Question

A material is 250 cm long, 8 cm wide and 10 cm deep. The material is subjected to an axial force of 4500kN resulting to an expansion of 3cm. calculate for stress, strain, and the young’s modulus of elasticity for the material

Answer 1

Given:  

Length of the material, L = 250 cm = 2.5 m  

width of the material, b = 8 cm = 0.08 m  

depth of the material, d = 10 cm = 0.1 m  

Axial force on the material, F = 4500 kN = 4500 × 1000 N

expansion, ΔL = 3 cm = 0.03 m  

To find :

The stress, strain, and the young’s modulus of elasticity for the material .

Solution :

Area of cross-section of face perpendicular to which force is applied.

Area of cross-section, A = b*d

                                  A = 0.08 × 0.1 = 0.008 m^2

Stress =  axial force / area of cross-section

=> stress = 4500 * 1000 / 0.008

=> stress = 5.625 * 10 ^ 8 Pa

Strain = 0.03 / 2.5 = 0.012

Young’s modulus of elasticity =  5.625 * 10 ^ 8 / 0.012 = 4.6875 * 10 ^ 10 Pa

The stress is 5.625 * 10 ^ 8 Pa , strain = 0.03 / 2.5 = 0.012 , and the young’s modulus of elasticity =  5.625 * 10 ^ 8 / 0.012 = 4.6875 * 10 ^ 10 Pa