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Stress Due to Thermal Gradient

  1. Feb 5, 2014 #1
    I am trying to understand stresses that are induced by thermal gradients. Now, I can think of a hundred different questions to ask, but I want to take baby steps to get there. Let's just talk about a simple cantilever beam in the x-y plane where the x-axis is the beam's longitudinal axis and the y-axis is parallel to its height. Z points 'into the page' along its thickness.

    Let's say that there is a temperature gradient in the x-direction. The beam will want to elongate due to thermal expansion, but the gradient will cause it to want to elongate by different amounts. I am assuming that this will only induce an axial stress/strain.

    What if I want to do a simple back-of-envelope calculation to approximate the equivalent axial load required to induce this stress? I was assuming that I can just take the worst ΔT and use:

    ΔL/L = ε = αΔT
    σ = P/A = εE
    so that:
    P/(AE) = αΔT
    ∴ P = AEαΔT​

    Any thoughts or issues with this approach?
     
  2. jcsd
  3. Feb 5, 2014 #2

    AlephZero

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    Your math is OK if the temperature of the whole beam changes by ΔT.

    If "thermal gradient" means different parts of the beam are at different temperatures, you need to work out the total change in length by integrating the thermal strains αΔT along the length of the beam.
     
  4. Feb 5, 2014 #3
    Hi Aleph. OK. So what I am looking to do is to find a way to hand-calculate this to just get a rough number. Something to let me sanity check some ANSYS results.

    Let's assume now that a cantilevered beam starts at Tref. In the real world, if thermal BCs are applied, it is likely that the beam will end up with a continuous temperature distribution. But to simplify things, let's just say that the beam is length 2L and the resulting temperature field is such that the first half of the beam (0≤x<L) is at T1 and the second half of the beam (L<x≤2L) is at T2. This is obviously not possible, but I think it will illustrate the idea. Let T2>T1.

    Let:
    ΔT1=T1-Tref
    ΔT2=T2-Tref

    We know that the first half of the beam wants to expand by Δx1=LαΔT1 and the second by LαΔT2.

    Now right at x=L there is an issue. Immediately to the left at x=L- the beam wants to expand by Δx1 and immediately to the right at x=L+ it wants to expand by Δx2.

    So the right half wants to grow more than the left half. The left half tries to keep up, but it can't and falls behind by Δx=Δx2-Δx1. So we have:

    Δx=Δx2-Δx1
    →Δx=LαΔT2 - LαΔT1
    →Δx=Lα(T2 - Tref) - Lα(T1 - Tref)
    →Δx=Lα(T2 - T1).

    This would be the net expansion of the beam. So the net strain could be used to find an equivalent load using the logic from the OP.

    I know this all assumes constant α's and E's, but as a 1st order approximation, I would think this could give me a worst-case equivalent load if I let T2 be the highest temp from the actual distribution and let T1 be the lowest temp from the actual distribution.

    Thoughts?
     
  5. Feb 5, 2014 #4
    Here is a secondary question that I think is really more important for me to understand. If a bar is completely unrestrained and we heat it up from uniform T1 to uniform T2 it will have a thermal "strain" associated with. But is it really a "strain" in the sense that there is an associated stress? I think the answer is "no" since it is unrestrained.

    Now I ask the same exact question except that instead we go from uniform T1 to a non-uniform temperature distribution (i.e. we have a gradient). Even though the bar is externally non-restrained, I think that we do end up with an induced stress since it cannot all expand the same. So there are restraints in that sense.

    Does this sound right?
     
  6. Feb 5, 2014 #5

    SteamKing

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    Look at it this way: if you were to apply heat to the top of the bar all along its length, while keeping the bottom of the bar at the reference temperature, the elongation of the fibers in the top of the bar, relative to the length of the fibers in the bottom of the bar, are going to warp the bar into a curved shape. Now, if the top of the bar were restrained so that the length of the fibers could not increase due to the change in temperature, then compressive stresses must build up in the top fibers of the bar to counteract the increase in length due to thermal expansion.
     
  7. Feb 5, 2014 #6

    AlephZero

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    I think you are confusing strain with displacement. Assuming the end at 2L is free, the strain in the left half is αΔT1 and the axial displacement at a distance x along the left half of the beam is u = αΔT1x. In other words the beam "grows" in a linear manner from the fixed end.

    The discontinuity in strain doesn't matter. That is no different to joining two rods of different materials end to end, or two rods of the same material but different areas.

    For the right half, the strain is αΔT2 and the displacement in the right half will be
    αΔT1L + αΔT2(x-L).
    In other words the right half just "grows" from the center point, but at a different rate.

    If the right hand end is restrained or has an axial force applied to it, then there is an elastic strain ε as well as the thermal strains. If Young's modulus is constant (independent of temperature), ε will be constant for the whole length of the beam. In this case the displacement in the left half would be u = (αΔT1 - ε)x and the displacement in the right half would be
    (αΔT1 - ε)L + (αΔT2 - ε)(x-L). (Eq. 1)

    If there is an axial force on the end of the beam, you can find the elastic stress in the beam = F/A and the elastic strain ε = F/(AE).

    If there is a constraint on the displacement at the right hand end of the beam, Eq. 1 gives the displacement at the end and you can solve that to find ε, and then the axial stress = εE and the axial reaction force at the end of the beam is εEA.
     
    Last edited: Feb 5, 2014
  8. Feb 5, 2014 #7

    AlephZero

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    Yes it is a strain, because the definition of strain is just (change in length)/(original length). It doesn't matter why the length changes.

    But it doesn't have a stress associated with it, because it is not an elastic strain.

    There may be other types of strain in the structure as well. e.g. plastic strain, creep strain, etc.

    The bottom line is that configuration of the bar has to satisfy several conditions:
    1. For the stress field, every small volume has to be in equilibrium between forces the on its boundaries (from the elastic stresses, and the applied loads) and any internal forces acting on the volume (e.g. its weight).
    2. The strain field has to satisfy the compatibility equations, which basically means the material has to "fit together" properly without any internal holes or other geometrical discontinuities.
    3. The stresses and strains have to satisfy the constitutive equations for the material (e..g for a linear isotropic material, the relations between the stresses and strains, and Young's modulus and Poisson's ratio)

    The thermal strains are similar to prescribed displacements at the boundary - they only depend on the temperature field, not on what happens to the structure.

    So in general, you are right, to satisfy conditions 1, 2, and 3, there will be some elastic strains, and therefore some stresses, as well as the thermal strains.

    If the compatibility equations (condition 2) are satisfied, you can integrate the strains to get the displacements, without any problems about getting inconsistent results depending what path you take through the structure from one position to another.

    Wow, that's a complete course in continuum mechanics, in one post - everything else is just details :smile:
     
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