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I want to calculate stress matrix in a cube with two faces parallel to x axis and perpendicular to z axis (sorry I don't know how can I put a picture in this post).

There are two force uniform distributions (that we'll indicate with p) over this two surfaces: the superior is in the x direction, the inferior in the -x direction.

So, I'll have only shear stress, and a shear-modulus $\mu$ dipendence.

We suppose uniform stress tensor in the cube, because every infinitesimal dV of medium is in statical equilibrium with +pdS force with x direction by the superior infinitesimal dV and -pdS force by the inferior infintesimal dV for 3rd Newton's law.

Remembering that any stress over a surface is $t_{ij}n{j}$, where $n_j$ is the normal versor to the surface, we must write:

$T_{ij}n_1=0$ because we haven't any force on the surfaces perpendicular to the x-axis; so the first column is composed from three 0;

$T_{ij}n_2=0$ because we haven't any force on the surfaces perpendicular to the y-axis;so the second column is composed from three 0 too;

$T_{ij}n_3=p n_1$, because we have the force-distribution p over the surfaces that are perpendicular to z axis.

$n_1$ is $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $n_2$ is $\begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix}$, $n_3$ is $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. So the matrix $T_{ij}$ becomes:

$$\begin{pmatrix} 0 & 0 & p\\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$

But this haven't any sense, because stress-tensor must be simmetric for conservation of angular momentum. Where is the mistake?

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You most probably do not need an answer anymore. However I stumbled in just the same kind of inconsistency and I found this post, so this comment might help others like me. I understand this is the configuration you have in mind... very commonly used to introduce shear stress:

enter image description here

You are right: on the one hand, you can't have any force on a free surface (like the vertical ones on the left and right side of the cube) in a stationary configuration; on the other one you need a symmetric stress tensor otherwise you get problems with the torque of surface forces... no doubts about that... but in this way you also get a non-zero force on those vertical surfaces. Good point, it does not make any sense: I agree.

Well, today I struggled for hours trying to understand what I was not getting here and... Boy - I was kind of shocked in the beginning - the problem is in a stupid assumption, i.e. that what is show in the figure above is REALLY what happens if you pull a cube laterally from the top surface and at the same time prevent the bottom surface from sliding on the floor. The problem is that you simply won't get that nice uniform diamond-like deformation... but something quite more complicate and definitely non-uniform. So in the end the reason of the inconsistency you mention is that the "solution" in the figure is not a solution of the mechanical problem.

In particular (beam theory can help here) I would say that:

  1. If you consider the bottom surface, it really has to be glued to the ground because the left edge will be pulled up and the right one will be pushed against the floor. Lateral friction is not enough to prevent the cube from rotating: try with a foam cube and you will see...

  2. The bottom surface will exert a complex force profile with various components perpendicular to the ground (both positive and negative depending on the position). THIS will balance the torque of the force on the top surface if external forces just pull the top surface of the cube.

  3. Obviously the strain is not at all uniform here... there must be some sort of non-trivial bending, in particular at the base of the cube.

  4. Lateral surfaces are free and, you are right, they have to exert zero force! Be sure that force on those faces is indeed zero in the correct solution of this problem.

In short, once you properly glue the bottom of the cube to the floor, you can expect something more similar to what visible here

enter image description here

rather than in the picture above.

It is just funny that sketches like the one above are so widely used to explain shear stress. I think it is kind of risky and can lead to inconsistencies and (legitimate) doubts in students, I wonder how many persons are aware.

I have to say that some texts indicate the presence of vertical external forces also on those two lateral surfaces (then it all works and the uniform deformation above is correct)... but more often this is neglected.

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  • $\begingroup$ Dear Stefano, this is the very first true answer to this question of many years ago! Thanks for that. I realised the same after a while: the uniform configuration of the stress tensor is an ideal configuration of an infinite medium where you select a (say) cube at rest and see that for the various continuity of stresses at the border of this surface you obtain a uniform stress tensor. Of course, as you said, in reality you have discontinuity of stress between the surfaces parallel to the ground and the perpendicular ones: this creates the asymmetries that you pointed out. Cheers! SB $\endgroup$ – Boy Simone Aug 31 '17 at 10:25
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Boy, the right stress tensor for similar static situations is symmetric, indeed. It's not hard to see why: the stress tensor knows about the density of forces and an asymmetry would destroy the equilibrium. See

http://en.wikipedia.org/wiki/Stress_(mechanics)#Equilibrium_equations_and_symmetry_of_the_stress_tensor

The last paragraph of the section above contains a proof why the stress tensor is symmetric. You may say that the antisymmetric part would act as a torque.

