Question

a 5cm cube has its upper face displaced by 0.2 cm by a tangential force of 8N. calculate the shearing strain. shearing stress and mod of rigidity of the material.

Answer 1

According to question,

Displacement of cube    x  = 0.2 Cm
Side of Cube    a = 5 Cm
Force     F  = 8 N


We know,    
Strain =  Extension / Length
           = x  / a
           = 0.2 / 5
           = 0..04

Stress  = F / A
             = 8 / 5 * 5 
            = 8 / 25
              = 0.032

Answer 2

Answer:

Explanation:

Given,

L=5cm

Delta L=0.2cm

Surface area A=LxL=25cm^2

F=8N=8x10^5 dyne

Therefore, Shearing Stress=F/A

=8x10^5 /25

=32x10^3 dyne/cm^2

Again, Shearing strain=delta L/L

=0.2/5

= 0.04

Now,

Modulus of rigidity G=FL /A delta L

=8x10^5 dyne/cm2

Hope,These answers will help you.