13
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Permeability and Seepage -2
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Conditions favourable for the formation
quick sand
Quick sand is not a type of sand but a flow condition
occurring within a cohesion-less soil when its effective stress
is reduced to zero due to upward flow of water.
Quick sand occurs in nature when water is being forced
upward under pressurized conditions.
In this case, the pressure of the escaping water exceeds
the weight of the soil and the sand grains are forced apart.
The result is that the soil has no capability to support a load.
Why does quick condition or boiling occurs
mostly in fine sands or silts?
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Some practical examples of quick conditions
Excavations
in
granular
cofferdams alongside rivers
materials
behind
Any place where artesian pressures exist (i.e. where
head of water is greater than the usual static water
pressure).
-- When a pervious underground structure is
continuous and connected to a place where head is
higher.
Behind river embankments to protect floods
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Some practical examples of quick conditions
Contrary to popular belief, it is not possible to drown
in quick sand, because the density of quick sand is
much greater than that of water.
ρ quicksand >> ρ water
Consequently, it is literally impossible for a person to
be sucked into quicksand and disappear. So, a
person walking into quicksand would sink to about
waist depth and then float.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
•
•
Example
The sand layer of the
soil profile shown in figure
is under artesian pressure.
A trench is to be
excavated in the clay up
to a depth of 4m.
Determine the depth of
water h to avoid boiling.
6m
Boiling condition
4m
Water level
h
stiff clay γ = 18kN/m3
3m
P
Sand
Rock
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
5m
Boiling condition
• At P: u = 5γw
• At P: σ’ = hγw + 2γclay - 5γw
• σ’ = 0 (boiling condition)
• hγw + 2γclay - 5γw = 0
• h = (5γw - 2γclay )/γw
•
= (5 x 9.81 – 2x18)/9.81
• h = 1.33m
6m
• Solution:
• At P: σ = hγw + (6-4)γclay
•
= hγw + 2γclay
4m
Water level
h = 1.33 m
stiff clay γ = 18kN/m3
3m
P
Sand
Rock
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
5m
B
Seepage forces
Water flowing past a
soil particle exerts a drag
force on the particle in
the direction of flow.
h
The drag force is
caused
by
pressure
gradient and by viscous
drag.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Wd
L
(Fs)c
Seepage forces
Direction
of flow
B
W
Wd
(Fs)c
FBD of grain
FBD of soil
At critical condition; h = hc
 Gs + e 

γ w AL = (hc + L )Aγ w
 1+ e 
hc  Gs − 1 
ic = = 

L  1+ e 
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Considering FBD of grain in the direction of flow at i = ic
Wd = (Fs )c + B
(Fs )c = Wd − B
(Fs )c
(Fs )c
(Fs )c
 Gs γ w 
=
 AL − Vsγ w
 1+ e 
(
V − VV )
 Gs γ w 
Vγ w
=
 AL −
V
 1+ e 
 Gs γ w 
=
 AL − (1 − n) ALγ w
 1+ e 
 Gs − 1 
( Fs ) c = 
γ wV
 1+ e 
( Fs ) c = icγ wV
Seepage pressure
ps = icγw
If h < hc then seepage force Js is iγwV
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Seepage force
Area per unit width of section = ∆l x 1
Volume affected by
seepage force = ∆l2 x 1
h1 > h2
h1
∆h
h2
Direction of
flow
Force applied to sand
particles = γwh1 -γwh2 (∆l x 1)
∆l
∆l
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Seepage force
Force applied to sand particles = γw (∆h) (∆l x 1)
= γw (∆h/∆l) (∆l2 x 1)
J
Seepage pressure ps
= i γw V
= Seepage force per
unit volume
= i γw
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Critical hydraulic gradient – Quick condition
The quick condition occurs at a critical upward
hydraulic gradient ic, when the seepage force just
balances the buoyant weight of an element of soil.
(Shear stresses on the sides of the element are
neglected.)
