Stress • Introduction • The stress tensor • Cauchy formula • Principal ...

Stress

•  Introduction •  The stress tensor •  Cauchy formula •  Principal stresses •  Mohr circle •  Pore pressure •  Mohr-Coulomb criterion •  Power-law creep •  The state of stress in the lithosphere

Recommended reading is: •  Means, Stress and strain: Basic concepts of continuum mechanics for geologists •  Turcotte and Schubert, Geodynamics •  Pollard and Fletcher, Fundamentals of structural geology

Stress: Introduction

Geologists and geophysicists use stress rather than forces. Why? Consider two blocks of different cross-sections. Intuitively, the blocks whose cross-section is smaller is going to deform a lot more than the other.

For this reason, it makes sense to work with stresses.

STRESS = FORCE/AREA

Pascal=Newton/m2

Fig. from Twiss and Moores

Stress: Introduction.

•  A stress of 1Pa is very small. For example, the load due to 1m of water is about 104Pa

•  So it is better to use MPa •  Other units are bars and atm:

1MPa = 106Pa = 10bars = 9.8692atm


•  The traction is a vector quantity that acts at a point on an imaginary or real surface of arbitrary orientation.

•  The traction at a point on a surface is equal and opposite to the traction that at that same point for the same surface with opposite outward unit normal vector.


In general, a stress acting on a plane may be expressed as a sum of shear and normal stresses.

•  Normal stress: The component of stress acting perpendicular to the plane

•  Shear stress: The component of stress acting parallel to the plane


Consider a small cubic element of rock extracted from the Earth. The stress acting on this element may be visualized as follows:

•  The first index indicates the plane in question

•  The second index indicates the direction at which the stress acts


The sign convention adopted by geophysicist differs from that adopted by engineers: •  In engineering, the stress is positive if it acts in the positive

direction on the positive plane. In other words, the stress is positive in tension and negative in compression.

•  In geophysics, the stress is positive if it acts in the negative direction on the positive plane. In other words, the stress is positive in compression and negative in tension. [This is because the stresses in the Earth are compressive (although locally tensional stresses are possible too)].

Stress: The stress tensor

It is convenient to “pack” the stress components into a tensor form. In 3D: In 2D (that is, when ) :

σ ij =

σ11 σ12 σ13σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

!

"

####

$

%

&&&&

σ ij =σ11 σ12σ 21 σ 22

!

"

##

$

%

&&

σ13 =σ 31 =σ 23 =σ 32 =σ 33 = 0

Stress: The stress tensor

The stress tensor is symmetric: Unless the torque is equal to zero, the block will rotate. In the example below, the condition of zero torque may be written as: And we get that: . So the number of independent parameters is equal to 3.

− σ xyδyδz( )δx / 2+ σ yxδxδz( )δy / 2 = 0σ xy =σ yx

Similarly, in 3D we get that: , and the number of independent stress components is equal to 6

σ xy =σ yx, σ xz =σ zx and σ yz =σ zy

Stress: Cauchy formula

Consider a small cubic element of rock extracted from the earth, and imagine a plane boundary with an outward normal, n, and an area, δA cutting through this element - so it is reduced to a triangular element with sides 1 and 2.


The force components acting on sides 1 and 2 are: 1. Note that: 2. Replacing 2 in 1 gives: 3.

€

f1x = −σ xxδAcosγf1y = −σ xyδAcosγf2x = −σ yxδAsinγf2y = −σ yyδAsinγ .

€

cosγ = nxsinγ = ny.

€

f1x = −σ xxδAnxf1y = −σ xyδAnxf2x = −σ yxδAnyf2y = −σ yyδAny .


Force balance leads to: Rearranging the above: This is equivalent to: where tj is the traction acting on ni.

€

fx∑ = txδA −σ xxδAnx −σ yxδAny = 0

fy∑ = tyδA −σ xyδAnx −σ yyδAny = 0.

€

tx =σ xxnx +σ yxny

ty =σ xynx +σ yyny .

€

t j =σ ijni,

Stress: Principal stresses

We have learned that the stress tensor is symmetric. A property of symmetric matrices is that they may be diagonalized. The transformation from the non-diagonal to the diagonal tensor requires transformation of the coordinate system. The axes of the new coordinate system are the principal axes, and the diagonal elements of the tensor are referred to as the principal stresses. Note that the shear stresses along the principal axes are equal to zero.

€

σ ij =

σ1 0 00 σ 2 00 0 σ 3

#

$

% % %

&

'

( ( ( .

