Question

A shaft of length \(L\) is made of two materials—an inner core and an outer rim—perfectly bonded with no slip. The inner core has diameter \(d_i\), the outer diameter is \(d_o\), and the rigidities are \(G_i\) and \(G_o\) respectively. Given: \( d_o = 2 d_i \) and \( G_i = 3 G_o \). Under an applied torque, the maximum shear stresses in the outer rim and inner core are \( \tau_o \) and \( \tau_i \). Find the ratio \( \tau_i / \tau_o \) (round off to 2 decimals).

Hint:

In composite torsion problems, use compatibility of twist:
\(\tau / (G r)\) must be equal for all bonded materials, not the polar moment.

Answer 1

Correct Answer: 1.48

For a composite circular shaft with perfect bonding, the angle of twist per unit length is the same in both materials: \[ \frac{\tau_i}{G_i r_i} = \frac{\tau_o}{G_o r_o} \] Here, \[ r_i = \frac{d_i}{2}, \qquad r_o = \frac{d_o}{2} \] Given: \[ d_o = 2 d_i \quad \Rightarrow \quad r_o = 2 r_i \] \[ G_i = 3 G_o \] Substitute into the compatibility condition: \[ \frac{\tau_i}{3 G_o r_i} = \frac{\tau_o}{G_o (2 r_i)} \] Cancel \( G_o \) and \( r_i \): \[ \frac{\tau_i}{3} = \frac{\tau_o}{2} \] Thus, \[ \tau_i = \frac{3}{2} \tau_o = 1.5\,\tau_o \] Therefore, \[ \frac{\tau_i}{\tau_o} = 1.5 \] Rounded to 2 decimal places, the ratio is \(1.50\).