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

Tmax = 20KN-m = 20 × 103 N-m

d = 7cm = 0.07m

ιmax = ?

θ = ?

L = unit length say 1m

G = 105 N/mm2 = 1011 N/m2

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

Or ; Tmax = (π/16) ιmax.D3 ...(i)

= (ιD4/32) .G.θmax/L ...(ii)

Where; Tmax = Maximum torque

θmax = Maximum angle of twist

ιmax = Maximum shear stress

Using equation (i)

20 × 103 = (π/16) ιmax.(0.07)3

ιmax = 296965491.5 N/m2

Using equation (ii)

Tmax = (πD4/32) .G.θmax/L

20 × 103 = (π(0.07)4/32) .1011.θmax/1

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

θmax = 4.86