# Why does stress need to be described with a tensor?

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

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

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

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?

• 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. – probably_someone Jun 25 '17 at 1:07
• 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. – probably_someone Jun 25 '17 at 1:10
• 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. – JMac Jun 25 '17 at 1:47
• @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? – Stan Shunpike Jun 25 '17 at 2:29
• @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. – Chet Miller Jun 25 '17 at 3:00