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Homework Help: Finding axial extension in steel bar

  1. May 3, 2010 #1
    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


    [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] = 40x103/((.12x pi)/4) = 50265482.46

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

    therefore, e = 50265482.46/ 210x109 = 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
  2. jcsd
  3. May 3, 2010 #2


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    Homework Helper

    First of all, the area equals D^2 x PI / 4, where D is the diameter. You calculated as if it were the radius.
  4. May 3, 2010 #3
    Thanks for the quick reply

    I did new calculations and still am getting a wrong answer,

    I found the [tex]\sigma[/tex] = 1273239.545

    the strain = 6.063045452x10-6

    and the extension I found to be .0303 mm

    I reckon it has to do with the value I am using for E but it wasn't given in the question so I am not sure, any thoughts?
  5. May 3, 2010 #4


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    Homework Helper

    Make sure your units are consistent - use milimeters for length and newtons for force (i.e. megapascals for stress), and convert the assumed steel "Gpa" to "Mpa".
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