Question

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.

Answer 1

Answer:

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)

Answer 2

Answer:

The resultant strain in the aluminum specimen is $4.14 \times {10^{ - 3}}$

Explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

$A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2$

Tension acting on object, T = 35000 N

The elastic modulus for aluminum is,$E=69\ GPa=69\times 10^9\ Pa$

The stress acting on material is proportional to the strain. Its formula is given by :

$\epsilon=\dfrac{\sigma}{E}$

$\sigma$ is the stress

$\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}$

Thus, The resultant strain in the aluminum specimen is $4.14 \times {10^{ - 3}}$