Question

Solve a hollow steel shaft has an outside diameter of 100mm and inside diameter of 75 mm.the shaft is subjected to a pure torque of 7.5kn m and is 2m long.the modulus of rigidity for steel 80 gpa *determine the maximum sharing stress in shaft
*the magnitude of the angle of twist
*if the shaft rotates at 200 rev/min determined the power generated by the shaft

Answer 1

• The maximum shearing stress in hollow shaft is 66.125 MPa.
• The magnitude of angle of twist is 0.0279 rad.
• The power generated by the shaft is  157 kW.

What is power ?

Power in a rotating body like shaft is the product of Torque and the angular speed of that shaft.

P = Tω

The Torque is related to the shearing stress is

T = π/16 xτ(Do³ -Di³)

* Given is the Torque T= 7.5 kN.m, the inner diameter Di = 0.075m, outer diameter Do= 0.1m

Then, the maximum shearing stress will be

7.5 x 10³ =  π/16 xτ x (0.1³ - 0.075³)

τ = 66.125 MPa

Thus, the shearing stress is 66.125 MPa.

* From the Torque equation

T/J = Gθ/l

Where, J = π/32 (Do⁴ -Di⁴) = 0.00000671 m⁴

l = length of shaft = 2m, modulus of rigidity G = 80GPa = 80x 10⁹ Pa.

Substitute the values, we get the angle of twist

7.5 x10³ /0.00000671 = 80x 10⁹xθ/2

θ = 0.0279 rad.

Thus, the angle of twist is0.0279 rad.

* Power generated by the shaft is

P = 2πNT/60

where N = revolution/ minute = 200

Put the values, we get

P =  2π x 200 x 7500/60

P = 157 kW.

Thus, the power generated is 157 kW.

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