SOIL MECHANICSSOIL MECHANICS(BCE 3303)(BCE 3303)

Seepage and FlownetSeepage and Flownet

Lecture

Mdm Nur SyazwaniMdm Nur Syazwani

LINTON UNIVERSITY COLLEGELINTON UNIVERSITY COLLEGESCHOOL OF CIVIL ENGINEERINGSCHOOL OF CIVIL ENGINEERING

INTRODUCTION

DAMWater

Seepage

Vx

Vz

Water

Vx

Vz

Two-dimensional flow

PERMEABLE SOIL

INTRODUCTION

• Pore spaces between soil particles are interconnected and water is free to flow within the soil mass

• Flow of water through soils is called seepage. • Seepage takes place when there is difference in

water levels on the two sides of the structure• The vertical and horizontal velocity components

vary from point to point within the cross-section of the soil mass

• If we know the permeability of the soil, how do we compute the discharge through the soil?

• How do we compute the pore water pressures at various locations in the flow region or assess the uplift loading on the bottom of the concrete dam?

• Is there any problem with hydraulic gradient being too high within the soil?

To address all these, let’s look at some fundamentals in flow……

FLOW THROUGH SOILS

• Bernoulli’s equation for steady flow of non-viscous incompressible flow:

Total head =

• When water flow through soil, the seepage velocity is often very small and negligible, Bernoulli’s equation becomes:

g

vz

g

p

w 2

2

Pressure HeadElevation Head

Velocity Head

zg

p

w

Total head =

DARCY’S LAW

• In saturated condition, 1-D flow is governed by Darcy’s Law:

• The quantity flowing is therefore given by:

L

hkkiv

Flow Velocity

Permeability of SoilHydraulic Gradient

Difference in Total Head

Flow Path Length

AkiAvq Area through which flow is taking place

TWO-DIMENSIONAL FLOW

Seepage taking place around water retaining structure (sheet pile, dams etc.) and embankments is 2-D i.e. vx & vz vary from point to point

A Flow Net is a graphical solution to the Laplace equation for two-dimensional flow in a homogenous and isotropic (kh = kv) soil mass

Two sets of derived orthogonal curves :

a)Equipotential Lines

b)Flow Lines

FLOW NET(GRAPHICAL PROPERTIES)

FLOW NET(GRAPHICAL PROPERTIES)

a) Flow lines and Equipotential lines are perpendicular

b) Grids are curvilinear squares, where diagonals cross at right angles

c) Impermeable boundary is a flow line

d) The quantity of seepage,

(flow interval)

(equipotential drop)

e) At any point,

Total head = Elevation Head + Pressure Head

d

f

N

NkHq

pet hhh

FLOW NET (CONSTRUCTION RULES)

a) Draw to scale the cross sections of the structure, water elevations, and aquifer profiles

b) Establish boundary conditions, and draw one or two flow lines and equipotential lines near the boundaries

c) Sketch intermediate flow lines and equipotential lines by smooth curves adhering to right-angle intersections and square grids. Where flow direction is a straight line, flow lines are an equal distance apart and parallel

FLOW NET (CONSTRUCTION RULES)

d) Continue sketching until a problem develops. Each problem will indicate changes to be made in the entire net. Successive trials will result in a reasonably consistent flow net

e) In most cases, 5 to 10 flow lines are usually sufficient. Depending on the number of flow lines selected, the number of equipotential lines will automatically be fixed by geometry and grid layout

INSTABILITY (‘PIPING’)

• ‘Piping’ effect – an unstable condition cause by the vertical component of seepage pressure (upward direction) exceeds the weight of the soil (downward direction)

• Piping failure can lead to the collapse of a water-retaining structure

• Factor of safety against piping = Downward Weight

Upward Seepage Force

Example 1

Impermeable Stratum

Sand

8m

3.5m0.5m

Water Level

Water Level

k = 6.5 x 10-4 m/s

e = 0.68

G = 2.6214m

Example 1

Impermeable Stratum

8m

3.5m0.5m

Water Level

Water Level

14m

k = 6.5 x 10-4 m/s

e = 0.68

G = 2.62

Example 1 – Sheet Piles

Impermeable Stratum

8m

3.5m0.5m

Water Level

Water Level

14m

k = 6.5 x 10-4 m/s

e = 0.68

G = 2.62

A

Impermeable Stratum

Solution 1a: 

Flow Interval, Nf = 4.3Equipotential Drop, Nd = 11

 

The quantity of seepage beneath the sheet pile,

= 2744.18 l/h per m  

3600100011

3.43105.6 4

d

f

N

NkHq

Solution 1b:

 Taking the impermeable stratum as datum,

hT @ upstream = 17.5mhT @ downstream = 14.5m

hT loss = 0.273m (per square)

11

3

dN

H

Point ht (m) he hp u (kPa)

A 17.5 17.5 0 0

B 17.5 - (2 x 0.273) = 16.954 10 6.954 68.22

C(TOE) 17.5 - (5 x 0.273) = 16.135 6 10.135 99.42

D 17.5 - (8 x 0.273) = 15.316 8 7.316 71.77

E 17.5 - (10 x 0.273) = 14.77 12 2.77 27.17

Solution 1c:Check again piping:

19.27 kN/m3

 

Effective weight of soil,

302.72 kN/m run

Upward seepage force,

397.68 kN/m run  Piping failure will occur due to the upward seepage force is (397.68

kN/m) larger than the effective weight of the soil (302.72 kN/m)

81.968.01

68.062.2

1 ws

sat e

eG

2

8881.927.19W

2

842.99

2

DuF TOE

Example 1 (continued)

Impermeable Stratum

11m

3.5m0.5m

Water Level

Water Level

14m

k = 6.5 x 10-4 m/s

e = 0.68

G = 2.62

Increase the sheet pile to a depth of 11m

Flow Interval, Nf = 6.3Equipotential Drop, Nd = 16

Flow Interval, Nf = 4.5Equipotential Drop, Nd = 16

3600100016

3.63105.6 4

d

f

N

NkHq 36001000

16

5.43105.6 4

d

f

N

NkHq

2764.13 l/h per m 1974.38 l/h per m

hT loss = 0.1875 (per square)

16

3

dN

H

Point ht (m) he hp u (kPa)

Point ht (m) he hp u (kPa)

Example 2

A river has a water depth of 3m above the clayey sand base. The clayey sand layer is 14m thick which in turn overlies impermeable rock.

Laboratory tests indicate that the average permeability of the clayey sand is 4 x 10-3 m/s. The void ratio of the clayey sand is 0.6 and the specific gravity of the grains is 2.65.

It is required to excavate a 35m long, 5m wide and 6m deep trench across the river. To facilitate this work, a cofferdam is to be constructed by driving 2 lines of sheet piles to a depth of 10m below the clayey sand.

 

Example 2 – Cofferdam

Example 3 – Concrete Dam

Example

Impermeable Stratum

Sand

12m

5.0m1.0m

Water Level

Water Level

k = 1 x 10-4 m/s

21mγsat = 16.5 kN/m3

Example

Impermeable Stratum

12m

5.0m1.0m

Water Level

Water Level

k = 1 x 10-4 m/s

21mγsat = 16.5 kN/m3

Example

Impermeable Stratum

12m

5.0m1.0m

Water Level

Water Level

k = 1 x 10-4 m/s

21mγsat = 16.5 kN/m3