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Thickness of object to halve the stress

  1. May 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Untitled_zps6etin2kv.png

    The diagram represents a steel tube with wall thickness w which is small in comparison with the diameter of the tube. The tube is under tension, caused by a force T, parallel to the axis of the tube. To reduce the stress in the material of the tube, it is proposed to thicken the wall. The tube diameter and the tension being constant, which wall thickness gives half the stress?

    2. Relevant equations
    stress = Force / Area

    3. The attempt at a solution
    (The answer to this question is 2w)

    From formula stress = force / area, to halve the stress we have to double the area. But I don't really understand the effect of changing the thickness.
    If we double the thickness, will the diameter change? In my opinion the diameter won't change because the wall thickness w is small in comparison with the diameter so changing the thickness won't change the stress

    Thanks
     
  2. jcsd
  3. May 22, 2016 #2

    Merlin3189

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    What area do you need to change?
     
  4. May 22, 2016 #3
    Area of circle
     
  5. May 22, 2016 #4

    Merlin3189

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    Well, I don't think so. Not all of the circle is providing tension / resisting stress. The circle has two parts - the metal wall of thickness w and a lot of air (or maybe some other gas or liquid) inside.
     
  6. May 22, 2016 #5

    Merlin3189

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    I have to go out for a while now, so lets try to clarify a bit.
    Yes, the diameter won't change. That is , as you say, implied by their statement that D is large compared to w.
    But changing w does change the stress.
    The diameter is nearly all air or other insubstantial stuff incapable of providing tension. All of the tension is provided by the metal. If you change the area of the metal, you change the stress.
    The area of the metal is not πR2 nor πD2/4, because the metal is not a circle.
    Looking, as we are, at the cross-section of the pipe, the metal forms a ring or annulus. You can imagine it unwrapped to be just a strip, which would be approximately a rectangle.
     
  7. May 24, 2016 #6
    If we unwrapped the metal to be just a strip, the shape will be rectangle and the thickness will be the height while the circumference will be the length of rectangle. So doubling the thickness will double the area and halve the stress.

    So the tube in the question is open at both ends? The shape only consists of circular part of cylinder without base and top part?
     
  8. May 24, 2016 #7

    Merlin3189

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    Yes.
    But the ends would not matter in calculating the stress in the middle portion of the tube. (And it does say tube, not cylinder nor bar nor rod.) There the stress is all in the thin wall of the tube.
     
  9. May 24, 2016 #8
    Ok thanks a lot
     
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