01 PILRevetments General Intro

8/13/2019 01 PILRevetments General Intro

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Revetments

introduction


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Example of

structures

Granular materials

Artificial units

Sand


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Main function

Protection of the object (bank, dike, shore)

against loadings (waves, ship waves, currents,

mechanical damage, etc)


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Types of revetments

Materials:

Granular (rock)

Concrete

Asphalt

Geosynthetics

Wood

Steel

Vegetation

Combined

Other classifications:

- Permeable

- Semi-permeable

- Impermeable

- Statically stable

- Dynamic Stable

- Free blocks

- Grouted systems

- Interlocked

- Mats

- Slabs


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General types of revetments

Loose rock (riprap, dumped stones)

Placed/pitched blocks/stones

Mats

Asphalt

Grass Slabs

also Grass

Al i


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Alternatives

and selection


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Choice


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Riprap

on earth dams


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Alternative rock use


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Selection (choice of revetment)

Criteria:

Loads

Availability of materials Availability of space

Accessibility and/or equipment (construction)

Landscape

Maintenance

Cost

k d


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Wave attack and

Interactions with

structuress/tan=

L

H

gT

H2=s

2

andBreaker index

L=gT2/2=1.56T2

Llocal=T (gh)^0.5

h= local depth in front of structure


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Failure modes


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Surface erosion processes


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Geotechnical instability


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Design elements


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Design

D l i


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Developments in

Cover Layers


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Hydraulic

boundary

conditions


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F l ( ) b k fl i l i i


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For granular (non-cohesive ) banks, fluvial erosion is

modelled as for sediment transport

(with = bank angle):

Fluid Lift (FL)

Fluid Drag (FD)

Friction ()

Particle Weight (W)

Downslope component

of particle weight (Wd)

Normal component

of particle weight

(Wn)

Steve Darby


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Beginning of movement

of granular materials

)()(

*

2

*e

c

cr

cws

crcr Rf

gD

u

gD

CgUgRIu w //*

2


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)()(

*

2

*e

c

cr

cws

crcr Rf

gD

u

gD

CgUgRIu w //*


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Velocity distribution/profiles

-0 2


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us

k

h=K

-0.2

h

r

k

h1=K

-0.2

h

r

or


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27/92


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Stability criteria revetments


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Stability criteria revetments

rock - current attack

(a) K = f G, or

CDwUD/4=fD/6(s-w)g,providing:

U/(2g D) = (2/3)f/CD= Assuming = 42o(for rock), f = tan 42o = 0.90, and CD= 1.0, one obtains:U/(2g D) = = 0.60.

(b) The moment with respect to theturning point S gives the equation:

F b = G a, or(CF w UD/4)b=(D/6(s-w)g)a,

providing:

U/(2g D) = (2/3)(a/b)/CF=

Assuming a = b, CF = 1.0, then = 0.67. CF is a combination of coefficients for drag and liftforces.


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Isbash (1935) Ub2/2gD =

Ub= bottom velocity

= stability factor

g

UD b

ws

wc

2/1

2

or, in dimensionless form: c

b

Dg

U

2

2

(3a,b)

where Dc= characteristic size (usually, Dc = D50), = 1.4 for embedded stone, = 0.7

for exposed stone (conforming older results, Eq. 2), w = unit weight of water, s= unit

weight of stone, and g = acceleration of gravity. is the relative density defined as: = (s-w)/w = (s-w)/w (sand w, unit density of stones and water, respectively)

22

tan

tan1cos

sin

sin1

KsSlope factor

Comparison formulas


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Comparison formulas


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In conclusion, the modified Chinese formula (based on Isbash) should be:

ss K

r

rrgC

UDs

1

22

2

in which,22

tan

tan1cos

sin

sin1

Ks

=angle of slope; C2=

=angle of internal friction of material (for rock, 40 ).

C = 0.9 for normal riprap with turbulent flowC = 1.2 for low turbulence or for embedded (pitched) stones

When = , the rock revetment becomes unstable even without any hydraulicloading.Dn50= Ds/1.24

Because in Vietnam design codes follow the Chinese codes:


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Velocity profile

Measurements in UK rivers:

Ub= 0.74 to 0.9 Udepth-average

Ub measured at 10%of the water depth above the bed

???????

Under ice cover


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http://www.rsnz.org/publish/nzjmfr/1997/15

.pdf


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Designation of Design Flow

Velocity Bangladesh

U 2/2General approach to


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crh

cD

gU

2/2General approach tocurrent attack

- logarithmic velocity profile (Chezy),

r

hk

h

gg

C 121log

2

18

2

222

50

6

log75.5

12

log

18

D

h

k

h

gDg

U

crr

cr assuming kr= 2 D50.

Stricklers resistance formula for developed velocit

profile,

3/13/1

1322

625

rr

hk

h

k

h

g

- non-developed profile (Neill, 1967, Pilarczyk, 1995),

2.02.0

13232

rrh k

h

k

h

U 2/2General approach to


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crh

cD

gU

2/2General approach tocurrent attack

- logarithmic velocity profile (Chezy),

r

hk

h

gg

C 121log

2

18

2

222

50

6

log75.5

12

log

18

D

h

k

h

gDg

U

crr

cr assuming kr= 2 D50.

