Question

A hollow shaft of a motor is supported with bearings by the motor and extends horizontally 50mm out of the motor. The shaft has an outer diameter of 40mm and a thickness of 5mm. How much is the torque to produce a maximum shear stress of 200 MPa.

Ignoring the shafts own weight, how much is the load at the suspending end of the shaft (50mm away from the bearings) that produces 300 MPa maximum normal stress on the shaft.

Answer 1

The Torsion equation is given by,

Where T is the torque on the shaft

J is the polar moment of inertia

is the maximum shear stress on the shaft

r is the radius of the shaft

G is the modulus of rigidity

is the angle of twist

L is the length of Shaft

The polar moment of inertia of hollow shaft is

Where D is the outer diameter

d is the inner diameter

From the torsion equation, the Torque is given as

The given values are

To find the polar moment of inertia

Entering the values into the torque equation

The Torque on the shaft is 1040 Nm

In the second case, the shaft is at bending

The bending equation is given by

Where M is the bending moment acting on the shaft

   is the maximum bending stress or normal stress

I is the moment of inertia of the shaft

For hollow shaft

and y is the distance from the neutral axis

Form the bending equation

And The bending moment is equal to

Where F is the load and L is the length of the shaft

The Load at the end of the shaft is

The load at the end of the shaft is 15600 N or 15.6 kN