CONSTRUCTION OF CHECK DAM GOVERNMENT HALLA NEAR EJ BASAPURA , PAGADADINNI VILLAGE IN SINDHANUR TALUK, RAICHUR DISTRICT.
DESIGN DETAILS OF CHECK DAM I) Calculation for maximum flood discharge by Ryvey's formula Q = CM2/3 Where, Q = Discharge in cumecs.
6
M = Catchment area in Km2= 6 Sq.km
M
Q Say
C = Ryve’s Constant = = = Q=
9.75 32.19 33.00
9.75 x
Sq.km ( Pg No.23 of Minor Irrigation Manual ). 6
^ 2/3
Cumecs
II) Calculation for maximum flood discharge by Area- Velosity Method:Q=AV Where, Q = Discharge in cumecs. A = Area in Sq mt V =Velosity in mt/sec n = Co efficent of roughness= 0.033 R = Hydraulic Radian = A/P A= C/s 30.00 Sqm A= Average cross section Depth of flow
P= Wetted Perimeter= S= U/S Nala Bed Level D/S Nala Bed Level
0.65 60.100 99.434 98.783 0.651 280 0.00232
Difference Length Existing Bed Slope = R = Hydraulic Radian = A/P R= 0.499 V= (1/0.033) x 0.852 x 0.048 V= 30.30 x 0.852 x 0.048 V= 1.25 Mtr/sec Q= 30.00 x 1.25 Q= 37.36 Cumecs Say QSay Q=92 37.00 Cumecs Cumecs Compare with above two cases , consider Maximum Discharge Q=
37.00 Cumecs
III) To Find Length of check Dam Discharge over weir when top width is < 1.00m. Q =1.84 L H3/2 Where,Q = Discharge in cumecs. L = Length of Check Dam in mtrs H = Spillage over weir=
1.00 mts
1.84 x L x1 37.00 L= 20.1087 Say L= 21.00 mtr 3/2
As per site condition consider Length of weir L=
IV) Design of Weir Wall Discharge over weir when top width is < 1.00m. Q =1.84 L H3/2 Where,Q = Discharge in cumecs. L = Length of Check Dam in mtrs= H = Spillage over weir in mtrs H =. 0.97
21.00 mt
21.00 Mtr
V) Maximum Flood Discharge :Check Dam considered as a Submerged Type , Accordingly Maximum Discharge as shown below Q = 2/3 XCd1 X L X Squrt (2g) X (h1- h2) 3/2 + Cd2 X L X h2 X Sqrt (2g(h1-h2)) Where
h1 = Height of Water Upstream side above crest 0.97 h2 = Height of Water Downstream side above crest 0.3 L = Length of Weir Cd1 = 0.577 (As per Civil Engineer HANDBOOK By Sri Vazarani Page NO 437) Cd2 = 0.80
=( + Q=
0.67 x 0.577 X 21.00 X 4.429 X 0.8 x 21.00 X 0.300 X 3.63 ) 38.08 Cumecs is > 37.00 Cumecs There fore design structure is safe V) Calculation of maximum Scour depth Regime Scour Depth, Ds = 0.473 (Q/ Ksf) 1/3 Where, Q = Flood Discharge =
0.55
37.00 Cumecs.
Ksf = Lacy’s Silt factor = 1.00 ( for Medium sand as the type of Bed material, Pg No.197 of Engineer’sHand book) Therefore, Ds = 0.473 x [ 37.00 / 1.00 ] 1/3 Ds
=
1.56 Mt
Maximum Scour depth = 1.5 x Scour depth ( For Moderate Bend - Pg No.197 of Engineer’s Hand Book ).
Dm
=
Dm
=
Dm
1.5
=
1.5
2.34 Mt
X X
Actual Foundation Level = HFL – Dm
= 101.07 = 98.74 99.70
Ds.
