Question

A 6.1-m-long solid circular metal rod (E = 150 GPa) must not stretch more than 10 mm when a load of 109 kN is applied to it. Determine the smallest diameter rod that should be used. State your answer in mm.

Answer 1

To solve this problem we will apply the concepts related to the deformation of a body and the normal effort, from which we will obtain the area. From this area applied to the geometric concept of a circular bar we will find the diameter.

The deformation equation in a rod tells us that

$\delta = \frac{PL}{AE}$

Here,

P = Load

L = Length

A = Cross-sectional area

E = Elastic Modulus

Rearranging the Area,

$A = \frac{PL}{\delta E}$

Replacing we have that the area is,

$A = \frac{(109*10^{3})(6.1)}{(10*10^{-3})(150*10^9)}$

$A = 0.000443266m^2$

$A = 44.32*10^{-6}m^2$

Using the geometric expression for the Area we have,

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

$d = \sqrt{\frac{4A}{\pi}}$

$d = \sqrt{\frac{4(44.32)}{\pi}}$

$d = 7.51mm$

Therefore the smalles diameter rod is 7.51mm