Question

A circular rod with a gage length of 3.2 mm and a diameter of 2 cmcm is subjected to an axial load of 57 kNkN . If the modulus of elasticity is 200 GPaGPa , what is the change in length?

Answer 1

To solve this problem we will apply the concepts related to the change in length given by the following relation,

$\delta_l = \frac{Pl}{AE}$

Here the variables mean the following,

P = Load

l = Length

A = Area

E = Modulus of elasticity

Our values are,

$l = 3.2 m$

$\phi = 2cm = 0.02m$

$P = 57kN = 57*10^3N$

$E = 200Gpa$

We can obtain the value of the Area through the geometrical relation:

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

Replacing,

$A = \frac{\pi}{4} (0.02)^2$

$A = 3.14*10^{-4}m^2$

Using our first equation,

$\delta_l = \frac{Pl}{AE}$

$\delta_l = \frac{(57*10^3)(3.2)}{(3.14*10^{-2})(200*10^9)}$

$\delta_l = 0.000029044m$

$\delta_l = 0.029044mm$

Therefore the change in length is 0.029mm