Download LEC. 7: Stress I – Introduction to Dynamic Analysis

Lec. 7: Stress I: Introduction, Page 1 of 6
LEC. 7: Stress I – Introduction to Dynamic Analysis
The four categories of deformation are a rock’s response to stresses that are generated by
forces.
Units and Magnitude of Stress
A force is that which changes the state of rest/motion of a body. Remember Newton’s
First Law of Motion: an object at rest will remain at rest until acted on by an external
force. Newton’s Second Law of Motion: The acceleration of an object is proportional to
the force acted on it, and, inversely proportional to the mass of the object. The resulting
change of a rock due to an applied stress is something you’re now familiar with: strain.
Whether the rock strains or not under an applied stress, depends on the magnitude of the
stress, and the “strength” of the rock itself.
e.g., F = ma, where F = force, m = mass, and a = acceleration. For most of our
discussions in Structural Geology, we use the SI system of weights and measures (i.e.,
meters, kilograms, seconds). Therefore, force is measured in Newton’s (N), mass is
measured as kilograms (kg), acceleration is measured in m/second2;
e.g. 1N = kg*m*s-2.
Now, take for example, a tyrannosaurus rex standing on a concrete column 20 m high,
and 100 m in diameter. Chances are, the column will be able to support T Rex. On the
other hand, if we place T Rex on top of a concrete column 20 m, and 0.10 m (i.e., 10 cm)
in diameter, chances are that T Rex is going to crush the column. In both cases T Rex
has the same mass; in both cases the acceleration due to gravity is the same; the
difference is the area upon which T Rex sits. In the first case, T Rex’s force = mass *
acceleration due to gravity is applied over a column with a 100 m diameter. In the
second case, T Rex’s force is applied over a tiny, circular area, 0.10 m in diameter. We
can measure the stress, (s), exerted by T Rex on the columns which = Force/unit area. In
the textbook, Davis describes how the force of a bowling ball rolling down a lane is
~91.1 N.
Stress is measured in Pascals derived from the units N*m-2. So, 1N*m-2 = (1kg*m*s2
)*m-2 = 1Pa. Or, 105 Pa = 0.1 MPa (megapascal) = 1 bar. Atmospheric pressure at sea
level is 1 bar.
Generally, when we talk about the magnitude of stress we use either bars or pascals. So,
the stress that the bowling ball in the above example would exert upon hitting a pin with
a surface area of 0.50 m2 would be about 182 Pa.
We can establish the stress (or pressure) gradient in the subsurface in the following way.
What is the stress magnitude at one km depth beneath the earth’s surface?
Lec. 7: Stress I: Introduction, Page 2 of 6
First, we assume that the bulk composition of the crust is granite and then calculate the
force exerted by a 1 km cube of granite at 1 km depth. F = ma, where m (mass) = volume
* density: 1000m * 1000m * 1000m * 2700kg/m3 * 9.8 m/s2. The stress (s) created by
the weight of the cube of granite acting on the base of the cube (i.e., at 1 km depth) is
determined through s = force/unit area.
s = (1000m * 1000m * 1000m * 2700kg/m3 * 9.8 m/s2 )/ (1000 m * 1000 m)
s = 26,460,000 Pa, or 26.5 MPa at 1 km depth.
Another way to look at this is that for every 3.8 km, the lithostatic stress gradient
increases by 100 MPa (or 1 kb). Of course, this assumes that the entire overlying
material is compositionally analogous to granite with a uniform density. Is this a valid
assumption.
Another, simpler way to estimate the stress (pressure) at a given depth in a homogenous
media is s = rgh; where s = stress (MPa), r = density (kg/m3), g = acceleration due to
gravity (9.8 m/s2) and h = depth in m.
Stress: Magnitude and Direction
A more complete definition of force (and therefore stress) must include not only the
magnitude but also a direction in which the force is acting. To make matters more
complex; stress really refers to a whole collection of tractions (or force vectors) acting on
a single point at a given instant in time. A good way to visualize how stress works is to
consider a bulls-eye target in an archery contest. An arrow that hits the target represents
a traction acting on a plane – the bulls-eye. The magnitude of the traction represents the
length of the arrow that sticks out of the bulls-eye.
Now instead of the target being a planar disc, think of it as being a sphere. Within the
sphere there are an infinite number of disc-shaped cross sections that can be cut.
Therefore, there are an infinite number of orientations that an arrow could have that hits a
given cross section.
This is analogous to stress in nature: in a given rock we would like to know the
magnitude and orientation of all of the stresses acting on individual planes in all possible
orientations at a given point (P) within the rock. If we can do this for point P, then we
can describe what is called the stress tensor for point P. Now, if we can do this for every
point within the rock, then we can define the stress field, an entire explanation of all of
the stress tensors acting on a rock.
Stress on a Plane
Lec. 7: Stress I: Introduction, Page 3 of 6
The above figures shows a red (bold) arrow that is pointing obliquely into the x1x2 plane.
The stress on plane x1x2 exerted by the red (bold) arrow can be resolved into three
components: sn, ss2 and ss1. sn represents the normal stress component oriented parallel
to the x3 axis; ss2 and ss1 represent the shear stress component in both the x1 and x2
directions.
Stress at a Point
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In 3-D, there are nine tractions – vectors – which define the state of stress at a point.
• In the figure, s represents stress;
• the first subscript (s11) identifies the plane by indicating the axis which is
perpendicular to it;
• the second subscript (s11) shows which axis the traction vector is parallel to.
In the above example, the nine vectors (s11, s21, etc.) can be written in matrix form:
sij =
s11
s21
s12
s22
s13
s23
s31
s32
s33
sij represents the stress tensor. A tensor can relate two vector fields.
Lec. 7: Stress I: Introduction, Page 5 of 6
Principal Stresses
In the example of stress on a plane above, we could change the orientation of the plane
x1x2, such that it was perpendicular to the red (bold) arrow. When we do this, the shear
stress components go to zero and we are left with only the red (bold) arrow that is now
equal to the normal stresses on the plane.
x3
sn parallel to x 3
x2
x1
Now, if we extend this idea to the stress at a point concept, it turns out that there is one
orientation of the cube in which all of the shear stresses go to zero thus leaving each face
with only a normal stress acting on it.
Here, the remaining stresses, s 1,s 2, and s 3 represent the principal stresses,
where s 1 > s 2 > s 3
In this case, the Stress Tensor reduces to:
Lec. 7: Stress I: Introduction, Page 6 of 6
Another way to visualize the principal stresses is to imagine the stress ellipsoid which
consists of major, intermediate, and minor axes.