NAME OF PROJECT : DESIGN DATA : Culvert CH 500 with Glacis Fall : Chenderong Balai 1 Full supply discharge of distributory 2 Full supply level of distributory 3 Bed level of distributory 4 Bed width of distributory 5 Average bed level of parent channel 6 F.S.L. of parent channel 7 Safe exit gradient for canal bed material 8 Lacey's silt factor 9 Average N.S.L. (1) FIXATION OF CREST LEVEL AND WATERWAY : 8.000 6.900 6.500 3.000 6.800 7.900 0.200 0.700 Cumec m m m m m Generally the crest level
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  NAME OF PROJECT : Culvert CH 500 with Glacis Fall : Chenderong Balai DESIGN DATA : 1Full supply discharge of distributory 8.000 Cumec2Full supply level of distributory 6.900 m3Bed level of distributory 6.500 m4Bed width of distributory 3.000 m5Average bed level of parent channel 6.800 m6F.S.L. of parent channel 7.900 m7Safe exit gradient for canal bed material 0.200 8Lacey's silt factor  0.700 9Average N.S.L. (1)FIXATION OF CREST LEVEL AND WATERWAY : Generally the crest level of the distributory head regulator is kept .3 M higher than Bed level of Distributory.Adopt crest level of the regulator =6.800+ 0.300=7.100Waterway shall be worked out with the following formulaQ=Where,=7.900-6.900=1.000m=7.900-7.100=0.800m=0.800From fig. 6.5,Varshney vol-II, The ratio of modified coefficient to free discharge works out toCs=0.993CThus the coefficient of discharge =1.820X0.993=1.807(The value of 1.82 is for sharp crest free fall)Substituting the values,=6.187mThus clear waterway required=6.187mProvide 1 bays of  6.187 m giving a clear water way of6.187mHenceO.K. (2)LEVEL AND LENGTH OF DOWNSTREAM FLOOR The discharge and length of waterway is given by,Q=8.000Cumec,l =6.187m8.000Discharge intensity q =---------=1.293cumec/m6.187Head loss =u.s. F.S.L. -d.s. F.S.L.=7.900-6.900=1.000mC.B e H e3/2 h d  H e  h d  H e  B e  for,q =1.293cumec/m and1.000m1.254mCistern floor level required=D/s F.S.L.-=6.900-1.254=5.646mU/s specific energy ,+1.254+1.0002.254mfrom energy of flow curves=0.204m=1.195mLength of cistern floor required =4.952mThe downstream floor may be provided at R.L.5.646with a horizontal lengthof 5.000m. (from energy consideration) (4)Depth of sheet piles from scour considerations : U/S sheet pile : As per IS 6531 : 1994, on the upstream side of the head regulator, cut-off should be providedto the same depth as the cut-off stream of diversion work, if it exists or may be calculated asbelow :-Discharge intensity 'q'=1.293cumec/m=1.805mAnticipated scour = 1.25 R=2.256mR.L. of the bottom of scour hole=7.900-2.256=5.644mProvide U/S cut off depth = 1m minimum5.800Bottom R.L. of cut off ==5.644 D/S sheet pile : Discharge intensity 'q'=1.293cumec/m=1.805mAnticipated scour = 1.5 R=2.707mR.L. of the bottom of scour hole=6.900-2.707=4.193mMinimum depth of d/s cutoff below bed level =+0.5002.000=0.700mR.L. of the bottom of scour hole=5.800m(by minimum depth consideration)(Min. 1 m below cistern bed) =4.646mProvide sheet pile line down to elevation=4.193 (5)Total floor length and exit gradient The floor shall be subjected to the maximum static head when full supply is being maintained in theupstream for running the parent channel and there is no flow through the distributary head.Maximum static head =7.900-5.646=2.254mDepth of d.s. cut off =6.500-4.193=2.307mSince,H 1=---- ---------Hence,1-----------=-------------=0.205HH L =D/s specific energy E f2 (from Blench curves) =E f2  E f1 = E f2 H L  ==Prejump Depth D 1 corresponding to E f1 Postjump Depth D 2 corresponding to E f2 5 (D 2 - D 1 )Depth of scour 'R' = 1.35{q 2 /f} 1/3  Depth of scour 'R' = 1.35{q 2 /f} 1/3  y d  (where y d is water depth im m at d/s )G E  d π√λ  G E. d π√λ  3.705Hence requirement of total floor length b ==8.547mAdopt total floor length =9.000mTherefore,length of d/s floor=2/3 X total impervious length=6.000mThe floor length shall be provided as below:-Downstream horizontal floor=6.000mD/s glacis length with ( 2.000:1 slope)=2.908mCrest width=0.533mU/s glacis length with (1.000:1 slope)=0.300mUpstream floor=3.000mTotal=12.742m (6)Pressure calculations Let the floor thickness in the u/s be0.300 m and near the downstream cutoff be0.550m.(i) Upstream sheet pile Let the thickness of sheet pile=0.500md =6.500-5.644=0.856mb =12.742m1d0.856----- = ------ =--------=0.067 α b12.742From Khosla's Pressure curves7.9620.8390.769;0.069Correction for Depth =0.024 (+ ve)b'=11.7420.769+0.024=0.794(ii) Downstream sheet pile d =6.500-4.193=2.307mb =12.742m1d2.307------ = ------ =---------=0.181 α b12.742From Khosla's Pressure curves3.3070.3710.254;0.116Correction for Depth =0.028 (- ve)b'=11.7420.371-0.028=0.343Toe of Glacis Pressure =0.555 (7)Floor thickness Minimum Thickness of U/S F 0.300 Thickness under the crest =0.600mThickness under the toe of Glacis =1.009mProvide thickness for a distance 1.500from toe of Glacies1.01m Provided1.05 m% Pressure =0.5020.913m3.000from toe of Glacies0.91m Provided0.95 m% Pressure =0.4490.816m4.500from toe of Glacies0.82m Provided0.85 m% Pressure =0.3960.7206.000from toe of Glacies0.72m Provided0.55 From Khosla's exit gradient curve α = α. d λ = φ D =φ C =φ D − φ C =  φ C   corrected = λ = φ Ε1 =φ D1 =φ Ε1 − φ D1 =  φ Ε1   corrected =Provide the thickness for a distanceProvide the thickness for a distanceProvide the thickness for a distance


Dec 21, 2017

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Dec 21, 2017
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