Question

A specimen of some metal having a rectangular cross section 10.4 mm × 12.8 mm is pulled in tension with a force of 15900 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.

Answer 1

Answer:

$\sigma=0.00151$

Explanation:

We apply Hooke's Law as follows :

$\sigma= E\varepsilon$

where $\sigma$ is the applied stress

#The applied stress is also equal to :

$\sigma=\frac{F}{A_o}=\frac{F}{lw}$

Where $l$ and $w$ are the cross-sectional dimensions.

=>We equate the two stress equations:

$E\varepsilon=\frac{F}{lw}\\\\\varepsilon=\frac{F}{Elw}$

#We substitute our values in the equation as:

$\varepsilon=\frac{15900\ N}{(79\times 10^9\ N/m^2)(10.4\ m\times 10^{-3}\times 12.8\ m\times 10^{-3})}}\\\\\\=0.00151$

Hence, the resulting strain is 0.00151