PWD, WRO
Irrigation Design Handbook
8. DESIGN OF VERTICAL TYPE DROP ON PERMEABLE FOUNDATION
HYDRAULIC PARTICULARS
LOCATION LS 250 M
DISCHARGE REQUIRED 4.09 cumecs
U/S BED LEVEL 298.3 M
D/S BED LEVEL 296.9 M
LENGTH OF DROP 3 M
CONCENTRATION FACTOR 0 %
EXIT GRADIENT OF BED MATERIAL 0.2
FLOOR THICKNESS AT END (ASSUMED) 1 M
FlOOR THICKNESS FOR U/S APRON 0.6 M
THICKNESS OF CUTOFF 0.6 M
DESIGN OF CANAL SECTION
(UNLINED PORTION)
BED WIDTH 3 M
SIDE SLOPE 1 0.5
MANNINGS COEFFT N 0.025
BED SLOPE=1 IN 2500 0.0004
DEPTH OF WATER 1.5 M
1 1.5 1
0.5 0.5
3
Area "A" 5.625 Sq m
Wetted Perimeter "P" 6.354102 m
Hydraulic mean Depth "R"=A/P 0.885255
Velocity "V"=1/nR^2/3*S^0.5
=1/ 0.0X0.9^2/3X0.0^0.5
0.737568 m/s
Discharge "Q"=A X V 4.148821 Cumecs
> 4.09 CUMECS
Depth of water 1.5 m
D/S waterlevel=D/Sbed level+depth of water=296.9 +1.5
D/S waterlevel 298.4 m
Depth of u/s water 1.5 m
U/s water level = 298.3 + 1.5
U/s water level 299.8 m
Approach velocity 'v'=Q/A = 0.41/0.56
Approach velocity 0.727111 m/s
Velocity head =v /2g
2
= 0.7271^2 /2*9.81
Velocity head 0.026947 m
U/S TEL =U/S water level+velocity head= 299.80+0.02695
U/S TEL 299.827 m
Top width of drop (Assumed) 0.8 m
Q = 1.835 LH3/2(H/B)1/6 as per page 317 of Design of irrigation structures by Varshney
0.41=1.835X 3XH^3/2X(H/0.8)^1/6
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Head causing flow required for the design discharge 0.81825 m
say 0.82 m
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Crest level = U/s TEL-H 299.827 - 0.82
Crest level 299.00 m
d= Height of the crest in metre above d/s bed = 299.00 - 296.9
d= Height of the crest in metre above d/s bed 2.10
Width of crest = 0.55*d^0.5 as per page317 of Design of irrigation structures
=0.55*0.21^0.5 by R.S.Varshney
Width of crest 0.797026 m
say 0.8 m*
Provide a top width of 0.8 m
DESIGNING OF STILLING BASIN
Assuming depth of water cushion = 0.3 m
Basin level =296.90-0.3
Basin level 296.6 m
h= u/s water level-basinlevel =299.8-296.6
h = 3.2 m
v= (2gh)^.5 =(2*9.81*3.2)^0.5
Velocity of flow at the toe= 7.923636 m/sec
d1 =Q/(L/v) =0.41/3*0.79
Depth of flow at the toe d1= 0.172059 m
h= =3.2 -0.2
h= 3.027941 m
v = (2gh)^0.5 =2*9.81*0.30^0.5
v= 7.707672 m/s
d1 =Q/(L/v) =0.41/3*0.77
d1 0.17688 m
Hence initial depth d1 0.18 m
Froude's No. "F =v/(gd)^0.5 =0.77/(9.81x0.18)^0.5
Froude's No. "F" 5.800327
Sequent depth "d2" =d1/2(-1+(1+8F2)^0.5)
=0.18/2(-1+(1+8*0.58^2)^0.5)
Sequent depth "d2" 1.389262 m
Basin floor level RWL - Sequent Depth =298.4-0.14
297.0107 m
Depth of cistern = 1/4(H.HL)^2/3 as per page 634 Irrigation and hydraulic
structures by S.K. Garg
HL = U/S FSL - D/S FSL
= 299.8 - 298.4
HL 1.4 m
H 0.82 m
Depth of cistern DC =1/4*(0.83*1.4)^2/3
DC 0.273706 m
Provide a basin depth of 0.3 m
Basin floor level= 296.9 - 0.3 = 296.6
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Basin floor level based on hydraulic criteria = 297.0107 *
