Solucionário de Problemas de Engenharia Mecânica - Estática e Dinâmica

LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTA SIGUENOS EN VISITANOS PARA DESARGALOS GRATIS Chapter 1 Problems 11 through 14 are for student research 15 Consider force F at G reactions at B and D Extend lines of action for fullydeveloped friction DE and BE to find the point of concurrency at E for impending motion to the left The critical angle is θcr Resolve force F into components Facc and Fcr Facc is related to mass and acceleration Pin accelerates to left for any angle 0 θ θcr When θ θcr no magnitude of F will move the pin Consider force F at G reactions at A and C Extend lines of action for fullydeveloped friction AE and CE to find the point of concurrency at E for impending motion to the left The critical angle is θcr Resolve force F into components Facc and Fcr Facc is related to mass and acceleration Pin accelerates to right for any angle 0 θ θcr When θ θcr no magnitude of F will move the pin The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches The graphic approach accomplishes this quickly It is important to point out that this understanding enables a mathematical model to be constructed and that there are two of them This is the simplest problem in mechanical engineering Using it is a good way to begin a course What is the role of pin diameter d Yes changing the sense of F changes the response httplibrosysolucionariosnet 16 a ΣFy F fN cos θ N sin θ 0 1 ΣFx fN sin θ N cos θ Tr 0 F Nsin θ f cos θ Ans T Nrf sin θ cos θ Combining T Fr 1 f tan θ tan θ f KFr Ans 2 b If T detent selflocking tan θ f 0 θcr tan¹ f Ans Friction is fully developed Check If F 10 lbf f 020 θ 45 r 2 in N 10 020 cos 45 sin 45 1768 lbf Tr 1728020 sin 45 cos 45 15 lbf fN 0201728 354 lbf θcr tan¹ f tan¹020 1131 1131 θ 90 17 a F F₀ k0 F₀ T₁ F₀r Ans b When teeth are about to clear F F₀ kx₂ From Prob 16 T₂ Fr f tan θ 1 tan θ f T₂ r F₀ kx₂f tan θ 1 tan θ f Ans 18 Given F 10 25x lbf r 2 in h 02 in θ 60 f 025 xᵢ 0 x𝑓 02 Fᵢ 10 lbf F𝑓 10 2502 105 lbf Ans httplibrosysolucionariosnet From Eq 1 of Prob 16 N F f cos θ sin θ Nᵢ 10 025 cos 60 sin 60 1349 lbf Ans N𝑓 10510 1349 1417 lbf Ans From Eq 2 of Prob 16 K 1 f tan θ tan θ f 1 025 tan 60 tan 60 025 0967 Ans Tᵢ 0967102 1933 lbf in T𝑓 09671052 2031 lbf in 19 a Point vehicles Q carshour vx 421v v²0324 Seek stationary point maximum dQdv 0 421 2v0324 v 2105 mph Q 4212105 2105²0324 13676 carsh Ans b Q vx l 0324v421 v² lv1 Maximize Q with l 105280 mi v Q 2218 1221431 2219 1221433 2220 1221435 2221 1221435 2222 1221434 loss of throughput 1368 12211221 12 Ans httplibrosysolucionariosnet Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design c increase in speed 222 21052105 55 Modest change in optimal speed Ans 110 This and the following problem may be the students first experience with a figure of merit Formulate fom to reflect larger figure of merit for larger merit Use a maximization optimization algorithm When one gets into computer implementation and answers are not known minimizing instead of maximizing is the largest error one can make FV F1 sin θ W 0 FH F1 cos θ F2 0 From which F1 Wsin θ F2 W cos θ sin θ fom S φγ volume φγ l1A1 l2A2 A1 F1 S W S sin θ l2 l1 cos θ A2 F2 S W cos θ S sin θ fom φγ l2 cos θ W S sin θ l2 W cos θ S sin θ φγ W l2 S 1 cos2 θ cos θ sin θ Set leading constant to unity θ fom 0 20 586 30 404 40 322 45 300 50 287 54736 2828 60 2886 θ 54736 Ans fom 2828 Alternative d dθ 1 cos2 θ cos θ sin θ 0 And solve resulting transcendental for θ Check second derivative to see if a maximum minimum or point of inflection has been found Or evaluate fom on either side of θ httplibrosysolucionariosnet Chapter 1 5 111 a x1 x2 X1 e1 X2 e2 error e x1 x2 X1 X2 e1 e2 Ans b x1 x2 X1 e1 X2 e2 e x1 x2 X1 X2 e1 e2 Ans c x1x2 X1 e1X2 e2 e x1x2 X1X2 X1e2 X2e1 e1e2 X1e2 X2e1 X1X2 e1 X1 e2 X2 Ans d x1 x2 X1 e1 X2 e2 X1 X2 1 e1 X1 1 e2 X2 1 e2 X21 1 e2 X2 and 1 e1 X11 e2 X2 1 e1 X1 e2 X2 e x1 x2 X1 X2 X1 X2 e1 X1 e2 X2 Ans 112 a x1 5 2236 067 977 5 X1 223 3correct digits x2 6 2449 487 742 78 X2 244 3correct digits x1 x2 5 6 4685 557 720 28 e1 x1 X1 5 223 0006 067 977 5 e2 x2 X2 6 244 0009 489 742 78 e e1 e2 5 223 6 244 0015 557 720 28 Sum x1 x2 X1 X2 e 223 244 0015 557 720 28 4685 557 720 28 Checks Ans b X1 224 X2 245 e1 5 224 0003 932 022 50 e2 6 245 0000 510 257 22 e e1 e2 0004 442 279 72 Sum X1 X2 e 224 245 0004 442 279 72 4685 557 720 28 Ans httplibrosysolucionariosnet 6 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 113 a σ 20689 1378 MPa b F 350445 1558 N 1558 kN c M 1200 lbf in 0113 1356 N m d A 24645 1548 mm2 e I 174 in4 2544 7242 cm4 f A 3616102 9332 km2 g E 211000689 14469103 MPa 1447 GPa h v 45 mih 161 7245 kmh i V 60 in3 2543 9832 cm3 0983 liter 114 a l 150305 4918 ft 5902 in b σ 600689 8696 kpsi c p 160689 2322 psi d Z 1841052543 1123 in3 e w 381175 0218 lbfin f δ 005254 000197 in g v 61200051 1200 ftmin h ϵ 00021 inin i V 3002543 1831 in3 115 a σ 200 153 131 MPa b σ 42103 61022 70106 Nm2 70 MPa c y 120080031033 3207641091024 1546102 m 155 mm d θ 1100250103 793π322541091034 9043102 rad 518 116 a σ 600 206 5 MPa b I 1 128243 9216 mm4 c I π 643241014 5147 cm4 d τ 1616 π2531033 5215106 Nm2 5215 MPa shi20396ch01qxd 6503 1211 PM Page 6 httplibrosysolucionariosnet Chapter 1 7 117 a τ 120103 π4202 382 MPa b σ 32800800103 π3231033 1989106 Nm2 1989 MPa c Z π 3236364 264 3334 mm3 d k 164 7931034109 81923321033 2868 Nm shi20396ch01qxd 6503 1211 PM Page 7 httplibrosysolucionariosnet Chapter 2 21 a b fNΔx f69 10 f690 x f fx fx2 fNΔx 60 2 120 7200 00029 70 1 70 4900 00015 80 3 240 19 200 00043 90 5 450 40 500 00072 100 8 800 80 000 00116 110 12 1320 145 200 00174 120 6 720 86 400 00087 130 10 1300 169 000 00145 140 8 1120 156 800 00116 150 5 750 112 500 00174 160 2 320 51 200 00029 170 3 510 86 700 00043 180 2 360 64 800 00029 190 1 190 36 100 00015 200 0 0 0 0 210 1 210 44 100 00015 69 8480 1 104 600 Eq 29 x 848069 1229 kcycles Eq 210 sx 1 104 600 8480269 69 1 12 303 kcycles Ans httplibrosysolucionariosnet 22 Data represents a 7class histogram with N 197 x f fx fx2 174 6 1044 181 656 182 9 1638 298 116 190 44 8360 1 588 400 198 67 13 266 2 626 688 206 53 10 918 2 249 108 214 12 2568 549 552 220 6 1320 290 400 197 39 114 7 789 900 x 39 114197 19855 kpsi Ans sx 7 783 900 39 1142197 197 112 955 kpsi Ans 23 Form a table x f fx fx2 64 2 128 8192 68 6 408 27 744 72 6 432 31 104 76 9 684 51 984 80 19 1520 121 600 84 10 840 70 560 88 4 352 30 976 92 2 184 16 928 58 4548 359 088 x 454858 784 kpsi sx 359 088 4548258 58 112 657 kpsi From Eq 214 fx 1 657 sqrt2π exp 12 x 7846572 httplibrosysolucionariosnet 24 a y f fy fy2 y fNw fy gy 5625 1 5625 3164063 5625 0072727 0001262 0000295 5875 0 0 0 5875 0 0008586 0004088 6125 0 0 0 6125 0 0042038 0031194 6375 3 19125 1219219 6375 0218182 0148106 0140262 6625 3 19875 1316719 6625 0218182 0375493 0393667 6875 6 4125 2835938 6875 0436364 0685057 0725002 7125 14 9975 7107188 7125 1018182 0899389 0915128 7375 15 110625 8158594 7375 1090909 0849697 0822462 7625 10 7625 5814063 7625 0727273 0577665 0544251 7875 2 1575 1240313 7875 0145455 0282608 0273138 8125 1 8125 6601563 8125 0072727 0099492 010672 55 396375 2866859 For a normal distribution ȳ 39637555 7207 sy 2866859 396375255 55 112 04358 fy 1 04358 sqrt2π exp 12 x 7207043582 For a lognormal distribution x ln 7206818 ln 1 00604742 19732 sx ln 1 00604742 00604 gy 1 x006042π exp 12 ln x 19732 006042 b Histogram httplibrosysolucionariosnet 25 Distribution is uniform in interval 05000 to 05008 in range numbers are a 05000 b 05008 in a Eq 222 μx a b 2 05000 05008 2 05004 Eq 223 σx b a 23 05008 05000 23 0000231 b PDF from Eq 220 fx 1250 if 05000 x 05008 in 0 otherwise c CDF from Eq 221 Fx 0 if x 05000 x 0500008 if 05000 x 05008 1 if x 05008 If all smaller diameters are removed by inspection a 05002 b 05008 μx 05002 05008 2 05005 in σx 05008 05002 23 0000173 in fx 16667 if 05002 x 05008 0 otherwise Fx 0 if x 05002 16667x 05002 if 05002 x 05008 1 if x 05008 26 Dimensions produced are due to tool dulling and wear When parts are mixed the distribution is uniform From Eqs 222 and 223 a μx 3 s 06241 30000581 06231 in b μx 3 s 06241 30000581 06251 in We suspect the dimension was 0623 0625 in Ans httplibrosysolucionariosnet 27 Fx0555x33 mm a Since Fx is linear the distribution is uniform at x a Fa 0 0555a33 a5946 mm Therefore at x b Fb 1 0555b33 b6126 mm Therefore Fx 0 x5946 mm 0555x33 5946 x 6126 mm 1 x 6126 mm The PDF is dFdx thus the range numbers are fx 0555 5946 x 6126 mm 0 otherwise Ans From the range numbers μx 5946 61262 6036 mm Ans σx 6126594623 0520 mm Ans b σ is an uncorrelated quotient F 3600 lbf A 0112 in² CF 30036000083 33 CA 000101120008 929 From Table 26 for σ σ μFμA 36000112 32 143 psi Ans σσ 32 143 0083 33² 0008 929²1 0008 929²12 2694 psi Ans Cσ 269432 143 00838 Ans Since F and A are lognormal division is closed and σ is lognormal too σ LN32 143 2694 psi Ans httplibrosysolucionariosnet 28 Cramers rule a₁ Σy Σx² Σx Σx² ΣyΣx³ ΣxyΣx² ΣxΣx³ Σx²² Ans Σxy Σx³ Σx² Σx³ a₂ Σx Σy Σx Σx² ΣxΣxy ΣyΣx² ΣxΣx³ Σx²² Ans Σx² Σxy Σx² Σx³ x y x² x³ xy 0 001 0 0 0 02 015 004 0008 0030 04 025 016 0064 0100 06 025 036 0216 0150 08 017 064 0512 0136 10 001 100 1000 0010 30 082 220 1800 0406 a₁1040 714 a₂ 1046 43 Ans Data Regression x y y 0 001 0 02 015 0166 286 04 025 0248 857 06 025 0247 714 08 017 0162 857 10 001 0005 71 httplibrosysolucionariosnet Data Regression Su Se Se2 SuSe 0 2035675 60 30 3908078 3 600 1 800 64 48 4032905 4 096 3 072 65 295 4064112 4 225 1 9175 82 45 4594626 6 724 3 690 101 51 5187554 10 201 5 151 119 50 5749275 14 161 5 950 120 48 5780481 14 400 5 760 130 67 6092548 16 900 8 710 134 60 6217375 17 956 8 040 145 64 6560649 21 025 9 280 180 84 7652884 32 400 15 120 195 78 8120985 38 025 15 210 205 96 8433052 42 025 19 680 207 87 8495466 42 849 18 009 210 87 8589086 44 100 18 270 213 75 8682706 45 369 15 975 225 99 9057187 50 625 22 275 225 87 9057187 50 625 19 575 227 116 91196 51 529 26 332 230 105 92132 2 52 900 24 150 238 109 9462874 56 644 25 942 242 106 9587701 58 564 25 652 265 105 103054 6 70 225 27 825 280 96 107735 6 78 400 26 880 295 99 112416 6 87 025 29 205 325 114 121778 6 105 625 37 050 325 117 121778 6 105 625 38 025 355 122 131140 6 126 025 43 310 5462 22745 1 251 868 501 8555 m0312 067 b20356 75 Ans httplibrosysolucionariosnet E Σy a₀ a₂x²² Ea₀ 2 Σy a₀ a₂x² 0 Σy na₀ a₂ Σx² 0 Σy na₀ a₂ Σx² Ea₂ 2 Σy a₀ a₂x²2x 0 Σxy a₀ Σx a₂ Σx³ Ans Cramers rule a₀ Σy Σx² n Σx² Σx³ Σy Σx² Σxy n Σx³ Σx Σx² Σxy Σx³ Σx Σx³ a₂ n Σy n Σx² n Σxy Σx Σy n Σx³ Σx Σx² Σx Σxy Σx Σx³ Data Regression x y y x² x³ xy 20 19 192 400 8000 380 40 17 168 1600 64000 680 60 13 128 3600 216000 780 80 7 72 6400 512000 560 200 56 12000 800000 2400 a₀ 80000056 120002400 4800000 20012000 20 a₂ 42400 20056 4800000 20012000 0002 Data Regression x y y x² y² xy x x x x² 02 71 7931803 004 5041 142 0633333 0401111111 04 103 9884918 016 10609 412 0433333 0187777778 06 121 11838032 036 14641 726 0233333 0054444444 08 138 13791147 064 19044 1104 0033333 0001111111 1 162 15744262 100 26244 1620 0166666 0027777778 2 252 25509836 400 63504 5040 1166666 1361111111 5 847 62 139083 9044 0 2033333333 m k 69044 5847 662 5² 97656 b Fᵢ 847 976565 6 59787 x 56 y 847 6 14117 sᵧₓ sqrt139083 59787847 976569044 6 2 0556 sb 0556 sqrt16 56² 20333 03964 lbf Fᵢ 59787 03964 lbf Ans b Eq 235 sm 0556 sqrt20333 03899 lbfin k 97656 03899 lbfin Ans 212 The expression ϵ δl is of the form xy Now δ 00015 0000092 in unspecified distribution l 2000 00081 in unspecified distribution Cx 0000092 00015 00613 Cy 00081 2000 000075 From Table 26 ε 00015 2000 000075 σϵ 000075 sqrt00613² 000405² 1 000405² 460710⁵ 0000046 We can predict ε and σϵ but not the distribution of ϵ 213 σ ϵE ϵ 00005 0000034 distribution unspecified E 295 0885 Mpsi distribution unspecified Cx 0000034 00005 0068 Cy 00885 295 0030 σ is of the form x y Table 26 σ εĒ 00005 295 10⁶ 14750 psi σσ 14750 sqrt0068² 0030² 0068² 0030²¹² 10967 psi Cσ 10967 14750 007435 214 δ Fl AE F 147 13 kip A 0226 0003 in² l 15 0004 in E 295 0885 Mpsi distributions unspecified CF 13 147 00884 CA 0003 0226 00133 Cl 0004 15 000267 CE 0885 295 003 Mean of δ δ Fl AE Fl1A1E From Table 26 δ F l 1A1E δ 1470015 10226 129510⁶ 0003 31 in Ans For the standard deviation using the firstorder terms in Table 26 σδ F l A E CF² Cl² CA² CE² 12 δ CF² Cl² CA² CE² 12 σδ 0003 31 00884² 000267² 00133² 003² 12 0000 313 in Ans COV Cδ 0000 313 0003 31 00945 Ans Force COV dominates There is no distributional information on δ 215 M 15000 1350 lbfin distribution unspecified d 200 0005 in distribution unspecified σ 32M π d³ CM 135015000 009 Cd 0005200 00025 σ is of the form xy Table 26 Mean σ 32M π d³ 32M π d³ 3215000 π2³ 19099 psi Ans Standard Deviation σσ σ CM² Cd³² 1 Cd³² 12 From Table 26 Cd³ 3Cd 300025 00075 σσ σ CM² 3Cd² 1 3Cd² 12 19099009² 00075²1 00075² 12 1725 psi Ans COV Cσ 172519099 00903 Ans Stress COV dominates No information of distribution of σ 216 Fraction discarded is α β The area under the PDF was unity Having discarded α β fraction the ordinates to the truncated PDF are multiplied by a a 1 1 α β New PDF gx is given by gx fx 1 α β for x₁ x x₂ 0 otherwise More formal proof gx has the property 1 x₁x₂ gx dx a x₁x₂ fx dx 1 a fx dx 0x₁ fx dx x₂ fx dx 1 a 1 Fx₁ 1 Fx₂ a 1 Fx₂ Fx₁ 1 1 β α 1 1 α β 217 a d U0748 0751 μd 0751 0748 2 07495 in σd 0751 0748 23 0000 866 in fx 1 b a 1 0751 0748 3333 in¹ Fx x 0748 0751 0748 3333 x 0748 b Fx₁ F0748 0 Fx₂ 0750 07483333 06667 If gx is truncated PDF becomes gx fx Fx₂ Fx₁ 3333 06667 0 500 in¹ μx a b 2 0748 0750 2 0749 in σx b a 23 0750 0748 23 0000 577 in 218 From Table A10 81 corresponds to z₁ 14 and 55 corresponds to z₂ 16 k₁ μ z₁ σ k₂ μ z₂ σ From which μ z₂ k₁ z₁ k₂ z₂ z₁ 169 1411 16 14 9933 σ k₂ k₁ z₂ z₁ 11 9 16 14 06667 The original density function is fk 1 06667 2π exp ½ k 9933 06667 ² Ans 219 From Prob 21 μ 1229 kcycles and σ 303 kcycles z₁₀ x₁₀ μ σ x₁₀ 1229 303 x₁₀ 1229 303 z₁₀ From Table A10 for 10 percent failure z₁₀ 1282 x₁₀ 1229 303 1282 841 kcycles Ans Chapter 2 21 220 x f fx f x2 x fNw fx 60 2 120 7200 60 0002899 0000399 70 1 70 4900 70 0001449 0001206 80 3 240 19200 80 0004348 0003009 90 5 450 40500 90 0007246 0006204 100 8 800 80000 100 0011594 0010567 110 12 1320 145200 110 0017391 0014871 120 6 720 86400 120 0008696 0017292 130 10 1300 169000 130 0014493 0016612 140 8 1120 156800 140 0011594 0013185 150 5 750 112500 150 0007246 0008647 160 2 320 51200 160 0002899 0004685 170 3 510 86700 170 0004348 0002097 180 2 360 64800 180 0002899 0000776 190 1 190 36100 190 0001449 0000237 200 0 0 0 200 0 598E05 210 1 210 44100 210 0001449 125E05 69 8480 x 1228986 sx 2288719 x fNw fx x fNw fx 55 0 0000214 145 0011594 0010935 55 0002899 0000214 145 0007246 0010935 65 0002899 0000711 155 0007246 0006518 65 0001449 0000711 155 0002899 0006518 75 0001449 0001951 165 0002899 000321 75 0004348 0001951 165 0004348 000321 85 0004348 0004425 175 0004348 0001306 85 0007246 0004425 175 0002899 0001306 95 0007246 0008292 185 0002899 0000439 95 0011594 0008292 185 0001449 0000439 105 0011594 0012839 195 0001449 0000122 105 0017391 0012839 195 0 0000122 115 0017391 0016423 205 0 28E05 115 0008696 0016423 205 0001499 28E05 125 0008696 0017357 215 0001499 531E06 125 0014493 0017357 215 0 531E06 135 0014493 0015157 135 0011594 0015157 shi20396ch02qxd 72103 328 PM Page 21 httplibrosysolucionariosnet 22 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 221 x f fx f x2 fNw f x 174 6 1044 181656 0003807 0001642 182 9 1638 298116 0005711 0009485 190 44 8360 1588400 0027919 0027742 198 67 13266 2626668 0042513 0041068 206 53 10918 2249108 0033629 0030773 214 12 2568 549552 0007614 0011671 222 6 1332 295704 0003807 0002241 1386 197 39126 7789204 x 1986091 sx 9695071 x fNw f x 170 0 0000529 170 0003807 0000529 178 0003807 0004297 178 0005711 0004297 186 0005711 0017663 186 0027919 0017663 194 0027919 0036752 194 0042513 0036752 202 0042513 0038708 202 0033629 0038708 210 0033629 0020635 210 0007614 0020635 218 0007614 0005568 218 0003807 0005568 226 0003807 000076 226 0 000076 Data PDF 0 0005 001 0015 002 0025 003 0035 004 0045 150 170 190 210 x 230 f Histogram PDF 0 0002 0004 0006 0008 001 0012 0014 0016 0018 f x 002 0 50 100 150 200 250 shi20396ch02qxd 72103 328 PM Page 22 httplibrosysolucionariosnet Chapter 2 23 222 x f fx f x2 fNw fx 64 2 128 8192 0008621 000548 68 6 408 27744 0025862 0017299 72 6 432 31104 0025862 0037705 76 9 684 51984 0038793 0056742 80 19 1520 121600 0081897 0058959 84 10 840 70560 0043103 0042298 88 4 352 30976 0017241 0020952 92 2 184 16928 0008621 0007165 624 58 4548 359088 x 7841379 sx 6572229 x fNw fx x fNw fx 62 0 0002684 82 0081897 0052305 62 0008621 0002684 82 0043103 0052305 66 0008621 0010197 86 0043103 003118 66 0025862 0010197 86 0017241 003118 70 0025862 0026749 90 0017241 0012833 70 0025862 0026749 90 0008621 0012833 74 0025862 0048446 94 0008621 0003647 74 0038793 0048446 94 0 0003647 78 0038793 0060581 78 0081897 0060581 223 σ 4 P πd2 440 π12 5093 kpsi ˆσσ 4 ˆσP πd2 485 π12 1082 kpsi ˆσsy 59 kpsi Data PDF x 0 001 002 003 004 005 006 007 008 009 60 70 80 90 100 f shi20396ch02qxd 72103 328 PM Page 23 httplibrosysolucionariosnet For no yield m Sy σ 0 z m μm σm 0 μm σm μm σm μm Sy σ 2747 kpsi σm σσ2 σSy212 1232 kpsi z 2747 1232 2230 From Table A10 pf 00129 R 1 pf 1 00129 0987 Ans 224 For a lognormal distribution Eq 218 μy ln μx ln 1 Cx2 Eq 219 σy ln1 Cx2 From Prob 223 μm Sy σ μx μy ln Sy ln 1 CSy2 ln σ ln 1 Cσ2 ln Sy σ 1 Cσ2 1 CSy2 σy ln 1 CSy2 ln 1 Cσ2 12 ln 1 CSy21 Cσ2 z μ σ ln Sy σ 1 Cσ2 1 CSy2 ln 1 CSy21 Cσ2 σ 4P π d2 430 π12 38197 kpsi σσ 4 σP π d2 451 π12 6494 kpsi Cσ 6494 38197 01700 CSy 381 496 007681 httplibrosysolucionariosnet ln 496 38197 1 01702 1 0076812 z ln 1 00768121 01702 1470 From Table A10 pf 00708 R 1 pf 0929 Ans 225 a a 1000 0001 in b 2000 0003 in c 3000 0005 in d 6020 0006 in w d a b c 6020 1 2 3 0020 in tw tall 0001 0003 0005 0006 0015 in w 0020 0015 in Ans b w 0020 σw σall2 000132 000332 000532 000632 000486 0005 in uniform w 0020 0005 in Ans 226 V ΔV a Δab Δbc Δc V ΔV abc bc Δa ac Δb ab Δc small higher order terms ΔV V Δa a Δb b Δc c Ans V a b c 1251875275 64453 in3 ΔV V 00011250 00021875 00032750 000296 ΔV ΔV V V 00029664453 00191 in3 Lower range number V ΔV 64453 00191 64262 in3 Ans Upper range number V ΔV 64453 00191 64644 in3 Ans httplibrosysolucionariosnet 227 a wmax 0014 in wmin 0004 in w 0014 00042 0009 in w 0009 0005 in w x y a b c 0009 a 0042 1000 a 1051 in tw tall 0005 ta 0002 0002 ta 0005 0002 0002 0001 in a 1051 0001 in Ans b σw σall2 σa2 σb2 σc2 σa2 σw2 σb2 σc2 0005 32 0002 32 0002 32 σa2 5667106 σa 5667106 000238 in a 1051 in σa 000238 in Ans 228 Choose 15 mm as basic size D d Table 28 fit is designated as 15H7h6 From Table A11 the tolerance grades are ΔD 0018 mm and Δd 0011 mm Hole Eq 238 Dmax D ΔD 15 0018 15018 mm Ans Dmin D 15000 mm Ans Shaft From Table A12 fundamental deviation δF 0 From Eq 239 dmax d δF 15000 0 15000 mm Ans dmin d δR Δd 15000 0 0011 14989 mm Ans 229 Choose 45 mm as basic size Table 28 designates fit as 45H7s6 From Table A11 the tolerance grades are ΔD 0025 mm and Δd 0016 mm Hole Eq 238 Dmax D ΔD 45000 0025 45025 mm Ans Dmin D 45000 mm Ans httplibrosysolucionariosnet Shaft From Table A12 fundamental deviation δF 0043 mm From Eq 240 dmin d δF 45000 0043 45043 mm Ans dmax d δF Δd 45000 0043 0016 45059 mm Ans 230 Choose 50 mm as basic size From Table 28 fit is 50H7g6 From Table A11 the tolerance grades are ΔD 0025 mm and Δd 0016 mm Hole Dmax D ΔD 50 0025 50025 mm Ans Dmin D 50000 mm Ans Shaft From Table A12 fundamental deviation 0009 mm dmax d δF 50000 0009 49991 mm Ans dmin d δF Δd 50000 0009 0016 49975 mm 231 Choose the basic size as 1000 in From Table 28 for 10 in the fit is H8f7 From Table A13 the tolerance grades are ΔD 00013 in and Δd 00008 in Hole Dmax D ΔDhole 1000 00013 10013 in Ans Dmin D 10000 in Ans Shaft From Table A14 Fundamental deviation 00008 in dmax d δF 10000 00008 09992 in Ans dmin d δF Δd 10000 00008 00008 09984 in Ans Alternatively dmin dmax Δd 09992 00008 09984 in Ans 232 Do W Di W Do W Di W 0139 3734 0139 4012 in tDo all tall 0004 0028 0004 0036 in Do 4012 0036 in Ans 233 Do Di 2W Do Di 2W 20892 2533 21958 mm tDo all t tDi 2tw 130 2013 156 mm Do 21958 156 mm Ans 234 Do Di 2W Do Di 2W 3734 20139 4012 mm tDo all t2 tDo2 2 tw212 00282 220004212 0029 in Do 4012 0029 in Ans 235 Do Di 2W Do Di 2W 20892 2533 21958 mm tDo all t2 1302 22013212 133 mm Do 21958 133 mm Ans 236 a w F W w F W 0106 0139 0033 in tw all t 0003 0004 tw 0007 in wmax w tw 0033 0007 0026 in wmin w tw 0033 0007 0040 in The minimum squeeze is 0026 in Ans b Ymax Do 4012 in Ymin max099Do Do 006 max39719 3952 3972 in Y 3992 0020 in Do w Y 0 w Y Do w Y Do 3992 4012 0020 in tw all t tY tDo 0020 0036 0056 in w 0020 0056 in wmax 0036 in wmin 0076 in Oring is more likely compressed than free prior to assembly of the end plate 237 a Figure defines w as gap w F W w F W 432 533 101 mm tw all t tF tW 013 013 026 mm wmax w tw 101 026 075 mm wmin w tw 101 026 127 mm The Oring is squeezed at least 075 mm b Ymax Do 21958 mm Ymin max099Do Do 152 max09921958 21958 152 21738 mm Y 21848 110 mm From the figure the stochastic equation is Do w Y or w Y Do w Y Do 21848 21958 110 mm tw all t tY tDo 110 034 144 mm wmax w tw 110 144 034 mm wmin w tw 110 144 254 mm The Oring is more likely to be circumferentially compressed than free prior to assembly of the end plate 238 wmax 0020 in wmin 0040 in w 12 0020 0040 0030 in tw 12 0020 0040 0010 in b 0750 0001 in c 0120 0005 in d 0875 0001 in w ā b c d 0030 ā 0875 0120 0750 ā 0875 0120 0750 0030 ā 1715 in Absolute tw Σall t 0010 ta 0001 0005 0001 ta 0010 0001 0005 0001 0003 in a 1715 0003 in Ans Statistical For a normal distribution of dimensions tw2 Σall t2 ta2 tb2 tc2 td2 ta tw2 tb2 tc2 td212 00102 00012 00052 0001212 00085 a 1715 00085 in Ans 239 x n nx nx2 93 19 1767 164311 95 25 2375 225625 97 38 3685 357542 99 17 1683 166617 101 12 1212 122412 103 10 1030 106090 105 5 525 55125 107 4 428 45796 109 4 436 47524 111 2 222 24624 136 13364 1315704 x 13364136 9826 kpsi sx 1315704 133642136 13512 430 kpsi Under normal hypothesis z001 x001 9826430 x001 9826 430 z001 9826 43023267 8826 883 kpsi Ans 240 From Prob 239 μx 9826 kpsi and σx 430 kpsi Cx σx μx 430 9826 004376 From Eqs 218 and 219 μy ln9826 0043 7622 4587 σy ln1 004376212 004374 For a yield strength exceeded by 99 of the population z001 ln x001 μyσy ln x001 μy σy z001 From Table A10 for 1 failure z001 2326 Thus ln x001 4587 0043742326 4485 x001 887 kpsi Ans The normal PDF is given by Eq 214 as fx 1 430 sqrt2π exp12 x 9826 4302 For the lognormal distribution from Eq 217 defining gx gx 1 x 004374 sqrt2π exp12 ln x 45870043742 x kpsi fNw fx gx x kpsi fNw fx gx 92 000000 003215 003263 102 003676 006356 006134 92 006985 003215 003263 104 003676 003806 003708 94 006985 005680 005890 104 001838 003806 003708 94 009191 005680 005890 106 001838 001836 001869 96 009191 008081 008308 106 001471 001836 001869 96 013971 008081 008308 108 001471 000713 000793 98 013971 009261 009297 108 001471 000713 000793 98 006250 009261 009297 110 001471 000223 000286 100 006250 008548 008367 110 000735 000223 000286 100 004412 008548 008367 112 000735 000056 000089 102 004412 006356 006134 112 000000 000056 000089 Note rows are repeated to draw histogram The normal and lognormal are almost the same However the data is quite skewed and perhaps a Weibull distribution should be explored For a method of establishing the Weibull parameters see Shigley JE and CR Mischke Mechanical Engineering Design McGrawHill 5th ed 1989 Sec 412 241 Let x Sfe104 x0 79 kpsi θ 862 kpsi b 26 Eq 228 x x0 θ x0 Γ1 1b x 79 862 79 Γ1 126 79 72 Γ138 From Table A34 Γ138 088854 x 79 72 088854 854 kpsi Ans Eq 229 σx θ x0Γ1 2b Γ21 1b12 862 79Γ1 226 Γ21 12612 72 092376 088854212 264 kpsi Ans Cx σx x 264 854 0031 Ans 242 x Sut x0 277 θ 462 b 438 μx 277 462 277 Γ1 1438 277 185 Γ123 277 185 091075 4455 kpsi Ans σx 462 277Γ1 2438 Γ²1 143812 185Γ146 Γ²12312 18508856 0910 75²12 438 kpsi Ans Cx 4384455 0098 Ans From the Weibull survival equation R expx x₀θ x₀b 1 p R40 expx40 x₀θ x₀b 1 p40 exp40 277462 277438 0846 p40 1 R40 1 0846 0154 154 Ans 243 x Sut x₀ 1519 θ 1936 b 8 μx 1519 1936 1519Γ1 18 1519 417 Γ1125 1519 4170941 76 1912 kpsi Ans σx 1936 1519Γ1 28 Γ²1 1812 417Γ125 Γ²112512 4170906 40 0941 76²12 582 kpsi Ans Cx 5821912 0030 244 x Sut x₀ 476 θ 1256 b 1184 x 476 1256 476Γ1 11184 x 476 78 Γ108 476 780959 73 1225 kpsi σx 1256 476Γ1 21184 Γ²1 1118412 78Γ108 Γ²11712 780959 73 0936 70²12 224 kpsi httplibrosysolucionariosnet From Prob 242 p 1 expx x₀θ θ₀b 1 exp100 4761256 4761184 00090 Ans y Sy y₀ 641 θ 810 b 377 y 641 810 641Γ1 1377 641 169 Γ127 641 1690902 50 7935 kpsi σy 81 641Γ1 2377 Γ1 137712 σy 1690887 57 0902 50²12 457 kpsi p 1 expy y₀θ y₀377 p 1 exp70 64181 641377 0019 Ans 245 x Sut W1223 1346 364 kpsi px 120 1 100 since x₀ 120 kpsi px 133 exp133 12231346 1223364 0548 548 Ans 246 Using Eqs 228 and 229 and Table A34 μn n₀ θ n₀Γ1 1b 369 1336 369Γ1 1266 12285 kcycles σn θ n₀Γ1 2b Γ²1 1b 3479 kcycles For the Weibull density function Eq 227 fWn 2661336 369 n 3691336 369266 1 expn 3691336 369266 For the lognormal distribution Eqs 218 and 219 give μy ln12285 34791228522 4771 σy 1 3479122852 02778 httplibrosysolucionariosnet From Eq 217 the lognormal PDF is fLNn 102778 n 2π exp12 ln n 4771027782 We form a table of densities fWn and fLNn and plot n kcycles fWn fLNn 40 91E05 182E05 50 0000 991 0000 241 60 0002 498 0001 233 70 0004 380 0003 501 80 0006 401 0006 739 90 0008 301 0009 913 100 0009 822 0012 022 110 0010 750 0012 644 120 0010 965 0011 947 130 0010 459 0010 399 140 0009 346 0008 492 150 0007 827 0006 597 160 0006 139 0004 926 170 0004 507 0003 564 180 0003 092 0002 515 190 0001 979 0001 739 200 0001 180 0001 184 210 0000 654 0000 795 220 0000 336 0000 529 The Weibull L10 life comes from Eq 226 with a reliability of R 090 Thus n010 369 133 369ln10901266 781 kcycles Ans httplibrosysolucionariosnet The lognormal L10 life comes from the definition of the z variable That is ln n0 μy σy z or n0 expμy σy z From Table A10 for R 090 z 1282 Thus n0 exp4771 027781282 827 kcycles Ans 247 Form a table x gx i L105 fi fix105 fix21010 105 1 305 3 915 279075 00557 2 355 7 2485 882175 01474 3 405 11 4455 1804275 02514 4 455 16 7280 33124 03168 5 505 21 10605 5355525 03216 6 555 13 7215 4004325 02789 7 605 13 7865 4758325 02151 8 655 6 3930 257415 01517 9 705 2 1410 99405 01000 10 755 0 0 0 00625 11 805 4 3220 25921 00375 12 855 3 2565 2193075 00218 13 905 0 0 0 00124 14 955 0 0 0 00069 15 1005 1 1005 1010025 00038 100 52950 297595 x 5295105100 5295105 cycles Ans sx 2975951010 52951052 100 100 1 12 1319105 cycles Ans Cx sx 13195295 0249 μy ln 5295105 02492 2 13149 σy ln1 02492 0245 gx 1 x σy 2π exp12 ln x μy σy 2 gx 1628 x exp12 ln x 13149 0245 2 Superposed histogram and PDF 248 x Su W703 844 201 Eq 228 μx 703 844 703Γ1 1201 703 844 703Γ1498 703 844 703 0886 17 828 kpsi Ans Eq 229 σx 844 703Γ1 2201 Γ21 120112 σx 1410997 91 0886 17212 6502 kpsi Cx 6502 828 0079 Ans 249 Take the Weibull equation for the standard deviation σx θ x0Γ1 2b Γ21 1b12 and the mean equation solved for x x0 x x0 θ x0Γ1 1b Dividing the first by the second σx x x0 Γ1 2b Γ21 1b12 Γ1 1b 42 49 338 Γ1 2b Γ21 1b 1 R 02763 Make a table and solve for b iteratively b 1 2b 1 1b Γ1 2b Γ1 1b 3 167 133 0903 30 0893 38 0363 4 15 125 0886 23 0906 40 0280 41 149 124 0885 95 0908 52 0271 b 4068 Using MathCad Ans x x0 θ x0 Γ1 1b 338 49 338 Γ1 14068 498 kpsi Ans 250 x Sy W347 39 293 kpsi x 347 39 347Γ1 1293 347 43Γ134 347 430892 22 385 kpsi σx 39 347Γ1 2293 Γ21 129312 43Γ168 Γ213412 430905 00 0892 22212 142 kpsi Ans Cx 142 385 0037 Ans 251 x Mrev f fx fx2 1 11 11 11 2 22 44 88 3 38 114 342 4 57 228 912 5 31 155 775 6 19 114 684 7 15 105 735 8 12 96 768 9 11 99 891 10 9 90 900 11 7 77 847 12 5 60 720 Sum 78 237 1193 7673 μx 1193106237 5034106 cycles σx 76731012 11931062 237 237 1 2658106 cycles Cx 2658 5034 0528 From Eqs 218 and 219 μy ln5034106 0528²2 15292 σy sqrtln1 0528² 0496 From Eq 217 defining gx gx 1 x0496 sqrt2π exp12 ln x 15292 0496 ² x Mrev fNw gx 106 05 000000 000011 05 004641 000011 15 004641 005204 15 009283 005204 25 009283 016992 25 016034 016992 35 016034 020754 35 024051 020754 45 024051 017848 45 013080 017848 55 013080 013158 55 008017 013158 65 008017 009011 65 006329 009011 75 006329 005953 75 005063 005953 85 005063 003869 85 004641 003869 95 004641 002501 95 003797 002501 105 003797 001618 105 002954 001618 115 002954 001051 115 002110 001051 125 002110 000687 125 000000 000687 z ln x μy σy ln x μy σy z 15292 0496 z L10 life where 10 of bearings fail from Table A10 z 1282 Thus ln x 15292 04961282 1466 x 232 106 rev Ans 31 From Table A20 Sut 470 MPa 68 kpsi Sy 390 MPa 57 kpsi Ans 32 From Table A20 Sut 620 MPa 90 kpsi Sy 340 MPa 495 kpsi Ans 33 Comparison of yield strengths Sut of G10500 HR is 620 470 132 times larger than SAE1020 CD Ans Syt of SAE1020 CD is 390 340 115 times larger than G10500 HR Ans From Table A20 the ductilities reduction in areas show SAE1020 CD is 40 35 114 times larger than G10500 Ans The stiffness values of these materials are identical Ans Table A20 Table A5 Sut Sy Ductility Stiffness MPa kpsi MPa kpsi R GPa Mpsi SAE1020 CD 47068 390 57 40 20730 UNS10500 HR 62090 340495 35 20730 34 From Table A21 1040 QT Sy 593 86 MPa kpsi at 205C 400F Ans 35 From Table A21 1040 QT R 65 at 650C 1200F Ans 36 Using Table A5 the specific strengths are UNS G10350 HR steel Sy W 395103 0282 140105 in Ans 2024 T4 aluminum Sy W 43103 0098 439105 in Ans Ti6Al4V titanium Sy W 140103 016 875105 in Ans ASTM 30 gray cast iron has no yield strength Ans Chapter 3 shi20396ch03qxd 81803 1018 AM Page 40 httplibrosysolucionariosnet 37 The specific moduli are UNS G10350 HR steel EW 301060282 106108 in Ans 2024 T4 aluminum EW 1031060098 105108 in Ans Ti6Al4V titanium EW 165106016 103108 in Ans Gray cast iron EW 145106026 558107 in Ans 38 2G1 ν E ν E 2G2G From Table A5 Steel ν 30 21152115 0304 Ans Aluminum ν 104 23902390 0333 Ans Beryllium copper ν 18 2727 0286 Ans Gray cast iron ν 145 2626 0208 Ans 39 Su 855 kpsi Ans Sy 455 kpsi Ans E 900003 30 000 kpsi Ans R A0 AFA0 01987 0107701987 100 458 Ans ε Δll0 l l0l0 l0A0 1 310 To plot σtrue vs ε the following equations are applied to the data A0 π0503²4 01987 in² Eq 34 ε lnll0 for 0 ΔL 00028 in ε lnA0A for ΔL 00028 in σtrue PA The results are summarized in the table below and plotted on the next page The last 5 points of data are used to plot log σ vs log ε The curve fit gives m 02306 log σ0 51852 σ0 1532 kpsi Ans For 20 cold work Eq 310 and Eq 313 give A A01 W 019871 02 01590 in² ε lnA0A ln0198701590 02231 Eq 314 Sy σ0 εm 15320223102306 1084 kpsi Ans Eq 315 with Su 855 kpsi from Prob 39 Su Su1 W 8551 02 1069 kpsi Ans Table with columns P ΔL A ε σtrue log ε log σtrue 0 0 0198 713 0 0 1 000 00004 0198 713 00002 5032388 3699 01 3701 774 2 000 00006 0198 713 00003 1006478 3522 94 4002 804 3 000 00010 0198 713 00005 1509717 3301 14 4178 895 4 000 00013 0198 713 000065 2012955 3187 23 4303 834 7 000 00023 0198 713 0001149 3522672 2939 55 4546 872 8 400 00028 0198 713 0001 399 4227206 2854 18 4626 053 8 800 00036 0198 4 0001 575 4435484 2802 61 4646 941 9 200 00089 0197 8 0004 604 4651163 2336 85 4667 562 9 100 0196 3 0012 216 4635762 1913 05 4666 121 13 200 0192 4 0032 284 6860707 1491 01 4836 369 15 200 0187 5 0058 082 8106667 1235 96 4908 842 17 000 0156 3 0240 083 1087652 0619 64 5036 49 16 400 0130 7 0418 956 1254782 0377 83 5098 568 14 800 0107 7 0612 511 1374188 0212 89 5138 046 Tangent modulus at σ 0 is E0 ΔσΔε 5000 002103 0 25106 psi At σ 20 kpsi E20 26 1910315 1103 140106 psi Ans Let x represent ε103 and y represent σ kpsi table of values for x y x2 x3 xy Substituting a1 2972632362 874668523621142632362 6852362 2099367 a2 21148746 29768523621142632362 6852362 214242 The tangent modulus is dydx dσdε 2099367 2214242x 2099367 428483x At σ 0 E0 2099 Mpsi Ans At σ 20 kpsi 20 2099367x 214242x2 x 1069 873 Taking the first root ε 1069 and the tangent modulus is E20 2099367 4284831069 1641 Mpsi Ans Determine the equation for the 01 percent offset line y 2099x b at y 0 x 1 b 2099 