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Axial stress

  • Thread starter SenseAO
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  • #1
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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.
 

Answers and Replies

  • #2
vela
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Start by looking up the definition of stress.
 
  • #3
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Yes i have and i get the picture in my head. But i just cant place where the 16mm goes.
 
  • #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.
 
  • #5
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F/A?

Force being 25kn and the area being 16mmx16mm?
 
  • #6
vela
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Good. What precisely does A stand for? It's an area, but the area of what?
 
  • #7
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The cross sectional area of the beam?
 
  • #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?
 
  • #9
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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?
 
  • #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.
 
  • #11
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The 16mm corresponds to the bar
 
  • #12
vela
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The length of the bar?
 
  • #13
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yes, uniformly accross the bar
 
  • #14
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Figured it out.
 

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