Question

A helical compression spring made of a wire of circular cross-section is subjected to a compressive load. The maximum shear stress induced in the cross-section of the wire is 24 MPa. For the same compressive load, if both the wire diameter and the mean coil diameter are doubled, the maximum shear stress (in MPa) induced in the cross-section of the wire is ________.

Answer 1

Concept:

For a helical spring, maximum shear stress developed in the wire, 

\({τ _{shear}} = \frac{{8WD}}{{\pi {d^3}}} {{\;\;\;\;\;}} \ldots \left( 1 \right)\)

Where, 

D =  Mean coil diameter, d =  Wire diameter, W  = Axial load

Calculation:

Given:

τ_max1 = 24 MPa, d₂ = 2d₁,  D₂ = 2D₁

By using equation (1),

\(\Rightarrow \tau_{shear2} = \frac{{8WD_2}}{{\pi {{\left( {d_2} \right)}^3}}} \)

\(\Rightarrow \frac{τ _{2}}{\tau_1} = \frac{{8WD_2}}{{\pi {d_2^3}}}\times\frac{{\pi {d_1^3}}}{{8WD_1}} \)

\(\Rightarrow \frac{τ _{2}}{\tau_1} = \frac{{D_2}}{{ {D_1}}}\times\left(\frac{{{d_1}}}{{d_2}} \right)^3\)

\(\Rightarrow \frac{τ _{2}}{\tau_1} = \frac{{2D_1}}{{ {D_1}}}\times\left(\frac{{{d_1}}}{{2d_1}} \right)^3=\frac{1}{4}\)

\(\Rightarrow \tau_{shear2} = \frac{\tau_{shear1}}{4} = \frac{24}{4} = 6 \ MPa\;\)