Question

A shaft subjected to torsion experiences a pure shear stress τ on the surface. The maximum principal stress on the surface which is at 45° to the axis will have a value 
1. τ cos 45°
2. 2τ cos 45°
3. 2τ sin 45°cos 45° 
4. τ cos² 45° 

Answer 1

Correct Answer - Option 3 : 2τ sin 45°cos 45° 

Concept:

For biaxial stress accompanied by simple shear stress at any plane then,

Maximum principal stress is

\({{\rm{σ }}_{\rm{p}}} = \frac{{{{\rm{σ }}_{\rm{x}}} + {{\rm{σ }}_{\rm{y}}}}}{2} - \frac{{{{\rm{σ }}_{\rm{x}}} - {{\rm{σ }}_{\rm{y}}}}}{2}\cos 2{\rm{θ }} - {{\rm{τ }}_{{\rm{xy}}}}\sin 2{\rm{θ }}\)

when there is the normal stress in a plane which is inclined at an angle θ with the horizontal axis;

σ_x and σ_y are the normal stress in the direction of the X-axis and Y-axis respectively;

τ_xy is the shear stress acting in the XY plane.

Tensile normal stress is taken as positive, compressive normal stress is taken as negative.

Calculations:

Given:

It is a pure shear stress case

  σx = σx = 0 and τxy = τ and θ = 45° 

Maximum principal stress:

\({{\rm{σ }}_{\rm{p}}} = \left(\frac{{0 + 0}}{2}\right) - \left(\frac{{0 - 0}}{2}\right)\cos (2 \times 45) - {\rm{τ }} \times \sin (2 \times 45^{\circ}) \)

σ_p = τ sin2 45° .....................(Clockwise shear stress is taken as positive along the x-x axis)

σ_p = 2τ sin 45°cos 45° ........(sin 2θ = 2 sinθcosθ) 

∴ The resulting normal stress (tensile or compressive) on planes inclined at 45° to the direction of the shear stresses is equal to τ.