Question

A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T that may be applied to the shaft if the allowable shear stress is equal to 100 MPa.

Answer 1

Answer:

Explanation:

Using equation of pure torsion

$\frac{T}{I_{polar} }=\frac{t}{r}$

where

T is the applied Torque

$I_{polar}$ is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

$D_{0} =$ Outer Diameter

$D_{i} =$ Inner Diameter

$I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}$

Applying values in the above equation we get

$I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\I_{polar}= 1.74$ x $10^{-7} m^{4}$

Thus from the equation of torsion we get

$T=\frac{I_{polar} t}{r}$

Applying values we get

$T=\frac{1.74X10^{-7}X100X10^{6} }{.021}$

T =829.97Nm