A steel specimen 150 mm^2 in X-section stretches 0.05 mm over a 50 mm gauge length under an axial load of 30 kN. Calculate the strain energy stored in the specimen at this stage. If the load at the elastic limit for the specimen is 50 kN, calculate the elongation at the elastic limit and proof resilience.

Data:

$A=150 mm^2$

$\delta l=0.05mm$

$l=50mm$

$P=30KN$

$\delta l=\frac{pl}{E}$ and stress $p=\frac{P}{A}=\frac{30*10^3}{150}$

$p=200N/mm^2$

$0.05=\frac{200*50}{E}$

$E=2*10^5 N/mm^2$

Strain energy=$u=\frac{p^2}{2E}*Al=\frac{200^2}{2*2*10^5}*150*50$

$u=750 Nmm$

For 30KN $\delta l=0.05mm$

For 50KN, $\delta l'=?$

$\delta l'=\frac{0.05}{30}*50$

$\delta l'=0.0833mm$

Strain energy at elastic limit=proof resilience=

$=\frac{\frac{50*10^3}{150}^2*150*50}{2*2*10^5}$

$=2083.33Nmm$