22-Mec-A2 Kinematics and Dynamics of Machines · December 2017
Question 6 of 7: Base-excited vibration of a mono-cycle model
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — December 2017 · 16-Mec-A2 Kinematics and Dynamics of Machines · 3 hours, open book · Answer five questions, at least one from Part B (Part A mechanisms & machine dynamics Q1–5, Part B mechanical vibration Q6–7). Marks: 20 each. All seven questions are solved here.
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (instant centres & Kennedy’s rule §6.3–6.4, cams Ch. 8, epicyclic trains §9.7–9.9, balancing Ch. 13); Uicker, Pennock & Shigley, Theory of Machines and Mechanisms, 5th ed.; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery, 3rd ed.; S. S. Rao, Mechanical Vibrations, 6th ed. (Part B: Ch. 3 base excitation, Ch. 6 & Ch. 9 multi-DOF / torsional).
Check (measured geometry). Questions 1 and 2 carry no printed dimensions — the exam instructs the candidate to scale every length off the drawings (1:6 for Q1, 1:10 for Q2). The joint coordinates used below were read from the printed figure of the figures; the resulting linear quantities are therefore accurate to roughly ±3–5 %. Scale-free results (time ratio, mobility, angular-velocity ratios, mode shapes) are unaffected.
Question 6: Base-excited vibration of a mono-cycle model (20 marks)
Given. Single-DOF base excitation: $m=100$ kg, $k=10000$ N/m, $c=100$ N·s/m; road amplitude $Y=b=0.025$ m, wavelength $a=2.75$ m; forcing frequency $\omega_f=2\pi v/a$ set to $0.95,1.0,1.15$ times $\omega_n$.
Find. Steady-state mass amplitude $X$ at each frequency ratio.
Base-excited $m$–$c$–$k$ model rolling on a sinusoidal road, and its displacement transmissibility $X/Y$ vs frequency ratio $r$; the three requested ratios are marked (peak near $r=1$).
Approach. The road drives the base through the spring and damper, so this is displacement transmissibility: compute $\omega_n$ and $\zeta$, form the frequency ratio $r$, and evaluate the standard base-excitation amplitude ratio.
System parameters.
$$\omega_n=\sqrt{k/m}=\sqrt{10000/100}=10\ \text{rad/s},\qquad \zeta=\frac{c}{2\sqrt{km}}=\frac{100}{2\sqrt{10^6}}=0.05$$
Transmissibility. For base motion $y_b=Y\sin\omega_f t$ the steady-state amplitude is
$$\frac{X}{Y}=\sqrt{\frac{1+(2\zeta r)^2}{(1-r^2)^2+(2\zeta r)^2}},\qquad r=\frac{\omega_f}{\omega_n}$$
$r=0.95$. $X/Y=7.38\Rightarrow \boxed{X=0.025(7.38)=0.185\ \text{m}}$ (185 mm). Here $v=r\,\omega_n a/(2\pi)=4.16$ m/s.
$r=1.0$ (resonance). $X/Y=10.05\Rightarrow \boxed{X=0.251\ \text{m}}$ (251 mm) — the largest, limited only by damping. $v=4.38$ m/s.
$r=1.15$. $X/Y=2.94\Rightarrow \boxed{X=0.0735\ \text{m}}$ (73.5 mm); past resonance the response falls rapidly. $v=5.03$ m/s.
