44025203 Design of a Barrage MS

Design of Barrage Hydraulic Structures

Muhammad Azhar Saleem2003/II-MS-C-STRU-01

Design of Barrage

Input Design DataMaximum Discharge, Q max 500000 cusecsMinimum Discharge, Qmin 12000 cusecsRiver Bed Level, RBL 582 ftHigh Flood Level, HFL 600 ftLowest water level, LWL 587 ftNumbers of canals on left side 1Numbers of canals on right side 2Maximum Discharge of one Canal 3500 cusecsSlope of river 1 ft/mileLacey's Looseness Coefficient, LLC 1.8

1- Minimum Stable Wetted PerimeterWetted perimeter, Pw = 2.67√ Qmax 1888 ftWidth between abutment, Wa = LLC x Pw 3398 ftNumber of bays 50Bay width 60 ftNumber of fish ladder 1Width of one fish ladder 26 ftNumber of divide walls 2Width of on divide wall 15 ftwidth of one pier 7 ftTotal number of piers 47Total width of bays 3000 ftTotal width of piers 329 ftWidth between abutment, Wa 3385 ft



147.71 cusecs/ft166.67 cusecs/ft

2- Calculation of Lacey's Silt FactorS = (1/1844) x f**5/3 / Q**1/6Lacey's silt factor, f 2.04

3- Fixation of Crest LevelAfflux 3 ftHeight of crest above river bed, P 6 ftScour depth, R = 0.9(qabt**2 / f)**1/3 19.82 ftDepth of water above crest, Ho = R- P 13.82 ftApproach velocity, Vo = qabt / R 7.45 ft/sEnergy head, ho = Vo**2 / 2g 0.86 ftEo = Ho + ho 14.69 ftDo = HFL - RBL 18.00 ftE1 = Do + ho + Afflux 21.86 ftLevel of E1 = RBL + E1 603.86 ftCrest level = Level of E1 - Eo 589.18 ftMaximum d/s water level 600 fth = d/s WL - Crest Level 10.82 ftUsing Gibson Curveh / Eo 0.74C' / C 0.84C 3.8 fpsC' = (C'/C) x C 3.19

Q = C' x W clear x Eo**3/2 538948 cusecs O.K

Discharge between abutments, qabt

Discharge over weir, q weir



4- Design of Undersluices

Difference between undersluices & main weir 3 ftNumber of undersluices (N1) 2Number of bays for one undersluices (N2) 5Flow through undesluices as % of main weir 120 %Crest level of undersluices 586.18 ftb1 = N1 x Bay width 300 ft

200 cusecs/ftScour depth, R = 0.9(qus**2 / f)**1/3 24.26 ftDo, (may be Do = R) 24.26 ft

8.24 ft/secEnergy head, ho = Vo**2 / 2g 1.06 ftMaximum U/S E.L = HFL + Afflux + ho 604.06 ftEo = U/S E.L - Crest Level 17.88 fth = (U/S E.L - Afflux) - Crest level 14.88 fth / Eo 0.83Using Gibson CurveC' / C 0.76C' = (C'/C) x C 2.89

131000 cusecs431158 cusecs562158 cusecs O.K

%water through undersluices=(Q1+Q3)/Qmain weir*100 30.4 % 1

HenceCrest Level of Undersluices 586.18 ftNumber of Bays on Each Side 5

qus = % flow x q weir

Approach velocity, Vo = qus / R

Q1 & Q3, ( Q = C' x Wclear x Eo**3/2)Q main weir = C' x (Wclear(bays) - Wclear( us) )x Eo**3/2

Total Discharge = Q1 + Q3 + Q main weir



5- Determination of Water Levels and Energy Levels

5.1 Check for main weir

Q DSWL USWL D Vo ho h Ho Eo h/Eo C'/C C' Q(USWL-RBL) (Vo**2/2g) (DSWL-CL) (USWL-CL) (USWL+ho-CL) (Gibson)

(cusecs) (ft) (ft) (ft) (ft/s) (ft) (ft) (ft) (ft) (cusecs/ft) (cusecs)

For normal state 600000 601.5 604.0 22.0 9.1 1.28 12.32 14.82 16.11 0.765 0.820 3.12 201.4 604304.51 O.K

500000 600 602.5 20.5 8.1 1.03 10.82 13.32 14.35 0.754 0.815 3.10 168.4 505080.27 O.K

250000 597 598.5 16.5 5.1 0.396 7.82 9.32 9.72 0.805 0.780 2.96 89.8 269469.2 O.K

125000 592 594.5 12.5 3.3 0.17 2.82 5.32 5.50 0.514 0.940 3.57 46.0 138093.92 O.K

For retrogressed state600000 595.5 602.0 20.0 10.0 1.55 6.32 12.82 14.38 0.440 0.960 3.65 198.9 596585.232 NOT O.K

500000 596 601.0 19.0 8.8 1.19 6.82 11.82 13.02 0.524 0.930 3.53 166.0 498022.699 NOT O.K

250000 591 597.0 15.0 5.6 0.48 1.82 7.82 8.30 0.220 0.970 3.69 88.2 264578.81 O.K

For accreted state600000 604 606.0 24.0 8.3 1.08 14.82 16.82 17.90 0.828 0.765 2.91 220.2 660591.93 O.K

500000 602.5 604.0 22.0 7.6 0.89 13.32 14.82 15.72 0.848 0.740 2.81 175.2 525555.63 O.K

250000 601.5 602.0 20.0 4.2 0.27 12.32 12.82 13.09 0.941 0.520 1.98 93.6 280866.24 O.K

5.2 Check fo undersluices

Increase in flow 20 %Concentration of flow 157199 cusecs

Q DSWL USWL D Vo ho h Ho Eo h/Eo C'/C C' Q

qclear

qclear/D

qclear



(USWL-RBL) (Vo**2/2g) (DSWL-CL) (USWL-CL) (USWL+ho-CL) (Gibson)

(cusecs) (ft) (ft) (ft) (ft/s) (ft) (ft) (ft) (ft) (cusecs/ft) (cusecs)

For normal state157199 601.5 603.5 21.5 12.2 2.31 15.32 17.32 19.63 0.781 0.82 3.12 271.00 162602.65 O.K

For Retrogressed state157199 595.5 601.5 19.5 13.4 2.80 9.32 15.32 18.13 0.514 0.94 3.57 275.68 165409 O.K

For accreted state157199 604.0 605.5 23.5 11.1 1.93 17.82 19.32 21.25 0.839 0.77 2.93 286.71 172025 O.K

6- Fixation of d/s Floor Levels and Length of d/s Glacis and d/s Floor

6.1 Fixation of d/s floor levels for normal weir section using blench curves

Q USEL DSEL DSFL(USWL+ho) (DSWL +ho) (USEL-DSEL) (blench curve)

(cusecs) (cusecs/ft) (ft) (ft) (ft) (ft) (ft)

Normal state of river600000 201.4 605.28 602.78 2.50 19.1 583.68500000 168.4 603.53 601.03 2.50 17.3 583.73250000 89.8 598.90 597.40 1.50 11.2 586.20

For Retrogressed state of river600000 198.9 603.55 597.05 6.50 21 576.05500000 166.0 602.19 597.19 5.00 18.3 578.89250000 88.2 597.48 591.48 6.00 13 578.48

For accreted state of river600000 220.2 607.08 605.08 2.00 19.6 585.48500000 175.2 604.89 603.39 1.50 16.6 586.79250000 93.6 602.27 601.77 0.50 10.4 591.37

qclear/D

qclear hL E2

(DSEL - E2)



Hence d/s Floor level 576.00 ft

6.2 Fixation of d/s floor levels for undersluices using blench curves



Normal state of river162603 271.00 605.81 603.81 2.00 22.3 581.51

For Retrogressed state of river165409 275.68 604.30 598.30 6.00 25.1 573.20

For accreted state of river172025 286.71 607.43 605.93 1.50 22.6 583.33

Hence d/s Floor level for underslu 573.00 ft

7- Fixation of d/s floor level for normal barrage section usingCrump's method and determination of floor length

Q 500000 cusecsMaximum DSWL 602.5 ftUSWL 604.0 ftUSEL 604.89 ftDSFL 576.00 ftRBL 582 ftCrest level 589.18 ftDpool (Max. DSWL - DSFL) 26.5 ft

qclear hL E2

(DSEL - E2)



d/s Velocity (Q/(Dpool x Wa) 5.57 ft/sec

0.48 ftDSEL (DSWL + velocity head) 602.98 ftK (USEL - Crest Level ) 15.71 ftL (USEL - DSEL ) 1.91 ftq (Q / Total width of bays) 166.67 cusecs/ft

9.52 ftL/C 0.20(K+F)/C ,(from crumps curve) 1.92F, [(K+F)/C x C - K] 2.56Level of intersection of jump with glacis= Crest level - F 586.61 ft

16.37 ft10.61 ft

Slope of d/s glacis 1: 3Length of glacis d/s of jump, (slope x submergency) 31.84 ft

73.66 ft41.83 ft

Say 42 ft

d/s velocity head (V2/2g)

Critical Depth, C, (q2 / g)1/3

E2, ( DSEL - Level of intersection of jump)Submergency of jump, (Level of intersection of jump - DSFL )

Length of stilling pool, (4.5 x E2)Length of d/s floor, (Length of stilling pool -Length of glacis d/s of jump)

L K

y F x



b)Q 500000 cusecsMinimum DSWL 596 ftUSWL 601 ftUSEL 602.19 ftDSFL 576 ftRBL 582 ftCrest level 589.18 ftDpool (Min. DSWL - DSFL) 20 ftd/s Velocity (Q/(Dpool x Wa) 7.39 ft/sec

0.85 ftDSEL, (DSWL + velocity head) 596.85 ftK (USEL - Crest Level ) 13.01 ftL (USEL - DSEL ) 5.34 ftq (Q / Total width of bays) 166.67 cusecs/ft

9.52 ftL/C 0.56(K+F)/C ,(from crumps curve) 2.48F, [(K+F)/C x C - K] 10.59 ftLevel of intersection of jump with glacis= Crest level - F 578.58 ft

18.27 ft2.58 ft

Slope of d/s glacis 1: 3




Fig:1 Various Parameters for using Crump's Curve



Length of glacis d/s of jump, (slope x submergency) 7.74 ft

82.19 ft74.45 ft

Say 75.00 ft

8 - Fixation of d/s floor length for undersluicesa)

Q 157199 cusecsMaximum DSWL 604 ftUSWL 605.5 ftUSEL 607.43 ftDSFL 573 ftRBL 582 ftCrest level 586.18 ftDpool (Max. DSWL - DSFL) 31 ftd/s Velocity (Q/(Dpool x Wa) 8.5 ft/sec



22.39 ft




E2, ( DSEL - Level of intersection of jump)



9.72 ftSlope of d/s glacis 1: 3Length of glacis d/s of jump, (slope x submergency) 29.16 ft

101 ft71.60 ft

Say 72.00 ftb)

Q 157199 cusecsMinimum DSWL 595.5 ftUSWL 601.5 ftUSEL 604.3 ftDSFL 573 ftRBL 582 ftCrest level 586.18 ftDpool (Min. DSWL - DSFL) 22.5 ftd/s Velocity (Q/(Dpool x Wa) 11.64 ft/sec



23.16 ft

Submergency of jump, (Level of intersection of jump - DSFL )




E2, ( DSEL - Level of intersection of jump)



1.44 ftSlope of d/s glacis 1: 3Length of glacis d/s of jump, (slope x submergency) 4.32 ft

104 ft99.91 ft

Say 100 ft

Hence we shall provide d/s floor length = 100 ft

9- Check for Adequacy for d/s floor levels using conjugate depth method.

9.1 For normal weir section

Ф 1.00Floor level of stilling pool 576.00 ft

Discharge in river, Q (cusecs) 500000 250000

400000 200000max. min. max min

USEL (ft) 604.89 602.19 602.27 597.48DSWL (ft) 602.5 596.0 601.5 591E = USEL - DSFL 28.89 26.19 26.27 21.48

166.67 166.67 83.33 83.33

26.50 20.00 25.50 15.00

1.073 1.243 0.619 0.837

Conjugate depth coefficientsz 0.145 0.170 0.080 0.110z' 0.635 0.671 0.504 0.573

Submergency of jump, (Level of intersection of jump - DSFL )


Discharge through main weir, Q1= 0.8Q (cusecs)

Intensity of flow on d/s floor, q = Q1/width of main weir

Depth in stilling pool, Dpool = DSWL - DSFL

f(z) = q/E3/2



Conjugate depths4.19 4.45 2.10 2.36

18.36 17.57 13.24 12.31

8.14 2.43 12.26 2.69Remarks O.K O.K O.K O.K

9.2 For undersluices section


Discharge in river, Q (cusecs) 500000Discharge through U.S with 20% concentration, (1.2 x (Q1 + Q2)) 157199

Max. Min.USEL (ft) 607.43 604.30DSWL (ft) 604.00 595.5E = USEL - DSFL (ft) 34.43 31.30

262 262

31.00 22.50

1.30 1.50

Conjugate depth coefficientsz 0.180 0.217z' 0.684 0.695

Conjugate depths6.20 6.79

23.54 21.75

7.46 0.75Remarks O.K O.K

10 - Scour Protection

d1 = z x E

d2 = z' x E

Jump submergency = Dpool - d2

Intensity of flow on d/s floor, q = Q1/Total width of all U/S


f(z) = q/E3/2

d1 = z x E

d2 = z' x E

Jump submergency = Dpool - d2 (ft)



