Question

A specimen of aluminum having a rectangular cross section 10 mm x 12.7 mm is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain.

Answer 1

The resulting strain is 4.05 x 10⁻³

Step-by-step explanation:

Given;

dimension of the specimen = 10 mm x 12.7 mm

Cross sectional area of the aluminum specimen = 10 mm x 12.7 mm = 127 mm² = 1.27 x 10⁻⁴ m²

applied force, F = 35,500 N

Young's modulus is given by;

`E = (Stress)/(Strain)\\\\E = (F)/(A(strain))\\\\E = (F)/(A(\epsilon))\\\\\epsilon = (F)/(A E)\\\\`

Where;

ε is the resulting strain

E is Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

`\epsilon = (35500)/((1.27*10^(-4)) (69*10^(9)))\\\\ \epsilon = 4.05*10^(-3)`

Therefore, the resulting strain is 4.05 x 10⁻³