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 370 mm (14.57 in.), what must be the diameter to allow an elongation of 0.53 mm (0.02087 in.)?

Answer 1

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

$\Delta L=\dfrac{PL}{AE}$

$A=\dfrac{PL}{\Delta LE}$

$A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}$

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

Answer 2

Answer:

$d=7.32\ mm$

Explanation:

Given:

• Young's modulus, $E=110\times 10^3\ MPa$

• yield strength, $\sigma_y=240\ MPa$

• load applied, $F=6640\ N$

• initial length of rod, $l=370\ mm$

• elongation allowed, $\Delta l=0.53$

We know,

Stress:

$\sigma=\frac{F}{A}$

where: A = cross sectional area

Strain:

$\epsilon = \frac{\Delta l}{l}$

& by Hooke's Law within the elastic limits:

$E=\frac{\sigma}{\epsilon}$

$\therefore 110\times 10^3=\frac{F}{A}\div \frac{\Delta l}{l}$

$\therefore 110\times 10^3=\frac{6640\times 4}{\pi.d^2}\div \frac{0.53}{370}$

where: d = diameter of the copper rod

$d=7.32\ mm$