Trabalho Global - Vibrações

Chapter 1 Fundamentals of Vibration 11 Head upper torso hips and legs Head and upper torso Head neck upper torso spinal column Hips and legs Arms spinal column shoulder Hips and legs legs legs i ii iii 12 Seat Restraint belts Windshield Instrument panel Slant footboard Floor Impact force 13 unbalanced forces Mass of frame reciprocating engine foundation block soil damping stiffness a one degree of freedom model unbalanced forces Mass of frame reciprocating engine damping stiffness of bolts and elastic pad Mass of foundation block damping stiffness of soil b Two degree of freedom model x1 x1 x2 14 mass of body stiffness of wheels damping of suspension i mass of body suspension mass of front rear wheels wheels ii mass of passengers plus seats damping stiffness of seats mass moment of inertia of body damping stiffness of front suspension damping stiffness of front wheels front wheels damping stiffness of rear suspension rear wheels damping stiffness of rear wheels iii 15 Radiator sheet metal m body m engine Bumper Driveline 16 Shock absorber Plow 1keq 12k1 1k2 12k3 keq 2 k1 k2 k3 k2 k3 2 k1 k3 k1 k2 b New deflection mgkeq δ12 keq 2 mgδ1 2 k1 k k1 112 Let x vertical displacement of mass M xs resulting deformation of each inclined spring 113 k23 k2 k3k2 k3 k4 Aρg πd24 ρg From kinetic energy 12 m1l1 θ2 12 m2 ml3 θ2 12 Jeq θ2 From potential energy 12 k1 l1 θ2 12 k23 l2 θ2 12 kt θ2 12 k4 l3 θ2 12 keq θ2 Jeq m1 l12 m2 m l32 keq k1 l12 k23 l22 kt k4 l32 114 k2 EAl2 πEtdtl2 k1 πEDd4l k2 k1 gives l2 4td tDd 115 a F Fx0 dFdxx0 x x0 500x 2x3x10 500 6x2x 10x10 1100x 4000 b at x 9 mm Exact F9 500 x 9 293 5958 N Approximate F9 1100 x 9 4000 5900 N Error 09735 c at x 11 mm Exact F11 500 x 11 2113 8162 N Approximate F11 1100 x 11 4000 8100 N Error 07596 116 p v γ constant E1 Differentiation of E1 gives dp vγ p γ vγ1 dv 0 dp pγv dv E2 change in volume when mass moves by dx dv Adx E3 Eqs E2 and E3 give dp pγ Av dx Force due to pressure change dF dpA p γ A2v dx spring constant of air spring k dFdx p γ A2v 117 Equivalent spring constants in different directions are ke1 k5 k6 k7k5 k6 k5 k7 k6 k7 ke2 k8 k9k8 k9 ke3 k1 k2k1 k2 ke4 k3 k4k3 k4 If the force P moves by x spring located at θi undergoes a displacement of xi x cos θi derivation as in problem 112 Equivalence of potential energy gives 12 keq x2 12 Σi14 kei xi2 keq Σi14 kei cos2 θi 118 From problem 116 k p γ A2v with γ 14 for air Let p 14 MPa k 13000 Nm 14 10614 A2v A2v 66326 103 Let diameter of a piston d 5 cm A π45 1022 19635 103 m2 v A66326 103 58127 104 m3 Let h 5 cm π4 D2 5 102 v D 012166 m 119 F a x b x3 2 104 x 4 107 x3 Around x Fx Fx dFdxx x x When x 102 m Fx 2 104 102 4 107 106 240 N dFdxx a 3 b x2 2 104 3 4 107 104 32000 Hence Fx 240 32000 x 001 32000 x 80 N Since the linearized spring constant is given by Fx keq x we have keq 32000 Nm 120 Fi aixi bixi3 i12 Springs in series W a1δ1 b1δ13 1 W a2δ2 b2δ23 2 W keqδst 3 δst δ1 δ2 4 Solve Eqs1 and 2 for δ1 and δ2 respectively Substitute the result in Eq4 and then in Eq3 to find keq Springs in parallel W F1 F2 a1δst b1δst3 a2δst b2δst3 keq δst keq a1 b1 δst2 a2 b2 δst2 121 k G d4 8 D3 N 8x106 Nm Dd 6 N 10 W π D N ρ π d24 where ρ weight per unit volume f1 12 kgW 12 G d2 g 2 π2 D4 N2 ρ 04 Hz Using G 731 x 109 Nm2 ρ 76000 Nm3 g 981 msec2 Dd 6 8 10 N 10 15 20 d 04 06 values of k and f1 are computed Combination of Dd 6 N 10 and d 20 m corresponding to k 34606 x 106 Nm and f1 04801 Hz can be taken as an acceptable design 122 Total elongation strain is same in each material ϵs ϵa xl 1 where x is the total elongation Equation 1 can be expressed as σsEs σaEa xl 2 or σs Es xl 3 σa Ea xl 4 Total axial force is F Fs Fa σs As σa Aa 5 where Fs and Fa denote the axial forces acting on steel and aluminum respectively and As and Aa represent the crosssectional areas of the two materials Equating F to keq x where keq denotes the equivalent spring constant of the bimetallic bar we obtain from Eqs 3 to 5 F keq x Es xl As Ea xl Aa or keq Es Asl Ea Aal 6 123 J π2 r4 area polar moment of inertia at section x 15708 01 005 x4 m4 Knowing that the angle of twist θ between the ends of a uniform shaft of length l under a torque T is given by θ T l GJ the angle of twist for an element of length dx can be expressed as dθ T dx GJ T dx 80 109 15708 01 005 x4 1 The total angle of twist can be determined by integrating Eq 1 from x0 to 1 as θ from 0 to 1 T dx 125664 1010 01 005 x4 T 125664 1010 from 0 to 1 dx 01 005 x4 2 But from 0 to 1 dx 01 005 x4 1 005 from 0 to 1 005 dx 01 005 x4 20 from 0 to 005 dy 01 y4 46667 104 where y 005 x Hence θ T 46667 104 125664 1010 T 03714 106 rad This gives kt Tθ 26925 106 Nmrad 124 The steel and aluminum hollow shafts can be treated as two torsional springs in parallel For a hollow shaft kt π G 32 l D4 d4 For the steel shaft G 80 109 Pa l 5 m D 025 m d 015 m and hence kt1 π 8 1010 32 5 0254 0154 534072 106 Nmrad a For the aluminum shaft G 26 109 Pa l 5 m D 015 m d 01 m and hence kt2 π 26 109 32 5 0154 0104 0207395 106 Nmrad keq kt1 kt2 534072 106 020739 106 554811 106 Nmrad b With G 26 109 Pa l 5 m D 015 m and d 005 m kt2 π26109325 0154 0054 0255255 106 Nmrad keq k t1 kt2 534072 106 0255255 106 5595975 106 Nmrad 125 For Helical spring k G d4 64 n R3 Spring 1 k1 82 109 5 1024 64 10 15 1023 2372685 kNm Spring 2 k2 27 109 25 1024 64 10 125 1023 84375 kNm a Spring 2 inside spring 1 parallel keq k1 k2 2457060 Nm b Spring 2 on top of spring 1 series 1 keq 1 k1 1 k2 k2 k1 k1 k2 which gives keq 814776 Nm 126 For a helical spring k G d4 64 n R3 k1 82 109 25 1024 64 10 15 1023 1482928 Nm k2 27 109 125 1024 64 10 125 1023 52734 Nm a Spring 2 inside spring 1 keq k1 k2 1535663 Nm b Spring 2 on top of spring 1 1 keq 1 k1 1 k2 or keq k1 k2 k1 k2 1482928 52734 1482928 52734 50923 Nm 127 x2 y sin 30 x1 y cos 30 Equivalence of strain energies 12 keq y2 12 k2 x22 12 k1 x12 12 k1 y2 cos2 30 12 k2 y2 sin2 30 ie keq 34 k1 14 k2 with k1 A1 E1 l1 π4 0252 0242 200 109 25 3078761 kNm and k2 A2 E2 l2 π4 0182 0172 200 109 19 2893572 kNm keq 34 k1 14 k2 3032464 kNm Similarly the equivalent damping constant can be found as using equivalence of kinetic energy Ceq 34 c1 14 c2 655 Nsm 128 Outer diameter 75 mm Inner diameter 73 mm Length 1 m Solution stainless steel E 200 GPa G 83 GPa for each tube D 75 mm d 73 mm ls 1 m Axial stiffness A E l π4 D2 d2 E l π4 000752 000732 200 109 1 464956 Nm Ka Torsional stiffness π G 32 l D4 d4 π 83 109 32 1 000754 000734 26421 Nmrad Kt for