Question

A $0.05m$ cube has its upper face displaced by $0.2cm$ by a tangential force of $8N.$Calculate the shearing stain, shearing stress and modulus of rigidity of the material of the cube.

Answer 1

Shearing strain is the ratio of small change in length to the original length. Shearing stress is the ratio of change in force applied to the area. Modulus of rigidity is the ratio of shearing stress to the shearing strain.

Complete step by step answer:
It is given in the question that,
A cube has dimension, $l = 0.05m$ . . . (1)
Displacement of the upper face of the cube, $\Delta l = 0.2cm.$ . . . (2)
The tangential force$,F = 8N.$ . . . (3)
We have to find:
shearing strain, shearing stress and the modulus of rigidity.
Now, we have from the above equation or data,
$l = 0.05m = 5 \times {10^{ - 2}}m$
$\Delta l = 0.2cm = 0.2 \times {10^{ - 2}}m,$
Tangential Force$(F) = 8N.$
By using the definition of shearing strain, we get
Shearing strain$= \dfrac{{\Delta l}}{l}$
$= \dfrac{{0.2}}{5}$
$= 0.04$
By using the definition of shearing stress, we get
Shearing stress$= \dfrac{F}{{{l^2}}} = \dfrac{8}{{{{(5 \times {{10}^{ - 2}})}^2}}} = 3200N/{m^2}$
In the same way, by using the definition of modulus of rigidity, we get
Modulus of Rigidity, $\eta$=shearing stress shearing strain
$= \dfrac{{3200}}{{0.04}} = 80000N/{m^2}$

Note: This is a very simple question, if you know the required formula. So it is important to know the formulae and definitions. Shearing strain, Shearing stress and rigidity are important factors required to study elasticity properties of any body.