Design of check dam

April 4, 2018 | Author: Naveen Yadav | Category: Dam, Hydraulics, Earth & Life Sciences, Earth Sciences, Chemical Engineering


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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 00 ] 1/3 Ds = 1.III) To Find Length of check Dam Discharge over weir when top width is < 1.473 x [ 37.1087 Say L= 21.56 Mt Maximum Scour depth = 1.00 Mtr V) Maximum Flood Discharge :Check Dam considered as a Submerged Type .3 L = Length of Weir Cd1 = 0. .00 ( for Medium sand as the type of Bed material.00 / 1.473 (Q/ Ksf) 1/3 Where.97 h2 = Height of Water Downstream side above crest 0.577 X 21.55 37.Q = Discharge in cumecs. Q =1.197 of Engineer’s Hand Book ).00 Cumecs There fore design structure is safe V) Calculation of maximum Scour depth Regime Scour Depth.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 < 1.97 21.Q = Discharge in cumecs.80 =( + Q= 0.577 (As per Civil Engineer HANDBOOK By Sri Vazarani Page NO 437) Cd2 = 0.00 X 0.00 mt 21.00 mts 1.08 Cumecs is > 37.00 L= 20.300 X 3.8 x 21.h2) 3/2 + Cd2 X L X h2 X Sqrt (2g(h1-h2)) Where h1 = Height of Water Upstream side above crest 0.00 Cumecs. Pg No.63 ) 38.67 x 0. Ds = 0. Q = Flood Discharge = 0.197 of Engineer’sHand book) Therefore. L = Length of Check Dam in mtrs H = Spillage over weir= 1. Ksf = Lacy’s Silt factor = 1. Ds = 0.84 x L x1 37.84 L H3/2 Where.00 X 4. L = Length of Check Dam in mtrs= H = Spillage over weir in mtrs H =. Accordingly Maximum Discharge as shown below Q = 2/3 XCd1 X L X Squrt (2g) X (h1.429 X 0.84 L H3/2 Where.Pg No.00m. Q =1. 0.5 x Scour depth ( For Moderate Bend .00m. 56 - 2.97 5. For Coarse sand (south Indian rivers)=12 HL =Maximum Head Loss.5 d = 1.00 d = 3.44 98.80 m Bottom width of Weir = 0. Height of check dam = 3.00 kgs 3.07 = 98.674 So Provide 1.87 M 2. i. There Fore the required foundation levelis below than the Nala bed level.00 3.7 6.82 10.07 G)      D/S Cutoff Wall U/S Scour Depth =2.00 m V) Design of Bodywall:Top Width of the Weir m( Assumed ).8 1.5 = 1.674 1.60 4.70 m = Bligh's Creep Co-efficent .5 2.77 2671.e Height of weir above low water .80 = 96.70mts VI) Design of Stilling Basin and Aprons:Calculation of Creep Length: Creep Length L= C*HL Where.25 97.8m below NBL i. 1. 0.80 Mt Foundation below Nala Bed Required Foundation Level = 98.e. Hence OK.04 Provide 0.13 kgs U/S Scour Depth =1.25 mtr Height of Cutoff Wall = 101.07 Where as we provide 1.60 0.70 Ds.85 2.20 m 1.74 99.80 x h + Top width.60 C) Abut ments : Height of Abutment= Surface area for steel= Steel required= F)      U/S Cutoff Wall 2. C 96.Dm = Dm = Dm 1.85 Provide 1.20 mts Surface area for steel= Steel required= 693.34 Mt X X Actual Foundation Level = HFL – Dm = 101.63 3.44 mtr Height of Cutoff Wall = 101. 3.34 Crest Level of Check Dam Which is below the lowest Nala Bed Level 98.87 Provide @ 1. 76 H 3/2 1.69 = 98.76 = 1. Pg No. Dm = 1.5 x Scour depth ( For Moderate Bend .63 Mt Maximum Scour depth = 1.225 and crest level 100.658 m Length of downstream floor is L2= 2.197 of Engineer’sHand book) 2 Therefore.44 .00 x 0.K Garge Where.21* C *Sqrty (HL/13) = 2.97 L= 11.35 1.63 x mt (=/1)^0.21 x 12 x = 7.89 m Then Ksf = Lacy’s Silt factor = 1. q = Flood Discharge = 37.63 2.25 m But actually we provide 5.99 mt Crest level= 99.44 Mt Level of Bottom of U/s cut off wall = 100.35 x [q2 / f ] 1/3 Ds = = 1.00 mt Length Solid cushion To calculate Head over the weir when High Flood discharge is passing 0.00 m q= Q/L = 1. f=1 page 554) Ds = 1.69 Total Height = 4.Pg No.197 of Engineer’s Hand Book ).5 X = 2.35 x [q / f ] 1/3 ( from Text Book of Irrigation and Hydraulic structures by S.35 ( for Medium sand as the type of Bed material.33 = 1. L= Length of Weir= 21.7 U/S HFL (assuming Nala Bed Level is 99. Dm Dm = 1. Ds = 1.2734 q= 1.5 X Ds.70 (H) 3/2 Where.99 m Head over the crest 0.24 mt 1.8 x H= 0.00 Cumecs.The Length of creep required including creep along cutt-off L= C*HL = 12. 31 L1= 1.658 10.E.25 mt L3 L3 = 9.20m thick M-20cc 0.5 m mt VII)      Thickness of Apron:Rate of Change of Uplift Pressure = = 0.93 0. th.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. M15 and 0.36 kscm (a)    Uplift Pressure at A = 1.67 = 10. th.425/75)) = 9. However Provide Provide 0.E. Total Creep Length except U/S floor = 10.39 m However Provide 0. Sub Division SINDHANUR Executive Engineer P. M20 for cushion bed.24/13)*(2.20mtr.31 + 0.12 But L2 7.25 mtr. (b)   Uplift Pressure at B = Thickness of Concrete Required = .97 0.DIVISION RAICHUR .67 mt Total Creep Length provided = 10.25 3.25 m Section officer Assistant Executive Engineer P.35 mt So provide 0.25 m thick M-15cc and 0.31 Mtr L1= Creep Length -Total Creep Length provided except U/S floor = 11.45 m However 0.5 m 2.Now.20 0.86 Thickness of Concrete Required = 0.12 = 1.R.48 0.R.92 - 7.87 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.
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