heat exchanger with 6 tubes Axial stiffness 6Ka 2789736 Nm Torsional stiffness 6Kt 158524 Nmrad 129 Assume small angles θ1 and θ2 θ2 p1 p2 θ1 x1 horizontal displacement of cg of mass m1 θ1 r1 x2 vertical displacement of cg of mass m2 θ2 r2 p1 θ1 r2 p2 y1 horizontal displacement of springs k1 and k2 θ1 r1 l1 y2 vertical displacement of springs k3 and k4 θ2 l2 p1 l1 θ1 p2 Equivalence of kinetic energies gives 12 Jeqθ12 12 J1 θ12 12 J2 θ22 12 m1 x12 12 m2 x22 Jeq J1 J2 p1 p22 m1 r12 m2 r22 p1 p22 Equivalence of potential energies gives 12 keq θ12 12 k12 y12 12 k34 y22 12 kt1 θ12 12 kt2 θ22 with k12 k1 k2 k34 k3 k4 k3 k4 y1 θ1 r1 l1 y2 p1 l2 θ1 p2 and θ2 p1 θ1 p2 keq k1 k2 r1 l12 k3 k4 k3 k4 p12 l22 p22 kt1 kt2 p12 p22 130 θ x b x1 x α b From equivalence of kinetic energies 12 meq x2 12 m1 x12 12 m2 x2 12 J0 θ2 meq m1 α b2 m2 J0 1 b2 Let θᵢ angular velocity of the motor input Angular velocities of different gear sets are Jmotor J₁ J₂ J₃ J₄ J₅ J₂N Jload Equivalence of kinetic energies gives 12 Jeq θᵢ² 12 Jmotor θᵢ² 12 Σ k1 to 2N Jₖ θₖ² 12 Jload θload² Jeq Jmotor J₁ J₂ J₃ n₁n₂² J₄ J₅n₁n₂ n₃n₄² J₂N Jloadn₁n₂ n₃n₄ n₂N1n₂N² 132 Equivalence of kinetic energies gives 12 Jeq θ₁² 12 J₁ θ₁² 12 J₂ θ₂² where θ₂ θ₁ n₁n₂ Jeq J₁ J₂ n₁n₂² 133 When point A moves by distance x xₕ the walking beam rotates by the angle θb xₕℓ₃ This corresponds to a linear motion of point B xB θb ℓ₂ xₕ ℓ₂ℓ₃ and the angular rotation of crank can be found from the relation xB rc sin θc ℓ₄ cos φ rc sin θc ℓ₄ 1 rc²ℓ₄² sin²θc For large values of ℓ₄ compared to rc and for small values of x and θc we have xB rc sin θc rc θc or θc xBrc xₕ ℓ₂ℓ₃ rc The kinetic energy of the system can be expressed as T 12 mₕ xₕ² 12 Jb θb² 12 Jc θc² Equating this to T 12 meq xₕ² we obtain meq mₕ Jbℓ₃² Jcℓ₃ rc² 134 When mass m is displaced by x the bell crank lever rotates by the angle θb xℓ₁ This makes the center of the sphere displace by xs θb ℓ₂ Since the sphere rotates without slip it rotates by an angle θs xsrs θb ℓ₂rs x ℓ₂ℓ₁ rs The kinetic energy of the system can be expressed as T 12 m x² 12 J₀ θ² 12 Js θs² 12 ms xs² 12 m x² 12 J₀ xℓ₁² 12 25 ms rs² x² ℓ₂²ℓ₁² rs² 12 ms x ℓ₂ℓ₁² since for a sphere Js 25 ms rs² Equating this to T 12 meq x² we obtain meq m J₀ℓ₁² 75 ms ℓ₂²ℓ₁² 135 a Fi ci F₂ c₂ F₃ c₃ Fi damping force of ci ci x₂ x₁ i123 Feq damping force of ceq ceq x₂ x₁ F₁ F₂ F₃ ceq c₁ c₂ c₃ b c₁ c₂ c₃ x₁ x₂ x₃ x₄ F₁ c₁ x₂ x₁ F₂ c₂ x₃ x₂ F₃ c₃ x₄ x₃ x₄ x₁ x₄ x₃ x₃ x₂ x₂ x₁ Feqceq F₃c₃ F₂c₂ F₁c₁ Since Feq F₁ F₂ F₃ 1ceq 1c₁ 1c₂ 1c₃ c Equating the energies dissipated in a cycle π ceq ω x₁² π c₁ ω x₁² π c₂ ω x₂² π c₃ ω x₃² where x₁ θ₁ ℓ₁ x₂ θ₂ ℓ₂ and x₃ θ₃ ℓ₃ ceq c₁ c₂ ℓ₂ℓ₁² c₃ ℓ₃ℓ₁² d Equating the energies dissipated in a cycle π cteq ω θ₁² π ct₁ ω θ₁² π ct₂ ω θ₂² π ct₃ ω θ₃² where θ₂ θ₁ n₁n₂ and θ₃ θ₁ n₁n₃ cteq ct₁ ct₂ n₁n₂² ct₃ n₁n₃² 136 Damping constant desired C 180 Nsm viscosity of the fluid μ 30 10³ Nsm² c μ 3 π D³ ℓ 1 2dD 4 d³ Eq 1 Assuming x Dd as the unknown with ℓ 5 cm Eq 1 can be written as c μ 3 π ℓ 3 4 1 2x or 180 30 10³ 3 π 5 10² 4 x³ 1 2x This gives x³ 2x² 5092958 0 Using a trial and error procedure the solution of this cubic equation can be found as x 3641 Using D 75 cm we get d 75h3641 0206 cm 137 Damping constant 20 10⁶ Nsm SAE 30 at 21C diameter of piston 65 mm Solution c μ 3 π D³ ℓ 4 d³ 1 2dD μ 031 Pas D diameter of piston d radial clearance ℓ axial length of piston Let d 002 mm D 60 mm and above equation gives 20 10⁶ 031 3πℓ60³4002³ 1 2 00260 ℓ 10135 10³ m Tangential velocity of inner cylinder D2 ω For small d rate of change of velocity of fluid is dvdr D2 ωd shear stress between cylinders is τ μ dvdr μ D ω2 d and shear force is F τ Area τ π D lh π μ D² ω lh2 d Torque developed M t1 F D2 For small h rate of change of velocity of fluid in vertical direction is dvdy r ωh Shear stress is τ μ dvdy μ r ωh Force on area dA dF τ dA Torque between bottom surfaces of cylinders is M t2 dm t2 dA where dm t2 dF r μ r³ ωh dr dθ ie M t2 μ ωh r0 to D2 θ0 to 2π r³ dr dθ μ ω π D⁴ 64 h Total torque M t M t1 M t2 π μ D³ ω lh4 d π μ ω D⁴ 64 h Expressing M t as c t v c t ω D2 we get damping constant c t π μ D² lh2 d π μ D³32 h F a x b x² 5 x 02 x² Fx Fx0 dFdx x0 x x0 At x0 5 ms Fx0 5 5 02 25 30 N dFdx x0 5 04 x 5 7 and hence Fx 30 7 x 5 7 x 5 Thus the linearized damping constant is given by Fx 7 x c eq x or c eq 7 Nsm 140 Damping constant due to skin friction drag is c 100 μ ℓ² d 1 Damping constant of a platetype damper is c p μ A h 2 where A area of plates and h distance between the plates If the area of plates A in Fig 135 is taken to be same as the area of the plate shown in Fig 184 we have A ℓ d Equating 1 and 2 gives 100 μ ℓ² d μ ℓ dh 3 from which the clearance between the plates can be determined as h 1100 ℓ 141 c 6 π μ ℓ h³ a h2² r² a² r² a h2 h When μ 03445 Pas ℓ 01 m h 0001 m a 002 m and r 0005 m c 6 π 03445 01 10³³ 002 00005² 0005² 002² 0005² 002 00005 0001 42056394 Nsm 142 c 6 π μ ℓ h³ a h2² r² a² r²a h2 h Basic data ℓ 10 cm h 01 cm a 2 cm r 05 cm μ 03445 Damping constant with basic data c 42056230 Nsm a r changed to 1 cm new c 26177920 Nsm b h changed to 005 cm new c 350608910 Nsm c a changed to 4 cm new c 387545860 Nsm 143 x 5 2 i A ei θ A cos θ i A sin θ A cos θ 5 A sin θ 2 A A cos θ² A sin θ² 5² 2² 53852 θ tan¹ A sin θ A cos θ tan¹ 25 218014 144 x 1 1 2 i a₁ a₂ i x 2 3 4 i b₁ b₂ i x x 1 x 2 a₁ b₁ i a₂ b₂ 4 2 i A ei θ A cos θ i A sin θ A 4² 2² 44721 θ tan¹ 24 265651 145 z₁ 3 4 i z₂ 1 2 i z z₁ z₂ 3 4 i 1 2 i 2 6 i A ei θ where A 2² 6² 63246 and θ tan¹ 62 tan¹ 3 12490 rad 146 z₁ 1 2 i z₂ 3 4 i z z₁ z₂ 1 2 i3 4 i 11 2 i A ei θ where A 11² 2² 111803 and θ tan¹ 211 01798 rad 147 z z₁z₂ 1 2 i 3 4 i 1 2 i3 4 i 3 4 i3 4 i 5 10 i25 02 04 i A ei θ where A 02² 04² 04472 and θ tan¹ 0402 tan¹ 2 11071 rad 148 xt X cos ω t yt Y cos ω t ϕ a x²X² cos² ω t y²Y² cos² ω t ϕ 2 xyXY cos ϕ 2 cos ω t cos ω t ϕ cos ϕ x²X² y²Y² 2xyXY cos ϕ cos² ω t cos² ω t ϕ 2 cos ω t cos ϕ cos ω t ϕ 1 Noting that cos² α 12 1 cos 2 α Eq 1 can be rewritten as x²X² y²Y² 2 xyXY cos ϕ 12 12 cos 2 ω t 12 12 cos 2 ω t 2 ϕ 2 cos ω t cos ϕ cos ω t ϕ 1 12 2 cos 2ωt2ωt2ϕ2 cos 2ωt2ωt2ϕ2 2 cos ω t cos ϕ cos ω t ϕ 1 cos 2 ω t ϕ cos ϕ 2 cos ω t cos ϕ cos ω t ϕ 1 cos 2 ω t ϕ cos ϕ 2 cos ϕ 12 cos ω t ϕ ω t cos ω t ϕ ω t 1 cos ϕ cos 2 ω t ϕ cos ϕ cos ϕ cos 2 ω t ϕ 1 cos² ϕ sin² ϕ 2 b When ϕ 0 Eq 2 reduces to x²X² y²Y² 2 xyXY xX yY² 0 which gives X XY y This indicates that the locus of the resultant motion is a straight line When ϕ π2 Eq 2 reduces to x²X² y²Y² 1 which