Question 7 of 7: Lateral vibration of a two-mass shaft (Part B)

Question 7: Lateral vibration of a two-mass shaft (Part B) (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. Clamped–clamped shaft of total length $3L$ with equal masses $m$ at the third-points $x=L$ and $x=2L$; $d=25\ \text{mm}$, $L=0.25\ \text{m}$, $E=210\ \text{GPa}$. Second moment $I=\pi d^4/64=1.917\times10^{-8}\ \text{m}^4$, so $EI=210\times10^{9}\cdot I=4.03\times10^{3}\ \text{N}\cdot\text{m}^2$.

Find. (i) the equations of motion; (ii) $\omega_1,\omega_2$ and their mode shapes.

Approach. Take the two lateral deflections $x_1,x_2$ at the masses as coordinates. Build the flexibility (influence-coefficient) matrix of the clamped–clamped beam at those two points, invert to get stiffness, then solve the standard eigenproblem $(\mathbf K-\omega^2\mathbf M)\boldsymbol\phi=\mathbf 0$ with $\mathbf M=m\,\mathbf I$.

1. Coordinates and equations of motion. With $x_1,x_2$ the transverse displacements at the two masses, $$m\ddot x_1+k_{11}x_1+k_{12}x_2=0,\qquad m\ddot x_2+k_{21}x_1+k_{22}x_2=0,$$ i.e. $\mathbf M\ddot{\mathbf x}+\mathbf K\mathbf x=\mathbf 0$ with $\mathbf M=m\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $\mathbf K=\mathbf a^{-1}$.
2. Flexibility coefficients (clamped–clamped, load points at $L$ and $2L$). The direct term is $a_{11}=a^3b^3/(3\,\ell^3 EI)$ with $\ell=3L,\ a=L,\ b=2L$; by symmetry $a_{22}=a_{11}$. Evaluating (and the cross term from the beam Green’s function): $$a_{11}=a_{22}=3.83\times10^{-7}\ \text{m/N},\qquad a_{12}=a_{21}=2.64\times10^{-7}\ \text{m/N}.$$
3. Eigenvalues. For $\mathbf M=m\mathbf I$ the modal equation $\det(\mathbf a\,m\,\omega^2-\mathbf I)=0$ gives $\omega^2=1/(m\lambda)$ with $\lambda$ the eigenvalues of $\mathbf a$. By symmetry the eigenvectors are the symmetric $(1,1)$ and antisymmetric $(1,-1)$ combinations: $$\lambda_1=a_{11}+a_{12}=6.47\times10^{-7}\ \ (\text{mode }(1,1)),\qquad \lambda_2=a_{11}-a_{12}=1.20\times10^{-7}\ \ (\text{mode }(1,-1)).$$
4. Natural frequencies and mode shapes. $$\boxed{\omega_1=\frac{1}{\sqrt{m\,\lambda_1}}=\frac{1243}{\sqrt{m}}\ \text{rad/s}},\quad \boldsymbol\phi_1=\begin{Bmatrix}1\\1\end{Bmatrix}\ \text{(symmetric)};$$ $$\boxed{\omega_2=\frac{1}{\sqrt{m\,\lambda_2}}=\frac{2890}{\sqrt{m}}\ \text{rad/s}},\quad \boldsymbol\phi_2=\begin{Bmatrix}1\\-1\end{Bmatrix}\ \text{(antisymmetric)}.$$ The lower mode has both gears deflecting the same way (softer, larger flexibility); the higher mode has them opposed, with a node at mid-span.