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A Linear elasticity, strutural physics, pressure load

  1. Mar 12, 2017 #1
    I have been studying structural physics lately, in particular the linear elastic eqns, governed by the equation of motion. https://en.wikipedia.org/wiki/Linear_elasticity

    I am trying to solve these equations numerically. I was thinking of a scenario of a rod being compressed on one end by some pressure load, say 1kPa. Does this 1kPa load correspond to the external force term in the equation of motion or does it translate to a normal stress, which would be the divergence term in the equation of motion?
     
  2. jcsd
  3. Mar 13, 2017 #2
    There is not a lot here to work with for what you are looking for. 1 kPa, generally speaking, is a pressure or a stress. Pressures and stresses = Force/Area. As stated in your question, 1 kPa is a static pressure (or stress can't tell from what is given). In order to apply any dynamics to the question you have to specify the forcing function. There is no guidance that can be provided as the question is currently posed.
     
  4. Mar 13, 2017 #3
    Thank you for the reply. Sorry for the lack of details. Let me provide some more:

    (1) I am trying to numerical solve the linear elastic equations https://en.wikipedia.org/wiki/Linear_elasticity#Mathematical_formulation
    In the above link there is a forcing term F, a time derivative, and a divergence term.
    (2) I am solving the equation of motion using the finite volume method.
    (3) The pressure, 1kPa, was just a hypothetical example, although I would like to test my code on some compression problems, e.g., a column under compression with some constant pressure/force load on one end.

    I guess one of my questions is, is this pressure load actually a boundary condition and/or is it just a force term?
     
  5. Mar 13, 2017 #4
    It's a boundary condition on the stress.
     
  6. Mar 13, 2017 #5
    But this boundary condition is translated to the force term or am I not understanding this? x
     
  7. Mar 13, 2017 #6
    You are trying to solve the stress equilibrium equations for a Hookean solid, for the displacements,correct?
     
  8. Mar 13, 2017 #7
    I am not too familiar with solid mechanics terminology, but I believe equilibrium equations mean steady state? If so, then no, I am solving the transient equation (the equation of motion in this link https://en.wikipedia.org/wiki/Linear_elasticity#Mathematical_formulation)

    And yes, I am solving for the displacements in my formulation.
     
  9. Mar 13, 2017 #8
    OK. So, as I said before, the pressure comes in as a boundary condition. In particular, if you are using the usual convention that tensile stresses are positive and compressive stresses are negative, the normal component of the stress vector on the portion of the boundary acted upon by the pressure is equal to - p. And the normal component of the stress vector would be expressed in terms of the strains at the boundary, which, in turn would be expressed in terms of the displacement gradients at the boundary.
     
  10. Mar 13, 2017 #9
    I see thank you. Does the force term not come in to play in this case then?
     
  11. Mar 13, 2017 #10
    The force term in those equations is a body force, not a force on the boundary.
     
  12. Mar 13, 2017 #11
    Oh I see. What exactly is a body force? I thought external loads were "body" forces, but it seems I do not understand what this is.
     
  13. Mar 13, 2017 #12
    They are forces like gravity, magnetic force, electrostatic force. They act in the interior.
     
    Last edited: Mar 13, 2017
  14. Mar 13, 2017 #13
    Ahhh I see.


    So when I solve the equation, I am solving it in the displacement formulation, where I convert the stress tensor term to a displacement term (using Hooke's law and the strain-displacement relationship). So stress is not explicitly computed for. Does this change at all how the pressure is translated to a stress term?
     
  15. Mar 13, 2017 #14
    Yes.
    What you do is express the boundary condition in terms of the displacements and their derivatives at the boundary. So the strain-displacement relations come in also. This is analogous to a transient heat transfer problem where the heat flux at the boundary is expressed in terms of the normal derivative of temperature.
     
  16. Mar 13, 2017 #15
    Thanks chestermiller. Very helpful analogy. I am a fluids/heat transfer guy, and new to structures.

    So in structural BCs, are the following BCs all the same: stress BC, traction BC, pressure load BC, force load BC?
     
  17. Mar 16, 2017 #16
    If I am given a stress/pressure BC, I need to compute the respective displacement. To do this, I use 2 relationships:
    Hooke's Law
    (1) ##\boldsymbol{\sigma} = 2\mu\boldsymbol{\epsilon} + \lambda tr(\boldsymbol{\epsilon})I ##
    and the
    Strain-displacement relationship
    (2)
    ##\boldsymbol{\epsilon} = \frac{1}{2} [\boldsymbol{\nabla {u}} + (\boldsymbol{\nabla{u}})^T]##

    Substituting (2) in (1) yields
    ##\boldsymbol{\sigma} = \mu\boldsymbol{\nabla{u}} + \mu\boldsymbol{(\nabla{u})}^T+ \lambda I tr(\boldsymbol{\nabla{u}}) ##

    in 3-D the LHS and RHS are 3x3 tensors.
    in 1-D the LHS and RHS are 1-D 3 element arrays.

    Let the displacement vector be ##\boldsymbol{u} = [u_x, u_y, u_z]##.
    Let the symmetric stress tensor be

    ##
    \begin{bmatrix}
    \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
    \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\
    \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \\
    \end{bmatrix}


    ##

    For now, consider 1-D in the x direction. Meaning that ##u_y = u_z= \frac{\partial}{\partial y} = \frac{\partial}{\partial z} = 0##
    Equation (2) simplifies to:
    ##\sigma_{xx} = \mu\frac{\partial u_x}{\partial x} + \mu\frac{\partial u_x}{\partial x}+ \lambda \frac{\partial u_x}{\partial x} ##

    Given the normal stress in the x-direction and the lame parameters, this equation can be easily solved for ##\frac{\partial u_x}{\partial x}## and the solution is unique.

    However, in the 3-D sense, it seems ##\boldsymbol{\nabla u}## cannot be analytically solved and the solution is not unique?
     
  18. Mar 16, 2017 #17
    oh wait a minute..... My boundary condition for stress is typically only going to be a normal stress in 1-D direction. In this case, I believe the 3-D tensor equation will reduce to a scalar equation.

    Edit: It would only reduce to a scalar equation if the normal stress (i.e., the face experiencing this normal stress is normal to one of the axes) was oriented along 1 of the 3 axes.
     
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