denotes an ellipse with its major and minor axes along x and y directions respectively When ϕ π Eq 2 reduces to that of a straight line as in the case of ϕ 0 149 Equation for resultant motion x²X² y²Y² 2 xyXY cos² ϕ sin² ϕ 1 When y 0 Eq 1 reduces to x²X² sin² ϕ and hence x X sin ϕ 62 OS in figure 2 When x 0 Eq 1 reduces to y²Y² sin² ϕ and hence y Y sin ϕ 60 OT in figure 3 It can be seen that OR X cos ϕ 76 in figure 4 OSOR X sin ϕX cos ϕ tan ϕ 6276 08158 or ϕ 392072 5 From Eqs 2 and 4 we find X X sin ϕ² X cos ϕ² 62² 76² 98082 mm Equations 3 and 5 give Y 60sin ϕ 60sin 392072 94918 mm 150 a xt A1000 cos 50 t α m where A is in mm E₁ x0 A1000 cos α 0003 A cos α 3 E₂ ẍ0 50 A1000 sin α 1 A sin α 20 E₃ A A cos α² A sin α²12 202237 mm α tan¹ A sin αA cos α tan¹ 66667 814692 14219 rad xt 202237 cos 50 t 14219 mm b cos A B cos A cos B sin A sin B EqE₁ can be expressed as xt A cos 50tcos α A sin 50tsin α A₁ cos ωt A₂ sin ωt where ω 50 A₁ A cos α A₂ A sin α xt 3 cos 50 t 20 sin 50 t mm 151 xt A₁ cos ωt A₂ sin ωt dxdt t A₁ ω sin ωt A₂ ω cos ωt d²xdt² A₁ ω² cos ωt A₂ ω² sin ωt d²xdt² ω² xt where ω² is a constant Hence xt is a simple harmonic motion 152 a Using trigonometric relations x₁t 5 cos 3t cos 1 sin 1t sin 1 x₂t 10 cos 3t cos 2 sin 3t sin 2 xt x₁t x₂t cos 3t 5 cos 1 10 cos 2 sin 3t 5 sin 1 10 sin 2 If xt A cos ωt α A cos ωt cos α A sin ωt sin α ω 3 A cos α 5 cos 1 10 cos 2 14599 A sin α 5 sin 1 10 sin 2 133003 A A cos α² A sin α² 133802 α tan¹ A sin αA cos α tan¹ 91104 962640 168 rad Angle between x₁t and xt is 962640 573 38964 b Using vector addition For an arbitrary value of ωt 1 harmonic motions x₁t and x₂t can be shown as in the figure From vector addition we find xt 1338 cosωt 168 c Using complex numbers x₁t Re A₁ eiωt1 Re 5 eiωt1 x₂t Re A₂ eiωt2 Re 10 eiωt2 If xt Re A eiωtα A cos 3tα A₁ cos 3t1 A₂ cos 3t2 ie A cos 3t cos α sin 3t sin α 5 cos 3t cos 1 sin 3t sin 1 10 cos 3t cos 2 sin 3t sin 2 ie A cos α 5 cos 1 10 cos 2 A sin α 5 sin 1 10 sin 2 A 133802 α 168 rad xt Re 133802 ei3t168 153 xt 10 sin ωt 60 x₁t x₂t where x₁t 5 sin ωt 30 and x₂t A sin ωt α 10 sin ωt cos 60 cos ωt sin 60 5 sin ωt cos 30 cos ωt sin 30 A sin ωt cos α cos ωt sin α 10 cos 60 5 cos 30 A cos α A cos α 06699 10 sin 60 5 sin 30 A sin α A sin α 61603 A 06699² 61603² 61966 α tan1 6160306699 837938 x₂t 61966 sin ωt 837938 154 xt ½ cos π2 t sin πt ½ cos π2 t 1 4 sin π2 t From the nature of the graph of xt it can be seen that xt is periodic with a time period of τ 4 Graph 155 If xt is harmonic xt ω² xt Here xt 2 cos 2t cos 3t xt 8 cos 2t 9 cos 3t constant times xt xt is not harmonic 156 xt ½ cos π2 t cos πt xt π²8 cos π2 t π² cos πt constant times xt xt is not harmonic 157 xt x₁t x₂t 3 sin 30t 3 sin 29t since sin A sin B 2 sin AB2 cos AB2 xt 6 cos t2 sin 592 t xt sin 592 t 6 cos t2 This equation shows that the amplitude 6 cos t2 varies with time between a maximum value of 6 and a minimum value of 0 The frequency of this oscillation beat frequency is ωB 1 Note Beat frequency is twice the frequency of the term 6 cos t2 since two peaks pass in each cycle of 6 cos t2 158 The resultant motion of two harmonic motions having identical amplitudes X but slightly different frequencies ω and ω δω is given by Eq 167 xt 2X cos ω t δω t2 cos δ ω t2 Thus the maximum amplitude of the resultant motion is equal to 2X and the beat frequency is equal to δω From Fig 188 we find that 2X 5 mm or X 25 mm and δω2 2πτbeat 2πτlarger 2π126 42 0374 radsec or δω 0748 radsec and ω δω2 2πτsmaller 2π1 62832 radsec Hence ω 62832 03740 59092 radsec Thus the amplitudes of the two motions X 25 mm and their frequencies are ω 59092 radsec and ω δω 59092 07480 66572 radsec 159 A 005 m ω 10 Hz 62832 radsec period τ 2πω 2π62832 01 sec maximum velocity A ω 005 62832 31416 ms maximum acceleration A ω² 005 62832² 197393 ms² 160 ω 15 cps 94248 radsec xmax 05 g 05 981 4905 ms² A ω² A amplitude 490594248² 00005522 m xmax max velocity A ω 005204 ms 161 x A cos ωt xmax A 025 mm x ω²A cos ωt xmax A ω² 04 g 3924 mms² ω² 2924A 15696 rads² operating speed of pump ω 1252837 rads 199395 rpm 162 xt X sin 2πtτ xrms 1τ 0τ X² sin² 2πtτ dt12 Using sin² 2πtτ 1 cos 4πtτ2 we obtain xrms X²τ 0τ 12 12 cos 4πtτ dt12 X²τ t2 12 τ4π sin 4πtτ 0τ 12 X²τ τ2 τ8π sin 4π 0 012 X2 163 xt A tτ 0 t τ xrms 1τ 0τ A²τ² t² dt12 1τ A²τ² t³30 12 A²τ³ τ³312 A²312 A3 164 For even functions xt xt From Eq 173 bn 2τ 0τ xt sin nωt dt 2τ τ2τ2 xt sin nωt dt 2τ τ20 xt sin nωt dt 0τ2 xt sin nωt dt E1 Since sin nωt sin nωt odd function of t the product of xt and sin nωt is an odd function Further for an odd function ft ft ft and aa ft dt a0 ft dt 0a ft dt aa ft dt a0 ft dt 0a ft dt 0a ft dt 0 E2 Equations E1 and E2 lead to bn 0 Also since cos nωt is an even function we get an 2τ τ2τ2 xt cos nωt dt 4τ 0τ2 xt cos nωt dt For odd functions xt xt From Eq 172 an 2τ 0τ xt cos nωt dt 2τ τ2τ2 xt cos nωt dt Since cos nωt is an even function cosnωt cos nωt the product of xt and cos nωt is an odd function Hence an 0 Further since sin nωt is an odd function xt sin nωt is an even function and hence bn 4τ 0τ2 xt sin nωt dt 165 xt A 0 t τ2 A τ2 t τ a xt A 0 t 3τ4 A τ4 t 3τ4 A 3τ4 t τ b xt 0 0 t τ2 2A τ2 t τ c xt 2A 0 t τ4 0 τ4 t 3τ4 2A 3τ4 t τ d a xt xt odd function hence a0 an 0 bn 2τ 0τ xt sin nωt dt 2τ A 0τ2 sin nωt dt A τ2τ sin nωt dt 2Aτ cos nωtnω 0πω 2Aτ cos nωtnωπωτ2 2Aτ nω 2 cos nπ cos 0 cos 2nπ xt 4Aπ 12n1 sin2n1ωt b xt xt even function hence bn 0 a0 2τ 0τ xt dt 2τ At0τ4 Atτ43τ4 At3τ4τ 0 an 2τ 0τ xt cos nωt dt 2Anπω sin nωt 0τ4 sin nωt τ43τ4 sin nωt 3τ4τ Anπ 2 sin nπ2 2 sin 3nπ2 sin 2nπ 4Anπ for n159 4Anπ for n3711 xt 4Aπ 1n12n1 cos 2πn1tτ c a0 2τ 0τ xt dt 2τ 0 2A tτ2τ 2A an 2τ 0τ xt cos nωt dt 4Anωτ sin nωtτ2τ 0 bn 2τ 0τ xt sin nωt dt 4Anωτ cos nωtτ2τ 4Anωτ cos 2πn cos nπ xt 4Aπ 12n1 sin 2n1ωt with ω 2πτ d xt xt even function hence bn 0 a0 2τ 0τ xt dt 2τ 2A τ4 0 2A τ 3τ4 2A an 2τ 0τ xt cos nωt dt 4Anωτ sin nωtτ40 sin nωtτ3τ4 4Anωτ sin nπ2 sin 2nπ sin 3nπ2 xt 4Aπτ 1n12n1 cos 2π2n1tτ with ω 2πτ 166 xt A sin 2πtτ 0 t τ2 0 τ2 t τ a0 2τ 0τ xt dt 2τ 0τ2 sin 2πtτ dt 2Aτ τ22π cos 2πtττ20 2Aπ an 2τ 0τ xt cos nωt dt 2Aτ 0τ2 sin 2πtτ cos nωt dt E1 Using the relation sin mωt cos nωt sin mnωt sin mnωt 2 Eq E1 can be rewritten as an Aτ 0πω sin 1n ωt sin 1n ωt dt when n1 a1 Aτ 0πω sin 2ωt dt 0 when n2 3 4 an Aτ cos 1n ωt1n ω cos 1n ωt1n ωπω0 A2τ 1 cos 1n π1n 1 cos 1n π1n 0 if n is odd 2A n1n1π if n is even Similarly bn 2τ 0τ xt sin nωt dt 2Aτ 0τ2 sin 2πtτ cos nωt dt Aτ 0τ2 cos 1n ωt cos 1n ωt dt When n1 b1 