Question

20 kNm Torque is applied to a shaft of 7 cm diameter. Calculate: 

(i) Maximum shear stress in the shaft. 

(ii) Angle of twist per unit length of shaft. Take G = 10⁵ N/mm².

Answer 1

T_max = 20KN-m = 20 × 10³ N-m 

d = 7cm = 0.07m 

ι_max = ? 

θ = ? 

L = unit length say 1m 

G = 10⁵ N/mm² = 10¹¹ N/m² 

Using torsion equation ; T/J = ι/R = G.θ/L 

Or ; T_max = (π/16) ι_max.D³ ...(i) 

= (ιD⁴/32) .G.θ_max/L ...(ii) 

Where; T_max = Maximum torque 

θ_max = Maximum angle of twist 

ι_max = Maximum shear stress 

Using equation (i) 

20 × 10³ = (π/16) ι_max.(0.07)³ 

ι_max = 296965491.5 N/m² 

Using equation (ii) 

T_max = (πD⁴/32) .G.θ_max/L 

20 × 10³ = (π(0.07)⁴/32) .10¹¹.θ_max/1 

θ_max = 0.08484 radian = 0.08484 × (180/π) 

θ_max = 4.86º