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 

ι_max = ? 

θ = ? 

G = 80 GPa = 80 × 10⁹ N/m² 

Using the relation, T_max = (π/16) τ_max.D³ 

1000 = (π/16) τ_max.(0.04)³ 

τ_max = 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 = 8.56º