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Enthalpy under uniaxial compression of solid

  1. Jun 13, 2013 #1

    I have found several references (I can send you the links if you want) that point to the fact that the enthalpy under uniaxial compression along z of a solid is simply given by:


    This appears to make sense as during uniaxial compression the Pz pressure component is the only one doing work (zero transverse strain). If you do a second subsequent uniaxial compression (along y say), then by the same logic the enthalpy should be given by U+PyV. However at the starting point of this second uniaxial compression the enthalpy is then given simultaneously by U+PzV and U+PyV, which can't possibly be true for general non-hydrostatic stress conditions. Hence my question is: is this formulation of the enthalpy valid only starting from equilibrium and doing only one uniaxial compression, or is it valid for general non-hydrostatic conditions?

    Many thanks,

  2. jcsd
  3. Jun 13, 2013 #2
    Further to my previous post, the reason why I'm interested in this is beacuse I would like to be able to calculate the transition parameters of a pressure-induced phase transition under non-hydrostatic stress conditions. Basically I have an energy surface of the material calculated with DFT, and I would like to be able to calculate the enthalpy at any point on this surface including the points away from the hydrostat, in order to find the enthalpy curves across the two phases involved in the phase transition and from their intersection the transition pressure. However I have the impression from the literature that the theory of thermodynamics under non-hydrostatic stress is still a source of confusion
  4. Jun 13, 2013 #3
    Are you talking about elastic solids, that are capable of storing deformational energy? If so, are you confining the discussion to small strains?
  5. Jun 13, 2013 #4
    Yes, I am indeed talking about elastic solids (no plasticity involved), and ideally I would like to derive a well-defined function of state equivalent to the enthalpy which is capable of determining phase equilibrium under non-hydrostatic stress and that reduces to the conventional definition U+pv when conditions are set back to hydrostatic.

    I have attached a copy of the cold energy surface as a function of lattice geometry that I'm working with. The hydrostat is the dotted line and there the enthalpy is given trivially by U+pV, but everywhere else conditions are non-hydrostatic. The standard definition of enthalpy under such non-hydrostatic conditions given in textbooks like Landau and Lifgarbagez is:

    H=U+ V_0 \sum_{i} \sigma_i \epsilon_i

    where V_0 is the equilibrium volume and sigma and epsilon are the stress and strain tensors respectively. This is only valid for small strains, but the main problem is that it is incomplete: it is of the form U+P \Delta V, and hence a term equivalent to V_0 \Delta P is missing, but it is not trivial what form this term should have under non-hydrostatic conditions. This made me think that paths of uniaxial compression could provide the key for finding a well-defined function of state at each point on the surface

    Attached Files:

  6. Jun 13, 2013 #5
    I forgot to mention: the strains I'm dealing with are finite and anisotropic, as you can see in the energy surfce
  7. Jun 13, 2013 #6
    Sorry, g_mogni. I feel like I'm in a little over my head on this one. If you are considering phase transitions, the usual thermodynamic function to work with the the Gibbs free energy. As you probably know, during a phase transition, this is the same (per unit mass) for both phases. If no phase transition were involved, then there would be a strain energy function that would characterize the deformational mechanics, even at large deformations. The strain energy is a function of the three invariants of the Cauchy Green tensor.

    Last edited: Jun 13, 2013
  8. Jun 14, 2013 #7
    I know that the usual function to find phase equilibrium is the Gibbs free energy, but in my case the energy surface is calculated at 0K, hence G reduces to the enthalpy. On paper the problem is simple: how to find a way to generalise the PV term to non-hydrostatic stress, but it could be that there is no solution as I haven't been able to find any clear answers in the literature yet

    Anyway, thanks for the help
  9. Jun 14, 2013 #8
    I have attached another version of the energy surface that shows the trajectory of the line of uniaxial compression along z (those along y and x of course lie outside the plane of the plot) in the form of the red line. The dotted lines are contours of constant shear stress. I'm fairly certain that if you start from the equlibrium position (red dot), the enthalpy along the uniaxial line is given simply by U+pzV, but of course being able to calculate the enthalpy along just one line outside of the hydrostat doesn't help much.

    Hence I was wondering if one could in theory reach any point on the surface by considering the line of uniaxial compression crossing it, which in general will intersect the hydorstat at positions away from equilibrium, and use the same formula U+pzV. In this case the problem of how to find the enthalpy everywhere on the surface would be solved. I was wondering if anyone had ever come across a similar problem

    Attached Files:

  10. Jun 14, 2013 #9
    Check out the Continuum Mechanics literature. Continuum Mechanics and Thermodynamics. I'm very confident that they have successfully analyzed this.

  11. Jun 14, 2013 #10


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    In general you have to generalize H=U+pV to [itex]H=U+sum_ij \sigma_{ij} \epsilon_{ij} [/itex], where σ is the stress tensor and ε the strain tensor.
  12. Jun 14, 2013 #11
    Yes, I have come across that formula in Landau and Lifgarbagez "Theory of Elasticity". The problem with it is that only writing the stress-strain term leaves it incomplete: if you revert the conditions back to hydrostatic, you are left with $U+P \Delta V$, since $V_0 (\varepsilon_x+\varepsilon_y+\varepsilon_z)=\Delta V$. Hence there is a term of the form $PV_0$ which is missing to obtain $U+PV$ again, but it's not clear what form this term should have under non-hydristatic conditions. Landau himself admitted this in the footnote on page 9 of his book on elasticity
  13. Jun 15, 2013 #12
    Maybe this will help a little.

    For an incompressible isotropic elastic solid, the strain energy per unit volume W can be expressed in terms of the 3 principal stretches for a given deformation λ1, λ2, and λ3 by
    The principal stresses are related to the strain energy function by:
    [tex]σ_j=\frac{\partial W}{\partial \lnλ_j}[/tex]
  14. Jun 15, 2013 #13
    Do you happen to know an expression for the strain energy density under finite strains?
  15. Jun 15, 2013 #14
    The strain energy density for finite strains is a function of the three invariants of the Cauchy Green tensor, or, equivalently, the three principal stretch ratios.
  16. Jun 15, 2013 #15
    Sorry I'm not very familiar with the theory of elasticity for finite strains, but I just read that the three invariants that you mentioned are given by:

    [itex] I=tr(C)=\lambda_1^2 +\lambda_2^2 +\lambda_3^2 [/itex]

    [itex] II=\lambda_1^2 \lambda_2^2 +\lambda_1^2 \lambda_3^2 +\lambda_2^2 \lambda_3^2 [/itex]

    [itex] III=det(C)=\lambda_1 \lambda_2 \lambda_3 [/itex]

    Is this correct?

  17. Jun 15, 2013 #16
    Actually I could just use the small-strain approximation for the strain energy:

    [itex]U=\frac{1}{2} \sigma_{ij} \varepsilon_{ij} [/itex]

    and integrate the formula numerically across small deformational steps
  18. Jun 15, 2013 #17
  19. Jun 15, 2013 #18
    Only if you know the stress at finite strains.
  20. Jun 15, 2013 #19
    See Mechanical Properties of Solid Polymers, Second Edition, I. M. Ward, Macmillan, pages 44 - 47.

  21. Jun 16, 2013 #20
    Thanks, I will have a look at that book in the library ASAP
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