Week 7
Joints and Shear Fractures I. Read pages 204-256 in Chapter
5: Joints and Shear Fractures.
You are expected to read all the
sections listed below. Information from the sections in italics
will be discussed in class. You are expected to read the other
sections and you may be called on in class to answer questions
based on that material.
- Definitions and distinctions p.204-214
- A Detailed Look at Individual Joint Surfaces p.214-221
- A Close Look at Propagation Relationships Among Joint
Surfaces p.221-226
- Creation of Joints and Shear Fractures in the Laboratory
p.226-245
- The Influence of Pore Fluid Pressure p.245-251
- A Microscopic Look at the Mechanics of Fracturing
p.252-256
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You should become familiar with the
following terms during this weeks lectures and readings:
| abnormal fluid pressure |
angle of internal friction |
angle of sliding friction |
antitaxial veins |
| asperities |
Byerlee's law |
cohesive strength |
Coulomb envelope |
| Coulomb law of failure |
crack-seal vein |
crystal fibers |
en echelon joints |
| effective stress |
fluid pressure ratio |
Griffith cracks |
hackles |
| hydraulic fracture |
hydrostatic pressure |
joint |
joint bands |
| joint fringe |
joint intersections |
joint system |
opening (Mode I) fractures |
| origin |
plume axis |
plumose structure |
process zone |
| ribs |
ridge-in-groove lineations |
scissors (Mode III) fractures |
shear fractures |
| slickenlines |
sliding (Mode II) fractures |
structural domains |
syntaxial vein |
| systematic joint |
tensile strength |
vein |
von Mises criterion |
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You should be able to answer the questions
below following this week:
- Describe the ornamentation associated with a joint
surface.
- How can joint intersections be used to determine the
relative ages of joints?
- What is the approximate ratio of critical shear stress to
normal stress needed to cause brittle failure?
- What is the relative orientation of shear fractures
predicted by the Coulomb law of failure?
- Use a copy of the Mohr diagram in Figure 5.42 to answer
the following question. What is the maximum compressive
stress needed to cause failure if the confining pressure
is zero. What is the critical shear stress at failure?
- Use Figure 5.44 to estimate the angle of shear fractures
during failure of: a) Westerly granite; b) Carrera
marble; c) Berea sandstone.
- What are the differences between syntaxial and antitaxial
veins?
- How does fluid (pore) pressure influence the stress
necessary to cause faulting in both undeformed and
previously faulted rocks. Use relevant equations in your
answer.
- What is the Mohr-Coulomb law of failure and how is it
related to the Mohr circle for stress?
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Joints and Shear Fractures
Fracture classification
- Mode I (opening) fractures: extension
perpendicular to fracture surface (joints)
- Mode II (sliding) fractures: shear parallel
to the fracture surface and perpendicular to the fracture
front (shear fractures)
- Mode III (scissors) fractures: shear
parallel to the fracture surface and parallel to the
fracture front (shear fractures)
Joint surface characteristics
- main joint face bordered by fringes (= joint termination)
- joint ornamentation
- origin - point where joint propagated
- hackles - linear or curved markings on
joint face (collectively = plumose structure)
- ribs - curved steps in joint face, oriented
perpendicular to hackles, old joint terminations
Joint intersections and terminations
- joint appearance varies with level of exposure
- individual en echelon joints at surface
- single continuous surface at deeper levels
- hook patterns form at joint terminations,
common in proximity to other joints (parking lot)
Creation of Joints and Shear Fractures in the Lab
multiple experiments with brittle deformation yield several
Mohr circles that can be used to define the envelope of
failure. Failure stress can therefore be predicted for a
range of conditions.
- envelope is parabolic in tensile field of Mohr diagram
and linear in compressive field
- tensile strength is less than compressive strength (at
least 1:2 ratio)
- failure envelope (Coulomb envelope) has a slope of f,
(the angle of internal friction)
Coulomb law of failure
critical shear stress = cohesive strength + coefficient
of internal friction (normal stress)
for unfaulted rocks where:
- cohesive strength of rock = point where failure envelope
intersects y-axis)
- coefficient of internal friction = tangent of slope of
failure envelope
We can use the equation to determine the:
approximate ratio of shear stress to normal stress for
failure;
approximate orientation of the fault formed by brittle
failure
The Mohr stress diagram with the Coulomb failure envelope can
be used to illustrate the relative magnitudes of the principal
stresses needed for failure
With additional confining pressure the rock will deform with
plastic behavior and the von Mises criterion will govern
deformation.
- The failure envelope will become concave-downward and may
flatten
- A Mohr circle constructed for this part of the failure
envelope will intesect the envelope with a failure angle
(2q) of 90 degrees
- indicating a shear zone oriented at 45 degrees to maximum
principal stress.
Coulomb failure criterion deals with fracture at a point in a
homogeneous isotropic material that is homogeneously stressed.
However, in nature, most faulting occurs on pre-existing
faults.
Failure of pre-fractured rocks
Thus we must consider the implications for failure along
faults that will not necessarily be oriented at the optimum angle
to the maximum principal stress.
- for pre-existing faults, cohesive strength no longer
applies
- the coefficient of internal friction (tanf) is replaced
by the coefficient of sliding friction
Frictional resistance to sliding is greater for rough surfaces
which contain asperites - irregularities on the sliding
surface.
Mohr circle for stress plotted for experiments with
pre-fractured rocks shows a linear relationship between shear
stress and normal stress.
- the angle of the slope is ff
- the "new" failure criterion for pre-fractured
rocks is critical shear stress = tanff (normal stress), Byerlee's
law
Under normal conditions, pre-existing fractures with a range
of orientations may fail prior to the formation of a new fracture
oriented at 30o to maximum principal stress. (see Fig. 5.48)
Influence of Pore Fluid Pressure
Hubbert & Rubey showed that high pore pressures can
decrease the effect of normal stress
- effective stress = (normal stress - pore pressure[Pf])
- critical shear stress = cohesive strength + tanf (normal
stress) becomes
In both cases, critical shear stress, is substantially reduced
fluid pressure ratio = fluid pressure/lithostatic
(normal) pressure fpr = Pf/Pl [Pf = fpr(Pl) = fpr (normal
stress)]
- Pf = fluid density x depth x gravity
- Pl (= normal stress) = rock denisty x depth x gravity
- fpr = fluid density/rock density ~ 1000/2500 = 0.4
The equation, critical shear stress = tanff (normal stress -
Pf) can be rewritten as
- critical shear stress = tanff x normal stress(1 - fpr)
- for unfractured rocks, sc = s0 + tanf sn(1 - l) , with
elevated fluid pressures
Vein Formation
Elevated fluid pressures can explain the formation of joints
at substantial depths (Fig. 5.49).
As joints open, they may be filled with vein material to form crystal
fiber veins, which may preserve a record of the veins
opening.
- crystals grow in opening direction (direction of least
principal stress)
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