Question

1) (35 points) A model of a vehicle suspension system is shown below. The of a 500 kg vehicle is connected to the wheels through a suspension system that is modeled as a spring in parallel with a viscous damper. The wheels are assumed to be rigid and follow the road contour which is also shown below. If the vehicle travels at a constant speed of 52 m/s, what is the acceleration amplitude of the vehicle? m = 500 kg m ZE k = 4 x 10 Uc = 3000 N-s v = 52. m 2.5 m 0.01 m y(5) = 0.01 sin 2πξ 2.5

Answer 1

Given, m=jookg. k= 4x1050 c = 3000 N/m V = 52 m/s 4X105 Natural frequency (wn) - = V 800 = 28.3 rad srce 3000 Damping ratio (E)= 2min 26500) (28.3) -0.106 2.5 Ginen, mathematical description of road contour is y C6 ) = o-olsin 27 = 0.01sin (0-8X) 9(&) - 0.0i sin(0-8xv+) [:: €=vt] wheel follows road contour, a harmonie base displacement Since, the word it acts as w = 0.800

0.xf52) rad/see - 130.7 130.7 ratio =r=w : frequency 28.3 wn 4.62 The steady-state amplitude of absolute displacement is given by X= T (2, 4) 1 +(284) Vc1-x2)+(2&M) У AXT 0:49. >> X = YT (4•62,0-106) - (0.01) 1+(260-106)(4-62)) V[1-64-6236)++ [2(0-106) 4-67 = 6-87x10 m The wehicle acceleration amplitude is А = 3-х 7 A = (130. 7) * (6.87*104) A 11.7 m/s2