Question

The solid rod AB has a diameter dAB 5 60 mm. The pipe CD has an outer diameter of 90 mm and a wall thickness of 6 mm. Knowing that both the rod and the pipe are made of steel for which the allowable shearing stress is 75 MPa, determine the largest torque T that can be applied at A.

Answer 1

Answer:

3180.86 Nm

Explanation:

Moment of inertia for shaft AB, $I_{AB}= \frac {\pi d^{4}}{32}=\frac {\pi 0.06^{4}}{32}=1.27235\times 10^{-6} m^{4}$

Torque in solid shaft AB will be given by

$T_{AB}=\frac {\tau I_{AB}}{r}=\frac {75\times 10^{6} \times 1.27235\times 10^{-6}}{0.03}=3180.862562 Nm\approx 3180.86 Nm$

Where $\tau$ is shear stress, $I_{AB}$ is polar moment of inertia for shaft AB, r is the radius of shaft B

The inner diameter of pipe CD can found considering that the thickness of pipe is 0.006 m hence diameter= 0.09-(2*0.006)= 0.078 m

Moment of inertia for shaft CD will be

$I_{CD}=\frac {\pi (0.09^{4}-0.078^{4})}{32}=2.8073\times 10^{-6} m^{4}$

Torque for shaft CD will be

$T_{CD} =\frac {\tau I_{CD}}{r}$ and here r = 0.045 m

$T_{CD}}=\frac {75\times 10^{6} \times 2.8073\times 10^{-6}}{0.045}=4678.837 Nm\approx 4678.84 Nm$

The minimum of the two torques is the largest torque that can be applied. Therefore, the torque to apply equals 3180.86 Nm