W L Mmax 0300 kN m c 05d σmax Mmax c I η σmax 2332 MPa Ans Since σmax σY 250 MPa the above analysis is valid Problem 1461 If the maximum allowable bending stress for the W250 X 22 structural A36 steel beam is σallow 140 MPa determine the maximum height h from which a 250N 25kg weight can be released from rest and strike the center of the beam Given L 15m W 250N E 200GPa σallow 140MPa Solution Use W 250x22 I 288 106 mm4 d 254mm Static Displacement st W L3 3E I st 004883 mm From Appendix C Maximum Stress Impact factor η 1 1 2 h st Maximum moment occurs at midspan where Mmax W L c 05d σmax Mmax c I η σmax σallow σallow W L d 2I 1 1 2 h st h st 2 2 I σallow W L d 1 2 1 h 0171 m Ans Problem 1462 A 200N 20kg weight is dropped from a height of h 06 m onto the center of the cantilevered A36 steel beam If the beam is a W250 X 22 determine the vertical displacement of its end B due to the impact Given L 15m h 06m W 200N E 200GPa σY 250MPa Solution Use W 250x22 I 288 106 mm4 d 254mm Static Displacement From Appendix C At midspan st W L3 3E I st 003906 mm θst W L2 2E I θst 000003906 rad At end B st st θst L st 009766 mm Maximum Displacement Impact factor η 1 1 2 h st η 17627 Bmax st η Bmax 1721 mm Ans v 05 m s Problem 1463 The steel beam AB acts to stop the oncoming railroad car which has a mass of 10 Mg and is coasting towards it at v 05 ms Determine the maximum stress developed in the beam if it is struck at its center by the car The beam is simply supported and only horizontal forces occur at A and B Assume that the railroad car and the supporting framework for the beam remains rigid Also compute the maximum deflection of the beam Est 200 GPa σY 250 MPa Given L 2m mo 10 103 kg E 200GPa do 200mm σY 250MPa d 153mm Solution Weight W mo g Section Properties I do 4 12 Static Displacement At midspan From Appendix C st W L3 48E I st 061292 mm Equivalent Spring Stiffness k W st k 160000 kN m Maximum Displacement max st v2 g max 3953 mm Maximum Force Pmax k max Pmax 63246 kN Maximum Stress c 05do Maximum moment occurs at midspan Mmax 025Pmax L σmax Mmax c I σmax 2372 MPa Ans σY 250 MPa OK Problem 1464 The tugboat has a weight of 600 kN 60tonne and is traveling forward at 06 ms when it strikes the 300mmdiameter fender post AB used to protect a bridge pier If the post is made from treated white spruce and is assumed fixed at the river bed determine the maximum horizontal distance the top of the post will move due to the impact Assume the tugboat is rigid and neglect the effect of the water Given LBC 36m LCA 09m Amax 4108 mm E 965GPa v 06 m s Ans Solution Section Properties I π do 4 64 Maximum Displacement From Appendix C At point C Pmax 3 E I LBC 3 Cmax 1 Conservation of Energy 1 2 m v2 1 2 Pmax Cmax 1 2 W g v2 1 2 3 E I LBC 3 Cmax Cmax Cmax W v2 LBC 3 3gE I Cmax 29879 mm From Eq1 Pmax 3 E I LBC 3 Cmax Pmax 7372 kN From Appendix C At point C θCmax Pmax LBC 2 2E I θCmax 012450 rad At end A Amax Cmax θCmax LCA do 300mm W 600kN Problem 1465 The W250 X18 beam is made from A36 steel and is cantilevered from the wall at B The spring mounted on the beam has a stiffness of k 200 kNm If a weight of 40 N 4 kg is dropped onto the spring from a height of 1 m determine the maximum bending stress developed in the beam Given L 25m h 1m W 40N E 200GPa ksp 200kN m σY 250MPa Solution Use W 250x18 I 225 106 mm4 d 251mm Eq Spring Constants kb 3E I L3 From Appendix C kb 864 kN m Equilibrium requires Fsp Fb Hence ksp sp kb b sp kb ksp b 1 Conservation of Energy Ui Ur W h max 1 2 ksp sp 2 1 2 kb b 2 2 However max b sp 3 Substituting Eqs1 and 3 into 2 Given W h b kb ksp b 1 2 ksp kb ksp b 2 1 2 kb b 2 4 Solving Eq 4 Guess b 10mm b Find b b 42184 mm Maximum Stress c 05d Pmax kb b Pmax 364 kN Mmax Pmax L Mmax 911 kN m σmax Mmax c I σmax 5082 MPa Ans σY 250 MPa OK Problem 1466 The 375N 375kg block has a downward velocity of 06 ms when it is 09 m from the top of the wooden beam Determine the maximum bending stress in the beam due to the impact and compute the maximum deflection of its end D Ew 14 GPa Assume the material will not yield Given d 300mm L 30m h 09m LCD 15m E 14GPa W 375N v 06 m s Solution Section Property I d4 12 Eq Spring Constants kb 48E I L3 From Appendix C kb 16800 kN m Conservation of Energy Ui Ur 1 2 W g v2 W h max 1 2 kb b 2 1 However b max 2 Substituting Eq2 into 1 Given 1 2 W g v2 W h max 1 2 kb max 2 3 Solving Eq 3 Guess max 10mm max Find max max 6425 mm Maximum Stress c 05d b max Pmax kb b Pmax 10795 kN Maximum moment occurs at midspan Mmax Pmax L 4 Mmax 8096 kN m σmax Mmax c I σmax 1799 MPa Ans Maximum Displacement From Appendix C At point C θCmax Pmax L2 16E I θCmax 0006425 rad At end D Dmax θCmax LCD Dmax 9638 mm Ans Problem 1467 The 375N 375kg block has a downward velocity of 06 ms when it is 09 m from the top of the wood beam Determine the maximum bending stress in the beam due to the impact and compute the maximum deflection of point B Ew 14 GPa Given d 300mm L 30m h 09m LCD 15m E 14GPa W 375N v 06 m s Solution Section Property I d4 12 Eq Spring Constants kb 48E I L3 From Appendix C kb 16800 kN m Conservation of Energy Ui Ur 1 2 W g v2 W h max 1 2 kb b 2 1 However b max 2 Substituting Eq2 into 1 Given 1 2 W g v2 W h max 1 2 kb max 2 3 Solving Eq 3 Guess max 10mm max Find max max 6425 mm Maximum Stress c 05d b max Pmax kb b Pmax 10795 kN Maximum moment occurs at midspan Mmax Pmax L 4 Mmax 8096 kN m σmax Mmax c I σmax 1799 MPa Ans Problem 1468 Determine the maximum height h from which an 400 N 40 kg weight can be dropped onto the end of the A36 steel W150 X 18 beam without exceeding the maximum elastic stress Given L 3m W 04kN E 200GPa σY 250MPa Solution Use W 150x18 I 919 106 mm4 d 153mm Static Displacement at B Support Reactions ΣFy0 A C W 0 1 ΣΜC0 A L W L 0 2 Solving Eqs 1 and 2 A W A 04 kN C W A C 08 kN Internal Moment Function M1 A x1 M2 M1 L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 2 0 L x1 M1 2 d Ui 2 2E I 0 L x1 A x1 2 d Ui 0783 J External Work The external work done by the weight W is Ue 1 2 W Bst Conservation of Energy Ui Ue Ui 1 2 W Bst Bst 2 Ui W Bst 3917 mm Eq Spring Constant kB W Bst kB 10211 kN m Maximum Force c 05d Maximum moment occurs at midsupport Mmax Pmax L σmax Mmax c I σY Pmax L d 2I Pmax 2IσY L d Pmax 1001 kN Maximum Displacement at B max Pmax kB max 9804 mm Conservation of Energy Ui Ur W h max 1 2 kB max 2 h 1 2W kB max 2 max h 1129 m Ans Problem 1469 The 400 N 40 kg weight is dropped from rest at a height of h 1125 m onto the end of the A36 steel W150 X 18 beam Determine the maximum bending stress developed in the beam Given L 3m W 04kN h 1125m E 200GPa σY 250MPa Solution Use W 150x18 I 919 106 mm4 d 153mm Static Displacement at B Support Reactions ΣFy0 A C W 0 1 ΣΜC0 A L W L 0 2 Solving Eqs 1 and 2 A W A 04 kN C W A C 08 kN Internal Moment Function M1 A x1 M2 M1 L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 2 0 L x1 M1 2 d Ui 2 2E I 0 L x1 A x1 2 d Ui 0783 J External Work The external work done by the weight W is Ue 1 2 W Bst Conservation of Energy Ui Ue Ui 1 2 W Bst Bst 2 Ui W Bst 3917 mm Eq Spring Constant kB W Bst kB 10211 kN m Maximum Displacement at B B max Conservation of Energy Ui Ur Given W h max 1 2 kB max 2 1 Solving Eq 1 Guess max 10mm max Find max max 9788 mm Maximum Stress c 05d B max Pmax kB B Pmax 999 kN Maximum moment occurs at midsupport Mmax Pmax L σmax Mmax c I σmax 24960 MPa Ans σY 250 MPa OK 1 2 mo v2 1 2 kb beam 2 1 4 kb 2 ksp beam 2 Given L 18m c 75mm mo 18 103 kg I 300 106 mm4 E 2GPa σY 30MPa ksp 1500 kN m v 075 m s Solution Equilibrium This requires Fsp 1 2 Pbeam Then ksp sp 1 2 kb beam sp 1 2 kb ksp beam 1 Weight W mo g Static Displacement At midspan From Appendix C st W L3 48E I st 3575 mm Equivalent Spring Stiffness kb W st kb 493827 kN m Conservation of Energy Ui Ur 1 2 mo v2 1 2 kb beam 2 2 1 2 ksp sp 2 2 Substituting Eq1 into 2 Given Problem 1470 The car bumper is made of polycarbonatepolybutylene terephthalate If E 20 GPa determine the maximum deflection and maximum stress in the bumper if it strikes the rigid post when the car is coasting at v 075 ms The car has a mass of 180 Mg and the bumper can be considered simply supported on two spring supports connected to the rigid frame of the car For the bumper take I 300106 mm4 c 75 mm σY 30 MPa and k 15 MNm 3 Solving Eq 3 Guess beam 10mm beam Find beam beam 8803 mm Maximum Displacement From Eq1 sp 1 2 kb ksp beam sp 14490 mm Hence max sp beam max 2329 mm Ans Maximum Stress Pmax kb beam Pmax 4347 kN Maximum moment occurs at midspan Mmax Pmax L 4 Mmax 1956 kN m σmax Mmax c I σmax 489 MPa Ans σY 30 MPa OK θ 30deg Given LBC 