Provide basin floor level as 296.6 m
Length of basin 5x(H*HL)^0.5 As per page 629 of Irrigation engineering and
hydraulic structures by S.K. Garg
=5X(0.83*1.4)^0.5
5.351534 m
Say 5.40 m
Provide a basin length of 5.5 m
DESIGN OF PERMEABLE FOUNDATION
Bottom width of drop body wall (from drop wall stabilit 2.6 m
Scour depth and depth of cutoff
Scour depth = 0.475(Q/f)^1/3
taking silt factor "f" =1.0
Scour depth = "R" =0.475*(0.41/1)^1/3
Scour depth "R" 0.759629 m
U/S Scour level= U/S MFL -1.25 R 298.8505 m
U/s Floor level (taken = basin level unless specified 296.6 m
If u/s floor level is different type in cell F119 296.6 m
Depth of cutoff required from scour considerations =U/S floor level-scour level
=296.60-0.2989
Depth of cutoff required from scour considerations -2.25046 m
Minimum depth of cutoff =(u/s water depth/3)+0.6 1.1 m
Provide a u/s cutoff depth of 1.1 m
However provide a depth of 1.5 m
Depth of u/s cutoff 1.5 m
Bottom of u/s cutoff= =296.60-1.5 295.1 m
D/s Scour level D/S FSL - 1.5R 297.2606 m
Depth of Cutoff required from scour considerations -0.36056 m
Minimum depth of cutoff =(D/s water depth/2)+0.6 1.35 m
Provide a d/s cutoff depth of 1.4 m
However provide a depth of 2m
Depth of d/s cutoff 2m
Bottom of d/s cutoff =296.9-2.0 294.9 m
Floor Length Required From Exit Gradient Considerations
Exit gradient 0.2
Taking head upto bed level only
H = U/S Bed level - D/S Bed level =298.3-296.9
H = U/S Bed level - D/S Bed level 1.4 m
d = Depth of d/scutoff 2m
Ge = H 1
d PÖl
l = 1.240863
l = (1+(1+a2)^0.5)/2
a = Ö (2 l-1)2 -1 1.09 m
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Length of floor required from exit gradient 2.19 m
Length of floor required from hydraulic criteria =length of basin+width of bodywall+
2*cut-off thick
Length of floor required from hydraulic criteria = 5.50+2.600+2*0.60
Length of floor required from hydraulic criteria 9.3 m
Maximum say 9.5 m
Length of floor provided 9.5 m
Calculation of pressures
u/s pile
= b/d '= 6.333
1/a = 0.158 Say 0.16
From table in pg 91 of CBIP pub. no. 12
For 1/a = 0.16
fD = 23.953
fE = 34.779
fE1 =100 - 0 = 100 %
fD1 =100 - 23.953 = 76.04737 %
fc1 =100 - 34.779 = 65.22105 %
Correction for floor thickness
Assumed floor thickness for U/S = 0.6 m
CorrectionCf =( ( fD1-fC1)/d)*thickness
CorrectionCf = 4.330526 +ve
Correction for mutual interference of pile
Ci = 19x ÖD/b1x (d+D)/b
Where b1 = centre to centre distance of piles
b = total floor length provided
d = depth of pile =1.5-0.6 0.9
D = depthof pile below the floor 1.1
of the first pile whose inter-
ference has to be determined
Ci = 1.406246
corrected fc1 =0.652+0.43+0.14
corrected fc1 = 70.95783 %
D/S Pile
= b/d '= 4.75
1/a = 0.21052632 Say 0.21
From table in pg 91 of CBIP pub. no. 12
For 1/a = 0.21
fD = 27.1263
fE = 39.747
fE2 = 39.74737 %
fD2 = 27.12632 %
(III) Typical Design of Various Irrigation Structures - 8.Vertical drop ( Design) Page No.
