A bar having a length of 5 in. and cross-sectional area of 0.7 i n . 2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear elastic behavior.

E=1.969 × 10¹¹ Pa

Explanation:

The formula to apply is;

E=F*L/A*ΔL

where

E=Young modulus of elasticity

F=Force in newtons

L=Original length in meters,m

A=area in square meters m²

ΔL= Change in length in meters,m

Given

F= 8000 lb = 8000*4.448 =35584 N

L= 5 in = 0.127 m

A= 0.7 in² =0.0004516 m²

ΔL = 0.002 in = 5.08e-5 m

Applying the formula

E=(35584 * 0.127)/(0.0004516*5.08e-5 )

E=1.969 × 10¹¹ Pa

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