15m P 4kN LAB LBC sin θ E 200GPa A 1250mm2 Solution Member Lengths Ans LAB 3000 m Member Real Forces NAB P NBC 0 Member Virtual Forces Apply Virtual Force F 1kN nAB F cos θ nAB 11547 kN nBC nAB sin θ nBC 05774 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 Sum p1 p2 nNLEA 1 Virtual Work Equation F Bh Sum E A Bh Sum E A 1 F Bh 00554 mm Problem 1471 Determine the horizontal displacement of joint B on the twomember frame Each A36 steel member has a crosssectional area of 1250 mm2 Problem 1472 Determine the horizontal displacement of joint B Each A36 steel member has a crosssectional area of 1250 mm2 Given LBC 1m LAC 15m P 3kN E 200GPa A 1250mm2 Solution θ atan LAC LBC θ 56310 deg Member Lengths LAB LBC 2 LAC 2 LAB 1803 m Member Real Forces NAB P sin θ NAB 36056 kN NBC NAB cos θ NBC 20000 kN Member Virtual Forces Apply Virtual Force F 1kN nAB 0 nAB 00000 kN nBC F nBC 10000 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 Sum p1 p2 nNLEA 1 Virtual Work Equation F Bh Sum E A Bh Sum E A 1 F Bh 800 10 3 mm Ans Problem 1473 Determine the vertical displacement of joint B Each A36 steel member has a crosssectional area of 1250 mm2 Given LBC 1m LAC 15m P 3kN E 200GPa A 1250mm2 Solution θ atan LAC LBC θ 56310 deg Member Lengths LAB LBC 2 LAC 2 LAB 1803 m Member Real Forces NAB P sin θ NAB 36056 kN NBC NAB cos θ NBC 20000 kN Member Virtual Forces Apply Virtual Force F 1kN nAB F sin θ nAB 12019 kN nBC nAB cos θ nBC 06667 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 Sum p1 p2 nNLEA 1 Virtual Work Equation F Bv Sum E A Bv Sum E A 1 F Bv 3658 10 3 mm Ans Problem 1474 Determine the horizontal displacement of point B Each A36 steel member has a crosssectional area of 1250 mm2 Given b 18m h 24m P 1kN E 200GPa A 1250mm2 Solution θ atan h b θ 53130 deg Member Lengths LAB b2 h2 LAB 3000 m LBC b2 h2 LBC 3000 m LAC 2 b LAC 3600 m Member Real Forces NAB P 2cos θ NAB 08333 kN NBC NAB NBC 08333 kN NAC NAB cos θ NAC 05000 kN Member Virtual Forces Apply Virtual Force F 1kN nAB F 2cos θ nAB 08333 kN nBC nAB nBC 08333 kN nAC nAB cos θ nAC 05000 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 AC n3 nAC N3 NAC L3 LAC p3 n3 N3 L3 Sum p1 p2 p3 nNLEA 1 Virtual Work Equation F Bh Sum E A Bh Sum E A 1 F Bh 2027 10 3 mm Ans Problem 1475 Determine the vertical displacement of point B Each A36 steel member has a crosssectional area of 1250 mm2 Given b 18m h 24m P 1kN E 200GPa A 1250mm2 Solution θ atan h b θ 53130 deg Member Lengths LAB b2 h2 LAB 3000 m LBC b2 h2 LBC 3000 m LAC 2 b LAC 3600 m Member Real Forces NAB P 2cos θ NAB 08333 kN NBC NAB NBC 08333 kN NAC NAB cos θ NAC 05000 kN Member Virtual Forces Apply Virtual Force F 1kN nAB F 2sin θ nAB 06250 kN nBC nAB nBC 06250 kN nAC nAB cos θ nAC 03750 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 AC n3 nAC N3 NAC L3 LAC p3 n3 N3 L3 Sum p1 p2 p3 nNLEA 1 Virtual Work Equation F Bv Sum E A Bv Sum E A 1 F Bv 270 10 3 mm Ans Problem 1476 Determine the horizontal displacement of point C Each A36 steel member has a crosssectional area of 400 mm2 Given b 15m h 2m P1 5kN E 200GPa A 400mm2 P2 10kN Solution θ atan h b θ 53130 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NDA 0 NCD P2 NCD 10kN NAC P2 sin θ NAC 125 kN NBC P1 NAC cos θ NBC 125 kN NAB NAC sin θ NAB 10kN Member Virtual Forces Apply Virtual Force F 1kN nDA 0 nCD 0 nAC 0 nBC F nAB 0 Member Virtual n Real N Length L Product p DA n1 nDA N1 NDA L1 LDA p1 n1 N1 L1 CD n2 nCD N2 NCD L2 LCD p2 n2 N2 L2 AC n3 nAC N3 NAC L3 LAC p3 n3 N3 L3 BC n4 nBC N4 NBC L4 LBC p4 n4 N4 L4 AB n5 nAB N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 nNLEA 1 Virtual Work Equation F Ch Sum E A Ch Sum E A 1 F Ch 0234 mm Ans Problem 1477 Determine the vertical displacement of point D Each A36 steel member has a crosssectional area of 400 mm2 Given b 15m h 2m P1 5kN E 200GPa A 400mm2 P2 10kN Solution θ atan h b θ 53130 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NDA 0 NCD P2 NCD 10kN NAC P2 sin θ NAC 125 kN NBC P1 NAC cos θ NBC 125 kN NAB NAC sin θ NAB 10kN Member Virtual Forces Apply Virual Force F 1kN nDA 0 nDA 0 nCD F nCD 1kN nAC F sin θ nAC 125 kN nBC nAC cos θ nBC 075 kN nAB nAC sin θ nAB 1kN Member Virtual n Real N Length L Product p DA n1 nDA N1 NDA L1 LDA p1 n1 N1 L1 CD n2 nCD N2 NCD L2 LCD p2 n2 N2 L2 AC n3 nAC N3 NAC L3 LAC p3 n3 N3 L3 BC n4 nBC N4 NBC L4 LBC p4 n4 N4 L4 AB n5 nAB N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 nNLEA 1 Virtual Work Equation F Dv Sum E A Dv Sum E A 1 F Dv 1164 mm Ans A 2800mm2 Given b 24m h 18m P 25kN E 200GPa LAE 3000 m Solution θ atan h b θ 36870 deg Member Lengths LBE h LAE b2 h2 LCE b2 h2 LCE 3000 m LAB b LEF b LAF h LBC b LDE b LCD h Member Real Forces NBE P NAE P 2sin θ NAE 208333 kN NCE P 2sin θ NCE 208333 kN NAB NAE cos θ NAB 166667 kN NBC NCE cos θ NBC 166667 kN NEF 0 NDE 0 NAF 0 NCD 0 Member Virtual Forces Apply Virtual Force F 1kN nBE F nAE F 2sin θ nAE 08333 kN nCE F 2sin θ nCE 08333 kN nAB nAE cos θ nAB 06667 kN nBC nCE cos θ nBC 06667 kN nEF 0 nDE 0 nAF 0 nCD 0 Problem 1478 Determine the vertical displacement of point B Each A36 steel member has a crosssectional area of 2800 mm2 Est 200 GPa Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 EF n3 nEF N3 NEF L3 LEF p3 n3 N3 L3 DE n4 nDE N4 NDE L4 LDE p4 n4 N4 L4 AF n5 nAF N5 NAF L5 LAF p5 n5 N5 L5 CD n6 nCD N6 NCD L6 LCD p6 n6 N6 L6 AE n7 nAE N7 NAE L7 LAE p7 n7 N7 L7 CE n8 nCE N8 NCE L8 LCE p8 n8 N8 L8 BE n9 nBE N9 NBE L9 LBE p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 nNLEA 1 Virtual Work Equation F Bv Sum E A Bv Sum E A 1 F Bv 03616 mm Ans A 2800mm2 Given b 24m h 18m P 25kN E 200GPa LAE 3000 m Solution θ atan h b θ 36870 deg Member Lengths LBE h LAE b2 h2 nCD 0 LCE b2 h2 LCE 3000 m LAB b LEF b LAF h LBC b LDE b LCD h Member Real Forces NBE P NAE P 2sin θ NAE 208333 kN NCE P 2sin θ NCE 208333 kN NAB NAE cos θ NAB 166667 kN NBC NCE cos θ NBC 166667 kN NEF 0 NDE 0 NAF 0 NCD 0 Member Virtual Forces Apply Virtual Force F 1kN nBE 0 nAE F 2sin θ nAE 08333 kN nCE F 2sin θ nCE 08333 kN nAB nAE cos θ nAB 06667 kN nBC nCE cos θ nBC 06667 kN nEF 0 nDE 0 nAF 0 Problem 1479 Determine the vertical displacement of point E Each A36 steel member has a crosssectional area of 2800 mm2 Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 EF n3 nEF N3 NEF L3 LEF p3 n3 N3 L3 DE n4 nDE N4 NDE L4 LDE p4 n4 N4 L4 AF n5 nAF N5 NAF L5 LAF p5 n5 N5 L5 CD n6 nCD N6 NCD L6 LCD p6 n6 N6 L6 AE n7 nAE N7 NAE L7 LAE p7 n7 N7 L7 CE n8 nCE N8 NCE L8 LCE p8 n8 N8 L8 BE n9 nBE N9 NBE L9 LBE p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 nNLEA 1 Virtual Work Equation F Bv Sum E A Bv Sum E A 1 F Bv 02813 mm Ans Problem 1480 Determine the horizontal displacement of point D Each A36 steel member has a crosssectional area of 300 mm2 Given h 3m b 3m P1 4kN E 200GPa A 300mm2 P2 2kN Solution θ atan 2 h b θ 63435 deg Member Lengths c 05 b 2 h2 LAE h LED h LCD c LEC 05 b LBC c LAC c Member Real Forces NCD P1 cos θ NCD 89443 kN NED NCD sin θ NED 8kN NAE NED NAE 8kN NEC P2 NEC 2 kN NAC 05 NEC cos θ NAC 22361 kN NBC NCD 05 NEC cos θ NBC 111803 kN Member Virtual Forces Apply Virtual Force F 1kN nCD F cos θ nCD 22361 kN nED nCD sin θ nED 2kN nAE nED nAE 2kN nEC 0 nEC 0kN nAC 0 nAC 0kN nBC nCD nBC 22361 kN Member Virtual n Real N Length L Product p CD n1 nCD N1 NCD L1 LCD p1 n1 N1 L1 ED n2 nED N2 NED L2 LED p2 n2 N2 L2 AE n3 nAE N3 NAE L3 LAE p3 n3 N3 L3 EC n4 nEC N4 NEC L4 LEC p4 n4 N4 L4 AC n5 nAC N5 NAC L5 LAC p5 n5 N5 L5 BC n6 nBC N6 NBC L6 LBC p6 n6 N6 L6 Sum p1 p2 p3 p4 p5 p6 nNLEA 1 Virtual Work Equation F Dh Sum E A Dh Sum E A 1 F Dh 4116 mm Ans Problem 1481 Determine the horizontal displacement of point E Each A36 steel member has a crosssectional area of 300 mm2 Given h 3m b 3m P1 4kN E 200GPa A 300mm2 P2 2kN Solution θ atan 2 h b θ 63435 deg Member Lengths c 05 b 2 h2 LAE h LED h LCD c LEC 05 b LBC c LAC c Member Real Forces NCD P1 cos θ NCD 89443 kN NED NCD sin θ NED 8kN NAE NED NAE 8kN NEC P2 NEC 2 kN NAC 05 NEC cos θ NAC 22361 kN NBC NCD 05 NEC cos θ NBC 111803 kN Member Virtual Forces Apply Virtual Force F 1kN nCD 0 nCD 0kN nED 0 nED 0kN nAE 0 nAE 0kN nEC F nEC 1 kN nAC 05 nEC cos θ nAC 1118 kN nBC 05 nEC cos θ nBC 1118 kN Member Virtual n Real N Length L Product p CD n1 nCD N1 NCD L1 LCD p1 n1 N1 L1 ED n2 nED N2 NED L2 LED p2 n2 N2 L2 AE n3 nAE N3 NAE L3 LAE p3 n3 N3 L3 EC n4 nEC N4 NEC L4 LEC p4 n4 N4 L4 AC n5 nAC N5 NAC L5 LAC p5 n5 N5 L5 BC n6 nBC N6 NBC L6 LBC p6 n6 N6 L6 Sum p1 p2 p3 p4 p5 p6 nNLEA 1 Virtual Work Equation F Dh Sum E A Dh Sum E A 1 F Dh 08885 mm Ans Problem 1482 Determine the horizontal displacement of joint B of the truss Each A36 steel member has a crosssectional area of 400 mm2 Given h 15m b 2m P2 5kN E 200GPa A 400mm2 P1 4kN Solution θ atan h b θ 36870 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NAB 0 NBC P1 NBC 4kN NAC P1 cos θ NAC 5 kN NCD P2 NAC sin θ NCD 2 kN NDA NAC cos θ NDA 4kN Member Virtual Forces Apply Virtual Force F 1kN nAB 0 nBC F nBC 1kN nAC F cos θ nAC 125 kN nCD nAC sin θ nCD 075 kN nDA nAC cos θ nDA 1kN Member Virtual n Real N Length L Product p DA n1 nDA N1 NDA L1 LDA p1 n1 N1 L1 CD n2 nCD N2 NCD L2 LCD p2 n2 N2 L2 AC n3 nAC N3 NAC L3 LAC p3 n3 N3 L3 BC n4 nBC N4 NBC L4 LBC p4 n4 N4 L4 AB n5 nAB N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 nNLEA 1 Virtual Work Equation F Bh Sum E A Bh Sum E A 1 F Bh 0367 mm Ans Problem 1483 Determine the vertical displacement of joint C of the truss Each A36 steel member has a crosssectional area of 400 mm2 Given h 15m b 2m P2 5kN E 200GPa A 400mm2 P1 4kN Solution θ atan h b θ 36870 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NAB 0 NBC P1 NBC 4kN NAC P1 cos θ NAC 5 kN NCD P2 NAC sin θ NCD 2 kN NDA NAC cos θ NDA 4kN Member Virtual Forces Apply Virtual Force F 1kN nAB 0 nBC 0 nAC 0 nCD F nDA 0 Member Virtual n Real N Length L Product p DA n1 nDA N1 NDA L1 LDA p1 n1 N1 L1 CD n2 nCD N2 NCD L2 LCD p2 n2 N2 L2 AC n3 nAC N3 NAC L3 LAC p3 n3 N3 L3 BC n4 nBC N4 NBC L4 LBC p4 n4 N4 L4 AB n5 nAB N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 nNLEA 1 Virtual Work Equation F Cv Sum E A Cv Sum E A 1 F Cv 00375 mm Ans Problem 1484 Determine the vertical displacement of joint A Each A36 steel member has a crosssectional area of 400 mm2 Given h 2m b1 15m b2 3m E 200GPa A 400mm2 P1 40kN P2 60kN Solution θ1 atan h b1 θ1 53130 deg θ2 atan h b2 θ2 33690 deg Support Reactions Cv 0 ΣFy0 Dv P1 P2 0 1 ΣFx0 Ch Dh 0 2 ΣΜD0 P1 b1 b2 P2 b2 Ch h 0 3 Solving Eqs 2 and 3 Ch P1 b1 b2 P2 b2 h Ch 180kN Dh Ch Dh 180 kN Solving Eq 1 Dv P1 P2 Dv 100kN Member Lengths c1 b1 2 h2 c2 b2 2 h2 LAE b1 LED b2 LBE h LAB c1 LBC b2 LBD c2 Member Virtual Forces Member Real Forces Apply Virtual Force F 1kN NAB P1 sin θ1 NAB 50kN nAB F sin θ1 nAB 125 kN NAE NAB cos θ1 NAE 30 kN nAE nAB cos θ1 nAE 075 kN NBE P2 NBE 60kN nBE 0 nBE 0kN NED NAE NED 30 kN nED nAE nED 075 kN NBC Ch NBC 180kN nBC F b1 b2 h nBC 225 kN NBD Dv sin θ2 NBD 1803 kN nBD F sin θ2 nBD 180 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 AE n2 nAE N2 NAE L2 LAE p2 n2 N2 L2 BE n3 nBE N3 NBE L3 LBE p3 n3 N3 L3 ED n4 nED N4 NED L4 LED p4 n4 N4 L4 BC n5 nBC N5 NBC L5 LBC p5 n5 N5 L5 BD n6 nBD N6 NBD L6 LBD p6 n6 N6 L6 Sum p1 p2 p3 p4 p5 p6 nNLEA 1 Virtual Work Equation F Av Sum E A Av Sum E A 1 F Av 3305 mm Ans Problem 1485 DetermiNe the vertical displacemeNt of joiNt C Each A36 steel member has a crosssectioNal area of 2800 mm2 Given b 4m h 3m E 200GPa A 2800mm2 P1 30kN P2 40kN Solution θ atan h b θ 36870 deg Support Reactions By symmetry AER ΣFy0 2R 2P1 P2 0 R 05 2P1 P2 R 50kN Member Lengths c b2 h2 c 5m LCH h LAI c LBH c LDH LBH LEG LAI LAB b LBC b LCD LBC LDE LAB LAJ h LBI h LDG LBI LEF LAJ LJI b LIH b LHG LIH LGF LJI Member Real Forces NCH P2 NAI R sin θ NBH P2 2sin θ NDH NBH NEG NAI NAB NAI cos θ NBC NAB NBH cos θ NCD NBC NDE NAB NAJ 0 NBI NAI sin θ NDG NBI NEF NAJ NJI 0 NIH NAI cos θ NHG NIH NGF NJI Member Virtual Forces Apply Virtual Force F 1kN Support Reactions By symmetry AER ΣFy0 2R F 0 R 05 F R 05 kN nCH F nAI R sin θ nBH F 2sin θ nDH nBH nEG nAI nAB nAI cos θ nBC nAB nBH cos θ nCD nBC nDE nAB nAJ 0 nBI nAI sin θ nDG nBI nEF nAJ nJI 0 nIH nAI cos θ nHG nIH nGF nJI Member Virtual N Real N Length L Product p CH n1 nCH N1 NCH L1 LCH p1 n1 N1 L1 AI n2 nAI N2 NAI L2 LAI p2 n2 N2 L2 AB n3 nAB N3 NAB L3 LAB p3 n3 N3 L3 AJ n4 nAJ N4 NAJ L4 LAJ p4 n4 N4 L4 JI n5 nJI N5 NJI L5 LJI p5 n5 N5 L5 BH n6 nBH N6 NBH L6 LBH p6 n6 N6 L6 BC n7 nBC N7 NBC L7 LBC p7 n7 N7 L7 BI n8 nBI N8 NBI L8 LBI p8 n8 N8 L8 IH n9 nIH N9 NIH L9 LIH p9 n9 N9 L9 DH n10 nDH N10 NDH L10 LDH p10 n10 N10 L10 CD n11 nCD N11 NCD L11 LCD p11 n11 N11 L11 DG n12 nDG N12 NDG L12 LDG p12 n12 N12 L12 HG n13 nHG N13 NHG L13 LHG p13 n13 N13 L13 EG n14 nEG N14 NEG L14 LEG p14 n14 N14 L14 DE n15 nDE N15 NDE L15 LDE p15 n15 N15 L15 EF n16 nEF N16 NEF L16 LEF p16 n16 N16 L16 GF n17 nGF N17 NGF L17 LGF p17 n17 N17 L17 Sum1 p1 p2 p3 p4 p5 p6 p7 p8 p9 Sum2 p10 p11 p12 p13 p14 p15 p16 p17 Sum Sum1 Sum2 nNLEA 1 Virtual Work Equation F Cv Sum E A Cv Sum E A 1 F Cv 5266 mm Ans Problem 1486 Determine the vertical displacement of joint H Each A36 steel member has a crosssectional area of 2800 mm2 Given b 4m h 3m E 200GPa A 2800mm2 P1 30kN P2 40kN Solution θ atan h b θ 36870 deg Support Reactions By symmetry AER ΣFy0 2R 2P1 P2 0 R 05 2P1 P2 R 50kN Member Lengths c b2 h2 c 5m LCH h LAI c LBH c LDH LBH LEG LAI LAB b LBC b LCD LBC LDE LAB LAJ h LBI h LDG LBI LEF LAJ LJI b LIH b LHG LIH LGF LJI Member Real Forces NCH P2 NAI R sin θ NBH P2 2sin θ NDH NBH NEG NAI NAB NAI cos θ NBC NAB NBH cos θ NCD NBC NDE NAB NAJ 0 NBI NAI sin θ NDG NBI NEF NAJ NJI 0 NIH NAI cos θ NHG NIH NGF NJI Member Virtual Forces Apply Virtual Force F 1kN Support Reactions By symmetry AER ΣFy0 2R F 0 R 05 F R 05 kN nCH 0 nAI R sin θ nBH F 2sin θ nDH nBH nEG nAI nAB nAI cos θ nBC nAB nBH cos θ nCD nBC nDE nAB nAJ 0 nBI nAI sin θ nDG nBI nEF nAJ nJI 0 nIH nAI cos θ nHG nIH nGF nJI Member Virtual N Real N Length L Product p CH n1 nCH N1 NCH L1 LCH p1 n1 N1 L1 AI n2 nAI N2 NAI L2 LAI p2 n2 N2 L2 AB n3 nAB N3 NAB L3 LAB p3 n3 N3 L3 AJ n4 nAJ N4 NAJ L4 LAJ p4 n4 N4 L4 JI n5 nJI N5 NJI L5 LJI p5 n5 N5 L5 BH n6 nBH N6 NBH L6 LBH p6 n6 N6 L6 BC n7 nBC N7 NBC L7 LBC p7 n7 N7 L7 BI n8 nBI N8 NBI L8 LBI p8 n8 N8 L8 IH n9 nIH N9 NIH L9 LIH p9 n9 N9 L9 DH n10 nDH N10 NDH L10 LDH p10 n10 N10 L10 CD n11 nCD N11 NCD L11 LCD p11 n11 N11 L11 DG n12 nDG N12 NDG L12 LDG p12 n12 N12 L12 HG n13 nHG N13 NHG L13 LHG p13 n13 N13 L13 EG n14 nEG N14 NEG L14 LEG p14 n14 N14 L14 DE n15 nDE N15 NDE L15 LDE p15 n15 N15 L15 EF n16 nEF N16 NEF L16 LEF p16 n16 N16 L16 GF n17 nGF N17 NGF L17 LGF p17 n17 N17 L17 Sum1 p1 p2 p3 p4 p5 p6 p7 p8 p9 Sum2 p10 p11 p12 p13 p14 p15 p16 p17 Sum Sum1 Sum2 nNLEA 1 Virtual Work Equation F Cv Sum E A Cv Sum E A 1 F Cv 5052 mm Ans Problem 1487 Determine the displacement of point C and the slope at point B EI is constant Problem 1488 Determine the displacement at point C EI is constant Problem 1489 Determine the slope at point C EI is constant Problem 1490 Determine the slope at point A EI is constant Problem 1491 Determine the displacement of point C of the beam made from A36 steel and having a moment of inertia of I 21106 mm4 Given a 15m b 3m P 40kN E 200GPa I 21 106 mm4 Solution Virtual Force F 1kN Real Support Reactions Support Reactions due to Virtual Force at C ΣFy0 A B P 0 1 ΣFy0 A B F 0 1 ΣΜB0 A b P a b 0 2 ΣΜB0 A b F a 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P a b b B P A A F a b B F A A 60kN B 20 kN A 05 kN B 15 kN Internal Moment Functions OA BA CB Real M1 P x1 M2 B x2 M3 0 Virtual m1 0 m2 A b x2 m3 F x3 L 0 mMEIdx 1 Virtual Work Equation F C 1 E I 0 a m1 M1 x1 d 0 b m2 M2 x2 d 0 a m3 M3 x3 d C 1 E I 1 F 0 0 b x2 A b x2 B x2 d 0 C 1071 mm Ans I 21 106 mm4 Given a 15m b 3m P 40kN E 200GPa Support Reactions due to Virtual Moment at B Solution Virtual Moment m 1kN m Real Support Reactions Ans ΣFy0 A B P 0 1 ΣFy0 A B 0 1 ΣΜB0 A b P a b 0 2 ΣΜB0 A b m 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P a b b B P A A m b B A A 60kN B 20 kN A 033 kN B 033 kN Internal Moment Functions OA BA CB Real M1 P x1 M2 B x2 M3 0 Virtual m1 0 m2 A b x2 m3 0 L 0 mMEIdx 1 θ Virtual Work Equation m θB 1 E I 0 a m1 M1 x1 d 0 b m2 M2 x2 d 0 a m3 M3 x3 d θB 1 E I 1 m 0 0 b x2 A b x2 B x2 d 0 θB 7143 10 3 rad θB 0409 deg clockwise Problem 1492 Determine the slope at B of the beam made from A36 steel and having a moment of inertia of I 21106 mm4 Problem 1493 Determine the displacement of point C of the W360 X 39 beam made from A36 steel Given a 15m b 3m P 40kN E 200GPa Solution Virtual Force F 1kN Use W 360x39 I 102 106 mm4 Real Support Reactions Support Reactions due to Virtual Force at C ΣFy0 A B 2P 0 1 ΣFy0 A B F 0 1 Due to symmetry A B 2 Due to symmetry A B 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P B P A 05F B 05F A 40kN B 40kN A 05 kN B 05 kN Internal Moment Functions OA BA DB Real M1 P x1 M2 P a M3 P x3 Virtual m1 0 m2 B x2 m3 0 for x2 05b L 0 mMEIdx 1 Virtual Work Equation F C 1 E I 0 a m1 M1 x1 d 2 0 05 b m2 M2 x2 d 0 a m3 M3 x3 d C 1 E I 1 F 0 2 0 05 b x2 B x2 P a d 0 C 331 mm Ans Problem 1494 Determine the slope at A of the W360 X 39 beam made from A36 steel Given a 15m b 3m P 40kN E 200GPa Solution Virtual Moment m 1kN m Use W 360x39 I 102 106 mm4 Real Support Reactions Support Reactions due to Virtual Moment at A ΣFy0 A B 2P 0 1 ΣFy0 A B 0 1 Due to symmetry A B 2 ΣΜB0 A b m 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P B P A m b B A A 40kN B 40kN A 033 kN B 033 kN Internal Moment Functions OA BA DB Real M1 P x1 M2 P a M3 P x3 Virtual m1 0 m2 B x2 m3 0 L 0 mMEIdx 1 θ Virtual Work Equation m θA 1 E I 0 a m1 M1 x1 d 0 b m2 M2 x2 d 0 a m3 M3 x3 d θA 1 E I 1 m 0 0 b x2 P a B x2 d 0 θA 4412 10 3 rad θA 0253 deg anticlockwise Ans do 38mm Given a 06m b 16m c 045m PD 14kN F 1kN PE 32kN E 200GPa L a b c Solution Virtual Force m3 C x3 Section Property I π do 4 64 I 10235 103 mm4 Real Support Reactions ΣFy0 A C PD PE 0 1 C PD PE A ΣΜC0 A L PD b c PE c 0 2 Solving Eqs 1 and 2 A 1 L PD b c PE c A 1626 kN A C F 0 C 2974 kN Support Reactions due to Virtual Force at B ΣFy0 1 ΣΜC0 A L F b a c 0 2 Solving Eqs 1 and 2 A 1 L F b a c C F A A 0547 kN C 0453 kN Internal Moment Functions AD DB EB CE Real M1 A x1 M2 A a x2 PD x2 M4 C c x4 PE x4 M3 C x3 Virtual m1 A x1 m2 A a x2 m4 C c x4 Problem 1495 Determine the displacement at B of the 38mmdiameter A36 steel shaft L 0 mMEIdx 1 Virtual Work Equation F B 1 E I 0 a m1 M1 x1 d 0 a m2 M2 x2 d 0 b a m4 M4 x4 d 0 c m3 M3 x3 d I1 0 a m1 M1 x1 d I1 0 a x1 A x1 A x1 d I1 006407 kN2 m3 I2 0 a m2 M2 x2 d I2 0 a x2 A a x2 A a x2 PD x2 d I2 031064 kN2 m3 I4 0 b a m4 M4 x4 d I4 0 b a x4 C c x4 C c x4 PE x4 d I4 051840 kN2 m3 I3 0 c m3 M3 x3 d I3 0 c x3 C x3 C x3 d I3 004090 kN2 m3 B 1 E I 1 F I1 I2 I4 I3 B 4563 mm Ans Problem 1496 Determine the slope of the 30mmdiameter A36 steel shaft at the bearing support A Given a 06m b 16m c 045m PD 14kN do 38mm PE 32kN E 200GPa L a b c Solution Virtual Moment m 1kN m Section Property I π do 4 64 I 10235 103 mm4 Real Support Reactions ΣFy0 A C PD PE 0 1 ΣΜC0 A L PD b c PE c 0 2 Solving Eqs 1 and 2 A 1 L PD b c PE c A 1626 kN C PD PE A C 2974 kN Support Reactions due to Virtual Moment at A ΣFy0 A C 0 1 ΣΜC0 A L m 0 2 Solving Eqs 1 and 2 A m L C A A 0377 kN C 0377 kN Internal Moment Functions AD ED CE Real M1 A x1 M4 C c x4 PE x4 M3 C x3 Virtual m1 m A x1 m4 C c x4 m3 C x3 L 0 mMEIdx 1 θ Virtual Work Equation m θA 1 E I 0 a m1 M1 x1 d 0 b m4 M4 x4 d 0 c m3 M3 x3 d I1 0 a m1 M1 x1 d I1 0 a x1 m A x1 A x1 d I1 024857 kN2 m3 I4 0 b m4 M4 x4 d I4 0 b x4 C c x4 C c x4 PE x4 d I4 084403 kN2 m3 I3 0 c m3 M3 x3 d I3 0 c x3 C x3 C x3 d I3 003408 kN2 m3 θA 1 E I 1 m I1 I4 I3 θA 55038 10 3 rad θA 3153 deg clockwise Ans Problem 1497 Determine the displacement at pulley B The A36 steel shaft has a diameter of 30 mm Given a 04m b 03m PD 4kN do 30mm PE 3kN E 200GPa PC 2kN L 2a 2 b Solution Virtual Force F 1kN 1 Section Property 2 I π do 4 64 I 3976 103 mm4 Real Support Reactions ΣFy0 A C PD PE PC 0 1 ΣΜC0 A L PD L a PE b PC 2 b 0 2 Solving Eqs 1 and 2 A 1 L PD L a PE b PC 2 b A 4357 kN C PD PE PC A C 4643 kN Support Reactions due to Virtual Force at B ΣFy0 A C F 1 ΣΜC0 A L F 2 b 0 2 Solving Eqs 1 and 2 A 2 b F L C F A A 04286 kN C 05714 kN Internal Moment Functions AD DB ED CE Real M1 A x1 M2 A a x2 PD x2 M4 C b x4 PE x4 M3 C x3 Virtual m1 A x1 m2 A a x2 m4 C b x4 m3 C x3 L 0 mMEIdx 1 Virtual Work Equation F B 1 E I 0 a m1 M1 x1 d 0 a m2 M2 x1 d 0 b m4 M4 x4 d 0 b m3 M3 x3 d I1 0 a m1 M1 x1 d I1 0 a x1 A x1 A x1 d I1 003984 kN2 m3 I2 0 a m2 M2 x2 d I2 0 a x2 A a x2 A a x2 PD x2 d I2 018743 kN2 m3 I4 0 b m4 M4 x4 d I4 0 b x4 C b x4 C b x4 PE x4 d I4 012857 kN2 m3 I3 0 b m3 M3 x3 d I3 0 b x3 C x3 C x3 d I3 002388 kN2 m3 B 1 E I 1 F I1 I2 I4 I3 B 4775 mm Ans Problem 1498 The simply supported beam having a square cross section is subjected to a uniform load w Determine the maximum deflection of the beam caused only by bending and caused by bending and shear Take E 3G Problem 1499 Determine the displacement at point C EI is constant Problem 14100 Determine theslope at point B EI is constant Problem 14101 The A36 steel beam has a moment of inertia of I 125106 mm4 Determine the displacement at D Given a 4m Mo 18kN m b 3m E 200GPa I 125 106 mm4 Solution Virtual Force F 1kN Real Support Reactions ΣFy0 B C 0 1 Due to symmetry B C 2 Solving Eqs 1 and 2 B 0 C 0 Support Reactions due to Virtual Force at D ΣFy0 B C F 0 1 Due to symmetry B C 2 Solving Eqs 1 and 2 B 05F C 05F B 05 kN C 05 kN Internal Moment Functions AB BD CD EC Real M1 Mo M2 Mo M3 Mo M4 Mo Virtual m1 0 m2 B x2 m3 C x3 m4 0 for x2 b L 0 mMEIdx 1 Virtual Work Equation F D 1 E I 0 a m1 M1 x1 d 0 b m2 M2 x2 d 0 b m3 M3 x3 d 0 a m4 M4 x4 d D 1 E I 1 F 0 0 b x2 B x2 Mo d 0 b x3 C x3 Mo d 0 D 324 mm Ans Problem 14102 The A36 steel beam has a moment of inertia of I 125106 mm4 Determine the slope at A Given a 4m Mo 18kN m b 3m E 200GPa I 125 106 mm4 Solution Virtual Moment m 1kN m Real Support Reactions ΣFy0 B C 0 1 Due to symmetry B C 2 Solving Eqs 1 and 2 B 0 C 0 Support Reactions due to Virtual Moment at A ΣFy0 B C 0 1 ΣΜC0 B 2 b m 0 2 Solving Eqs 1 and 2 B m 2 b C B B 01667 kN C 01667 kN Internal Moment Functions AB CB EC Real M1 Mo M3 Mo M4 Mo Virtual m1 m m3 C x3 m4 0 for x3 2b L 0 mMEIdx 1 θ Virtual Work Equation m θA 1 E I 0 a m1 M1 x1 d 0 2 b m3 M3 x3 d 0 a m4 M4 x4 d θA 1 E I 1 m 0 a m Mo x1 d 0 2 b x3 C x3 Mo d 0 θA 5040 10 3 rad θA 0289 deg anticlockwise Ans Problem 14103 The A36 structural steel beam has a moment of inertia of I 125106 mm4 Determine the slope of the beam at B Given a 4m Mo 18kN m b 3m E 200GPa I 125 106 mm4 Solution Virtual Moment m 1kN m Real Support Reactions ΣFy0 B C 0 1 Due to symmetry B C 2 Solving Eqs 1 and 2 B 0 C 0 Support Reactions due to Virtual Moment at B ΣFy0 B C 0 1 ΣΜC0 B 2 b m 0 2 Solving Eqs 1 and 2 B m 2 b C B B 01667 kN C 01667 kN Internal Moment Functions AB BC EC Real M1 Mo M2 Mo M4 Mo Virtual m1 0 m2 m B x2 m4 0 for x2 2b L 0 mMEIdx 1 θ Virtual Work Equation m θA 1 E I 0 a m1 M1 x1 d 0 2 b m2 M2 x2 d 0 a m4 M4 x4 d θB 1 E I 1 m 0 0 2 b x2 m B x2 Mo d 0 θB 2160 10 3 rad θB 0124 deg anticlockwise Ans Problem 14104 Determine the slope at A EI is constant Problem 14105 Determine the displacement at C EI is constant Problem 14106 Determine the slope at B EI is constant Problem 14107 The beam is made of oak for which Eo 11 GPa Determine the slope and displacement at A Given L 3m E 11GPa h 400mm wo 4kN m t 200mm Solution Section Property I t h3 12 a Applying Virtual Force at A F 1kN Internal Moment Functions clockwise Ans Real M1 x1 2 wo x1 L x1 3 M2 wo 2 L L 3 x2 wo 2 x2 2 Virtual m1 F x1 m2 F L x2 L 0 mMEIdx 1 Virtual Work Equation F A 1 E I 0 L m1 M1 x1 d 0 L m2 M2 x2 d A 1 E I 0 L x1 x1 x1 2 wo x1 L x1 3 d 0 L x2 L x2 wo 2 L L 3 x2 wo 2 x2 2 d A 2738 mm Ans b Applying Virtual Moment at A m 1kN m Internal Moment Functions Real M1 x1 2 wo x1 L x1 3 M2 wo 2 L L 3 x2 wo 2 x2 2 Virtual m1 m m2 m L 0 mMEIdx 1 θ Virtual Work Equation m θA 1 E I 0 L m1 M1 x1 d 0 L m2 M2 x2 d θA 1 E I 0 L x1 x1 2 wo x1 L x1 3 d 0 L x2 wo 2 L L 3 x2 wo 2 x2 2 d θA 5753 10 3 rad CB AC CB AC Problem 14108 Determine the displacement at B EI is constant Problem 14109 Determine the slope and displacement at point C EI is constant Problem 14110 Bar ABC has a rectangular cross section of 300mm by 100 mm Attached rod DB has a diameter of 20 mm If both members are made of A36 steel determine the vertical displacement of point C due to the loading Consider only the effect of bending in ABC and axial force in DB Given a 4m h 300mm P 20kN E 200GPa b 3m t 100mm do 20mm Solution Virtual Force F 1kN Section Properties Lo a2 b2 Ao π do 2 4 I t h3 12 Virtual Support Reactions ΣFy0 Av Bv F 0 1 ΣΜB0 Av b F b 0 2 Solving Eqs 1 and 2 Av F Av 1 F Bv F Av Bv 2F Axial Force in the Rod BD nBD Bv Lo a nBD 25 F Real Support Reactions Similarly Av 1 P Bv 2 P NBD 25 P Internal Moment and Force Functions AB BC BD Real M1 Av x1 M2 P x2 N3 NBD Virtual m1 Av x1 m2 F x2 n3 nBD nNLEA mMEIdx 1 L 0 Σ Virtual Work Equation F C 1 E I 0 b m1 M1 x1 d 0 b m2 M2 x2 d n3 N3 Lo E A C 1 E I 1 F 0 b x1 Av x1 Av x1 d 0 b x2 F x2 P x2 d 25 F 25 P Lo E Ao F C 1795 mm Ans 2 Given a 4m h 300mm P 20kN E 200GPa b 3m t 100mm do 20mm Solution Virtual Moment m 1kN m Section Properties Lo a2 b2 Ao π do 2 4 I t h3 12 Real Support Reactions ΣFy0 Av Bv P 0 1 ΣΜB0 Av b P b 0 2 Solving Eqs 1 and 2 Av P Av 1 P Bv P Av Bv 2P Axial Force in the Rod BD NBD Bv Lo a NBD 25 P Virtual Support Reactions ΣFy0 Av Bv 0 1 ΣΜB0 Av b m 0 Problem 14111 Bar ABC has a rectangular cross section of 300mm by 100 mm Attached rod DB has a diameter of 20 mm If both members are made of A36 steel determine the slope at A due to the loading Consider only the effect of bending in ABC and axial force in DB