Question

A solid shaft AB, of 100 mm diameter, is rigidly connected at itsends to two hollow shafts ACand ED (both of external diameter 138 mm and internal diameter 106 mm, At Cand Dtwo masses of 130 kg and 250 kg, and of radius of gyration 175 mm and 375 mm respectively, are attached to the ends of the hollow shafts. Determine from first principles the frequency of free torsional vibrations and the position of the node 600mn 400mm -5m

Answer 1

Please refer the diagram below:

We assume steel shaft

G=80x10⁹ GPa

In the diagram above, k₂ is torsional stiffness of shaft AB, k₁ is that of shaft AC and k₃ is that of shaft BD.

$k_2=\frac{GJ_2}{l_2}=\frac{80\times10^9(\pi D^4/32)}{1.5}=\frac{80\times10^9(\pi 0.1^4/32)}{1.5}=0.5236\times10^6\: Nm/rad$

$k_1=\frac{GJ_1}{l_1}=\frac{80\times10^9[\pi (D_o^4-D_i^4)/32]}{0.6}=\frac{80\times10^9[\pi (0.138^4-0.106^4)/32]}{1.5}=3.095\times10^6\: Nm/rad$

$k_3=\frac{GJ_3}{l_3}=\frac{80\times10^9[\pi (D_o^4-D_i^4)/32]}{0.4}=\frac{80\times10^9[\pi (0.138^4-0.106^4)/32]}{0.4}=4.642\times10^6\: Nm/rad$

The equivalent spring stiffness is given by

$\frac{1}{k_e}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}=\frac{1}{0.5236\times10^6}+\frac{1}{3.095\times10^6}+\frac{1}{4.642\times10^6}=0.4084\times10^6\: Nm/rad$

Moments of Inertia

The equivalent system is shown below:

Refer free body diagram below:

ke(-) 好レ/

The equations of motion are

$I_1\ddot \theta_1=-k_e(\theta_1-\theta_1)$

$I_2\ddot \theta_2=k_e(\theta_1-\theta_1)$

In Matrix form

$\begin{bmatrix} I_1 &0 \\ 0 & I_2 \end{bmatrix}\begin{Bmatrix} \ddot\theta_1\\ \ddot\theta_2 \end{Bmatrix}+\begin{bmatrix} k_e &-k_e \\ -k_e & k_e \end{bmatrix}\begin{Bmatrix} \theta_1\\ \theta_2 \end{Bmatrix}=0$

We assume solution in the form

$\theta_1(t)=\Theta _1sin(\omega t+\phi)$

$\theta_2(t)=\Theta _2sin(\omega t+\phi)$

Substituting in the matrix equation above

$\begin{bmatrix} -I_1\omega^2+k_e &-k_e \\ -k_e & -I_2\omega^2+k_e \end{bmatrix}\begin{bmatrix} \Theta_1\\ \Theta_2 \end{bmatrix}=0$

For non trivial solution

$det\begin{bmatrix} -I_1\omega^2+k_e &-k_e \\ -k_e & -I_2\omega^2+k_e \end{bmatrix}=0$

Substituting values

$det\begin{bmatrix} -4.1344\omega^2+0.4084\times10^6 &-0.4084\times10^6 \\ -0.4084\times10^6 & -35.156\omega^2+0.4084\times10^6 \end{bmatrix}=0$

The characteristics equation is

$(-4.1344\omega^2+0.4084\times10^6)(-35.156\omega^2+0.4084\times10^6)-(0.4084\times10^6)^2=0$

Solving the above the lowest value of $\omega$ is 0

The other value of $\omega$

$\omega=332.26\: rad/s$

Mode shapes

For $\omega$=0

From Matrix equation

$k_e\Theta_1-k_e\Theta_2=0$

$\Theta_1=\Theta_2$

Mode shape is

$\phi_1=\begin{Bmatrix} 1\\ 1 \end{Bmatrix}$

For $\omega$=332.26 rad/s

$[-4.1344(332.26^2)+0.4084\times10^6]\Theta_1-0.4084\times10^6\Theta_2=0$

$\Theta_2=-0.1176\Theta_1$

Or

$\phi_2=\begin{Bmatrix} 1\\ -0.1176 \end{Bmatrix}$