PWD, WRO
Irrigation Design Handbook
fc2 = 0%
(III) Typical Design of Various Irrigation Structures - 8.Vertical drop ( Design) Page No.
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Irrigation Design Handbook
Correction for floor thickness
Assumed floor thickness for D/S = 1m
CorrectionCf = 6.310526
Correction for mutual interference of pile
Ci = 19x ÖD/b1x (d+D)/b
Where
d = depth of pile 1.110738
D = depthof pile below the floor 0.910738
of the first pile whose inter-
ference has to be determined
Ci = 1.293302
corrected f E2 = 32.14354 %
s.no. f pile1 pile2
1 fE 100 32.14
2 fD 76.05 27.13
3 fC 70.96 0.00
Elevation of HGL
U/S WL D/S WL Head U/S Pile D/S Pile
fE1 fD1 fC1 fE2 fD2 fC2
100 76.04737 70.96 32.14 27.13 0.00
Static head
299.00 296.9 2.1 2.1 1.596995 1.490114 0.675014 0.569653 0
299 298.497 298.3901 297.575 297.4697 296.9
Dynamic head
299.8 298.4 1.4 1.4 1.064663 0.99341 0.45001 0.379768 0
299.8 299.4647 299.3934 298.85 298.7798 298.4
HGL Line
299.3934
298.85
298.3901
297.575
296.9
8.9
(III) Typical Design of Various Irrigation Structures - 8.Vertical drop ( Design) Page No.
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FLOOR THICKNESS
Assuming that there is water in the basin upto d/S bed level, immedieately after the flow ceases,
the head is calculated from d/s bed level. 5.5
(a) Unbalanced head at 1.3 m from 0.6
centre of d/s pile 0.794074
Thickness required=H/(G-1) 0.567196
Provide a thickness 0.6 m
(b) Unbalanced head at 3.3 m from
centre of d/s pile 0.977242
Thickness required 0.69803
Provide a thickness of 0.7 m
(c) Unbalanced head at 5.8 m from
centre of d/s pile ie at toe 1.206203
Thickness required 0.861574
Provide a thickness of 0.9 m
Dynamic condition
Check for thickness of drop at toe
Assuming the jump forms at toe itself for worst condition
Elevaton of HGL at toe ie 5.8 m from 299.2041
Centre of D/s pile
Unbalanced head at 5.8 m from 2.304135
centre of d/s pile ie at toe
Taking 2/3 rd the head , Head = 1.53609
Thickness required 1.097207
Provide a thickness of 1.1 m
UPSTREAM PROTECTION WORKS
As per pg 304 of Theory and Design of Irrigation structures by R.S Varshney
Pitching for a length equal to u/s water depth shall be provided with a slope of 1:10
U/S water depth 1.5 m
Provide stone protection for length = 3m
DOWNSTREAM PROTECTION WORKS
As per pg 305 of Theory and Design of Irrigation structures by R.S Varshney
Pitching for a minimum length
equal to 3 times d/s water depth shall be provided
D/S water depth 1.5 m
Length of pitching required 4.5
Provide stone pitching for a length of 5m
(III) Typical Design of Various Irrigation Structures - 8.Vertical drop ( Design) Page No.
