Let us begin by calculating the stress acting on a single plane passing through point P and oriented q degrees with respect to the z-axis (Figure 7).
Moreover, let the area of this plane be so small that its units are measured in millionths of a meter and let q = 50o. As shown in Figure 7, our plane represents one face of a right triangular prism. Moreover, we can view the trace of our hypothetical plane as the hypotenuse (XZ) of a right triangle whose legs are represented by line segments ZO and OX which are parallel to the x and z-axes of our coordinate system respectively (Figure 8).
XZ is n x 10-6 m long, while ZO and OX are 10-6 m long. Thus, (1) the area of the face of the triangular prism that is perpendicular to the z axis is labeled Areaox and is equal to n x 10-12 m x sin(50o), (2) the area of the face of the triangular prism that is perpendicular to the x-axis is labeled Areaoz and is equal to n x 10-12 m x cos(50o), and (3) the area of our plane of interest is labeled Areazx and is equal to n x 10-12 m2 .
We now assume that all of the forces acting on our right triangular prism are balanced. Recalling that stress is force divided by area we write:
and:
where Fox is the force acting in the OX direction, Fxo is the force acting in the XO direction, Areaoz = n x 10 –12 m2 x cos(50o), Areaxz = n x 10 –12 m2, and sx = 20 MPa. If we are in a condition of static equilibrium, then:
Rearranging, canceling similar terms, and isolating Sx results in:
Solving for Sz involves a similar set of steps. First we we write:
and:
where Fzo is the force acting in the ZO direction, Foz is the force acting in the OZ direction, and Areaox = n x10 –12 m2 x sin(50o), Areaxz = n x 10 –12 m2, and sz = 40 MPa. If we are in a condition of static equilibrium, then:
Again rearranging, canceling similar terms, and isolating Sz results in:
Sz and Sx are the components of the stress vector or traction, sxz. To solve for the magnitude of this traction we use the Pythagorean theorem, hence:
As shown in Figure 9, b is the angle that sxz makes with the z-axis, and sin(b) = Sx/sxz = 0.3869.
Taking the arcsine of 0.3869 indicates that sxz makes an angle of 22.76o with respect to the z-axis. Thus, sxz is the stress vector or traction on a plane oriented at 50o to the z-axis produced by a vertical normal stress of 40MPa and a horizontal tectonic stress of 20MPa. The traction sxz has a magnitude of 32.23MPa and is oriented 22.76o clockwise with respect to the z-axis, but what would the magnitudes and directions of other tractions be on all of the other possible planes running through point P? In other words what is the stress tensor?