Question

A 7.3-mm-diameter rod of 3003-H14 aluminum alloy with an elastic modulus of 51MPa and a poisson's ratio of 0.3 is subjected to a 4-KN tensile load. Calculate the resulting rod diameter to 5 significant figures. Answer:

Answer 1

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

Area of cross-section = A = ?

Diameter of rod = d = 7.3mm = 0.0073m

(0.0073)2

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

Tensile stress = $\sigma$ = ?

Tensile load = P = 4 kN = 4000 N

$\sigma = \frac{4000}{4.1854*10^{-05}}$

$\sigma = 95.5703 * 10^{6} Pa = 95.5703 MPa$

$E = \frac {\sigma}{Longitudinal strain}$

$Longitudinal strain = \frac {\sigma}{E}$

Elastic modulus = E = 51 GPa = 51000 MPa

$Longitudinal strain = \frac {95.5703}{51000}$

$\mu = \frac {Lateral strain}{Longitudinal strain}$

$Lateral strain = \mu* Longitudinal strain$

Poisson's ratio = $\mu$ = 0.3

Lateral strain = (Change in diameter)/d

Change in diameter = Lateral strain * d

Change in diameter = (5.6217*10^(-4))*7.3

Change in diameter = 4.1038*10^(-3)mm

New diameter = d + Change in diameter

New diameter = 7.3 - 4.1038*10^(-3)

New diameter = 7.2959 mm