For main weirAssume flow concentration 20 %q = Qmax /(total width of bays) x 1.2 200 cusecs/ft

24.26 ft

10.1 - d/s scour protection for main weir

Safety factor for d/s floor critical condition 1.75Depth, R' = safety factor x R 42.46 ft

596 ftd/s appron (floor) level, (DSFL) 576.00 ftDepth of water on apron (Min DSWL - DSFL) 20.00 ftIncrease in depth due to concentration 0.50 ftDepth of water with concentration, D' 20.50 ftDepth of scour below apron = R' - D' 21.96 ftSlope of protection 1: 3

69.44 ftLength of d/s stone apron in horizontal position

= length of apron x (1.25t/1.75t) 50 ft

R = 0.9 (q2/f)1/3

Minimum DSWL for Qmax

Length of apron to cover surface of scour = Sqrt (12+32)x(R'-D')

D

R'

t

1:3

2.5(R-D)Bed Level DSFL

MIN. W.L

Deepest Possible Scour



10.2 U/s Scour Protection for main weir

Safety factor for u/s floor critical condition 1.25R' = Safety factor x R 30.33 ft

601.00 ftU/s apron level, (RBL) 582 ftDepth of water on apron = USWL -RBL 19.00 ftIncrease in depth due to concentration 0.50 ftTotal depth with concentration, D' 19.50 ftDepth of scour below apron = R' - D' 10.83 ftSlope of protection 1: 3

34.24 ftLength of u/s stone apron in horizontal position

25 ft

For undersluices

Assume flow concentration 20 %

262 cusecs/ft

29.05 ft

10.3 - d/s scour protection for undersluices

Minimum USWL for Qmax


= length of apron x (1.25t/1.75t)

q = (Q1+Q3)/Total width of undersluicesx 1.2

R = 0.9 (q2/f)1/3

t

3(R-D)


Fig:2 Scour Protection




595.5 ftd/s appron (floor) level, (DSFL) 573.00 ftDepth of water on apron (Min DSWL - DSFL) 22.50 ftIncrease in depth due to concentration 0.5 ftDepth of water with concentration, D' 23.00 ftDepth of scour below apron = R' - D' 27.83 ftSlope of protection 1: 3


63 ft

10.4 U/s Scour Protection for undersluices


601.5 ftU/s apron level, (RBL) 582 ftDepth of water on apron = USWL -RBL 19.5 ftIncrease in depth due to concentration 0.5 ftTotal depth with concentration, D' 20 ftDepth of scour below apron = R' - D' 16.31 ftSlope of protection 1: 3


37 ft

Minimum DSWL for Q1 + Q3



Minimum USWL for Q1 + Q3





10.5 - Thickness of Aprons

The following table gives the required valuves of "t" (Fig:2) for protection of various grades of sand and slope of rivers.

Fall in inches/mile 3 9 12 18 24Sand classification Thickness of stone pitching in inches

Very coarse 16 19 22 25 28Coarse 22 25 28 31 34Medium 28 31 34 37 40

Fine 34 37 40 43 46Very fine 40 43 45 49 52

Type of soil Medium sandSlope of river 12 in/mileThickness if stone pitching, t 34 in

5 ftSize of concrete blocks over filter 4 ft cubeSummary

Total Length of d/s stone apron 50 ft4 ft Thick bloke apron = 1/3 x total length 16 ft (block= 4'x4'x4')5 ft Thick stone apron 34 ft

Total length of u/s apron 25 ft4 ft Thick bloke apron = 1/3 x total length 8 ft (block= 4'x4'x4')5 ft Thick stone apron 17 ft

Thickness of stone apron in horizontal position = 1.75xt/slope



11 - Inverted Filter Design

Size of Concrete blocks 4 ft cubeThickness of shingle (3' - 6") 9 inThickness of coarse shingle (3/4" - 3") 9 inThickness of fine shingle (3/16" - 3/4") 6 inSpacing b/w conc. Blocks filled with fine shingle 2 in

12- Design of guide banks

Concrete Blocks(4'x4'x4')

9" Gravel

9" Coarse sand

6" Sand

Spacing /Jhries (2")

Fig: 3 Inverted Filter



i) Length of each guid bank measured in straight line along

5078 ft

ii) 677 ft

iii) For the nose of the u/s guide bank and the full length of d/sguide bank use Lacey's depth = 1.75 x R 42.46 ft

For remaining u/s guide bank lacey's depth = 1.25 x R 30.33 ftiv) Possible slope of scour 1: 3v) Free board u/s 7 ft above HFL

Free board d/s 6 ft above HFLThese free boards also include allowance for accretion.

vi) Top of guide bank 10 ftvii) Side slope of guide bank 1: 3viii) Minimum apron thickness 4 ft

Length of barrage, Wa 3385 ftLength of u/s guide bank 5078 ftLength of d/s guide bank 677 ftRadius of u/s curved part 600 ftRadius of d/s curved part 400 ft

Maximum u/s angle protected

Maximum d/s angle protected

12.1 Determination of levels of guide banksMerrimen's backwater formula

the barrage u/s , Lu/s = 1.5 x Wa

Length of each guid bank d/s of barrage, Ld/s = 0.2 x Wa

140o

57o - 80o

L=d1−d2

S+D( 1

S−C2

g )[Φ( d1

D )−Φ ( d2

D )]



L = length of back water curve

Chezy's Coefficient, C 71 ( max for earthen channels)

Bed slope of river, S 1/ 5000RBL 582 ftD/s HFL with accretion 602.5 ftD = D/s HFL with accretion - RBL 20.5 ftU/s HFL with accretion 604.0 ft

22.0 ft

21.78 ft

1.073

1.062

0.7870

0.8287L 5240 ftLength of guide bank 5078

Comments O.K

d1 = U/s HFL with accretion - RBL

Assume d2 (in between d1 and D)

d1/D

d2/D

Ф (d1/D) (from Bresse back water function table)

Ф (d2/D)

d1d2 D

L

Bresse Backwater Function

S

L=d1−d2

S+D( 1

S−C2

g )[Φ( d1

D )−Φ ( d2

D )]



Rise in RBL = Length of guide bank / slope 1.02 from barrage level

Water level along h/w axis at 5078 ft u/s of barrage= RBL + Rise in RBL + d2 604.80 ft

i) Level at the nose of u/s guide bank = W/L + free board 611.80 ftii) Level at the barrage = HFL + free board 607 ftiii) Water level d/s of barrage 602.5 ft

