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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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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
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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
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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):
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
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:
):
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
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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20
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