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.

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

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .