Question

A solid shaft is replaced by a hollow one. The external diameter of which is 5/4 times the internal diameter. Allowing the same intensity of torsional stress in each, compare the weight and the stiffness of the solid with that of the hollow shaft.

Answer 1

Diameter of solid shaft = d 

Outer Diameter of hollow shaft = d_O 

Inner Diameter of hollow shaft = d_i 

Given that d_O = 1.25d_i

Same intensity of torsional stress i.e ; ι_S = ι_H

W_S/W_H = ? 

K_S/K_H = ? 

Since ; ι_S = 16T/πd³ ...(i) 

and ι_H = 16T/[π(d_O⁴ - d_i⁴)/d_O] ...(ii) 

Equate equation (i) and (ii)

16T/πd³ = 16T/[π(d_O⁴ - d_i⁴)/d_O], putting the value d_O = 1.25d_i 

we get d = 1.048 d_i or 0.838 d_O 

Now for ratio of weight 

Since ρ_S = ρ_H, L_S = L_H 

W_S = ρ_S.g(volume)_S = ρ_S.g.A_S.L_S 

W_H = ρ_H.g(volume)_H = ρ_H.g.A_H.L_H 

Now ; W_S/W_H = ρ_S.g.A_S.L_S /ρ_H.g.A_H.L_H = A_S/A_H = (ρ/4)d² /[(π/4)(d_O² – d_i²)] Now putting the value d_i = d/1.048 ; d_O = d/0.838 ; we get 

W_S/W_H =1.9525

Now for ratio of torsional stiffness 

K = T/θ = G.J/L 

K_S = G_S.J_S/L_S 

K_H = G_H.J_H/L_H 

K_S/K_H = G_S/G_H[{(π/4)(d⁴)} / {(π/4)(d_O⁴ – d_i⁴)}] 

If G_H = G_S and d_i = d/1.048 ; d_O = d/0.838, than 

K_S/K_H = 0.837