D/s free board 6 ftLevel of guide bank d/s = W/L + Free board 608.5 ft

13 - Design of Guide Bank Apron

Working on same lines as in section 10,

Length of unlanched horizontal apron = 2.5(R' - D') 27.07 ft

W.L

T=1.07t

2.5 (R' - D')D

RT

t


1 :3



Length of launched apron at 1:3 slope = 3.16(R' - D') 34.22 ft34 inches

say 3 ft

Volume of stones in apron = t x length launched apron 102.65Minimum thickness of unlaunched apron = 1.07t 3.2 ftMean thickness of unlaunched apron = volume/ 2.5(R' - D') 3.8 ft

5.0 ft

14 - Design of Marginal Bunds

i) Top width 20 ftii) Top level above estimated HFL after allowing 1.5ft accretion 5 ftiii) 1: 3iv) Back slope to be such as to provide minimum cover of 2 ft,

over hydraulic gradient of 1:6v) U/s water level at nose of guide bank 611.80 ft

Free board of marginal bund 5 ftHence level of marginal bund 616.80 ft

Calculation of length of backwater curve:

Merrimen's equation can be used to calculate backwater length

Thickness of stone apron, t (as calculated previously)

ft3/unit width

Maximum thickness of unlaunched apron = 2tmean - tmin

Front slope of marginal bunds (not pitched)

L=d1−d2

S+D( 1

S−C2

g )[Φ( d1

D )−Φ ( d2

D )]



604.0 ftRBL 582 ftNormal W.L without weir 600 ft

22.0 ftD = Normal W.L - RBL 18 ftSlope 1: 5000Table for length of backwater curve

D (5)x(10) Lx(1) =(11)+(4)

1 2 3 4 5 6 7 8 9 10 11 12

18 22.0 21.5 2500 4843.4 1.222 1.194 0.3162 0.3397 0.0235 2049 454918 21.5 21.0 2500 4843.4 1.194 1.167 0.3397 0.3636 0.0239 2084 458418 21.0 20.5 2500 4843.4 1.167 1.139 0.3636 0.3908 0.0272 2371 487118 20.5 20.0 2500 4843.4 1.139 1.111 0.3908 0.4198 0.029 2528 502818 20.0 19.5 2500 4843.4 1.111 1.083 0.4198 0.457 0.0372 3243 574318 19.5 19.0 2500 4843.4 1.083 1.056 0.457 0.4993 0.0423 3688 618818 19.0 18.5 2500 4843.4 1.056 1.028 0.4993 0.5511 0.0518 4516 701618 18.5 18.1 2000 4843.4 1.028 1.006 0.5511 0.6207 0.0696 6068 8068

Total 46047 ft8.73 miles

Maximum USWL at Qmax

d1 = Maximum USWL - RBL

d1 d2

d1d2 D

L


d1−d2

ST 1=

d1

DT 2=

d2

D Φ1( d1

D ) Φ2( d2

D )1S−C2

gΦ2−Φ1



Hence length of backwater cure = 8.73 miles

Part IIDesign of barrage profile for sub surface flow condition

15 - Fixing of Depth of Sheet Piles

Scour depth, R 19.82 ftDepth of u/s sheet pile from HFL = 1.5 R 30 ft

604.0 ftRL of bottomo fu/s sheet pile = Max. USWL - 1.5R 574.0 ftDepth of d/s sheet pile below HFL = 2R 40 ftRL of bottom of intermediate sheet pile = Max. USWL - 2R 564.0 ftLet RL of bottom of d/s sheet pile 550 ft

604.0 ft

589.18 ft

582.0 1: 4

A K 1: 3

576.00B L N P R

574.0 75.00

M

564.0 Q

1.5H = 48 28.70 6 39.53 550.0

Max. USWL for Qmax



16 - Calculation of Exit Gradient

Let the water be headed up to Max. accreted level u/s 604.0 ftand no flow d/s.Retogression 4 ftDSFL 576.0 ft

32.0 ft26.0 ft

Total length of concrete floor = b 197.23 ft

α = b/d 7.59form α ~ curve 0.152

Differential head causing seepeage, H = Max. u/s WL - (DSFL - Retrogression)

Depth of d/s sheet pile, d = DSFL - RL of bottom of d/s sheet pile

b

db1

H

E

D

C

α = b/d

Parmeters of Khosla's Curve

GE=Hdx

1π √λ

1π √ λ

1π √ λ



0.187 SAFE

17 - Calculation of Uplift Pressure After Applying Correction

17.1 U/s pile line:

48 ftTotal length of concrete floor, b 197.23 ftDepth of u/s sheet pile,d 8.0 ft

2.5 ft0.040624.65

0.243

0.757= From khosla's curve 100 - 33 67 %= From khosla's curve 100 - 31 69 %= From khosla's curve 64 %

i) Correction for floor thickness

Correction in 0.938 %

Correction in -ve 0.625 %

ii) Correction for interface of sheet pile

Correction in due to second pile =

GE

Length of concrete floor upto u/s sheet pile, b1

Assume tf u/s floor thickness1/α = d/bα = b/d

b1/b

1 - b1/b

ΦA

ΦB

ΦK

ΦEΦc

ΦD

ΦK=t fd

(ΦB−ΦK )

ΦA=t fd

(ΦA−ΦB )

ΦK 19[ d+Db ]√ D

b'



Depth of u/s sheet pile,d 8.0 ftD = RBL - RL of bottom of second pile 18.0 ftTotal length of concrete floor, b 197.23 ftDistance between two piles, b' 74.23 ftCorrection in +ve 1.23 %

iii) Slope correction for

Correction for

For 1:4 slope, Fs (from slope correction curve) 3.3

74.23 ft28.70 ft

Correction for -1.28Hence Correted 68.375 %Corrected 67 %Corrected 64.895 %

17.2 Intermediate sheet pile at toe of d/s glacis:

Assume floor thickness 10 ftDSFL 576.00 ftRL of intermediate sheet pile 564.00 ftd = DSFL - RL of Intermediate sheet pile 12.00 ftTotal length of concrete floor, b 197.23 ft

122.23 ft

0.620

Distance between two piles, b1

Horizontal projection of u/s glacis, bs = (crest level - RBL) x 1/slope

Length of concrete floor up to sheet pile, b1

b1/b

ΦK

ΦK

ΦK=−FS

bsb1

ΦK

ΦA

ΦB

ΦK



0.38016.44

From khosla's curve 100 - 55 45 %From khosla's curve 100 - 58 42 %From khosla's curve 36.5 %

i) Correction due to floor thickness

Correction in -2.5 %


ii) Correction due to interference of pile

Correction in due to u/s sheet pile

Depth of Intermediate sheet pile,d 12.0 ftD = RBL - RL of bottom of u/s sheet pile 18.00 ftTotal length of concrete floor, b 197.23 ftDistance between two piles, b' 74.23 ft

Correction in = -1.42 %

Correction in due to d/s sheet pile =Depth of Intermediate sheet pile,d 12.00 ftD = DSFL - RL of bottom of d/s sheet pile 26.00 ftTotal length of concrete floor, b 197.23 ftDistance between two piles, b' 75.00 ft

