Question

A bar of cross-section area A is subjected two equal and opposite tensile forces at its ends as shown in figure. Consider a plane $BB\prime$ making an angle $\theta$ with the length.

The ratio of tensile stress to the shearing stress on the plane $BB\prime$ is

• A. $tan\theta$
• B. $sec\theta$
• C. $cot\theta$
• D. $cos\theta$

Answer 1

Answer: (A)

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

Component of force F along the plane,
$\therefore F_p=Fcos\theta$
Component of force F normal to the plane,
$F_N=Fcos(90^{°}-\theta )=Fsin\theta$
Let the area of the face BB′ be A′. Then
$\frac{A}{A^{\prime }}=sin\theta$
$\therefore A^{\prime }=\frac{A}{sin\theta }$
$\therefore$ Tensile stress $=\frac{Fsin\theta }{A^{\prime }}=\frac{F}{A}si{n}^2\theta$
Shearing stress $=\frac{Fcos\theta }{A^{\prime }}$
$=\frac{F}{A}cos\theta sin\theta =\frac{Fsin2\theta }{2A}$
Their corresponding ratio is
$\frac{Tensilestress}{Shearingstress}=\frac{F}{A}si{n}^2\theta \times \frac{A}{Fsin\theta cos\theta }=tan\theta$