A specimen is subjected to a pure shear stress regime of intensit

| A specimen is subjected to a pure shear stress regime of intensity τ. The resulting tensile and compressive stresses σ, which occur on planes inclined at 45° to the direction of the shear stresses, would be

A. τ 

B. \(\frac{{\rm{\tau }}}{2}\)

C. \(\sqrt 2 {\rm{\;\tau }}\)

D. \(\frac{{\rm{\tau }}}{{\sqrt 2 }}\)

Please scroll down to see the correct answer and solution guide.

Right Answer is: A

SOLUTION

Concept:

For a bi-axial state of stress normal stress at any plane is given by,

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

Where, \({{\rm{σ }}_{\rm{n}}}\) is the normal stress in a plane which is inclined at an angle θ with the horizontal axis;

\({{\rm{σ }}_{\rm{x}}}{\rm{\;and\;}}{{\rm{σ }}_{\rm{y}}}\) are the normal stress in the direction of X axis and Y axis respectively;

\({{\rm{τ }}_{{\rm{xy}}}}\) is the shear stress acting in XY plane.

Tensile normal stress is taken as positive, compressive normal stress is taken as negative and anticlockwise shear stress is taken as positive.

Calculations:

Given, it a pure shear case.

∴ σ_x = σ_x = 0 and τ_xy = τ and θ = 45°

\(∴ {{\rm{σ }}_{\rm{n}}} = \frac{{0 + 0}}{2} + \frac{{0 - 0}}{2}\cos 2 \times 45 + {\rm{τ }} \times \sin 2 \times 45 = {\rm{τ }}\)

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