The document describes the design of a Sarda type fall. Key elements include a trapezoidal crest wall, cistern, impervious floor designed using Bligh's theory, downstream wings and protection works. An example problem is included where a Sarda fall is designed for a given canal reach with a discharge of 60 cumecs and drop of 1.5m. The crest wall, cistern, impervious floor length of 15m, downstream wings and protections works are designed and friction blocks are used as energy dissipators.

1. UNIT-V Canal Falls & Canal regulation works
• Canal Falls - Types of falls and their
location, Design principles of Notch
Fall and Sarada type Fall.
• Canal regulation works, principles of
design of distributary and head
regulators, Canal Cross Regulators -
canal outlets, types of canal modules,
proportionality, sensitivity and
flexibility. Cross Drainage works: types,
selection of site,
Dr GK Viswanadh Prof. of Civil Engineering & Director UGC-HRDC JNTUH 1/3/2022 1

2. Design procedure of Sarda fall
• Simple vertical drop fall or sarda fall consists, a vertical drop which
allows the upstream water to fall with sudden impact on downstream.
• The downstream acts like cushion for the upstream water and dissipate
extra energy.
• This type of fall was first introduced on the Sarda canal system in U.P.
based on the experiments carried out at Bahadarabad Research station,
U.P. Therefore, it is also called Sarda Fall.
• This is a vertical drop weir with rectangular crest for discharges less than
14m3/s.
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Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
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3. 1/3/2022
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4. • The various elements of the Sarda type fall are
(i) crest wall (body wall)
(ii) Cistern
(iii) Impervious floor
(iv) downstream protection and
(v) Upstream approach.
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5. Sarda type falls
• The sarda type fall is a modification of vertical drop fall.
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6. Design procedure of Sarda fall…
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7. Design criteria
The design criteria of the various elements are as follows:
Crest Wall:
• The crest wall may be designed either for free flow conditions or for submerged
flow condition.
• The discharge equations adopted for rectangular and trapezoidal crests are as
follows.
Rectangular crest weir with free flow:𝑸 = 𝟏. 𝟖𝟑𝟓 𝑳𝑯
𝟑
𝟐
𝑯
𝑩
𝟏
𝟔
here Q = discharge in m3/s
L= Length of crest wall(m)
H= height of water above the crest wall on the u/s side (m)
B= top width of crest(m)
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8. Trapezoidal weir with free flow: 𝑄 = 1.99 𝐿𝐻
3
2
𝐻
𝐵
1
6
For drowned condition of both types of weirs, neglecting velocity of
approach with a coefficient of discharge Cd for free and drowned
portions,
Where HL = drop in the water surface , m
and hd = height of the water surface on the d/s side above the crest of the
weir, in m.
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9. Width of the crest wall; B:
for rectangular crest wall
Where B= top width of crest wall in m B1 = base width of the crest wall (m)
H= height of water surface above the top of crest wall u/s
s= specific gravity of the material of crest wall
And for trapezoidal crest wall,
with side slope of u/s face as 1 in 3 and
side slope of d/s face as 1 in 8.
The stability of the crest wall shall be checked for fall over 1.5m.
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10. Crest level:
Crest level = u/s F.S.L.- H for free overfall condition
If the height of crest wall above u/s bed is h, and
u/s full supply depth = D1,
h= D1-H
for drowned condition, h=D1-(HL + hd)
f
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11. Cistern : Length of cistern =
Where E is the depth of crest below the u/s T.E.L.
Impervious floor: Total length of impervious floor shall be designed based
on Bligh’s or Khosla’s theory.
Minimum length of d/s impervious floor
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14. • The d/s floor should be made thick enough to resist the uplift pressures,
the minimum thickness should be limited to 45cm on concrete under
35cm of brick masonry.
• A sufficient depth of cutoff below the floor should be provided at the
d/s end of floor for safety against exit gradient.
Minimum depth of u/s cutoff = where D1 = u/s F.S.D.
Minimum depth of d/s cutoff = , where D2= d/s F.S.D.
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15. Energy Dissipators:
• For large discharges, two rows of friction blocks in the cistern and two
rows of chute blocks on the impervious floor at the d/s end shall be
provided as additional energy dissipators.
• Both friction blocks and chute blocks are staggered and the size and
position of these blocks are follows:
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16. Friction blocks: Length = 2 dc
Width = dc
Height = dc
Distance of the first row from the d/s toe of crest wall = 1.5 dc.
Spacing between two rows = dc
spacing between the block in the same row = 2dc
where dc = critical depth =
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18. Chute Blocks:
Length= width = Height =
Spacing between two rows=
Spacing between blocks in same row=
• Where D2 is the full supply depth downstream.
• One row of chute blocks is provided just at the d/s end of
impervious floor and the other upstream of the above row.
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19. Protection works:
• Bed protection u/s of crest wall: Brick pitching is laid on the
channel bed for 2 to 4m, length adjacent to the crest wall on
the u/s side with a downward slope of 1 in 10 towards the
crest wall.
• At the level of brick pitching a few drain holes are provided in
the crest wall for draining purpose during canal closure.
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20. Upstream Wings: are kept segmental with a radius of 6HL, subtending an
angle of 60° at the centre and continued thereafter tangentially into the
bank by a minimum of 1m.
Bed Protection d/s : The d/s bed pitching is kept horizontal for a length of
6m and sloped at 1 in 10 thereafter for a length of 5 to 15m for falls
varying from 0.75m to 1.5m.
The d/s wing walls are kept vertical for a length of 5 to 8 times from
the crest and then flared or wraped from vertical to 1:1 or : 1 at end of
flooring
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21. 1/3/2022 Dr GK Viswanadh prof. of Civil engineering & Director UGC HRDC JNTUH 21

