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.

3180.86 Nm

Explanation:

Moment of inertia for shaft AB,

Torque in solid shaft AB will be given by

Where is shear stress, 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

Torque for shaft CD will be

and here r = 0.045 m

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

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