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STABILITY ANALYSIS OF DROP WALL:
T.B.L 300.3 m
F.T.L 299.00 m
Maximum flood discharge 4.09 cumecs
U/S. bed level 298.3 m
D/S bed level 296.9 m
Depth of water = 1.5 m
U/S waterlevel= 299.8 m
D/S waterlevel= 298.4 m
Top level of wingwall= 299.00 m
Foundation level= 296.6 m
Height of wing wall= 2.4 m
Width of platform= B= 0.8 m
Rear slope= 1: 0.75
Rear batter 1.8 m
Front slope= 1: 0
Front batter= 0 m
b3= Base width= 2.6 m
Height= 2.4 m
Density= 2 t/m3
f= 22 o
d= f/2= 11 o
a o= 0
ah= B*I*ao 0
av= 1/2*ah= 0
l= tan a(h)/(1±a(v)=
-1
00 or 0
a= 36.8699 o
i= 0
Increment= 0%
Unit weight of masonry= 2.3 t/m3
Unit weight of water= 1 t/m3
STRESS DUE TO EARTH PRESSURE ACTIVE EARTH PRESSURE:
cosl 1
Cos^2(f-l-a)= 0.9341
Cos^2a= 0.64
Cos(d+l+a)= 0.6708
Sin(f+d)= 0.5446
Sin(f-i-l)= 0.3746
Cos(a-i)= 0.8
Cos(d+l+a)= 0.6708
1+ Sin(f+d)*Sin(f-i-l) 0.5
= 1.6166
Cos(a-i)*Cos(d+a+l)
1 2
1+ Sin(f+d)*Sin(f-i-l) 0.5 = 0.38265
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Cos(a-i)*Cos(d+a+l)
Cos^2( f-d-a) = 2.17586
cosl* Cos^2a* f-a-l)
(1+aV)= 1
Ca= (1+aV)* cos^2 ( f-d-a) 1 2
cosl* Cos^2( Cos(d+a) 1+ Sin(f+d)*Sin(f-i) ^0.5
Cos(a-i)*Cos(d+a)
Ca = 0.83259
Ps=1/2*w*h *Ca=
2
Earth pressure: 4.79572 t
Vertical component= 3.55662 t
Horizontal component= 3.21705 t
0.8 299.00
0.75
1 2.4
a b
O 296.6
0.8
1.8
2.6
Condition1: No Flow Condition
Sl.no FORCES
Coefficient L B H Unit wt. Verticl HorizontaL.A MOMENT
I .Weight of masonry +VE -VE
a. 0.5 1 1.8 2.4 2.3 4.968 1.4 6.9552
b. 1 1 0.8 2.4 2.3 4.416 0.4 1.7664
III . Static earth pressure:
Vertical component: 3.55662 2 7.113235
Horizontal component: 3.21705 0.8 2.573637
V= 12.9406 3.21705 15.83483 2.573637
X= 1.02477 m
e= b/2-x= 0.27523 13.2612
6e/b= 0.63514
Maximum stress= P/A*(1+6e/b)= 8.1383 t/m2
Minimum stress= P/A*(1-6e/b)= 1.816 t/m2
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Condition2: Maximum Flow Condition
U/S waterlevel= 299.8 m
Top level of wingwall= 299.00 m
Foundation level= 296.6 m
Height of wing wall= 2.4 m
Width of platform= 0.8 m
Front slope= b1= : 0.75
Front batter 1 1.8 m
Rear slope= : 0
Rear batter= 1 0 m
b3= Base width= 2.6 m
Height= 2.4 m
Density= 1 t/m3
f= 22 o
d= f/4= 5.5 o
a o= 0
ah= B*I*ao 0
av= 1/2*ah= 0
l= tan a(h)/(1±a(v)=
-1
00 or 0
a= 36.8699 o
i= 0
Increment= 0%
Unit weight of masonry= 2.3 t/m3
Unit weight of water= 1 t/m3
STRESS DUE TO EARTH PRESSURE ACTIVE EARTH PRESSURE:
