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

Answer:

7.65 mm

Explanation:

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

Also, $\sigma=E\times \frac {\triangle L}{L}$

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

Therefore,

$\frac {F}{A}= E\times \frac {\triangle L}{L}$

Making A the subject of the formula then

$A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}$

Since $A=\frac {\pi d^{2}}{4}$ then  

$d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm$