Craig's Soil Mechanics 7th Edition
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Craig's Soil Mechanics 7th Edition

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is calculated at the points of intersection of the equipoten-
tials with the base of the dam. The total head at each point is obtained from the flow
net and the elevation head from the section. The calculations are shown in Table 2.3
and the pressure diagram is plotted in Figure 2.9.

Example 2.3

A river bed consists of a layer of sand 8.25m thick overlying impermeable rock; the
depth of water is 2.50m. A long cofferdam 5.50m wide is formed by driving two lines
of sheet piling to a depth of 6.00m below the level of the river bed and excavation to a
depth of 2.00m below bed level is carried out within the cofferdam. The water level
within the cofferdam is kept at excavation level by pumping. If the flow of water into

Table 2.2

Level z
(m)

hb
(m)

ub/�w
(m)

hf
(m)

uf /�w
(m)

ub � uf
(kN/m2)

1 0 2.30 2.30 0 0 22.6
2 �2.70 2.10 4.80 0 2.70 20.6
3 �4.00 2.00 6.00 0 4.00 19.6
4 �5.50 1.83 7.33 0.21 5.71 15.9
5 �7.10 1.68 8.78 0.50 7.60 11.6
6 �8.30 1.51 9.81 0.84 9.14 6.6
7 �8.70 1.25 9.95 1.04 9.74 2.1

Flow nets 47

the cofferdam is 0.25m3/h per unit length, what is the coefficient of permeability of the
sand? What is the hydraulic gradient immediately below the excavated surface?
The section and flow net appear in Figure 2.10. In the flow net there are 6.0 flow

channels and 10 equipotential drops. The total head loss is 4.50m. The coefficient of
permeability is given by

k ¼ q
h Nf=Ndð Þ

Table 2.3

Point h
(m)

z
(m)

h� z
(m)

u ¼ �w(h� z)
(kN/m2)

1 0.27 �1:80 2.07 20.3
2 0.53 �1:80 2.33 22.9
3 0.80 �1:80 2.60 25.5
4 1.07 �2:10 3.17 31.1
5 1.33 �2:40 3.73 36.6
6 1.60 �2:40 4.00 39.2
7 1.87 �2:40 4.27 41.9
7 1
2

2.00 �2:40 4.40 43.1

Figure 2.9 Example 2.2.

48 Seepage

¼ 0:25
4:50� 6=10� 602 ¼ 2:6� 10

�5 m=s

The distance (�s) between the last two equipotentials is measured as 0.9m. The
required hydraulic gradient is given by

i ¼ �h
�s

¼ 4:50
10� 0:9 ¼ 0:50

2.5 ANISOTROPIC SOIL CONDITIONS

It will now be assumed that the soil, although homogeneous, is anisotropic with
respect to permeability. Most natural soil deposits are anisotropic, with the coefficient
of permeability having a maximum value in the direction of stratification and a
minimum value in the direction normal to that of stratification; these directions are
denoted by x and z, respectively, i.e.

kx ¼ kmax and kz ¼ kmin

Figure 2.10 Example 2.3.

Anisotropic soil conditions 49

In this case the generalized form of Darcy’s law is

vx ¼ kxix ¼ �kx @h
@x

ð2:18aÞ

vz ¼ kziz ¼ �kz @h
@z

ð2:18bÞ

Also, in any direction s, inclined at angle
 to the x direction, the coefficient of
permeability is defined by the equation

vs ¼ �ks @h
@s

Now

@h

@s
¼ @h
@x

@x

@s
þ @h
@z

@z

@s

i.e.

vs

ks
¼ vx
kx

cos
þ vz
kz
sin

The components of discharge velocity are also related as follows:

vx ¼ vs cos
vz ¼ vs sin

Hence

1

ks
¼ cos

2

kx
þ sin

2

kz

or

s2

ks
¼ x

2

kx
þ z

2

kz
ð2:19Þ

The directional variation of permeability is thus described by Equation 2.19 which
represents the ellipse shown in Figure 2.11.
Given the generalized form of Darcy’s law (Equation 2.18) the equation of con-

tinuity (2.6) can be written:

kx
@2h

@x2
þ kz @

2h

@z2
¼ 0 ð2:20Þ

50 Seepage

or

@2h

ðkz=kxÞ@x2 þ
@2h

@z2
¼ 0

Substituting

xt ¼ x
ffiffiffiffiffi
kz

kx

s
ð2:21Þ

the equation of continuity becomes

@2h

@x2t
þ @

2h

@z2
¼ 0 ð2:22Þ

which is the equation of continuity for an isotropic soil in an xt�z plane.
Thus, Equation 2.21 defines a scale factor which can be applied in the x direction to