Stress: Mohr circle

Given a set of principal stresses, one can compute the shear and normal stresses on any plane. Adding vectors in directions parallel and normal to the plane in question:

€

FN = F1 cosθ + F3 sinθ

FS = F1 sinθ − F3 cosθ.

Stress: Mohr circle

This is equivalent to: Substituting the following trigonometric identities: gives:

€

σN =σ1 cos2θ +σ 3 sin

2θ

σ S = (σ1 −σ 3)sinθ cosθ.

€

sin2θ = (1− cos2θ) /2

sinθ cosθ = sin2θ /2.

€

σN =σ1 +σ 3

2+σ1 −σ 3

2cos2θ

σ S =σ1 −σ 3

2sin2θ.

Stress: Mohr circle This equation describes a circle on a σS vs. σN space, with a center on the horizontal axis at (σ1 + σ 3)/2, and a radius that is equal to (σ 1 - σ 3)/2. (σ1 + σ 3)/2 is the mean stress. (σ1 - σ 3)/2 is the differential stress. θ is the angle between σ1 and the normal to the plane - positive when measured counter-clockwise from σ1 .

€

σN =σ1 +σ 3

2+σ1 −σ 3

2cos2θ

σ S =σ1 −σ 3

2sin2θ.

σ1 ≥σ 2 ≥σ 3

A matter of convention:

Stress: Mohr circle

Note that for a given stress tensor, the mean stress is independent of the plane in question, that is: We can thus write the stress tensor as a sum of the mean stress field and the deviatoric stress field:

€

σmean =σ1 +σ 2

2=σ11 +σ 22

2.

€

σ11 σ12 σ13σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

#

$

% % %

&

'

( ( (

=

σmean 0 00 σmean 00 0 σmean

#

$

% % %

&

'

( ( (

+

σ11 −σmean σ12 σ13σ 21 σ 22 −σmean σ 23

σ 31 σ 32 σ 33 −σmean

#

$

% % %

&

'

( ( (

Stress: Mohr circle

Note that: •  Shear stresses equal to zero at θ=0 and 90 degrees.

•  Maximum shear stress is equal to (σ 1 - σ 3)/2 at θ=45 degrees.

•  The shear stresses along the principal directions are equal to zero.

•  The principal axes are orthogonal.

Stress: Mohr circle

Mohr circle in 3D

σ ij =90 0 00 60 00 0 30

!

"

###

$

%

&&&

Stress: Mohr circle

Mohr circle in 3D

σ ij =90 0 00 60 00 0 30

!

"

###

$

%

&&&

Stress: Mohr circle

Mohr circle in 3D

σ ij =90 0 00 60 00 0 30

!

"

###

$

%

&&&

Stress: Mohr circle

Mohr circle in 3D

σ ij =90 0 00 60 00 0 30

!

"

###

$

%

&&&

Stress: Mohr circle

Mohr circle in 3D A single Mohr circle describes the variation of shear and normal stress along a principal plane (a plane that contains 2 principal axes). The representation of a 3D state of stress is obtained by the superposition of three Mohr circles, as follows: The state of stress on planes that are not perpendicular to a principal plane fall within the shaded area.

Stress: The state of stress

Uniaxial stress: Only one non-zero principal stress. For example: Biaxial stress: One principal stress equals zero, the other two do not. For example:

Stress: The state of stress

Triaxial stress: All principal stresses are non-zero. For example: Axial stress: Two of the three principal stresses are equal. For example:

Stress: Pore pressure and effective stress

Pressure is a special state of stress in which the shear stress is equal to zero, i.e.: σ1= σ2= σ3. Question: How does this state of stress plot on a Mohr diagram? It is useful to consider two pressures: Lithostatic and hydrostatic. Lithostatic pressure: The stress equals the weight of the overlying column of rock. In the absence of tectonic forces or fluids, the state of stress would be lithostatic.

€

Plithostatic = g ρ(z)dzz∫ .


Hydrostatic pressure: The stress equals the weight of a column of water.

€

Phydrostatic = gρwaterz.Pore fluid: Is the fluid within the pores. Pore pressure: Is the pressure within the pore fluid. Usually the fluid is water, but it can also be oil or gas. In a granular medium, the pore pressure acts to reduce the contact between the grains.

fluid presses out equally in all directions in a saturated sediment the weight of all the water above produces a pore pressure that tends to push the grains apart


Effective stress: The effective stress tensor is: Question: Is pressure a vector or a scalar?

