Question

A bar of cross-sectional area $A$ is subjected two equal and opposite tensile forces at its ends as shown in figure. Consider a plane BB' making an angle $\theta$ with the length. For what value of $\theta$, shearing stress is maximum?

• A. $0^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${45}^o$

Consider the equilibrium of the plane BB'.

A force $F$ must be acting on this plane making an angle $({90}^o-\theta )$ with the normal ON. Resolving $F$ into two components, along the plane and normal to the plane.

Component of force $F$ along the plane, $F_P=F\cos \theta$

Component of force $F$ normal to the plane,

$F_N=F\cos ({90}^o-\theta )=F\sin \theta$

Let the area of the face BB' be A'. Then

$\frac{A}{A^{\prime }}=\sin \theta$

$\therefore A^{\prime }=\frac{A}{\sin \theta }$

Shearing stress $=\frac{F\cos \theta }{A^{\prime }}=\frac{F}{A}\cos \theta \sin \theta =\frac{F\sin 2\theta }{2A}$

Shearing stress$=\frac{F_p\sin 2\theta }{2A}$

Shearing stress is maximum when $\sin 2\theta =1$

or $\sin 2\theta =\sin {90}^{o}or2\theta ={90}^o$

or $\theta ={45}^o$