transform a given anisotropic flow region into a fictitious isotropic flow region in
which the Laplace equation is valid. Once the flow net (representing the solution of the
Laplace equation) has been drawn for the transformed section the flow net for the
natural section can be obtained by applying the inverse of the scaling factor. Essential
data, however, can normally be obtained from the transformed section. The necessary
transformation could also be made in the z direction.
The value of coefficient of permeability applying to the transformed section,

referred to as the equivalent isotropic coefficient, is

k0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
ðkxkzÞ

p
ð2:23Þ

A formal proof of Equation 2.23 has been given by Vreedenburgh [9]. The validity of
Equation 2.23 can be demonstrated by considering an elemental flow net field through

Figure 2.11 Permeability ellipse.

Anisotropic soil conditions 51

which flow is in the x direction. The flow net field is drawn to the transformed and natural
scales in Figure 2.12, the transformation being in the x direction. The discharge velocity vx
can be expressed in terms of either k0 (transformed section) or kx (natural section), i.e.

vx ¼ �k0 @h
@xt

¼ �kx @h
@x

where

@h

@xt
¼ @hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðkz=kxÞp @x

Thus

k0 ¼ kx
ffiffiffiffiffi
kz

kx

s
¼

ffiffiffiffiffiffiffiffiffiffiffiffiffi
ðkxkzÞ

p

2.6 NON-HOMOGENEOUS SOIL CONDITIONS

Two isotropic soil layers of thicknesses H1 and H2 are shown in Figure 2.13, the
respective coefficients of permeability being k1 and k2; the boundary between the
layers is horizontal. (If the layers are anisotropic, k1 and k2 represent the equivalent
isotropic coefficients for the layers.) The two layers can be considered as a single
homogeneous anisotropic layer of thickness (H1 þH2) in which the coefficients in the
directions parallel and normal to that of stratification are kx and kz, respectively.
For one-dimensional seepage in the horizontal direction, the equipotentials in each

layer are vertical. If h1 and h2 represent total head at any point in the respective layers,
then for a common point on the boundary h1 ¼ h2. Therefore, any vertical line
through the two layers represents a common equipotential. Thus, the hydraulic
gradients in the two layers, and in the equivalent single layer, are equal; the equal
hydraulic gradients are denoted by ix.

Figure 2.12 Elemental flow net field.

52 Seepage

The total horizontal flow per unit time is given by

qx ¼ ðH1 þH2Þkxix ¼ ðH1k1 þH2k2Þix
; kx ¼ H1k1 þH2k2

H1 þH2 ð2:24Þ

For one-dimensional seepage in the vertical direction, the discharge velocities in
each layer, and in the equivalent single layer, must be equal if the requirement of
continuity is to be satisfied. Thus

vz ¼ kziz ¼ k1i1 ¼ k2i2
where iz is the average hydraulic gradient over the depth (H1 þH2). Therefore

i1 ¼ kz
k1
iz and i2 ¼ kz

k2
iz

Now the loss in total head over the depth (H1 þH2) is equal to the sum of the losses in
total head in the individual layers, i.e.

izðH1 þH2Þ ¼ i1H1 þ i2H2
¼ kziz H1

k1
þH2

k2

� �

; kz ¼ H1 þH2
H1

k1

� �
þ H2

k2

� � ð2:25Þ

Similar expressions for kx and kz apply in the case of any number of soil layers. It can be
shown that kx must always be greater than kz, i.e. seepage can occur more readily in the
direction parallel to stratification than in the direction perpendicular to stratification.

2.7 TRANSFER CONDITION

Consideration is now given to the condition which must be satisfied when seepage
takes place diagonally across the boundary between two isotropic soils 1 and 2 having

Figure 2.13 Non-homogeneous soil conditions.

Transfer condition 53

coefficients of permeability k1 and k2, respectively. The direction of seepage approach-
ing a point B on the boundary ABC is at angle
1 to the normal at B, as shown in
Figure 2.14; the discharge velocity approaching B is v1. The components of v1 along
the boundary and normal to the boundary are v1s and v1n respectively. The direction of
seepage leaving point B is at angle
2 to the normal, as shown; the discharge velocity
leaving B is v2. The components of v2 are v2s and v2n.
For soils 1 and 2 respectively