Read 16.7 for Today. Read 16.12 for Friday.
The stress tensor maps maps a given force to a set of stresses along a particular set of orthogonal axes. We write this mapping as
The elements of the stress tensor indicate the amount of force exerted by a medium in particular direction at a particular surface. For example, choose a surface perpendicular to the x-axis so that dA = n dA = dA (1, 0, 0). Then the force normal to the surface is σ11 dA, and the force parallel to the surface in the y-direction is σ12 dA.
The stress tensor is symmetric. The argument is similar to the proof that pressure in a static fluid is isotropic, but relies on the fact that the angular momentum of a volume element scales as λ5 (L = Iω, and I ∝ mr²) while the torque is proportional to λ³ (Γ ∝ Fl = Σ l dA).
Write down the stress tensor in a static fluid at a point where the pressure is p.
The force is everywhere normal to the chosen surface, pointing inward with magnitude p, so
negative since dA points outward. This is the result for a diagonal stress tensor, where each diagonal element has magnitude -p,
where 1 is the 3×3 unit matrix (1's on the diagonal, 0's off the diagonal).
At a certain point P in a continuous medium the stress tensor is
| Σ = | -1 | 2 | 0 |
| 2 | -2 | 0 | |
| 0 | 0 | 1 |
A small surface element at P has area dA and is parallel to the plane x + y + z = 0. What is the force on this surface element and what is the angle between this force and the normal to the surface element? To be definite, take P to be in the positive octant (x, y, z all positive), and assume that the outside of the surface is the side away from the origin.
The elements of Σ have units of pressure, even though they were not specified in the problem.
To find the force on the surface, we need to find the normal to the surface n. This may be most easily done by noting that the constraint for the surface is a function, f(x,y,z), and the gradient of the function is normal to it, so
Using this, the force is
The angle between the force and the normal to the surface is
This value of cosine corresponds to the angle 35.3° = 0.615 radians.