ENV-2E1Y: Fluvial Geomorphology:
2004 - 5
Slope Stability and Geotechnics
Landslide Hazards
River Bank Stability
Section 2 - Water Flow in Soils
N.K. Tovey
Н.К.Тови М.А., д-р технических наук
Lecture 5
Lecture 6
Landslide on Main Highway at km 365 west of Sao Paulo: August 2002
2.1 Introduction
• Three component parts to the water pressure:-
– pressure arising from a static head (wZ)
– excess pore water pressure (pressure head differential
which actually causes water flow. (u )
wv
– a velocity head =
2g
•total pore water pressure (pwp) =
H 
total =
head
Z

u
w

position + pressure +
head
head
2
v2
2g
velocity
head
 w v2
 w Z u 
2g
2.2
Hydraulic Gradient
Pressure at A is w h1
and at B is w h2
Standpipes
h1 h2
hydraulic gradient = i 


P1
h1
h2
Water
In
more generally
P2
Water
Out
B
A
Soil
Sample
i
dh
ds
NOTE: The hydraulic gradient as
defined above is dimensionless
(i.e. has no units).
some other disciplines
i
 w ( h1 h2 )
.....(kNm-3)

Fig. 2.1 Flow of water in a simple channel section
2.3
The Permeameter - (Constant Head)
Water IN
h
va 
Q
At
Q - flow rate
At - cross section area
Darcy’s Law
z
va  k i   k
dh
ds
k - coefficient
of permeability
2.5
Results from Permeameter
Loose Dense
e=0.744
Quicksand
occurs
k=5.89 mm/s
Medium Dense
e=0.620
k=2.93 mm/s
2.5
Results from Permeameter
m - total mass of sand
A - cross section of sand column
L - length (height) of column of sand
Volume occupied = A.L
Volume of Sand grains =
e
AL  m
m
Gs  w
Gs  w
m
Gs w
ALG s  w

1
m
Further Comments about Permeability
• Falling Head Permeameter is used for clays
- in constant head permeameter, flow rate is far to small to get
meaningful readings
•
Formation of a Quicksand - Piping
•
occurs when upward seepage force
= downward force from self weight
icrit
hcrit
A t  ' Gs  1



z
At  w
1e
2.10 Flow of Water in Soils
• Analogies in Heat Flow and Electricity
– In HEAT FLOW - (ENV-2D02)
kA
Q
(1  2 )

Where Q is the heat flow rate
1 is the internal temperature
2 is the external temperature
A is the cross-section area
k is the thermal conductivity
 is the path length
– In the FLOW of ELECTRICTY
kA
I 
( E1  E 2 )

Where I is the current
E1 is the inlet voltage
E2 is the outlet voltage
A is the cross-section area
k is the electrical conductivity
 is the path length
2.10 Flow of Water in Soils (continued)
• In the FLOW of WATER in SOILS
kA
Q
( h1 h2 )

1)
Where Q is the water flow rate
h1 is the inlet head
h2 is the outlet head
A is the cross-section area
k is the permeability
 is the path length
Mathematical solutions
a) exact solutions for certain simple situations
b) solutions by successive approximate - e.g. relaxation methods
2)
Graphical solutions
3)
Solutions using the electrical analogue
4)
Solutions using models
Only graphical methods will be used in this course
2.12 Graphical Solutions - Flow Nets
Flow Lines
1) flow lines and equipotentials
are at right angles to one
another.
Equipotentials
2) the cylinder walls are also flow
lines.
3) distances between the
equipotentials are equal
Water IN
head drops between the
equipotentials are also equal.
2.12 Asymetric Flow
2.12 Asymetric Flow
C
A
B
• Intersections are at right angles
• approximate to curvilinear square
D
2.12 Asymetric Flow
C
a

A
B
• Intersections are at right angles
• approximate to curvilinear square
D
2.12 Asymetric Flow (continued)
pressure drop between AB and CD is H
and let there be nd pressure drops and nf flow lines.
kH
By Darcy' s Law : v  ki 
nd 
kHa
q f  kia 
nd 
and a  
where qf is the flow per unit cross-section and
a x 1 is the cross- section between flow lines.
kH
qf 
nd
the total seepage =
nf qf 
n f kH
nd
h
H
i 

nd 
Summary of Flow Nets
Solutions are relatively straightforward.
1) draw the appropriate flow net
2) count the number of pressure drops in the flow net
(over the relevant distance)
3) count the number of flow lines
4) do a simple calculation
–
–
–
work out total flow
work out pressure at any given point
etc.
2.13
Seepage around an obstruction
H
B
A
2.13
Seepage around an obstruction
upward seepage force
=
downward force of the soil =
A quicksand will occur if
N ab  w H
nd
' 
N ab  w H
 ' 
nd
or
N ab H
'

nd 
w
actual downward force of the soil
Factor of safety = ----------------------------------------------------------downwards force required to resist seepage force
'  nd
Fs 
 w H N ab
In the above example,
but very approximately
nd = 10 and
' = w so
Nab ~ 3.5
10l
Fs 
3.5H
i.e. the distance must exceed 0.35 times the difference in head of water.
2.14
Flow nets Summary
Rules for drawing flow nets:1) All impervious boundaries are flow lines.
2) All permeable boundaries are equipotentials
3) Phreatic surface - pressure is atmospheric, i.e. excess pressure is zero.
Change in head between adjacent equipotentials equals the vertical
distance between the points on the phreatic surface.
Water
4) All equipotentials are at right angles to flow lines
table
h
5) All parts of the flow net must have the same geometric proportions
h
(e.g. square or similarly shaped rectangles).
h
6) Goodh approximations can be obtained with 4 - 6 flow channels. More
accurate results are possible with higher numbers of flow channels, but the
h
time taken goes up in proportion to the number of channels.
h
The extra precision is usually not worth the extra effort.
2.17
Uplift on Obstructions
Uplift arises the total water pressure exerted on the base.
Static head (constant for flat based obstruction)
excess head.
4m
6m
Head of Water (m)
3m
4
3
2
1
0
0
1
2
3
Distance under obstruction (m)
2
2.17
Uplift on Obstructions
If total uplift force > the self weight downward
object will be displaced downstream.
 Draw flow net
 Plot graph of uplift pressure (Y –axis) against distance
along base (X-axis).
Uplift pressure is estimate from
flownet
 head at the upstream head is ~0.75 of total head
 head at the down stream end it is ~0.25 of the total head.
2.17
Uplift on Obstructions
• Base of the obstruction is 2m below the surface
• uplift force from the static head is 2w multiplied by width
(i.e. 6w kN per metre length).
• the upward force is the area under the curve multiplied by w.
In this example upward force = 6w kN per metre length,
i.e. in this case it equals the static head uplift.
total uplift = 12w kN m-1.
Uplift reduces ability of the obstruction to resist
movement through the pressure of water
 potential boulder blockages in a river
 man-made drop structure built in river engineering works to
dissipate energy (see RDH's part of the Course).
 quicksand might form at the down stream end of the
obstruction.
2.3
The Permeameter - (Constant Head)
Water IN
h
z