Question

2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 and E 5 200 GPa, determine (a) the elongation of the rod in a 200-mm gage length, (b) the change in diameter of the rod.

Answer 1

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

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

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

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

$\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}$

$\delta = 0.1973mm$

Therefore the elongaton of the rod in a 200mm gage length is $0.1973mm$

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

$\upsilon = \frac{l_{a}}{l_{s}}$

Where,

$\upsilon =$Poission's ratio

$l_{a}$= Lateral strain

$l_{s}$= Linear strain

$\upsilon = \frac{\delta/d}{\delta/l}$

$0.3 = \frac{\delta/22}{0.1973/200}$

$0.3(\frac{0.1973}{200}) = \frac{\delta}{22}$

$\delta = 6.5109*10^{-3}mm$

Therefore the change in diameter of the rod is $6.5109*10^{-3}mm$