Question

4.46 Consider the model of a vehicle given in Problem 4.15 illustrated in Figure P4.15 defined by 2000 0 0 50 1000 -1000 -1000 11,000 Suppose that the tire rolls over a bump modeled as the initial conditions of x(0) [0 0.01 and x(0) 0. Use modal analysis to calculate the response of the car xi(t). Plot the response for three cycles

Answer 1

A slightly more sophisisticated model of a vehicle suspension system

The natural frequencies for k1 = 10^3 N/m, k2 = 10^4N/m, m2 = 50kg, and m1 = 2000kg

Calculate the mass-normalized stiffness matrix: K MKM-2 Here, M is the mass matrix and Kis the stiffness 2000 0 0 50 1000 -1000 -1000 11000 Substitute for M and for K 2000 0 1000 1000 2000 (0 -1000 1000050 0 50 0.0224 0 1000 -1000 0.02240 0.1414-1000 1000 0 0.1414 0.0224 0 22.4-141.4 0.141422. 1555.4 0.502 - 3.167 3.167 219.94

Calculate the eigenvalues: Here, 2 is the eigenvalue and 1 is the identity matrix. .502-3.167 3.167 219.94 Substitute forI 219.94for K ana 0.502-3.167 3.167219.9401 0.502-λ -3.167 21 9.94- (0.502-1)(219.94-2)-3.1672-0 3.167 λ-0.456, 220

Calculate the natural frequencies: 0.456 0.675 rad/s

a, = =./220 14.832 rad/s

For λ-0.456 Calculate eigenvectors Here, v, is the vector 0.502 -3.167 Substitute for 1, 0.456 for λ and I 'll l for 219.94for K ana 12 0.502 -3.167 219.94 0.4560 3.167 Vi 0.046- 3.167219.484vi2 0.046Vi-3.167Vi2- 3.167vii +219.484Vi, 68.85v 0 12 Therefore, eigenvector is v, -68.85

Calculate normalized eigenvectors 12 (1+68.8s)-1 12-0.0144 Substituteequation (1), 0.9999 0.0144

For λ= 220 Calculate eigenvectors Here, v is the vector 0.502 -3.167 -3.167 219.94 Substitutee for 1 , 220 for λ and | | for 0.502 -3.167 3.167 219.94 219.5 -3.167 2 3.167 -0.06 V22 210 -219.5v2l-3.167v220 3.167v2 -0.06V22 ½,--0.01 44m Therefore, eigenvector is -0.0144 V2 -V221

Calculate normalized eigenvectors 2122 (1 +0.01442)ย์" 22 -0.9999 Substitute 0.9999 for V22 in equation (2), -0.0144 v,-| 0.9999

Calculate the matrixP 0.9999-0.0144 0.0144 0.9999

Calculate the matrix of mode shapes: S = M-1/2 P 0.02240 0.9999-0.0144 0.0144 0.9999 0.1414or M-2and 0.9999-0.0144 0.14140.0144 0.9999 Substitute 「0.0224 0.022 0.0003 0.002 0.141

Calculate s- 0.9999 0.0144 Substitute for pT and -0.01440.9999 .9999or p and 44.72 7.07 orM2 -1 「0.9999 0.0 144T4472 -0.0144 0.9999 0 7.07 44.7160.102 0.644 7.069

Calculate the modal initial conditions r(o)-S x. 44.7160.102 0.644 7.069 44.716 0.102 0 -0.644 7.069 0.01 Substitute for x s and 0.01 r(0)- 0.001 0.0707 Similarly r(o)-S 44.7160.102 0.644 7.069 44.716 0.1021[0 Substitute 0r Sanfor r(0)- -0.644 7.0690

Calculate the modal solutions: ro-GZZsin(-..tan씬 10. 악 10ノ Substitute 0.675 rad/s for a,, 0.001 for o and 0 for o 10 2) r (t)-0.001 sin! 0.675+ 0.001 cos(0.6751)

5(,).m sin(w.tm" a) 20 Substitute 14.832 rad/s for o,, 0.0707 for o and ofor r (t 0.0707 sin 14.832t- 0.0707 cos(14.832/) 0.0707 cos(14.832)

Calculate the solution for the response: 0.022 -0.0003 0.002 0.141 0.00 l cos(0.6751) |forrώ 0.0707 cos(14.8321) Substitute for sand 0.0220.000370.001 cos (0.6751) 0.002 0.1410.0707 cos(14.8321) 0.000022 cos(0.675t)-0.000021 cos (14.8321) 0.000002 cos(0.6751)+0.01 cos (14.832) Therefore, the response000022cos(0.6751)-0.000021cos(14.832/) Therefore, the response is x- 0.000002 cos(0.6751)+0.01cos (14.832/)

Plot of x, (t)-0.000022 cos (0.6751)-0.000021cos(14.8321) mm with

Plot of x2(1)0.000002 cos(0.675t)+0.01 cos (14.8321) mm with