## Tuesday, 18 April 2017

Today we will discuss strain energy stored in a body due to shear stress with the help of this post. Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box which is provided below this post.

Let us consider that we have one rectangular block ABCD of length l, height h and breadth b as displayed in following figure. Now we need to produce here the condition of shear loading of rectangular block in order to develop the shear stress and shear strain.

Therefore, let us consider that bottom face AB of the rectangular block ABCD is fixed and we have applied a shear force P gradually over the top face CD of the rectangular block as displayed in following figure.
As we can see from above figure that top face CD of the rectangular block ABCD will move by CC1 or DD1 due to the application of shear force P which is applied gradually over the top face CD of the rectangular block.

We have following information from above diagram for a body which is subjected with shear loading.

Ï„ = Shear stress developed in the body due to shear force P
Ï• = Shear strain = CC1/CB = DD1/DA
G = Modulus of rigidity of the material of the body
l= Length of the body i.e. rectangular block ABCD
b= Breadth of the body i.e. rectangular block ABCD
h= Height of the body i.e. rectangular block ABCD
A= Cross sectional area of the body = b x l
V= Volume of the body = l x b x h
CC1 or DD1 = Shear deformation of the body
U = Strain energy stored in the body i.e. rectangular block ABCD due to shear stress

As we have already discussed that when a body will be loaded within its elastic limit, the work done by the load in deforming the body will be equal to the strain energy stored in the body.

Strain energy stored in the body i.e. rectangular block ABCD = Work done by the load in deforming the body.

Strain energy stored in the body i.e. rectangular block ABCD = Average load x distance

As we have seen above that shear force P is applied gradually over the top face CD of the rectangular block and therefore average load will be P/2.

Therefore, we will have
Strain energy stored in the body i.e. rectangular block ABCD = P/2 x CC1

Let us recall the concept of shear stress and shear strain and we will find out here that value of shear force P and shear deformation CC1 and we will use above equation to put the value of shear force P and shear deformation CC1.

Shear stress = Shear load/ Cross sectional area
Ï„ = P/ (b x l)
P = Ï„ x b x l

Shear strain = Shear stress / Modulus of rigidity
Ï• = Ï„ / G
CC1/CB = Ï„ / G
CC1/ h = Ï„ / G
CC1 = Ï„ x h / G

Now we have values of shear force P and distance CC1 and we will use above equation of strain energy stored in the body due to shear stress to put the value of shear force P and shear deformation CC1.

Strain energy stored in the body i.e. rectangular block ABCD = (Ï„ x b x l)/2 x (Ï„ x h / G)
U = Ï„2 x (b x l x h) /2G
U = Ï„2 x V /2G

### Reference:

Strength of material, By R. K. Bansal