€

effective stress = normal stress - pore pressure

€

σ11 − P σ12 σ13σ 21 σ 22 − P σ 23

σ 31 σ 32 σ 33 − P

$

%

& & &

'

(

) ) )

=

σ11 σ12 σ13σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

$

%

& & &

'

(

) ) ) −

P 0 00 P 00 0 P

$

%

& & &

'

(

) ) )


The effect of pore pressure increase (for example, due to water pumping) is to lower the effective stress. Graphically, this may be illustrated as follows:

P

Stress: Frictional resistance

Question: Given that all objects shown below are of equal mass and identical shape, in which case the frictional force is greater?


Leonardo Da Vinci (1452-1519) showed that the friction force is independent of the geometrical area of contact.

Movie from: http://movies.nano-world.org/movies/frictionmodule/en/


Amontons' first law: The frictional force is independent of the geometrical contact area. Amontons' second law: Friction, FS, is proportional to the normal force, FN:

Movie from: http://movies.nano-world.org/movies/frictionmodule/en/

€

FS = µFN


Byerlee law: The coefficient of sliding friction for most rock types was experimentally determined to be 0.6-0.8.

Friction measured at maximum stress


The best fit line in the diagram defines the failure envelope used in Mohr Circle analysis. Mohr Circle equations and the Coulomb equation are both used to determine the ideal angle between maximum stress and the failure plane, which is referred to as φ. For a coefficient of friction of 0.7, φ is equal to 30 degrees. Question: for φ equals 30 degrees, compute θ. €

σ s = S0 +σ n tanφ = S0 + µσ n ,

Stress: Coulomb-Mohr criterion

Figure from Structural Geology by Fossen

Show animation from: http://folk.uib.no/nglhe/e-modules/Chapter%207/07%20Brittle%20deformation.swf


Figure from Structural Geology by Fossen


Normal stress

Shear stress

Coulomb equation

Question: What’s the effect of increasing the pore pressure?

So it is the effective normal stress that we should plug into the Coulomb-Mohr criterion:


Normal stress

Shear stress

Coulomb equation

P

σ s = S0 +σ n tanφ = S0 +µ σ n −P( ) .


Normal stress

Shear stress

Coulomb equation The set of two possible

solutions is referred to as the conjugate set


Normal stress

Shear stress



Imperial Valley (CA)

Figure from Scholz [1990]


Izu peninsula (Japan)

Figure from Scholz [1990]

Stress: Anderson model

Since the shear stress acting on the free surface is zero, the free surface is a principal plane. Thus, either σ1 or σ2 or σ3 is perpendicular to the free surface.

Stress: Anderson model

Vertical σ1 gives rise to normal faults dipping at about 60 degrees. Vertical σ3 gives rise to thrust faults dipping at about 30 degrees. Vertical σ2 gives rise to strike-slip faults dipping at about 90 degrees.

Stress: Fault mechanics

We assume x and y are principal directions of the stress tensor, and σyy is equal to the lithostatic pressure. Then: and: for a thrust: for a normal fault:

σ yy = ρgh

Δσ > 0

σ xx = ρgh+Δσ

Δσ < 0


the effective normal stress is: 1. and the shear stress is: 2. plugging 1 and 2 into the Coulomb criterion we get: 3.

σ n = ρgh−P +Δσ21+ cos2θ( )

τ = ±Δσ2sin2θ

Δσ =2µ ρgh−P( )

±sin2θ −µ 1+ cos2θ( ).

Stress: Fault mechanics The crust contains many preexisting fractures. These zones of weakness can be reactivated at an angle that minimizes the tectonic stress. The angle that gives the minimum value of stress is found via solution of: which yields: 4. and therefore:

tan2θ = 1µ

,

dΔσdθ

= 0,

tan2β = ± 1µ

.


Δσ

thrust normal Δσ =

±2µ ρgy−P( )1+µ 2( )

12 µ

.

Replacing 4 in 3 gives:

If the rock is already fractured, faults with certain orientation may be reactivated (Sibson, 1985).

fractured

unfractured τ

σ


Brittle deformation: Consequences of pre-fractured rocks

What are the conditions for fault reactivation (Sibson, 1985)? which may be rewritten as: This can be reduced to:

€

τ = µσ *,

€

σ1* −σ3

*( )sin2θ = µ σ1* +σ3

*( ) − σ1* −σ3

*( )cos2θ[ ].

€

R = σ1*

σ3*

#

$ %

&

' ( = 1+ µcotθ( ) 1− µ tanθ( ).


