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Inverse bulk modulus / compression or shear?

  1. Aug 9, 2017 #1
    IMG_2144.jpg IMG_2145.jpg




    Hi, I'm afraid I not very good at these questions just yet and would like a walk through a bit better than the one I was given by my tutors.

    Thank you, please refer to the inline image.

     
  2. jcsd
  3. Aug 9, 2017 #2

    scottdave

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    What is your question? Who did the handwritten working out - you or the tutor?
    What they gave you (Compressive Strength) is the maximum amount of stress that the concrete can handle. Obviously the maximum stress will occur at the bottom, with the weight of all the concrete above it pushing down.
     
  4. Aug 9, 2017 #3
    The tutor wrote that out
    I think the compressive strength is the inverse of the bulk modulus?
    Could someone walk me through the problem solution? A few steps are skipped and the reasoning is not explicitly stated.
     
  5. Aug 10, 2017 #4

    scottdave

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    The bulk modulus would allow you to calculate the amount volume strain, resulting from a given change in pressure ( ΔP ). The volume strain is dimensionless - think of it like a percentage change. If you draw a stress strain curve, think of the modulus as the slope of the curve in the linear region, and the Compressive strength as the point on the curve where it starts to fail.
     
  6. Aug 10, 2017 #5
    Sorry to ask but could you please do the work they did or annotate it? It would help
     
  7. Aug 10, 2017 #6

    scottdave

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    Look at a flat square meter at the bottom. How much weight os needed to cause the maximum pressure (maximum strength)? Using g find the amount of mass necessary. Use density to find the height of a column of concrete.
     
  8. Aug 10, 2017 #7
    Neither the bulk modulus nor the shear modulus nor the Young's (tensile) modulus determine the failure behavior of a material. The failure behavior of a material is not the same thing as its stress-strain behavior. They are entirely different concepts.
     
  9. Aug 10, 2017 #8

    scottdave

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    Good point. I should have stated that to let the OP know that those are not needed to solve the problem. The Compressive Strength is the maximum pressure that it is designed withstand before you are getting into risk of failure.
     
  10. Aug 10, 2017 #9
    Actually, it is not the pressure (which is the isotropic part of the stress tensor) that causes failure. It is the anisotropic part (i.e., unequal principal stresses) that causes failure (along shear planes at an angle to the principal stresses). When we talk about compressive strength, we are talking about unixial loading of a bar or column, with zero principle stresses in the transverse directions. If we somehow placed the bar or column under isotropic pressure loading (like lowering it to the bottom of the ocean or placing it in a liquid high compressive chamber), it would not fail until a much higher compressive stress than the uniaxial compressive strength.
     
  11. Aug 10, 2017 #10

    scottdave

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    Thanks @Chestermiller . Yes it is starting to come back to me. It has been around 20 years, since I took that materials course. But I do remember when I first was introduced to tensors. I had a similar feeling to the first time that I was introduced to imaginary numbers. :woot:
     
  12. Aug 10, 2017 #11
    Once I learned dyadic notation, all my problems with tensors vanished.
     
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