Question

A $2m$ long axially loaded mild steel rod of $8mm$ diameter exhibits the load-displacement $(P-\delta )$ behaviour as shown in the figure.


Assume the yield stress of steel as $250MPa.$ The complementary strain energy (in N-mm) stored in the bar up to its linear elastic behaviour will be

• A. 15707.96

Answer 1

The correct option is A 15707.96

The area enclosed by the inclined line and the vertical axis is called complementary strain energy


$ϵ=\frac{\delta }{l}=\frac{2.5}{2000}=\frac{1}{800}$

${\sigma }_y=250MPa$

Complementry strain energy = strain energy

$=\frac{1}{2}\times {\sigma }_y\times \epsilon \times \text{vol of bar}$

$=\frac{1}{2}\times 250\times \frac{1}{800}\times \frac{\pi }{4}\times 8^2\times 2000$

$=15707.96Nmm$