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)

$tan\theta$

Answer 2

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 $\frac{Tensile \,stress}{Shearing \,stress}=\frac{F}{A}sin^{2}\,\theta\times\frac{A}{F\,sin\,\theta\,cos\,\theta}=tan\,\theta$