Basically, what I need to know guys is that when we divide a vector by a scalar, we get a vector. Then what is different in the case of stress?? I mean, WHY IS IT STILL A SCALAR?

  • $\begingroup$ Have you noticed that pressure is also a scalar? You might have an easier time finding an explanation of that. $\endgroup$ – dmckee Feb 8 '18 at 19:22
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    $\begingroup$ Why do you say stress is a scalar? It would help to know the context of the question. Stress is not a scalar. But sometimes we are interested in only one component of stress, so we see only one real number. But that number is not a scalar, it's a component of a rank 2 tensor, as @Chemomechanics points out in his answer (which I recommend you select as the accepted answer). $\endgroup$ – garyp Feb 9 '18 at 16:18

Stress is a rank-two tensor that couples one 3-D vector (a direction associated with an area) to another (a force). The tensor can hold 3×3=9 values according to the various directional combinations, but for static problems, only six values can be independent.

In symbolic form, the stress $\boldsymbol\sigma$ is

$$\boldsymbol\sigma=\begin{bmatrix}\sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz}\\\sigma_{zx} & \sigma_{zy} & \sigma_{zz}\end{bmatrix}=\begin{bmatrix}\sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ & \sigma_{yy} & \sigma_{yz}\\& & \sigma_{zz}\end{bmatrix}$$

where the missing elements are the nonindependent ones for a nonaccelerating infinitesimal cube. (A free-body diagram would show you that if $\sigma_{xy}\neq\sigma_{yx}$, then the cube must start to rotate.) One of the indices (it doesn't matter which due to the symmetry) is the direction of the surface normal vector, and the other is the force direction. Again, in the static case, $\sigma_{xx}=\sigma_{(-x)(-x)}$.

Pressure, a scalar, is one-third the negative of the trace of the stress tensor written in matrix form: $P=-\frac{1}{3}(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})$. This is one of three invariants (i.e., coordinate-system-independent parameters) of the stress tensor.

If you were considering a single scalar value of stress, it's because you'd already implicitly chosen the surface orientation and force direction.


The state of stress at a given location within a material is described by the stress tensor, which is a 2nd order tensor. The stress tensor helps us determine the force per unit area (called either the traction vector or the stress vector) on a surface of arbitrary orientation at the given location. The traction vector on a surface is obtained from the stress tensor by applying the Cauchy Stress Relationship: if $\mathbf{n}$ represents a unit normal vector to the surface (thus establishing its spatial orientation), the traction vector on the surface is obtained by taking the dot product of the stress tensor with the unit normal. The traction vector thus obtained has both magnitude and direction, and, it is not necessarily perpendicular to the surface. The component of the traction vector normal to the surface is called the normal stress; the component tangent to the surface is called the shear stress.

The isotropic part of the stress tensor is called the pressure (actually, minus the pressure), and, when dotted with the unit normal to any specified internal surface gives a traction vector of magnitude equal to -p, and direction normal to the surface; there is no shear stress component to the portion of the traction vector contributed by the (isotropic) pressure portion.


Good question. Well, stress is a scalar quantity, once the direction is fixed.

In general you have nine possible stresses, that can be gruoped in a matrix. ($F_x, \ F_y \ \text{and} \ F_x$ in each plane $XY, \ YZ, \ XZ) $.

But I guess this is confusing you, so let's focus on only one of them: horizontal force through on a cube's side.

Yes, force is a vector, and surface can be seen as a vector as well, but when you say "stress", it means "magnitude (scalar) of the force perpendicular to that element of area.

So the direction is previously fixed, and you only need a number to complete the information. That's what happens with pressure.

You can have a force on $y$ axis on the $XY$ plane, that's one stress $\sigma_{xy}$.

When you have force on $z$ axis over the $XY$ plane, that's force perpendicular to that plane. It is a scalar because it is "total force" (magnitude) over a surface whose orientation has already been described. That's why pressure is a scalar, and so are the other stresses.

If your surface does not coincide with those planes, you must take the projection of the force on each plane, and then compute the scalar stresses.

  • $\begingroup$ According to your description pressure is not a scalar. You say it varies with the orientation of the surface. Pressure does not vary with orientation. $\endgroup$ – garyp Feb 8 '18 at 20:21
  • $\begingroup$ In a solid different components of the stress tensor can have different values. Pressure is a scalar and does not change with direction. Indeed, it is defined differently in a solid compared to a fluid, as @Chemomechanics reminds me. $\endgroup$ – garyp Feb 9 '18 at 3:24
  • $\begingroup$ It's true, I mixed two terms. I'll change it. $\endgroup$ – FGSUZ Feb 9 '18 at 14:08
  • $\begingroup$ One final thing (I hope!) regarding terminology. A single number by itself does not a scalar make. If $\vec{a}=a_x \hat{i} + a_y \hat{j}$, $a_x$ is not a scalar. A scalar does not change its value as the coordinate system changes. $\sigma_{xx}$ is not a scalar. It's a real number that can only be described as "component of a rank 2 tensor". So one needs to be careful about the word scalar. In particular you say " pressure is a scalar, and so are the other stresses", but "the other stresses" are not scalars, they are components of a rank 2 tensor. $\endgroup$ – garyp Feb 9 '18 at 16:14
  • $\begingroup$ That's right, in fact I was about to include that but I finally didn't. Thanks anyways. $\endgroup$ – FGSUZ Feb 9 '18 at 17:45

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