Transcript Loading joints

Distribution of Microcracks
in Rocks

Uniform


As in igneous rocks where microcrack
density is not related to local structures but
rather to a pervasive process such as cooling.
Variable

as a function of distance with respect to local
structure such as joints, shear fractures, and
larger faults.
Mechanical Processes and Microcracking



The tensile stress necessary for propagation of
microcracks arise either from thermal or mechanical
processes.
Mechanical processes include:
 stress concentrations at grain boundaries and around
cavities.
 elastic mismatch-induced cracking
 kink band and deformation lamellae-associated
microcracking.
 twin-induced microcracking.
Mechanically-induced microcracks tend to have a nonuniform distribution in association with local
structures or discontinuities.
Thermal Stresses and Microcracking



Arise from differential and incompatible expansion or
contraction between:
 Grains of different thermoelastic properties
 Similar but mis-aligned anisotropic grains.
Experimental work shows that a:
DT < 50oC can generate thermally-induced microcracks.
 Such changes can occur during burial and uplift; so
they should be common within uplifted rocks.
Thermally-induced microcracks are more likely to have a
uniform distribution in large rock volume.
Loading History

Loading refers to history of applied forces (over
time), displacement, and temperature, which
control the net deformation of a rock.

There are three types of loading:
 Gravitational
 Thermal
 Tectonic
Gravitational Loading






Is the most important, and varies with time by erosion.
Is due to the weight of the overlying rocks: sz = rg z
It affects horizontal stress because rocks are elastic and
tend to expand transverse to compressive vertical stress
Poisson ratio: n = et/el where n = 0.25 for most rocks when
material is free to expand laterally.
If rock is not free to expand (et = 0), a transverse stress st
is created which is given by:
st = n /(1- n) sz
which for a n = 0.25 becomes:
st = sz/3 = rgz/3
This means the horizontal stress, induced by gravitational
load, is 1/3 the vertical load.
Uplift and Denudation

Many joints form by the release of tensile
stresses generated during uplift and denudation.

These stresses change the pre-existing state of
stress (Ds) in three ways:
1. Horizontal stretching through the geometry
of uplift, Dss
2. Expansion through the release of the
gravitational load, Dsg
3. Contraction through cooling, Dst
Uplift and Denudation, cont’d



If a body of rock with mean density, r, is lithified at some
depth (h) in the earth, it is subjected to an isotropic stress,
P:
P = sz= sx = sy = rgh
The state of stress after a given amount of erosion
(negative Dz) is:
sz = rg(h+Dz)
The new horizontal stress is determined by the sum of the
effects due to :
 horizontal stretching
 expansion by the release of the gravitational load.
 contraction through cooling.
sx= sy = rgh + D ss + D sg + D st
Horizontal Stretching

When a segment of the crust is uplifted; it
may stretch horizontally.

The amount of stretching is inversely
proportional to the radius of curvature of the
flexure, i.e.:

The strongly curved parts of folds formed
under brittle upper crustal conditions
commonly display more concentrations of
joints and small faults.
Release of the Gravitational Load

Release of gravitational load by erosion
causes expansion of compressed rock.

Since rocks are not free to expand
horizontally, much of the horizontal
compressive stress remains after erosion.
Thermal Contraction



A change in horizontal stress is produced during
erosion as a result of cooling.
Strain and temperature are related by:
e = a DT
where a is the coefficient of thermal expansion.
If, however, the rock is not free to expand or
contract (i.e., et = 0), a stress, called thermal
loading, is set up, given by:
Dst = -aE /1-n DT
Example

The surface of a lava flow is in tension because
it cools faster than inside it; so it fractures.

If an uplift leads to a DT=100oC in a confined rock,
we get:
st = -aE /1-n DT
st = (10-6/oC) (10 5 Mpa) / (1-0.25) 100oC
Which gives a st = -13 Mpa; equal to the tensile
strength of most rocks, so joints will form.

