Geotechnical Engineering

DHARM

226 GEOTECHNICAL ENGINEERING

on application of the pressure increment, pore water starts to flow towards the drainage faces.

Let ∂h be the head lost between the two faces of this elementary layer, corresponding to a

decrease of hydrostatic excess pressure ∂u.

Equation 6.2, for flow of water through soil, holds here also,

k

h

x

k h

z e

e s

t

S e

xz t

.

()

∂ ..

∂

+ ∂

∂

=

+

∂

∂

+ ∂

∂

L

NM

O

QP

2

2

2

2

1

1 ...(Eq. 6.2)

For one-dimensional flow situation, this reduces to:

k

h

z e

e s

t

S e

z t

.

()

∂ ..

∂

=

+

∂

∂

+ ∂

∂

L

NM

O

QP

2

2

1

1

During the process of consolidation, the degree of saturation is taken to remain con-

stant at 100%, while void ratio changes causing reduction in volume and dissipation of excess

hydrostatic pressure through expulsion of pore water; that is,

S = 100% or unity, and

∂

∂

S

t

= 0.

∴ (^) k h
z e
e
tt
e
z e
.
()
.
∂
∂
=−

• ∂
  ∂
  =−
  ∂
  ∂+
  L
  N
  M
  O
  Q
  P
  2
  2
  1
  11
  ,
  negative sign denoting decrease of e for increase of h.
  Since volume decrease can be due to a decrease in the void ratio only as the pore water and soil
  grains are virtually incompressible,
  ∂
  ∂+
  F
  HG
  I
  t KJ
  e
  1 e represents time-rate of volume change per unit
  volume.
  The flow is only due to the hydrostatic excess pressure,
  h =
  u
  γw
  , where γw = unit weight of water.
  ∴
  ku
  z
  V
  γw t
  .
  ∂
  ∂
  =−
  ∂
  ∂
  2
  2 ...(Eq. 7.17)
  (This can also be considered as the continuity equation for a non-zero net out-flow,
  while Laplace’s equation represents inflow being equal to out-flow).
  Here k is the permeability of soil in the direction of flow, and ∂V represents the change
  in volume per unit volume. The change in hydrostatic excess pressure, ∂u, changes the
  intergranular or effective stress by the same magnitude, the total stress remaining constant.
  The change in volume per unit volume, ∂V, may be written, as per the definition of the
  modulus of volume change, mv;
  ∂V = mv.∂σ = – mv.∂u, since an increase ∂σ represents a decrease ∂u.
  Differentiating both sides with respect to time,
  ∂
  ∂
  =−
  ∂
  ∂
  V
  t
  m
  u
  v t

. ...(Eq. 7.18)
From Eqs. 7.17 and 7.18, we have:
∂
∂

=

∂

∂

u

t

k

m

u

γwv. z

.

2

2