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Axial stress

  1. Oct 24, 2011 #1
    A steel bar that is 16mm in diameter is resisting a force of 25kN. What is the normal axial stress in the bar(MPa)


    Since MPa = N sqmm would it be 25kn/(0.016m x 0.016m)

    Im more after the process on how to get the answer than the answer itself.
     
  2. jcsd
  3. Oct 24, 2011 #2

    vela

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    Start by looking up the definition of stress.
     
  4. Oct 24, 2011 #3
    Yes i have and i get the picture in my head. But i just cant place where the 16mm goes.
     
  5. Oct 24, 2011 #4

    vela

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    What quantities goes into calculating the stress?

    Perhaps you're simply missing the implication of the use of the word diameter.
     
  6. Oct 24, 2011 #5
    F/A?

    Force being 25kn and the area being 16mmx16mm?
     
  7. Oct 24, 2011 #6

    vela

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    Good. What precisely does A stand for? It's an area, but the area of what?
     
  8. Oct 24, 2011 #7
    The cross sectional area of the beam?
     
  9. Oct 24, 2011 #8

    vela

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    Right. The cross-sectional area depends on the shape of the cross section. A=0.016m x 0.016m would work if the cross section were square, but is that the case here?
     
  10. Oct 24, 2011 #9
    Does not specify. So than we would just do 25kn/0.016m = to give us 1562.5kNm converting this to Nmm would give us 1562 x 10^6 Nmm

    am i somewhat right?
     
  11. Oct 24, 2011 #10

    vela

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    No, that's not correct. For one thing, when you divide, the units divide as well, so you end up with kN/m, not kN m.

    What does the 16-mm given correspond to? The problem statement implies what the shape is.
     
  12. Oct 24, 2011 #11
    The 16mm corresponds to the bar
     
  13. Oct 24, 2011 #12

    vela

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    The length of the bar?
     
  14. Oct 24, 2011 #13
    yes, uniformly accross the bar
     
  15. Oct 24, 2011 #14
    Figured it out.
     
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