y 2099x 2099 2099367x 214242x2 214242x2 2099 0 x 3130 Sy0001 2099313 21423132 447 kpsi Ans The roots are N R1 1 hR12 The sign being significant N R1 hR12 1 Ans Substitute for N in ε0 lnRhRN Gives ε0 lnRhRR1 hR12 R ln1 hR12 Ans These constitute a useful pair of equations in coldforming situations allowing the surface strains to be found so that coldworking strength enhancement can be estimated 314 τ 16Tπd3 16Tπ1253 1061033 26076T MPa γ θπ180rL θπ180125350 62333104θ For G take the first 10 data points for the linear part of the curve table with T θdeg γ103 τMPa γ103x τMPay x2 xy and their respective values y mx b τ y γ x where m is the shear modulus G m NΣxy ΣxΣy NΣx2 Σx2 773 MPa103 773 GPa Ans b Σy mΣx N 1462 MPa From curve Sys 200 MPa Ans Note since τ is not uniform the offset yield does not apply so we are using the elastic limit as an approximation 315 x f fx fx2 385 2 770 296450 395 9 3555 1404225 405 30 12150 4920750 415 65 26975 11194630 425 101 42925 18243130 435 112 48720 21193200 445 90 40050 17822250 455 54 24570 11179350 465 25 11625 5405625 475 9 4275 2030625 485 2 970 470450 495 1 495 245025 Σ 5280 500 217080 94405700 x 21708500 43416 σx 944057 217082 500 500 1 17808 Cx 17808 43416 004102 y ln 43416 ln1 0041022 37691 httplibrosysolucionariosnet σy ln1 0041022 00410 gx 1 x004102π exp12ln x 37691004102 x fNw gx x fNw gx 38 0 0001488 45 0180 0142268 38 0004 0001488 45 0108 0142268 39 0004 0009057 46 0108 0073814 39 0018 0009057 46 0050 0073814 40 0018 0035793 47 0050 0029410 40 0060 0035793 47 0018 0029410 41 0060 0094704 48 0018 0009152 41 0130 0094704 48 0004 0009152 42 0130 0172538 49 0004 0002259 42 0202 0172538 49 0002 0002259 43 0202 0222074 50 0002 0000449 43 0224 0222074 50 0 0000449 44 0224 0206748 44 0180 0206748 Sy LN4342 1781 kpsi Ans 316 From Table A22 AISI 1212 Sy 280 kpsi σf 106 kpsi Sut 615 kpsi σ0 110 kpsi m 024 εf 085 From Eq 312 εu m 024 Eq 310 A0 Ai 11 W 11 02 125 Eq 313 εi ln 125 02231 εi εu httplibrosysolucionariosnet Eq 314 Sy σ0 εim 11002231024 767 kpsi Ans Eq 315 Su Su 1 W 615 1 02 769 kpsi Ans 317 For HB 250 Eq 317 Su 0495 250 124 kpsi 341 250 853 MPa Ans 318 For the data given Σ HB 2530 Σ HB2 640226 HB 2530 10 253 σHB 640226 25302 10 9 3887 Eq 317 Su 0495253 1252 kpsi Ans σsu 0495 3887 192 kpsi Ans 319 From Prob 318 HB 253 and σHB 3887 Eq 318 Su 023253 125 457 kpsi Ans σsu 0233887 0894 kpsi Ans 320 a uR 4552 230 345 in lbfin3 Ans b P ΔL A A0A 1 ε σ PA0 0 0 0 0 1000 00004 00002 503239 2000 00006 00003 1006478 3000 00010 00005 1509717 4000 00013 000065 2012955 7000 00023 000115 3522672 8400 00028 00014 4227206 8800 00036 00018 4428502 9200 00089 000445 4629797 9100 01963 0012291 4579473 13200 01924 0032811 6642753 15200 01875 0059802 7649230 17000 01563 0271355 8555060 16400 01307 0520373 8253117 14800 01077 0845059 7447935 httplibrosysolucionariosnet uT i15Ai 12 43 0000001 545 0000004 45 0001 512 45 00076 5000059 8 0004 4581 00004 0059 880 0000845 04 667103 inlbfin3 Ans Chapter 4 41 1 RC RA RB RD C A B W D 1 2 3 RB RA W RB RC RA 2 1 W RA RBx RBx RBy RBy RB 2 1 1 Scale of corner magnified W A B e f d W A RA RB B 1 2 W A RA RB B 1 1 2 a b c shi20396ch04qxd 81803 1035 AM Page 50 httplibrosysolucionariosnet RA 2 sin 60 1732 kN Ans RB 2 sin 30 1 kN Ans S 06 m α tan¹ 060406 3096 RAsin 135 800sin 3096 RA 1100 N Ans ROsin 1404 800sin 3096 RO 377 N Ans RO 12tan 30 2078 kN Ans RA 12sin 30 24 kN Ans h 45tan 30 7794 m MA 0 9RE 7794400 cos 30 45400 sin 30 0 RE 400 N Ans Fx 0 RAx 400 cos 30 0 RAx 3464 N Fy 0 RAy 400 400 sin 30 0 RAy 200 N RA 34642 2002 400 N Ans Step 2 Find components of RC on link 4 and RD MC 0 40045 7794 19RD 0 RD 3054 N Ans Fx 0 RCx4 3054 N Fy 0 RCy4 400 N Step 3 Find components of RC on link 2 Fx 0 RCx2 3054 3464 0 RCx2 41 N Fy 0 RCy2 200 N Ans M0 0 1860 14R2 830 440 0 R2 7143 lbf Fy 0 R1 40 30 7143 60 0 R1 143 lbf M1 1434 572 lbfin M2 572 41434 17144 lbfin M3 17144 11436 240 lbf in M4 240 604 0 checks b Fy 0 R0 2 40150 26 kN M0 0 M0 200002 400001500425 655 Nm M1 655 260002 135 Nm M2 135 6000150 45 Nm M3 45 12 6000150 0 checks c M0 0 10R2 61000 0 R2 600 lbf Fy 0 R1 1000 600 0 R1 400 lbf M1 4006 2400 lbfft M2 2400 6004 0 checks d MC 0 10R1 22000 81000 0 R1 1200 lbf Fy 0 1200 1000 2000 R2 0 R2 1800 lbf M1 12002 2400 lbfft M2 2400 2006 3600 lbfft M3 3600 18002 0 checks e MB 0 7R1 3400 3800 0 R1 1714 lbf Fy 0 1714 400 R2 800 0 R2 13714 lbf M1 17144 6857 lbfft M2 6857 57143 2400 lbfft M3 2400 8003 0 checks f Break at A R1 VA 12 408 160 lbf MD 0 12160 10R2 3205 0 R2 352 lbf Fy 0 160 352 320 R3 0 R3 128 lbf M1 12 1604 320 lbfin M2 320 12 1604 0 checks hinge M3 0 1602 320 lbfin M4 320 1925 640 lbfin M5 640 1285 0 checks Chapter 4 55 44 a q R1x1 40x 41 30x 81 R2x 141 60x 181 V R1 40x 40 30x 80 R2x 140 60x 180 1 M R1x 40x 41 30x 81 R2x 141 60x 181 2 for x 18 V 0 and M 0 Eqs 1 and 2 give 0 R1 40 30 R2 60 R1 R2 70 3 0 R118 4014 3010 4R2 9R1 2R2 130 4 Solve 3 and 4 simultaneously to get R1 143 lbf R2 7143 lbf Ans From Eqs 1 and 2 at x 0 V R1 143 lbf M 0 x 4 V 143 40 4143 M 143x x 8 V 143 40 30 1143 M 1438 408 41 17144 x 14 V 143 40 30 7143 60 M 14314 4014 4 3014 8 240 x 18 V 0 M 0 See curves of V and M in Prob 43 solution b q R0x1 M0x2 2000x 021 4000x 0350 4000x 050 V R0 M0x1 2000x 020 4000x 0351 4000x 051 1 M R0x M0 2000x 021 2000x 0352 2000x 052 2 at x 05m V M 0 Eqs 1 and 2 give R0 2000 400005 035 0 R1 2600 N 26 kN Ans R005 M0 200005 02 200005 0352 0 with R0 2600 N M0 655 N m Ans With R0 and M0 Eqs 1 and 2 give the same V and M curves as Prob 43 note for V M0x1 has no physical meaning c q R1x1 1000x 61 R2x 101 V R1 1000x 60 R2x 100 1 M R1x 1000x 61 R2x 101 2 at x 10 ft V M 0 Eqs 1 and 2 give R1 1000 R2 0 R1 R2 1000 10R1 100010 6 0 R1 400 lbf R2 1000 400 600 lbf 0 x 6 V 400 lbf M 400x 6 x 10 V 400 1000x 60 600 lbf M 400x 1000x 6 6000 600x See curves of Prob 43 solution shi20396ch04qxd 81803 1035 AM Page 55 httplibrosysolucionariosnet 56 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design d q R1x1 1000x 21 2000x 81 R2x 101 V R1 1000x 20 2000x 80 R2x 100 1 M R1x 1000x 21 2000x 81 R2x 101 2 At x 10 V M 0 from Eqs 1 and 2 R1 1000 2000 R2 0 R1 R2 3000 10R1 100010 2 200010 8 0 R1 1200 lbf R2 3000 1200 1800 lbf 0 x 2 V 1200 lbf M 1200x lbf ft 2 x 8 V 1200 1000 200 lbf M 1200x 1000x 2 200x 2000 lbf ft 8 x 10 V 1200 1000 2000 1800 lbf M 1200x 1000x 2 2000x 8 1800x 18 000 lbf ft Plots are the same as in Prob 43 e q R1x1 400x 41 R2x 71 800x 101 V R1 400x 40 R2x 70 800x 100 1 M R1x 400x 41 R2x 71 800x 101 2 at x 10 V M 0 R1 400 R2 800 0 R1 R2 1200 3 10R1 4006 R23 0 10R1 3R2 2400 4 Solve Eqs 3 and 4 simultaneously R1 1714 lbf R2 13714 lbf 0 x 4 V 1714 lbf M 1714x lbf ft 4 x 7 V 1714 400 5714 lbf M 1714x 400x 4 lbf ft 5714x 1600 7 x 10 V 1714 400 13714 800 lbf M 1714x 400x 4 13714x 7 800x 8000 lbf ft Plots are the same as in Prob 43 f q R1x1 40x0 40x 80 R2x 101 320x 151 R3x 20 V R1 40x 40x 81 R2x 100 320x 150 R3x 200 1 M R1x 20x2 20x 82 R2x 101 320x 151 R3x 201 2 M 0 at x 8 in 8R1 2082 0 R1 160 lbf at x 20 V and M 0 160 4020 4012 R2 320 R3 0 R2 R3 480 16020 20202 20122 10R2 3205 0 R2 352 lbf R3 480 352 128 lbf 0 x 8 V 160 40x lbf M 160x 20x2 lbf in 8 x 10 V 160 40x 40x 8 160 lbf M 160x 20x2 20x 82 1280 160x lbf in shi20396ch04qxd 81803 1035 AM Page 56 httplibrosysolucionariosnet 10 x 15 V 160 40x 40x 8 352 192 lbf M 160x 20x² 20x 8 352x 10 192x 2240 15 x 20 V 160 40x 40x 8 352 320 128 lbf M 160x 20x² 20x 8 352x 10 320x 15 128x 2560 Plots of V and M are the same as in Prob 43 45 Solution depends upon the beam selected 46 a Moment at center xc l 2a2 Mc w2l2l 2a l2² wl2 l4 a At reaction Mr wa²2 a 225 l 10 in w 100 lbfin Mc 100102 104 225 125 lbfin Mr 100225²2 2531 lbfin Ans b Minimum occurs when Mc Mr wl2 l4 a wa²2 a² al 025l² 0 Taking the positive root a 12l l² 4025l² l2 2 1 02071l Ans for l 10 in and w 100 lbf Mmin 10020207110² 2145 lbfin 47 For the ith wire from bottom from summing forces vertically a Ti i 1W From summing moments about point a Ma Wl xi iWxi 0 Giving xi li 1 So W l 1 1 l 2 x l 2 1 l 3 y l 3 1 l 4 z l 4 1 l 5 b With straight rigid wires the mobile is not stable Any perturbation can lead to all wires becoming collinear Consider a wire of length l bent at its string support Mₐ 0 Mₐ i W l i 1 cos α i l W i 1 cos β 0 i W l i 1 cos α cos β 0 Moment vanishes when α β for any wire Consider a ccw rotation angle β which makes α α β and β α β Mₐ i W l i 1 cosα β cosα β 2 i W l i 1 sin α sin β 2 i W l β i 1 sin α There exists a correcting moment of opposite sense to arbitrary rotation β An equation for an upward bend can be found by changing the sign of W The moment will no longer be correcting A curved convexupward bend of wire will produce stable equilibrium too but the equation would change somewhat φₚ 12 tan¹43 266 cw τ₁ R 5 φₛ 45 266 184 ccw C 9 16 2 125 CD 16 9 2 35 R 5² 35² 610 σ₁ 61 125 186 φₚ 12 tan¹535 275 ccw σ₂ 125 61 64 τ₁ R 610 φₛ 45 275 175 cw C 24 10 2 17 CD 24 10 2 7 R 7² 6² 922 σ₁ 17 922 2622 σ₂ 17 922 778 φₚ 12 90 tan¹76 697 ccw τ₁ R 922 φₛ 697 45 247 ccw C 9 19 2 14 CD 19 9 2 5 R 5² 8² 9434 σ₁ 14 943 2343 σ₂ 14 943 457 φₚ 12 90 tan¹58 610 cw τ₁ R 9434 φₛ 61 45 16 cw 49 a C 12424 CD 12428 R sqrt82 72 1063 sigma1 410631463 sigma2 41063 663 phip 12 90 tan1 87 694 ccw tau1 R 1063 phis 694 45 244 ccw b C 65205 CD 65255 R sqrt552 82 971 sigma105 971 1021 sigma205 971 921 phip 12 tan1 855 2775 ccw tau1 R 971 phis 45 2775 1725 cw c C 8 72 05 CD 8 72 75 R sqrt752 62 960 sigma1960 05 910 sigma2 05 96 101 phip 12 90 tan1 756 7067 cw tau1 R 960 phis 7067 45 2567 cw d C 96215 CD 96275 R sqrt752 32 8078 sigma1 15 8078 958 sigma2 15 8078 658 phip 12 tan1 375 109 cw tau1 R 8078 phis 45 109 341 ccw 410 a C 201025 CD 2010215 R sqrt152 82 17 sigma1 5 17 22 sigma2 5 17 12 phip 12 tan1 815 1404 cw tau1 R 17 phis 45 1404 3096 ccw b C 30 102 10 CD 30 102 20 R 20² 10² 2236 σ₁ 10 2236 3236 σ₂ 10 2236 1236 ϕp 12 tan¹1020 1328 ccw τ₁ R 2236 ϕs 45 1328 3172 cw c C 10 182 4 CD 10 182 14 R 14² 9² 1664 σ₁ 4 1664 2064 σ₂ 4 1664 1264 ϕp 12 90 tan¹ 149 7363 cw τ₁ R 1664 ϕs 7363 45 2863 cw d C 12 222 5 CD 12 222 17 R 17² 12² 2081 σ₁ 5 2081 2581 σ₂ 5 2081 1581 ϕp 12 90 tan¹ 1712 7239 cw τ₁ R 2081 ϕs 7239 45 2739 cw 411 a b C 0 102 5 CD 10 02 5 R 5² 4² 640 σ₁ 5 640 1140 σ₂ 0 σ₃ 5 640 140 τ₁₃ R 640 τ₁₂ 11402 570 τ₂₃ 1402 070 c C 2 82 5 CD 8 22 3 R 3² 4² 5 σ₁ 5 5 0 σ₂ 0 σ₃ 5 5 10 τ₁₃ 102 5 τ₁₂ 0 τ₂₃ 5 d C 10 302 10 CD 10 302 20 R 20² 10² 2236 σ₁ 10 2236 1236 σ₂ 0 σ₃ 10 2236 3236 τ₁₃ 2236 τ₁₂ 12362 618 τ₂₃ 32362 1618 412 a C 80 302 55 CD 80 302 25 R 25² 20² 3202 σ₁ 0 σ₂ 55 3202 2298 230 σ₃ 55 320 870 τ₁₂ 232 115 τ₂₃ 320 τ₁₃ 872 435 b C 30 60 2 15 CD 60 30 2 45 R 45² 30² 541 σ₁ 15 541 391 σ₂ 0 σ₃ 15 541 691 τ₁₃ 391 691 2 541 τ₁₂ 391 2 196 τ₂₃ 691 2 346 c C 40 0 2 20 CD 40 0 2 20 R 20² 20² 283 σ₁ 