Could you please tell me how to prove that elastic energy density is equal to 1/2 x strain?

Asked by  | 11th Jan, 2013, 08:03: PM

Expert Answer:

When a wire is put under a tensile stress,work is done against the inter-atomic forces.

This work is stored in the wire in the form of elastic potential energy.

 When a wire of original length L and area of cross-section A is subjected to a deforming force F along the length of the wire.

 let the length of the wire is elongated by l.

now from Hooke’s law , we know that

Y = stress / strain.

Also here stress = Force(F)/area (A)

And strain = l/L

Hence Y = (F/A)/(l/L)

 F= YA  (l/L). Here Y is the Young’s modulus of the material of the wire. Now for a further

elongation of infinitesimal small length dl, then the work done:

 dW  is  F * dl or

  Y A ld l /L.

Therefore the amount of work done (W) in increasing the length of the wire from

 l to L +l, that is from l = 0 to l = l is

 

   = (YA/2) * l2/L

  = ½ * Y *(l/L)2 *AL

= ½ * young’s modulus * strain2 * volume of the wire

= ½ * stress *strain *volume of the wire

This work is stored in the wire in the form of elastic potential energy (U). Therefore the

elastic potential energy per unit volume of the wire (U) is

U = 1/2 x Stess x Strain.

 U = ½ ? ?

Answered by  | 12th Jan, 2013, 09:04: PM

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