Stress physics: Encyclopedia II - Stress physics - Stress tensor

Stress physics - Stress tensor

Because the behavior of a body does not depend on the coordinates used to measure it, stress can be described by a tensor. The stress tensor is symmetric and can always be resolved into the sum of two symmetric tensors:

• a mean or hydrostatic stress tensor, involving only pure tension and compression; and
• a shear stress tensor, involving only shear stress.

In the case of a fluid, Pascal's law shows that the hydrostatic stress is the same in all directions, at least to a first approximation, so can be captured by the scalar quantity pressure. Thus, in the case of a solid, the hydrostatic (or isostatic) pressure p is defined as one third of the trace of the tensor, i.e., the mean of the diagonal terms.

Stress physics - Generalized notation

In the generalized stress tensor notation, the tensor components are written σ_ij, where i and j are in {1;2;3}.

The first step is to number the sides of the cube. When the lines are parallel to a vector base , then:

• the sides perpendicular to are called j and -j; and
• from the center of the cube, points toward the j side, while the -j side is at the opposite.

σ_ij is then the component along the i axis that applies on the j side of the cube. (Or in books in the English language, σ_ij is the stress on the i face acting in the j direction -- the transpose of the subscript notation herein. But transposing the subscript notation produces the same stress tensor, since a symmetric matrix is equal to its transpose.)

This generalized notation allows an easy writing of equations of the continuum mechanics, such as the generalized Hooke's law:

The correspondence with the former notation is thus:

Stress physics - Why is stress a symmetric tensor?

The fact that the stress is a symmetric tensor follows from some simple considerations. The force on a small volume element will be the sum of all the stress forces over the surface area of that element. Suppose we have a volume element in the form of a long bar with a triangular cross section, where the triangle is a right triangle. We can neglect the forces on the ends of the bar, because they are small compared to the faces of the bar. Let be the (vector) area of one face of the bar, be the area of the other, and be the area of the "hypotenuse face" of the bar. It can be seen that

Lets say is the force on area and likewise for the other faces. Since the stress is by definition the force per unit area, it is clear that

The total force on the volume element will be:

Let's suppose that the volume element contains mass, at a constant density. The important point is that if we make the volume smaller, say by halving all lengths, the area will decrease by a factor of four, while the volume will decrease by a factor of eight. As the size of the volume element goes to zero, the ratio of area to volume will become infinite. The total stress force on the element is proportional to its area, and so as the volume of the element goes to zero, the force/mass (i.e. acceleration) will also become infinite, unless the total force is zero. In other words:

This, along with the second equation above, proves that the σ function is a linear vector operator (i.e. a tensor).

By an entirely analogous argument, we can show that the total torque on the volume element (due to stress forces) must be zero, and that it follws from this restriction that the stress tensor must be symmetric.

Adapted from the Wikipedia article "Stress tensor", under the G.N U Free Docmentation License. Please also see http://en.wikipedia.org/wiki