Is tension a component of the stress tensor?

[Chestermiller]

Ok. All clear. Thanks again. Looking forward to your explanation about the stress tensor being symmetric (or not).

To be clear, the stress tensor is symmetric (meaning that the components of the dyadics with corresponding different indices are equal). So the way you represented it, it was not symmetric.

Throughout this discussion, the stress tensor was homogeneous but not symmetric. Is the stress tensor always symmetric in the case of fluids and can be symmetric or asymmetric in the case of solid materials?

For the vast majority of practical materials, the stress tensor is symmetric for both solids and liquids.

When the stress tensor is homogenous (not a function of position), it means that the internal stress is the same at every point in the material. What are a couple of simple examples of a material described by a homogeneous stress tensor and a couple of examples of a material whose stress tensor is instead inhomogeneous?

Homogeneous: A rod under tension with the load distributed uniformly at its ends. A fluid under hydrostatic pressure.
Inhomogeneous: A rod under tension with the load distributed non-uniformly at its ends. A beam being bent. Most objects encountered in practice.

For the stress tensor to be homogeneous, does the material need to have a constant density, be homogeneous and isotropic, and does the external force need to be uniformly applied to material?

Pretty much yes. When you solve some problems, you will get the idea.

 

[Chestermiller]

A 2nd order tensor is said to be symmetric if it is equal to its "transpose." The transpose of a tensor is obtained by switching the order of the unit vectors in each of its dyadics.

 

[fog37]

Thanks. I see my mistake in post #29: I assumed that being symmetric meant that the terms $i_{x^{'}}i_{y^{'}}$ and $i_{y^{'}}i_{x^{'}}$ were the same. Symmetry instead means that the components of these two dyadics are the same...