Aτ 0πω dt cos 2ω t dt A2 When n2 3 4 bn Aτ sin 1n ωt1n ω sin 1n ωt1n ωπω0 0 xt Aπ A2 sin ωt 2Aπ cos nωtn2 1 n246 167 xt 2Atτ 0 t τ2 2Atτ 2A τ2 t τ a0 2τ ₀τ xt dt 2τ ₀τ2 2Atτ dt τ2τ 2Atτ 2A dt 2τ 2Aτ t²2 ₀τ2 2Aτ t²2 τ2τ 2A t τ2τ 2τ Aτ4 3Aτ4 Aτ A an 2τ ₀τ xt cos nωt dt 2τ ₀τ2 2Aτ t cos nωt dt τ2τ 2Aτ t 2A cos nωt dt 2τ 2Aτ t sin nωt nω cos nωt n²ω² 0τ2 2Aτ t sin nωt nω cos nωt n²ω² τ2τ 2A sin nωt nω τ2τ As τ 2πω an ωπ Aωπ n² ω² cos nπ Aωπ n² ω² Aωπ n² ω² cos 2π n Aωπ n² ω² cos nπ 2A n² π² cos nπ 1 4A n² π² n135 0 n246 bn 2τ ₀τ xt sin nωt dt 2τ ₀τ2 2Aτ t sin nωt dt τ2τ 2Aτ t 2A sin nωt dt 2τ 2Aτ t cos nωt nω sin nωt n² ω² 0τ2 2Aτ t cos nωt nω sin nωt n² ω² τ2τ 2A cos nωt nω τ2τ ωπ Anω cos nπ 2Anω cos 2nπ Anω cos nπ 2Anω cos 2nπ 2Anω cos nπ 0 xt A2 4Aπ² n135 1n² cos nωt 168 xt 4Atτ 0 t τ4 4Atτ 2A τ4 t 3τ4 4Atτ 4A 3τ4 t τ a0 2τ ₀τ xt dt 2τ 4Aτ t²2 0τ4 4Aτ t²2 2A t τ43τ4 4Aτ t²2 4A t 3τ4τ 0 an 2τ ₀τ xt cos nωt dt 2τ 4Aτ ₀τ4 t cos nωt dt 4Aτ τ43τ4 t cos nωt dt 2A τ43τ4 cos nωt dt 4Aτ 3τ4τ t cos nωt dt 4A 3τ4τ cos nωt dt 2τ 4Aτ t sin nωt nω cos nωt n² ω² 0τ4 4Aτ t sin nωt nω cos nωt n² ω² τ43τ4 2A sin nωt nω τ43τ4 ωπ sin nπ2 Anω Anω 2Anω cos nπ2 2Aπ n² ω² 2Aπ n² ω sin 3nπ2 3Anω 2Anω 3Anω 4Anω cos 3nπ2 2Aπ n² ω 2Aπ n² ω cos 2π n 2Aπ n² ω cos 0 2Aπ n² ω 0 bn 2τ ₀τ xt sin nωt dt 2τ 4Aτ ₀τ4 t sin nωt dt 4Aτ τ43τ4 t sin nωt dt 2A τ43τ4 sin nωt dt 4Aτ 3τ4τ t sin nωt dt 4A 3τ4τ sin nωt dt 2τ 4Aτ 1 n² ω² sin nωt tnω cos nωt 0τ4 4Aτ 1 n² ω² sin nωt tnω cos nωt τ43τ4 2A cos nωt nω τ43τ4 4Aπ² n² sin nπ2 sin 3nπ2 0 if n is even 8Aπ² n² 1n12 if n is odd xt 8Aπ² n135 1n12 sin nωt n² 169 xt A 1 tτ 0 t τ a0 2τ ₀τ xt dt 2Aτ ₀τ 1 tτ dt 2Aτ t t²2τ 0τ A an 2τ ₀τ xt cos nωt dt 2Aτ sin nωt nω tτ sin nωt nω cos nωt τ n² ω² 02πω 0 bn 2τ ₀τ xt sin nωt dt 2Aτ cos nωt nω tτ cos nωt nω sin nωt τ n² ω² 02πω A π n xt A2 Aπ n1 sin nωt n 170 The truncated series of k terms can be denoted as xt a₀2 n1k an cos n ω t n1k bn sin n ω t 1 with xt denoting an approximation to the exact xt given by Eq 170 The error to be minimized is given by E πωπω e²t dt 2 where et xt xt 3 and xt is the exact value with infinite series on the right hand side of Eq 1 Treating E as a function of the unknowns an and bn it can be minimized by setting Ean 2 πωπω xt xt cos n ω t dt 0 4 Ebn 2 πωπω xt xt sin n ω t dt 0 5 Rearranging Eq 4 gives from πω to πω xt cos n ω t dt from πω to πω xt cos n ω t dt 6 Using orthogonality property the right hand side of Eq 6 can be expressed as from πω to πω xt cos n ω t dt 0 for m n āₙ πω for m n 7 This leads to from πω to πω xt cos n ω t dt āₙ πω 8 or āₙ ωπ from πω to πω xt cos n ω t dt n0 1 2 k 9 In a similar manner we can derive bₙ ωπ from πω to πω xt sin n ω t dt n1 2 k 10 It can be observed that Eqs 9 and 10 are similar to those of Eqs E3 and E4 171 i ti xi n1 n2 n3 xi cos 2π ti 032 xi sin 2π ti 032 xi cos 4π ti 032 xi sin 4π ti 032 xi cos 6π ti 032 xi sin 6π ti 032 1 002 9 83149 34442 63639 63640 34441 83149 2 004 13 91924 91924 00000 130000 91924 91923 3 006 17 65056 157060 120209 120208 157059 65057 4 008 29 00000 290000 290000 00000 00000 290000 5 010 43 164556 397267 304053 304059 397271 164548 6 012 59 417195 417191 00000 590000 417187 417199 7 014 63 582045 241087 445482 445472 241101 582040 8 016 57 570000 00000 570000 00000 570000 00000 9 018 49 452700 187518 346477 346487 187505 452705 10 020 35 247485 247489 00000 350000 247493 247482 11 022 35 133936 323359 247493 247482 323354 133950 12 024 41 00000 410000 410000 00000 00000 410000 13 026 47 179866 434221 332333 332347 434229 179847 14 028 41 289917 289911 00000 410000 289905 289923 15 030 13 120105 49747 91927 91921 49755 120102 16 032 7 70000 00000 70000 00000 70000 00000 34 from i1 to 16 558 1667897 313278 116552 915984 432234 268281 18 from i1 to 16 6975 203487 39160 14569 114498 54029 33535 172 Speed 100 rpm In a minute a point will be subjected to the maximum pressure A pmax 700 kPa 100 4 400 times Hence period τ 60400 015 sec pt A 0 t τ4 0 τ4 t τ a₀ 2τ from 0 to τ pt dt 2τAt₀τ4 A2 350 kPa am 2τ from 0 to τ pt cos m ω t dt 2 Aτ sin m ω t m ω ₀τ4 Aπm sin m π2 bm 2τ from 0 to τ pt sin m ω t dt 2 Aτ sin m ω t m ω ₀τ4 Aπm cos m π2 1 Evaluation of am and bm m1 a₁ Aπ sin π2 Aπ 222817 kPa b₁ Aπ cos π2 1 222817 kPa m2 a₂ A2π sin π 0 b₂ A2π cos π 1 222817 kPa m3 a₃ A3π sin 3π2 74272 kPa b₃ A3π cos 3π2 1 74272 kPa pt a₀2 from m1 to am cos m ω t bm sin m ω t kPa 35 173 Speed 200 rpm In a minute a point will be subjected to the maximum pressure A pmax 700 kPa 200 6 1200 times Hence period τ 601200 005 sec pt A 0 t τ4 0 τ4 t τ a₀ 2τ from 0 to τ pt dt 2τ At₀τ4 A2 350 kPa am 2τ from 0 to τ pt cos m ω t dt 2Aτ sin m ω t m ω₀τ4 Aπm sin m π2 bm 2τ from 0 to τ pt sin m ω t dt 2Aτ sin m ω t m ω₀τ4 Aπm cos m π2 1 Evaluation of am and bm m1 a₁ Aπ sin π2 Aπ 222817 kPa b₁ Aπ cos π2 1 222817 kPa m2 a₂ A2π sin π 0 b₂ A2π cos π 1 222817 kPa m3 a₃ A3π sin 3π2 74272 kPa b₃ A3π cos 3π2 1 74272 kPa pt a₀2 from m1 to am cos m ω t bm sin m ω t kPa 174 i tᵢ Mᵢ n1 n2 n3 Mₜᵢ cos 2π tᵢ 0012 Mₜᵢ sin 2π tᵢ 0012 Mₜᵢ cos 4π tᵢ 0012 Mₜᵢ sin 4π tᵢ 0012 Mₜᵢ cos 6π tᵢ 0012 Mₜᵢ sin 6π tᵢ 0012 1 00005 770 7437627 1992912 6668391 3850010 5444712 5444731 2 00010 810 7014802 4050007 4049988 7014912 00000 8100000 3 00015 850 6010798 6010417 00000 8500000 6010442 6010273 4 00020 910 4549978 7880845 4550041 7880808 9100000 00000 36 5 00025 1010 2614043 9755859 874689 504995 714171 714184 6 00030 1170 00000 11700000 1170000 0000 0000 1170000 7 00035 1370 3545874 13233169 1186449 685010 968748 968725 8 00040 1610 8050073 13942966 8049877 1394309 1610000 0000 9 00045 1890 13364407 13364229 0000 1890000 1336410 1336454 10 00050 1750 15155491 8747922 875019 1515534 0000 1750000 11 00055 1630 15744619 4218647 1411634 814979 1152608 1152560 12 00060 1510 15100000 0000 1510000 0000 1510000 0000 13 00065 1390 13426345 3597671 1203767 695014 982858 982898 14 00070 1290 11171677 6450088 644982 1117183 0000 1290000 15 00075 1190 8414492 8414648 0000 1190000 841479 841435 16 00080 1110 5549897 9612942 555021 961276 1110000 0000 17 00085 1050 2717498 10142249 809337 524982 742440 742485 18 00090 990 00000 9900000 990000 0000 0000 990000 19 00095 930 2407123 8983081 805393 465018 657633 657586 20 00100 890 4450095 7707571 444981 770773 890000 0000 21 00105 850 6010478 6010337 0000 850000 601022 601060 22 00110 810 7014868 4049895 405022 701468 0000 810000 23 00115 770 7437659 1992798 666851 384980 544500 544444 24 00120 750 7500000 00000 750000 00000 750000 0000 24 Σ 27300 49793242 