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 6640 N (1493 lbf). If the length of the rod is 380 mm (14.96 in.), what must be the diameter to allow an elongation of 0.50 mm (0.01969 in.)?

Answer 1

Answer:

The diameter is 7.64 mm.

Explanation:

Given that,

Force = 6640 N

Length = 380 mm

Elongation = 0.50 mm

Modulus of elasticity =  110 GPa

We need to calculate the area

Using formula of stress

$\sigma=\dfrac{F}{A}$....(I)

$\sigma=\dfrac{E\Delta l}{l}$....(II)

From equation (I) and (II)

$\dfrac{F}{A}=\dfrac{E\Delta l}{l}$

$A=\dfrac{Fl}{E\Delta l}$

Put the value into the formula

$A=\dfrac{6640\times380\times10^{-3}}{110\times10^{9}\times0.50\times10^{-3}}$

$A=4.587\times10^{-5}\ m^2$

We need to calculate the diameter

$A=\dfrac{\pi d^2}{4}$

$4.587\times10^{-5}=\dfrac{\pi d^2}{4}$

$d=\sqrt{\dfrac{4\times A}{\pi}}$

$d=\sqrt{\dfrac{4\times4.587\times10^{-5}}{\pi}}$

$d=7.64\ mm$

Hence, The diameter is 7.64 mm.