Solving Eqs 1 and 2 Av m b Av 03333 kN Bv Av Bv 03333 kN Axial Force in the Rod BD nBD Bv Lo a nBD 0417 kN Internal Moment and Force Functions AB BC BD Real M1 Av x1 M2 P x2 N3 NBD Virtual m1 m Av x1 m2 0 n3 nBD nNLEA mMEIdx 1 L 0 Σ θ Virtual Work Equation m θA 1 E I 0 b m1 M1 x1 d 0 b m2 M2 x2 d n3 N3 Lo E A θA 1 E I 1 m 0 b x1 m Av x1 Av x1 d 0 nBD 25 P Lo E Ao m θA 99120 10 6 rad anticlockwise Ans Problem 14112 Determine the vertical displacement of point A on the angle bracket due to the concentrated force P The bracket is fixed connected to its support EI is constant Consider only the effect of bending Problem 14113 The Lshaped frame is made from two segments each of length L and flexural stiffness EI If it is subjected to the uniform distributed load determine the horizontal displacement of the end C Problem 14114 The Lshaped frame is made from two segments each of length L and flexural stiffness EI If it is subjected to the uniform distributed load determine the vertical displacement of the point B Problem 14115 Determine the horizontal displacement of point C EI is constant There is a fixed support at A Consider only the effect of bending Given a 16m b 15m c 15m Pv 3kN Ph 4kN Mo 18kN m Solution Virtual Force F 1kN Internal Moment Functions CB BE EA Real M1 Mo Pv x1 M2 Mo Pv a M3 Mo Pv a Ph x3 Virtual m1 0 m2 F x2 m3 F b x3 L 0 mMEIdx 1 Virtual Work Equation Set EI 1kN m2 F C 1 E I 0 a m1 M1 x1 d 0 b m2 M2 x2 d 0 c m3 M3 x3 d C 1 EI 1 F 0 0 b x2 F x2 Mo Pv a d 0 c x3 F b x3 Mo Pv a Ph x3 d C 4095 m C 4095 EI m Ans Problem 14116 The ring rests on the rigid surface and is subjected to the vertical load P Determine the vertical displacement at B EI is constant Problem 14117 Solve Prob 1471 using Castiglianos theorem Given LBC 15m P 4kN θ 30deg E 200GPa A 1250mm2 Solution Member Lengths LAB LBC sin θ LAB 3000 m Member Real Forces NAB P NBC 0 Member Virtual Forces Apply Virtual Force P nAB P cos θ nAB P sec θ nBC nAB sin θ nBC P tan θ n P Member Set P0 Real N Length L Product p AB n1 sec θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 tan θ N2 NBC L2 LBC p2 n2 N2 L2 Sum p1 p2 PNLEA n Castiglianos Theorem Bh Sum E A Bh 00554 mm Ans Problem 14118 Solve Prob 1473 using Castiglianos theorem Given LBC 1m LAC 15m P 3kN E 200GPa A 1250mm2 Solution θ atan LAC LBC θ 56310 deg Member Lengths LAB LBC 2 LAC 2 LAB 1803 m Member Real Forces NAB P sin θ NAB 36056 kN NBC NAB cos θ NBC 20000 kN Member Virtual Forces Apply Virtual Force P nAB P sin θ nAB P csc θ nBC P cot θ nBC nAB cos θ n P Member Set P0 Real N Length L Product p AB n1 csc θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 cot θ N2 NBC L2 LBC p2 n2 N2 L2 Sum p1 p2 PNLEA n Castiglianos Theorem Bv Sum E A Bv 3658 10 3 mm Ans Problem 14119 Solve Prob 1474 using Castiglianos theorem Given b 18m h 24m P 1kN E 200GPa A 1250mm2 Solution θ atan h b θ 53130 deg Member Lengths LAB b2 h2 LAB 3000 m LBC b2 h2 LBC 3000 m LAC 2 b LAC 3600 m Member Real Forces NAB P 2cos θ NAB 08333 kN NBC NAB NBC 08333 kN NAC NAB cos θ NAC 05000 kN Member Virtual Forces Apply Virtual Force P nAB P 2cos θ nAB 05 P sec θ nBC nAB nBC 05 P sec θ nAC nAB cos θ nAC 05 P n P Member Set P0 Real N Length L Product p AB n1 05 sec θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 05 sec θ N2 NBC L2 LBC p2 n2 N2 L2 AC n3 05 N3 NAC L3 LAC p3 n3 N3 L3 Sum p1 p2 p3 PNLEA n Castiglianos Theorem Bh Sum E A Bh 2027 10 3 mm Ans Problem 14120 Solve Prob 1472 using Castiglianos theorem Given LBC 1m LAC 15m P 3kN E 200GPa A 1250mm2 Solution θ atan LAC LBC θ 56310 deg Member Lengths LAB LBC 2 LAC 2 LAB 1803 m Member Real Forces NAB P sin θ NAB 36056 kN NBC NAB cos θ NBC 20000 kN Member Virtual Forces Apply Virtual Force P nAB 0 nBC P n P Member Set P0 Real N Length L Product p AB n1 0 N1 NAB L1 LAB p1 n1 N1 L1 BC n2 1 N2 NBC L2 LBC p2 n2 N2 L2 Sum p1 p2 PNLEA n Castiglianos Theorem Bh Sum E A Bh 800 10 3 mm Ans Problem 14121 Solve Prob 1475 using Castiglianos theorem Given b 18m h 24m P 1kN E 200GPa A 1250mm2 Solution θ atan h b θ 53130 deg Member Lengths LAB b2 h2 LAB 3000 m LBC b2 h2 LBC 3000 m LAC 2 b LAC 3600 m Member Real Forces NAB P 2cos θ NAB 08333 kN NBC NAB NBC 08333 kN NAC NAB cos θ NAC 05000 kN Member Virtual Forces Apply Virtual Force P nAB P 2sin θ nAB 05 Pcsc θ nBC nAB nBC 05 P csc θ nAC nAB cos θ nAC 05 P cot θ n P Member Set P0 Real N Length L Product p AB n1 05 csc θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 05 csc θ N2 NBC L2 LBC p2 n2 N2 L2 AC n3 05 cot θ N3 NAC L3 LAC p3 n3 N3 L3 Sum p1 p2 p3 PNLEA n Castiglianos Theorem Bv Sum E A Bv 270 10 3 mm Ans Problem 14122 Solve Prob 1476 using Castiglianos theorem Given b 15m h 2m P1 5kN E 200GPa A 400mm2 P2 10kN Solution θ atan h b θ 53130 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NDA 0 NCD P2 NCD 10kN NAC P2 sin θ NAC 125 kN NBC P1 NAC cos θ NBC 125 kN NAB NAC sin θ NAB 10kN Member Virtual Forces Apply Virtual Force P nDA 0 nCD 0 nAC 0 nBC P nAB 0 n P Member Set P0 Real N Length L Product p DA n1 0 N1 NDA L1 LDA p1 n1 N1 L1 CD n2 0 N2 NCD L2 LCD p2 n2 N2 L2 AC n3 0 N3 NAC L3 LAC p3 n3 N3 L3 BC n4 1 N4 NBC L4 LBC p4 n4 N4 L4 AB n5 0 N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 PNLEA n Castiglianos Theorem Ch Sum E A Ch 0234 mm Ans Problem 14123 Solve Prob 1477 using Castiglianos theorem Given b 15m h 2m P1 5kN E 200GPa A 400mm2 P2 10kN Solution θ atan h b θ 53130 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NDA 0 NCD P2 NCD 10kN NAC P2 sin θ NAC 125 kN NBC P1 NAC cos θ NBC 125 kN NAB NAC sin θ NAB 10kN Member Virtual Forces Apply Virual Force P nDA 0 nCD P nAC P sin θ nAC P csc θ nBC nAC cos θ nBC P cot θ nAB nAC sin θ nAB P n P Member Set P0 Real N Length L Product p DA n1 0 N1 NDA L1 LDA p1 n1 N1 L1 CD n2 1 N2 NCD L2 LCD p2 n2 N2 L2 AC n3 csc θ N3 NAC L3 LAC p3 n3 N3 L3 BC n4 cot θ N4 NBC L4 LBC p4 n4 N4 L4 AB n5 1 N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 PNLEA n Castiglianos Theorem Dv Sum E A Dv 1164 mm Ans Problem 14124 Solve Prob 1480 using Castiglianos theorem Given h 3m b 3m P1 4kN E 200GPa A 300mm2 P2 2kN Solution θ atan 2 h b θ 63435 deg Member Lengths c 05 b 2 h2 LAE h LED h LCD c LEC 05 b LBC c LAC c Member Real Forces NCD P1 cos θ NCD 89443 kN NED NCD sin θ NED 8kN NAE NED NAE 8kN NEC P2 NEC 2 kN NAC 05 NEC cos θ NAC 22361 kN NBC NCD 05 NEC cos θ NBC 111803 kN Member Virtual Forces Apply Virtual Force P nCD P cos θ nCD P sec θ nED nCD sin θ nED P tan θ nAE nED nAE P tan θ nEC 0 nAC 0 nBC nCD nBC P sec θ n P Real N Length L Product p Member Set P0 CD n1 sec θ N1 NCD L1 LCD p1 n1 N1 L1 ED n2 tan θ N2 NED L2 LED p2 n2 N2 L2 AE n3 tan θ N3 NAE L3 LAE p3 n3 N3 L3 EC n4 0 N4 NEC L4 LEC p4 n4 N4 L4 AC n5 0 N5 NAC L5 LAC p5 n5 N5 L5 BC n6 sec θ N6 NBC L6 LBC p6 n6 N6 L6 Sum p1 p2 p3 p4 p5 p6 PNLEA n Castiglianos Theorem Dh Sum E A Dh 4116 mm Ans Problem 14125 Solve Prob 1478 using Castiglianos theorem Given b 24m h 18m P 25kN E 200GPa A 2800mm2 Solution θ atan h b θ 36870 deg Member Lengths LBE h LAE b2 h2 LAE 3000 m LCE b2 h2 LCE 3000 m LAB b LEF b LAF h LBC b LDE b LCD h Member Real Forces NBE P NAE P 2sin θ NAE 208333 kN NCE P 2sin θ NCE 208333 kN NAB NAE cos θ NAB 166667 kN NBC NCE cos θ NBC 166667 kN NEF 0 NDE 0 NAF 0 NCD 0 Member Virtual Forces Apply Virtual Force P nBE P nAE P 2sin θ nAE 05 Pcsc θ nCE P 2sin θ nCE 05 Pcsc θ nAB nAE cos θ nAB 05P cot θ nBC nCE cos θ nBC 05P cot θ nEF 0 nDE 0 nAF 0 nCD 0 n P Member Set P0 Real N Length L Product p AB n1 05 cot θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 05 cot θ N2 NBC L2 LBC p2 n2 N2 L2 EF n3 0 N3 NEF L3 LEF p3 n3 N3 L3 DE n4 0 N4 NDE L4 LDE p4 n4 N4 L4 AF n5 0 N5 NAF L5 LAF p5 n5 N5 L5 CD n6 0 N6 NCD L6 LCD p6 n6 N6 L6 AE n7 05 csc θ N7 NAE L7 LAE p7 n7 N7 L7 CE n8 05 csc θ N8 NCE L8 LCE p8 n8 N8 L8 BE n9 1 N9 NBE L9 LBE p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 PNLEA n Castiglianos Theorem Bv Sum E A Bv 03616 mm Ans Problem 14126 Solve Prob 1479 using Castiglianos theorem Given b 24m h 18m P 25kN E 200GPa A 2800mm2 Solution θ atan h b θ 36870 deg Member Lengths LBE h LAE b2 h2 LAE 3000 m LCE b2 h2 LCE 3000 m LAB b LEF b LAF h LBC b LDE b LCD h Member Real Forces NBE P NAE P 2sin θ NAE 208333 kN NCE P 2sin θ NCE 208333 kN NAB NAE cos θ NAB 166667 kN NBC NCE cos θ NBC 166667 kN NEF 0 NDE 0 NAF 0 NCD 0 Member Virtual Forces Apply Virtual Force P nBE 0 nAE P 2sin θ nAE 05 Pcsc θ nCE P 2sin θ nCE 05 Pcsc θ nAB nAE cos θ nAB 05P cot θ nBC nCE cos θ nBC 05P cot θ nEF 0 nDE 0 nAF 0 nCD 0 n P Member Set P0 Real N Length L Product p AB n1 05 cot θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 05 cot θ N2 NBC