 Gs − 1 
ic = 

 υ 
 The critical hydraulic gradient is typically around 1.0 for many
soils. Fluidized beds in chemical engineering systems rely on
deliberate generation of quick conditions to ensure that the
chemical process can occur most efficiently.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Example problem
Determine and plot total
stress, PWP and effective
stress diagrams if: (i) h = 1 m;
(ii) h = 4 m; and (iii) h = 2 m.
h
i =
2
Case – I; i = 0.5;
Case – II; i = 2;
Case – III; i = 1
h
a
2m
b
c
d
1m
γsat =20 kN/m3
1m
Datum
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Case - I
Point
i = 0.5
PH [m]
EH [m]
TH [m]
a
0
4
4
b
2
2
4
d
5
0
5
c
3.5
1
4.5
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Case - I
h=1m
σ
i = 0.5
σ´
u
a
2m
b
c
d
1m
1m
Datum
20
20
γsat =20 kN/m3
40
60
35
50
5
10
Stress units are in kN/m2
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Case - II
Point
i=2
PH [m]
EH [m]
TH [m]
a
0
4
4
b
2
2
4
d
8
0
8
c
5
1
6
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Case - II
h=4m
Stress units are in kN/m2
σ
σ´
u
a
2m
b
c
d
1m
1m
Datum
20
20
γsat =20 kN/m3
-10
40
50
-20
60
80
Quick sand condition would
have already occurred…
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Case - III
i=1
Point
PH [m]
EH [m]
TH [m]
a
0
4
4
b
2
2
4
d
6
0
6
c
4
1
5
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Case - III
h=2m
Stress units are in kN/m2
σ
σ´
u
a
0
2m
b
c
d
1m
1m
Datum
20
20
γsat =20 kN/m3
40
40
0
0
0
60
60
Just subjected to
Quick sand condition…
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Example problem
A large open excavation was made in a stratum of clay with a
saturated unit weight of 17.6 kN/m3. When the depth of the
excavation reached 7.5 m, the bottom rose, gradually cracked
and was flooded from below by a mixture of sand and water.
Subsequent borings showed that the clay was underlain by a
bed of sand with its surface at a depth of 11 m.
Compute the elevation for which the water would have risen
from the sand surface into a drill hole before the excavation was
started.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Solution
11m
γsat = 17.6 kN/m3
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Solution
Elevation for which the
water would have risen
from the sand surface
into a drill hole before
the excavation was
started.
h = 6.16 m
7.5 m
h
A
For σ´= 0 at A: σ´= 17.6 x (11 – 7.5) - 10 x h = 0
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Example problem
Determine and plot the total vertical stress, pore water
and effective vertical stress distribution at levels A, B
and C.
4m
A
6m
B
3m
Silty clay; Gs = 2.70; w = 45.2 %
C
3m
Coarse sand; Gs = 2.65; e = 0.833
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Example problem
An artesian pressure
exists in the lower
sand layer.
1m
1m
γd = 18.5 kN/m3
Determine σ, u, and
σ ′ at A, B, and C.
2m
γsat = 19 kN/m3
(TH)A = 2 + 3 = 5m
1.33 m
(TH)B = 7 + 0 = 7m
2m
A
2m
1m
C γclay = 17 kN/m3
B
DATUM
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Solution
At point B:
σ = 18.5 x 1 + 19 x 2 + 17 x 3 = 107.5 kN/m2
u = 10 x 7
= 70 kN/m2
σ′ = 107.5 – 70
= 37.5 kN/m2
At point A:
σ = 18.5 x 1 + 19 x 2 = 56.5 kN/m2
u = 10 x 2
= 20 kN/m2
σ′ = 56.5 – 20
= 36.5 kN/m2
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Solution
At point c:
σ = 18.5 x 1 + 19 x 2 + 17 x 2 = 90.5 kN/m2
u = 10 x (PH)c = 10 x 5.33
= 53.3 kN/m2
σ′ = 107.5 – 53.3
= 38.2 kN/m2
(TH)C = (TH)B - i Z
= 7 – (2/3) x 1 = 6.33 m
(PH)C = (TH)C – Zc = 6.33 – 1 = 5.33 m
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
1m
1m
2m
A
56.5
20
36.5
53.3
38.2
70
37.5
2m
1m
C
B
90.5
DATUM
107.5
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Measurement of soil permeabilities
The rate of flow of water q (volume/time) through
cross-sectional area A is found to be proportional to
hydraulic gradient i according to Darcy’s law:
q
v =
= ki
A
h
i =
L
where v is flow velocity and k is coefficient of permeability with
dimensions of velocity (length/time).
