Structural Geology
Force and Stress –
Normal and Shear Stress
Lecture 5 – Spring 2016
1
Rocks and Force
• Rocks constantly experience the force of gravity
• They may also experience a variety of other
forces, including tectonic forces and forces
associated with impact
• Previously, we saw that force is defined by the
following equation:
 F = mA
 where F is the force vector, m is mass, and A is the
acceleration vector
2
Responses to
Force
Figure 3.2 in text
• Rocks respond to applied forces in one of two ways:
 A. Movement – Newtonian mechanics
 B. Distortion – continuum mechanics
3
Types of Force
• Body forces
 Fb ∝ m
• Surface forces
 Fs ∝ area
4
Definition of Stress
• We have previously seen that stress is an internal
force set up as the result of external forces acting
on a body
• Stress is usually represented by the Greek letter
sigma, σ
 σ = F/Area
 Defined in this manner, stress is really an intensity of
force, which measures how concentrated the force is
• Subscripts are often attached to σ to add additional
information
5
Stress in Different Dimensions
• In two dimensional problems, stress is a
vector quantity, and is sometimes called
traction
• In three dimensions, stress is a second-order
tensor, which will be discussed shortly
6
Traction
• Stress in an arbitrary
direction may be resolved into
components
• A. Normal stress,
denoted σn
• B. Shear stress, denoted
σs or τ (tau)
• Figure 3.3 in text
7
Resolution of the Stress Vector
• Figure 3_4a illustrates the
principle of stress resolution
• A plane face is ABCD in the
drawing
• Note: The section through a cube
implies a plane, i.e. two
dimensions
8
Force and Stress
Figure 3.4a
& b in text
• A force, F, is applied along rib AB
• Line EF in the drawing is the trace of a plane which makes
an angle θ with the top and bottom surfaces of ABCD
• The force can be resolved into components Fn perpendicular
to the plane, and Fs parallel to the plane
9
Fn and Fs
•
•
•
•
•
σ = F/AB
(Note: F = σAB)
Fn = F cos θ = σAB cos θ
Since AB = EF cos θ,
Fn = σEF cos2θ
Fs = F sin θ = σAB sin θ = σ EF sin θ cos θ
10
Trigonometry Identity
• We can use the following trigonometric
identity to simplify Fs
 sin θ cos θ = ½ (sin 2θ)
• Fs = σEF ½ (sin 2θ)
11
Normal and Shear Stress
• σn = Fn/EF = σ cos2θ
• σs = Fs/EF = σ ½ (sin 2θ)
12
Stress Vector Resolution
• Thus, the stress vector acting on a plane can
be resolved into vector components normal
and parallel to the plane
• Their magnitudes vary as a function of the
orientation of the plane
13
Normal Force and Stress vs. θ
• Plot of the normalized values
of normal force and the
normal stress versus theta
• The curves have a slightly
different shape, but in both
cases the normalized values
decrease and go to zero at θ =
90º
Figure 3.4c in text
14
Shear Force and Stress vs. θ
• The curves in this case are
nearly identical until θ = 25º,
then the shear force increases
faster than the shear stress
• After 45º, the shear force
continues to increase, but the
shear stress again goes to zero
at 90 º
Figure 3.4d in text
15
Three Dimensional Stress
• 3D analysis is more complicated than 2D, because stress in
three dimensions is a second order tensor
• To reduce the complexity somewhat, we can stipulate that
the rock being discussed is at rest
 According to Newton’s Third Law of Motion, bodies at rest are
acted upon by balanced forces, otherwise they would start to move
• We can consider the stress at a point within the body
 A point is the intersection of an infinite number of planes, in every
possible orientation
 Using a point allows us to discuss force and stress components in
any plane
16
Stress Ellipse
• Figure 3. 5a shows a plane cut by four other planes (a
through d)
• The stresses on each plane are plotted, and are perpendicular
to their respective planes
• Since the body is at rest, every stress is opposed by an equal
an opposite stress
• We can connect the endpoints of the two dimensional stress
vectors with a smooth curve, generating the ellipse shown
17
Stress Ellipsoid
• If we were to draw similar ellipses in the two
additional, mutually perpendicular, planes, we could
then combine the data to generate a three dimensional
ellipsoid, as shown in figure 3.5b
• This is known as the stress ellipsoid
18
Principal Stresses
• Ellipsoids are characterized by three principal axes
• In the stress ellipsoid, these axes are known as the
