Question 5 of 6: Base-excited quarter-car — steady-state amplitudes

Question 5: Base-excited quarter-car — steady-state amplitudes (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

The wheel follows the road, so the base displacement is $y(t)=b\sin\omega t$ with forcing frequency $\omega=2\pi v/a$.

Find. the absolute steady-state amplitude $X$ of the mass at $\omega=0.8\omega_n,\ 1.0\omega_n,\ 1.1\omega_n$.

Model and transmissibility

The equation of motion for base excitation is $m\ddot x+c(\dot x-\dot y)+k(x-y)=0$, i.e. $m\ddot x+c\dot x+kx=c\dot y+ky$. For $y=b\sin\omega t$ the steady-state absolute amplitude is $X=b\,T_d$, with the displacement transmissibility

$$T_d=\frac{X}{b}=\sqrt{\frac{1+(2\zeta r)^2}{(1-r^2)^2+(2\zeta r)^2}},\qquad r=\frac{\omega}{\omega_n}.$$

System constants:

$$\omega_n=\sqrt{k/m}=\sqrt{10000/100}=10\ \text{rad/s},\qquad \zeta=\frac{c}{2\sqrt{km}}=\frac{100}{2\sqrt{10^{6}}}=0.05.$$

The road speed for each case, $v=\dfrac{\omega a}{2\pi}=\dfrac{r\,\omega_n a}{2\pi}$, is listed for reference.

Amplitudes at the three speeds

Sample check at $r=1$: $T_d=\sqrt{\dfrac{1+(2\cdot0.05\cdot1)^2}{0+(0.1)^2}}=\sqrt{\dfrac{1.01}{0.01}}=10.05$, so $X=0.05(10.05)=\boxed{0.503\ \text{m}}$ — a tenfold amplification of the 50 mm road bump. Just above resonance ($r=1.1$) the amplitude has already dropped to 0.21 m, and by $r=0.8$ to 0.14 m; a light-damped bounce system is worst exactly at the resonant road speed.