Question

Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

Answer 1

`strain = 1.4 * 10^(-3)`

Step-by-step explanation:

As we know by the formula of elasticity that

`E = (stress)/(strain)`

now we have

`E = 110 GPA`

`E = 110 * 10^9 Pa`

Area = 15.2 mm x 19.1 mm

`A = 290.3 * 10^(-6)`

now we also know that force is given as

`F = 44500 N`

here we have

stress = Force / Area

`stress = (44500)/(290.3 * 10^(-6))`

`stress = 1.53 * 10^8 N/m^2`

now from above formula we have

`strain = (stress)/(E)`

`strain = (1.53 * 10^8)/(110 * 10^9)`

`strain = 1.4 * 10^(-3)`