# Pore Pressure Coefficients by tZ865u

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```									in equilibrium under principal stresses:
with a pore a V and Porosity
Consider soil element:
whose volume =water pressure: = n
Pore Pressure Coefficients
σ1

u0        σ3

V, n

σ2
Pore Pressure Coefficient, B
Then, the element is subjected to an increase in
stress in all 3 directions of σ3 which produces an
increase in pore pressure of u3
σ1 + σ3

u0 +u3
u0               σ3 + σ3

V, n

σ2 + σ3
Pore Pressure Coefficient, B
As a result, the effective stress in each
This increase in effective stress reduces the
direction skeleton by the pore
volume of the soilincreases andσ3 -u3 space.
σ1 + σ3

σ3 - u3

σ +
σ3 -3u3σ3
u0 +u3

V, n
σ3 - u3
σ2 + σ3
Pore Pressure Coefficient, B
σ3 - u3

Soil Skeleton
σ3 - u3

V, n
σ3 - u3                    ΔVss  CsV Δσ 3  Δu3 
where:

Vss = the change in volume of the soil skeleton caused by an
increase in the cell pressure, σ3

Cs = the compressibility of the soil skeleton under an isotropic
effective stress increment;
i.e., the fraction of volume reduction per kPa increase in cell
pressure
Pore Pressure Coefficient, B
σ3 - u3

Pore Space (Voids)
σ3 - u3

V, n
ΔVPS  CVVV Δu3
σ3 - u3
where:

VPS = the volume reduction in pore space caused by a change
in the pore pressure, u3
CV = the compressibility of the pore fluid; i.e., the fraction of
volume reduction per kPa increase in pore pressure

Since: VV  nV            Then:   ΔVPS  CVnVu3
Pore Pressure Coefficient, B
σ3 - u3
Assuming:
1. the soil particles are incompressible
2. no drainage of the pore fluid
σ3 - u3
Therefore, the reduction in soil
V, n
σ3 - u3
skeleton volume must equal the
reduction in volume of pore space

Therefore:           CSV Δσ 3  Δu3   CVnVΔu3
             
             
      1      
or:                  Δu3  Δσ3              
 1  n CV  
 
      C 
       S 
Pore Pressure Coefficient, B
σ3 - u3                                       
             
The value:         1      
             
σ3 - u3
 CV  
 1  n 
      C 
V, n
       S 
σ3 - u3
is called the pore pressure coefficient, B
So, u3 = B σ3
If the void space is completely saturated, Cv = 0 and B = 1
When soils are partially is estimated > 0 and B < 1
In an undrained triaxial test, Bsaturated, Cv by increasing the
cell pressure by σ3 and measuring the resulting change in pore
This is illustrated on Figure 4.26 in the text.
pressure, u3 so that:
Δu3
B
Δσ 3
Pore Pressure Coefficient, A
What happens if the element is subjected to an
increase in axial (major principal) stress of σ1 which
produces an increase in pore pressure of u1

σ1 + σ1

u0 +u1
u0               σ3 - u1

V, n

σ2 - u1
Pore Pressure Coefficients
As change in effective stress also in each the
This a result, the effective stress changes of
the minor soil skeleton and the by -u1
volume of the directions increasespore space.
σ1 + σ1

σ1 - u1

-σ3 -u1
u1
u0 +u1

V, n
-u1
σ2 - u1
Pore Pressure Coefficients
σ1 - u1
If we assume for a minute that soil is an elastic
material, then the Volume change of the soil
skeleton can be expressed from elastic theory:
- u1
ΔVss  3 CsV Δσ 1  3Δu1 
1
V, n
- u3

As before, the change in volume of the pore space:
ΔVPS  CVnVu1
Again, if the soil particles are incompressible and no drainage
of the pore fluid, then:
1
3   CSV Δσ 1  3Δu1   CVnVΔu1
Pore Pressure Coefficient, A
σ1 - u1                                      
             
or: Δu  1
      1      
Δσ1  1 BΔ 1
1
3              
 CV        3
- u1               1  n 
      C 
V, n
       S 
- u3
Since soils are NOT elastic, this is rewritten as:
u1 = AB σ1 or Δu1  A Δσ 1 where A  AB
For different values of saturated soil test, determined by
A value is A for pressure during the to be determined by
where A of a pore a fully σ1coefficient canbeu1 is measured,
although the values at failure are of particular interest:
experiment
measuring the pore water pressure during the application of
the deviatorΔustress in an undrained triaxial test
A        1
, B  1 for saturation   
Δσ 1
Pore Pressure Coefficient, A
Normally
Consolidated
σ1 - u1
Figure 4.28 in the text illustrates the variation of
A with OCR (Overconsolidation Ratio).
Lightly Over-
- u1                                Consolidated

V, n                                         Heavily Over-
- u3
Consolidated

In highly overconsolidated (normally consolidated clays), A
For heavilycompressible soils clays, A may lie <between -0.5 & 0
For lightly overconsolidated clays, 0 A < 0.5
ranges between 0.5 and 1.0
maximum historical effective stress
OCR 
current effective stress
Pore Pressure Coefficient, B
The third pore pressure coefficient is determined
from the response, u to a combination of the
effects of increasing both the cell pressure, σ3
and the axial stress (σ1 -σ3) or deviator stress.

If we divide through by σ1
From the two previous effects: u = u3 + u1
u3 = Bσ3
{
u3 + u1 = u = B[σ3+A(σ1-σ3)]
Δu

Δu
 B


 Δσ
{
 Δσ3
Δσ1 u1 1= BA(σ1
 Δσ3 
 Δσ -σ )
 A 1 

 Δσ3 

1   3
Δu
or:           B1  (1  A)  1 
 Δσ  or:
           B
Δσ1                      1      Δσ 1
Pore Pressure Coefficient, B
Δu         The third pore pressure coefficient is not a
B
Δσ 1          constant but depends on σ3 and σ1

With no movement of water (undrained) and no change in
water table level during subsequent consolidation, u =
initial excess pore water pressure in fully saturated soils.

Testing under Back Pressure
When a sample of the calculation extracted pressure
This process allowssaturated clay isof the pore from the
coefficient, B.
ground, it can swell thereby decreasing Sr as it breathes in air
The pore pressure can be raised artificially (in sync with σ3)
Values of B  0.95 are considered to represent
to a datum value for excess pore water pressure and then the
to consolidate
sample can be allowedsaturation. back to the in situ
conditions (saturation, pore water pressure).

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