Thermal Contraction …

For slow uplift, the change in temperature is
determined by the equilibrium thermal gradient.
Dst = dT/dz Dz

For very rapid uplift the temperature change may be
substantially higher

In this case, given a DT= dT/dz Dz, the maximum
thermal estimate for stress is:
Dst = aE /1-n dT/dz Dz
The Net Effect of Uplift

The stress generated during uplift and erosion
is the net effect of the horizontal stretching,
expansion by the release of the gravitational
load , and contraction through cooling.
sx= rgh + Dss + Dsg + Dst

The horizontal stress, sx , at a depth z =(h+D z)
after some erosion (negative Dz) is given by:
sx = rg[ z -(1-2n )/(1-n) Dz]+aE/1-n dT/dz Dz
The Net Effect of Uplift

Notice that the resulting stress is a function of
several material properties (r, n, a, E ) and the
thermal gradient (dT/dz).

Thus, under constant thermal stress, some
rocks with different properties may form
joints.
 Joints form by the release of tensile stresses
generated by uplift and denudation.
Residual Stress

A stress system that acts in a body of rock at
equilibrium when external forces and temperature
gradients do not exist.

Builds up in rocks normal to compressive stresses
when rocks cannot accommodate strain in
transverse direction
Note: n = et/el , et = 0
Residual Stress

Caused by change in T and P in homogeneous
rocks with different thermoelastic properties or
their spatial relationship (e.g., granite surrounded
by schist).

Residual stresses generated by pressure release,
and cooling during uplift and erosion, play an
important role in the formation of some joints.
Differential Stress for Jointing



The shape and size of the Mohr circle places an upper
limit for the deviatoric stress and hence depth of joint
formation.
The maximum deviatoric stress
(s*1 - s*3) for a tensile effective normal stress ( s*n < 0) is
about 6 times the tensile strength (To).
(s*1-s*3) < -6 To
This is even smaller (4 times the tensile strength ) for true
joints:
(s*1-s*3) < |4To|
Depth of Jointing

Joints require tensile effective normal stress (s*n
<0)

But the vertical stress due to gravitational load is
compressive (sz = rgz)

Therefore, formation of joints at deep levels
require both small deviatoric stress and high fluid
pressure (l)

For transitional joint , the estimate for the
maximum depth, z, is given:
z = | (2+2o2)To / rg(1-l) |

For true tensile joints, the maximum effective
stress (s*1) is limited to 3 times the tensile
strength (s*1 < |3To|).
The maximum depth for true joints is given by
(is 60% of z for transitional joints):
z < | 3To / rg(1-l) |

Orientation of Joints

Fluid-solid interfaces such as earth surface and
open fluid-filled joints are principal planes of
stress (since fluid cannot support shear stress).

Thus, the principal stress must be
perpendicular to these interfaces.
Orientation of Joints

Since joints form perpendicular to the least principal
stress (a = 0o; q = 90o),
we expect joints to be either:
 parallel (s3 vertical) or
 perpendicular (s1 or s2 vertical) to the earth
surface (dikes, veins, sills).

Tensile joints propagating near a preexisting open joint
will bend into perpendicular or parallel orientation with
respect to the preexisting joint.
Joints in Bedded Sedimentary Rocks

Assume a multi-layer of different rock type (1, 2, …, n)
with their own individual Young’s modulus (E), tensile
strength (T), and thickness (d): F and e are the
horizontal stress & strain.
_____________________________
E1, T1, d1,
F1 = E1e1 layer 1
_____________________________
E2, T2, d2,
F2 = E2e2 layer 2
_____________________________
En, Tn, dn,
Fn = Enen layer n
____________________________
Joints vs. Rock type in Bedded Rocks

The new horizontal stress in each of the layers after a
uniform horizontal stretching (ex; e.g., due to
epeirogenic warping) is:
F1 = E1 (e1 - ex)
F2 = E2 (e2 - ex)
F3 = E3 (e3 - ex)
Fn = En (en - ex)


Notice that even though the ex is uniform, the horizontal
stresses in each layer will be different. Thus some layers
will fracture first if their tensile strength is reached.
Therefore, fracture spacing is a function of rocks type (T,
E).
Joint Spacing and Bed Thickness






Cracks release significant stress only within a radius of
about one crack length.
Thus, a fracture has little effect on the state of stress in
its own bed more than one bed thickness away.
The rest of the bed is very close to the stress for tensile
failure.
With the small increase in strain, other joints will
form, each only relaxing the stress within one bed
thickness spacing.
Thus, fracture spacing is related to bed thickness.
With continued strain, the entire bed will be jointed with
spacing equal to bed thickness.