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 $BB′$ making an angle $\theta$ with the length.

The ratio of tensile stress to the shearing stress on the plane $BB′$ 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,
$\therefore F_p=F\,cos\theta$
Component of force F normal to the plane,
$F_N = F\,cos (90^° - \theta) = F\, sin\theta$
Let the area of the face BB′ be A′. Then
$\frac{A}{A'}=sin\,\theta$
$\therefore A'=\frac{A}{sin\,\theta }$
$\therefore $ Tensile stress $=\frac{F\,sin\,\theta }{A'}=\frac{F}{A} sin^{2}\,\theta$
Shearing stress $=\frac{F\,cos\,\theta}{A'}$
$=\frac{F}{A}cos\,\theta \,sin\,\theta =\frac{F\,sin\,2\theta}{2\,A}$
Their corresponding ratio is
$\frac{Tensile \,stress}{Shearing \,stress}=\frac{F}{A}sin^{2}\,\theta\times\frac{A}{F\,sin\,\theta\,cos\,\theta}=tan\,\theta$