an aluminum bar 125mm and having a scuare cross section 16.5mmon an edge is pulled in tenstion with a load of 66700N and expects an elongeation of 0.43mm . Assuming the deformation is entirely elastic . calculate the modulus of elasticity of the aluminium

The modulus of elasticity, also known as Young's modulus, can be calculated using the formula:

E = (F/A) / (ΔL/L)

where: E is the modulus of elasticity, F is the force applied, A is the cross-sectional area, ΔL is the change in length, and L is the original length.

Given in the problem: F = 66700 N, ΔL = 0.43 mm = 0.43 x 10^-3 m (converting mm to m), L = 125 mm = 125 x 10^-3 m (converting mm to m), A = (16.5 mm)^2 = 16.5 x 10^-3 m x 16.5 x 10^-3 m = 272.25 x 10^-6 m^2 (converting mm to m and then squaring).

Substituting these values into the formula gives:

E = (66700 N / 272.25 x 10^-6 m^2) / (0.43 x 10^-3 m / 125 x 10^-3 m) E = (245.3 x 10^6 Pa) / 0.00344 E = 71.3 x 10^9 Pa

So, the modulus of elasticity of the aluminum is approximately 71.3 GPa.