design of drops d-22-3 (1)
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KM10.150-2 DESIGN OF U/S WING WALLS AND STABILITY ANALYSIS FOR
DROP AT
KM10.150-D221.90206.157w32.723w1w4PH3.173w2PvA0.601.30203.434X0.250.25c
=2.40.450w5202.984e =2.1B2.40.5( B )Kv =0.04Taking moments about '
A 'Kh =0.158S.No.ForceMagnitude in TonnesL.A in 'M'Moment in
T.M.1w10.6X2.723X2.43.921-0.3001.1762w20.5X1.3X2.723X2.44.248-1.0334.3893w30.5X1.3X2.723X2.13.717-1.4675.4514Pv0.04X2.1X2.723X2.7230.623---5PH0.158X2.1X2.723X2.723-2.4601.089-2.680Total
Load(W) =12.509TTotal Moment=8.338T.M.Lever Arm=Total Moment/ Total
load =0.667Meccentricity = 0.50 X base width - Leveer arm
=(e)=0.283MB/6 =0.3170.283Hence O KStresses in Concrete = W/B(1
(6e)/B) =8.905T/M2(Max)and3.474T/M2(Min)Taking moments about ' B
'S.No.ForceMagnitude in TonnesL.A in 'M'Moment in
T.M.1w10.6X2.723X2.43.921-0.5502.1572w20.5X1.3X2.723X2.44.248-1.2835.4513w30.5X1.3X2.723X2.13.717-1.7176.3814w40.25X2.723X2.11.4302.2753.2525w52.4X0.45X2.42.5921.2003.1106Pv0.04X2.1X3.173X3.1730.846--7PH0.158X2.1X3.173X3.173-3.3411.269-4.240Total
Load(W) =16.753TTotal Moment=16.112T.M.Lever Arm=Total Moment/
Total load =0.962Meccentricity = 0.50 X base width - Leveer arm
=(e)=0.238MB/6 =0.4000.238Hence O KStresses in Soils = W/B(1
(6e)/B) =11.139T/M2(Max)and2.822T/M2(Min)DESIGN OF D/S WING WALLS
AND STABILITY ANALYSIS FOR DROP AT KM10.150
D-221.1205.157w31.550w1w4PH2.000w2PvA0.600.50203.607X0.250.25c
=2.40.450w5203.157e =2.1B1.60.5( B )Kv =0.04Taking moments about '
A 'Kh =0.158S.No.ForceMagnitude in TonnesL.A in 'M'Moment in
T.M.1w10.6X1.55X2.42.232-0.3000.6702w20.5X0.5X1.55X2.40.930-0.7670.7133w30.5X0.5X1.55X2.10.814-0.9330.7604Pv0.04X2.1X1.55X1.550.202---5PH0.158X2.1X1.55X1.55-0.7970.620-0.494Total
Load(W) =4.178TTotal Moment=1.648T.M.Lever Arm=Total Moment/ Total
load =0.394Meccentricity = 0.50 X base width - Leveer arm
=(e)=0.156MB/6 =0.1830.156Hence O KStresses in Concrete = W/B(1
(6e)/B) =4.134T/M2(Max)and1.030T/M2(Min)Taking moments about ' B
'S.No.ForceMagnitude in TonnesL.A in 'M'Moment in
T.M.1w10.6X1.55X2.42.232-0.5501.2282w20.5X0.5X1.55X2.40.930-1.0170.9463w30.5X0.5X1.55X2.10.814-1.1830.9634w40.25X1.55X2.10.8141.4751.2005w51.6X0.45X2.41.7280.8001.3826Pv0.04X2.1X2X20.336--7PH0.158X2.1X2X2-1.3270.800-1.062Total
Load(W) =6.854TTotal Moment=4.657T.M.Lever Arm=Total Moment/ Total
load =0.680Meccentricity = 0.50 X base width - Leveer arm
=(e)=0.120MB/6 =0.2670.120Hence O KStresses in Soils = W/B(1
(6e)/B) =6.219T/M2(Max)and2.348T/M2(Min)
km10.150-1 DESIGN OF1.00M DROP AT KM 10.150 OF D-22 OF AMRP
PRROJECTDATAU / SD / S1. Discharge1.08Cumecs1.08Cumecs2.
