Question

A compound shaft 1.5m long fixed at one end is subjected to a torque of 15KN-m at the free end and of 20KNm at the Junction point as shown in fig. Determine 

(a) The maximum shearing in each portion of the shaft. 

(b) The angle of twist at the junction of the two section and at the free ends. Take G = 0.82 × 10⁵ N/mm².

Answer 1

Torque in BC = 15 KN-m 

Torque in AB = 35 KN-m 

Shaft are in series i.e.; θ = θ₁ + θ₂ 

θ = (T.L/G.J)_AB + (T.L/G,J)_BC 

= (35 × 10⁶ × 750)/{0.82 × 10⁵ × (π/32)(120)⁴} + (15 × 10⁶ × 750)/ {0.82 × 10⁵ x (π/32)(90)⁴} 

θ = 0.037 rad = 2.12º

Now from torque equation T/J = ι/R ; ι = T.R/J = 16T/πd³ 

ι_BC = (16 × 15 × 10⁶)/π(90)³ 

ι_BC = 104.85 N/mm² 

ι_AB = (16 × 35 × 10⁶)/π(120)³ 

ι_AB = 103.21 N/mm²