1.56 -
2.34
Crest Level of Check Dam Which is below the lowest Nala Bed Level 98.674 So Provide 1.80 Mt Foundation below Nala Bed Required Foundation Level = 98.674 1.80 = 96.87 Provide @ 1.8m below NBL i.e. There Fore the required foundation levelis below than the Nala bed level, Hence OK. Height of check dam = 3.00 m V) Design of Bodywall:Top Width of the Weir m( Assumed ). 0.80 m Bottom width of Weir = 0.80 x h + Top width. 3.20 mts
Surface area for steel= Steel required=
693.00 kgs
3.00
3.60
C) Abut ments : Height of Abutment= Surface area for steel= Steel required= F) U/S Cutoff Wall
2.44 mtr
Height of Cutoff Wall =
101.07
G) D/S Cutoff Wall
U/S Scour Depth =2.00 d =
3.25 mtr
Height of Cutoff Wall =
101.07
Where as we provide 1.70mts VI) Design of Stilling Basin and Aprons:Calculation of Creep Length: Creep Length L= C*HL Where,
C
96.7
6.60
4.87 M
2.77 2671.13 kgs
U/S Scour Depth =1.5 d =
1.8
1.97
5.85
2.44
98.63
3.25
97.82
10.60
0.04 Provide
0.85 Provide
1.20 m
1.70 m
= Bligh's Creep Co-efficent , For Coarse sand (south Indian rivers)=12
HL =Maximum Head Loss, i.e Height of weir above low water
The Length of creep required including creep along cutt-off L= C*HL = 12.00 x 0.97 L= 11.658 m Length of downstream floor is L2= 2.21* C *Sqrty (HL/13) = 2.21 x 12 x = 7.25 m But actually we provide 5.00 mt Length Solid cushion To calculate Head over the weir when High Flood discharge is passing
H 3/2 1.76 = 1.8 x H= 0.99 m Head over the crest 0.99 mt Crest level= 99.7 U/S HFL (assuming Nala Bed Level is 99.225 and crest level 100.69 Total Height = 4.89 m Then
Ksf = Lacy’s Silt factor = 1.35 ( for Medium sand as the type of Bed material, Pg No.197 of Engineer’sHand book) 2 Therefore, Ds = 1.35 x [q / f ] 1/3 ( from Text Book of Irrigation and Hydraulic structures by S.K Garge Where, f=1 page 554) Ds = 1.35 x [q2 / f ] 1/3
Ds
= =
1.35 1.63
x mt
(=/1)^0.33
= 1.63 Mt Maximum Scour depth = 1.5 x Scour depth ( For Moderate Bend - Pg No.197 of Engineer’s Hand Book ). Dm = 1.5 X Ds. Dm
Dm
=
1.5
X
= 2.44 Mt Level of Bottom of U/s cut off wall = 100.69 = 98.24 mt
1.63
2.44
Now, Total Creep Length except U/S floor = 10.31 Mtr L1= Creep Length -Total Creep Length provided except U/S floor = 11.658 10.31 L1= 1.35 mt So provide 0.67 mt Total Creep Length provided = 10.31 + 0.67 = 10.98 mt Provide Length is More Than Creep Length L2+L3= 18 *C* SQRT( (HL/13)*(Q/75)) Where L3
=
Length of Loose Talus in D/S
L2+L3= 18 x 12 sqrt ((1.24/13)*(2.425/75)) = 9.12 But L2 7.25 mt L3 L3
=
9.12
= 1.87 mt AS Per site codition the solid apron is And Provide Loose talus in U/s say Total Length of creep provided= L1+L2 +L3 +L4 = 13.92
-
7.25 3.5 m 2.5 m
mt
VII) Thickness of Apron:Rate of Change of Uplift Pressure = = 0.36 kscm (a) Uplift Pressure at A = 1.86 Thickness of Concrete Required = 0.39 m However Provide 0.45 m However 0.25 mtr. th. M15 and 0.20mtr. th. M20 for cushion bed. (b) Uplift Pressure at B = Thickness of Concrete Required = . However Provide Provide 0.25 m thick M-15cc and 0.20m thick M-20cc
0.93
0.97 0.48 0.20 0.25 m
Section officer Assistant Executive Engineer P.R.E. Sub Division SINDHANUR
