Geotechnical Engineering

(Jeff_L) #1
DHARM

SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 155


We know,

k =

qrr
zz

log ( / )e
()

21
2

2
1
π−^2

=

2 303 2 70 15 100
60 31 3 29

10
22

.log(/)
(. )

××
×−π

cm/s

= 1.18 × 10–1 mm/s.

Example 5.13. Determine the order of magnitude of the composite shape factor in the Poiseulle’s
equation adapted for flow of water through uniform sands that have spherical grains and a
void ratio of 0.9, basing this determination on Hazen’s approximate expression for permeabil-
ity.


Poiseulle’s equation adapted for the flow of water through soil is:

k = D e
e
s^2 C

3
1

..
()

γ.
μ+
with the usual notation, C being the composite shape factor.
By Allen Hazen’s relationship,
k = 100 D 102
D 10 is the same as the diameter of grains, Ds, for uniform sands.
∴ k = 100 Ds^2. Here Ds is in cm while k is in cm/s.
Substituting –

100 Ds^2 = D

e
e
s^2 C

3
1

..
()

.

γ
μ+

C =^100

1
.. 3

μ ()
γ

+e
e

; here μ is in N-Sec/cm^2 and γ is in N/cm^3.

C =

100 10 10 1 9
981 10 09

76
33

×××
××

−.
.(.)
since μ = 10–3 N-sec/cm^2 (at 20°C) and γ = 9.81 kN/m^3
= 0.002657.

Example 5.14. A cohesionless soil has a permeability of 0.036 cm per second at a void ratio of
0.36. Make predictions of the permeability of this soil when at a void ratio of 0.45 according to
the two functions of void ratio that are proposed.


k 1 : k 2 =

e
e

e
e

1

3

1

2

3


































































() (^112)
++()
0.036 : k 2 = (. )
.
:
(. )
.
036
136
045
145
33
= 0.546 : 1
∴ k 2 =
1
0546.
× 0.36 mm/s = 6.60 × 10–1 mm/s
Also, k 1 : k 2 = e 12 : e 22
0.036 : k 2 = (0.36)^2 : (0.45)^2
= 0.1296 : 0.2025
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