1 - b1/bα = b/dΦL=ΦEΦM=ΦDΦN=Φ

C

ΦL

ΦL=t fd

(Φ L−ΦM )

ΦN=t fd

(ΦM−ΦN )

19[ d+Db ]√ D

b'

ΦL 19[ d+Db ]√ D

b'

ΦN 19[ d+Db ]√ D

b'

ΦN 19[ d+Db ]√ D

b'



Correction in = 2.16 %


4.5

39.53 ft

74.23 ft


Hence Corrected 43.47 %Corrected 42 %Corrected 43.24 %

17.3 D/s sheet pile at the end of impervious floor

Assume floor thickness 7 ftDepth of d/s sheet pile, d 26.00 ftTotal length of concrete floor, b 197.23 ft

0.132From khosla's curve 100 - 68 32 %From khosla's curve 100 - 78 22 %From khosla's curve 0 %


Correction in -2.69 %

for '1:3 slope, Fs

bs = (crest level - DSFL) x 1/slope


1/α = d/b

ΦN 19[ d+Db ]√ D

b'

ΦL

ΦL=F S×bs

b1

ΦLΦM

ΦN

ΦP=ΦE

ΦQ=ΦDΦR=ΦC

ΦP




ii) Correction due to interface of piles

Correction in Depth of d/s sheet pile,d 26.00 ftD = DSFL - RL of bottom of intermediate sheet pile 12.00 ftTotal length of concrete floor, b 197.23 ftDistance between two piles, b' 75.00 ft


HenceCorrected 27.84 %Corrected 22 %Corrected 5.92 %

Table: Uplift pressure at E, D, C and along the sheet piles

Symbol used in u/s Pile Intermediate d/s PileKhosla cueve line Line Line

= 68.38% = 43.47% = 27.84%= 67% = 42% = 22%= 64.89% = 43.24% = 5.92%

18 - Calculation For Floor Thickness:

ΦPΦR

ΦP

19[ d+Db ]√ D

b'ΦP

ΦP

ΦR

ΦQ

ΦEΦD

ΦC

ΦA

ΦB

ΦK

ΦL

ΦM

ΦN

ΦP

ΦQ

ΦR

t f=Φ

100 (G−1 )×H



where

H = Maximum differential head causing seepageG = Specific gravity of concrete 2.4

a) Thickness of floor at A

Assumend thickness 2.5 ftH 32 ftThickness from uplift pressure 15.63 ft

say 16.00 ftb) Thickness of floor at L

Assumend thickness 10 ftThickness from uplift pressure 9.94 ft

say 10.00 ftc) Thickness of floor at N


say 10.00 ftd) Thickness of floor at P


say 7.00 fte) Thickness of floor at crest

tf = Thickness of floor in ft = % Uplift pressure

t f=Φ

100 (G−1 )×H

Φ



Uplift pressure at crest 54.88 %HenceThickness of floor at crest d/s of gate 12.544 ft

say 13 ft

K

L

Pressure at crest

x

y

=ΦL+ΦK−ΦL

y( x )

ΦK

ΦL

Design of Barrage

Input Design DataMaximum Discharge, Q max 540000 cusecsMinimum Discharge, Qmin 12000 cusecsRiver Bed Level, RBL 582 ftHigh Flood Level, HFL 600 ftLowest water level, LWL 587 ftNumbers of canals on left side 1Numbers of canals on right side 2Maximum Discharge of one Canal 3500 cusecsSlope of river 1 ft/mileLacey's Looseness Coefficient, LLC 1.8

1- Minimum Stable Wetted PerimeterWetted perimeter, Pw = 2.67√ Qmax 1962 ftWidth between abutment, Wa = LLC x Pw 3532 ftNumber of bays 55Bay width 60 ftNumber of fish ladder 1Width of one fish ladder 25 ftNumber of divide walls 2Width of one divide wall 15 ftwidth of one pier 7 ftTotal number of piers 52Total width of bays 3300 ftTotal width of piers 364 ftWidth between abutment, Wa 3719 ft

145.20 cusecs/ft163.64 cusecs/ft

2- Calculation of Lacey's Silt FactorS = (1/1844) x f**5/3 / Q**1/6Lacey's silt factor, f 2.06

3- Fixation of Crest LevelAfflux 3 ftHeight of crest above river bed, P 6 ftScour depth, R = 0.9(qabt**2 / f)**1/3 19.55 ftDepth of water above crest, Ho = R- P 13.55 ftApproach velocity, Vo = qabt / R 7.43 ft/sEnergy head, ho = Vo**2 / 2g 0.86 ftEo = Ho + ho 14.41 ft

Discharge between abutments, qabt

Discharge over weir, q weir

Do = HFL - RBL 18.00 ftE1 = Do + ho + Afflux 21.86 ftLevel of E1 = RBL + E1 603.86 ftCrest level = Level of E1 - Eo 589.45 ftMaximum d/s water level 600 fth = d/s WL - Crest Level 10.55 ftUsing Gibson Curveh / Eo 0.73C' / C 0.86C 3.8 fpsC' = (C'/C) x C 3.27

Q = C' x W clear x Eo**3/2 589630 cusecsO.K

4- Design of Undersluices

Difference between undersluices & main weir 3 ftNumber of undersluices (N1) 2Number of bays for one undersluices (N2) 5Flow through undesluices as % of main weir 120 %Crest level of undersluices 586.45 ftb1 = N1 x Bay width 300 ft

196.364 cusecs/ftScour depth, R = 0.9(qus**2 / f)**1/3 23.91 ftDo, (may be Do = R) 23.91 ft

8.21 ft/secEnergy head, ho = Vo**2 / 2g 1.05 ftMaximum U/S E.L = HFL + Afflux + ho 604.05 ftEo = U/S E.L - Crest Level 17.60 fth = (U/S E.L - Afflux) - Crest level 14.60 fth / Eo 0.83Using Gibson CurveC' / C 0.77C' = (C'/C) x C 2.93

129586 cusecs482425 cusecs

O.K 612010 cusecs%water through undersluices=(Q1+Q3)/Qmain weir*100 1 26.9 %HenceCrest Level of Undersluices 586.45 ftNumber of Bays on Each Side 5

5- Determination of Water Levels and Energy Levels

qus = % flow x q weir

Approach velocity, Vo = qus / R

Q1 & Q3, ( Q = C' x Wclear x Eo**3/2)Q main weir = C' x (Wclear(bays) - Wclear( us) )x Eo**3/2

Total Discharge = Q1 + Q3 + Q main weir

5.1 Check for main weir

Q DSWL USWL D Vo ho h Ho(USWL-RBL) (Vo**2/2g) (DSWL-CL) (USWL-CL)

(cusecs) (ft) (ft) (ft) (ft/s) (ft) (ft) (ft)

