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I Infinitesimal cube and the stress tensor

  1. Jan 2, 2017 #1


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    The Cauchy stress tensor at a material point is usually visualized using an infinitesimal cube. The stress vectors (traction vectors) on opposite sides of the cube are equal in magnitude and opposite in direction. As a result, the infinitesimal cube is in equilibrium.

    However, when we derive the equilibrium equation for the domain, we have to consider the spatial change in stress along the sides of the cube. We then get [tex]\mathrm{div}{\matrix{\sigma}}=\vec{0},[/tex] where ##\matrix{\sigma}## is the Cauchy stress tensor. It is assumed that there are no body forces, such as gravity.

    My question is:
    What is the mathematical justification for the fact that the infinitesimal cube has the same volume in both cases (##\mathrm{d}V=\mathrm{d}x\hspace{1pt}\mathrm{d}y\hspace{1pt}\mathrm{d}z##), but in the first case the spatial change in stress is excluded and in the second case it is included?

    I must say my knowledge in differential geometry is not very deep. I understand that the spatial change in stress approaches zero as the volume of the cube approaches zero, but this does not really explain the issue for me.
    Last edited: Jan 2, 2017
  2. jcsd
  3. Jan 2, 2017 #2
    The normal components of the stress vectors on the leading and trailing faces of the cube can be different because the shear stresses on the sides of the cube can contribute to the force balances in the directions under consideration. In this way, the cube can still be in equilibrium. In terms of the question about volume change, we assume that the strains are small, so that the shape of the cube is not affected in the differential force balance, to linear terms in the stresses and deformations. It is possible for more complicated materials that don't obey Hooke's law to carry out a large deformation analysis, and include all the kinematic changes in the force balance, including changes in volume.
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