Question

A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6660 N (1500 lbf). If the length of the rod is 380 mm (15.0 in.), what must be the diameter to allow an elongation of 0.50 mm (0.020 in.)?

Answer 1

7.65 mm

Step-by-step explanation:

Stress,
`\sigma=\frac {F}{A}` where F is the force and A is the area

Also,
`\sigma=E* \frac {\triangle L}{L}`

Where E is Young’s modulus, L is the length and
`\triangle L` is the elongation

Therefore,

`\frac {F}{A}= E* \frac {\triangle L}{L}`

Making A the subject of the formula then

`A=\frac {FL}{E\triangle L}=\frac {6660* 380}{110* 10^(9)* 0.5}=4.60145* 10^(-5) m^(2)`

Since
`A=\frac {\pi d^(2)}{4}` then

`d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4* 4.60145* 10^(-5)}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm`