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 = (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 = (Pl)/(AE)`

`\delta = ((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 = (l_(a))/(l_(s))`

Where,

`\upsilon =`Poission's ratio

`l_(a)`= Lateral strain

`l_(s)`= Linear strain

`\upsilon = (\delta/d)/(\delta/l)`

`0.3 = (\delta/22)/(0.1973/200)`

`0.3((0.1973)/(200)) = (\delta)/(22)`

`\delta = 6.5109*10^(-3)mm`

Therefore the change in diameter of the rod is
`6.5109*10^(-3)mm`