L2 LBC p2 n2 N2 L2 EF n3 0 N3 NEF L3 LEF p3 n3 N3 L3 DE n4 0 N4 NDE L4 LDE p4 n4 N4 L4 AF n5 0 N5 NAF L5 LAF p5 n5 N5 L5 CD n6 0 N6 NCD L6 LCD p6 n6 N6 L6 AE n7 05 csc θ N7 NAE L7 LAE p7 n7 N7 L7 CE n8 05 csc θ N8 NCE L8 LCE p8 n8 N8 L8 BE n9 0 N9 NBE L9 LBE p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 PNLEA n Castiglianos Theorem Bv Sum E A Bv 02813 mm Ans Problem 14127 Solve Prob 1481 using Castiglianos theorem Given h 3m b 3m P1 4kN E 200GPa A 300mm2 P2 2kN Solution θ atan 2 h b θ 63435 deg Member Lengths c 05 b 2 h2 LAE h LED h LCD c LEC 05 b LBC c LAC c Member Real Forces NCD P1 cos θ NCD 89443 kN NED NCD sin θ NED 8kN NAE NED NAE 8kN NEC P2 NEC 2 kN NAC 05 NEC cos θ NAC 22361 kN NBC NCD 05 NEC cos θ NBC 111803 kN Member Virtual Forces Apply Virtual Force P nCD 0 nED 0 nAE 0 nEC P nAC 05 nEC cos θ nAC 05P sec θ nBC 05 nEC cos θ nBC 05 P sec θ n P Member Set P0 Real N Length L Product p CD n1 0 N1 NCD L1 LCD p1 n1 N1 L1 ED n2 0 N2 NED L2 LED p2 n2 N2 L2 AE n3 0 N3 NAE L3 LAE p3 n3 N3 L3 EC n4 1 N4 NEC L4 LEC p4 n4 N4 L4 AC n5 05 sec θ N5 NAC L5 LAC p5 n5 N5 L5 BC n6 05 sec θ N6 NBC L6 LBC p6 n6 N6 L6 Sum p1 p2 p3 p4 p5 p6 PNLEA n Castiglianos Theorem Dh Sum E A Dh 08885 mm Ans Problem 14128 Solve Prob 1484 using Castiglianos theorem Given h 2m b1 15m b2 3m E 200GPa A 400mm2 P1 40kN P2 60kN Solution θ1 atan h b1 θ1 53130 deg θ2 atan h b2 θ2 33690 deg Support Reactions Cv 0 ΣFy0 Dv P1 P2 0 1 ΣFx0 Ch Dh 0 2 ΣΜD0 P1 b1 b2 P2 b2 Ch h 0 3 Solving Eqs 2 and 3 Ch P1 b1 b2 P2 b2 h Ch 180kN Dh Ch Dh 180 kN Solving Eq 1 Dv P1 P2 Dv 100kN Member Lengths c1 b1 2 h2 c2 b2 2 h2 LAE b1 LED b2 LBE h LAB c1 LBC b2 LBD c2 Member Virtual Forces Member Real Forces Apply Virtual Force P NAB P1 sin θ1 NAB 50kN nAB P sin θ1 nAB P csc θ1 NAE NAB cos θ1 NAE 30 kN nAE nAB cos θ1 nAE P cot θ1 NBE P2 NBE 60kN nBE 0 NED NAE NED 30 kN nED nAE nED P cot θ1 NBC Ch NBC 180kN nBC P b1 b2 h NBD Dv sin θ2 NBD 1803 kN nBD P sin θ2 nBD P csc θ2 n P Member Set P0 Real N Length L Product p AB n1 csc θ1 N1 NAB L1 LAB p1 n1 N1 L1 AE n2 cot θ1 N2 NAE L2 LAE p2 n2 N2 L2 BE n3 0 N3 NBE L3 LBE p3 n3 N3 L3 ED n4 cot θ1 N4 NED L4 LED p4 n4 N4 L4 BC n5 b1 b2 h N5 NBC L5 LBC p5 n5 N5 L5 BD n6 csc θ2 N6 NBD L6 LBD p6 n6 N6 L6 Sum p1 p2 p3 p4 p5 p6 PNLEA n Castiglianos Theorem Av Sum E A Av 3305 mm Ans Problem 14129 Solve Prob 1482 using Castiglianos theorem Given h 15m b 2m P2 5kN E 200GPa A 400mm2 P1 4kN Solution θ atan h b θ 36870 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NAB 0 NBC P1 NBC 4kN NAC P1 cos θ NAC 5 kN NCD P2 NAC sin θ NCD 2 kN NDA NAC cos θ NDA 4kN Member Virtual Forces Apply Virtual Force P nAB 0 nBC P nAC P cos θ nAC P sec θ nCD nAC sin θ nCD P tan θ nDA nAC cos θ nDA P n P Member Set P0 Real N Length L Product p DA n1 1 N1 NDA L1 LDA p1 n1 N1 L1 CD n2 tan θ N2 NCD L2 LCD p2 n2 N2 L2 AC n3 sec θ N3 NAC L3 LAC p3 n3 N3 L3 BC n4 1 N4 NBC L4 LBC p4 n4 N4 L4 AB n5 0 N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 PNLEA n Castiglianos Theorem Bh Sum E A Bh 0367 mm Ans Problem 14130 Solve Prob 1483 using Castiglianos theorem Given h 15m b 2m P2 5kN E 200GPa A 400mm2 P1 4kN Solution θ atan h b θ 36870 deg Member Lengths LCD h LAB h LAC b2 h2 LAC 25 m LDA b LBC b Member Real Forces NAB 0 NBC P1 NBC 4kN NAC P1 cos θ NAC 5 kN NCD P2 NAC sin θ NCD 2 kN NDA NAC cos θ NDA 4kN Member Virtual Forces Apply Virtual Force P nAB 0 nBC 0 nAC 0 nCD P nDA 0 n P Member Set P0 Real N Length L Product p DA n1 0 N1 NDA L1 LDA p1 n1 N1 L1 CD n2 1 N2 NCD L2 LCD p2 n2 N2 L2 AC n3 0 N3 NAC L3 LAC p3 n3 N3 L3 BC n4 0 N4 NBC L4 LBC p4 n4 N4 L4 AB n5 0 N5 NAB L5 LAB p5 n5 N5 L5 Sum p1 p2 p3 p4 p5 PNLEA n Castiglianos Theorem Cv Sum E A Cv 00375 mm Ans Problem 14131 Solve Prob 1485 using Castiglianos theorem Given b 4m h 3m E 200GPa A 2800mm2 P1 30kN P2 40kN Solution θ atan h b θ 36870 deg Support Reactions By symmetry AER ΣFy0 2R 2P1 P2 0 R 05 2P1 P2 R 50kN Member Lengths c b2 h2 c 5m LCH h LAI c LBH c LDH LBH LEG LAI LAB b LBC b LCD LBC LDE LAB LAJ h LBI h LDG LBI LEF LAJ LJI b LIH b LHG LIH LGF LJI Member Real Forces NCH P2 NAI R sin θ NBH P2 2sin θ NDH NBH NEG NAI NAB NAI cos θ NBC NAB NBH cos θ NCD NBC NDE NAB NAJ 0 NBI NAI sin θ NDG NBI NEF NAJ NJI 0 NIH NAI cos θ NHG NIH NGF NJI Member Virtual Forces Apply Virtual Force P Support Reactions By symmetry AER ΣFy0 2R F 0 R 05 P nCH P nAI R sin θ nBH P 2sin θ nDH nBH nEG nAI nAB nAI cos θ nBC nAB nBH cos θ nCD nBC nDE nAB nAJ 0 nBI nAI sin θ nDG nBI nEF nAJ nJI 0 nIH nAI cos θ nHG nIH nGF nJI nCH P nAI 05 P csc θ nBH 05 P csc θ nDH nBH nEG nAI nAB 05 P cot θ nBC P cot θ nCD nBC nDE nAB nAJ 0 nBI 05 P nDG nBI nEF nAJ nJI 0 nIH 05 P cot θ nHG nIH nGF nJI n P Member Set P0 Real N Length L Product p CH n1 1 N1 NCH L1 LCH p1 n1 N1 L1 AI n2 05 csc θ N2 NAI L2 LAI p2 n2 N2 L2 AB n3 05 cot θ N3 NAB L3 LAB p3 n3 N3 L3 AJ n4 0 N4 NAJ L4 LAJ p4 n4 N4 L4 JI n5 0 N5 NJI L5 LJI p5 n5 N5 L5 BH n6 05 csc θ N6 NBH L6 LBH p6 n6 N6 L6 BC n7 cot θ N7 NBC L7 LBC p7 n7 N7 L7 BI n8 05 N8 NBI L8 LBI p8 n8 N8 L8 IH n9 05 cot θ N9 NIH L9 LIH p9 n9 N9 L9 DH n10 05 csc θ N10 NDH L10 LDH p10 n10 N10 L10 CD n11 cot θ N11 NCD L11 LCD p11 n11 N11 L11 DG n12 05 N12 NDG L12 LDG p12 n12 N12 L12 HG n13 05 cot θ N13 NHG L13 LHG p13 n13 N13 L13 EG n14 05 csc θ N14 NEG L14 LEG p14 n14 N14 L14 DE n15 05 cot θ N15 NDE L15 LDE p15 n15 N15 L15 EF n16 0 N16 NEF L16 LEF p16 n16 N16 L16 GF n17 0 N17 NGF L17 LGF p17 n17 N17 L17 Sum1 p1 p2 p3 p4 p5 p6 p7 p8 p9 Sum2 p10 p11 p12 p13 p14 p15 p16 p17 Sum Sum1 Sum2 PNLEA n Castiglianos Theorem Cv Sum E A Cv 5266 mm Ans Problem 14132 Solve Prob 1486 using Castiglianos theorem Given b 4m h 3m E 200GPa A 2800mm2 P1 30kN P2 40kN Solution θ atan h b θ 36870 deg Support Reactions By symmetry AER ΣFy0 2R 2P1 P2 0 R 05 2P1 P2 R 50kN Member Lengths c b2 h2 c 5m LCH h LAI c LBH c LDH LBH LEG LAI LAB b LBC b LCD LBC LDE LAB LAJ h LBI h LDG LBI LEF LAJ LJI b LIH b LHG LIH LGF LJI Member Real Forces NCH P2 NAI R sin θ NBH P2 2sin θ NDH NBH NEG NAI NAB NAI cos θ NBC NAB NBH cos θ NCD NBC NDE NAB NAJ 0 NBI NAI sin θ NDG NBI NEF NAJ NJI 0 NIH NAI cos θ NHG NIH NGF NJI Member Virtual Forces Apply Virtual Force P Support Reactions By symmetry AER ΣFy0 2R F 0 R 05 P nCH 0 nAI R sin θ nBH P 2sin θ nDH nBH nEG nAI nAB nAI cos θ nBC nAB nBH cos θ nCD nBC nDE nAB nAJ 0 nBI nAI sin θ nDG nBI nEF nAJ nJI 0 nIH nAI cos θ nHG nIH nGF nJI nCH 0 nAI 05 P csc θ nBH 05 P csc θ nDH nBH nEG nAI nAB 05 P cot θ nBC P cot θ nCD nBC nDE nAB nAJ 0 nBI 05 P nDG nBI nEF nAJ nJI 0 nIH 05 P cot θ nHG nIH nGF nJI n P Member Set P0 Real N Length L Product p CH n1 0 N1 NCH L1 LCH p1 n1 N1 L1 AI n2 05 csc θ N2 NAI L2 LAI p2 n2 N2 L2 AB n3 05 cot θ N3 NAB L3 LAB p3 n3 N3 L3 AJ n4 0 N4 NAJ L4 LAJ p4 n4 N4 L4 JI n5 0 N5 NJI L5 LJI p5 n5 N5 L5 BH n6 05 csc θ N6 NBH L6 LBH p6 n6 N6 L6 BC n7 cot θ N7 NBC L7 LBC p7 n7 N7 L7 BI n8 05 N8 NBI L8 LBI p8 n8 N8 L8 IH n9 05 cot θ N9 NIH L9 LIH p9 n9 N9 L9 DH n10 05 csc θ N10 NDH L10 LDH p10 n10 N10 L10 CD n11 cot θ N11 NCD L11 LCD p11 n11 N11 L11 DG n12 05 N12 NDG L12 LDG p12 n12 N12 L12 HG n13 05 cot θ N13 NHG L13 LHG p13 n13 N13 L13 EG n14 05 csc θ N14 NEG L14 LEG p14 n14 N14 L14 DE n15 05 cot θ N15 NDE L15 LDE p15 n15 N15 L15 EF n16 0 N16 NEF L16 LEF p16 n16 N16 L16 GF n17 0 N17 NGF L17 LGF p17 n17 N17 L17 Sum1 p1 p2 p3 p4 p5 p6 p7 p8 p9 Sum2 p10 p11 p12 p13 p14 p15 p16 p17 Sum Sum1 Sum2 PNLEA n Castiglianos Theorem Cv Sum E A Cv 5052 mm Ans Problem 14133 Solve Prob 1487 using Castiglianos theorem Problem 14134 Solve Prob 1489 using Castiglianos theorem Problem 14135 Solve Prob 1490 using Castiglianos theorem Problem 14136 Solve Prob 1488 using Castiglianos theorem Problem 14137 Solve Prob 1491 using Castiglianos theorem Given a 15m b 3m P 40kN E 200GPa I 21 106 mm4 Solution Apply a Virtual Force P at C Real Support Reactions Support Reactions due to Virtual Force at C ΣFy0 A B P 0 1 ΣFy0 A B P 0 1 ΣΜB0 A b P a b 0 2 ΣΜB0 A b P a 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P a b b B P A A P a b B P a b b A 60kN B 20 kN Internal Moment Functions OA BA CB Real M1 P x1 M2 B x2 M3 0 Virtual m1 0 m2 A b x2 m3 P x3 m P 0 a b b x2 x3 Set P0 0 a b b x2 x3 L 0 m PMEI dx Castiglianos Theorem C 1 E I 0 a x1 P m1 d d M1 d 0 b x2 P m2 d d M2 d 0 a x3 P m3 d d M3 d C 1 E I 0 0 b x2 a b b x2 B x2 d 0 C 1071 mm Ans Problem 14138 Solve Prob 1493 using Castiglianos theorem Given a 15m b 3m P 40kN E 200GPa Solution Apply a Virtual Force P at C Use W 360x39 I 102 106 mm4 Real Support Reactions Support Reactions due to Virtual Force at C ΣFy0 A B 2P 0 1 ΣFy0 A B P 0 1 Due to symmetry A B 2 Due to symmetry A B 