20 283 483 σ₂ 20 283 83 σ₃ σz 30 τ₁₃ 483 30 2 391 τ₁₂ 283 τ₂₃ 30 83 2 109 d C 50 2 25 CD 50 2 25 R 25² 30² 391 σ₁ 25 391 641 σ₂ 25 391 141 σ₃ σz 20 τ₁₃ 641 20 2 421 τ₁₂ 391 τ₂₃ 20 141 2 295 413 σ F A 2000 π 405² 10 190 psi 1019 kpsi Ans δ FL AE σ L E 10 190 72 3010⁶ 002446 in Ans ε₁ δ L 002446 72 34010⁶ 340μ Ans From Table A5 v 0292 ε₂ vε₁ 0292340 993μ Ans Δd ε₂ d 99310⁶05 49610⁶ in Ans 414 From Table A5 E 717 GPa δ σ L E 13510⁶ 3 71710⁹ 56510³ m 565 mm Ans 415 From Table 42 biaxial case From Table A5 E 207 GPa and v 0292 σₓ Eεₓ vεy 1 v² 20710⁹00021 0292000067 1 0292²10⁶ 431 MPa Ans σᵧ 20710⁹000067 029200021 1 0292²10⁶ 129 MPa Ans 416 The engineer has assumed the stress to be uniform That is ΣFₜ F cosθ τA 0 τ F A cosθ When failure occurs in shear Ssu F A cosθ The uniform stress assumption is common practice but is not exact If interested in the details see p 570 of 6th edition 417 From Eq 415 σ³ 2 6 4σ² 26 24 64 3² 2² 5²σ 264 2325 22² 65² 43² 0 σ³ 66σ 118 0 Roots are 7012 189 8903 kpsi Ans τ₁₂ 7012 189 2 256 kpsi τ₂₃ 8903 189 2 540 kpsi τmax τ₁₃ 8903 7012 2 796 kpsi Ans Note For Probs 417 to 419 one can also find the eigenvalues of the matrix σ σₓ τₓᵧ τ𝓏ₓ τₓᵧ σᵧ τᵧ𝓏 τ𝓏ₓ τᵧ𝓏 σz for the principal stresses 418 From Eq 415 σ³ 10 0 10σ² 100 1010 010 20² 102² 0²σ 10010 2201020 10102² 00² 1020² 0 σ³ 20σ² 500σ 6000 0 Roots are 30 10 20 MPa Ans τ₁₂ 30 10 2 10 MPa τ₂₃ 10 20 2 15 MPa τmax τ₁₃ 30 20 2 25 MPa Ans 419 From Eq 415 σ³ 1 4 4σ² 14 14 44 2² 4² 2²σ 144 2242 14² 42² 42² 0 σ³ 9σ² 0 156 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 68 See Prob 67 for plot a For all methods n O B O A 155 103 15 b BCM n O D OC 14 08 175 All other methods n OE OC 155 08 19 c For all methods n OL OK 52 068 76 d MNS n O J OF 512 082 62 BCM n OG OF 285 082 35 M1M n O H OF 33 082 40 M2M n OI OF 382 082 47 69 Given Sy 42 kpsi Sut 662 kpsi εf 090 Since εf 005 the material is ductile and thus we may follow convention by setting Syc Syt Use DE theory for analytical solution For σ use Eq 613 or 615 for plane stress and Eq 612 or 614 for general 3D a σ 92 95 5212 1229 kpsi n 42 1229 342 Ans b σ 122 33212 1308 kpsi n 42 1308 321 Ans c σ 42 49 92 35212 1166 kpsi n 42 1166 360 Ans d σ 112 114 42 31212 9798 n 42 9798 429 Ans shi20396ch06qxd 81803 1222 PM Page 156 httplibrosysolucionariosnet 184 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design The remaining Marin factors are ka 5775700718 0606 kc kd ke kf 1 Eq 717 Se 060608882873 MPa 1546 MPa Eq 713 a 095702 1546 1702 Eq 714 b 1 3 log 09570 1546 0173 64 Eq 712 Sf aN b 17021040173 64 3439 MPa n Sf σa or σa Sf n 6800 103 b3 3439 15 b 276 mm Check values for kb Se etc kb 127142760107 0891 Se 060608912873 1551 MPa a 095702 1551 1697 b 1 3 log 09570 1551 0173 17 Sf 16971040173 17 3444 MPa 6800 103 b3 3444 15 b 275 mm Ans 710 Table A20 Sut 440 MPa Sy 370 MPa S e 0504440 2218 MPa Table 74 ka 4514400265 0899 kb 1 axial loading Eq 725 kc 085 Se 089910852218 1695 MPa Table A151 dw 1260 02 Kt 25 12 Fa Fa 10 60 1018 shi20396ch07qxd 81803 1235 PM Page 184 httplibrosysolucionariosnet Chapter 8 215 d LG H 1109 04375 15465 in This would be rounded to 175 in per Table A17 The bolt is long enough Ans e ld L LT 175 125 0500 in Ans lt LG ld 1109 0500 0609 in Ans These lengths are needed to estimate bolt spring rate kb Note In an analysis problem you need not know the fasteners length at the outset although you can certainly check if appropriate 812 a LT 2D 6 214 6 34 mm Ans b From Table A33 the maximum washer thickness is 35 mm Thus the grip is LG 14 14 35 315 mm Ans c From Table A31 H 128 mm d LG H 315 128 443 mm This would be rounded to L 50 mm The bolt is long enough Ans e ld L LT 50 34 16 mm Ans lt LG ld 315 16 155 mm Ans These lengths are needed to estimate the bolt spring rate kb 813 a LT 2D 1 4 205 025 125 in Ans b L G h d 2 t1 d 2 0875 05 2 1125 in Ans c L h 15d t1 15d 0875 1505 1625 in From Table A17 this rounds to 175 in The cap screw is long enough Ans d ld L LT 175 125 0500 in Ans lt L G ld 1125 05 0625 in Ans 814 a LT 212 6 30 mm Ans b L G h d 2 t1 d 2 20 12 2 26 mm Ans c L h 15d t1 15d 20 1512 38 mm This rounds to 40 mm Table A17 The fastener is long enough Ans d ld L LT 40 30 10 mm Ans lT L G ld 26 10 16 mm Ans shi20396ch08qxd 81803 1242 PM Page 215 httplibrosysolucionariosnet Chapter 8 221 From Eq 820 Top frustum D 18 t 13 E 207 GPa k1 5316 MNm Midfrustum t 7 E 207 GPa D 249 mm k2 15 620 MNm Bottom frustum D 18 t 6 E 100 GPa k3 3887 MNm km 1 15316 155 620 13887 2158 MNm Ans C 744 744 2158 0256 Ans From Prob 822 Fi 379 kN n Sp At Fi C P 60000843 379 0256848 584 Ans 824 Calculation of bolt stiffness H 716 in LT 212 14 1 14 in LG 12 58 0095 122 in L 1125 716 0095 166 in Use L 175 in ld L LT 175 125 0500 in lt 1125 0095 0500 072 in Ad π05024 01963 in2 At 01419 in2 UNC kt At E lt 0141930 072 59125 Mlbfin kd Ad E ld 0196330 0500 11778 Mlbfin kb 1 159125 111778 3936 Mlbfin Ans 5 8 1 2 0095 3 4 1454 1327 3 4 0860 122 061 shi20396ch08qxd 81803 1242 PM Page 221 httplibrosysolucionariosnet Chapter 8 223 826 Refer to Prob 824 and its solutionAdditional information A 35 in Ds 425 in static pressure 1500 psi Db 6 inC joint constant 0267 ten SAE grade 5 bolts P 1 10 π4252 4 1500 2128 lbf From Tables 82 and 89 At 01419 in2 Sp 85 000 psi Fi 0750141985 9046 kip From Eq 828 n Sp At Fi C P 8501419 9046 02672128 531 Ans 827 From Fig 821 t1 025 in h 025 0065 0315 in l h d2 0315 316 05025 in D1 150375 057705025 08524 in D2 150375 05625 in l2 050252 0251 25 in Frustum 1 Washer E 30 Mpsi t 0065 in D 05625 in k 7857 Mlbfin by computer Frustum 2 Cap portion E 14 Mpsi t 0186 25 in D 05625 200650577 06375 in k 2346 Mlbfin by computer Frustum 3 Frame and Cap E 14 Mpsi t 0251 25 in D 05625 in k 1431 Mlbfin by computer km 1 17857 12346 11431 799 Mlbfin Ans 08524 05625 025125 08524 06375 018625 05625 06375 0065 shi20396ch08qxd 81803 1242 PM Page 223 httplibrosysolucionariosnet Chapter 8 225 b Washerst 34 mm d 20 mm D 30 mm E 207 GPa k1 42 175 MNm Cast iron t 20 mm d 20 mm D 30 234 tan 30 3393 mm E 135 GPa k2 7885 MNm Steel t 20 mm d 20 mm D 3393 mm E 207 GPa k3 12 090 MNm km 242 175 17885 112 0901 3892 MNm Bolt LG 468 mm Nut H 18 mm L 468 18 648 mm Use L 80 mm LT 220 6 46 mm ld 80 46 34 mm lt 468 34 128 mm At 245 mm2 Ad π2024 3142 mm2 kb Ad At E Adlt Atld 3142245207 3142128 24534 1290 MNm C 12901290 3892 02489 Sp 600 MPa Fi 1323 kN n Sp At Fi CPN 6000245 1323 02489154 157 Ans Bolts are a bit oversized for the load 830 a ISO M 20 25 grade 88 coarse pitch bolts lubricated Table 82 At 245 mm2 Table 811 Sp 600 MPa Ad π2024 3142 mm2 Fp 2450600 147 kN Fi 090Fp 090147 1323 kN T 018132320 476 N m Ans b L LG H 48 18 66 mm Therefore set L 80 mm per Table A17 LT 2D 6 220 6 46 mm ld L LT 80 46 34 mm lt LG ld 48 34 14 mm 14 Not to scale 80 48 grip 46 34 shi20396ch08qxd 81803 1242 PM Page 225 httplibrosysolucionariosnet Chapter 8 239 Bearing on member Ab td 1010 100 mm2 σb 7103 100 70 MPa n Sy σ 370 70 529 Strength of member At A M 14200 280 N m IA 1 1210503 10103 1033103 mm4 σA Mc IA 28025 1033103103 6776 MPa n Sy σA 370 6776 546 At C M 14350 490 N m IC 1 1210503 1042103 mm4 σC 49025 1042103103 11756 MPa n Sy σC 370 11756 315 546 C more critical n min272 529 315 272 Ans 845 Fs 3000 lbf P 30003 7 1286 lbf H 7 16 in l LG 1 2 1 2 0095 1095 in L LG H 1095 716 1532 in 1 2 1 2 13 4 l 3000 lbf Fs P O 3 7 3 Pivot about this point shi20396ch08qxd 81803 1242 PM Page 239 httplibrosysolucionariosnet Chapter 9 91 Eq 93 F 0707hlτ 0707516420 177 kip Ans 92 Table 96 τall 210 kpsi f 1485h kipin 1485516 464 kipin F f l 4644 1856 kip Ans 93 Table A20 1018 HR Sut 58 kpsi Sy 32 kpsi 1018 CR Sut 64 kpsi Sy 54 kpsi Coldrolled properties degrade to hotrolled properties in the neighborhood of the weld Table 94 τall min030Sut 040Sy min03058 04032 min174 128 128 kpsi for both materials Eq 93 F 0707hlτall F 07075164128 113 kip Ans 94 Eq 93 τ 2F hl 232 51642 181 kpsi Ans 95 b d 2 in a Primary shear Table 91 τ y V A F 14145162 113F kpsi F 7 1414 shi20396ch09qxd 81903 930 AM Page 246 httplibrosysolucionariosnet Chapter 10 281 a Spring over a Rod b Spring in a Hole Source Parameter Values Source Parameter Values d 0075 008 0085 d 0075 008 0085 D 0875 088 0885 D 0875 0870 0865 ID 0800 0800 0800 ID 0800 0790 0780 OD 0950 0960 0970 OD 0950 0950 0950 Eq 102 C 11667 11000 10412 Eq 102 C 11667 10875 10176 Eq 109 Na 6937 8828 11061 Eq 109 Na 6937 9136 11846 Table 101 Nt 8937 10828 13061 Table 101 Nt 8937 11136 13846 Table 101 Ls 0670 0866 1110 Table 101 Ls 0670 0891 1177 115y Ls L0 2970 3166 3410 115y Ls L0 2970 3191 3477 Eq 1013 L0cr 4603 4629 4655 Eq 1013 L0cr 4603 4576 4550 Table 104 A 201000 201000 201000 Table 104 A 201000 201000 201000 Table 104 m 0145 0145 0145 Table 104 m 0145 0145 0145 Eq 1014 Sut 292626 289900 287363 Eq 1014 Sut 292626 289900 287363 Table 106 Ssy 131681 130455 129313 Table 106 Ssy 131681 130455 129313 Eq 106 KB 1115 1122 1129 Eq 106 KB 1115 1123 1133 Eq 103 n 0973 1155 1357 Eq 103 n 0973 1167 1384 Eq 1022 fom 0282 0391 0536 Eq 1022 fom 0282 0398 0555 For ns 12 the optimal size is d 0085 in for both cases 1019 From the figure L0 120 mm OD 50 mm and d 34 mm Thus D OD d 50 34 466 mm a By counting Nt 125 turns Since the ends are squared along 14 turn on each end Na 125 05 12 turns Ans p 12012 10 mm Ans The solid stack is 13 diameters across the top and 12 across the bottom Ls 1334 442 mm Ans b d 34254 01339 in and from Table 105 G 786 GPa k d4G 8D3Na 344786109 8466312 103 1080 Nm Ans c Fs kL0 Ls 1080120 442103 819 N Ans d C Dd 46634 1371 KB 41371 2 41371 3 1096 τs 8KB Fs D πd3 81096819466 π343 271 MPa Ans 1020 One approach is to select A22747 HD steel for its low cost Then for y1 38 at F1 10 lbf k 100375 2667 lbfin Try d 0080 in 14 gauge shi20396ch10qxd 81103 440 PM Page 281 httplibrosysolucionariosnet Chapter 10 285 The shaded areas depict conditions outside the recommended design conditions Thus one spring is satisfactoryA313 as wound unpeened squared and ground d 00915 in OD 0879 0092 0971 in Nt 1584 turns 1024 The steps are the same as in Prob 1023 except that the GerberZimmerli criterion is replaced with GoodmanZimmerli Sse Ssa 1 SsmSsu The problem then proceeds as in Prob 1023 The results for the wire sizes are shown below see solution to Prob 1023 for additional details Iteration of d for the first trial d1 d2 d3 d4 d1 d2 d3 d4 d 0080 00915 01055 01205 d 0080 00915 01055 01205 m 0146 0146 0263 0263 KB 1151 1108 1078 1058 A 169000 169000 128000 128000 τa 29008 29040 29090 29127 Sut 244363 239618 231257 223311 n f 1500 1500 1500 1500 Ssu 163723 160544 154942 149618 Na 14444 6572 2951 1429 Ssy 85527 83866 80940 78159 Nt 16444 8572 4951 3429 Sse 52706 53239 54261 55345 Ls 1316 0784 0522 0413 Ssa 43513 43560 43634 43691 ymax 2875 2875 