Question

The tangential or shear stress on an oblique plane at an angle $\theta $ to the cross-section of a body which is subjected to a direct tensile stress $\left( \sigma \right)$ is equal to

A) $\dfrac{\sigma }{2}\sin 2\theta $

B) $\sigma \cos \theta $

C) $\sigma {\cos ^2}\theta $

D) $\sigma {\sin ^2}\theta $

Answer 1

Hint: Shear stress, force that appears to induce deformation of the material by sliding down a plane or plane parallel to the stress applied. The ensuing shear is of great significance in nature, being intimately connected to the down slope movement of earth materials and earthquakes.

Complete step by step solution:

When an external force acts on an object, it is deformed. When the position of the force is parallel to the plane of the object. The deformation is going to be along the plane. The tension of the object here is shear stress or tangential stress.

It happens when the force vector components are parallel to the cross-section region of the material. In the case of normal / longitudinal tension, the force vectors would be perpendicular to the cross-section field in which they are acting.

Shear force diagrams display the overall shear strength of each cross-section of the structural component over the length of the beam or structural component. However, this force is not uniformly distributed across the individual cross-section of the beam or structural portion. The maximum shear stress is the maximum localised shear pressure in a limited area.

Let us consider the force to be $F$ and area to be $A$

We know that shear stress is equal to force over an area.

$\sigma = \dfrac{F}{A}$

Now, taking the horizontal component of force and area in an oblique plane we have,

$\dfrac{{F\sin \theta }}

{{\dfrac{A}

{{\cos \theta }}}} = \dfrac{\sigma }

{2}\sin 2\theta $

Hence, option (A) is the correct answer. 

Note: Here we have to be careful while multiplying otherwise we may write $\cos \theta $ instead of $\sin \theta $. Also, we have to remember the formula of shear stress. Tangential or shear stress is the stress which is parallel to the concerned plane so if we calculate the shear stress for an oblique plane then we have to consider the component of force which is parallel to the oblique plane.