(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A cylindrical steel bar, 20-mm in diameter and 5.0-m in length, is subjected to an axial tensile load of 40-kN applied at the ends of the steel bar. Determine the axial extension, in millimeters, of the steel bar under this loading condition?

2. Relevant equations

E=[tex]\sigma[/tex]/e

[tex]\sigma[/tex] = F/A

e= change in L / Lo

3. The attempt at a solution

To find the extension we need to find (change in Lo)

[tex]\sigma[/tex] = 40x10^{3}/((.1^{2}x pi)/4) = 50265482.46

Even though not given in the question I am presuming E for steel = 210Gpa

therefore, e = 50265482.46/ 210x10^{9}= 2.393594403x10-4

as e = (change in L/Lo), the change in lenght = (2.393594403x10-4 x 5) = 1.2x10-3 m and 1.2 mm

This is a multiple choice question and this is not one of the answers, can anyone see where I am going wrong?

ANSWERS: 5x10-5 , 0.05, 3.2x10-3, 5.0, 3.2

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# Homework Help: Finding axial extension in steel bar

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