cxlass 29 Design of Sarada type Fall.pdf

Civil Engineering

1. UNIT-V Canal Falls & Canal regulation works
•CanalFalls-Typesoffallsandtheir
location,DesignprinciplesofNotch
FallandSaradatypeFall.
•Canalregulationworks,principlesof
designofdistributaryandhead
regulators,CanalCrossRegulators-
canaloutlets,typesofcanalmodules,
proportionality,sensitivityand
flexibility.CrossDrainageworks:types,
selectionofsite,
Dr GK Viswanadh Prof. of Civil Engineering & Director UGC-HRDC JNTUH 1/3/2022 1

2. Design procedure of Sarda fall
•Simpleverticaldropfallorsardafallconsists,averticaldropwhich
allowstheupstreamwatertofallwithsuddenimpactondownstream.
•Thedownstreamactslikecushionfortheupstreamwateranddissipate
extraenergy.
•ThistypeoffallwasfirstintroducedontheSardacanalsysteminU.P.
basedontheexperimentscarriedoutatBahadarabadResearchstation,
U.P.Therefore,itisalsocalledSardaFall.
•Thisisaverticaldropweirwithrectangularcrestfordischargeslessthan
14m
3
/s.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
2

3. 1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
3

4. •ThevariouselementsoftheSardatypefallare
(i)crestwall(bodywall)
(ii)Cistern
(iii) Impervious floor
(iv) downstream protection and
(v) Upstream approach.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
4

5. Sarda type falls
•The sarda type fall is a modification of vertical drop fall.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
5

6. Design procedure of Sarda fall…
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
6

7. Design criteria
The design criteria of the various elements are as follows:
CrestWall:
•Thecrestwallmaybedesignedeitherforfreeflowconditionsorforsubmerged
flowcondition.
•Thedischargeequationsadoptedforrectangularandtrapezoidalcrestsareas
follows.
Rectangularcrestweirwithfreeflow:??????=�.??????��????????????
�
�
??????
??????
�
�
hereQ=dischargeinm
3
/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)
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
7

8. Trapezoidal weir with free flow: ??????=1.99????????????
3
2
??????
??????
1
6
Fordrownedconditionofbothtypesofweirs,neglectingvelocityof
approachwithacoefficientofdischargeC
dforfreeanddrowned
portions,
Where H
L= drop in the water surface , m
and h
d= height of the water surface on the d/s side above the crest of the
weir, in m.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
8

9. Width of the crest wall; B:
for rectangular crest wall
Where B= top width of crest wall in m B
1= 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.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
9

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 = D
1,
h= D
1-H
for drowned condition, h=D
1-(H
L+ h
d)
f
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
10

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
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
11

12. 1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
12

13. 1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
13

14. •Thed/sfloorshouldbemadethickenoughtoresisttheupliftpressures,
theminimumthicknessshouldbelimitedto45cmonconcreteunder
35cmofbrickmasonry.
•Asufficientdepthofcutoffbelowthefloorshouldbeprovidedatthe
d/sendoffloorforsafetyagainstexitgradient.
Minimum depth of u/s cutoff = where D
1= u/s F.S.D.
Minimum depth of d/s cutoff = , where D
2= d/s F.S.D.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
14

15. EnergyDissipators:
•Forlargedischarges,tworowsoffrictionblocksinthecisternandtwo
rowsofchuteblocksontheimperviousflooratthed/sendshallbe
providedasadditionalenergydissipators.
•Bothfrictionblocksandchuteblocksarestaggeredandthesizeand
positionoftheseblocksarefollows:
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
15

16. Friction blocks:Length = 2 d
c
Width = d
c
Height = d
c
Distance of the first row from the d/s toe of crest wall = 1.5 d
c.
Spacing between two rows = d
c
spacing between the block in the same row = 2d
c
where d
c= critical depth =
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
16

