Strain Energy

Strain Energy

Strain Energy of the member is defined as the internal work done in defoming the body by the action of externally applied forces. This energy in elastic bodies is known as elastic strain energy :

Strain Energy in uniaxial Loading

Fig .1

Let as consider an infinitesimal element of dimensions as shown in Fig .1. Let the element be subjected to normal stress s_x.

The forces acting on the face of this element is s_x. dy. dz

where

dydz = Area of the element due to the application of forces, the element deforms to an amount = Î_x dx

  Î_x = strain in the material in x – direction

       

Assuming the element material to be as linearly elastic the stress is directly proportional to strain as shown in Fig . 2.

Fig .2

\ From Fig .2 the force that acts on the element increases linearly from zero until it attains its full value.

Hence average force on the element is equal to ½ s_x . dy. dz.

\ Therefore the workdone by the above force

Force = average force x deformed length

           = ½ s_x. dydz . Î_x . dx

For a perfectly elastic body the above work done is the internal strain energy “du”.

where dv = dxdydz

   = Volume of the element

By rearranging the above equation we can write

The equation (4) represents the strain energy in elastic body per unit volume of the material its strain energy – density ‘u_o' .

From Hook's Law for elastic bodies, it may be recalled that

In the case of a rod of uniform cross – section subjected at its ends an equal and opposite forces of magnitude P as shown in the Fig .3.

Fig .3

ILLUSTRATIVE PROBLEMS

1. Three round bars having the same length ‘L' but different shapes are shown in fig below. The first bar has a diameter ‘d' over its entire length, the second had this diameter over one – fourth of its length, and the third has this diameter over one eighth of its length. All three bars are subjected to the same load P. Compare the amounts of strain energy stored in the bars, assuming the linear elastic behavior.

Solution :

From the above results it may be observed that the strain energy decreases as the volume of the bar increases.

2. Suppose a rod AB must acquire an elastic strain energy of 13.6 N.m using E = 200 GPa. Determine the required yield strength of steel. If the factor of safety w.r.t. permanent deformation is equal to 5.

Solution :

Factor of safety = 5

Therefore, the strain energy of the rod should be u = 5 [13.6] = 68 N.m

Strain Energy density

The volume of the rod is

Yield Strength :

As we know that the modulus of resilience is equal to the strain energy density when maximum stress is equal to s_x .

It is important to note that, since energy loads are not linearly related to the stress they produce, factor of safety associated with energy loads should be applied to the energy loads and not to the stresses.