A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 35000 N (7868 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.

Strain = 4.139 x 10^(-3)

Explanation:

We are given;

Dimension; 9.5 mm by 12.9 mm = 0.0095m by 0.0129m

Elastic Modulus; E = 69GPa = 69 x 10^(9)N/m²

Force = 35,000N

Now, Elastic modulus is given by;

E = σ/ε

Where

E is elastic modulus

σ is stress

ε is strain

Now, stress is given by the formula;

σ = F/A

Area = 0.0095m x 0.0129m = 0.00012255 m²

Thus, σ = 35000/0.00012255 = 285597715.218 N/m²

Now, we are looking for strain.

Let's make ε the subject;

E = σ/ε, Thus, ε = σ/E = 285597715.218/69 x 10^(9) = 0.00413909732 = 4.139 x 10^(-3)

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2