17. 1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
17

18. Chute Blocks:
Length= width = Height =
Spacing between two rows=
Spacing between blocks in same row=
•Where D
2is the full supply depth downstream.
•Onerowofchuteblocksisprovidedjustatthed/sendof
imperviousfloorandtheotherupstreamoftheaboverow.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
18

19. Protection works:
•Bedprotectionu/sofcrestwall:Brickpitchingislaidonthe
channelbedfor2to4m,lengthadjacenttothecrestwallon
theu/ssidewithadownwardslopeof1in10towardsthe
crestwall.
•Atthelevelofbrickpitchingafewdrainholesareprovidedin
thecrestwallfordrainingpurposeduringcanalclosure.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
19

20. UpstreamWings:arekeptsegmentalwitharadiusof6H
L,subtendingan
angleof60°atthecentreandcontinuedthereaftertangentiallyintothe
bankbyaminimumof1m.
BedProtectiond/s:Thed/sbedpitchingiskepthorizontalforalengthof
6mandslopedat1in10thereafterforalengthof5to15mforfalls
varyingfrom0.75mto1.5m.
Thed/swingwallsarekeptverticalforalengthof5to8times from
thecrestandthenflaredorwrapedfromverticalto1:1or:1atendof
flooring
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
20

21. 1/3/2022 Dr GK Viswanadh prof. of Civil engineering & Director UGC HRDC JNTUH 21

22. Thesidepitchingwith1:1or :1slopeisprovidedafterthereturn
wingonthed/s.
Atoewallshouldbeprovidedbetweenbedpitchingandsidepitching
toprovideafirmsupporttothelatter.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
22

23. Problem on Sarda type fall
Design a Sarda type fall with the following data.
Full supply discharge=60m
3
/s.
Full supply levelu/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
DesigntheflooronthebasisofBligh’stheoryassumingthecreep
coefficienttobe8.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
23

24. 1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
24

25. Problem on Sarda type fall…
Solution:
1.Calculation of H and d:
Sincetrapezoidalcrestistobeprovidedfordischargegreaterthan14
m
3
/s.
Full supply depth u/s D
1= 1.80 m
H+d= D
1+drop in bed level = 1.8+ (146.50 -145.00 =) 1.5 =3.3m
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
25

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=D
1-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
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
26

27. Problem on Sarda type fall…
Assuming the side slopes of the channel to be 1:1, velocity of approach, Va=Q/A
�??????���??????????????????�.�.??????.=�??????���??????????????????�.�.??????.+
??????
??????
2
2??????
=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
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
27

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
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
28

29. Problem on Sarda type fall…
4. Design of impervious floor:
Seepage head, H
s= d=2.4m
Bligh’s creep coefficient =8
Provide u/s cut off of depth d
1=1.0m
d/s cut off of depth d
2=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 = l
d= 2(D
1+1.2)+H
L
= 2(1.8+1.2)+1.5=7.5m
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
29

30. Problem on Sarda type fall…
Provide l
d=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.cfloor 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.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
30

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+freeboard.
= 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.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
31

32. Problem on Sarda type fall…
6. D/S Protection:
(i)Bed pitching: Provide 200mm thick dry pitching
Length of bed pitching=9+2H
L= 9+2X1.5=12m
This should be horizontal uptothe 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.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
32

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,
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
33

34. Problem on Sarda type fall…
Friction blocks: Length = 2 d
c Width = d
c Height = d
c
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.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
34

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
•Providetworowsofstaggeredcubeblocksofsize0.2x0.2x0.2mon
imperviousflooratd/sendwithaspacingof0.2m
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
35

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
•Provideu/swingshavingsegmentalportionofradius5.5mand
subtending60°atthecentrefromtheu/sedgeofcrestwall.
•Thereaftertheyaretakenstraightandembeddedinthebankby1.0m
roomtheF.S.L.line.
1/3/2022
Dr GK Viswanadh prof. of Civil engineering & Director UGC
HRDC JNTUH
36