Two-rotor torsional system frequency

Model

  [ A ]=========shaft (l, GJ)=========[ B ]
   I_A         k_t = GJ/l             I_B
   theta_A                            theta_B

Let $\theta_A$ and $\theta_B$ be the angular displacements of the rotors, both measured in the same sense. (In the vibrating mode they turn in opposite directions, so the result will show $\theta_A$ and $\theta_B$ with opposite signs.)

Equations of motion

The shaft torque is $k_t(\theta_A - \theta_B)$ (relative twist):

$$I_A\ddot\theta_A + k_t(\theta_A - \theta_B) = 0, \qquad I_B\ddot\theta_B - k_t(\theta_A - \theta_B) = 0$$

Take harmonic motion $\theta = \Theta\sin\omega t$:

$$(k_t - I_A\omega^2)\Theta_A - k_t\Theta_B = 0, \qquad -k_t\Theta_A + (k_t - I_B\omega^2)\Theta_B = 0$$

For a non-trivial solution, the determinant is zero:

$$(k_t - I_A\omega^2)(k_t - I_B\omega^2) - k_t^2 = 0 \;\Rightarrow\; \omega^2\left[I_AI_B\omega^2 - k_t(I_A+I_B)\right] = 0$$

The root $\omega = 0$ is rigid-body rotation. The other root is the natural frequency:

$$\omega_n = \sqrt{\frac{k_t(I_A+I_B)}{I_AI_B}} = \sqrt{\frac{GJ}{l}\cdot\frac{I_A+I_B}{I_AI_B}}, \qquad f_n = \frac{\omega_n}{2\pi}$$

Position of the node

At the node the shaft does not twist (its angular displacement is zero). From the first equation, $\dfrac{\Theta_B}{\Theta_A} = \dfrac{k_t - I_A\omega_n^2}{k_t} = -\dfrac{I_A}{I_B}$, so the amplitudes are inversely proportional to the inertias and are of opposite sign.

Let $l_A$ and $l_B$ be the distances of the node from A and B ($l_A + l_B = l$). The twist per unit length is uniform, so $|\Theta_A|/l_A = |\Theta_B|/l_B$:

$$l_A = \frac{I_B}{I_A+I_B}\,l, \qquad l_B = \frac{I_A}{I_A+I_B}\,l$$

The node is nearer the rotor with the larger inertia. Each rotor vibrates as if fixed to a shaft of length $l_A$ or $l_B$, with $\omega_n = \sqrt{\dfrac{GJ}{l_A I_A}} = \sqrt{\dfrac{GJ}{l_B I_B}}$.