A bar having a length of 5 in. and cross-sectional area of 0.7 in.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.

Answer: 2,857,142.5lb/in²

Explanation:

The modulus of elasticity is defined as the ratio of tensile stress to tensile strain of a material.

Mathematically,

Modulus of elasticity = Tensile stress/tensile strain

Tensile stress is the ratio of the force acting on a body to its unit cross sectional area.

Stress = Force (F) / Area (A)

Given force = 8000lb

Area = 0.7in²

Stress = 8000/0.7

Stress = 11,428.57lb/in²

Tensile strain is the ratio of the extension of a material to its original length.

Strain = extension (e) / original length (Lo)

Given extension = 0.002in

Original length of the bar = 5in

Strain = 0.002in/5in

Strain = 0.004 (it has no unit)

Therefore the modulus of elasticity = 11428.57/0.004

= 2,857,142.5lb/in²