principal stresses, which have two properties
 They are mutually perpendicular
 They are perpendicular to three planes which have no
shear stresses
 The three planes are known as the Principal Planes of
Stress
 Each principal stress is a vector
19
Stress Using Cartesian
Coordinates
Figure 3.6 in text
• Stress can be visualized in
another manner
• Using a standard three
dimensional Cartesian
coordinate system, and we
picture a point cube, as shown
in figure 3.6
• We can resolve the stress
acting on each face of the cube
into three components
20
Stress
Notation
• The face normal to the x-axis has a component σxx
• First subscript refers to the plane, in this case the one normal
to the x-axis
• Second subscript refers to the component along axis x
• In addition, we have two shear stresses, σxy and σxz, which
lie along the y and z axes within the plane under consideration
21
Table of Stress Components
Doing the same for the other principal stress axes,
we generate a table
Stress on face
normal to:
x
y
z
x
σxx
σxy
σxz
y
σyx
σyy
σyz
z
σzx
σzy
σzz
X
Y
Z
In the direction of:
22
Equivalence of Shear Stress
Components
• Since the object is at rest, three of the six
shear stress components must be equivalent
to the other three (otherwise the object
would move)
 σxy = σyx, σxz = σzx, and σyz = σzy
23
Independent Stress Components
• This leaves six independent components:
Stress on face
normal to:
x
y
z
x
σxx
σxy
σxz
y
σxy
σyy
σyz
z
σxz
σyz
σzz
X
Y
Z
In the direction of:
24
Principal Stress Table
Stress on face
normal to:
x
y
z
x
σxx
0
0
y
0
σyy
0
z
0
0
σzz
X
Y
Z
In the direction of:
Thus oriented, the axes are known as the principal axes
of stress, and the planes perpendicular are the principal
25
planes of stress
Isotropic Stress
• It is possible that the three principal stresses
will be equal in magnitude
• If this condition is met, the stress is said to
be isotropic
• The stress ellipsoid becomes a sphere
26
Anisotropic Stress
• When the principal stresses are unequal,
they are said to be anisotropic
• We then introduce another convention:
 σ1  σ2  σ 3
• σ1 is called the maximum principal stress
• σ2 is the intermediate principal stress
• σ3 is the minimum principal stress
27
Types of Stress
• General Triaxial Stress
σ1 ≥ σ 2 ≥ σ3 ≠ 0
• Biaxial stress, with one axis at zero
 σ1 ≥ 0 ≥ σ3 or σ1 ≥ σ2 ≥ 0
• Uniaxial tension
 σ1 = σ2 = 0; σ3 < 0
28
Uniaxial Stress
• Uniaxial compression
 σ1 > 0; σ2 = σ3 = 0
• Hydrostatic or lithostatic pressure
 σ 1 = σ 2 = σ3
29
Gabriel Auguste Daubrée
• Daubrée (1814-1896) was an early
experimenter in many aspects of the
geological sciences
• He taught mineralogy at the French
School of Mines
• He introduced synthesis techniques
and extended these to general
experimental work
30
Daubrée
Experiment
• For example, Figure 3-7a
shows a picture of wax
placed between two wooden
plates, an experiment first
performed by Daubrée
• He reported some of his
results at the first
International Geological
Conference in 1878 (Paris)
31
Diagram of Daubrée Experiment
• Plane AB is arbitrary, and
it makes an angle θ with
σ3
• We can make two other
simplifying assumptions:
 Line AB has unit length
 The plane represented by
AB within the block has
unit area
32
Forces in Balance
• The forces parallel and perpendicular to AB
must balance
• We resolve force z AB into the
component of the force z BC (parallel to
σ1) along CD plus the component of the
force z AC along CD (parallel to σ3)
• Note that ACD = θ
33
Resolving Forces
• Various forces may be resolved as:




forceBC = σ1cosθ (Force = stress C
forceAC = σ3sinθ
AreaBC = 1 C (cos θ)
AreaAC = 1 C (sin θ)
area)
 On the AB surface, there is a normal stress,
σn and a shear stress, σs
34
Normal Stress
• The normal stress is the same as the stress
along CD:
 σn = σ1cosθCcosθ + σ3sinθCsinθ = σ1cos2θ +
σ3sin2θ
• Since cos2θ = ½ (1 + cos2θ) and sin2θ = ½
(1 - cos2θ) we get
 σn = ½ (σ1 + σ3) + ½ (σ1 - σ3) C cos2θ
35
Shear Stress
• We resolve force 2 AB into the component
of the force z BC along AB plus the
component of the force z AC along AB
 σs = σ1cosθCsinθ - σ3sinθCcosθ =
(σ1 - σ3) sinθCcosθ
36
Simplification of Shear Stress
• Substituting sinθCcosθ = ½ sin2θ gives
σs = ½(σ1 - σ3)sin2θ
• The planes of maximum normal stress are at
θ = 0E relative to σ3, because cos2θ = 1 at
θ = 0E
• The planes of maximum shear stress are at
45E relative to σ3, because sin2θ = 1 at
45E
37