22-Mec-A2 Kinematics and Dynamics of Machines · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (mobility Ch. 2, cams Ch. 8, epicyclic/compound trains §9.6–9.9, 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 multi-DOF torsional).
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. Shaft diameter $d=48$ mm; four equal segments $L=100$ mm between the fixed ends and the three equal disks; $G=70$ GPa; $J_1=J_2=J_3=0.01\ \text{kg}\cdot\text{m}^2$. Coordinates: absolute twist angles $\theta_1,\theta_2,\theta_3$ of the three disks.
Find. the equations of motion, and the natural frequencies with mode shapes.
$$J_p=\frac{\pi d^4}{32}=\frac{\pi(0.048)^4}{32}=5.211\times10^{-7}\ \text{m}^4,\qquad k_t=\frac{GJ_p}{L}=\frac{70\times10^{9}(5.211\times10^{-7})}{0.10}=3.648\times10^{5}\ \text{N}\cdot\text{m/rad}.$$
All four segments share this $k_t$ (equal length), and all three disks share $J=0.01\ \text{kg}\cdot\text{m}^2$.
Each disk is linked to its neighbours (or a wall) by a segment of stiffness $k_t$. Newton/Lagrange on the three twist angles gives $\mathbf{J}\ddot{\boldsymbol\theta}+\mathbf{K}\boldsymbol\theta=\mathbf 0$ with
$$\mathbf J=J\,\mathbf I_3,\qquad \mathbf K=k_t\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}.$$
(The diagonal $2k_t$ on disks 1 and 3 comes from a wall spring plus a neighbour spring; the off-diagonals are the shared inter-disk springs.)
For the uniform fixed–fixed chain the eigenvalues of the tridiagonal $[2,-1]$ matrix are $\lambda_j=2-2\cos\!\dfrac{j\pi}{4}$, so
$$\omega_j=\sqrt{\frac{k_t}{J}\Big(2-2\cos\frac{j\pi}{4}\Big)},\qquad \phi_i^{(j)}=\sin\frac{ij\pi}{4},\quad i,j=1,2,3,$$
with $k_t/J=3.648\times10^{7}\ \text{s}^{-2}$. Evaluating:
| Mode $j$ | $\lambda_j=2-2\cos(j\pi/4)$ | $\omega_j$ (rad/s) | $f_j$ (Hz) | Mode shape $(\theta_1,\theta_2,\theta_3)$ |
|---|---|---|---|---|
| 1 | $2-\sqrt2=0.586$ | 4623 | 735.7 | $(1,\ \sqrt2,\ 1)$ — all in phase |
| 2 | $2$ | 8542 | 1359.5 | $(1,\ 0,\ -1)$ — centre node still |
| 3 | $2+\sqrt2=3.414$ | 11160 | 1776.2 | $(1,\ -\sqrt2,\ 1)$ — centre opposed |
Taking the fundamental as the requested single answer:
$$\boxed{\ \omega_1=4.62\times10^{3}\ \text{rad/s}\ (735.7\ \text{Hz}),\quad \boldsymbol\phi_1=(1,\ \sqrt2,\ 1)\ }$$
— the symmetric mode in which all three disks twist the same way, the middle disk with the largest amplitude.
| Quantity | Value |
|---|---|
| Segment torsional stiffness $k_t$ | $3.648\times10^{5}\ \text{N}\cdot\text{m/rad}$ |
| $\omega_1$ / $f_1$ (fundamental) | 4623 rad/s / 735.7 Hz, shape $(1,\sqrt2,1)$ |
| $\omega_2$ / $f_2$ | 8542 rad/s / 1359.5 Hz, shape $(1,0,-1)$ |
| $\omega_3$ / $f_3$ | 11160 rad/s / 1776.2 Hz, shape $(1,-\sqrt2,1)$ |
