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Calculating deformation energy

  1. Jun 18, 2014 #1


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    Hi all,

    I've been trying to understand how to calculate the elastic energy of a deformed object. For example, if I have a box, and I push on it and it deforms, how can I calculate this deformation energy?

    I don't know much about elasticity, but I have read a little about strain energy and find the following equation promising:

    However, I keep coming across different types of strain tensors, and principle strain, and shear strain, and all kinds of things, and I just don't know how to put it together. I would imagine that you simply take a lot of dot products between forces and displacements, but I don't know how to take care of this with these tensors and such.

    So for the box example, could you help me understand how I could go about calculating this energy? Thanks!
  2. jcsd
  3. Jun 18, 2014 #2

    Andrew Mason

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    I don't think there is any simple way to determine a formula based on first principles in order to calculate the deformation energy of a body.

    Generally speaking, the energy used in deforming the body is converted to heat. The internal energy of the deformed body is the same as the undeformed body. So the work done ends up as heat (using the first law of thermodynamics: Q = ΔU + W → Q = W if ΔU=0).

    The easiest way to measure the energy of deformation (ie. the work done in achieving deformation) might be to measure the heat flow.

  4. Jun 18, 2014 #3


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    A thermodynamicist might want to count it all as heat energy, but a mech engineer would prefer to work with stresses and strains.

    There are two basic ways to find the elastic internal energy. One is to find the work done to deform the body, which is the OP's
    That doesn't involve any tensors - forces and displacements are vectors.

    If the material behavior is linear (i.e. there is no permanent plastic deformation, etc), the other way is to integrate the strain energy density over the volume of the body. The SED is either ##\frac 1 2 \sigma_{ij}\epsilon_{ij}## in tensor notation, or ##\frac 1 2 \sigma^T \epsilon## writing stresses and strains the engineer's way, as vectors of length 6. That fact that "engineers' shear strain" is twice "tensor shear strain" conveniently makes the two formulas give the same answer. (The engineer's formula includes each shear strain once, but the tensor formula includes the two equal off-diagonal terms separately).

    If you ask a specific question about what is confusing you, you will probably get some help here, but that statement is too vague to answer apart from the general advice to find one good source of teaching material on continuum mechanics and work through it systematically, rather than jumping around from one website to another.
  5. Jun 19, 2014 #4


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    Actually, this is specific enough for me already! Thanks very much to both of you!
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