To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as
[tex]\delta = \frac{Pl}{AE}[/tex]
Where,
P = Tensile Force
L= Length
A = Cross sectional Area
E = Young's modulus
PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then
[tex]\delta = \frac{Pl}{AE}[/tex]
[tex]\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}[/tex]
[tex]\delta = 0.1973mm[/tex]
Therefore the elongaton of the rod in a 200mm gage length is [tex]0.1973mm[/tex]
PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,
[tex]\upsilon = \frac{l_{a}}{l_{s}}[/tex]
Where,
[tex]\upsilon =[/tex]Poission's ratio
[tex]l_{a}[/tex]= Lateral strain
[tex]l_{s}[/tex]= Linear strain
[tex]\upsilon = \frac{\delta/d}{\delta/l}[/tex]
[tex]0.3 = \frac{\delta/22}{0.1973/200}[/tex]
[tex]0.3(\frac{0.1973}{200}) = \frac{\delta}{22}[/tex]
[tex]\delta = 6.5109*10^{-3}mm[/tex]
Therefore the change in diameter of the rod is [tex]6.5109*10^{-3}mm[/tex]
