BED, BANK & SHORE BED, BANK & SHORE PROTECTION - CHAPTER 3 pps

Chapter 3
Flow - stability
Introduction
• focus on stability of loose non-cohesive grains
• rock: important material for protection
• grains may vary in size from μc (sand) to m (rock)
Uniform flow – Horizontal bed
Forces on a grain in flow
1
2
1
2
1
2
2
DD
D
w
222
SS
F
ww
2
LL
L
w
Drag force : =
Cu
FA
Shear force : = F
Cu ud

FA
Lift force : =
Cu
FA
ρ
ρρ
ρ
⎫
⎪
⎪
⎪
≈
⎬
⎪
⎪
⎪
⎭
Balance equations
d g K =
u
d g = d g
-

u
2
c
w
ws
2
c

Δ→Δ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∝
ρ
ρρ
d
g ) - (
d

u

dO W = dO
F
: 0 = M
W =
F
: 0 = V
FW x f =
F
: 0 = H
3
ws
22

c
w
SD,
L
F
SD,
ρρρ
∝
⎪
⎭
⎪
⎬
⎫
⋅⋅Σ
Σ
=Σ
)()(
Relation between load and strength
Isbash (1930)
g 2
u
0.7 = d or 1.7 =
d g
u
or d g 2 1.2 =
u
2
cc
c
Δ

Δ
Δ
used for first approximation when:
• relation between velocity and waterdepth not clear
(e.g a jet entering a body of water)
()
()
2
**
*
Re
cc c
c
sw
uud
ff
gd gd
τ
ψ
ρ
ρυ
⎛⎞
====
⎜⎟
−Δ
⎝⎠
Shields (1936)
*
*
Re

c
ud
υ
=
:
c
ψ
Shields parameter (stability parameter)
Note:
:
ψ
Mobility parameter (when actual u used)
*
Re Re≠
Critical shear stress
Shields Van Rijn
()
()
⎟
⎠
⎞
⎜
⎝
⎛
==
Δ
=
−
=
υρρ

τ
ψ
du
ff
dg
u
dg
cc
ws
c
c
*
*
2
*
Re
Example
1500
1033.1
002.01
Re
6
*
*
=
×
×
==
−
ν

du
c
What is u
*c
for sand with d = 2 mm?
•
Ψ
c
= 0.055
•
smgdu
gd
u
cc
c
c
/042.0002.081.965.1055.0
*
*
=×××=Δ=→
Δ
=
ψ
ψ
04.063
1033.1
002.0042.0
Re
6
*

*
=→=
×
×
==
−
c
c
du
ψ
ν
smgdu
cc
/036.0002.081.965.104.0
*
=
×
×
×
=Δ=
ψ
Guess: u
*c
= 1 m/s
Shields
Example (cont.)
42
)1033.1(
81.965.1
002.0

3
26
3
2
*
=
×
×
=
Δ
=
ν
g
dd
d = 2 mm
0.04
c
Ψ=
•
smgdu
cc
/036.0002.081.965.104.0
*
=
×
×
×
=Δ=
ψ
Van Rijn

Relative protrusion
Load and strength distribution
0. no movement at all
1. occasional movement at some locations
2. frequent movement at some locations
3. frequent movement at several locations
4. frequent movement at many locations
5. frequent movement at all locations
6. continuous movement at all locations
7. general transport of the grains
Shields
Videos on stability of rock on a bed
with current only
u = 0.70 m/s, Ψ = 0.04
u = 0.83 m/s, Ψ = 0.05
u = 0.92 m/s, Ψ = 0.06
u = 0.97 m/s, Ψ = 0.07
u = 0.60 m/s, Ψ = 0.03
Threshold of motion
Shields Paintal
(Extrapolation of transport to zero)
*1816
*
* 2.5
3
6.56 10 (for 0.05)
with
13 (for 0.05)
s
s

s
s
q
q
q
q
gd
ψψ
ψψ
⎫
=⋅ <
⎪
=
⎬
=>
Δ
⎪
⎭
Stone dimension
dVM
n
==
3
3
/
ρ
50 50n
dd
≠
Nominal diameter:

50
50
0.84
n
d
d
≈
Influence of waterdepth
*
g
uu
C
=
50
c
c
n
C
u
g
d
g
ψ
=
Δ
1.7
ic
u
gd
=

Δ
Isbash:
12
18log
r
R
C
k
=
roughness
k
r
= 2*d
50
or k
r
= 3*d
50
Attention:
cic
uu≠
Uniform flow:
2
*c
c
u
gd
ψ
=
Δ

Influence waterdepth on critical
velocity
Practical application
u
gd
C
g
d
u
C
c
n
c
n
c
c
Δ
Δ
50
50
2
2
=→=
ψ
ψ
Roughness and threshold of motion
Choose Ψ:
do we select Ψ on the safe side or do we
use the expected value of Ψ ??
Lammers, 1997

d
n50
=0.146m
Angles of repose for non-cohesive
materials
1:1
1:1.5
1:2
1:3
Influence of slope on stability
Case b: slope parallel to flow
Case c: slope perpendicular to flow
φ = 40
ο
Slope parallel to current
φ
α
φ
α
α
α
=
W
W - W
=
F(0)
)F(
= )K(
tan
sintancos

//
//
φ
α
φ
φ
α
φ
α
φ
α

) - (
=
sin
sin
sin
sincoscossin
=
−
Slope perpendicular to current
φ
αφαα
α
tan
sintancos
2
222
=
-

=
F(0)
)F(
= )K(
φ
α
φ
α
α
sin
sin
tan
tan
cos
2
2
2
2
- 1 = - 1 =
Stability on top of sill
use velocity on top of the sill
Experiments: first damage at downstream crest
Stability and head difference
Shields is useless here because
Shields contains waterdepth
waterlevel downstream is below
the top of the dam
Shields for flow over sill
)h - (h g 2 )
d

h
0.04 + (0.5 = )h - (h g 2 =
u
du
n
d
du
2
2
1
50
μ
discharge coefficient
Vertical constriction
Stability with flow under weir
Shields in horizontal constriction
50
gap
c
n
u
= C
g
gd
ψ
sin
sin
2
4
2

1 -
α
φ
Δ
General formula
(horizontal closure with trucks)
Correction
α slope of construction
ϕ angle of repose
(internal stability)
α = 30
o
; φ = 40
o
50
3
4.5log
c
n
h
d
ψ
⎛⎞
=
⎜⎟
⎝⎠
50 50
12 1/2 3
18log 18log
2

nn
hh
C
dd
×
==
Damage at half depth: