A bar has a length of 8 in. and cross-sectional area of 12 in2. Part A Determine the modulus of elasticity of the material if it is subjected to an axial tensile load of 10 kip and stretches 0.006 in. The material has linear-elastic behavior.

Answer: modulus elasticity = 4.9 * 10^6 N/in²

Explanation: modulus elasticity = tensile stress / tensile strain.

Tensile stress = force / area.

Force = 10kip = 4.44 * 10^4N (note that 1kip = 4448.22N), area = 12 in²

Tensile stress = 4.44 * 10^4 / 12 = 3.7 * 10³ N/in²

Tensile strain = extension / length

extension = 0.006in length = 8 in

Tensile strain = 0.006 / 8 = 7.5 * 10^-4.

Modulus = stress / strain = 3.7 * 10³ / 7.5 * 10^-4

Modulus = 0.49 * 10^7

modulus = 4.9 * 10^6 N/in²