Drop structure lecture 4

Trepezoidal Sarda type fall
Straight glacis fall
3/20/2014 1
PREPARED BY VIDHI H. KHOKHANI
ASST. PROF. DIET

u/s wing walls :
D/s wing walls :
D/s *bed pitching :
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Radius = 5 to 6 times H
= 5 x 0.83 to 6 x 0.83
= 4.15 to 4.98 m
Say R = 4.50 m
Provide u/s wing walls having
segmental portion with radius 4.5 m
and subtending an angle of 60°, at the
centre and starting from the u/s edge
of the crest wall.
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ASST. PROF. DIET

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ASST. PROF. DIET

Length of d/s wing wall with
vertical face
= 6 H • HL
= 6-7 0.83 x 1.5
= 6.70 m
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ASST. PROF. DIET

Length of warped wing with a splay
of 1 in 3 = height * splay = 2.5 * 3
= 7.5 m
Total length of d/s wing wall = 6.70
+ 7.5 = 14.2 m.
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ASST. PROF. DIET
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Length of d/s bed pitching
= 9 + 2Hl
= 9 + 2 x 1.5
= 12 m
Provide horizontal pitching up to
the end of masonry wing walls for a
length of 6.3 m and then sloping at 1
in 10 for length of 6 m.
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ASST. PROF. DIET
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Wingwall
103.77
12 M
PITCHING
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PREPARED BY V.H.KHOKHANI
ASSISTANT PROFESSOR, DIET

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ASST. PROF. DIET
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FLUMED
FALL
UNFLUMED
FALL
METER
FALL
NONMETER
FALL
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1. Crest design
2. Cistern design
3. Impervious floor
4. u/s wing walls
5. d/s wing walls
6. d/s side protection
7. Roughening devices
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Length of crest = width of bed (for
unflumed fall)
L = 35.0 m
For non-meter fall, narrow crest is
provided.
Discharge Q is given by :
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H = 0.85 m
R.L. of crest = u/s FSL – H
= 203.5 – 0.85
= 202.65 M
height of crest above u/s bed,
= R.L.of crest – U/S bed level
= 202.65 - 201.5 = 1.15
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Crest width for non meter fall
= (2/3) H
= (2/3) 0.85
= 0.57
The d/s glacis is provided in a slope of
2 : 1
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Compare it with d/s bed level
because as we know cistern is
always below d/s bed level.
Ef2 can be obtained from blench
curve from available q and Hl
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d/s bed level
u/s bed level
cistern level
0.25 ef2

d/s TEL = d/s FSL + Ha =202 + 0.02
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d/s bed RL is 200 m which is lower
than obtained cistern level.
So provide minimum depth of
cistern = 0.25 Ef2 = 0.25 (1.40) =
0.35 m.
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Length of cistern = 5 ef2 to 6 ef2
= 5(1.40) to 6 (1.40)
= 7.0 to 8.4 m
Provide length of cistern 8 .0 m
The cistern is joined to the d/s bed of
the channel with a slope of 5 : 1.
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d/s bed level
u/s bed level
cistern level
0. 35 m
8.0 m
5 in 1

Assuming 0.6 m additional cutoff
depth.
u/s cutoff depth = D1/3 + 0.6
D/S cutoff depth = D2/2 + 0.6
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d/s bed level
u/s bed level
cistern level
0. 35 m
8.0 m
5 in 1
0.2 m
Deflector
wall
1.6 m
1.40 m

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Provide total horizontal length of
impervious flow = 20.60 m
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d/s bed level = 200
u/s bed level
= 201.5
cistern level =
199.65
8.0 m
1.6 m
1.40 m
crest level =
202.65
0.5 IN 1
L = b
Lu = total length – d/s glacis – u/s glacis – crest width

Provide total horizontal length of
impervious flow = 20.60 m
crest width = 0.55 m
horizontal length of u/s glacis =
(202.65 - 201.5) x 0.5 = 0.58 m
horizontal length of d/s glacis =
(202.65 - 199.65) x 2 = 6.0 m
length of cistern = 8 m
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(a) u/s cutoff wall
d{ = 1.30 m
b = 20.60 m
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Assume a thickness of 0.4 m at the
end of u/s floor
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(b) d/s cutoff
d2 = 1.60 m
b = 20.6 m
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Assume thickness of 0.5 m at the
end of d/s floor.
Neglecting effect of mutual
interference.
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ib) At 3 m from the toe of glacis :
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(2) D/s wing walls :
Provide d/s wing walls upto the end of the
d/s floor and then join them to the return
walls.
Top level of wing wall = 202 + 0.5 = 202.5
m.
d/s Bed Protection :
No bed protection is required as a deflector
has been provided,
d/s Side Protection :
Length of side protection = 3D2 = 3 x 2= 6
m
Provide 20 cm thick brick side pitching
over 10 cm ballast for a length of
6 m at a slope of 1 : 1 beyond the d/s
impervious floor.
3/20/2014
ASST. PROF. DIET
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Drop structure lecture 4

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Drop structure lecture 4