Consider a soil element:whose volume = V and Porosity = n

V, n

in equilibrium under principal stresses:σ1

σ3

σ2

with a pore water pressure:

u0

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

V, n

σ1

σ3

σ2

u0

σ1 + σ3

σ2 + σ3

σ3 + σ3 u0 +u3

As a result, the effective stress in each direction increases by σ3 -u3

V, n

σ1 + σ3

σ2 + σ3

σ3 + σ3 u0 +u3

σ3 - u3

σ3 - u3

σ3 - u3

This increase in effective stress reduces the volume of the soil skeleton and the

pore space.

Soil Skeleton

V, n

σ3 - u3

σ3 - u3

σ3 - u3

33sss ΔuΔσVCΔV 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 Space (Voids)

V, n

σ3 - u3

σ3 - u3

σ3 - u3

3VVPS ΔuVCΔV 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: nVVV Then: 3u nVCΔV VPS

Assuming:

V, n

σ3 - u3

σ3 - u3

σ3 - u3

1. the soil particles are incompressible2. no drainage of the pore fluid

Therefore, the reduction in soil skeleton volume must equal the

reduction in volume of pore space

or:

Therefore: 3V33S nVΔCΔuΔσVC u

S

V

33

CC

n1

1ΔσΔu

The value:

V, n

σ3 - u3

σ3 - u3

σ3 - u3

is called the pore pressure coefficient, B

If the void space is completely saturated, Cv = 0 and B = 1

This is illustrated on Figure 4.26 in the text.

So, u3 = B σ3

S

V

CC

n1

1

When soils are partially saturated, Cv > 0 and B < 1In an undrained triaxial test, B is estimated by increasing the cell pressure by σ3 and measuring the resulting

change in pore pressure, u3 so that:

3

3

ΔσΔu

B

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

V, n

σ1

σ3

σ2

u0

σ1 + σ1

σ2 - u1

σ3 - u1 u0 +u1

As a result, the effective stress in each of the minor directions increases by -u1

V, n

σ1 + σ1

σ2 - u1

σ3 -u1 u0 +u1

σ1 - u1

-u1

- u1

This change in effective stress also changes the volume of the soil skeleton and the pore

space.

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:

V, n

- u1

- u3

σ1 - u1

11s31

ss 3ΔΔσVCΔV u

As before, the change in volume of the pore space:

Again, if the soil particles are incompressible and no drainage of the pore fluid, then:

1u nVCΔV VPS

1V11S31 nVΔC3ΔΔσVC uu

Since soils are NOT elastic, this is rewritten as:

where A is a pore pressure coefficient to be determined by experiment

u1 = AB σ1 or

A value of A for a fully saturated soil can be determined by measuring the pore water pressure during the

application of the deviator stress in an undrained triaxial test

For different values of σ1 during the test, u1 is measured, although the values at failure are of particular

interest: saturation for 1 BΔσΔu

A1

1 ,

V, n

- u1

- u3

σ1 - u1

11

S

V

1 BΔ31Δσ

CC

n1

131Δu

or:

ABA where ΔσAΔu 11

For lightly overconsolidated clays, 0 < A < 0.5

Figure 4.28 in the text illustrates the variation of A with OCR (Overconsolidation Ratio).

In highly compressible soils (normally consolidated clays), A ranges between 0.5 and 1.0

For heavily overconsolidated clays, A may lie between -0.5 & 0

V, n

- u1

- u3

σ1 - u1

Normally Consolidated

Lightly Over-Consolidated

stress effective currentstress effective historical maximum

OCR

Heavily Over-Consolidated

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.From the two previous effects: u = u3 + u1

u = B[σ3+A(σ1-σ3)]

or:

If we divide through by σ1

u3 + u1

=

u3 = Bσ3

u1 = BA(σ1-σ3){

1

3

1

3

1 ΔσΔσ

1AΔσΔσ

BΔσΔu

1

3

1 ΔσΔσ

1A)(11BΔσΔu

or: BΔσΔu

1

The third pore pressure coefficient is not a 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

BΔσΔu

1

When a sample of saturated clay is extracted from the ground, it can swell thereby decreasing Sr as it breathes

in airThe pore pressure can be raised artificially (in sync with σ3) to a datum value for excess pore water pressure and

then the sample can be allowed to consolidate back to the in situ conditions (saturation, pore water pressure).

This process allows the calculation of the pore pressure coefficient, B.

Values of B 0.95 are considered to represent saturation.