18037673 343270 1754047 428734 661855 i1 12 24 Σ 2275 4149436 1503139 28606 146171 35728 55155 i1 175 i ti xi n1 n2 n3 xi cos 2π ti06 xi sin 2π ti06 xi cos 4π ti06 xi sin 4π ti06 xi cos 6π ti06 xi sin 6π ti06 1 0025 900 869 233 779 450 636 636 2 0050 1700 1472 850 850 1472 000 1700 3 0075 2300 1626 1626 000 2300 1626 1626 4 0100 2500 1250 2165 1250 2165 2500 000 5 0125 2600 673 2511 2252 1300 1838 1838 6 0150 2800 000 2800 2800 000 000 2800 7 0175 3300 854 3188 2858 1650 2333 2333 8 0200 3500 1750 3031 1750 3031 3500 000 9 0225 3400 2404 2404 000 3400 2404 2404 10 0250 2900 2511 1450 1450 2511 000 2900 11 075 2400 2318 621 2078 1200 1697 1697 12 0300 2600 2600 000 2600 000 2600 000 13 0325 3200 3091 828 2771 1600 2263 2263 14 0350 4000 3464 2000 2000 3464 000 4000 15 0375 1800 1273 1273 000 1800 1273 1273 16 0400 800 400 693 400 693 800 000 17 0425 500 129 483 433 250 354 354 18 0450 1400 000 1400 1400 000 000 1400 19 0475 2800 725 2705 2425 1400 1980 1980 20 0500 3700 1850 3204 1850 3204 3700 000 21 0525 3300 2333 2333 000 3300 2333 2334 22 0550 2900 2511 1450 1450 2511 000 2900 23 0575 2200 2125 569 1905 1100 1556 1556 24 0600 000 000 000 000 000 000 000 24 Σ 23900 24190 28230 3972 14718 4526 488 i1 12 24 Σ 1992 2016 2353 331 1226 377 041 i1 176 Program1m Program for calling the subroutine FORIER Run Program1m in MATLAB Command Window Program1m and forierm should be in the same file folder and set the path to this folder Following 6 lines contain problemdependent data n16 m3 time032 x9 13 17 29 43 59 63 57 49 35 35 41 47 41 13 7 t002002032 end of problemdependent data Following line calls subroutine forierm azeroabxsinxcosforiernmtimext following outputs data fprintfFourier series expansion of the function xt fprintfData fprintfNumber of data points in one cycle 30f n fprintf fprintfNumber of Fourier Coefficients required 30f m fprintf fprintfTime period 86e time fprintfStation i fprintfTime at station i ti fprintfxi at ti for i1n fprintf 8d256e276e itixi end fprintf fprintfResults of Fourier analysis fprintfazero86e azero fprintfvalues of i ai bi for i1m fprintf100g 86e206e iaibi end Subroutine forierm function azeroabxsinxcosforiernmtimext pi31416 sumz00 for i1n sumzsumzxi end azero20sumzn for ii1m sums00 sumc00 for i1n theta20pitiiitime xcosixicostheta xsinixisintheta sumssumsxsini sumcsumcxcosi end aii20sumcn bii20sumsn end program1 Fourier series expansion of the function xt Data Number of data points in one cycle 16 Number of Fourier Coefficients required 3 Time period 3200000e001 Station i Time at station i ti xi at ti 1 2000000e002 9000000e000 2 4000000e002 1300000e001 3 6000000e002 1700000e001 4 8000000e002 2900000e001 5 1000000e001 4300000e001 6 1200000e001 5900000e001 7 1400000e001 6300000e001 8 1600000e001 5700000e001 9 1800000e001 4900000e001 10 2000000e001 3500000e001 11 2200000e001 3500000e001 12 2400000e001 4100000e001 13 2600000e001 4700000e001 14 2800000e001 4100000e001 15 3000000e001 1300000e001 16 3200000e001 7000000e000 Results of Fourier analysis azero6975000e001 values of i ai bi 1 2084870e001 3915985e000 2 1456887e000 1144979e001 3 5402900e000 3353473e000 Ex177m for i 1101 ti 032i1100 xi 34875 208487cos19635ti 39160sin19635ti 14569cos3927ti 114498sin3927ti 54029cos58905ti 33535sin58905ti end plottx xlabelt ylabelxt Ex178m u 03445 l 10 h0 01 a0 2 r0 05 First case r changes for i 1101 ri 05 i105100 c1i 6piu1h03 a0 h022 ri2 a02ri2a0h02 h0 end Second case h changes for i 1101 hi 005 i1005100 c2i 6piu1hi3 a0 hi22 r02 a02r02a0hi2 hi end Third case a changes for i 1101 ai 2 i12100 c3i 6piu1h03 ai h022 r02 ai2r02aih02 h0 end subplot311 plotrc1 xlabelr ylabelcr subplot312 plothc2 xlabelh ylabelch subplot313 plotac3 xlabela ylabelca Ex179m for i 1101 xi i14100 kai 1000xi 100xi2 kbi 500 500 xi2 end plotxka hold on plotxkb xlabelx ylabelka solid line kb dash line Ex180m for i 1201 ti i130200 x1i 3sin30ti x2i 3sin29ti xi x1i x2i end plottx xlabelt ylabelx Results of Ex181 Please input the data Please input n 24 Please input m 3 Please input time 0012 Please input the value of xi i 0 n1 770 810 850 910 1010 1170 1370 1610 1890 1750 1630 1510 1390 1290 1190 1110 1050 990 930 890 850 810 770 750 Please input the value of ti i 0 n1 00005 0001 00015 0002 00025 0003 00035 0004 00045 0005 00055 0006 00065 0007 00075 0008 00085 0009 00095 0010 00105 0011 00115 0012 FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 24 NUMBER OF FOURIER COEFFICIENTS REQUIRED 3 TIME PERIOD 1200000e002 TIME AT VARIOUS STATIONS TI 5000000e004 1000000e003 1500000e003 2000000e003 2500000e003 3000000e003 3500000e003 4000000e003 4500000e003 5000000e003 5500000e003 6000000e003 6500000e003 7000000e003 7500000e003 8000000e003 8500000e003 9000000e003 9500000e003 1000000e002 1050000e002 1100000e002 1150000e002 1200000e002 KNOWN VALUES OF XI AT TI 7700000e002 8100000e002 8500000e002 9100000e002 1010000e003 1170000e003 1370000e003 1610000e003 1890000e003 1750000e003 1630000e003 1510000e003 1390000e003 1290000e003 1190000e003 1050000e003 9900000e002 9300000e002 8900000e002 8500000e002 8100000e002 7700000e002 7500000e002 RESULTS OF FOURIER ANALYSIS AZERO 2275000e003 VALUES OF I AI AND BI ARE 1 4149436e002 1503138e002 2 2860518e001 1461703e002 3 3572730e001 5515471e001 Results of Ex182 Please input the data Please input n 16 Please input m 3 Please input time 032 Please input the value of xi i 0 n1 9 13 17 29 43 59 63 57 49 35 35 41 47 41 13 7 Please input the value of ti i 0 n1 002 004 006 008 010 012 014 016 018 020 022 024 026 028 030 032 FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 16 NUMBER OF FOURIER COEFFICIENTS REQUIRED 3 TIME PERIOD 3200000e001 TIME AT VARIOUS STATIONS TI 2000000e002 4000000e002 6000000e002 8000000e002 1000000e001 1200000e001 1400000e001 1600000e001 1800000e001 2000000e001 2200000e001 2400000e001 2600000e001 2800000e001 3000000e001 3200000e001 KNOWN VALUES OF XI AT TI 9000000e000 1300000e001 1700000e001 2900000e001 4300000e001 5900000e001 6300000e001 5700000e001 4900000e001 3500000e001 3500000e001 4100000e001 4700000e001 4100000e001 1300000e001 7000000e000 RESULTS OF FOURIER ANALYSIS AZERO 6975000e001 VALUES OF I AI AND BI ARE 1 2084870e001 3915985e000 2 1456887e000 1144979e001 3 5402900e000 3353473e000 Results of Ex183 Please input the data Please input n 24 Please input m 6 Please input time 06 Please input the value of xi i 0 n1 9 17 23 25 26 28 33 35 34 29 24 26 32 40 18 8 5 14 28 37 33 29 22 0 Please input the value of ti i 0 n1 0025 