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.

The ratio of tensile stress to the shearing stress on the plane BB’ is

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

Answer 1

Answer: (A)

Answer 2

: Consider the equilibrium of the plane BB’ A force F must be acting on this plane making an angle with the normal ON. Resolving F into two components, along the plane and normal to the plane.

Component of force F along the plane,

Component of force F normal to the plane,

$\mathrm { F } _ { \mathrm { N } } = \mathrm { F } \cos \left( 90 ^ { \circ } - \theta \right) = \mathrm { F } \sin \theta$

Let the area of the face BB’ be . Then

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

$\therefore$Tensile stress

Shearing stress

Their corresponding ratio is

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