A plot of the stress ratio for fault reactivation versus the reactivation angle for µ=0.75 [Sibson, 1985]:

• The optimum angle for reactivation as a shear fracture is θ* with R=4.

• R increases to infinity as θ approaches 0 and 2θ*.

• R<0 for θ>2θ*, which requires σ3<0.


A further limitation on the allowable stress states for fault reactivation is that they must not induce failure of the surrounding rock. The stress circle is constrained by the the failure criteria of the intact rock.


The role of pre-existing foliation for constant σ3

Figure from Fossen

Brittle deformation: Consequences of pre-fractured rocks •  Under constant stress (strain) field, faults rotate.

•  The range of angles beyond which fault reactivation is impossible is referred to as the lock-up angles. •  Once faults orientation reaches the lock-up angle, they become inactive and a new set forms.

(the idea that strain accommodation in the brittle crust results in fault rotation is attributed to Freund [1974].)

Figure from Ron et al. [2000]





•  Note that exactly the same fault configuration may result from rotation of the stress field.

Figure from Ron et al. [2000]

Question: suggest ways to distinguish fault rotation from stress rotation?

Brittle deformation: Stress trajectories and stress concentration

Stress trajectory: A line indicating the continuous change in the orientation of a principal stress throughout a body. Although trajectories may curve, their intersections with other principal stresses remain perpendicular.


Superposition of a tectonic horizontal compressive stress on the standard state of stress


Adding a horizontal shear stress that increases with depth and has a constant magnitude along the base of the block


Example for stress concentrators

discontinuities inclusion cavity

E1

E2



Brittle deformation: Theoretical vs. actual strength

The modifying effect of a cavity on the distribution of stress in solid (Iglis, 1913):

Elliptical cavity: €

σ c =σ a 1+ 2c b( ).


Stress concentration for c=3b:

•  Schematic Illustration of local stress concentrations in a material with a circular and an elliptical hole.

•  A material with elliptical hole will be easier to pull apart.

Figure from Fossen

€

σ c =σ a 1+ 2c b( ).


Stress concentration for c=3b:

Thus, the local stress about a sharp notch or corner could rise to a level several times that of the applied stress, and even submicroscopic flaws my act as stress concentrators that weaken the solid.

€

σ c =σ a 1+ 2c b( ).


The figures above show the change in the fault-parallel shear stress and fault-perpendicular normal stress, due to right-lateral slip along a dislocation embedded in an infinite elastic medium

Isochromatics - contours of equal shear stress



Brittle deformation: Stress determination

World stress map

1. Faulting and focal mechanism


•  According to Anderson’s

hypothesis, dikes intrude perpendicular to the least compressive stress

•  Dikes that intrude a region where the stress field is homogeneous are expected to follow straight path, and are expected to follow a curved path if the stress is heterogeneous.

2. Igneous dikes


Map of Spanish Peaks region of southeastern Colorado with traces of dikes and trajectories of maximum compressive principal stress

Muller and Pollard, 1977

2. Igneous dikes


3. Hydraulic fracturing

•  Increase the fluid pressure until the wellbore fractures •  Record the fluid pressure and fracture orientation •  The stress vs. time record is used for the inference of the least

horizontal stress •  The magnitude of the least horizontal stress is equal to the ISIP

(The Instantaneous Shut-in Pressure (ISIP) is just sufficient to hold the fracture open)


•  Measurements of the minimum horizontal stress as a function of depth in the 3.5 km deep Cajon Pass borehole near the San Andreas fault

•  A series of ISIP measurements were carried out using both hydro-fractures and pre-existing fractures

60% lithostatic

lithostatic

(Zoback and Healy, 1992)

3. Hydraulic fracturing


4. Wellbore breakouts

•  Although the borehole drill cut is cylindrical, some wellbores have systematic increase in radii along two diametrically opposed zones

•  This fracturing is caused by the stress concentration induced by the borehole

•  The interpretation is that the breakouts align with the direction of the least horizontal stress


Map of Western Canada with directions of least horizontal compressive stress inferred from wellbore breakouts

4. Wellbore breakouts


•  Drill a hole in an intact rock •  Install strain gauges at 3

perpendicular directions •  Overcore the first hole •  The strains on the strain gauges

are used to determine the original state of stress.