For normal state 648000 602 604.5 22.5 8.7 1.18 12.55 15.05540000 600.5 603.0 21.0 7.8 0.94 11.05 13.55270000 597.5 598.5 16.5 5.0 0.382 8.05 9.05135000 593 594.5 12.5 3.3 0.17 3.55 5.05

For retrogressed state648000 596 602.5 20.5 9.6 1.42 6.55 13.05540000 596.5 601.5 19.5 8.4 1.09 7.05 12.05270000 592 597.0 15.0 5.5 0.46 2.55 7.55

For accreted state648000 604.5 605.5 23.5 8.4 1.08 15.05 16.05540000 603 604.0 22.0 7.4 0.86 13.55 14.55270000 602 602.5 20.5 4.0 0.25 12.55 13.05

5.2 Check for undersluices

Increase in flow 20 %Concentration of flow 2E+05 cusecs

Q DSWL USWL D Vo ho h Ho(USWL-RBL) (Vo**2/2g) (DSWL-CL) (USWL-CL)

(cusecs) (ft) (ft) (ft) (ft/s) (ft) (ft) (ft)

For normal state155503 594 601.5 19.5 13.3 2.74 7.55 15.05

For Retrogressed state155503 590 601 19 13.6 2.89 3.55 14.55

For accreted state155503 598.0 602.5 20.5 12.6 2.48 11.55 16.05




qclear/D

qclear/D

qclear hL E2

(DSEL - E2)


Normal state of river648000 201.3 605.68 603.18 2.50 18.6 584.58540000 170.8 603.94 601.44 2.50 17.6 583.84270000 83.6 598.88 597.88 1.00 10 587.88

For Retrogressed state of river648000 198.8 603.92 597.42 6.50 20.8 576.62540000 168.4 602.59 597.59 5.00 17.3 580.29270000 84.4 597.46 592.46 5.00 12 580.46

For accreted state of river648000 196.7 606.58 605.58 1.00 17 588.58540000 167.8 604.86 603.86 1.00 15.4 588.46270000 86.6 602.75 602.25 0.50 10.5 591.75

Hence d/s Floor level ### ft

6.2 Fixation of d/s floor levels for undersluices using blench curves



Normal state of river157414 262.36 604.24 596.74 7.50 27.5 569.24

For Retrogressed state of river156064 260.11 603.89 592.89 11.00 26.5 566.39

For accreted state of river165503 275.84 604.98 600.48 4.50 28.4 572.08

Hence d/s Floor level for undersluices 566.00 ft

7- Fixation of d/s floor level for normal barrage section usingCrump's method and determination of floor length

Q 540000 cusecsMaximum DSWL 603.00 ftUSWL 604.00 ftUSEL 604.86 ftDSFL 577.00 ftRBL 582 ftCrest level 589.45 ftDpool (Max. DSWL - DSFL) 26 ftd/s Velocity (Q/(Dpool x Wa) 5.58 ft/sec

0.48 ft

qclear hL E2

(DSEL - E2)


DSEL (DSWL + velocity head) 603.48 ftK (USEL - Crest Level ) 15.41 ftL (USEL - DSEL ) 1.38 ftq (Q / Total width of bays) 163.64 cusecs/ft

9.40 ftL/C 0.15(K+F)/C ,(from crumps curve) 1.92F, [(K+F)/C x C - K] 2.65Level of intersection of jump with glacis= Crest level - F 586.80 ft

16.68 ft9.80 ft


75.06 ft45.64 ft

Say 46 ft

b)Q 540000 cusecsMinimum DSWL 596.5 ftUSWL 601.5 ftUSEL 602.59 ftDSFL 577.00 ftRBL 582 ftCrest level 589.45 ftDpool (Min. DSWL - DSFL) 19.5 ftd/s Velocity (Q/(Dpool x Wa) 7.45 ft/sec

0.86 ftDSEL, (DSWL + velocity head) 597.36 ftK (USEL - Crest Level ) 13.14 ft





L K

y F x

Fig:1 Various Parameters for using Crump's Curve

L (USEL - DSEL ) 5.23 ftq (Q / Total width of bays) 163.64 cusecs/ft


18.09 ft2.27 ft


81.42 ft74.61 ft

Say 75.00 ft

8 - Fixation of d/s floor length for undersluicesa)

Q 155503 cusecsMaximum DSWL 598 ftUSWL 602.5 ftUSEL 604.98 ftDSFL 566.00 ftRBL 582 ftCrest level 586.45 ftDpool (Max. DSWL - DSFL) 32 ftd/s Velocity (Q/(Dpool x Wa) 8.1 ft/sec



22.15 ft10.87 ft








100 ft67.06 ft

Say 68.00 ftb)

Q 155503 cusecsMinimum DSWL 590 ftUSWL 601 ftUSEL 603.89 ftDSFL 566.00 ftRBL 582 ftCrest level 586.45 ftDpool (Min. DSWL - DSFL) 24 ftd/s Velocity (Q/(Dpool x Wa) 10.80 ft/sec



24.97 ft0.84 ft


112 ft109.88 ft

Say 110 ft

Hence we shall provide d/s floor length = 110 ft


9.1 For normal weir section


Discharge in river, Q (cusecs) 540000






432000max. min.

USEL (ft) 604.86 602.59DSWL (ft) 603 596.5E = USEL - DSFL 27.86 25.59

160.00 160.00

26.00 19.50

1.088 1.236

Conjugate depth coefficientsz 0.150 0.170z' 0.643 0.671

Conjugate depths4.18 4.35

17.92 17.17

8.08 2.33Remarks O.K O.K

9.2 For undersluices section

Ф 1.00Floor level of stilling pool 566.00

Discharge in river, Q (cusecs) 540000Discharge through U.S with 20% concentration, (1.2 x (Q1 + Q2)) 155503

Max.USEL (ft) 604.98DSWL (ft) 598.00E = USEL - DSFL (ft) 38.98

259

32.00

1.06

Conjugate depth coefficientsz 0.145z' 0.635

Conjugate depths5.65

24.75

7.25Remarks O.K

10 - Scour ProtectionFor main weirAssume flow concentration 20 %q = Qmax /(total width of bays) x 1.2 196.364 cusecs/ft

23.91 ft

Discharge through main weir, Q1= 0.8Q (cusecs)

Intensity of flow on d/s floor, q = Q1/width of main weir


f(z) = q/E3/2

d1 = z x E

d2 = z' x E

Jump submergency = Dpool - d2

Intensity of flow on d/s floor, q = Q1/Total width of all U/S


f(z) = q/E3/2

d1 = z x E

d2 = z' x E

Jump submergency = Dpool - d2 (ft)

