Virtual Labs

Torsional System as Two Degree of Freedom System

A torsional system consists of two discs mounted on a shaft, as shown in Fig. 1.

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Fig 1. Two DOF Torsional System

The shaft can be divided into three segments, having rotational spring constants as $k_{t1}$ , $k_{t2}$ , and $k_{t3}$ , as shown in Figure 1.

Mass moment of inertia of the discs are $J_1$ and $J_2$ , and $M_{t1}$ and $M_{t2}$ are the applied torque and the rotational degree of freedoms are $\theta_1$ and $\theta_2$ .

The differential equations of rotational motion for the discs can be written as,

$J₁θ̈₁ + (k_{t1} + k_{t2})θ₁ - k_{t2}θ₂ = M_{t1}$
$J₂θ̈₂ - k_{t2}θ₁ + (k_{t2} + k_{t3})θ₂ = M_{t2}$

Consider a harmonic oscillation where $\theta_i = \Theta_i \sin\omega t$ . The frequencies can be obtained by equating the determinant to zero,

$\det \begin{bmatrix} -J_1\omega^2 + (k_{t_1} + k_{t_2}) & -k_{t_2} \\ -k_{t_2} & -J_2\omega^2 + (k_{t_2} + k_{t_3}) \end{bmatrix} = 0$

And,

$J_1J_2\omega^4 - \left\{ (k_{t_1} + k_{t_2})J_2 + (k_{t_2} + k_{t_3})J_1 \right\} \omega^2 + \left\{ (k_{t_1} + k_{t_2})(k_{t_2} + k_{t_3}) - k_{t_2}^2 \right\} = 0$

Which is the characteristic equation. The roots of the equations are,

$\omega_2^2, \omega_1^2 = \frac{1}{2}\left\{ \frac{(k_{t_1} + k_{t_2})J_2 + (k_{t_2} + k_{t_3})J_1}{J_1J_2} \right\} \pm \frac{1}{2}\left[ \left\{ \frac{(k_{t_1} + k_{t_2})J_2 + (k_{t_2} + k_{t_3})J_1}{J_1J_2} \right\}^2 - 4\left\{ \frac{(k_{t_1} + k_{t_2})(k_{t_2} + k_{t_3}) - k_{t_2}^2}{J_1J_2} \right\} \right]^{\frac{1}{2}}$

where $\omega_1$ and $\omega_2$ are the natural frequencies of the system.