This technique rests on the assumption that the overcoring completely relieves the stress in the isolated block

5. Overcoring

Brittle deformation: Stress maps

Fig. from Heidback et al., 2007


Fig. from Heidback et al., 2007


Fig. from Townend and Zoback, 2004 Breakout: inward-pointing arrows Hydraulic fracturing: stars Focal mechanism: circles


Fig. from Townend and Zoback, 2004

Breakout: inward-pointing arrows Hydraulic fracturing: stars Focal mechanism: circles


•  Stress orientations are consistent on regional scale •  Boundaries between provinces of uniform stress direction

generally coincide with major structural boundaries •  In many regions, the orientation of the maximum compressive

horizontal stress generally coincides with directions of plate motion

•  In central California, the axis of maximum horizontal compressive stress lies at a high angle to the SAF, implying low friction (this is highly debated matter)

Brittle deformation: Does Anderson theory hold for all faults?

Question: What is the nature of the contact?

Figure from Vapnik and Avigad [2004]

Eclogite-Blueschist

Greenschist

Geological map of Tinos Island (Cyclades, Greece)



Figure from Zeffren et al. [2005]

•  These low angle normal faults are widespread all over the Cycladic massif (Greece).

•  Such tectonic contacts are commonly referred to as detachments.


Very low angle reverse faults are referred to as low-angle overthrusts or thrust sheets.

•  Allochthon: A package of rocks which has been moved a long way from their original place of deposition.

•  Autochthon: Rocks that have moved little from their place of formation.

•  Klippe: An isolated remnant of allochthon resting on a lower plate.

•  Tectonic window: A ''hole'' through the thrust sheet that exposes isolated area of rocks that lie beneath the thrust.


Chief Mountain (Montana) is an example of a klippe. It consists of a Precambrian block which rests directly above younger Cretaceous gray shales. The surrounding portion of the thrust sheet has be removed by erosion leaving behind this isolated block of Proterozoic rock.


Find the tectonic window

Brittle deformation: Mechanics of low angle faults

The problem is: how to force a very thin sheet to slide as a (almost) rigid body, without causing internal faulting.

Brittle deformation: Coulomb failure criterion in terms of principal stresses

Recall that the Coulomb failure criterion in terms of normal and shear stress is: where: •  µ is the coefficient of friction (equals tanφ) •  c is cohesion The Coulomb failure criterion in terms of the principal stresses is: where:

€

σ s = c + µσ n ,

€

σ1 = C0 + Kσ 3 ,

€

K =1+ sinφ1− sinφ

and C0 = 2c cosφ1- sinφ

.


The problem: is how to force a very thin sheet to slide as a (almost) rigid body without causing internal faulting.

€

σ xxdz0

z

∫ = σ zx0

x

∫ dx .

Force balance along the x direction requires that:


Let’s first evaluate the right-hand side of the force balance equation. Since σzx is the frictional resistance to sliding, we write: where µ is the coefficient of friction. The normal stress is just the lithospheric stress: with ρ being the rock density and g being the acceleration of gravity. So:

€

σ zx = µσ n = µσ zz ,

€

σ zz = ρgz ,

€

σ zx0

x

∫ dx = µρgz0

x

∫ dx = µρgzx .


Next we evaluate the left-hand side of the force balance equation. We are seeking an expression for the largest σxx that the block can support without breaking. So we'll use the expression for Coulomb criterion. We identify: We thus get: Equation the right and left hand sides gives:

€

σ xx =σ1 and σ zz =σ 3 .

€

σ xxdz0

z

∫ = C0 + Kρgz( )0

z

∫ dz = C0z + Kρgz2

2 .

€

C0z + Kρgz2

2= µρgzx .


Thus, given a thrust sheet of thickness z, the largest x is given by: •  We plug in realistic numbers: φ = 300 (i.e., µ = 0.6), S0 = 20 MPa, and ρ = 2300 Kg/m3.

•  With these values we get: x = 5.4 km + 2.6 z.

Often, however, it is observed that x >> 5.4 km + 2.6 z

€

x =C0

µρg+ Kz

2µ .


There are several ways out of this paradox: •  Rheology is incorrect.

•  Elevated pore pressure.

•  My favorite solution for this ''paradox'' is that thrust sheets do not move as rigid bodies, instead they slide like a wrinkle of a rug!


The wrinkle model:


The wrinkle model:

Figure taken from: http://ic.ucsc.edu/~casey/eart150/Lectures/DefMech/14deformationmechanisms.htm


Insight from seismology, the 1992 Landers example and the carpet wrinkle analogy:

Wald and Heaton, 1994

Stress • Introduction • The stress tensor • Cauchy formula • Principal ...

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