R = 0.9 (q2/f)1/3

10.1 - d/s scour protection for main weir


596.5 ftd/s appron (floor) level, (DSFL) 577.00 ftDepth of water on apron (Min DSWL - DSFL) 19.50 ftIncrease in depth due to concentration 0.50 ftDepth of water with concentration, D' 20.00 ftDepth of scour below apron = R' - D' 21.84 ftSlope of protection 1: 3



10.2 U/s Scour Protection for main weir


601.50 ftU/s apron level, (RBL) 582 ftDepth of water on apron = USWL -RBL 19.50 ftIncrease in depth due to concentration 0.50 ftTotal depth with concentration, D' 20.00 ftDepth of scour below apron = R' - D' 9.88 ftSlope of protection 1: 3


Minimum DSWL for Qmax


Minimum USWL for Qmax


D

R'

t

1:3

3(R-D)

2.5(R-D)Bed Level DSFL

MIN. W.L


Fig:2 Scour Protection


For undersluices

Assume flow concentration 20 %

259 cusecs/ft

28.76 ft

10.3 - d/s scour protection for undersluices


590 ftd/s appron (floor) level, (DSFL) 566.00 ftDepth of water on apron (Min DSWL - DSFL) 24.00 ftIncrease in depth due to concentration 0.5 ftDepth of water with concentration, D' 24.50 ftDepth of scour below apron = R' - D' 25.84 ftSlope of protection 1: 3



10.4 U/s Scour Protection for undersluices


601 ftU/s apron level, (RBL) 582 ftDepth of water on apron = USWL -RBL 19 ftIncrease in depth due to concentration 0.5 ftTotal depth with concentration, D' 19.5 ftDepth of scour below apron = R' - D' 16.46 ftSlope of protection 1: 3



10.5 - Thickness of Aprons

The following table gives the required valuves of "t" (Fig:2) for protection of various grades of sand and slope of rivers.

q = (Q1+Q3)/Total width of undersluicesx 1.2

R = 0.9 (q2/f)1/3

Minimum DSWL for Q1 + Q3


Minimum USWL for Q1 + Q3


Fall in inches/mile 3 9 12 18 24Sand classification Thickness of stone pitching in inches

Very coarse 16 19 22 25 28Coarse 22 25 28 31 34Medium 28 31 34 37 40

Fine 34 37 40 43 46Very fine 40 43 45 49 52

Type of soil Medium sandSlope of river 12 in/mileThickness if stone pitching, t 34 inThickness of stone apron in horizontal position = 1.75xt/slo 5 ftSize of concrete blocks over filter 4 ft cubeSummary

Total Length of d/s stone apron 50 ft4 ft Thick bloke apron = 1/3 x total length 16 ft (block= 4'x4'x4')5 ft Thick stone apron 34 ft

Total length of u/s apron 23 ft4 ft Thick bloke apron = 1/3 x total length 7 ft (block= 4'x4'x4')5 ft Thick stone apron 16 ft

11 - Inverted Filter Design

Size of Concrete blocks 4 ft cubeThickness of shingle (3' - 6") 9 inThickness of coarse shingle (3/4" - 3") 9 inThickness of fine shingle (3/16" - 3/4") 6 inSpacing b/w conc. Blocks filled with fine shingle 2 in


9" Gravel

9" Coarse sand

6" Sand

Spacing /Jhries (2")

12- Design of guide banks

i) Length of each guid bank measured in straight line along

5579 ft

ii) 743.8 ft

iii) For the nose of the u/s guide bank and the full length of d/sguide bank use Lacey's depth = 1.75 x R 41.84 ft

For remaining u/s guide bank lacey's depth = 1.25 x R 29.88 ftiv) Possible slope of scour 1: 3v) Free board u/s 7 ft above HFL

Free board d/s 6 ft above HFLThese free boards also include allowance for accretion.

vi) Top of guide bank 10 ftvii) Side slope of guide bank 1: 3viii) Minimum apron thickness 4 ft

Length of barrage, Wa 3719 ftLength of u/s guide bank 5579 ftLength of d/s guide bank 743.8 ftRadius of u/s curved part 600 ftRadius of d/s curved part 400 ft

Maximum u/s angle protected

Maximum d/s angle protected

12.1 Determination of levels of guide banksMerrimen's backwater formula

L = length of back water curve

the barrage u/s , Lu/s = 1.5 x Wa

Length of each guid bank d/s of barrage, Ld/s = 0.2 x Wa

140o

57o - 80o


9" Gravel

9" Coarse sand

6" Sand

Fig: 3 Inverted Filter

d1d2 D

L=d1−d2

S+D( 1

S−C2

g )[Φ( d1

D )−Φ ( d2

D )]

Chezy's Coefficient, C 71 ( max for earthen channels)

Bed slope of river, S 1/ 5000RBL 582 ftD/s HFL with accretion 603 ftD = D/s HFL with accretion - RBL 21 ftU/s HFL with accretion 604.0 ft

22.0 ft

21.8 ft

1.048

1.038

0.8999

0.958L 6909 ftLength of guide bank 5579

Comments O.K

Rise in RBL = Length of guide bank / slope 1.12 from barrage level

Water level along h/w axis at 5579 ft u/s of barrage= RBL + Rise in RBL + d2 604.92 ft

i) Level at the nose of u/s guide bank = W/L + free board 611.92 ftii) Level at the barrage = HFL + free board 607.5 ftiii) Water level d/s of barrage 603.0 ft

D/s free board 6 ftLevel of guide bank d/s = W/L + Free board 609.0 ft

13 - Design of Guide Bank Apron

d1 = U/s HFL with accretion - RBL

Assume d2 (in between d1 and D)

d1/D

d2/D

Ф (d1/D) (from Bresse back water function table)

Ф (d2/D)

W.L

T=1.07t

2.5 (R' - D')D

RT

1 :3

d1d2 D

L


S

Working on same lines as in section 10,

Length of unlanched horizontal apron = 2.5(R' - D') 24.71 ftLength of launched apron at 1:3 slope = 3.16(R' - D') 31.23 ft

34 inchessay 3 ft

Volume of stones in apron = t x length launched apron 93.69Minimum thickness of unlaunched apron = 1.07t 3.2 ftMean thickness of unlaunched apron = volume/ 2.5(R' - D') 3.8 ft

5.0 ft

14 - Design of Marginal Bunds

i) Top width 20 ftii) Top level above estimated HFL after allowing 1.5ft accretio 5 ftiii) 1: 3iv) Back slope to be such as to provide minimum cover of 2 ft,

over hydraulic gradient of 1:6v) U/s water level at nose of guide bank 611.92 ft

Free board of marginal bund 5 ftHence level of marginal bund 616.92 ft

Calculation of length of backwater curve:

Merrimen's equation can be used to calculate backwater length

Thickness of stone apron, t (as calculated previously)

ft3/unit width

Maximum thickness of unlaunched apron = 2tmean - tmin

Front slope of marginal bunds (not pitched)

t


d1d2 D

L


L=d1−d2

S+D( 1

S−C2

g )[Φ( d1

D )−Φ ( d2

D )]

