Question

In the figure, the suspension system is composed of a damper and spring having the damping and stiffness coefficients of \(\mathrm{C}=60[\mathrm{Ns} / \mathrm{m}]\) and \(\mathrm{K}=7200[\mathrm{~N} / \mathrm{m}]\)

respectively. The test rig shown in the figure is used to obtain the oscillation characteristics of the suspension system in the \(y\) -direction. The input of the system is given as the road profile as a sine function and it is given by velocity \(V\) in \(x\) -direction. The tire attached at point \(O\) has stiffness of \(K_{7}=10000[\mathrm{~N} / \mathrm{m}]\) and the lumped mass of the whole system at point \(O\) is \(2[\mathrm{~kg}]\)

(a) As a velocity \(\mathrm{V}=6 \mathrm{~m} / \mathrm{s}\) is given as road profile, find the force applied to the suspension system at point \(\mathrm{O}\). (10pts)

(b) Find the steady state oscillations of the system due to the acting force that is computed at part (a). (10pts)

(c) Find the velocity \(V\) which will make the system oscillate at resonance frequency. (5pts)

Hint: \(T \omega=2 \pi\)

Answer 1

Velocity \(=\) frequency \(\times\) wavelength

$$ v=\frac{\omega}{2 \pi} \times \lambda $$

from figure \(\lambda=120 \mathrm{~mm}\)

For velocity of \(6 \mathrm{~m} / \mathrm{s}\)

$$ \begin{array}{c} 6=\frac{\omega}{2 \pi} \times 0.12 \quad=>\omega=314.16 \mathrm{rad} / \mathrm{sec} \\ \omega_{\mathrm{n}}=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}=\sqrt{\frac{10000}{2}}=70.7106 \mathrm{rad} / \mathrm{sec} \\ \mathrm{C}_{\mathrm{c}}=2 \mathrm{~m} \omega_{\mathrm{n}}=2 \times 2 \times 70.7106=282.84 \mathrm{Ns} / \mathrm{m} \\ \xi=\frac{\mathrm{C}}{\mathrm{C}_{\mathrm{c}}}=\frac{60}{282.84}=0.21213 \\ \frac{\mathrm{X}}{\mathrm{Y}}=\frac{1+\left(2 \xi \frac{\omega}{\omega_{\mathrm{n}}}\right)^{2}}{\sqrt{\left(1-\left(\frac{\omega}{\omega_{\mathrm{n}}}\right)^{2}\right)^{2}+\left(2 \xi \frac{\omega}{\omega_{\mathrm{n}}}\right)^{2}}} \end{array} $$

from figure \(\quad Y=12 \mathrm{~mm}=0.012 \mathrm{~m}\)

By substituting all values to the above equation, we get

$$ X=0.00136 \mathrm{~m} $$

a)

Force \(\mathrm{F}=\mathrm{X} \sqrt{\mathrm{K}^{2}+\mathrm{C}^{2} \omega^{2}}\)

By substituting all values to the above equation, we get

$$ \mathbf{F}=29.0513 \mathrm{~N} $$

b)

Steady state amplitude of vibration \(\mathbf{X}=\mathbf{1 . 3 6} \mathbf{~ m m}\)

c)

Velocity \(\mathrm{V}\) which will make the system oscillate at resonance frequency.

$$ \begin{array}{c} \mathrm{v}=\frac{\omega_{\mathrm{n}}}{2 \pi} \times \lambda \\ \mathrm{v}=\frac{70.7106}{2 \pi} \times 0.12 \\ \mathrm{v}=1.3504 \mathrm{~m} / \mathrm{s} \end{array} $$