Flow through Soils II (ch8). 2D flow Vel. vectors are confined to a single plane 1D flow vel vectors parallel 2D flow vel vectors not necessarily parallel.

Flow through Soils II(ch8)

2D flow

Vel. vectors are confined to a single plane

1D flow vel vectors parallel

2D flow vel vectors not necessarily parallel

wall

soil

Laplace’s Equation

dxx

vv xx

zv

xv

1D flow - Darcy’s law 2D flow - Laplace’s equation

dzz

vv zz

x

z

dz

dx L

Rate of in vel in z-dir

Laplace’s Equation

Laplace’s equation

“represents energy loss through any resistive medium”

Assumptions

Darcy is valid

S = 100%

is of constant volume

Isotropic kx = kz

Homogeneous k = same throughout

Remember main point = get Q

Examples

Q beneath a dam Q in excavations

Solving

Mathematics

02

2

2

2

z

h

x

h

Graphical

Flownets

Flownets

Flowline

wall

impervious

Equipotential line

Flowpath – “channel” between two flowlines

Equipotential line – along any eqpl, the total head is the same

Flownets

1st eqpl: starts in inlet

last eqpl: ends at outlet

impervious

1st eqpl

Of soil

last eqpl

Flownets

vel

Flowline

Flowline

node

eqpl

Requirements:

Perpendicular crossings at nodes

Maintain “squareosity”

Flownets

X

Mistake – redraw!

Flownets - Example

10m

14m

2m

L = 100 m

k = .1 cm/s

2m

A

B

Flownets - Example

h = 14 – 2 = 12 m

h at first eq = 24 m

Q = h L k (NFP/NED)

Q = 0.3 m3/sec

NFP = 3

NED = 12

h at last eq = 12 m

Flownets - Example

h at first eq = 24 m

Head loss per drop = h/NED = 12 / 12 = 1 m

he at A = 8 m

h at A = 24 – 8 = 16 m

hp at A = h – he = 16 – 8 = 8 m

uA = (hp)(w) = 78.5 kPa

Flownets - Example

h at first eq = 24 m

Head loss per drop = h/NED = 12 / 12 = 1 m

he at B = 10 m

h at B = 24 – 9 = 15 m

hp at B = h – he = 15 – 10 = 5 m

uB = (hp)(w) = 49 kPa

Uplift pressures

Concrete dam

uuplift

Concrete dam

= W/A

If uuplift ~ , structure can float away!

Uplift Pressures

Case 1 (no flow)

u = (hp)(w) = uhydrostatic

Draw flownet

Compute h at several points along the structure’s base

Determine hp at these points

Case 2 (with flow)

Find u at these points by u = (hp)(w)

Filters

Problem

- water pressure

- soil filter

- Soil particles migrate out

Sol.

- drains

Problem

Sol.

Filters

2 filter purposes

Allow adequate drainage

Soil retained is called base soil

Disallow particle migration

Filter soil

Filters

2 filter failure types

Clogging: base clogs filter pores – (k decreases)

Piping: base migrates through filter

D

dmax

D

dmax

Simple Cubic Packing

D/dmax = 2.4

Tetrahedral Packing

D/dmax = 6.5

Interstice SizesFilters - principles

Arching Bridging

Filters - principles

To disallow particle migration

To allow adequate drainage (maintain proper k)

Terzaghi’s Filter criteria

5485

15 tod

D 54

15

15 tod

D

See table 8.2 in book… Example:

Sherard’s Filter criteria

Example (see notes (pad (lined sheet))…

Example: Filter selection