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P B P A 05P B 05P A 40kN B 40kN Internal Moment Functions OA BA DB Real M1 P x1 M2 P a M3 P x3 Virtual m1 0 m2 B x2 m3 0 for x2 05b m P 0 05x2 0 Set P0 0 05x2 0 L 0 m PMEI dx Castiglianos Theorem C 1 E I 0 a x1 P m1 d d M1 d 2 0 05 b x2 P m2 d d M2 d 0 a x3 P m3 d d M3 d C 1 E I 0 2 0 05 b x2 05x2 P a d 0 C 331 mm Ans Problem 14139 Solve Prob 1494 using Castiglianos theorem Given a 15m b 3m P 40kN E 200GPa Solution Apply a Virtual Moment M at A Use W 360x39 I 102 106 mm4 Real Support Reactions Support Reactions due to Virtual Moment at A ΣFy0 A B 2P 0 1 ΣFy0 A B 0 1 Due to symmetry A B 2 ΣΜB0 A b M 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P B P A M b B M b A 40kN B 40kN Internal Moment Functions OA BA DB Real M1 P x1 M2 P a M3 P x3 Virtual m1 0 m2 B x2 m3 0 m M 0 x2 b 0 Set M0 0 x2 b 0 L 0 m MMEIdx θ Castiglianos Theorem θA 1 E I 0 a x1 M m1 d d M1 d 0 b x2 M m2 d d M2 d 0 a x3 M m3 d d M3 d θA 1 E I 0 0 b x2 x2 b P a d 0 θA 4412 10 3 rad θA 0253 deg anticlockwise Ans Problem 14140 Solve Prob 1492 using Castiglianos theorem Given a 15m b 3m P 40kN E 200GPa I 21 106 mm4 Solution Apply a Virtual Moment M at B Real Support Reactions Support Reactions due to Virtual Moment at B ΣFy0 A B P 0 1 ΣFy0 A B 0 ΣΜB0 A b P a b 0 2 ΣΜB0 A b M 0 Solving Eqs 1 and 2 Solving Eqs 1 and 2 A P a b b B P A A M b B M b A 60kN B 20 kN Internal Moment Functions OA BA CB Real M1 P x1 M2 B x2 M3 0 Virtual m1 0 m2 A b x2 m3 0 m M 0 b x2 b 0 Set M0 0 b x2 b 0 L 0 m MMEIdx θ Castiglianos Theorem θB 1 E I 0 a x1 M m1 d d M1 d 0 b x2 M m2 d d M2 d 0 a x3 M m3 d d M3 d θB 1 E I 0 0 b x2 b x2 b B x2 d 0 θB 7143 10 3 rad θB 0409 deg clockwise Ans Problem 14141 Solve Prob 1495 using Castiglianos theorem Given a 06m b 16m c 045m PD 14kN do 38mm PE 32kN E 200GPa L a b c Solution Apply a Virtual Force P at B Section Property I π do 4 64 I 10235 103 mm4 Real Support Reactions ΣFy0 A C PD PE 0 1 ΣΜC0 A L PD b c PE c 0 2 Solving Eqs 1 and 2 A 1 L PD b c PE c A 1626 kN C PD PE A C 2974 kN Support Reactions due to Virtual Force at B ΣFy0 A C P 0 1 ΣΜC0 A L P b a c 0 2 Solving Eqs 1 and 2 A P L 2a L C P 2a L Internal Moment Functions AD DB EB CE Real M1 A x1 M2 A a x2 PD x2 M4 C c x4 PE x4 M3 C x3 Virtual m1 A x1 m2 A a x2 m4 C c x4 m3 C x3 m P L 2a L x1 L 2a L a x2 2a L c x4 2a L x3 Set P0 L 2a L x1 L 2a L a x2 2a L c x4 2a L x3 L 0 m PMEI dx Castiglianos Theorem B 1 E I I1 I2 I4 I3 I1 0 a x1 P m1 d d M1 d I1 0 a x1 L 2a L x1 A x1 d I1 006407 kN m3 I2 0 a x2 P m2 d d M2 d I2 0 a x2 L 2a L a x2 A a x2 PD x2 d I2 031064 kN m3 I4 0 b a x4 P m4 d d M4 d I4 0 b a x4 2a L c x4 C c x4 PE x4 d I4 051840 kN m3 I3 0 c x3 P m3 d d M3 d I3 0 c x3 2a L x3 C x3 d I3 004090 kN m3 B 1 E I I1 I2 I4 I3 B 4563 mm Ans Problem 14142 Solve Prob 1497 using Castiglianos theorem Given a 04m b 03m PD 4kN do 30mm PE 3kN E 200GPa PC 2kN L 2a 2 b Solution Apply a Virtual Force P at B 1 Section Property 2 I π do 4 64 I 3976 103 mm4 Real Support Reactions ΣFy0 A C PD PE PC 0 1 ΣΜC0 A L PD L a PE b PC 2 b 0 2 Solving Eqs 1 and 2 A 1 L PD L a PE b PC 2 b A 4357 kN C PD PE PC A C 4643 kN Support Reactions due to Virtual Force at B ΣFy0 A C P 1 ΣΜC0 A L P 2 b 0 2 Solving Eqs 1 and 2 A 2 b P L C L 2 b L P Internal Moment Functions AD DB ED CE Real M1 A x1 M2 A a x2 PD x2 M4 C b x4 PE x4 M3 C x3 Virtual m1 A x1 m2 A a x2 m4 C b x4 m3 C x3 m P 2 b L x1 2 b L a x2 L 2 b L b x4 L 2 b L x3 Set P0 2 b L x1 2 b L a x2 L 2 b L b x4 L 2 b L x3 L 0 m PMEI dx Castiglianos Theorem B 1 E I I1 I2 I4 I3 I1 0 a x1 P m1 d d M1 d I1 003984 kN m3 I2 0 a x2 P m2 d d M2 d I2 018743 kN m3 I4 0 b x4 P m4 d d M4 d I4 012857 kN m3 I3 0 b x3 P m3 d d M3 d I3 002388 kN m3 B 1 E I I1 I2 I4 I3 B 4775 mm Ans I1 0 a x1 2 b L x1 A x1 d I2 0 a x2 2 b L a x2 A a x2 PD x2 d I4 0 b x4 L 2 b L b x4 C b x4 PE x4 d I3 0 b x3 L 2 b L x3 C x3 d Problem 14143 Solve Prob 1499 using Castiglianos theorem Problem 14144 Solve Prob 1496 using Castiglianos theorem Given a 06m b 16m c 045m PD 14kN do 38mm PE 32kN E 200GPa L a b c Solution Apply a Virtual Moment M at A Section Property I π do 4 64 I 10235 103 mm4 Real Support Reactions ΣFy0 A C PD PE 0 1 ΣΜC0 A L PD b c PE c 0 2 Solving Eqs 1 and 2 A 1 L PD b c PE c A 1626 kN C PD PE A C 2974 kN ΣFy0 A C 0 1 ΣΜC0 A L M 0 2 Solving Eqs 1 and 2 A M L C M L Internal Moment Functions AD ED CE Real M1 A x1 M4 C c x4 PE x4 M3 C x3 Virtual m1 M A x1 m4 C c x4 m3 C x3 m M 1 1 L x1 1 L c x4 1 L x3 Set M0 1 1 L x1 1 L c x4 1 L x3 L 0 m MMEIdx θ Castiglianos Theorem θA 1 E I 0 a x1 M m1 d d M1 d 0 b x4 M m4 d d M4 d 0 c x3 M m3 d d M3 d I1 0 a x1 M m1 d d M1 d I1 0 a x1 1 1 L x1 A x1 d I1 024857 kN m2 I4 0 b x4 M m4 d d M4 d I4 0 b x4 1 L c x4 C c x4 PE x4 d I4 084403 kN m2 I3 0 c x3 M m3 d d M3 d I3 0 c x3 1 L x3 C x3 d I3 003408 kN m2 θA 1 E I I1 I4 I3 θA 55038 10 3 rad θA 3153 deg clockwise Ans Problem 14145 Solve Prob 14101 using Castiglianos theorem Given a 4m Mo 18kN m b 3m E 200GPa I 125 106 mm4 Solution Apply a Virtual Force P at D Real Support Reactions Support Reactions due to Virtual Force at D ΣFy0 B C 0 1 ΣFy0 B C F 0 1 Due to symmetry B C 2 Due to symmetry B C 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 B 0 C 0 B 05P C 05P Internal Moment Functions AB BD CD EC Real M1 Mo M2 Mo M3 Mo M4 Mo Virtual m1 0 m2 B x2 m3 C x3 m4 0 for x2 b m P 0 05 x2 05 x3 0 Set P0 0 05 x2 05 x3 0 L 0 m PMEI dx Castiglianos Theorem D 1 E I I1 I2 I3 I4 I1 0 a x1 P m1 d d M1 d I1 0 I2 0 b x2 05 x2 Mo d I2 0 b x2 P m2 d d M2 d I2 4050000 kN m3 I3 0 b x3 P m3 d d M3 d I3 0 b x3 05 x3 Mo d I3 4050000 kN m3 I4 0 b x4 P m4 d d M4 d I4 0 D 1 E I I1 I2 I4 I3 D 324 mm Ans Problem 14146 Solve Prob 14102 using Castiglianos theorem Given a 4m Mo 18kN m b 3m E 200GPa I 125 106 mm4 Solution Apply a Virtual Moment M at A Real Support Reactions Support Reactions due to Virtual Moment at A ΣFy0 B C 0 1 ΣFy0 B C 0 1 Due to symmetry B C 2 ΣΜC0 B 2 b M 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 B 0 C 0 B M 2 b C M 2 b Internal Moment Functions AB CB EC Real M1 Mo M3 Mo M4 Mo Virtual m1 M m3 C x3 m4 0 for x3 2b m M 1 1 2 b x3 0 Set M0 1 1 2 b x3 0 L 0 m MMEIdx θ Castiglianos Theorem θA 1 E I 0 a x1 M m1 d d M1 d 0 2 b x3 M m3 d d M3 d 0 a x4 M m4 d d M4 d θA 1 E I 0 a 1 Mo x1 d 0 2 b x3 1 2 b x3 Mo d 0 θA 5040 10 3 rad θA 0289 deg anticlockwise Ans Problem 14147 Solve Prob 14103 using Castiglianos theorem Given a 4m Mo 18kN m b 3m E 200GPa I 125 106 mm4 Solution Apply a Virtual Moment M at B Real Support Reactions Support Reactions due to Virtual Moment at B ΣFy0 B C 0 1 ΣFy0 B C 0 1 Due to symmetry B C 2 ΣΜC0 B 2 b M 0 2 Solving Eqs 1 and 2 Solving Eqs 1 and 2 B 0 C 0 B M 2 b C M 2 b Internal Moment Functions AB BC EC Real M1 Mo M2 Mo M4 Mo Virtual m1 0 m2 M B x2 m4 0 for x2 2b m M 0 1 1 2 b x2 0 Set M0 0 1 1 2 b x2 0 L 0 m MMEIdx θ Castiglianos Theorem θB 1 E I 0 a x1 M m1 d d M1 d 0 2 b x2 M m2 d d M2 d 0 a x4 M m4 d d M4 d θB 1 E I 0 0 2 b x2 1 1 2 b x2 Mo d 0 θB 2160 10 3 rad θB 0124 deg anticlockwise Ans Problem 14148 Solve Prob 14100 using Castiglianos theorem Problem 14149 Solve Prob 14105 using Castiglianos theorem Problem 14150 Solve Prob 14106 using Castiglianos theorem Problem 14151 Solve Prob 14107 using Castiglianos theorem Given L 3m E 11GPa h 400mm wo 4kN m t 200mm Solution Section Property I t h3 12 a Apply a Virtual Force P at A Internal Moment Functions clockwise Ans Real M1 x1 2 wo x1 L x1 3 M2 wo 2 L L 3 x2 wo 2 x2 2 Virtual m1 P x1 m2 P L x2 L 0 m PMEI dx Castiglianos Theorem A 1 E I 0 L x1 P m1 d d M1 d 0 L x2 P m2 d d M2 d A 1 E I 0 L x1 x1 x1 2 wo x1 L x1 3 d 0 L x2 L x2 wo 2 L L 3 x2 wo 2 x2 2 d A 2738 mm Ans b Apply a Virtual Moment M at A Internal Moment Functions Real M1 x1 2 wo x1 L x1 3 Virtual m1 M m2 M L 0 m MMEIdx θ Castiglianos Theorem θA 1 E I 0 L x1 M m1 d d M1 d 0 L x2 M m2 d d M2 d θA 1 E I 0 L x1 1 x1 2 wo x1 L x1 3 d 0 L x2 1 wo 2 L L 3 x2 wo 2 x2 2 d θA 5753 10 3 rad CB AC AC M2 wo 2 L L 3 x2 wo 2 x2 2 CB Problem 14152 Solve Prob 14104 using Castiglianos theorem Problem 14153 Solve Prob 14109 using Castiglianos theorem Problem 14154 Solve Prob 14113 using Castiglianos theorem Problem 14155 Solve Prob 14114 using Castiglianos theorem Problem 14156 Solve Prob 14108 using Castiglianos theorem Problem 14157 Solve Prob 14116 using Castiglianos theorem Problem 14158 Solve Prob 14115 using Castiglianos theorem Given a 16m b 15m c 15m Pv 3kN Ph 4kN Mo 18kN m Solution Apply a Virtual Force P at C Internal Moment Functions CB BE EA Real M1 Mo Pv x1 M2 Mo Pv a M3 Mo Pv a Ph x3 Virtual m1 0 m2 P x2 m3 P b x3 m P 0 x2 b x3 Set P0 0 x2 b x3 L 0 m PMEI