Steady-state response to a wavy road (base excitation): 22-Mec-A2 Kinematics and Dynamics of Machines, May 2016, Q5

Question 5: Steady-state response to a wavy road (base excitation) (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. Single-DOF base-excited system: $m=200$ kg on a parallel spring $k=16000$ N/m and damper $c=3220$ N·s/m; road profile $y_b=b\sin(2\pi x_b/a)$ with $b=0.02$ m, spatial period $a=1$ m; forward speed $v=1.35$ m/s.

Find. The steady-state absolute vertical motion $y(t)$ of the mass (amplitude and phase).

Figure 5.1 — Quarter-vehicle model. The wheel follows the sinusoidal road, imposing base displacement $y_b$ through the spring–damper onto the sprung mass $m$.

Approach. Riding at speed $v$ turns the spatial road wave into a harmonic base motion of frequency $\omega=2\pi v/a$. This is classic support excitation; the absolute-motion amplitude follows from the displacement-transmissibility formula.

Excitation frequency and system parameters. $$\omega=\frac{2\pi v}{a}=\frac{2\pi(1.35)}{1}=8.482\ \text{rad/s},\quad \omega_n=\sqrt{k/m}=\sqrt{80}=8.944\ \text{rad/s},$$ $$\zeta=\frac{c}{2\sqrt{km}}=\frac{3220}{2\sqrt{16000\cdot200}}=0.900,\quad r=\frac{\omega}{\omega_n}=0.948.$$
Governing equation and effective forcing. With $y$ the absolute displacement, $$m\ddot y+c\dot y+ky=c\dot y_b+ky_b=b\sqrt{k^2+(c\omega)^2}\,\sin(\omega t+\beta),\ \ \beta=\tan^{-1}\!\frac{c\omega}{k}.$$
Steady-state amplitude (transmissibility). $$\frac{Y}{b}=\sqrt{\frac{1+(2\zeta r)^2}{(1-r^2)^2+(2\zeta r)^2}}=1.157\ \Rightarrow\ \boxed{Y=1.157(0.02)=0.0231\ \text{m}=23.1\ \text{mm}}.$$
Phase. The mass lags the road by $\phi-\beta$ with $\phi=\tan^{-1}\!\dfrac{c\omega}{k-m\omega^2}=86.6^\circ$ and $\beta=59.7^\circ$, i.e. a net lag of $27.0^\circ$: $$y(t)=23.1\,\sin(8.482\,t-27.0^\circ)\ \text{mm}.$$

Quantity	Value
Excitation frequency $\omega$	8.482 rad/s
$\omega_n$, $\zeta$, $r$	8.944 rad/s, 0.90, 0.948
Steady-state amplitude $Y$	23.1 mm
Phase (lag behind road)	27.0°
Response	$y=23.1\sin(8.482t-27.0^\circ)$ mm

Topic: Harmonic base (support) excitation; displacement transmissibility.

Key relations: $\omega=2\pi v/a$; $|Y/b|=\sqrt{[1+(2\zeta r)^2]/[(1-r^2)^2+(2\zeta r)^2]}$; net phase $\phi-\beta$.

Why this works: Both the spring and the damper couple the base to the mass, so the effective forcing is the vector sum of a spring term ($ky_b$) and a damper term ($c\dot y_b$) — hence the $\sqrt{1+(2\zeta r)^2}$ numerator. Operating just below resonance ($r=0.95$) but with very heavy damping ($\zeta=0.9$) keeps amplification modest ($Y/b\approx1.16$) instead of the large peak a lightly-damped system would show.

Common pitfall: Using the force-transmissibility or the rotating-unbalance formula (wrong numerator), or dropping the damper term in the forcing. Converting speed to frequency wrongly ($\omega=2\pi v/a$, not $v/a$).

Source: Rao, Mechanical Vibrations §3.6 (base excitation, displacement transmissibility).

Question 5 of 6: Steady-state response to a wavy road (base excitation)

Question 5: Steady-state response to a wavy road (base excitation) (20 marks)