Question

Two circular solid shafts of diameters 4 cm and 6 cm respectively transmit same power at the same speed. The maximum shear stress in the 4 cm diameter shaft is 50 MPa. The maximum shear stress in the 6 cm diameter shaft will be approximately:
1. 50 MPa
2. 168.75 MPa
3. 15 MPa
4. None of these

Answer 1

Correct Answer - Option 3 : 15 MPa

Concept:

Using torsion equation:

\(\frac{T}{{{I_P}}} = \frac{{{{\rm{τ }}_{max}}}}{R} = \frac{{G\theta }}{L}\)

Where, T = applied torque, IP = Polar section modulus, τmax = Maximum shear stress in the shaft material,

R = Radius of the shaft, G = Modulus of Rigidity, θ = Angle of twist, L = Length of the shaft

\({I_P} = \frac{\pi }{{32}}{D^4}\)

R = D/2

From shear stress criteria:

\(\frac{{{{\rm{τ }}_{{\rm{max}}\left( {allowable} \right)}} × 2}}{D} = \frac{T}{{{I_P}}}\)

 

Substituting the values of \({I_P} \) and R we get 

Torque (T) = \(\frac{\pi }{{16}}τ {d^3}\)

The solid shaft transmits same power at the same speed i.e.  same torque at the same speed

Therefore, T1 = T2

\(\frac{\pi }{{16}}τ_1 {d_1^3}=\frac{\pi }{{16}}τ_2 {d_2^3}\)

\(\Rightarrow τ_1 {d_1^3}=τ_2 {d_2^3}\)

Calculation:

Given:

τ₁ = 50 MPa, d₁ = 4 cm, d₂ = 6 cm

50 × 4³ = τ2 × 63

τ2 = 14.81 MPa