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Stress Analysis

  1. Dec 19, 2008 #1
    An unloaded alumininium bar is 1.5m long and has a width and depth of 75mm and 25mm respectively, at room temperature. Determine the following when the bar is subjected to an axial tensile load of 180KN.

    a) The tensile stress
    b) The tensile strain
    c) The lateral strain
    d) The change in width and depth of the cross-section
    e) The change in length of the bar
    f) If the temperature of the unloaded bar is now raised by 200 degrees Celsius, determine its new length.

    Take Young's modulus, E = 65GPa, Poisson's ratio, v = 0.33,, Coefficient of thermal expansion, α = 23.4 x 10-6
     
  2. jcsd
  3. Dec 19, 2008 #2
    The answers i calculated were,
    a) 1.6MPa
    b) 2.46 x 10-5
    c) 8.118 x 10-6
    d) change in width = 1.845x10-6m
    Could not calculate depth
    e) 1.2177x10-5m
    f) 7.02 x 10-3m

    But most of the values I have calculated seem too small to be correct.
     
  4. Dec 19, 2008 #3

    Mapes

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    Hi jaymar023, welcome to PF. You're running into problems right away at (a). How do you calculate stress from load and cross-section area?
     
  5. Dec 19, 2008 #4
    Stress = Load/Area
     
  6. Dec 19, 2008 #5

    Mapes

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    Agreed. And the relevant area is?
     
  7. Dec 20, 2008 #6
    Width x Depth = 75 x 10-3 x 25 x 10-3
     
  8. Dec 20, 2008 #7
    So 180 x 103 / Area = 96MPa
     
  9. Dec 20, 2008 #8

    Mapes

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    Looks good.
     
  10. Dec 20, 2008 #9

    nvn

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    (a) Nice work, jaymar023. Alternately, leaving all units in N and mm (since 1 MPa = 1 N/mm^2), that would be sigma = P/A = (180 000 N)/[(75 mm)(25 mm)] = 96.0 MPa.

    By the way, there should always be a space between the numeric value and its following unit symbol. See international standard for writing units; i.e., ISO 31-0.

    Your approach looked correct on items b and c, so post b and c again using your corrected answer for item a. And post d, e, and f again using your new answers. Change in width and depth of the cross section would just be lateral strain times width or depth, right?
     
  11. Dec 22, 2008 #10
    a) 96 MPa
    b) 1.48 x 10-3
    c) 4.88 x 10-4
    d) Change in width = 4.88 x 10-4 x 25 x 10-3 = 1.22 x 10-5 m
    Change in depth = 4.88 x 10-4 x 75 x 10-3 = 3.66 x 10-5 m
    e) 2.22 x 10-3 m
    f) 7.02 x 10-3 m
     
  12. Dec 22, 2008 #11
    I am still not sure that d) and e) are correct
     
  13. Dec 22, 2008 #12

    nvn

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    jaymar023: Items a through e look correct, except c and d should be negative, because they are compressive strain and contraction. I didn't get the same answer as you got on item f. Notice question f is asking for the new length, not the change in length.
     
  14. Dec 23, 2008 #13
    So for f) the answer should be 8.52 m?
     
  15. Dec 23, 2008 #14

    nvn

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    No, that's not right. Keep trying.
     
  16. Dec 30, 2008 #15
    What about 1.50702 m ?
     
  17. Dec 30, 2008 #16

    nvn

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    That's correct. Nice work.
     
  18. Jan 6, 2009 #17
    thank you for you help, much appreciated.
     
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