2875 2875 α 29008 29040 29090 29127 L0 4191 3659 3397 3288 β 2785 2129 1602 1228 L0cr 3809 5924 9354 14219 C 9052 12309 16856 22433 τs 85782 85876 86022 86133 D 0724 1126 1778 2703 ns 0997 0977 0941 0907 f Hz 138806 144277 148617 151618 Without checking all of the design conditions it is obvious that none of the wire sizes satisfyns 12Also the Gerber line is closer to the yield line than the Goodman Setting n f 15 for Goodman makes it impossible to reach the yield line ns 1 The table below uses n f 2 Iteration of d for the second trial d1 d2 d3 d4 d1 d2 d3 d4 d 0080 00915 01055 01205 d 0080 00915 01055 01205 m 0146 0146 0263 0263 KB 1221 1154 1108 1079 A 169000 169000 128000 128000 τa 21756 21780 21817 21845 Sut 244363 239618 231257 223311 n f 2000 2000 2000 2000 Ssu 163723 160544 154942 149618 Na 40962 17594 7609 3602 Ssy 85527 83866 80940 78159 Nt 42962 19594 9609 5602 Sse 52706 53239 54261 55345 Ls 3437 1793 1014 0675 Ssa 43513 43560 43634 43691 ymax 2875 2875 2875 2875 α 21756 21780 21817 21845 L0 6312 4668 3889 3550 β 2785 2129 1602 1228 L0cr 2691 4266 6821 10449 C 6395 8864 12292 16485 τs 64336 64407 64517 64600 D 0512 0811 1297 1986 ns 1329 1302 1255 1210 f Hz 98065 103903 108376 111418 The satisfactory spring has design specifications of A313 as wound unpeened squared and ground d 00915 in OD 0811 0092 0903 in Nt 196 turns shi20396ch10qxd 81103 440 PM Page 285 httplibrosysolucionariosnet Chapter 10 293 nyB 732 429 171 fom 76π2d2Nb 2D 4 76π200812233 20398 4 1239 A tabulation of several wire sizes follow d 0081 0085 0092 0098 0105 012 Sut 243920 242210 239427 237229 234851 230317 Ssu 163427 162281 160416 158943 157350 154312 Sr 109764 108994 107742 106753 105683 103643 Se 57809 57403 56744 56223 55659 54585 Sa 42136 41841 41360 40980 40570 39786 C 4903 5484 6547 7510 8693 11451 D 0397 0466 0602 0736 0913 1374 OD 0478 0551 0694 0834 1018 1494 Fi calc 8572 7874 6798 5987 5141 3637 Fi rd 875 975 1075 1175 1275 1375 k 36000 36000 36000 36000 36000 36000 Na 2386 1790 1138 803 555 277 Nb 2350 1754 1102 768 519 242 L0 2617 2338 2127 2126 2266 2918 L18 lbf 2874 2567 2328 2300 2412 3036 σa A 21068 20920 20680 20490 20285 19893 n f A 2000 2000 2000 2000 2000 2000 KB 1301 1264 1216 1185 1157 1117 τabody 11141 10994 10775 10617 10457 10177 τmbody 33424 32982 32326 31852 31372 30532 Ssr 73176 72663 71828 71169 70455 69095 Sse 38519 38249 37809 37462 37087 36371 n f body 2531 2547 2569 2583 2596 2616 KB 1250 1250 1250 1250 1250 1250 τaB 10705 10872 11080 11200 11294 11391 τmB 32114 32615 33240 33601 33883 34173 SsrB 68298 67819 67040 66424 65758 64489 SseB 35708 35458 35050 34728 34380 33717 n f B 2519 2463 2388 2341 2298 2235 Sy 134156 133215 131685 130476 129168 126674 σAmax 84273 83682 82720 81961 81139 79573 nyA 1592 1592 1592 1592 1592 1592 τi 21663 23820 25741 27723 29629 31097 r 0945 1157 1444 1942 2906 4703 Ssybody 85372 84773 83800 83030 82198 80611 Ssay 30958 32688 34302 36507 39109 40832 nybody 2779 2973 3183 3438 3740 4012 SsyB 73176 72663 71828 71169 70455 69095 τBmax 42819 43486 44321 44801 45177 45564 nyB 1709 1671 1621 1589 1560 1516 fom 1246 1234 1245 1283 1357 1639 optimal fom The shaded areas show the conditions not satisfied shi20396ch10qxd 81103 440 PM Page 293 httplibrosysolucionariosnet Chapter 12 329 Now we can check the performance at cmin c and cmax Of immediate interest is the fom of the median clearance assembly 982 as compared to any other satisfactory bearing ensemble If a nominal 1875 in bearing is possible construct another table with tb 0 and td 0 c b d Tf Tmax ho Pst Tmax fos fom 00020 1879 1875 1572 19430 736 00030 1881 1875 1386 15710 864 00035 1882 1875 1335 14710 905 00040 1883 1875 1300 14010 932 00050 1885 1875 1257 13145 959 00055 1886 1875 1244 12880 963 00060 1887 1875 1234 12680 964 The range of clearance is 00030 c 00055 in That is enough room to fit in our de sign window d 18750000 0001 in b 18810003 0000 in The ensemble median assembly has fom 931 We just had room to fit in a design window based upon the homin constraint Further reduction in nominal diameter will preclude any smaller bearings A table constructed for a d 1750 in journal will prove this We choose the nominal 1875in bearing ensemble because it has the largest figure of merit Ans 1219 This is the same as Prob 1218 but uses design variables of nominal bushing bore b and radial clearance c The approach is similar to that of Prob 1218 and the tables will change slightly In the table for a nominal b 1875 in note that at c 0003 the constraints are loose Set b 1875 in d 1875 20003 1869 in For the ensemble b 18750003 0001 d 18690000 0001 Analyze at cmin 0003 c 0004 in and cmax 0005 in At cmin 0003 in Tf 1384 µ 3160 S 00297 Hloss 1035 Btuh and the Trumpler conditions are met At c 0004 in Tf 130F µ 3872 S 00205 Hloss 1106 Btuh fom 9246 and the Trumpler conditions are OK At cmax 0005 in Tf 12568F µ 4325 µreyn S 0014 66 Hloss 1129 Btuh and the Trumpler conditions are OK The ensemble figure of merit is slightly better this bearing is slightly smaller The lubri cant cooler has sufficient capacity shi20396ch12qxd 82903 222 PM Page 329 httplibrosysolucionariosnet 330 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 1220 From Table 121 Seireg and Dandage µ0 00141106 reyn and b 13600 µµreyn 00141 exp1360T 95 T in F 00141 exp136018C 127 C in C µMPa s 68900141 exp136018C 127 C in C For SAE 30 at 79C µ 68900141 exp13601879 127 152 MPa s Ans 1221 Originally d 20000000 0001 in b 20050003 0000 in Doubled d 40000000 0002 in b 40100006 0000 The radial load quadrupled to 3600 lbf when the analyses for parts a and b were carried out Some of the results are Trumpler Part c µ S Tf f rc Qs hoc ϵ Hloss ho ho f a 0007 3416 00310 1351 01612 656 01032 0897 9898 0000722 0000360 0005 67 b 00035 3416 00310 1351 01612 0870 01032 0897 1237 0000361 0000280 0005 67 The side flow Qs differs because there is a c3 term and consequently an 8fold increase Hloss is related by a 98981237 or an 8fold increase The existing ho is related by a 2fold increase Trumplers homin is related by a 1286fold increase fom 8237 for double size fom 10297 for original size an 8fold increase for doublesize 1222 From Table 128 K 061010 in3 minlbf ft h P 50011 500 psi V π DN12 π120012 524 ftmin Tables 1210 and 1211 f1 18 f2 1 Table 1212 PVmax 46 700 psi ftmin Pmax 3560 psi Vmax 100 ftmin Pmax 4 π F DL 4500 π11 637 psi 3560 psi OK P F DL 500 psi V 524 ftmin PV 500524 26 200 psi ftmin 46 700 psi ftmin OK shi20396ch12qxd 82903 222 PM Page 330 httplibrosysolucionariosnet 332 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design Trial 3 Try D 78 in L 15 in Pmax 42100 π7815 194 psi 3560 psi OK P 2100 7815 152 psi V 916 ftmin PV 152916 13 923 psi ftmin 46 700 psi ftmin OK D 78 in L 15 in is acceptable Ans Suggestion Try smaller sizes shi20396ch12qxd 82903 222 PM Page 332 httplibrosysolucionariosnet Chapter 13 131 dP 178 2125 in dG N2 N3 dP 1120 544 2125 4375 in NG PdG 84375 35 teeth Ans C 2125 43752 325 in Ans 132 nG 16001560 400 revmin Ans p πm 3π mm Ans C 315 602 1125 mm Ans 133 NG 20280 56 teeth Ans dG NGm 564 224 mm Ans dP NPm 204 80 mm Ans C 224 802 152 mm Ans 134 Mesh a 1P 13 03333 in Ans b 125P 1253 04167 in Ans c b a 00834 in Ans p πP π3 1047 in Ans t p2 10472 0523 in Ans Pinion BaseCircle d1 N1P 213 7 in d1b 7 cos 20 6578 in Ans Gear BaseCircle d2 N2P 283 9333 in d2b 9333 cos 20 8770 in Ans Base pitch pb pc cos φ π3 cos 20 0984 in Ans Contact Ratio mc Labpb 1530984 155 Ans See the next page for a drawing of the gears and the arc lengths shi20396ch13qxd 82903 1216 PM Page 333 httplibrosysolucionariosnet Chapter 13 351 1335 Sketch gear 2 pictorially Pt Pn cos ψ 4 cos 30 3464 teethin φt tan1 tan φn cos ψ tan1 tan 20 cos 30 2280 Sketch gear 3 pictorially dP 18 3464 5196 in Pinion Gear 2 Wr W t tan φt 800 tan 2280 336 lbf W a W t tan ψ 800 tan 30 462 lbf W 336i 462j 800k lbf Ans W 3362 4622 800212 983 lbf Ans Gear 3 W 336i 462j 800k lbf Ans W 983 lbf Ans dG 32 3464 9238 in TG W tr 8009238 7390 lbf in 1336 From Prob 1335 solution Notice that the idler shaft reaction contains a couple tending to turn the shaft endover end Also the idler teeth are bent both ways Idlers are more severely loaded than other gears belying their name Thus be cautious 800 336 462 4 800 800 336 336 462 3 462 800 2 336 462 W a TG Wr W t x 3 y z Wa Wr T Wt x y z 2 shi20396ch13qxd 82903 1216 PM Page 351 httplibrosysolucionariosnet 352 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 1337 Gear 3 Pt Pn cos ψ 7 cos 30 6062 teethin tan φt tan 20 cos 30 04203 φt 228 d3 54 6062 8908 in W t 500 lbf W a 500 tan 30 2887 lbf Wr 500 tan 228 2102 lbf W3 2102i 2887j 500k lbf Ans Gear 4 d4 14 6062 2309 in W t 500 8908 2309 1929 lbf W a 1929 tan 30 1114 lbf Wr 1929 tan 228 811 lbf W4 811i 1114j 1929k lbf Ans 1338 Pt 6 cos 30 5196 teethin d3 42 5196 8083 in φt 228 d2 16 5196 3079 in T2 63 02525 1720 916 lbf in W t T r 916 30792 595 lbf W a 595 tan 30 344 lbf Wr 595 tan 228 250 lbf W 344i 250j 595k lbf RDC 6i RDG 3i 404j T3 C A B D T2 y 3 2 x z y x W t Wr Wa Wt Wr Wa r4 r3 shi20396ch13qxd 82903 1216 PM Page 352 httplibrosysolucionariosnet 354 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design d2 167727 207 in d3 367727 466 in d4 287727 362 in T2 63 02575 1720 2748 lbf in W t 2748 2072 266 lbf Wr 266 tan 20 968 lbf W a 266 tan 15 713 lbf F2a 266i 968j 713k lbf Ans F3b 266 968i 266 968j 169i 169j lbf Ans F4c 968i 266j 713k lbf Ans 1340 d2 N Pn cos ψ 14 8 cos 30 2021 in d3 36 8 cos 30 5196 in d4 15 5 cos 15 3106 in d5 45 5 cos 15 9317 in C x y z b F t 23 Fr 23 Fa 23 Ft 54 Fa 54 Fr 54 D G H 3 2 3 26R 155R 4 31 2 F y D F x D F x C F y C F z D y Fr 43 Fx b3 Fy b3 F a 23 F r 23 Ft 23 F t 43 F a 43 3 Fb3 z x b y F t c4 Fr c4 F a c4 4 Fa 34 Fr 34 Ft 34 z x c shi20396ch13qxd 82903 1216 PM Page 354 httplibrosysolucionariosnet 358 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design Radial Fr D 1566j 1669k N Or Fr D 2289 N total radial Ft D 637i N thrust 1343 VW π15900 12 3534 ftmin W x WWt 33 00005 3534 4669 lbf pt px π 10 0314 16 in L 0314 162 0628 in λ tan1 0628 π15 759 W 467 cos 145 sin 759 005 cos 759 263 lbf W y 263 sin 145 658 lbf W z 263cos 145cos 759 005 sin 759 251 lbf So W 467i 658j 251k lbf Ans T 467075 35 lbf in Ans 1344 100101 Mesh dP 100 48 2083 33 in dG 101 48 2104 17 in x y z WWt G 075 T y z shi20396ch13qxd 82903 1216 PM Page 358 httplibrosysolucionariosnet Chapter 14 141 d N P 22 6 3667 in Table 142 Y 0331 V πdn 12 π36671200 12 1152 ftmin Eq 144b Kv 1200 1152 1200 196 W t T d2 63 025H nd2 63 02515 120036672 4297 lbf Eq 147 σ KvW t P FY 19642976 20331 7633 psi 763 kpsi Ans 142 d 16 12 1333 in Y 0296 V π1333700 12 2443 ftmin Eq 144b Kv 1200 2443 1200 1204 W t 63 025H nd2 63 02515 