Question

A solid circular shaft needs to be designed to transmit a torque of 55 N m. if the allowable shear stress of the material is 280 N/mm², find the diameter (in mm) of the shaft required to transmit the torque. (Assume, π = 22/7)
1. 5.62
2. 10
3. 31.62
4. 25.0

Answer 1

Correct Answer - Option 2 : 10

Concept:

Torsion equation

\(\frac{{{\tau }_{max}}}{r}=\frac{{{T}_{R}}}{J}=\frac{G\theta }{L}\)

\({{\tau }_{max}}=\frac{{{T}_{R}}\cdot r}{J}\)

τ_max = maximum allowable shear stress

T_R = Resistance torque

J = Polar moment of inertia

Calculation:

Given, T = 55 N.m = 55 × 10³ N.mm

τ_max = 280 N/mm² = 280 MPa

\({{\tau }_{max}}=\frac{T\cdot r}{J}\)

\(\Rightarrow 250=\frac{55\times {{10}^{3}}\times \text{D}}{2\times \frac{\pi }{32}\times {{\text{D}}^{4}}}\)

\(\Rightarrow {{D}^{3}}=\frac{55\times {{10}^{3}}\times 32}{2\times \frac{22}{7}\times 250}\)

⇒ D = 10 mm