20 Physics -- Elasticity

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Energy stored in a stretched wire

​When a body is deformed, work is done by the deforming forces against the intermolecular force of attraction between the molecules of the body. This work done is stored in the body in the form of elastic potential energy.​Consider a body of length L and of cross section area A. Suppose on applying force F, elongation x is produced in the body.​Then,  the young's modulus of elasticity  of the material of the body is given by,Y=FLAxY=FLAxF=YALx∴F=YAL

Then, small work done to produce an infinitesimally small extension dx is given by,dW=FdxdW=FdxIntegrating within limits from 0 to x,W=x0FdxW=∫0xFdxW=x0YALdxW=∫0xYALdxW=YALx0xdxW=YAL∫0xxdxW=YAL[x22]x0W=YAL[x22]0xW=12YALx²W=12⋅YAL⋅x²W=12YALxxW=12⋅YALx⋅xW=12FxW=12FxW=12×force×extensionW=12×force×extensionSince, work done = energy storedEnergy stored=12×force×extension∴Energy stored=12×force×extensionEnergy DensityEnergy density is defined as the elastic potential energy per unit volume of the deformed body.
​Thus, we have,Energy Density (U)=12FxVEnergy Density (U)=12⋅FxVhere, V = Volume of the deformed body andV=Area of cross section×lengthV=Area of cross section×lengthV=A×LV=A×LEnergy Density (U)=12FxAL∴Energy Density (U)=12FxALHere,FA=normal stressFA=normal stress
and,
xL=Longitudinal strainxL=Longitudinal strainEnergy density (U)=12×normal stress×longitudinal strain

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