70013332 2026 lbf Eq 147 σ KvW t P FY 1204202612 0750296 13 185 psi 132 kpsi Ans 143 d mN 12518 225 mm Y 0309 V π2251031800 60 2121 ms Eq 146b Kv 61 2121 61 1348 W t 60H πdn 6005103 π2251031800 2358 N Eq 148 σ KvW t FmY 13482358 121250309 686 MPa Ans shi20396ch14qxd 82003 1243 PM Page 360 httplibrosysolucionariosnet Chapter 14 361 144 d 515 75 mm Y 0290 V π75103200 60 07854 ms Assume steel and apply Eq 146b Kv 61 07854 61 1129 W t 60H πdn 605103 π75103200 6366 N Eq 148 σ KvW t FmY 11296366 6050290 826 MPa Ans 145 d 116 16 mm Y 0296 V π16103400 60 0335 ms Assume steel and apply Eq 146b Kv 61 0335 61 1055 W t 60H πdn 60015103 π16103400 4476 N Eq 148 F KvW t σmY 10554476 15010296 106 mm From Table A17 use F 11 mm Ans 146 d 1517 255 mm Y 0303 V π255103400 60 0534 ms Eq 146b Kv 61 0534 61 1088 W t 60H πdn 60025103 π255103400 468 N Eq 148 F KvW t σmY 1088468 75150303 149 mm Use F 15 mm Ans shi20396ch14qxd 82003 1243 PM Page 361 httplibrosysolucionariosnet 362 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 147 d 24 5 48 in Y 0337 V π4850 12 6283 ftmin Eq 144b Kv 1200 6283 1200 1052 W t 63 025H nd2 63 0256 50482 3151 lbf Eq 147 F KvW t P σY 105231515 201030337 246 in Use F 25 in Ans 148 d 16 5 32 in Y 0296 V π32600 12 5027 ftmin Eq 144b Kv 1200 5027 1200 1419 W t 63 02515 600322 9848 lbf Eq 147 F KvW t P σY 141998485 101030296 238 in Use F 25 in Ans 149 Try P 8 which gives d 188 225 in and Y 0309 V π225600 12 3534 ftmin Eq 144b Kv 1200 3534 1200 1295 W t 63 02525 6002252 2334 lbf Eq 147 F KvW t P σY 129523348 101030309 0783 in shi20396ch14qxd 82003 1243 PM Page 362 httplibrosysolucionariosnet Chapter 14 363 Using coarse integer pitches from Table 132 the following table is formed P d V Kv W t F 2 9000 1413717 2178 58356 0082 3 6000 942478 1785 87535 0152 4 4500 706858 1589 116713 0240 6 3000 471239 1393 175069 0473 8 2250 353429 1295 233426 0782 10 1800 282743 1236 291782 1167 12 1500 235619 1196 350139 1627 16 1125 176715 1147 466852 2773 Other considerations may dictate the selection Good candidates are P 8 F 78 in and P 10 F 125 in Ans 1410 Try m 2 mm which gives d 218 36 mm and Y 0309 V π36103900 60 1696 ms Eq 146b Kv 61 1696 61 1278 W t 6015103 π36103900 884 N Eq 148 F 1278884 7520309 244 mm Using the preferred module sizes from Table 132 m d V Kv W t F 100 180 0848 1139 1768388 86917 125 225 1060 1174 1414711 57324 150 270 1272 1209 1178926 40987 200 360 1696 1278 884194 24382 300 540 2545 1417 589463 12015 400 720 3393 1556 442097 7422 500 900 4241 1695 353678 5174 600 1080 5089 1834 294731 3888 800 1440 6786 2112 221049 2519 1000 1800 8482 2391 176839 1824 1200 2160 10179 2669 147366 1414 1600 2880 13572 3225 110524 0961 2000 3600 16965 3781 88419 0721 2500 4500 21206 4476 70736 0547 3200 5760 27143 5450 55262 0406 4000 7200 33929 6562 44210 0313 5000 9000 42412 7953 35368 0243 Other design considerations may dictate the size selection For the present design m 2 mm F 25 mm is a good selection Ans shi20396ch14qxd 82003 1243 PM Page 363 httplibrosysolucionariosnet 382 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design Gear wear Similarly W t 4 1182 lbf H4 590 hp Rating Hrated minH1 H2 H3 H4 min1575 1929 53 59 53 hp Ans Note differing capacities Can these be equalized 1425 From Prob 1424 W t 1 3151 lbf W t 2 3861 lbf W t 3 1061 lbf W t 4 1182 lbf W t 33 000KoH V 33 00012540 1649 1000 lbf Pinion bending The factor of safety based on load and stress is SFP W t 1 1000 3151 1000 315 Gear bending based on load and stress SFG W t 2 1000 3861 1000 386 Pinion wear based on load n3 W t 3 1000 1061 1000 106 based on stress SHP 106 103 Gear wear based on load n4 W t 4 1000 1182 1000 118 based on stress SHG 118 109 Factors of safety are used to assess the relative threat of loss of function 315 386 106 118 where the threat is from pinion wear By comparison the AGMA safety factors SFP SFG SHP SHG are 315 386 103 109 or 315 386 10612 11812 and the threat is again from pinion wear Depending on the magnitude of the numbers using SF and SH as defined by AGMA does not necessarily lead to the same conclusion concerning threat Therefore be cautious shi20396ch14qxd 82003 1243 PM Page 382 httplibrosysolucionariosnet Chapter 14 385 1429 n 1145 revmin Ko 125 NP 22T NG 60T mG 2727 dP 275 in dG 75 in YP 0331 YG 0422 JP 0335 JG 0405 P 8Tin F 1625 in HB 250 case and core both gears Cm 1 FdP 00591 Cf 00419 Cpm 1 Cma 0152 Ce 1 Km 11942 KT 1 Kβ 1 Ks 1 V 824 ftmin YNP 08318 YNG 0859 K R 1 I 0117 58 099St107 32 125 psi σallP 26 668 psi σallG 27 546 psi and it follows that W t 1 8793 lbf H1 2197 hp W t 2 1098 lbf H2 274 hp For wear W t 3 304 lbf H3 759 hp W t 4 340 lbf H4 850 hp Rating Hrated min2197 274 759 850 759 hp In Prob 1424 Hrated 53 hp Thus 759 530 01432 1 698 not 1 8 Ans The transmitted load rating is W t rated min8793 1098 304 340 304 lbf In Prob 1424 W t rated 1061 lbf Thus 304 1061 02865 1 349 not 1 4 Ans 1430 SP SH 1 Pd 4 JP 0345 JG 0410 Ko 125 Bending Table 144 099St107 13 000 psi σallP σallG 13 0001 111 13 000 psi W t 1 σallF JP KoKvKs Pd KmKβ 13 0003250345 1251534141241 1533 lbf H1 15331649 33 000 766 hp W t 2 W t 1 JGJP 153304100345 1822 lbf H2 H1JGJP 76604100345 910 hp shi20396ch14qxd 82003 1243 PM Page 385 httplibrosysolucionariosnet prerequisites and how quantitative it was The most important thing is to have the stu dent think about it The instructor can comment in class when students curiosity is heightened Options that will surface may include Select a throughhardening steel which will meet or exceed core hardness in the hot rolled condition then heattreating to gain the additional 86 points of Brinell hardness by bathquenching then tempering then generating the teeth in the blank Flame or induction hardening are possibilities The hardness goal for the case is sufficiently modest that carburizing and case harden ing may be too costly In this case the material selection will be different The initial step in a nitriding process brings the core hardness to 3338 Rockwell Cscale about 300350 Brinell which is too much Emphasize that development procedures are necessary in order to tune the BlackArt to the occasion Manufacturing personnel know what to do and the direction of adjust ments but how much is obtained by asking the gear or gear blank Refer your students to D W Dudley Gear Handbook library reference section for descriptions of heattreat ing processes 1512 Computer programs will vary 1513 A design program would ask the user to make the a priori decisions as indicated in Sec 155 p 794 MED7 The decision set can be organized as follows A priori decisions Function H Ko rpm mG temp NL R Design factor nd SF nd SH nd Tooth system Involute Straight Teeth Crowning φn Straddling Kmb Tooth count NPNG mG NP Design decisions Pitch and Face Pd F Quality number Qv Pinion hardness HB1 HB3 Gear hardness HB2 HB4 402 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design shi20396ch15qxd 82803 325 PM Page 402 httplibrosysolucionariosnet 406 1515 to 1522 Problem statement values of 25 hp 1125 revmin mG 10 Ka 125 nd 11 φn 20 ta 70F are not referenced in the table Parameters Selected 1515 1516 1517 1518 1519 1520 1521 1522 1 px 175 175 175 175 175 175 175 175 2 dW 360 360 360 360 360 410 360 360 3 FG 240 168 143 169 240 225 24 24 4 A 2000 2000 2000 2000 2000 2000 2500 2600 FAN FAN HW 382 382 382 382 382 380 412 412 HG 362 362 362 362 362 361 377 377 Hf 187 147 197 197 197 185 359 359 NW 3 3 3 3 3 3 3 3 NG 30 30 30 30 30 30 30 30 KW 125 80 50 115 185 Cs 607 854 1000 Cm 0759 0759 0759 Cv 0236 0236 0236 VG 492 492 492 492 492 563 492 492 W t G 2430 2430 2430 2430 2430 2120 2524 2524 W t W 1189 1189 1189 1189 1189 1038 1284 1284 f 00193 00193 00193 00193 00193 00183 0034A 0034A e 0948 0948 0948 0948 0948 0951 0913A 0913A PtG 1795 1795 1795 1795 1795 1571 1795 1795 Pn 1979 1979 1979 1979 1979 1732 1979 1979 CtoC 10156 10156 10156 10156 10156 116 10156 10156 ts 177 177 177 177 177 171 1796 1796 L 525 525 525 525 525 60 525 525 λ 249 249 249 249 249 2498 249 249 σG 5103 7290 8565 7247 5103 4158 5301 5301 dG 1671 1671 1671 1671 1671 19099 167 1671 shi20396ch15qxd 82803 325 PM Page 406 httplibrosysolucionariosnet Chapter 16 415 c The direction of brake pulley rotation affects the sense of Sy which has no effect on the brake shoe lever moment and hence no effect on Sx or the brake torque The brake shoe levers carry identical bending moments but the left lever carries a tension while the right carries compression column loading The right lever is de signed and used as a left lever producing interchangeable levers identical levers But do not infer from these identical loadings 1610 r 1352 675 in b 75 in θ2 45 From Table 163 for a rigid molded nonasbestos use a conservative estimate of pa 100 psi f 031 In Eq 1616 2θ2 sin 2θ2 2π4 sin 245 2571 From Prob 169 solution N Sx 4174P pabr 2 2571 1285pabr P 1285 417410075675 1560 lbf Ans Applying Eq 1618 for two shoes T 2af N 2742603141741560 29 980 lbf in Ans 1611 From Eq 1622 P1 pabD 2 90414 2 2520 lbf Ans f φ 025π270180 1178 Eq 1619 P2 P1 exp f φ 2520 exp1178 776 lbf Ans T P1 P2D 2 2520 77614 2 12 200 lbf in Ans Ans 1252P 2049P 4174P 268P Right shoe lever 2125P 1428P 2049P 4174P 1252P 168P Left shoe lever 2125P 0428P shi20396ch16qxd 82803 401 PM Page 415 httplibrosysolucionariosnet Chapter 16 417 Eq 1614 P2 P1 exp f φ 1680 exp0942 655 lbf T P1 P2 D 2 1680 65516 2 8200 lbf in Ans H Tn 63 025 8200200 63 025 260 hp Ans P 3P1 10 31680 10 504 lbf Ans b The radial load on the bearing pair is 1803 lbf If the bearing is straddle mounted with the drum at center span the bearing radial load is 18032 901 lbf c Eq 1622 p 2P bD pθ0 2P1 316 21680 316 70 psi Ans As it should be pθ270 2P2 316 2655 316 273 psi Ans 1615 Given φ 270 b2125 in f 020 T 150 lbf ft D825 in c2 225 in Notice that the pivoting rocker is not located on the vertical centerline of the drum a To have the band tighten for ccw rotation it is necessary to have c1 c2 When fric tion is fully developed P1 P2 exp f φ exp023π2 2566 Net torque on drum due to brake band T TP1 TP2 13 440 5240 8200 lbf in 1803 lbf 8200 lbfin 1680 lbf 655 lbf Force of shaft on the drum 1680 and 655 lbf TP1 16808 13 440 lbf in TP2 6558 5240 lbf in 1680 lbf 1803 lbf 655 lbf 13440 lbfin 5240 lbfin Force of belt on the drum R 16802 655212 1803 lbf 1680 lbf 655 lbf shi20396ch16qxd 82803 401 PM Page 417 httplibrosysolucionariosnet Chapter 16 419 c The torque ratio is 15012100 or 18fold P2 349 18 194 lbf P1 225P2 225194 436 lbf p 896 18 498 psi Ans Comment As the torque opposed by the locked brake increases P2 and P1 increase although ratio is still 225 then p follows The brake can selfdestruct Protection could be provided by a shear key 1616 a From Eq 1623 since F πpad 2 