We were discussingÂ Strain energy, Resilience, Proof
resilience and Modulus of resilienceÂ in our recent post and also we have
discussedÂ strain energy stored in a body when load will be applied graduallyÂ andÂ strain energy stored in a body when load will be applied suddenlyÂ during
our previous posts.Â

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 CC

_{1}or DD_{1}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.

P = Shear
load applied gradually

Ï„ = Shear
stress developed in the body due to shear force P

Ï• = Shear
strain = CC

_{1}/CB = DD_{1}/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

CC

_{1}or DD_{1}= 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 CC

_{1}
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 CC

_{1}and we will use above equation to put the value of shear force P and shear deformation CC_{1}.
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

CC

_{1}/CB = Ï„ / G
CC

_{1}/ h = Ï„ / G
CC

_{1}= Ï„ x h / G
Now we
have values of shear force P and distance CC

_{1}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 CC_{1}.
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
We will
discussÂ strain
energy stored in a body when body will be loaded with impact load.

###
**Reference:**

Strength
of material, By R. K. Bansal

Image
Courtesy: Google

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