I understand the idea that stress is basically force per unit area. So let's imagine a force on beam as follows:

enter image description here

Notice how the force only occurs in one dimension. But when I look up stress, I always find it defined using tensors as follows:

enter image description here

$$\mathbf{\sigma} = \left[{\begin{matrix} \sigma _x & \tau _{xy} & \tau _{xz} \\ \tau _{yx} & \sigma _y & \tau _{yz} \\ \tau _{zx} & \tau _{zy} & \sigma _z \\ \end{matrix}}\right]$$

This diagram and matrix makes it seem like stresses are being applied to multiple faces of the unit cube. But how is that possible if a force only pushes in one dimension? I am assuming in the second picture, all the vectors are coming from one force. So how come in the second picture there are 9 components required to describe single stress? In other words, why is the mechanical stress defined using a tensor if it is applied only on one side of the cube or object?

  • 1
    $\begingroup$ The fact that the bar bends should tell you that the force on an individual volume element isn't just in one direction. On the top of the bar, the surface is compressed, so an element feels inward force in the transverse direction. Similarly, on the bottom of the bar, the surface is stretched, so an element feels outward force in the transverse direction. $\endgroup$ – probably_someone Jun 25 '17 at 1:07
  • $\begingroup$ It's also worth noting that if you only apply force in one direction (and that direction is along one of your coordinate axes), then two of the three diagonal elements will be zero. For example, if you apply force in the $x$-direction, then $\sigma_y$ and $\sigma_z$ will be zero. However, there will still be non-zero off-diagonal elements, for the reasons given in the previous comment. $\endgroup$ – probably_someone Jun 25 '17 at 1:10
  • $\begingroup$ You don't necessarily need the stress tensor depending on the situation. The stress tensor is just able to fully describe the stress on an element. In some cases you can simplify it to 1D or 2D stress. A single force can have a complicated stress. $\endgroup$ – JMac Jun 25 '17 at 1:47
  • $\begingroup$ @probably_someone Regarding your first comment....but even if the forces occur in multiple directions, why would they occur on multiple faces of the "unit cube" (that's what I heard it called)? Additionally, you are discussing multiple forces on different sides of a large object. But does the unit cube behave this way? $\endgroup$ – Stan Shunpike Jun 25 '17 at 2:29
  • $\begingroup$ @probably_someone Regarding your 2nd comment, if the apply force in one direction and the stresses in the other two perpendicular directions are zero, these are the principal stresses, and the off-diagonal elements for this chosen coordinate system are zero. $\endgroup$ – Chet Miller Jun 25 '17 at 3:00

This diagram and matrix makes it seem like stresses are being applied to multiple faces of the unit cube. But how is that possible if a force only pushes in one dimension?

The idea that your second diagram above is trying to convey is that even with an infinitesimally sized cube, at the point where the cube is located one cannot associate a unique force acting in a single direction. IF one could, then there would be no need to represent the stresses by a (second rank) stress tensor field - one could simply represent the stresses by a simple vector field.

The idea that the second diagram is conveying is that the force acting at a specific point in a solid body due to internal stresses depends not just on the specific location in the body but also on the orientation of the (internal) surface that you are considering. That's why all those blue force vectors are pointing in all sorts of different directions even though they may refer to the exact same point in space. Those blue force vectors are associated with different surface orientations. So to represent this situation, we need to represent the stress state at a point in the body not as a vector but as a 2nd-rank tensor, an object that can be combined with a vector (a unit direction vector representing the orientation of the surface being considered) in order to give another vector (the force vector associated with that oriented surface).


Stan, in the example you gave, you are saying that you are applying a force along one of the coordinate directions, but the normal stresses and shear stresses with respect to the other coordinate directions are zero. So, in this coordinate system, only one component of the stress tensor is non-zero. However, if you held the body and the loading on the body fixed, but switched to a new coordinate system (obtained by rigidly rotating your Cartesian coordinate axes to a new orientation), more than one stress component would be non-zero for this new coordinate system (assuming that the rotation did not merely involve interchange of the coordinate directions). What the stress tensor allows you to do is to determine the normal traction and the shear traction on a surface of any specified orientation within your body. This is accomplished by applying the so-called Cauchy stress relationship, which uses the stress tensor to map a unit normal to a surface of specified orientation into the traction vector on that surface. So, basically, the stress tensor is a linear operator which maps a unit vector into a traction vector. Think of the stress tensor as a machine which has a vector entering and a different vector exiting.


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