Stricklers resistance formula for developed velocit

profile,

3/13/1

1322

625

rr

hk

h

k

h

g

- non-developed profile (Neill, 1967, Pilarczyk, 1995),

2.02.0

13232

rrh k

h

k

h

CK

u*K=d

2cs

c

22v

Schiereck


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In Schiereck 2001

CK

u*K=dn

2

cs

c

22v

50

gu=dor=

dguordgu

2

ccc

27.07.122.1

Isbash, 1930

Shields, 1936

duff

dg

u

dg

cc

ws

cc

**

2

* Re

u

g d

C

gd

u

C

c

n

c

nc

c 50

50

2

2

Practical relation;

Including velocity and

slope factor

d V Mn 3 3 /

22


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Schiereck 2001,

H03 final CK

u*K=d

2

cs

c

22v

sin

sintantancos

tansintancos 2

2

2

2

2

222

-1=-1=-

=F(0)

)F(=)K(

Influence of slope on stability

Case c); on slope

Kv=1 to1.6; usually 1.2


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For = 1.33*10-6 m2/s and= 2650 kg/m3, values of the grain size in mm areindicated on the graph.

Shields Diagram (D* =d50(g/2)1/3


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Example 3-1

What is u*c for sand (= 2650 kg/m3) with d = 2 mm?In the classical Shields-curve (Figure 3-2a) u*c appears

in both axes, so iteration is necessary. Suppose you dont

have the faintest idea how much u*c is and you make awild guess, say 1 m/s. Re* then becomes:

1*0.002/1.33*10-6 = 1500 c = 0.055 u*c =(1.65*9.81*0.002*0.055) = 0.042 m/sRe* = 63

u*c = 0.036 m/s. This is also the final value. Using Figure3-2b, you would have found directly d* =

0.002*(1.65*9.81/(1.33*10-6)2)1/3 = 42 c = 0.04 u*c = 0.036 m/s.


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Shields Diagram (D* =d50(g/2)1/3


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Transport

Paintal,1971

3

*

5.2*

1618*

with

)05.0for(13

)05.0for(1056.6

dg

qq

q

qs

s

s

s


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Example 3-2 transport

A value of 0.03 for c is considered a safe choice for thethreshold of motion. For stones with a characteristic

diameter of 0.4 m this gives qs = 6.56*1018*16*(gd3)

3*10-6 m3/m/s. This is equivalent with 86400*qs/d3 4stones per day per m width. The design velocity for anapron is usually a value which occurs only in exceptional

cases e.g. with a chance of 1% per year. This makes such a

transport quantity acceptable. Moreover, this is thetransport per m width, not per m2. This example alsoshows that there is always a chance of some damage and

inspection and maintenance is necessary for everystructure.


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Example 3-3

Given a (design) velocity of 4 m/s, a waterdepth of 10 m and =1.65, what stone size (dn50) is necessary to withstand this (uniform)

flow?

We take a c-value of 0.03 and a roughness of 2 times dn50. Since theroughness in equation depends on the still unknown diameter, wehave to iterate. So we have to start with a guess of either a value fordn50 or for C. We start with C=50 m/s (any guess is good). From

this we compute dn50 = 42/0.03*1.65*502 = 0.129 m. From there: C

= 18 log(12*10/2*0.129) = 48 m/s. From equation again: dn50 =42/0.03*1.65*482 = 0.14 m. And so on. Finally we find dn50 = 0.146

m. Using common sense, this iteration will converge in 3 or 4calculations. From appendix A we find that a stone size of 80/200 mm

will do.


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Rock specifications

Weight gradings and size relations for the standard light and heavy grading classes

1 3

1 0

/

,n

a

WD

g

1 2

1 24

/

,s

a

WD

g

0 806,n s

D DDs=equivalent sphere diameter

sD

)

1 3

5050

/

n

a

WD

g

50

50

0 84,nD

D


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Rock grading

range D50(cm) M50(kg) Dn50(cm)

Layerthickness 1.5Dn50(cm)

Minimal dumping quantity withlayer of 1.5 Dn50(kg/m

2)

30/60 mm40/100 mm50/150 mm80/200 mm5-40 kg10-60 kg40-200 kg60-300 kg

300-1000 kg1-3 ton3-6 ton6-10 ton

3.9-4.96.2-8.88.8-12.312.3-17.721-2626-3138-4445-51

71-77103-110136-143167-174

0.09-0.180.35-1.041.04-2.792.79-8.3112-2524-4384-131139-204

541-6921620-19803843-43927050-7790

3.76.38.912.619243541

6390118144

2020202029365362

95135177216

300300300300450550800950

1450205027003250

C t tt k Pil k


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Current attack_ Pilarczyk

g2u

KKK0.035=D

2

cr

s

hT

u

gD

K

K K

s

T h

2

0 035.