In your example, both $T_{xz}$ and $T_{zx}$ are equal to the shear stress you denoted $p$. In particular, it is not true that $T\cdot n_x=0$, as you wrote in the first step (with confusing extra indices), because the stress tensor measures the internal forces and not just forces that you actively add by your hand.

The internal forces respect the symmetry between the $x$ and $z$ axes. The only asymmetric tensor was one you wanted to "prescribe" to the system. But you can't prescribe arbitrary properties and behavior to the physical system: the objects' properties and behavior obey the laws of physics rather than your expectations. In particular, laws of physics guarantee that the shear stress will treat any pair of axes symmetrically.

You should imagine that the squares in an atomic grid in the $xz$ plane are deformed to rhombi. But the rhombic curvature exists relatively to both axes $x,z$. If you insert a probe into the solid that will measure the internal tension, be sure that you will get totally identical results when you push the $z=0$ plane in the $x$ direction as if you push the $x=0$ plane in the $z$ direction. The result of both things is to change the angle between the (former) $x=0$, $z=0$ planes in the solid. There's no difference.

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The stress tensor is only symmetric if either the spin density describing the density of internal angular momentum is zero. For most materials, this is the case, because we can't align most of the spins of the constituent atoms or molecules in the same direction.

Imagine the cube you mentioned embedded as a subsystem of a much larger material system. With an asymmetric stress-tensor, the forces will cause the cube to spin relative to the material it is embedded in, which for most materials, will be prevented by molecular binding forces between the molecules or atoms on both sides of the boundary. In other words, the cube can't spin relative to the material just outside the boundary. This is due to come counterforce due to molecular binding and shows up as a counterforce on the surfaces normal to the $x$-direction.

You also have to realize this stress analysis only applies in the limit as the cube size becomes infinitesimal. Otherwise, we can have unbalanced forces on the different surfaces, which only means the shape of the cube becomes deformed in time.

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  • $\begingroup$ Boy, I agree with QGR that at least in some definitions, the stress tensor doesn't have to be symmetric. However: It's very strange how you write your equations, such as $T_{ij}n_z=pn_x$. Because of the $ij$ indices, it looks like a set of 9 equations, but because of $n_z$ and $p_x$, it looks like the indices are fixed and it's only one equation. Moreover, it's strange to use the letter $p$ (like "pressure") for forces that are completely non-uniform and they behave as vectors. While the non-symmetricity of the stress tensor would be a contradiction, I wouldn't trust your derivation. $\endgroup$ – Luboš Motl Feb 10 '11 at 7:04
  • $\begingroup$ In particular, Boy, I don't trust your equation $T_{ij}n_x=0$. You just made it up, you don't have any evidence. I think that the right answer to your problem is a symmetric stress tensor. If you tilt a solid, deforming the grid in the $xz$ plane, then the squares in the grid become rhombi, and the rhombi feel the very same tension in the $xz$ double-direction as in the $zx$ double-direction. You need two indices. What's the problem if you conclude that both components $xz$ and $zx$ are equal to the shear stress? $\endgroup$ – Luboš Motl Feb 10 '11 at 7:09
  • $\begingroup$ @Luboš Motl I correct the simbology that you have told in the question.I put $T_{ij}n_x=0$ because my hands exercite forces only on the surfaces that are perpendicular to z axis, so surfaces that are parpendicular to x and y are free and the stress on them (that is fixed with boundary conditions) is $T_{ij}n_j=0$,with j=1,2 (as x, y).This was my derivation.I think my result it's wrong too, because the stress that I have obtained I asimmetric, that isn't possible in classical linear elasticity theory... $\endgroup$ – Boy Simone Feb 10 '11 at 11:57
  • $\begingroup$ ...I agree with you, there's a mistake, but I can't find it,I would see the correct derivation for stress tensor in this physical situation (that's statical). $\endgroup$ – Boy Simone Feb 10 '11 at 11:58

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