Transcript Document

Stress II
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, A cutting through this element - so it is reduced to a
triangular element with sides 1 and 2.
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The force components acting on sides 1 and 2 are:
1.
f1x   xxA cos
f1y   xyA cos
f 2x   yxA sin 
f 2y   yyA sin  .
Note that:
2.
cos  nx
sin   ny.

Replacing 2 in 1 gives:
3.

f1x   xxAnx
f1y   xyAnx
f 2x   yxAny
f 2y   yyAny .
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Force balance leads to:
f
f
x
 t xA   xxAnx   yxAny  0
y
 t yA   xyAnx   yyAny  0.
Rearranging the above:
t x   xx n x   yx n y

t y   xy n x   yy n y .
This is equivalent to:

t j   ij ni,
where tj is the traction acting on ni.

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Principal stresses
We have learned that the stress tensor is symmetric. A property of
symmetric matrices is that they may be diagonaliszd. 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.
1 0 0 


 ij  0  2 0 .


0
0


3 
Note that the shear stresses along the principal axes are equal to
zero.

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Using principal stresses to calculate shear and normal stresses on
a given plane
Adding vectors in directions parallel and normal to the plane in
question:
F  F cos  F sin 
N
1
3
FS  F1 sin  F3 cos .
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This is equivalent to:
 N  1 cos2    3 sin 2 
 S  (1   3 )sin  cos .
Substituting the following trigonometric identities:

sin2   (1 cos2 ) /2
sin  cos  sin2 /2.
gives:

N 
1   3
2
S 

1   3
1   3
2
2
cos2
sin2 .
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The above equation defines a circle with a center on the
horizontal axis at (1 +  3)/2, and a radius that is equal to ( 1 - 
3)/2.
(1 +  3)/2 is the mean stress.
(1 -  3)/2 is the deviatoric stress.
is the angle between 1 and the normal to the plane - positive
when measured counter-clockwise from  .
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Note that for a given stress tensor, the mean stress is
independent of the plane in question, that is:
 mean 
1   2
2

11   22
2
.
We can thus write the stress tensor as a sum of the mean stress
field and the deviatoric stress field:

11 12 13   mean

 



22
23   0
 21

 
 31  32  33   0
0
 mean
0
0  11   mean
 
0    21
 
 mean    31
12
13 

 22   mean
 23 

 32
 33   mean 
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Note that:
• Shear stresses equal to zero at =0 and 90 degrees.
• Maximum shear stress is equal to ( 1 -  3)/2 at =45 degrees.
• The shear stresses along the principal directions are equal to
zero.
• The principal axes are orthogonal.
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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.
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Mohr circles and 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 II
Mohr circles and 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:
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Pressure
This is a special state of stress in which the shear stress is equal
to zero, i.e.: 1= 2= 3.
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.
Plithostatic  g  (z)dz.
z
In the absence of tectonic forces or fluids, the state of stress
would be lithostatic.

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Pressure
Hydrostatic pressure:
The stress equals the weight of a column of water.
Phydrostatic  gwater z.

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Role of fluid pressure and effective stress
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.

Stress II
Effective stress:
effective stress
= normal stress - pore pressure
The effective stress tensor is:
11  P
12
13  11 12 13  P 0 0 

 
 

 22  P
 23   21  22  23  0 P 0 
  21

 
 





P



0
0
P
 31
  31

32
33
32
33  
Question: Is pressure a vector or a scalar?
Stress II
Effective stress:
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