Young’s modulus of elasticity is ratio between stress and strain. It can be expressed as:

\(Young’s\space\ Modulus=\frac{Stress}{Strain}\)\[E=\frac{f}{e}\]

#### Example

Find the young’s modulus of elasticity for the material which is 200 cm long, 7.5 cm wide and 15 cm deep. Tie material is subjected to axial force of 4200 KN. As a result material is stretched 2.67 cm.

##### Solution

Width of tie bar = b = 7.5 cm

Length of tie bar = d = 200 cm

Depth of tie bar = d = 15 cm

Axial Force = P = 4200 KN

Increase in length = 2.67 cm

Young’s Modulus of Elasticity = E = ?

Firstly find the cross sectional area of the material = A = b X d = 7.5 X 15

A = 112.5 centimeter square

\[Young's\space\ Modulus=\frac{Stress}{Strain}\]

\[E=\frac{\frac{P}{A}}{\frac{\delta l}{l}}\]

\[E=\frac{Pl}{A.\delta l}\]

\[E\space\ =\frac{4200\times 200}{112.5\times 2.67}\]

E = 2796.504 KN per centimeter square.