Symmetry of Strain and Stress

The strain tensor is symmetric by definition.  Each off-diagonal term is the sum of two terms, which are reversed when the tensor indices are, and the sum remains the same (a+b=b+a).  The two terms represent changes in length, or angle, for lines in two perpendicular planes, e.g. the change in a line in the x-direction embedded in a y-plane, plus the change in a line in the y-direction embedded in the x-plane.  (The nomenclature here identifies the plane by its normal vector)

In contrast, the off-diagonal terms of the stress tensor act in only one direction, in one plane.  E.g., the x-y component of the stress represents the force in the x-direction on a y-plane, and the y-x component represents the force in the y-direction on the x-plane.

In other words, the x-y component of the stress, acting only on the y-plane, produces strain in both the x and y planes, as does the y-x component of the stress.  This may lead us to a third reason for defining the strain tensor as it is.

Two questions are, 1) why are the x-y and y-x components of stress equal, and 2) why are they both linearly related to the x-y component of the strain tensor?

Mathematically, the second question is partly answered just by the symmetry of both tensors, but we may find a physical reason as well.  Another thing to note here, is that the strain tensor is dimensionless, but the stress tensor has units of force/area.  So stress and strain are different animals, so to speak, and any equation relating them linearly must have coefficients with the same units as the stress (or its inverse).

We start with the symmetry question, again considering a rectangle in the x-y plane.  Recall that the components of the stress tensor are to be interpreted as the forces acting on the corresponding planes (e.g., the i-kth component is the force in the i direction on the k plane).  Then we can label the drawing as shown below:

In equilibrium the rectangle is stationary, and the sum of the moments, or torques about the center of the rectangle must equal zero (conservation of angular momentum).  Since the stress components act on the areas, we let the rectangle represent one surface of a cube, whose length in the z direction is c.  Going counterclockwise from the right edge, we have

The stress components here must be taken as the average stresses over their respective areas, and are also assumed to be continuous functions of x,y and z.  Cancelling the area factor abc/2, and if the dimensions of the cube are small enough, the above equation, with some rearranging, becomes (to any desired degree of accuracy),

which shows the symmetry of the x-y components of the stress tensor.  The same thing can be done for the x-z and y-z components.

So the reason for the symmetry seems to be as follows:  the two x-y components are trying to rotate the cube counterclockwise, and the two y-x components are trying to rotate it clockwise. Then equilibrium is obtained by equating the clockwise and counterclockwise forces. As a and b become smaller, the difference between the two x-y (or y-x) components becomes small also, so that they are essentially equal.  Then the balance of forces produces the symmetry.