Stress = 50,000 / 625? = 25.46
Strain energy = ?²V/2E = 25.46×25.46×9817477 / (2×200000) = 15909.5 N-mm = 15.9 N-m.

Modulus of resilience is strain energy stored in material per unit volume.
u = U/v
= ( 100 x 1000 ) / ( 10 x 10x 5x 1000)
= 1/5 N-mm/mm³.

Concept:

For simply supported beam in concentrated load at mid-span:

Deflection at midpoint, δ:

\(\delta = \frac{{W{L^3}}}{{48EI}}\)

Calculation:

Strain energy, U:

\(U = \frac{1}{2}W\delta = \frac{1}{2}W \times \frac{{W{L^3}}}{{48EI}} = \frac{{{W^2}{L^3}}}{{96EI}}\)

Now,

W = 1 ? (point load)

∴ \(U = \frac{{{}{L^3}}}{{96EI}}\)

Instantaneous stress = E x instantaneous strain = E x ?L/L = 200,000x 2.1 / 3000 = 140N/mm².

Concept:

Strain energy (U)

U = Strain energy per unit volume (u) Ã— Volume of Member (V)

U = Area under load-deformation curve

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

Also,

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

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

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

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

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)

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

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

Calculation:

Given:

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

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

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

Stress = Load/ area = 60,000 / (?/4 D²) = 47.746 N/mm².

Area = ?/4 d2 = 625?
Load = 100kN = 100×1000 N
Stress = 2 x load / area = 2x100x1000 / (625?) = 101.86 N/mm².

Proof resilience is maximum stored energy at elastic limit. Resilience is ability of material to absorb energy when elastically deformed and to return it. Elasticity is property by which any body regain its original shape. Malleability is property by which any material can be beaten into thin sheets.

The strain energy stored in a body is equal to work done by applied load in stretching body.

Strain energy per unit volume for solid = q² / 4G.