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Axially Loaded Member and Tensile Stress oh so lost

  1. Dec 11, 2006 #1
    1. The problem statement, all variables and given/known data
    "The axial load for a given test sample carries 1590 N. Calculate tensile stress at sections (1) and (2) assuming the sample thickness is 5mm. (Rectangular cross section).

    http://img208.imageshack.us/img208/121/tensilestressby1.th.jpg [Broken]

    2. Relevant equations
    normal stress = P/A
    Pallow = (normal stress)allow * A

    3. The attempt at a solution
    This is a new one for me. All the previous homework problems involved a prismatic bar (a bar of uniform cross-section) and not this 'necked' bar.

    Since P are pulling forces, the bar is in tension.

    I know my first step would be to convert 5mm into meters = 0.005m

    But then I'm sorta lost.

    Do I just use the normal stress equation above for each section; 1 and 2? Keeping P = 1590N and then changing A for section 1 to 30mm = 0.03m, and for section 2 to 10mm = 0.01m?

    But then where does 5mm come into play?

    Thanks for all the help.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Dec 12, 2006 #2


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    Science Advisor

    You will use the cross sectional area at both locations to calculate the normal stresses. The 5mm is required to calculate the area. The 30 mm and 10 mm dimensions are not areas. How do you calculate the area of a rectangle?
  4. Dec 12, 2006 #3
    Area of a rectangle is length * width

    So at section 1, the area A would be 30mm*5mm = 150mm = 0.15m ?
    Section 2, the area A would be 10mm*5mm = 50mm = 0.05m ?

    Then I would just plug those areas into the normal stress = P/A equation for both sections?

    Thanks for the quick reply
  5. Dec 12, 2006 #4
    Those above areas should read m^2 and not just meters.
  6. Dec 12, 2006 #5
    normal stress of section 1 = 1590N / 0.15m^2 = 10600 N/m^2
    normal stress of section 2 = 1590N / 0.05m^2 = 31800 N/m^2

    Could it be that easy? Feel like I'm missing something still
  7. Dec 12, 2006 #6


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    It is in this case. You will never see an easier theoretical state of stress than that of a uniaxial tension test. There is no shear until you start dealing with transformations and Mohr's Circle stuff.
  8. Dec 12, 2006 #7
    Wow, well I expected that to be worse.

    Thank you very much
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