Page 63 - Physical Principles of Sedimentary Basin Analysis

3.5 Stress 45
σ yx
y

−σ xy dy x σ xy

dx
−σ yx

Figure 3.8. The shear stress that acts on the surface of an infinitesimal square element in the
xy-plane. (The box has unit height in the direction normal to the plane.)

The size of the normal stress on the plane is the scalar product of the stress vector S and
the normal vector n
S n = S i n i = n j σ ji n i , (3.23)
which yields the normal stress and tangential (or shear) stress on the plane

N = S n n and T = S − N, (3.24)
respectively.
The stress tensor turns out to be symmetric, a property that follows from the assumption
that any infinitesimal volume element of a continuum is torque-free. Recall that the torque
is a vector quantity defined as M = r × F. The torque M on an infinitesimal small element
in the xy-plane, as shown in Figure 3.8, is found by adding the contribution from all four
sides, and we get

dx dy
M = 2 (σ xy dy) − (σ yx dx) e z
2 2
= (σ xy − σ yx ) dx dy e z . (3.25)
There is one contribution from each side, where the first term is seen to be the shear force
F y = σ xy dy acting on the left side of the element with the distance dx/2 to the origin. A
torque-free element (M = 0)impliesthat σ xy = σ yx . The same argument can be applied
for the elements in the xz- and yz-planes as well. Therefore, it follows that the stress tensor
is symmetric, σ ij = σ ji , and that it has only six independent components. The symmetry
of the stress tensor allows the stress vector (3.21) to be written as the matrix–vector product

S i = σ ij n j . (3.26)
A later section shows that it is always possible to find a coordinate system where the stress
tensor becomes diagonal. It is often an advantage to work in such a coordinate system
because of simplifications.

Exercise 3.4 Show that the stress tensor σ ij = pδ ij gives the stress state in a fluid (at rest).
(Hint: The fluid acts on any plane with the normal stress p.)