ENV-2E1Y: Fluvial Geomorphology: 2004 - 5

Slope Stability and Geotechnics

Landslide Hazards

River Bank Stability

Section 2 - Water Flow in Soils

N.K. ToveyН.К.Тови М.А., д-р технических наук

Landslide on Main Highway at km 365 west of Sao Paulo: August 2002

Lecture 5 Lecture 6

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 )

g

vw

2

2

g2

vuZ

2w

w

g

vuZH

w 2

2

– a velocity head =

total = position + pressure + velocity

head head head head

•total pore water pressure (pwp) =

Fig. 2.1 Flow of water in a simple channel section

2.2 Hydraulic GradientPressure at A is w h1

and at B is w h2

hydraulic gradient =

Water In

Soil Sample

P1

h1

Standpipes

h2

Water Out

A B

21 hh

i

ds

dhi

)( 21 hh

i w

more generally

NOTE: The hydraulic gradient as defined above is dimensionless (i.e. has no units).

some other disciplines

.....(kNm-3)

P2

vQAat

Water IN

z

h

2.3 The Permeameter - (Constant Head)

k - coefficient of permeability

v k i kdhdsa

Q - flow rate

At - cross section area

Darcy’s Law

2.5 Results from Permeameter

Quicksand occurs

Medium Dense

e=0.620

k=2.93 mm/s

Loose Dense

e=0.744

k=5.89 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 = wsG

m

ws

ws

Gm

GmAL

e

1

m

ALG ws

• 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

Further Comments about Permeability

ih

zA

A

G

ecritcrit t

t w

s

' 1

1

• Analogies in Heat Flow and Electricity– In HEAT FLOW - (ENV-2D02)

2.10 Flow of Water in Soils

)( 21 Ak

Q

– In the FLOW of ELECTRICTY

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

)( 21 EEAk

I

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

• In the FLOW of WATER in SOILS

2.10 Flow of Water in Soils (continued)

)( 21 hhAk

Q

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

1) Mathematical solutionsa) exact solutions for certain simple situationsb) 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

1) flow lines and equipotentials are at right angles to one another.

2) the cylinder walls are also flow lines.

3) distances between the equipotentials are equal

head drops between the equipotentials are also equal.

2.12 Graphical Solutions - Flow Nets

Equi-potentials

Flow Lines

Water IN

2.12 Asymetric Flow

2.12 Asymetric Flow

• Intersections are at right angles

• approximate to curvilinear square

A

B

C

D

2.12 Asymetric Flow

• Intersections are at right angles

• approximate to curvilinear square

A

B

C

D

nd pressure drops

a

pressure drop between AB and CD is H and let there be nd pressure drops and nf flow lines. dn

Hhi

aand

n

kHakiaq

df

2.12 Asymetric Flow (continued)

where qf is the flow per unit cross-section and a x 1 is the cross- section between flow lines.

dn

kHkivLawsDarcyBy :'

the total seepage =

df n

kHq

d

fff n

kHnqn

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

A

B

upward seepage force =

2.13 Seepage around an obstruction d

wab

n

HN

'

'

d

wab

n

HN

ab

d

ws N

n

HF

'

downward force of the soil =

A quicksand will occur if

but very approximately ' = w so

actual downward force of the soilFactor of safety = ----------------------------------------------------------- downwards force required to resist seepage force

In the above example, nd = 10 and Nab ~ 3.5

i.e. the distance must exceed 0.35 times the difference in head of water.

H5.3

10Fs

l

wd

ab

n

HNor

'

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.

2.14 Flow nets Summary

h

h

h

h

h

h

Water table

Change in head between adjacent equipotentials equals the vertical distance between the points on the phreatic surface.

4) All equipotentials are at right angles to flow lines

5) All parts of the flow net must have the same geometric proportions

(e.g. square or similarly shaped rectangles).

6) Good approximations can be obtained with 4 - 6 flow channels. More accurate results are possible with higher numbers of flow channels, but the time taken goes up in proportion to the number of channels.

The extra precision is usually not worth the extra effort.

Uplift arises the total water pressure exerted on the base.

Static head (constant for flat based obstruction)excess head.

2.17 Uplift on Obstructions

0 1 2 3Distance under obstruction (m)

4

3

2

1

0H

ead

of

Wat

er (

m)

6 m

3 m

4 m

2

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 2w multiplied by width

(i.e. 6w kN per metre length).

• the upward force is the area under the curve multiplied by w.

In this example upward force = 6w kN per metre length, i.e. in this case it equals the static head uplift. total uplift = 12w 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.

2.17 Uplift on Obstructions

Water IN

z

h

2.3 The Permeameter - (Constant Head)