 The coefficient of permeability of a soil is a measure of the
conductance (i.e. the reciprocal of the resistance) that it
provides to the flow of water through its pores.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Measurement of soil permeabilities
The value of the coefficient of permeability k depends
on:
(i) Average size of the pores and is related to the
d = ed10
particle sizes and their packing,
(ii) Particle shape, and
d10
(iii) Soil structure.
d =
5
The ratio of permeabilities of typical sands/gravels to those of
typical clays is of the order of 106. A small proportion of fine
material in a coarse-grained soil can lead to a significant
reduction in permeability.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Measurement of soil permeabilities
A number of tests can be used to measure or estimate
the permeability of soils
Laboratory methods:
The constant head test (used for highly permeable soils)
The falling head test
(used for relatively impermeable soils)
Indirect methods: computation from grain size distribution
and During Oedometer test
Field methods:
1. Pumping tests and 2. Borehole tests
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Laboratory measurement of permeability
In determining permeability of coarse grained
material, large quantities of flow occur in short periods
of time and small quantities of flow occur over long
periods of time for fine grained soils.
Two aspects that need careful attention for all types
of soils are:
(i) To ensure that flow occurs only through the soil and not at the
interface between soil and the mould in which the soil is
contained.
(ii) The soil sample
observations
is
fully
saturated
before
recording
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Determination of coefficient of permeability in
the laboratory
Water
Constant
supply
water level
Constant Head test 
Q = Avt
h
i=
L
h
Q = Ak  t
L
QL
k=
Aht
A = area of
C/S of the
specimen
L
h
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Main features of constant head test
It is suitable for soils having a coefficient of
permeability in the range of 10-2 m/s to 10-5 m/s,
which applies to clean sand and sand-gravel
mixtures with less than 10 % fines.
It can be suitable for soils when used in their
completely disturbed or remolded states such as for
drainage materials and filters to confirm that their
performance will be adequate.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Falling head test or
Variable head test
dh
h1
h
h 2
Cross-sectional area of stand pipe = a
Cross-sectional area of specimen
=A
At time to head h1
L
At time t1 head h2
Let h be the head of water
at any time t. Let in time dt
the head drop by amount dh
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Falling head test
Quantity of water flowing through the sample in
time dt from Darcy’s law:
dQ = kiA( dt )
h
= k A( dt )
L
Quantity of discharge
can also be
dQ
=
expressed as:
kA t1
dt
− a ∫ dh =
∫
h1
L t =t0
 h1
aL
k = 2.303
log10 
At
 h2
h2



− a (dh)
As the time
increases
head
decreases!!
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Main features of constant head and
falling head tests
In the constant head test, permeability is computed on the
basis of fluid that passes through the soil sample.
While in the falling head
fluid flowing into the sample.
test, k is computed on the basis of
With the constant head test, time is required to accumulate
the fluid volume necessary to perform computation. Extreme
care would be required to prevent leaks in the apparatus and
evaporation of discharge water.
With the falling head test, the duration of the test is shortened
and care is required to prevent evaporation of water in the inlet
tube.
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
Problem
Determine and plot the total vertical stress, pore water
and effective vertical stress distribution at levels A, B
and C.
4m
A
6m
B
3m
Silty clay; Gs = 2.70; w = 45.2 %
C
3m
Coarse sand; Gs = 2.65; e = 0.833
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay