The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specifications require that the diameter of the shaft be uniform from A to D and that the angle of twist between A and D not exceed 1.45°. Knowing that τmax ≤ 60 MPa and G = 77.2 GPa, determine the minimum diameter shaft that can be used. (Round the final answer to one decimal place.)

The minimum diameter shaft that can be used is; 40.8 mm

We are given;

Torque(T) = 800 N·m

Angle of twist; φ = 1.45° = 1.45π/180 = 0.0253 rads

Maximum shear stress; τ = 60 MPa = 60 x 10^(6) Pa

G = 77.2 GPa = 77.2 x 10^(9) Pa

From the image attached;

length of shaft (L) = 0.4 + 0.6 + 0.3 = 1.3m

Now, the polar moment of inertia is calculated from,

J = πc⁴/2

where c is the radius of shaft

To solve this, we will find the diameter based on the angle of twist and also based on the shear stress and we will choose the smaller one.

Formula for angle twist is given as;

φ = TL/GJ

Now J = πc⁴/2

Thus, φ = 2TL/G(πc⁴)

c⁴ = 2TL/φGπ

c⁴ = [2 x 800 x 1.3]/(0.0253) x 77.2 x 10^(9) x π)

c⁴ = 0.00000033898

c = ∜0.00000033898

c = 0.024m

Formula for shear stress is given as;

τ = Tc/J

Putting πc⁴/2 for J, we have;

τ = 2T/πc³

c = ∛(2T/πτ)

c = ∛(2 x 800)/(π x 60 x 10^(6))

c = ∛0.00000848826

c = 0.0204m

So, comparing the two values of radius gotten, the one based on the shear stress is bigger and it's c = 0.0204m

Diameter = 2 x radius = 2 x 0.0204m = 0.0408m which is 40.8 mm

Read more about shear stress and torque at; link

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