0050 0075 0100 0125 0150 0175 0200 0225 0250 0275 0300 0 0325 0350 0375 0400 0425 0450 0475 0500 0525 0550 0575 060 0 0 FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 24 NUMBER OF FOURIER COEFFICIENTS REQUIRED 6 TIME PERIOD 6000000e001 TIME AT VARIOUS STATIONS TI 2500000e002 5000000e002 7500000e002 1000000e001 1250000e001 1500000e001 1750000e001 2000000e001 2250000e001 2500000e001 2750000e001 3000000e001 3250000e001 4250000e001 4500000e001 4750000e001 5000000e001 5250000e001 5500000e001 5750000e001 6000000e001 KNOWN VALUES OF XI AT TI 9000000e000 1700000e001 2300000e001 2500000e001 2600000e001 2800000e001 3300000e001 3500000e001 3400000e001 2900000e001 2400000e001 2600000e001 3200000e001 4000000e001 1800000e001 8000000e000 5000000e000 1400000e001 2800000e001 3700000e001 3300000e001 2900000e001 2200000e001 0000000e000 RESULTS OF FOURIER ANALYSIS AZERO 1991667e001 VALUES OF I AI AND BI ARE 1 2015867e001 2352528e001 2 3309933e000 1226463e001 3 3771938e000 4063822e001 4 9584350e001 3247643e000 5 1137750e000 1871580e000 6 1166755e000 1249975e000 Results of Ex184 Please input the data Please input n 14 Please input m 10 Please input time 035 Please input the value of xi i 0 n1 045 08 09 06 075 07 055 175 165 025 11 14 105 00 Please input the value of ti i 0 n1 0025 005 0075 01 0125 015 0175 02 0225 025 0275 03 0325 035 FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 14 NUMBER OF FOURIER COEFFICIENTS REQUIRED 10 TIME PERIOD 3500000e001 TIME AT VARIOUS STATIONS TI 2500000e002 5000000e002 7500000e002 1000000e001 1250000e001 1500000e001 1750000e001 2000000e001 2250000e001 2500000e001 2750000e001 3000000e001 3250000e001 3500000e001 2105353e314 3416413e312 KNOWN VALUES OF XI AT TI 4500000e001 8000000e001 9000000e001 6000000e001 7500000e001 7000000e001 5500000e001 1750000e000 1650000e000 2500000e001 1100000e000 1400000e000 1050000e000 0000000e000 2105353e314 1190440e311 RESULTS OF FOURIER ANALYSIS AZERO 3785714e001 VALUES OF I AI AND BI ARE 1 6195109e001 3504626e001 2 4624270e001 9240901e001 3 4149472e001 1712710e001 4 2227148e002 2469142e001 5 4543377e002 1043036e001 6 2043688e002 5600801e002 7 5000000e002 4775167e006 8 2043367e002 5599619e002 9 4542944e002 1042959e001 10 2228333e002 2469100e001 185 The main program and the output are shown below C C MAIN PROGRAM FOR CALLING THE SUBROUTINE FORIER C C C FOLLOWING 6 LINES NEED TO BE CHANGED FOR A DIFFERENT PROBLEM DIMENSION X24T24XSIN24XCOS24A5B5 DATA NMTIME 2450012 DATA X77081085091010101170137016101890175016301510 2 13901290119011101050990930870850810770750 DATA T 00050010015002002500300350040045005 2 00550060065007007500800850090095010105 3 0110115012 C END OF PROBLEMDEPENDENT DATA CALL FORIER NMTIMEXTAZEROABXSINXCOS PRINT 100 100 FORMAT 46H FOURIER SERIES EXPANSION OF THE FUNCTION XT PRINT 200 NMTIME 200 FORMAT 6H DATA37H NUMBER OF DATA POINTS IN ONE CYCLE 15 2 42H NUMBER OF FOURIER COEFFICIENTS REQURED 15 3 14H TIME PERIOD E158 PRINT 300 TII1N 300 FORMAT 33H TIME AT VARIOUS STATIONS TI 4E1581X PRINT 400 XII1N 400 FORMAT 31H KNOWN VALUES OF XI AT TI 4E1581X PRINT 500 500 FORMAT 29H RESULTS OF FOURIER ANALYSIS PRINT 600 AZERO 600 FORMAT 8H AZERO 2XE15831H VALUES OF I AI AND BI ARe 2 DO 700 I 1M 700 PRINT 800 I AI BI 800 FORMAT 152XE1582XE158 STOP END FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 24 NUMBER OF FOURIER COEFFICIENTS REQUIRED 5 TIME PERIOD 011999998E01 TIME AT VARIOUS STATIONS TI 050000008E03 099999993E03 015000000E02 020000001E02 024999999E02 030000000E02 035000001E02 040000007E02 045000017E02 049999990E02 055000000E02 060000010E02 064999983E02 069999993E02 075000003E02 080000013E02 084999986E02 089999996E02 095000006E02 010000002E01 010499999E01 011000000E01 011500001E01 011999998E01 KNOWN VALUES OF XI AT TI 077000000E03 081000000E03 085000000E03 091000000E03 010100000E04 011700000E04 013700000E04 016100000E04 018900000E04 017500000E04 016300000E04 015100000E04 013900000E04 012900000E04 011900000E04 011100000E04 010500000E04 099000000E03 093000000E03 089000000E03 085000000E03 081000000E03 077000000E03 075000000E03 RESULTS OF FOURIER ANALYSIS AZERO 022750000E04 VALUES OF I AI AND BI ARE 1 041494360E03 015031395E03 2 028605835E02 014617058E03 3 035727844E02 055154602E02 4 040830078E02 014440117E01 5 011577332E02 016659973E02 186 The main program and the output are given below C C MAIN PROGRAM FOR CALLING THE SUBROUTINE FORIER C C C FOLLOWING 6 LINES NEED TO BE CHANGED FOR A DIFFERENT PROBLEM DIMENSION X16T16XSIN16XCOS16A5B5 DATA NMTIME 165032 DATA X 913172943596357493535414741 2 137 DATA T 02040608112141619222242628 2 332 C END OF PROBLEMDEPENDENT DATA CALL FORIER NMTIMEXTAZEROABXSINXCOS PRINT 100 100 FORMAT 46H FOURIER SERIES EXPANSION OF THE FUNCTION XT PRINT 200 NMTIME 200 FORMAT 6H DATA37H NUMBER OF DATA POINTS IN ONE CYCLE 15 2 42H NUMBER OF FOURIER COEFFICIENTS REQUIRED 15 3 14H TIME PERIOD E158 PRINT 300 TII1N 300 FORMAT 33H TIME AT VARIOUS STATIONS TI 4E1581X PRINT 400 FORMAT 31H KNOWN VALUES OF XI AT TI 4E1581X PRINT 500 500 FORMAT 29H RESULTS OF FOURIER ANALYSIS PRINT 600 AZERO 600 FORMAT 8H AZERO 2XE15831H VALUES OF I AI AND BI ARE 2 DO 700 I 1M 700 PRINT 800 I AI BI 800 FORMAT 152XE1582XE158 STOP END FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 16 NUMBER OF FOURIER COEFFICIENTS REQUIRED 5 TIME PERIOD 031999999E00 TIME AT VARIOUS STATIONS TI 020000000E01 039999979E01 059999999E01 079999983E01 010000002E00 012000000E00 013999999E00 016000003E00 018000001E00 019999999E00 022000003E00 024000001E00 025999999E00 027999997E00 030000001E00 031999999E00 KNOWN VALUES OF XI AT TI 090000000E01 013000000E02 017000000E02 029000000E02 043000000E02 059000000E02 063000000E02 057000000E02 049000000E02 035000000E02 035000000E02 041000000E02 047000000E02 041000000E02 013000000E02 070000000E01 RESULTS OF FOURIER ANALYSIS AZERO 059750000E02 VALUES OF I AI AND BI ARE 1 020848709E02 039159737E01 2 014568996E01 011449797E02 3 054029312E01 033335175E01 4 017500381E01 024999523E01 5 026129245E01 010507224E01 The main program and the output are shown below 187 C C MAIN PROGRAM FOR CALLING THE SUBROUTINE FORIER C C C FOLLOWING 6 LINES NEED TO BE CHANGED FOR A DIFFERENT PROBLEM DIMENSION X24 T24 XSIN24 XCOS24A6B6 DATA NMTIME 24606 DATA 917232526283335342924263240188514 2 2837332922049 DATA T 025050751125151752225252753325 2 353754425454755525555756 C END OF PROBLEMDEPENDENT DATA CALL FORIER NMTIMEXTAZEROABXSINXCOS PRINT 100 100 FORMAT 46H FOURIER SERIES EXPANSION OF THE FUNCTION XT PRINT 