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

Strain = 4.139 x 10^(-3)

Step-by-step 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

The resultant strain in the aluminum specimen is
`4.14 * {10^( - 3)}`

Step-by-step explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

`A=9.5* 12.9=123.84* 10^(-6)\ mm^2A=122.55* 10^(-6)\ m^2`

Tension acting on object, T = 35000 N

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

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

`\epsilon=(\sigma)/(E)`

`\sigma` is the stress

`\epsilon=(F)/(EA)\\\epsilon=(35000)/(69* 10^9* 122.5* 10^(-6))\\\epsilon=4.14* 10^(-3)`

Thus, The resultant strain in the aluminum specimen is
`4.14 * {10^( - 3)}`