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Question: Strain energy per unit volume in a stretched string is \(A.\dfrac{1}{2}\times Stress\times Strain...

Strain energy per unit volume in a stretched string is
A.12×Stress×StrainA.\dfrac{1}{2}\times Stress\times Strain
B.Stress×StrainB.Stress\times Strain
C.(Stress×Strain)2C.{{\left( Stress\times Strain \right)}^{2}}
D.StressStrainD.\dfrac{Stress}{Strain}

Explanation

Solution

We know that the restoring force per unit area is called stress and strain is defined as the measure of deformation. Strain energy is the energy stored by a system when a system is undergoing deformation. We have to apply these concepts to find the strain energy per unit volume in relation to stress and strain.
Formula Used:
We will use the following formula to find the strain energy per unit volume:-
F=YAxLF=\dfrac{YAx}{L}.

Complete step-by-step solution:
Let Young’s modulus be YY, wire of length be LL, Area of cross-section be AA, the load is FF, the extension is xx, stress is σ\sigma and strain be ε\varepsilon and work done be WW and VV is the volume.
Y=FLAx\Rightarrow Y=\dfrac{FL}{Ax}………………. (i)(i)
From (i)(i) we can write
F=YAxLF=\dfrac{YAx}{L}……………… (ii)(ii)
W=F.dx\Rightarrow W=F.dx……………… (iii)(iii)
Where dxdx is the displacement
Putting (ii)(ii) in (iii)(iii) we get
W=YA(xL)dxW=YA\left( \dfrac{x}{L} \right)dx……………….(iv)(iv)
We have to integrate it to find the work done in elongation xx
W=YALxdxW=\dfrac{YA}{L}\int{xdx}………………….. (v)(v)
Integrating (v)(v) we get
W=12(YAx2L)\Rightarrow W=\dfrac{1}{2}\left( \dfrac{YA{{x}^{2}}}{L} \right)
This can be written as
W=12(YAxL)x\Rightarrow W=\dfrac{1}{2}\left( \dfrac{YAx}{L} \right)x
Using (ii)(ii) we get
W=12×F×x\Rightarrow W=\dfrac{1}{2}\times F\times x……………… (iv)(iv)
Work done is equal to strain energy, EE can be written as
E=12×F×x\Rightarrow E=\dfrac{1}{2}\times F\times x
Therefore, strain energy per unit volume can be written as
1×F×x2×V\dfrac{1\times F\times x}{2\times V}
1×F×x2×A×L\Rightarrow \dfrac{1\times F\times x}{2\times A\times L}………….. (v)(v)
This can be rearranged into
12×σ×ε\Rightarrow \dfrac{1}{2}\times \sigma \times \varepsilon
Because σ=FA\sigma =\dfrac{F}{A}and ε=xL\varepsilon =\dfrac{x}{L}.
Hence, option (A)(A) is the correct option.

Additional Information:
We should also have knowledge of Hooke’s law whenever we analyze stress, strain, and strain energy. Hooke’s law states that within proportional limit stress is directly proportional to strain produced. Mathematically it can be written as
σε\sigma \propto \varepsilon
σ=Eε\Rightarrow \sigma = E\varepsilon
σε=E\Rightarrow \dfrac{\sigma }{\varepsilon }= E
Where EE is proportionality constant.

Note: We should not get confused over the fact that strain and strain energies are different terms and their formulae are also different. We should also read the question carefully as it is said to find strain energy per unit volume not only strain energy. Strain energy stored in a material can be represented as work done on it.