200 NMTIME 200 FORMAT 6H DATA37H NUMBER OF DATA POINTS IN ONE CYCLE 2 42H NUMBER OF FOURIER COEFFICIENTS REQUIRED 15 3 14H TIME PERIOD E153 PRINT 300 TII1N 300 FORMAT 33H TIME AT VARIOUS STATIONS TI 4E1581X PRINT 400 XII1N 400 FORMAT 31H KNOWN VALUES OF XI AT TI 4E1581X PRINT 500 500 FORMAT 29H RESULTS OF FOURIER ANALYSIS PRINT 500 AZERO 600 FORMAT 8H AZERO 2XE15831H VALUES OF I AI AND BI ARE 2 DO 700 I 1M 700 PRINT 800 I AI BI 800 FORMAT 152XE1582XE158 STOP END FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 24 NUMBER OF FOURIER COEFFICIENTS REQUIRED 6 TIME PERIOD 060000002E00 TIME AT VARIOUS STATIONS TI 024999999E01 050000001E01 074999988E01 010000002E00 012500000E00 014999998E00 017500001E00 019999999E00 022500002E00 025000000E00 027499998E00 030000001E00 032499999E00 035000002E00 037500000E00 039999998E00 042500001E00 044999999E00 047500002E00 050000000E00 052499998E00 055000001E00 057499999E00 060000002E00 KNOWN VALUES OF XI AT TI 090000000E01 017000000E02 023000000E02 025000000E02 026000000E02 028000000E02 033000000E02 035000000E02 034000000E02 029000000E02 024000000E02 026000000E02 032000000E02 040000000E02 018000000E02 080000000E01 050000000E01014000000E02028000000E02037000000E02 033000000E02029000000E02022000000E02 000000000E00 RESULTS OF FOURIER ANALYSIS AZERO 019916672E0250 VALUES OF I AI AND BI ARE 1 020158676E02 023525284E02 2 033099222E01 012264638E02 3 037719278E01 040640426E00 4 095843577E00 032476425E01 5 011377630E01 018716125E01 6 011667604E01 012500324E01 The main program and the output are given below 188 C C MAIN PROGRAM FOR CALLING THE SUBROUTINE FORIER C C C FOLLOWING 6 LINES NEED TO BE CHANGED FOR A DIFFERENT PROBLEM DIMENSION X14 T14 XSIN14 XCOS14 A10 B10 DATA NMTIME 1410035 DATA X 45896757551751652511 2 1410500 DATA T 025050751125151752225252753325 2 35 C END OF PROBLEMDEPENDENT DATA CALL FORIER NMTIMEXTAZEROABXSINXCOS PRINT 100 100 FORMAT 46H FOURIER SERIES EXPANSION OF THE FUNCTION XT PRINT 200 NMTIME 200 FORMAT 6H DATA37H NUMBER OF DATA POINTS IN ONE CYCLE 15 2 42H NUMBER OF FOURIER COEFFICIENTS REQUIRED 15 3 14H TIME PERIOD E158 PRINT 300 TII1N 300 FORMAT 33H TIME AT VARIOUS STATIONS TI 4E1581X PRINT 400 XII1N 400 FORMAT 31H KNOWN VALUES OF XI AT TI 4E1581X PRINT 500 500 FORMAT 29H RESULTS OF FOURIER ANALYSIS PRINT 600 AZERO 600 FORMAT 8H AZERO 2XE15831H VALUES OF I AI AND BI ARE 2 DO 700 I 1M 700 PRINT 800 I AI BI 800 FORMAT 152XE1582XE158 STOP END FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 14 NUMBER OF FOURIER COEFFICIENTS REQUIRED 10 TIME PERIOD 035000002E0051 TIME AT VARIOUS STATIONS TI 024999999E01 050000001E01 074999988E01 010000002E00 012500000E00 014999998E00 017500001E00 019799999E00 022500002E00 025000000E00 027499998E00 030000001E00 032499999E00 035000002E00 KNOWN VALUES OF XI AT TI 044999999E00 080000001E00 089999998E00 060000002E00 075000000E00 069999999E00 055000001E00 017500000E01 016499996E01 025000000E00 011000004E01 013999996E01 010500002E01 000000000E00 RESULTS OF FOURIER ANALYSIS AZERO 037857169E00 VALUES OF I AI AND BI ARE 1 061951119E00 035046142E00 2 046242946E00 092408919E00 3 041494656E00 017127156E00 4 022273250E01 024691588E00 5 045435190E01 010430527E00 6 020434089E01 056009147E01 7 049997803E01 035924033E05 8 020436604E01 055996872E01 9 045427930E01 010429442E00 10 022274703E01 024690533E00 189 xp rl r cos θ l cos ϕ rl r cos ωt l1sin²ϕ E₁ But l sin ϕ r sin θ cos ϕ 1 r²l² sin² ωt12 E₂ Using E₂ in E₁ xp rl r cos ωt l 1 r²l² sin² ωt12 E₃ Let rl small 14 Using 1ε 1 ½ ε E₃ becomes xp r1 r2l rcos ωt r4l cos 2ωt E₄ a EqE₄ gives yp xp r 1 r2l r cos ωt ¼ rl cos 2ωt E₅ If rl is very small yp r cos ωt harmonic motion b To have amplitude of second harmonic smaller than that of first harmonic in EqE₅ we need to have 14 rl 125 ie rl 425 ie lr 625 once the amplitude of second harmonic is smaller by a factor of 25 the amplitudes of higher harmonics arising from the expansion of squarerootterm in E₃ are expected to be still smaller 190 Unbalanced force developed P 2 m ω² r cos ωt range of force 0 100 N range of frequency 25 50 Hz 15708 31416 radsec Parameters to be determined m r ω Let r 01 m To generate 100 N force at 25 Hz set Pmax 100 2 m 15708² 01 which gives m 100 215708² 01 00202641 kg 202641 g To generate 100 N force at 50 Hz set Pmax 100 2 m 31416² 01 which yields m 100 2 31416² 01 00050860 kg 50660 g 191 Goal Weight to be maintained at 5 005 kgmin Parameters to be determined Angular velocity of crank ω lengths of crank and connecting rod dimensions of the wedge dimensions of the orifice in the hopper dimensions of the actuating rod and dimensions of the lever arrangement Given Density of the material in the hopper Procedure Select ω based on available motor Determine the dimensions of the orifice in the hopper which delivers approximately 5 kgmin assuming continuous flow of material For trial dimensions of the wedge determine the increasedecrease in the size diameter of the orifice Choose the final dimensions of the wedge such that the material flow rate delivered by the orifice lies within the specified range 192 Force to be applied 1000 N frequency 50 Hz 31416 radsec Procedure 1 Select a motor that provides either directly or through a gear system the desired frequency Assume that it is connected to the cam 2 Setermine the sizes and dimensions of the plate cam and the roller 3 Choose the dimensions of the follower 4 Select the weight as 1000 N From thegeometry determine the range of displacement vertical motion of the weight 5 Determine the force exerted due to the falling weight 193 Considerations to be taken in the design of vibratory bowl feeders 1 Suitable design of the electromagnet and its coil 2 Radius of the bowl and the pitch of the spiral helical delivery track 3 Tooling to be fixed along the spiral track to reject the defective or outoftolerance or incorrectly oriented parts 4 Design of elastic supports 5 Size and location of the outlet 194 Axial spring constant of each tube k AEℓ Let diameter of each tube be 001 m 1 cm with thickness 0001 m 1 mm Then A π4 D² d² π4 001² 0008² 2827 10⁶ m² This gives k 2827 10⁶ 207 10¹¹ 2 2926 10⁵ Nm Since 76 tubes are in parallel we have the total axial stiffness as keq 76 k 76 2926 10⁵ 22238 10⁶ Nm The polar area moment of inertia of each tube is J π32 D⁴ d⁴ π32 001⁴ 0008⁴ 580 10⁸ m⁴ Torsional stiffness of each tube is given by GJℓ 796154 10⁹ 580 10⁸ 2 231 10³ Nmrad For 76 tubes in parallel equivalent torsional stiffness will be kteq 76 