Torsional vibration of a two-gear shaft: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2017, Q7

Question 7: Torsional vibration of a two-gear 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 fixed at both ends, three equal segments of length $L$ (wall–gear1–gear2–wall); each carries a gear of inertia $J=0.0075$ kg·m$^2$. $d=48$ mm, $L=0.15$ m, $G=70$ GPa.

Find. EOM in $\theta_1,\theta_2$; the two natural frequencies and mode shapes.

Fixed–fixed shaft with two equal torsional springs to ground and one coupling spring; the two mode shapes: in-phase $(1,1)$ and out-of-phase $(1,-1)$.

Approach. Model each shaft segment as a torsional spring $k_t=GI_p/L$; write Newton’s rotational equation for each gear, assemble the mass/stiffness matrices, and solve the eigenproblem.

Segment stiffness. Polar second moment $I_p=\dfrac{\pi d^4}{32}=\dfrac{\pi(0.048)^4}{32}=5.21\times10^{-7}$ m$^4$, so $$k_t=\frac{GI_p}{L}=\frac{(70\times10^9)(5.21\times10^{-7})}{0.15}=\boxed{2.432\times10^{5}\ \text{N}\cdot\text{m/rad}}$$
Equations of motion. Coordinates $\theta_1,\theta_2$ = gear rotations. Gear 1 is tied to the wall ($k_t$) and to gear 2 ($k_t$); gear 2 to gear 1 ($k_t$) and to the wall ($k_t$): $$J\ddot\theta_1+2k_t\theta_1-k_t\theta_2=0,\qquad J\ddot\theta_2-k_t\theta_1+2k_t\theta_2=0$$ i.e. $J\mathbf I\ddot{\boldsymbol\theta}+k_t\!\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\!\boldsymbol\theta=\mathbf 0$.
Eigenvalues. With $\boldsymbol\theta=\boldsymbol\Phi e^{i\omega t}$, $\det(k_t[\,{}^{\,2}_{-1}\,{}^{-1}_{\,2}]-\omega^2 J\mathbf I)=0$ gives $\omega^2=\dfrac{k_t}{J}(2\mp1)$: $$\omega_1=\sqrt{\frac{k_t}{J}}=\sqrt{\frac{2.432\times10^5}{0.0075}}=\boxed{5695\ \text{rad/s}}\ (906\ \text{Hz})$$ $$\omega_2=\sqrt{\frac{3k_t}{J}}=\boxed{9863\ \text{rad/s}}\ (1570\ \text{Hz})$$
Mode shapes. Substituting back: for $\omega_1$, $(2-1)\theta_1-\theta_2=0\Rightarrow\theta_2=\theta_1$ — both gears swing together, $\boldsymbol\Phi_1=(1,1)^{\!\top}$. For $\omega_2$, $(2-3)\theta_1-\theta_2=0\Rightarrow\theta_2=-\theta_1$ — opposed, $\boldsymbol\Phi_2=(1,-1)^{\!\top}$ (a node at the shaft mid-point).

Mode	Natural frequency	Shape $(\theta_1,\theta_2)$
1 (in-phase)	5695 rad/s (906 Hz)	$(1,\ 1)$
2 (out-of-phase)	9863 rad/s (1570 Hz)	$(1,\ -1)$

Topic. Two-DOF torsional vibration of a fixed–fixed shaft with two rotor inertias.

Key relations. $k_t=GI_p/L$, $I_p=\pi d^4/32$; $\mathbf M\ddot{\boldsymbol\theta}+\mathbf K\boldsymbol\theta=0$ with $\mathbf K=k_t[\,{}^{\,2}_{-1}\,{}^{-1}_{\,2}]$; $\omega^2=(k_t/J)(2\mp1)$.

Why this works. Equal segments and equal inertias make the system symmetric, so the modes split cleanly into a symmetric $(1,1)$ and an antisymmetric $(1,-1)$ pair; the antisymmetric mode has a stationary node at mid-shaft, doubling the effective stiffness and raising its frequency by $\sqrt3$.

Common pitfall. Using $I_p=\pi d^4/64$ (that is the bending second moment) — torsion needs the polar value $\pi d^4/32$; and omitting the wall springs, which each gear has in addition to the coupling spring.

Source. Rao, Mechanical Vibrations 6th ed., Ch. 5 (two-DOF systems) and §9 (torsional systems); shaft torsion $k_t=GI_p/L$.

Question 7 of 7: Torsional vibration of a two-gear shaft

Question 7: Torsional vibration of a two-gear shaft (20 marks)