cosl 1
Cos^2(f-l-a)= 0.9341
Cos^2a= 0.64
Cos(d+l+a)= 0.7388
Sin(f+d)= 0.4617
Sin(f-i-l)= 0.3746
Cos(a-i)= 0.8
Cos(d+l+a)= 0.7388
1+ Sin(f+d)*Sin(f-i-l 0.5= 1.54098
Cos(a-i)*Cos(d+a+l)
1 2
1+ Sin(f+d)*Sin(f-i-l) 0.5 = 0.42112
Cos(a-i)*Cos(d+a+l)
Cos^2( f-d-a) = 1.97561
cosl* Cos^2a* f-a-l)
(1+aV)= 1
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Ca= (1+aV)* f-a) 1 2
cosl* Cos^2( Cos(d+a) 1+ Sin(f+d)*Sin(f-i) ^0.5
Cos^2a* Cos(a-i)*Cos(d+a)
Ca = 0.83197
Pa=1/2*w*h *Ca=
2
Earth pressure: 2.39608 t
Vertical component= 1.61475 t
Horizontal component= 1.77025 t
0.8 299.00
0.75 0
1 1
2.4
a b c
o 296.6
0.8 0
1.8
2.6
Condition1: At maximum Flow Condition
Sl.no FORCES
Coefficient L B H Unit wt. Verticl HorizontaL.A MOMENT
I .Weight of masonry +VE -VE
a. 0.5 1 1.8 2.4 2.3 4.968 1.4 6.9552
b. 1 1 0.8 2.4 2.3 4.416 0.4 1.7664
c. 0.5 1 0 2.4 2.3 0 0
iv Weight of water 0 0
a 0.5 1 1.8 2.4 1 2.16 2 4.32
b 1 1 2.6 0.8 1 2.08 1.3 2.704
ii Earth pressure 0 0
Vertical component: 1.61475 2 3.229503
Horizontal component: 1.77025 0.8 1.416196
iii Water pressure
a 1 1 0.8 2.4 1 1.92 1.2 2.304
b 0.5 1 2.4 2.4 1 2.88 0.8 2.304
D/S Water pressure
0.5 1 1.5 1.5 1 1.125 0.5 0.5625
15.2388 7.69525 19.5376 6.024196
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Xbar= 0.88678
e= b/2-x= 0.41322
6e/b= 0.95359
Maximum stress= P/A*(1+6e/b)= 11.45 t/m2
Minimum stress= P/A*(1-6e/b)= 0.272 t/m2
DESIGN OF ABUTMENT AND UPSTREAM WING WALL
U/S waterlevel= 299.8 m
Top level ofabutment= 300.3 m
Foundation level= 296.6 m
Height of wing wall= 3.7 m
Width of platform= b1= 0.45 m
Front slope= 1: 0
Front batter 0m
Rear slope= 1: 0.7
Rear batter= 2.59 m
b3= Base width= 3.04 m
Height= 3.7 m
Density= 2 t/m3
f= 22 o
d= f/2= 11 o
a o= 0.02
ah= B*I*ao 0
av= 1/2*ah= 0
l= tan a(h)/(1±a(v)=
-1
00 or 0
a= 34.992 o
i= 0
Increment= 0%
Unit weight of masonry= 2.3 t/m3
Unit weight of water= 1 t/m3
STRESS DUE TO EARTH PRESSURE ACTIVE EARTH PRESSURE:
cosl 1
Cos^2(f-l-a)= 0.9495
Cos^2a= 0.6711
Cos(d+l+a)= 0.6948
Sin(f+d)= 0.5446
Sin(f-i-l)= 0.3746
Cos(a-i)= 0.8192
Cos(d+l+a)= 0.6948
1+ Sin(f+d)*Sin(f-i-l 0.5= 1.59872
Cos(a-i)*Cos(d+a+l)
1 2
1+ Sin(f+d)*Sin(f-i-l) 0.5 = 0.39125
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Cos(a-i)*Cos(d+a+l)
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Cos^2( f-d-a) = 2.03624
cosl* Cos^2a* f-a-l)
(1+aV)= 1
Ca= (1+aV)* f-a) 1 2
cosl* Cos^2( Cos(d+a) 1+ Sin(f+d)*Sin(f-i) ^0.5