231 10³ 1756 10⁶ Nmrad Chapter 2 Free Vibration of Single Degree of Freedom Systems 21 δst 5 x 103 m ωn gδst12 9815x10312 442945 radsec 70497 Hz 22 τn 021 sec 2π mk m 021k2π i τnnew 2π mknew 2π m15k 2π 021k2π 15k 01715 sec ii τnnew 2π mknew 2π m05k 2π 021k2π105k 02970 sec 23 ωn 62832 radsec km m k62832 when spring constant is reduced ωn decreases ωnnew 055 ωn 345576 radsec knewm k800m k800k x 62836 345576 k800k 055 k800k 0552 03025 k 11469534 Nm m k62832 m k628322 1146953439478602 m 02905 kg 24 k 100101000 10000 Nm ωn keqm 4km 4 x 1041012 632456 radsec τn 2πωn 62832632456 00993 sec 25 Unit weight 9000 N Natural frequency 5 rads 10 rads Solution m 900098 Let ωn 75 radsec ωn keqm keq mωn2 900098 752 516582 Nm 4 k where k is the stiffness of the air spring Thus k 5165824 129145 Nm 26 x A cos ωn t φ0 x ωn A sin ωn t φ0 x ωn2 A cos ωn t φ0 a ωn A 01 msec τn 2πωn 2 sec ωn 31416 radsec A 01ωn 003183 m d xo xt0 A cos φ0 002 m cosφ0 002A 06283 φ0 510724 b xo xt0 ωn A sinφ0 01 sin510724 007779 msec c xmax ωn2 A 314162 003183 0314151 msec2 27 For small angular rotation of bar PQ about P 12 k12eq θ l32 12 k1 θ l12 12 k2 θ l22 ie k12eq k1 l12 k2 l22 l32 Let keq overall spring constant at Q 1keq 1k12eq 1k3 keq k12eq k3 k12eq k3 k1 l1l32 k2 l2l32 k3 k1 l1l32 k2 l2l32 k3 ωn keqm k1 k2 l12 k2 k3 l22 m k1 l12 k2 l22 k3 l32 28 m 2000 kg δst 002 m ωn gδst12 98100212 221472 radsec 29 Let x be measured from the position of mass at which the springs are unstretched Equation of motion is m x k1 x δst k2x δst W sinθ E1 where δst k1 k2 W sin θ Thus Eq E1 becomes m x k1 k2 x 0 ωn k1 k2m 210 Cart weight 20 kN Steel Youngs modulus E 200 GPa 9m steel wire rope 1 mm diameter 75 m Solution k1 A1 E1 l1 π4 00012 200 109 9 174533 Nm k2 A2 E2 l2 174533 75 9 145444 Nm keq k1 k2 319977 Nm Let x be measured from the unstretched length of the springs The equation of motion is m ẍ k1 k2 x δst W sin θ where k1 k2 δst W sin θ ie m ẍ k1 k2x 0 Thus the natural frequency of vibration of the cart is given by ωn sqrtk1 k2m sqrt319977 98 20 103 39597 rads 211 Weight of electronic chassis 500 N To be able to use the unit in a vibratory environment with a frequency range of 0 5 Hz its natural frequency must be away from the frequency of the environment Let the natural frequency be ωn 10 Hz 62832 radsec Since ωn sqrtkeq m 62832 we have keq m ωn2 500 981628322 201857 104 Nm 4 k so that k spring constant of each spring 5046425 Nm For a helical spring k G d4 8 n D3 Assuming the material of springs as steel with G 80 109 Pa n 5 and d 0005 m we find k 5046425 80 109 00054 8 5 D3 This gives D3 1250 103 5046425 247700 109 or D 00291492 m 291492 cm 212 i with springs k1 and k2 Let ya yb yl be deflections of beam at distances a b l from fixed end 12 k12eg yl2 12 k1 ya2 12 k2 yb2 ie k12eg k1 ya yl2 k2 yb yl2 y F x2 6 E I 3 l x x a ya F a2 6 E I 3 l a x b yb F b2 6 E I 3 l b x l yl F l3 3 E I ωn sqrtk1 k3 ya yl2 k2 k3 yb yl2 m k1 ya yl2 k2 yb yl2 kbeam12 where kbeam 3 E I l3 k1 3 E I a4 3 l a2 k2 3 E I b4 3 l b2 m l3 k1 a4 3 l a2 k2 b4 3 l b2 12 E I l312 ii without springs k1 and k2 ωn sqrtkbeam m sqrt3 E I m l3 213 Let x1 x2 displacements of pulleys 1 2 x 2 x1 2 x2 E1 Let P tension in rope For equilibrium of pulley 1 2 P k1 x1 E2 For equilibrium of pulley 2 2 P k2 x2 E3 where 1 k1 1 4 k 1 4 k 1 2 k k1 2 k and k2 k k 2 k Combining Egs E1 to E3 x 2 x1 2 x2 2 2 P k1 2 2 P k2 4 P 1 2 k 1 2 k 4 P k Let keq equivalent spring constant of the system keq P x k 4 Equation of motion of mass m m ẍ keq x 0 ωn sqrtkeq m sqrtk 4 m 214 For a displacement of x of mass m pulleys 1 2 and 3 undergo displacements of 2x 4x and 8x respectively The equation of motion of mass m can be written as m ẍ F0 0 1 where F0 2 F1 4 F2 8 F3 as shown in figure Since F3 8 x k Eq 1 can be rewritten as m ẍ 8 F3 8 8 k 0 2 from which we can find ωn sqrt64 k m 8 sqrtk m 3 UNIVERSIDADE FEDERAL DE OURO PRETO ESCOLA DE MINAS DEMEC Curso de Engenharia Mecânica Disciplina MEC212 Vibrações de Sist Mecânicos Prof Gustavo Paulinelli Guimarães Trabalho Global Etapa 03 Nos experimentos realizados em laboratório um sistema com 2 graus de liberdade foi considerado composto por duas massas apoiadas em 4 vigas cada uma Na primeira massa foram instalados um pequeno motor com uma pequena massa desbalanceadora e um acelerômetro Na segunda massa foi instalado um acelerômetro Foram realizadas medições de aceleração em duas condições diferentes 1 Vibração livre com o motor desligado foram aplicadas condições iniciais fazendo o sistema vibrar livremente 2 Vibração forçada com o motor ligado em uma determinada rotação fixa o sistema foi colocado a vibrar de maneira forçada A medição de vibração forçada considerou uma rotação do motor para cada grupo Cada grupo realizou seu levantamento de informações bem como fotos e vídeos Os dados medidos estão disponibilizados junto a este documento Passo 1 Com os dados da medição livre faça a importação no Scilab Estime as frequências naturais utilizando os dados medidos Dica utilizar o algoritmo de expansão por séries de Fourier para encontrar as frequências Esses valores servirão de referência para validação do modelo analítico Implemente o cálculo dos coeficientes bn da expansão por séries de Fourier e estime as formas modais Passo 2 Com os dados da medição em vibração forçada faça a importação no Scilab Estime as frequências da força e a força máxima F0 utilizando os dados medidos Dica utilizar o algoritmo de expansão por séries de Fourier para encontrar a frequência Pesquise no livro texto da disciplina como calcular F0 e desenvolva como é possível a aplicação para este caso Passo 3 Calcular os parâmetros de massa equivalente e rigidez equivalente sem utilizar resultados de medições de aceleração Calcular a força máxima F0 a partir do valor da massa desbalanceadora e de sua excentricidade Dica basearse nos procedimentos do TG 2 Desenhe o modelo equivalente com as massas e rigidezes equivalentes encontradas Faça o diagrama de corpo livre e encontre as matrizes de massa e rigidez Calcule as frequências naturais e as formas modais utilizando a teoria para sistemas de vários GDL Demonstrar todos os cálculos Passo 4 Discutir todos os resultados Fazer uma análise comparativa clara e objetiva entre os valores de frequências naturais e formas modais obtidos a partir dos dados experimentais e os valores obtidos com cálculos analíticos Passo 5 Descrever o experimento resultados experimentais e analíticos através de uma apresentação em Powerpoint