We were discussing Strain energy, Resilience, Proof resilience and Modulus of resilience in our recent post and also we have
discussed strain energy stored in a body when load will be applied gradually
and strain energy stored in a body when
load will be applied suddenly during our previous posts.

Earlier we were discussing a body which is subjected with a sudden applied load, but in this case we have considered that body is subjected with an impact load.

As we know that strain energy stored in the body U will be provided by the following expression as mentioned here.
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Once we will have value of the stress (σ) induced in the vertical bar due to impact load, we will easily determine the value of strain energy stored in the vertical bar due to an impact load.

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Today we will
discuss strain energy stored in a body with impact load with the help of this
post.

Earlier we were discussing a body which is subjected with a sudden applied load, but in this case we have considered that body is subjected with an impact load.

Let us see
the following figure, where we can see one vertical bar which is fixed at the
upper end and there is collar at the lower end of the bar. Let us think that
one load is being dropped over the collar of the vertical bar from a height of
h as displayed in following figure.

Such a case
of loading could be considered as a body subjected with an impact load and let
us consider that there will be extension in the bar by x due to impact load P.

We will
find out here the stress induced in the body due to impact loading and
simultaneously we will also discuss for strain energy for this situation.

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.

We have
following information from above figure where a body is subjected with an impact
load.

σ= Stress
developed in the body due to impact load

E =
Young’s Modulus of elasticity of the material of the body

A= Cross
sectional area of the body

P = Impact
load

x =
Deformation or extension of the body i.e. vertical bar

L = Length
of the body i.e. vertical bar

V= Volume
of the body i.e. vertical bar = L.A

U = Strain
energy stored in the body i.e. vertical bar

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 vertical bar = Work done by the load in deforming the vertical
bar

Strain energy stored in the vertical bar = Load x Displacement

Strain
energy stored in the vertical bar = P. (h + x)

U = P. (h +
x)

As we know that strain energy stored in the body U will be provided by the following expression as mentioned here.

Now we
will secure the value of extension x in terms of Stress, Length of the body and
Young’s modulus of the body by using the concept of Hook’s Law.

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*According to Hook’s Law*

*According to Hook’s Law*

Within elastic limit, stress applied over an elastic
material will be directionally proportional to the strain produced due to
external loading and mathematically we can write above law as mentioned here.

Stress = E. Strain

Where E is Young’s Modulus of elasticity of the material

σ
= E. ε

σ
= E. (x/L)

x = σ.
L/ E

Let use the value of the extension or deformation “x”
in above equation and we will have.

Once we will have value of the stress (σ) induced in the vertical bar due to impact load, we will easily determine the value of strain energy stored in the vertical bar due to an impact load.

We will
discuss strain energy stored in a body due to shear stress in our next post.

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**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google