Transcript Slide 1

Lecture 8: Design of Erodible Channels
CEM001 Hydraulic Structures, Coastal and River Engineering
River Engineering Section
Dr Md Rowshon Kamal
[email protected]
H/P: 0126627589
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Hydraulic Parameters used in Design
of Unlined/Lined Channels
1. Channel must carry design flow/discharge (Qd).
2. Velocity in the channel must not be high to
cause scour.
3. Velocity in the channel must not be low to
cause deposition.
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Minimum Permissible Velocity
This is the lowest velocity which prevents both
sedimentation (deposition) and vegetation growth.
Recommendations by French (1985)
• Prevent from sedimentation – 0.61~0.91m/s
• Prevent from growth of vegetation – 0.76m/s
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1. Maximum Permissible Velocity Method
This is one of the methods we use to design a
channel.
Special committee on Irrigation Hydraulics (ASCE)
formed this method.
Design criteria:
Mean Flow Velocity < Max. Permissible Velocity
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Maximum Permissible Velocities
Maximum Permissible Velocity (m/s)
Material
Fine sand, non
colloidal
Alluvial silt, non
colloidal
Stiff clay, very
colloidal
Fine gravel
0.46
Water carrying
colloidal silts
0.76
0.61
1.07
1.14
1.52
0.76
1.52
Coarse gravel
1.22
1.83
Clear Water
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Maximum Permissible Velocities (con’t)
Above values will be changed if;
1. Reduce values by 25% for sinuous (meandering) channels.
2. Increase by 0.15 m/s for depths greater than 0.91m.
3. Reduce by 0.15 m/s if channel carries abrasive material.
4. Increase by 0.3 – 0.6 m/s for channels with high silt load.
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Example 02
Design a trapezoidal channel (side slope 1:2) to
carry 125.0 m3/s on a bed slope of 0.001. Use
Maximum Permissible Velocity Method.
Assume the following:Coarse gravel in water carrying colloidal silt
Depth to be greater than 1.0 m
Manning’s coefficient n = 0.025
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Answer Ex-02
From Table in slide no 5:
Corresponding allowable velocity = 1.83m/s
This may be increased by 0.15m/s because channel depth
assumed to be greater than 1.0m,
Modified allowable velocity = 1.98m/s
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Answer Ex-02
b
From Manning’s formula
1
y
z
R 2 / 3 S 01 / 2
Q
V 
A
n
 Vn 
R   1 / 2 
 S0 
3/ 2
 1.98  0.025


1/ 2
 0.001

3/ 2
Cross sectional area
A
Q 125

 63 .13m 2
V 1.98
Wetted perimeter
P
A 63.13

 32.31m
R 1.96
Cross sectional area
A  y(b  zy )
Wetted perimeter
P  b  2y 1 z2
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 1.96m
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Answer Ex-02
Substituting for the side slope,
z2
63.13  y(b  2 y)
32.31  b  2 5 y
Solve for y and b
2.472y 2  32.31y  63.13  0
y  2.4m
b  21 .5m
Design depth and width
y  10.6m
b  15.3m Negative value is
y  2.4m b  21.5m
not possible
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2. Permissible Tractive Force Method
(Shear Stress Method)
Most rational and widely used method.
Based on the consideration of equilibrium of particle
resting on the bed with drag and lifting forces
balanced by the submerged weight of particle.
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Permissible Tractive Forces by USBR
USBR recommends the following values for
boundary shear stresses:
 Course non-cohesive material (D > 5.0mm)
 bc  0.75D75
N / m2
D75 in mm
 Fine non-cohesive materials (Refer Example
2.4 – Next page please!)
 Cohesive sediments – Not covered by USBR
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For Fine Cohesive Materials
Graph of 0/g against d from Shields Diagram
0.01000
0/g = RS (m)
0.00100
USBR VALUES FOR FINE NONCOHESIVE SEDIMENTS
AMOUNT OF
SUSPENDED
SEDIMENT
HIGH
MED
SHIELDS' FUNCTION
LOW
0.00010
0.00001
0.01
0.10
1.00
Particle Diameter, d, (mm)
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10.00
100.00
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Adjustment for Sinuosity
Degree of Sinuosity
CS
Straight channels
1.00
Slightly sinuous
0.90
Moderately sinuous
0.75
Very sinuous
0.60
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Allowable Shear Stresses for a Trapezoidal
Channel
y
1
z
b
s = ks gyS0
τs and τb - Max Bottom & Side Shear Stresses
ks and kb - Depend on y, b, z
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b = kb gyS0
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Allowable Shear Stresses for a Trapezoidal
Channel (con’t)
ks and kb Factors
If
b
4
y
If
b
4
y
kb  0.97
Tables need to be used
Design Conditions
 b   bc or yb  ybc
 s   sc or ys  ysc
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ks  0.75
For bottom
For sides
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Bank Stability in a Trapezoidal Channel
Forces acting on the particle:
1. Drag force (FD)
2. Component from weight (Wsinθ)
3. Friction force opposing R
(Wcosθtanϕ)
F   s
D
Sand particle
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Q
Wsinq
q
Wcosq
Force normal
to the side
R  W 2 sin q   2 2
W
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Bank Stability in a Trapezoidal Channel
(con’t)
At incipient motion, resultant force, R will be
equal to friction force.
W cos q tan   W 2 sin 2 q   2 sc2
 sc 2 W 2 (cos2 q tan2   sin 2 q )
For side
1/ 2
 tan q 

 sc  cosq tan 1 
2

 tan  
W
2
(1)
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Bank Stability in a Trapezoidal Channel
(con’t)
For the bottom θ=0;
 bc 
W

tan 
(2)
Combining (1) and (2) gives;
1/ 2
 sc  sin q 

K
 1 
2
 bc  sin  
2
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Bank Stability in a Trapezoidal Channel
(con’t)
For finer materials, θ=0;
 sc
K
 cosq
 bc
i.e. Cohesive forces are much greater than the
gravity force.
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Example 03
Design a trapezoidal channel to carry 125.0m3/s
on a bed slope of 0.001. The channel is to be
excavated in coarse alluvium, containing
moderately angular stones with d75 of 50.0mm.
The angle of friction for this material is 40º, which
is also its angle of repose.
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Question 04
(i) Define the terms maximum and minimum permissible
velocities.
(ii) A river 30.0 m wide and 4.0 m deep and of a regular
rectangular cross-section carries a discharge of 350.0
m3/s through country with a bed slope of 0.0003. If the
bed material is coarse alluvium having a D50 size of
10.0 mm and specific gravity s = 2.65, estimate the total
transport load using the Ackers and White formula.
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Thank You
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