Question

A specimen of aluminum having a rectangular cross section 9.5 mm × 13.0 mm (0.3740 in. × 0.5118 in.) is pulled in tension with 35600 N (8003 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: 〖4.1x10〗^(-3)

Step-by-step explanation:

From the question, the parameters given are:

• Force= 35600N(8003lbf)

• Cross section (area)= 9.5mm x 13mm= 123.5 〖mm〗^2 = 0.0001235m^2 (after converting from 〖mm〗^2 to m^2 which is the standard unit for measuring length)

• Modulus of elasticity = 69 GPa = 69 x 〖10〗^9 N/m^2

We can solve this using the formula for elastic modulus ( E ) = stress/strain

making strain the subject of the formula the equation above becomes strain= stress/ elastic modulus

Where strain= change in length/ length

And

Stress= force/ area

now let’s calculate value for stress

Stress= 35600/0.0001235=288,259,109 N/m^2

substituting the value for stress into the equation strain= stress/ elastic modulus we have

Strain= 288,259,109 / 69 x 〖10〗^9 = 〖4.1x10〗^(-3)