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1/25
COULOMB STRESSCOULOMB STRESS
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cracks form cracks coalesce
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3
P
1
(1 + 3 )
(1 - 3 )
(1 - 3 )2 2
2
n
s
point P represents a plane that is degrees from 3plots 2 counterclockwise from 1 (note original diagram in upper right)
s
n
1
3
D
Mohr circle
A
BC
sP
nP
normal and shear stresses on point P are nP and sP, respectivelynP = 1/2(1 - 3 ) + 1/2(1 - 3 )cos 2sP = 1/2(1 - 3 ) sin 2
13
P: plane to left
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conduct triaxial loading experiments to determine applied stress at which sample breaks
c
a a= axial stress, 1c= confining stress, 3
first experiment set confining pressure low and increaseaxial load (stress) until sample breaks
second experiment set confining pressure higher and increaseaxial load (stress) until new sample breaks
keep repeating experiments
5/25
you will generate a series of pairs of confining stresses andassociated axial stresses at which samples break
40 540 500 MPa150 800 650 MPa400 1400 1000 MPa
ac
we use these as 1 and 3 and plot Mohr circles
get a sequence of circlesoffset from one another
a- c s
500
500
1000n
1500
diameters are stress difference;centers are stress sum/2
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s
n500
500
1000 1500
Mohr circles that definestress states wheresamples fracture
(critical stress states) together define the
failure envelopefor a particular rock
failure envelope
failure envelope is tangent to circles of all critical stress statesand is a straight line
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what does this straight line mean?
corresponds to Coulomb fracture criterion
Charles Coulomb in 18th century proposed that formation ofshear stress parallel to failure relates to normal stress by
s = C + tan (n) (empirical)
s = shear stress parallel to fracture at failureC = cohesion of rock (constant)n = normal stress across shear zone at instant of failuretan = = coefficient of internal friction (constant of proportionality)
this has form of y = mx + b (equation of a line)
y = s x = n b = intercept on s axis m = slope = tan =
so Coulomb criterion plots as straight line on n , s plot
8/25
suppose 1 is oriented east-west, horizontal, and equal to 40 MPa
3 is vertical and equal to 20 MPa
determine normal and shear stresses on a fault plane thatstrikes north-south and dips 55 west
5540 MPa
20 MPaangle, , between 3 and fault plane is 35;
2 = 70
1
3
but which direction on Mohr circle?
55
35
3
9/25
convention for shear stresses for Mohr circle diagram: sinistral is considered positive (+) dextral considered negative (-)
positive (sinistral) negative (dextral)
1
3
our example is dextral andthus, negative
10/25
to construct Mohr circle, we know
1 is equal to 40 MPa3 is equal to 20 MPa
2 = 70 and is negative
10
20 4070
10 n= 33.4 s = -9.4
P
n
s
1 + 32
distance from origin of center of circle(40+20)/2=30
1 - 3 diameter of circle: 40-20=20
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stress acts on every surface that passes through the pointwe can use three mutually perpendicular planes to describe
the stress state at the point, which we approximate as a cube
each of the three planeshas one normal component
& two shear components
therefore, 9 components necessaryto define stress at a point
3 normal6 shear
convention for describing: iji= plane on which component acts (defined by perpendicular)j = axis to which component is parallel
i = 1, 2, 3 j=1, 2, 3
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the nine components (six of which are independent) can be writtenin matrix form:
11 12 13
21 22 23
31 32 33
ij =
this is the stress tensor
describing stress as tensor allows relative ease of manipulation;can change reference frame (rotation; translation)
components on diagonal are normal stresses; off are shear stresses
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Earthquake ruptures before the Izmit earthquakeEarthquake ruptures before the Izmit earthquake
20/25
Coulomb stress field prior to the Izmit earthquakeCoulomb stress field prior to the Izmit earthquake
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Resolved stresses on the Izmit and Dzce faults before August 99Resolved stresses on the Izmit and Dzce faults before August 99
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Coulomb stress with varying coefficient of friction valuesCoulomb stress with varying coefficient of friction values
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Coulomb stress changes due to the Izmit earthquakeCoulomb stress changes due to the Izmit earthquake
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Stress changes on the Dzce fault due to the previous earthquakeStress changes on the Dzce fault due to the previous earthquakess
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Present day Coulomb stress changes due to the past earthquakesPresent day Coulomb stress changes due to the past earthquakes
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