Forces and dynamics - Physical Processes in Earth and Environmental Sciences - page 72

in or at the boundaries of the solid crust, whereas changes
in mass due to thermal effects are less common. The fact
that forces caused by changes in velocity or acceleration in
space deform the solid crust is witnessed by structures
formed in the rocky crust by them (Sections 4.14-4.16).

200 kg

200 kg

200 kg

3.13.1 Stress

A

B

F = mg

Stress, , is force F , per unit area, A , when acting over a sur-

face (Fig. 3.56). The units for stress are the Pascal (Pa)
which is Newton per square meter (Nm 2 ). More useful
units for the very large stresses relevant to tectonic studies
are the Kilopascal, kPa (10 3 Pa), Megapascal, MPa
(10 9 Pa), and Gigapascal, GPa (10 12 Pa). To illustrate the
physical difference between force and stress, imagine a ball
weighing 200 kg. The most important part of solid stress
is the force, the stress itself can be envisaged as the effect
or intensity of the force applied to a mass of rock and how
it is distributed or felt by the rock in every conceivable
direction of space. The force exerted by the ball in
Newtons (N) over any surface should be constant:
F

Surface A>>B

Fig. 3.57 Stress can be explained as the intensity of force over a
surface. The constant force exerted by the rock ball gives a higher
stress when it is put to rest over a column with a smaller surface
area: producing fractures if the internal resistance is exceeded.

The resulting stress over the smaller surface is bigger
even when the force has the same value in both cases.
In other words, the force is felt with more intensity over
the small surface and is thus able to produce deformation
easier than over the wide surface. The single vector corre-
sponding to the force per unit area which reflects the force
intensity over a surface F / A is called a traction and is only
one of the infinite components of the overall stress system
(Fig. 3.58). Because forces can change magnitude or
direction over a surface, we have to define different stress
values for every infinitesimal part of the surface or at any
point as d F /d A . If we consider a surface in a state of equi-
librium, any traction has to be balanced by an equal and
opposite traction. This pair of tractions constitutes the
surface stress (Fig. 3.59b), which can be resolved into nor-
mal and shear components (Fig. 3.59c). It is important to
remember that in a state of equilibrium the sum of the
forces acting over a surface equals zero.
We have so far considered only how the stress value is
distributed according to a single surface orientated per-
pendicular to the applied force. However, the orientation
of the surface in relation to the applied force is essential in
determining the resulting traction value and there should
be an orientation and magnitude for every possible surface
inclined at different angles. The stress tensor is composed
of all the individual surface stresses acting over a given
point. Stress analysis can be approached in two dimensions
(2D) or three dimensions (3D): in 2D orientations are
referred to an xy coordinate system, whereas in 3D an xyz
system is used, in which z is conventionally taken as the
vertical component. If all the tractions are equal in magni-
tude, as occurs in static fluids, the stress tensor has the
shape of a circle in 2D and of a sphere in 3D; we have seen
in our discussions of fluid stress (Section 1.16) that such a
state is called hydrostatic stress (Fig. 3.58b). In solids, when
the tractions usually have different magnitudes, the shape

1,960 N. In a simple
example, consider the 200 kg ball resting on the top sur-
face of a wide column, of area say 2 m 2 (Fig. 3.57a). The
force is distributed over all the surface and the column will
hold up depending on its material resistance. If the same
force is applied over a surface of a smaller column, for
example, 0.5 m 2 , of the same material, the identical force
of 1,960 N may lead the column to break (Fig. 3.57b).
Remembering that

m g 200 kg · 9.8 ms 2

F / A (Equation 1; Fig. 3.56) the
resulting stresses for both columns are 1,960 N/

2m 2 0.98 kPa and 1,960 N/0.5 m 2 3.9 kPa,

respectively.

F (N)

A

Area (m 2 )

Stress

s = F / A (Nm -2 = Pa)

(1)

Stress is force per unit area, measured in Pascals
(Pa) or Nm -2

Fig. 3.56 Revision: force and stress.