Torsional Vibration of a 3-Rotor Shaft: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2013, Q6

Question 6: Torsional Vibration of a 3-Rotor Shaft (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. Shaft diameter $d=48\ \text{mm}$, four equal spans $L=100\ \text{mm}$, shear modulus $G=70\ \text{GPa}$; three equal rotors $J=0.01\ \text{kg}\cdot ext{m}^2$. The shaft is fixed at both ends (walls in the figure), with the rotors between the four spans.

Find. The equations of motion, and the natural frequencies and mode shapes (one is required; all three are given).

Fig. 6 — fixed-fixed shaft carrying three rotors J1–J3 across four equal torsional springs (top), and the three torsional mode shapes with their natural frequencies (below).

Approach. Model each shaft span as a torsional spring $k_t=GJ_p/L$; take the three rotor rotations $\theta_1,\theta_2,\theta_3$ as coordinates; assemble $\mathbf M\ddot{\boldsymbol\theta}+\mathbf K\boldsymbol\theta=0$ and solve the eigenproblem.

Torsional stiffness of one span. Polar second moment $J_p=\dfrac{\pi d^4}{32}=\dfrac{\pi(0.048)^4}{32}=5.21\times10^{-7}\ \text{m}^4$, so $$k_t=\frac{GJ_p}{L}=\frac{70\times10^{9}(5.21\times10^{-7})}{0.1}=3.65\times10^{5}\ \text{N}\cdot ext{m/rad}.$$ All four spans are equal, $k_1=k_2=k_3=k_4=k_t\equiv k$.
Equations of motion (fixed ends, so springs 1 and 4 tie rotors 1 and 3 to ground): $$J\ddot\theta_1+(k_1+k_2)\theta_1-k_2\theta_2=0,$$ $$J\ddot\theta_2-k_2\theta_1+(k_2+k_3)\theta_2-k_3\theta_3=0,$$ $$J\ddot\theta_3-k_3\theta_2+(k_3+k_4)\theta_3=0.$$
Matrix form. With all $k$ and $J$ equal, $$\mathbf M=J\mathbf I,\qquad \mathbf K=k\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}.$$
Eigenvalues. Let $\lambda=\omega^2 J/k$. The tridiagonal matrix has $\lambda_j=2-2\cos\!\dfrac{j\pi}{4}$, $j=1,2,3$: $\lambda=0.586,\ 2,\ 3.414$. With $\sqrt{k/J}=6040\ \text{rad/s}$: $$\omega_1=\boxed{4623\ \text{rad/s}\ (736\ \text{Hz})},\quad \omega_2=8542,\quad \omega_3=11160\ \text{rad/s}.$$
Mode shapes. $\Phi_i^{(j)}=\sin\dfrac{ij\pi}{4}$ gives $$\boldsymbol\Phi_1=(1,\ \sqrt2,\ 1),\quad \boldsymbol\Phi_2=(1,\ 0,\ -1),\quad \boldsymbol\Phi_3=(1,\ -\sqrt2,\ 1).$$ Mode 1: all rotors in phase (fundamental); Mode 2: centre rotor stationary, ends opposed; Mode 3: adjacent rotors out of phase.

Mode	Natural frequency	Mode shape (θ₁,θ₂,θ₃)
1	4623 rad/s (736 Hz)	(1, 1.41, 1)
2	8542 rad/s (1360 Hz)	(1, 0, −1)
3	11160 rad/s (1776 Hz)	(1, −1.41, 1)

Topic. Multi-DOF torsional vibration — stiffness assembly and eigenanalysis.

Key relations. Torsional spring $k_t=GJ_p/L$, $J_p=\pi d^4/32$; eigenproblem $(\mathbf K-\omega^2\mathbf M)\boldsymbol\Phi=0$; for a uniform fixed-fixed chain $\omega_j^2=\dfrac{k}{J}\Big(2-2\cos\dfrac{j\pi}{N+1}\Big)$, $\Phi_i^{(j)}=\sin\dfrac{ij\pi}{N+1}$.

Why this works. Each rotor stores kinetic energy ($\tfrac12 J\dot\theta^2$) and each span stores torsional strain energy ($\tfrac12 k_t\Delta\theta^2$); Lagrange’s equations then give the symmetric $\mathbf M,\mathbf K$. Because the ends are fixed, springs 1 and 4 appear as ground stiffness on rotors 1 and 3, producing the classic $[2,-1,0;-1,2,-1;0,-1,2]$ pattern whose eigenpairs are known in closed form.

Common pitfall. Omitting the ground springs (treating it as free-free, which would give a rigid-body zero mode) — the figure clearly shows both ends built in. Also using $J_p=\pi d^4/64$ (that is the bending second moment $I$, half the polar value).

Source. Rao, Mechanical Vibrations, §6 (multidegree-of-freedom systems; torsional systems, eigenvalue problem).

Question 6 of 6: Torsional Vibration of a 3-Rotor Shaft

Question 6: Torsional Vibration of a 3-Rotor Shaft (20 marks)