F.S.L205.557M204.557M3. F.S.D0.8M0.8M4. Canal Bed Width1.9M1.9M5.
C.B.L.204.757M203.757M6. Drop Height1M-M7.
Velocity0.436M/Sec0.436M/Sec8. Surface Fall1in35001in35009. Side
Slopes ( I / O )1.5:1/2:11.5:1/2:110. T.B.L.206.157M205.157M11.
G.L.205.035M205.035M12. Top Width of Bank ( L / R
)2.70/5.00M2.70/5.00MU/S TEL205.567U/S FSL205.5570.70.48205.090U/S
CBL204.7570.333D/S
FSL204.55710.3203.757203.5840.7500.451.000.30.93.53.0202.7570.45Velocity
Head (ha) =v2/2g=0.010MU/S TEL = U/S FSL+ha =205.567MDischarge at
Km2.45(Q) =1.08CumecsAssume Bt as0.68MThroat Width of Drop(B)
=1.9MCanal Bed Width=1.9MFSD at U/S=0.8MHead over crest (D) = (
QX(Bt)1/6/ (1.835 X B ))3/5 =0.48MElevation of Crest = ( U/S TEL-D
)Height of Crest above B.L.= Crest level-U/S CBL=0.333M=205.090( D1
) = CrestLevel - D/S CBL=1.33MTop Width or Thickness of Crest
=0.55(D1)1/2 =0.64Bt = 0.55(Crestlevel-CisternLevel)0.50(Bt)or
Say0.70M=0.68(as per CWC manual)Cistern:H =Crest Level -Cistern
Level =1.51M( HL )=U/S TEL- D/S FSL=1.010MDepth of cistern
=(HLD)0.5/4=0.17MLength of Cistern = 5 (HLD)0.5=3.4673642276Mor
Say3.5MBase Width of Drop=0.90Mor Say0.9MTotal Length of Impervious
Floor on D/S =2(D1+1.30)+Drop=6.2667980421or Say6.5MBy Khosla's
Exit Gradiant Method :Depth of D/S Cut off wall (d) = D/S FSD/2
+0.60=1MGE =(H/d)X 1/() = b/d=7.4 = (1+(1+2))/2 =4.2336309405GE
=0.233004886 0.30Hence OK(1-E)AB0.903.53.00E7.40x = E
=(1/)cos-1((-1)/) =0.223Thickness of Apron =((D+D1)/0.3048)A
=0.91Pressure Head @ A = HXA =1.36=2.436657445ft. =Floor
Thickness=0.731M or Say0.750.7B =0.54Pressure Head @ B = HXB
=0.81MFloor Thickness=0.434M or Say0.45How ever provide floor
thicknesses as0.750and0.45mincluding 15 cm thick wearing coatThis
Drop is designed as unflumed Drop to avoid the unfavourable scour
on the d/s .Approach :The side walls shall be splayed straight at
an angle of 450 from the u/s edgeof the crest extending by 1.00 m
into the earthen bank from the FSL line.U/S protection:2.00m long
bed pitching shall be provided on u/s bed,sloping down towards
thecrest wall at 1:10 slope and two drain pipesof 100mm at u/s bed
level in thecrest wall so as to drain out the u/s bed during the
closure of the canalU/S curtain wall = u/s FSD/3 +0.60
=0.8666666667or say1mHowever provide 0.30m X 1.00m curtainwall300
mm thick rough stone dry packing shall be provided for u/s bed
pitchingand u/s side revetment.D/S Expansion:No d/s expansion is
contemplated because the drop is not flumed. The side wallsshall be
straight and parallel up to the end of the floor and shall be kept
vertical.D/S Protection:Length of bed Protection = 3 X d/s FSD
=3.000mSide Protection is same as Bed protection.300 mm thick rough
stone dry packing shall be provided for d/s bed pitchingand d/s
side revetment.D/S Curtain wall = d/s FSD/2 +0.60 =1.00mScour
Depth(R) =1.345 ( q2/f)1/3q = Q/b =0.5684210526f
=1.5R=0.8062541585Maximum ScourDepth =2.00R=1.61Maximum Scour Level
= d/s FSL - Max.Scour Depth =202.94
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