A strain, tension problem (materials of engineering)?

Question

A specimen of aluminum having a rectangular cross section 10mm x 12.7mm (.4in x .5in.) is pulled in tension with 35,500N (8000 lbs) force, producing only elastic deformation. Calculate the resulting strain.

Note: i've done stress problems but not strain..please help.

Answer 1

The strain in a member due to an applied load is equal to the Stress in the member divided by the Modulus of Elasticity of the material.

In your case, the stress is 8,000 pounds/0.4*0.5 square inches = 40,000psi.

There fore the strain is = 40,000/Modulus of Elasticity of aluminum.

I do not have a value of the elastic modulus of aluminum handy, but I think it is around 14,000,000
The Modulus of Elasticity of steel is 29,000,000

Answer 2

The youngs modulus for a material is given by:

E = Stress / Strain

Now, the youngs modulus of aluminium is 7x10^10. You should also know that stress is given by Stress=F/A, where F is the force applied (tentile or compressive) and A is the cross sectional area.

In this case, S=279527559.1 Nm^-2. Put that into the first equation, and then solve for the strain:

7x10^10 = 279527559.1 / Strain

Therefore, the strain is 0.00399