22. The side pitching with 1:1 or :1 slope is provided after the return
wing on the d/s.
A toe wall should be provided between bed pitching and side pitching
to provide a firm support to the latter.
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23. Problem on Sarda type fall
Design a Sarda type fall with the following data.
Full supply discharge=60m3/s.
Full supply level u/s=148.30m d/s=146.80m
F.S.D u/s=1.8m d/s=1.8m
Bed width u/s=35m d/s=35m
Bed level u/s=146.50 d/s=145.00
Drop = 1.5m
Design the floor on the basis of Bligh’s theory assuming the creep
coefficient to be 8.
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HRDC JNTUH
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25. Problem on Sarda type fall…
Solution:
1.Calculation of H and d:
Since trapezoidal crest is to be provided for discharge greater than 14
m3/s.
Full supply depth u/s D1 = 1.80 m
H+d = D1+drop in bed level = 1.8+ (146.50 -145.00 =) 1.5 =3.3m
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26. Problem on Sarda type fall…
H= height of water above the crest wall on the u/s side
d= (u/s FSL – d/s bed level ) – H
Height of crest above u/s bed level = h=D1-H=1.8-0.90=0.90m
2.Deign of crest wall : Top width of crest wall=B=1.0m
u/s slope = 1 in 3
d/s slope = 1 in 8
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27. Problem on Sarda type fall…
Assuming the side slopes of the channel to be 1:1, velocity of approach, Va=Q/A
𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑇. 𝐸. 𝐿. = 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝐹. 𝑆. 𝐿. +
𝑉
𝑎
2
2𝑔
= 148.30 + 0.04 = 148.34
Crest level =u/s F.S.L. – H = 148.30-0.90=147.40
u/s Specific energy E = u/s T.E.L.- crest level
= 148.34-147.40=0.94m
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28. Problem on Sarda type fall…
3. Design of cistern: Length of cistern
Provide a cistern of length 6.0m
R.L. of the bed of cistern=D/s bed level – x =145-0.31
= 144.69
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29. Problem on Sarda type fall…
4. Design of impervious floor:
Seepage head, Hs = d=2.4m
Bligh’s creep coefficient =8
Provide u/s cut off of depth d1=1.0m
d/s cut off of depth d2=1.5m
Vertical creep length=2(1.0+1.5)=5.0m
Length of horizontal impervious floor = 19.2-5.0=14.2m
Provide a length of impervious floor = 15m
Length of impervious floor d/s = ld = 2(D1+1.2)+HL
= 2(1.8+1.2)+1.5=7.5m
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30. Problem on Sarda type fall…
Provide ld=8.0m Length of impervious floor under the crest wall and
u/s = 15-8 = 7.0m
Total creep length = 15+2(1.0+1.5)=20m
Provide a nominal thick for the floor u/s of crest wall
= 2.4X0.55+0.31=1.63m
Thickness of floor say 1.4m
Provide 1.4 thick c.c floor over laid by 0.2m brick pitching:
Provide 1.4m thickness of 0.6m over laid by 0.2m thick brick pitching at
the d/s end of floor.
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31. Problem on Sarda type fall…
5. Design of d/s wings
Provide d/s wings vertical for a length of
The wings are then to be wrapped to 1:1 slope at a splay of 1 in 3.
Height of the top of d/s wing wall above bed = F.S.d+free board.
= 1.8+0.5=2.3m
Horizontal projection of the wrapped wing on 1:1 slope=2.3m
With a splay of 1 in 3, length of warped wing, measured along the centre
line of canal=2.3X3=6.9 0r 7.0m
Flooring and curtain walls to be checked for uplift and exit gradient by
Khosla’s theory.
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32. Problem on Sarda type fall…
6. D/S Protection:
(i) Bed pitching: Provide 200mm thick dry pitching
Length of bed pitching=9+2HL = 9+2X1.5=12m
This should be horizontal upto the end of wing walls and then slope at 1 in
10.
Warped wings commence 1m u/s of the d/s end of impervious floor.
Length of the sloping pitching = 12-6 = 6m
(ii) Curtain wall at the end of bed pitching : Thickness of curtain wall=0.4m
Depth of curtain wall =1m say.
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33. Problem on Sarda type fall…
(iii) Side pitching: Provide 20cm thick side pitching, warping from a slope
of 1:1 to and curtailed at an angle of 45°from the end of bed
pitching in plan.
(iv) Toe wall : Thickness = 0.4m
(v) Energy dissipator: Friction blocks and cube blocks are provided as
follows.
Critical depth,
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34. Problem on Sarda type fall…
Friction blocks: Length = 2 dc Width = dc Height = dc
Length = 2x0.7=1.4m
Width = 0.7m
Height = 0.7m
Distance of the first row from d/s toe of crest wall =1.5x0.7=1.05 or 1.0m
Spacing of blocks between two rows = 0.7m
Spacing of blocks in the same row=2x0.7=1.4m
Provide two rows of staggered friction blocks of size 1.4x0.7=0.7m in the
cistern.
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35. Problem on Sarda type fall…
Chute blocks : d/s water depth = 1.8m
Length = width = height=
Spacing between two rows=0.2m
Spacing between blocks in a row = 0.2m
• Provide two rows of staggered cube blocks of size 0.2x0.2x0.2m on
impervious floor at d/s end with a spacing of 0.2m
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36. Problem on Sarda type fall…
7. Design of u/s approach:
Radius of curved portion of u/s wings=5 to 6 times H
= 5 to 6(0.91) = 4.55 or 5.46 or 5.5m
• Provide u/s wings having segmental portion of radius 5.5m and
subtending 60° at the centre from the u/s edge of crest wall.
• There after they are taken straight and embedded in the bank by 1.0m
room the F.S.L. line.
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