D d then pa 2F πdD d and it follows that pa 25000 π225300 225 0189 Nmm2 or 189 000 Nm2 or 189 kPa Ans T F f 4 D d 5000025 4 300 225 164 043 N mm or 164 N m Ans b From Eq 1626 F πpa 4 D2 d2 pa 4F πD2 d2 45000 π3002 2252 0162 Nmm2 162 kPa Ans From Eq 1627 T π 12 f paD3 d3 π 120251621033003 22531033 166 N m Ans 1617 a Eq 1623 F πpad 2 D d π1204 2 65 4 1885 lbf Ans shi20396ch16qxd 82803 401 PM Page 419 httplibrosysolucionariosnet 424 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 1624 a The useful work performed in one revolution of the crank shaft is U 3520008015 84103 in lbf Accounting for friction the total work done in one revolution is U 841031 016 100103 in lbf Since 15 of the crank shaft stroke is 75 of a crank shaft revolution the energy fluctuation is E2 E1 84103 1001030075 765103 in lbf Ans b For the flywheel n 690 540 revmin ω 2πn 60 2π540 60 565 rads Since Cs 010 I E2 E1 Csω2 765103 0105652 2396 lbf in s2 Assuming all the mass is concentrated at the effective diameter d I md2 4 W 4gI d2 43862396 482 161 lbf Ans 1625 Use Ex 166 and Table 166 data for one cylinder of a 3cylinder engine Cs 030 n 2400 revmin or 251 rads Tm 33368 4π 804 in lbf Ans E2 E1 33531 10 590 in lbf I E2 E1 Csω2 10 590 0302512 0560 in lbf s2 Ans 1626 a 1 T21 F21rP T2 rG rP T2 n Ans rP rG T1 F12 F21 T2 shi20396ch16qxd 82803 401 PM Page 424 httplibrosysolucionariosnet 426 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design b For R constant nm Ie IM IP n2IP IP n2 R2IP n4 IL R2 Ans c For R 10 Ie n 0 0 2n1 21 n3 41021 n5 0 0 n6 n2 200 0 From which n 2430 Ans m 10 2430 4115 Ans Notice that nand m are independent of IL 1628 From Prob 1627 Ie IM IP n2IP IP n2 R2IP n4 IL R2 10 1 n21 1 n2 1001 n4 100 102 10 1 n2 1 n2 100 n4 1 n Ie 100 11400 150 3440 200 2250 243 2090 300 2230 400 2850 500 3720 600 4810 700 6110 800 7600 900 9300 1000 11202 Optimizing the partitioning of a double reduction lowered the geartrain inertia to 209112 0187 or to 19 of that of a single reduction This includes the two addi tional gears 1629 Figure 1629 applies t2 10 s t1 05 s t2 t1 t1 10 05 05 19 Ie n 0 1 2 243 4 6 8 10 100 209 shi20396ch16qxd 82803 401 PM Page 426 httplibrosysolucionariosnet 430 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design Scaling will affect do and di but the gear ratio changed I Scale up the flywheel in the Prob 1629 solution by a factor of 25 Thickness becomes 425 10 in d 3025 75 in do 75 102 80 in di 75 102 70 in W 838611 072 802 702 3026 lbf v 3026 026 11 638 in3 V π 4 l802 702 1178l l 11 638 1178 988 in Proportions can be varied The weight has increased 30261891 or about 16fold while the moment of inertia I increased 100fold The gear train transmits a steady 3 hp But the motor armature has its inertia magnified 100fold and during the punch there are decel eration stresses in the train With no motor armature information we cannot comment 1631 This can be the basis for a class discussion shi20396ch16qxd 82803 401 PM Page 430 httplibrosysolucionariosnet Chapter 17 437 If you only change the belt width the parameters in the following table change as shown Ex 171 This Problem b 600 413 w 0393 0271 Fc 256 176 F1a 420 289 F2 1724 42 Fi 2706 1477 f 033 080 dip 0139 0176 Friction underdeveloped Friction fully developed 176 The transmitted power is the same nFold b 6 in b 12 in Change Fc 2565 513 2 Fi 27035 6649 246 F1a 420 840 2 F2 1724 5924 344 Ha 2062 2062 1 nfs 11 11 1 f 0139 0125 090 dip 0328 0114 034 If we relax Fi to develop full friction f 080 and obtain longer life then nFold b 6 in b 12 in Change Fc 256 513 2 Fi 1481 1481 1 F1 2976 3232 109 F2 50 756 151 f 080 080 1 dip 0255 0503 2 shi20396ch17qxd 82803 358 PM Page 437 httplibrosysolucionariosnet Chapter 17 455 Evaluate K1 and K2 Eq 1738 Hd HnomKsnd Eq 1737 Ha K1K2Htab Equate Hd to Ha and solve for Htab Htab Ksnd Hnom K1K2 Table 1722 K1 1 Table 1723 K2 1 17 25 33 for 1 through 4 strands H tab 151125 1K2 4125 K2 Prepare a table to help with the design decisions Chain Lub Strands K2 H tab No Htab n f s Type 1 10 413 100 594 158 B 2 17 243 80 310 140 B 3 25 165 80 310 207 B 4 33 125 60 133 117 B Design Decisions We need a figure of merit to help with the choice If the best was 4 strands of No 60 chain then Decision 1 and 2 Choose four strand No 60 roller chain with n f s 117 n f s K1K2Htab Ks Hnom 133133 1525 117 Decision 3 Choose Type B lubrication Analysis Table 1720 Htab 133 hp Table 1719 p 075 in Try C 30 in in Eq 1734 L p 2C p N1 N2 2 N2 N12 4π2Cp 230075 17 84 2 84 172 4π230075 1333 134 From Eq 1735 with p 075 inC 3026 in Decision 4 Choose C 3026 in shi20396ch17qxd 82803 358 PM Page 455 httplibrosysolucionariosnet Chapter 17 459 Fatigue without bending n f Ff Ft 242 1176 206 Ans Fatigue with bending For a life of 01106 cycles from Fig 1721 pSu 41000 0004 Ff 0004240272 2 691 kip Eq 1750 n f 691 39 1176 256 Ans If we were to use the endurance strength at 106 cycles Ff 242 kip the factor of safety would be less than 1 indicating 106 cycle life impossible Comments There are a number of factors of safety used in wire rope analysis They are differ ent with different meanings There is no substitute for knowing exactly which fac tor of safety is written or spoken Static performance of a rope in tension is impressive In this problem at the drum we have a finite life The remedy for fatigue is the use of smaller diameter ropes with multiple ropes supporting the load See Ex 176 for the effectiveness of this approach It will also be used in Prob 1730 Remind students that wire ropes do not fail suddenly due to fatigue The outer wires gradually show wear and breaks such ropes should be retired Periodic in spections prevent fatigue failures by parting of the rope 1730 Since this is a design task a decision set is useful A priori decisions Function load height acceleration velocity life goal Design Factor nd Material IPS PS MPS or other Rope Lay number of strands number of wires per strand Decision variables Nominal wire size d Number of loadsupporting wires m From experience with Prob 1729 a 1in diameter rope is not likely to have much of a life so approach the problem with the d and m decisions open Function 5000 lbf load 90 foot liftacceleration 4 fts2velocity 2 fts life goal 105 cycles Design Factor nd 2 Material IPS Rope Regular lay 1in plowsteel 6 19 hoisting shi20396ch17qxd 82803 358 PM Page 459 httplibrosysolucionariosnet Chapter 17 463 From which d a 2cn1 Ans Substituting this result for d in Eq 1 gives m 4bcn1 a2 Ans 1734 Am 040d2 04022 16 in2 Er 12 Mpsi w 16d2 1622 64 lbfft wl 64480 3072 lbf Treat the rest of the system as rigid so that all of the stretch is due to the cage weighing 1000 lbf and the wires weight From Prob 56 δ1 Pl AE wll 2AE 100048012 1612106 307248012 21612106 03 0461 0761 in due to cage and wire The stretch due to the wire the cart and the cage is δ2 900048012 1612106 0761 3461 in Ans 1735 to 1738 Computer programs will vary shi20396ch17qxd 82803 358 PM Page 463 httplibrosysolucionariosnet Chapter 18 489 Boundary conditions y 0 at x 0 yields C2 0 y 0 at x 0315m yields C1 0295 25 Nm2 Equation 1 with C1 0295 25 provides the slopes at the bearings and gear The following table gives the results in the second column The third column gives the results from a similar finite element model The fourth column gives the result of a full model which models the 35 and 55 mm diameter steps x mm θ rad FE Model Full FE Model 0 00014260 00014270 00014160 140 00001466 00001467 00001646 315 00013120 00013280 00013150 The main discrepancy between the results is at the gear location x 140 mm The larger value in the full model is caused by the stiffer 55 mm diameter step As was stated earlier this step is not as stiff as modeling implicates so the exact answer is somewhere between the full model and the simplified model which in any event is a small value As expected modeling the 30 mm dia as 35 mm does not affect the results much It can be seen that the allowable slopes at the bearings are exceeded Thus either the load has to be reduced or the shaft beefed up If the allowable slope is 0001 rad then the maximum load should be Fmax 00010001 467 479 kN With a design factor this would be reduced further To increase the stiffness of the shaft increase the diameters by 0001 46000114 1097 Form a table Old d mm 2000 3000 3500 4000 4500 5500 New ideal d mm 2195 3292 3841 4389 4938 6035 Rounded up d mm 2200 3400 4000 4400 5000 6200 Repeating the full finite element model results in x 0 θ 930 104 rad x 140 mm θ 109 104 rad x 315 mm θ 865 104 rad Well within our goal Have the students try a goal of 00005 rad at the bearings Strength Due to stress concentrations and reduced shaft diameters there are a number of locations to look at A table of nominal stresses is given below Note that torsion is only to the right of the 7 kN load Using σ 32Mπd3 and τ 16Tπd3 x mm 0 15 40 100 110 140 210 275 300 330 σ MPa 0 220 370 619 478 609 520 396 176 0 τ MPa 0 0 0 0 0 6 85 127 202 681 σ MPa 0 220 370 619 478 618 531 453 392 1180 Table A20 for AISI 1020 CD steel Sut 470 MPa Sy 390 MPa shi20396ch18qxd 82803 417 PM Page 489 httplibrosysolucionariosnet The adequacy of their design must be demonstrated and possibly include a designers notebook Many of the fundaments of the course based on this text and this course are useful The student will find them useful and notice that heshe is doing it Dont let the students create a time sink for themselves Tell them how far you want them to go 1833 I used this task as a final exam when all of the students in the course had consistent test scores going into the final examination it was my expectation that they would not change things much by taking the examination This problem is a learning experience Following the task statement the following guidance was added Take the first half hour resisting the temptation of putting pencil to paper and decide what the problem really is Take another twenty minutes to list several possible remedies Pick one and show your instructor how you would implement it The students initial reaction is that heshe does not know much from the problem statement Then slowly the realization sets in that they do know some important things that the designer did not They knew how it failed where it failed and that the design wasnt good enough it was close though Also a fix at the bearing seat leadin could transfer the problem to the shoulder fillet and the problem may not be solved To many studentss credit they chose to keep the shaft geometry and selected a new material to realize about twice the Brinell hardness 496 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design shi20396ch18qxd 82803 417 PM Page 496 httplibrosysolucionariosnet