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Effect of thermal expansion on restraints

  1. Oct 5, 2009 #1
    When a beam undergoes thermal expansion and the restrains that support the beam are fixed so that no movement is allowed, how do you calculate the force reactions within the restraints? The restraints are two walls, and before expansion occurs no horizontal reaction forces are present.

    Since the amount by which the beam is deflected would be equal to the amount by which the beam thermally expands could it be determined with the deflection formula Delta=PL/AE?
     
  2. jcsd
  3. Oct 5, 2009 #2

    minger

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    Hint: think superposition

    Assume that there is intially only one wall, and then a force that pushes the beam back "into place".
     
  4. Oct 5, 2009 #3
    ok but do I use the formula i mentioned above to determine the Force that does that?
     
  5. Oct 6, 2009 #4

    minger

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    Yes, assume that the loading happens in two steps. The principle of superposition states that
    In layman's terms, it means that you can take the effect of each load component and add them up to get the sum. Since your bar is fully constrained, the sum effect is no displacement.

    So, use your formula to find the displacement that a certain delta T would cause. Then, using stress/strain equations, determine the force and stress required to "push" the beam back into place. That will be your second loading.
     
  6. Oct 7, 2009 #5
    Delta=PL/AE doesn't have anything to do with delta T though.
    Do you mean DeltaTx(Alpha)x(L)=Displacement?
    So I dont use Delta=PL/AE at all?

    For the stress strain equation would i just use
    y=mx
    where m = young's modulus
    x=percentage strain
    and y=stress?

    How would the method for doing this change if the first 1/3 of the Beam was made of a material with E=100,000 and the next 2/3 were made of a material with E=200,000?
     
    Last edited: Oct 7, 2009
  7. Oct 7, 2009 #6

    minger

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    I didn't check your equations, I was simply saying that the thermal expansion causes a change in length.
    [tex]
    \Delta x = l \alpha \Delta T[/tex]
    The constraints will then exert a force back on the beam with the magnitude that negates expansion. I'm not going to do it for you, but your PL/AE equation may come in handy.
     
  8. Oct 7, 2009 #7
    Would the method for doing this change if the first 1/3 was made of brass of a different cross sectional area to the 2/3 of the beam which is made of steel?
     
  9. Oct 8, 2009 #8

    minger

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    The method is going to stay the same, but the implementation would be a little trickier.
     
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