Cos^2a* Cos(a-i)*Cos(d+a)
Ca = 0.79668
Pa=1/2*w*h *Ca=
2
Earth pressure: 10.9066 t
Vertical component= 7.84448 t
Horizontal component= 7.57744 t
0.45 300.3
0 0.7
1 1
3.7
a b
296.6 o
0 0.45 2.59
0
3.04
Sl.no FORCES
Coefficient L B H Unit wt. Verticl HorizontaL.A MOMENT
I .Weight of masonry +VE -VE
a. 1 1 0.45 3.7 2.3 3.8295 2.815 10.78004
b. 0.5 1 2.59 3.7 2.3 11.0205 1.726667 19.02864
ii Earth pressure
Vertical component: 7.84448 0.863333 6.772399
Horizontal component: 7.57744 1.233333 9.345503
22.6944 7.57744 45.92659
2.02369
e= b/2-x= 0.50369
6e/b= 0.99413
Maximum stress= P/A*(1+6e/b)= 14.887 t/m2
Minimum stress= P/A*(1-6e/b)= 0.0438 t/m2
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DESIGN OF DOWNSTREAM WING WALL
D/S waterlevel= 298.4 m
Top level ofabutment= 298.9 m
Foundation level= 296.6 m
Height of wing wall= 2.3 m
Width of platform= b1= 0.45 m
Front slope= 1: 0
Front batter 0 m
Rear slope= 1: 0.7
Rear batter= 1.61 m
b3= Base width= 2.06 m
Height= 2.3 m
Density= 2 t/m3
f= 22 o
d= f/2= 11 o
a o= 0
ah= B*I*ao 0
av= 1/2*ah= 0
l= tan a(h)/(1±a(v)=
-1
00 or 0
a= 34.992 o
i= 0
Increment= 0%
Unit weight of masonry= 2.3 t/m3
Unit weight of water= 1 t/m3
STRESS DUE TO EARTH PRESSURE ACTIVE EARTH PRESSURE:
cosl 1
Cos^2(f-l-a)= 0.9495
Cos^2a= 0.6711
Cos(d+l+a)= 0.6948
Sin(f+d)= 0.5446
Sin(f-i-l)= 0.3746
Cos(a-i)= 0.8192
Cos(d+l+a)= 0.6948
1+ Sin(f+d)*Sin(f-i-l 0.5= 1.59872
Cos(a-i)*Cos(d+a+l)
1 2
1+ Sin(f+d)*Sin(f-i-l) 0.5 = 0.39125
Cos(a-i)*Cos(d+a+l)
Cos^2( f-d-a) = 2.03624
cosl* Cos^2a* f-a-l)
(1+aV)= 1
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Ca= (1+aV)* f-a) 1 2
cosl* Cos^2( Cos(d+a) 1+ Sin(f+d)*Sin(f-i) ^0.5
Cos^2a* Cos(a-i)*Cos(d+a)
Ca = 0.79668
Pa=1/2*w*h2*Ca=
Earth pressure: 4.21445 t
Vertical component= 3.03121 t
Horizontal component= 2.92802 t
0.45 298.9
0 0.7
1 1
2.3
a b
296.6 o
0.45 1.61
0
2.06
Sl.no FORCES
Coefficient L B H Unit wt. Verticl HorizontaL.A MOMENT
I .Weight of masonry +VE -VE
a. 1 1 0.45 2.3 2.3 2.3805 1.835 4.368217
b. 0.5 1 1.61 2.3 2.3 4.25845 1.073333 4.570736
ii Earth pressure
Vertical component: 3.03121 0.536667 1.62675
Horizontal component: 2.92802 0.766667 2.244817
9.67016 2.92802 12.81052
1.32475
e= b/2-x= 0.29475
6e/b= 0.85849
Maximum stress= P/A*(1+6e/b)= 8.7242 t/m2
Minimum stress= P/A*(1-6e/b)= 0.6643 t/m2
(III) Typical Design of Various Irrigation Structures - 8.Vertical Drop (Stability analysis of Drop wall) Page No.
