A rod of cross-sectional area 100 m² and length 5000 mm is subjected to a sudden load of 100 kN and Young's modulus is 200 MPa. What is the strain energy stored in it?

Concept:

Strain energy (U)

U = Strain energy per unit volume (u) × Volume of the Member (V)

U = Area under load-deformation curve

Strain energy per unit volume ⇒ \(u=\frac{1}{2}\times stress(\sigma)\times strain (\epsilon) =\frac{1}{2}\times \sigma \times\frac{\sigma}{E}=\frac{\sigma^2}{2E}\)

Also,

Load deformation curve for gradual and sudden loading is shown below:

In gradual loading, the loading starts from zero and increases gradually till the body is fully loaded, while in sudden loading, the load is suddenly applied on the body.

1. For Gradual Loading:

Strain Energy (U):

\(U=\frac{1}{2}\times P × δ L=\frac{1}{2}\times P ×\frac{\sigma\times L}{E}\)      ......(i)

\(U = u \times V=\frac{\sigma^2}{2E}\times A\times L\)      ......(ii)

From equation (i) and (ii)

\(??\therefore \frac{1}{2}\times P ×\frac{\sigma\times L}{E} =\frac{\sigma^2}{2E}\times A\times L\)

\({\sigma _{gradual}} = \frac{P}{A}\)

2. For sudden loading

Strain Energy (U):

\(U=P × δ L=P ×\frac{\sigma\times L}{E}\)      ......(i)

\(U = u \times V=\frac{\sigma^2}{2E}\times A\times L\)      ......(ii)

From equation (i) and (ii)

\(??\therefore P ×\frac{\sigma\times L}{E} =\frac{\sigma^2}{2E}\times A\times L\)

\({\sigma _{sudden}} = \frac{{2P}}{A}\)

Calculation:

Given:

A = 100 m2, L = 5000 mm = 5 m, P = 100 kN, E = 200 MPa

\(U=\frac{\sigma^2}{2E}× V=\frac{V}{2E}× \sigma^2=\frac{AL}{2E}× (\frac{2P}{A})^2=\frac{2P^2 L}{AE}\)

\(U=\frac{2P^2 L}{AE}=\frac{2 (100 \times 1000)^2 \times 5}{100 \times 200 \times 10^6} = \frac{2 \times10^{10} \times 5}{10^{10} \times 2}=5\; N.m\)