sin

sin-1=K

2

s

k

h=K

-0.2

h

r

k

h1=K

-0.2

h

r

or

E l bili f


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Example: stability of stone

CK

u*K

=d 2cs

c

22v

Schiereck 2001 H-03

T pes of protection:


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Types of protection: Revetment

Gravity structure, sloped protection

Falling apron, launching apron

Series of groynes, spurs

Single groyne

Series of solid, impermeable groynes

Series of permeable groynes

Temporary protections Bandal

Geotextile bags

Natural materials: bamboo structures

Eco-engineering

E l t t


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Waves , flow

Flow, turbulence

Example revetment

D i h l i


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Design channel cross-section):

1.1.1 Fig. 4.5-1: Sketch of a cross-section of an axi-symmetrical bend

In area I of Fig. 4.5-1 the bed profile is given by an empirical envelope curve of surveyed cross-

sections in the Jamuna river:

I II

h0

distance x (m), radius r (m)

Elevation z (m)

y r , y

5.5 h0

B

hch

Predicted bank line

Design water level

2

0

0 )5.5

1(h

yhz

)5.1log(.19.007.20 ch

c

ch BR

hh

Thorne:


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):




I II

h0


Elevation z (m)

y r , y

5.5 h0

B

chDesign bed level

The maximum water depth + maximum scour

depth:

Local scour by the protection structure

Confluence scour enhanced by the structure

Protrusion scour by the structure

Scour by channel narrowing How to combine different types of scour?

Bend scour is part of the maximum water depth


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Design toe flow velocity

Rc = bend radius (m)

B=width channel (m)

utoe = depth averaged flowvelocity above the toe(m/s)

Uch = cross sectionaveraged flow velocity(m/s)

Determine c9 and c10 frommonitoring data

Extra turbulence:

):




I II

h0


Elevation z (m)

y r , y

5.5 h0

B

ch

ch

c

ch

toe

B

Rcc

u

ulog109

,uuu )31()31(

special

normaltur

uuC

Utoe/Uavg= 1.750.5 log (Rc/Bch

)

Example USACE:

Two basic options for toe scour


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Two basic options for toe scourprotection

Maintenance

Deep scour

Slope 1:2 economic

Dredging trench 1:3 or 1:4

Sinking protection layer + small falling apron

Large falling apron:

Low waterlevel

predictederosion

alternatives


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a te at ves

Fill-in

Permeable groyne


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Permeable groyne The emperical width 2 h has a relation with the turbulent

eddies in the flow,

The permeability can vary with a change in water level Cross section projected perpendicular to the main

approach flow direction

h

About 2 h

h

About 2 h


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Bangladesh


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Example eco-engineering Catkin grass or Vetiver grass can protect

eroding bank and catch sediment when

flooded, Brahmaputra river

Flood plain with Robinia trees, bamboo

protect embankments against wave attack

Mangrove forests protect sea dykes againstwave attack in Red River delta.

Uncertainty about the degree of protection

Low cost protection

Success requires local experience

Effect varies in time

A relatively large space is required

roots

siltation h < 4 m

Examples under water protection:


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Fascine mattress on prepared slope (fill):

Geotextile, fascines and rip-rap toplayer Fixed position gives reliable protection

Falling apron

Only toplayer elements:

rip rap, rip rap in gabions, cc blocks,

geobags,

Flexible, dynamic, permeable for subsoil

Connection fixed protection and apron is a problem

Needs regular monitoring during falling process,

eventually also fixation after falling in finalposition (dumping gravel)

Economic

Launching apron: connected elements fall irregularly

p p

Bangladesh; example permeable groyne


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After the first flood:

Repair stone dumping was

necessary

Bangladesh; example permeable groyne

E l t t ti ( b t 5 )


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Example temporary protection (about 5 years)

Very economic

Local contractors

Present bank line will be maintained for a few years only,

Requires monitoring, sensitive for damage and it can attract

developing channels because of steep slopes

Conclusions on types of bank protection:

T b k Ri k f i i S i d


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Type bank

protection

Risk on functioning Space required

(width)

costs

Quay wall Strong foundation, bed

protection, fixed

bankline

Minimum space Expensive

Revetment Deep scouring due to

fallling apron, fixed

bankline

Moderate space Moderate costly

Solid groynes Deep scouring due to

fallling apron, fixed

bankline

Considerable space Economic

Bandal Strength is uncertain,

Sedimentation

Minimum space,

h < 8 m

Economic

Permeable groynes Moderate scour, flexible

bank line

Considerable space Economic

Temporary

protection

Deep scour,

maintenance

Moderate space

required

Very economic

Eco engineering Risk on bank erosion,

Maintenance effort,

Large areas

required

Very economic

Wave attack


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Wave attack

Wave characteristics

Wave definitions and wave height distribution

Wave prediction; examples


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Waveheight

Wave period

as a function of wind, depth

and fetch

Relative depth Shallow Water

1h

Transitional water depth

1h1

Deep Water

1h linear wave theory


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