Geotechnical Engineering

DHARM

SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 147

n

dz

dt

k

hz

z

..= ()c−

If z is zero when t is zero, the solution of this differential equation is:

t =

nh

k

z

h

c e zh

c

−− c

F

HG

I

KJ

−

L

N

M

M

O

Q

P

P

log 1 / ...(Eq. 5.42)

Here hc and k are assumed constants. However, as soon as z becomes larger than the
height of the bottom zone of partial capillary saturation the increasing air content leads to
variations in both hc and k. The permeability decreases and eventually becomes so small that
the tests required to determine the height of the zone of a partial capillary saturation needs a
very long period of time. The height to the top of the zone of partial saturation has no relation
to the constant capillary head hc, since the former depends on the size of the smaller pores, and
the letter acts only when there is maximum saturation and depends upon the size of the larger
pores.
It has been found from laboratory experiments that for values z less than about 20% of
hc, the saturation is relatively high, and hc and k are essentially constant.

For different sets of values of z and t, it is possible to solve for hc and k simultaneously.

5 .10 Illustrative Examples

Example 5.1. Determine the neutral and effective stress at a depth of 16 m below the ground

level for the following conditions: Water table is 3 m below ground level ; G = 2.68; e = 0.72;

average water content of the soil above water table is 8%.

(S.V.U.—B. Tech., (Part-time)—April, 1982)

The conditions are shown in Fig. 5.21:

GWT w=8%

G = 2.68

e = 0.72^13 13 mm

3 3mm

Fig. 5.21 Soil profile (example 5.1)

G = 2.68

e = 0.72

w = 8% for soil above water table.

γ =

Gw

e w

()

()

(^1).
1

• 
  


• 
  

  γ
  = 2.68 ×
  108
  172
  .
  .
  × 9.81 kN/m^3
  = 16.51 kN/m^3.