Sunday, 16 April 2017

STRAIN ENERGY STORED DUE TO IMPACT LOADING

STRAIN ENERGY STORED DUE TO IMPACT LOADING

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. 

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.

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.

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

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