Question

A torque of 1 KN-m is applied to a 40 mm diameter rod of 3 m length. Determine the maximum shearing stress induced and the twist produced. Take G = 80GPa.

Answer 1

T = 1 KN–m = 1000 N–m

d = 40 mm = 0.04 m

L = 3 m

ιmax = ?

θ = ?

G = 80 GPa = 80 × 109 N/m2

Using the relation, Tmax = (π/16) τmax.D3

1000 = (π/16) τmax.(0.04)3

τmax = 7.96 × 107 N/m2

Now using the relation, T = (πD4/32).G.θ/L

θ = T.L/[(πD4/32).G]

= (1000 × 3)/[{π(0.04)4/32}× 80 × 109]

θ = 0.1494 rad

= 0.1494 (180/π)

θ = 8.56