Question

A specimen of some metal having a rectangular cross section 10.5 mm x 13.7 mm is pulled in tension with a force of 2650 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.

Answer 1

Answer:

resulting strain will be 0.00322

Explanation:

given data

cross section = 10.5 mm × 13.7 mm

force = 2650 N

elastic modulus = 79 GPa

solution

we apply here elastic modulus that is express as

elastic modulus = $\frac{stress}{strain}$    ....................1

E = $\frac{\sigma }{\epsilon }$  

and here $\sigma$ = $\frac{force}{area}$    

$\epsilon = \frac{\sigma }{E}$  

$\epsilon = \frac{F }{AE}$

$\epsilon = \frac{34300}{10.8 \times 12.5 \times 10^{-6} \times 79 \times 10^9}$  

$\epsilon$ = 0.00322

so resulting strain will be 0.00322