Question

A specimen of aluminum having a rectangular cross section 9.9 mm × 12.4 mm (0.3898 in. × 0.4882 in.) is pulled in tension with 35900 N (8071 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.

Answer 1

Step-by-step explanation:

Given:

Dimensions:

Length, l = 9.9 mm

= 0.0099 m

Width, w = 12.4 mm

= 0.0124 m

Force, f = 35900 N

Young modulus, E = 69 GPa

= 6.9 × 10^10 Pa

Young modulus, E = stress/strain

Stress = force/area

Area of a rectangle = length, l × width, w

= 0.0099 × 0.0124

= 1.2276 × 10^-4 m^2

Therefore,

Strain = stress/young modulus

= (35900/1.2276 × 10^-4)/6.9 × 10^10

= 4.24 × 10^-3 m/m

= 0.00424 m/m