dx Castiglianos Theorem Set EI kN m2 EIo 1 C 1 EI 0 a x1 P m1 d d M1 d 0 b x2 P m2 d d M2 d 0 c x3 P m3 d d M3 d C 1 EI 0 0 b x2 x2 Mo Pv a d 0 c x3 b x3 Mo Pv a Ph x3 d C 4095 m EIo Ans Problem 14159 Determine the bending strain energy in the beam due to the loading shown EI is constant Problem 14160 The L2 steel bolt has a diameter of 5 mm and the link AB has a rectangular cross section that is 10 mm wide by 4 mm thick Determine the strain energy in the link AB due to bending and in the bolt due to axial force The bolt is tightened so that it has a tension of 1750 N Neglect the hole in the link Given L 160mm do 5mm E 200GPa a1 120mm a2 80mm P 1750N b 10mm h 4mm Solution Section Property I b h3 12 I 5333 100 mm4 Abolt π do 2 4 Abolt 1963 mm2 For Link AB Support Reactions ΣFy0 A B P 0 1 ΣΜB0 A a1 a2 P a2 0 2 Solving Eqs 1 and 2 A a2 a1 a2 P A 700N B P A B 1050N Internal Moment Functions M1 A x1 M2 B x2 L 0 2 i M 2EI dx U Bending Strain Energy Ubi 1 2E I 0 a1 x1 M1 2 d 0 a2 x2 M2 2 d Ubi 1 2E I 0 a1 x1 A x1 2 d 0 a2 x2 B x2 2 d Ubi 2205 J For Bolt L 0 2 i N 2EA dx U Axial Strain Energy Uai 1 2E Abolt 0 L y P2 d Uai 00624 J Problem 14161 The cantilevered beam is subjected to a couplemoment M0 applied at its end Determine the slope of the beam at B EI is constant Use the method of virtual work Problem 14162 Solve Prob 14161 using Castiglianos theorem Problem 14163 The cantilevered beam is subjected to a couple moment M0 applied at its end Determine the displacement of the beam at B EI is constant Use the method of virtual work Problem 14164 Solve Prob 14163 using Castiglianos theorem Problem 14165 Determine the vertical displacement of joint A Each bar is made of A36 steel and has a crosssectional area of 600 mm2 Use the conservation of energy Given b 15m h 2m A 600mm2 E 200GPa P 5kN Solution c h2 b2 Member Forces Applying the method of joints to Joint A Given ΣFy0 FAB h c P 0 1 ΣFx0 FAD FAB b c 0 2 Solving Eqs 1 and 2 Guess FAB 1kN FAD 1kN T FAB FAD Find FAB FAD FAB FAD 625 375 kN Ans Applying the method of joints to Joint B Given ΣFy0 FBD h c FAB h c 0 3 ΣFx0 FBC FBD b c FAB b c 0 4 Solving Eqs 3 and 4 Guess FBD 1kN FBC 1kN C FBD FBC Find FBD FBC FBD FBC 625 750 kN T N L2EA U 2 i Axial Strain Energy Ui 1 2 E A FAB 2 c FAD 2 2b FBD 2 c FBC 2 b Ui 1341 J External Work The external work done by the force P is Ue 1 2 P Ah Conservation of Energy Ui Ue Ui 1 2 P Av Av 2 Ui P Av 0536 mm C Problem 14166 Determine the displacement of point B on the aluminum beam Eal 75 GPa Use the conservation of energy Given a 4m bf 150mm dw 150mm E 75GPa P 15kN df 25mm tw 25mm Solution L 2a Section Property h df dw yc Σ yi Ai Σ Ai yc bf df 05df tw dw 05dw df bf df tw dw yc 5625 mm I 1 12 bf df 3 bf df 05df yc 2 1 12 tw dw 3 tw dw 05dw df yc 2 I 2158203125 mm4 Support Reactions ΣFy0 A C P 0 1 ΣΜC0 A L P a 0 2 Solving Eqs 1 and 2 A P a L A 75 kN C P A C 75 kN Internal Moment Function M1 A x1 M2 C x2 L 0 2 i M 2EI dx U Bending Strain Energy Ui 1 2E I 0 a x1 M1 2 d 0 a x2 M2 2 d Ui 1 2E I 0 a x1 A x1 2 d 0 a x2 C x2 2 d Ui 74136 J External Work The external work done by the force P is Ue 1 2 P B Conservation of Energy Ui Ue Ui 1 2 P B B 2 Ui P B 9885 mm Ans Problem 14167 A 100N 10kg weight is dropped from a height of 12 m onto the end of a cantilevered A36 steel beam If the beam is a W310 X 74 determine the maximum stress developed in the beam Given L 36m h 12m W 100N E 200GPa σY 250MPa Solution Use W 310x74 I 165 106 mm4 d 310mm Static Displacement st W L3 3E I From Appendix C st 004713 mm Maximum Stress Impact factor η 1 1 2 h st η 22667 Maximum moment occurs at support A where Mmax W L Mmax 0360 kN m c 05d σmax Mmax c I η σmax 767 MPa Ans Since σmax σY 250 MPa the above analysis is valid NBD NCD sin θ A 400mm2 Given b 2m P 45kN h 15m E 200GPa Solution θ atan h b θ 36870 deg Support Reactions By symmetry ACR ΣFy0 2R P 0 R 05P Member Lengths c h2 b2 c 25 m LAF h LBE h LCD h LAE c LBD c LEF b LDE b LAB b LBC b Member Real Forces By inspection NAF 0 NEF 0 NBC 0 NCD 225 kN Problem 14168 Determine the vertical displacement of joint B For each member A 400 mm2 E 200 GPa Use the method of virtual work NBD 375 kN NDE NBD cos θ NDE 30 kN NBE P NBD sin θ NBE 225 kN NAE NBE sin θ NAE 375 kN NAB NAE cos θ NAB 30kN Member Virtual Forces Apply Virtual Force at B F 1kN Similarly Support Reactions R 05F nAF 0 nEF 0 nBC 0 nCD R nCD 05 kN nBD nCD sin θ nBD 0833 kN nDE nBD cos θ nDE 0667 kN nBE F nBD sin θ nBE 05 kN nAE nBE sin θ nAE 0833 kN nAB nAE cos θ nAB 0667 kN NCD R Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 CD n3 nCD N3 NCD L3 LCD p3 n3 N3 L3 DE n4 nDE N4 NDE L4 LDE p4 n4 N4 L4 EF n5 nEF N5 NEF L5 LEF p5 n5 N5 L5 AF n6 nAF N6 NAF L6 LAF p6 n6 N6 L6 AE n7 nAE N7 NAE L7 LAE p7 n7 N7 L7 BE n8 nBE N8 NBE L8 LBE p8 n8 N8 L8 BD n9 nBD N9 NBD L9 LBD p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 nNLEA 1 Virtual Work Equation F Bv Sum E A Bv Sum E A 1 F Bv 3375 mm Ans Problem 14169 Solve Prob 14168 using Castiglianos theorem Given b 2m A 400mm2 P 45kN h 15m E 200GPa Solution θ atan h b θ 36870 deg Support Reactions By symmetry ACR ΣFy0 2R P 0 R 05P Member Lengths c h2 b2 c 25 m LAF h LBE h LCD h LAE c LBD c LEF b LDE b LAB b LBC b Member Real Forces By inspection NAF 0 NEF 0 NBC 0 NCD R NCD 225 kN NBD NCD sin θ NBD 375 kN NDE NBD cos θ NDE 30 kN NBE P NBD sin θ NBE 225 kN NAE NBE sin θ NAE 375 kN NAB NAE cos θ NAB 30kN Member Virtual Forces Apply Virtual Force P at B Similarly Support Reactions R 05P nAF 0 nEF 0 nBC 0 nCD R nCD 05 P nBD nCD sin θ nBD 05P csc θ nDE nBD cos θ nDE 05 P cot θ nBE P nBD sin θ nBE 05P nAE nBE sin θ nAE 05 P csc θ nAB nAE cos θ nAB 05P cot θ n P Member Set P0 Real N Length L Product p AB n1 05 cot θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 0 N2 NBC L2 LBC p2 n2 N2 L2 CD n3 05 N3 NCD L3 LCD p3 n3 N3 L3 DE n4 05 cot θ N4 NDE L4 LDE p4 n4 N4 L4 EF n5 0 N5 NEF L5 LEF p5 n5 N5 L5 AF n6 0 N6 NAF L6 LAF p6 n6 N6 L6 AE n7 05 csc θ N7 NAE L7 LAE p7 n7 N7 L7 BE n8 05 N8 NBE L8 LBE p8 n8 N8 L8 BD n9 05 csc θ N9 NBD L9 LBD p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 PNLEA n Castiglianos Theorem Bv Sum E A Bv 3375 mm Ans Problem 14170 Determine the vertical displacement of joint E For each member A 400 mm2 E 200 GPa Use the method of virtual work Given b 2m A 400mm2 P 45kN h 15m E 200GPa Solution θ atan h b θ 36870 deg Support Reactions By symmetry ACR ΣFy0 2R P 0 R 05P Member Lengths c h2 b2 c 25 m LAF h LBE h LCD h LAE c LBD c LEF b LDE b LAB b LBC b Member Real Forces By inspection NAF 0 NEF 0 NBC 0 NCD R NCD 225 kN NBD NCD sin θ NBD 375 kN NDE NBD cos θ NDE 30 kN NBE P NBD sin θ NBE 225 kN NAE NBE sin θ NAE 375 kN NAB NAE cos θ NAB 30kN Member Virtual Forces Apply Virtual Force at E F 1kN Similarly Support Reactions R 05F nAF 0 nEF 0 nBC 0 nCD R nCD 05 kN nBD nCD sin θ nBD 0833 kN nDE nBD cos θ nDE 0667 kN nBE nBD sin θ nBE 05 kN nAE F nBE sin θ nAE 0833 kN nAB nAE cos θ nAB 0667 kN Member Virtual n Real N Length L Product p AB n1 nAB N1 NAB L1 LAB p1 n1 N1 L1 BC n2 nBC N2 NBC L2 LBC p2 n2 N2 L2 CD n3 nCD N3 NCD L3 LCD p3 n3 N3 L3 DE n4 nDE N4 NDE L4 LDE p4 n4 N4 L4 EF n5 nEF N5 NEF L5 LEF p5 n5 N5 L5 AF n6 nAF N6 NAF L6 LAF p6 n6 N6 L6 AE n7 nAE N7 NAE L7 LAE p7 n7 N7 L7 BE n8 nBE N8 NBE L8 LBE p8 n8 N8 L8 BD n9 nBD N9 NBD L9 LBD p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 nNLEA 1 Virtual Work Equation F Bv Sum E A Bv Sum E A 1 F Bv 2953 mm Ans Problem 14171 Solve Prob 14170 using Castiglianos theorem Given b 2m A 400mm2 P 45kN h 15m E 200GPa Solution θ atan h b θ 36870 deg Support Reactions By symmetry ACR ΣFy0 2R P 0 R 05P Member Lengths c h2 b2 c 25 m LAF h LBE h LCD h LAE c LBD c LEF b LDE b LAB b LBC b Member Real Forces By inspection NAF 0 NEF 0 NBC 0 NCD R NCD 225 kN NBD NCD sin θ NBD 375 kN NDE NBD cos θ NDE 30 kN NBE P NBD sin θ NBE 225 kN NAE NBE sin θ NAE 375 kN NAB NAE cos θ NAB 30kN Member Virtual Forces Apply Virtual Force P at E Similarly Support Reactions R 05P nAF 0 nEF 0 nBC 0 nCD R nCD 05 P nBD 05P csc θ nBD nCD sin θ nDE 05 P cot θ nDE nBD cos θ nBE nBD sin θ nBE 05 P nAE 05 P csc θ nAE P nBE sin θ nAB 05P cot θ nAB nAE cos θ n P Member Set P0 Real N Length L Product p AB n1 05 cot θ N1 NAB L1 LAB p1 n1 N1 L1 BC n2 0 N2 NBC L2 LBC p2 n2 N2 L2 CD n3 05 N3 NCD L3 LCD p3 n3 N3 L3 DE n4 05 cot θ N4 NDE L4 LDE p4 n4 N4 L4 EF n5 0 N5 NEF L5 LEF p5 n5 N5 L5 AF n6 0 N6 NAF L6 LAF p6 n6 N6 L6 AE n7 05 csc θ N7 NAE L7 LAE p7 n7 N7 L7 BE n8 05 N8 NBE L8 LBE p8 n8 N8 L8 BD n9 05 csc θ N9 NBD L9 LBD p9 n9 N9 L9 Sum p1 p2 p3 p4 p5 p6 p7 p8 p9 PNLEA n Castiglianos Theorem Bv Sum E A Bv 2953 mm Ans