Question

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain.

Answer 1

0.00119

Step-by-step explanation:

F = Force = 44500 N

A = Area =
`0.0152* 0.0191\ m^2`

E = Young's modulus of copper =
`128* 10^(9)\ Pa`

Stress is given by

`\sigma=(F)/(A)`

Strain is given by

`\epsilon=(\sigma)/(E)\\\Rightarrow \eplison=((F)/(A))/(E)\\\Rightarrow \epsilon=((44500)/(0.0152* 0.0191))/(128* 10^(9))\\\Rightarrow \epsilon=0.00119`

The resulting strain is 0.00119