604.0 ftRBL 582 ftNormal W.L without weir 600 ft

22.0 ftD = Normal W.L - RBL 18 ftSlope 1: 5000Table for length of backwater curve

D

1 2 3 4 5 6 7 8

18 22.0 21.5 2500 4843.4 1.222 1.194 0.455318 21.5 21.0 2500 4843.4 1.194 1.167 0.479918 21.0 20.5 2500 4843.4 1.167 1.139 0.534518 20.5 20.0 2500 4843.4 1.139 1.111 0.625318 20.0 19.5 2500 4843.4 1.111 1.083 0.645318 19.5 19.0 2500 4843.4 1.083 1.056 0.795618 19.0 18.5 2500 4843.4 1.056 1.028 0.856418 18.5 18.1 2000 4843.4 1.028 1.006 1.155

Hence length of backwater cure = 22.4 miles


15 - Fixing of Depth of Sheet Piles

Scour depth, R 19.55 ftDepth of u/s sheet pile from HFL = 1.5 R 30 ft

604.0 ftRL of bottomo fu/s sheet pile = Max. USWL - 1.5R 574.0 ftDepth of d/s sheet pile below HFL = 2R 40 ftRL of bottom of intermediate sheet pile = Max. USWL - 2R 564.0 ftLet RL of bottom of d/s sheet pile 548 ft

604.0 ft

589.45 ft

582.0 1: 4

Maximum USWL at Qmax

d1 = Maximum USWL - RBL

d1 d2

Max. USWL for Qmax


d1−d2

ST 1=

d1

DT 2=

d2

D Φ1( d1

D )1S−C2

g

A K 1: 3

577.00B L N P R

574.0 75.00

M

564.0 Q

1.5H = 46.5 29.81 6 37.35 548.0

16 - Calculation of Exit Gradient

Let the water be headed up to Max. accreted level u/s 604.0 ftand no flow d/s.Retogression 4 ftDSFL 577.0 ft

31.0 ft29.0 ft

Total length of concrete floor = b 194.66 ft

α = b/d 6.71form α ~ curve 0.153

0.164 SAFE


17.1 U/s pile line:

46.5 ftTotal length of concrete floor, b 194.66 ft

Differential head causing seepeage, H = Max. u/s WL - (DSFL - Retrogression)

Depth of d/s sheet pile, d = DSFL - RL of bottom of d/s sheet pile

GE

Length of concrete floor upto u/s sheet pile, b1

b

db1

H

E

D

C

α = b/d

Parmeters of Khosla's Curve

GE=Hd

x1

π √λ

1π √ λ

1π √ λ

Depth of u/s sheet pile,d 8.0 ft

2.5 ft0.041124.33

0.239

0.761= From khosla's curve 100 - 34 66 %= From khosla's curve 100 - 33 67 %= From khosla's curve 61 %

i) Correction for floor thickness


Correction in -ve 0.3125 %

ii) Correction for interface of sheet pile

Correction in due to second pile =

Depth of u/s sheet pile,d 8.0 ftD = RBL - RL of bottom of second pile 18.0 ftTotal length of concrete floor, b 194.66 ftDistance between two piles, b' 73.16 ftCorrection in +ve 1.26 %


Correction for

For 1:4 slope, Fs (from slope correction curve) 3.3

73.16 ft29.81 ft

Correction for -1.34Hence Correted 66.6875 %Corrected 66 %Corrected 62.477 %

17.2 Intermediate sheet pile at toe of d/s glacis:

Assume floor thickness 10 ftDSFL 577.00 ft

Assume tf u/s floor thickness1/α = d/bα = b/d

b1/b

1 - b1/b


Horizontal projection of u/s glacis, bs = (crest level - RBL) x 1/slope

ΦA

ΦB

ΦK

ΦEΦc

ΦD

ΦK=t fd

(ΦB−ΦK )

ΦA=t fd

(ΦA−ΦB )

ΦK 19[ d+Db ]√ D

b'

ΦK

ΦK

ΦK=−FS

bsb1

ΦK

ΦA

ΦB

ΦK

RL of intermediate sheet pile 564.00 ftd = DSFL - RL of Intermediate sheet pile 13.00 ftTotal length of concrete floor, b 194.66 ft

119.66 ft

0.615

0.38514.97

From khosla's curve 100 - 48 52 %From khosla's curve 100 - 46 54 %From khosla's curve 36.5 %




ii) Correction due to interference of pile

Correction in due to u/s sheet pile

Depth of Intermediate sheet pile,d 13.0 ftD = RBL - RL of bottom of u/s sheet pile 18.00 ftTotal length of concrete floor, b 194.66 ftDistance between two piles, b' 73.16 ft


Correction in due to d/s sheet pile =Depth of Intermediate sheet pile,d 13.00 ftD = DSFL - RL of bottom of d/s sheet pile 29.00 ftTotal length of concrete floor, b 194.66 ftDistance between two piles, b' 75.00 ft

Correction in = 2.55 %


4.5

37.35 ft

73.16 ft


Length of concrete floor up to sheet pile, b1

b1/b

1 - b1/bα = b/d

for '1:3 slope, Fs

bs = (crest level - DSFL) x 1/slope


ΦL=ΦEΦM=ΦDΦN=Φ

C

ΦL

ΦL=t fd

(Φ L−ΦM )

ΦN=t fd

(ΦM−ΦN )

19[ d+Db ]√ D

b'

ΦL 19[ d+Db ]√ D

b'

ΦN 19[ d+Db ]√ D

b'

ΦN 19[ d+Db ]√ D

b'

ΦL

ΦL=F S×bs

b1

Hence Corrected 54.34 %Corrected 54 %Corrected 52.51 %

17.3 D/s sheet pile at the end of impervious floor

Assume floor thickness 7 ftDepth of d/s sheet pile, d 29.00 ftTotal length of concrete floor, b 194.66 ft

0.149From khosla's curve 100 - 68 32 %From khosla's curve 100 - 78 22 %From khosla's curve 0 %


Correction in -2.41 %Correction in 5.31 %

ii) Correction due to interface of piles

Correction in Depth of d/s sheet pile,d 29.00 ftD = DSFL - RL of bottom of intermediate sheet pile 13.00 ftTotal length of concrete floor, b 194.66 ftDistance between two piles, b' 75.00 ft


HenceCorrected 27.88 %Corrected 22 %Corrected 5.31 %

Table: Uplift pressure at E, D, C and along the sheet piles

Symbol used in u/s Pile Intermediate d/s PileKhosla cueve line Line Line

= 66.69% = 54.34% = 27.88%= 66% = 54% = 22%

1/α = d/b

ΦL=F S×bs

b1

ΦLΦM

ΦN

ΦP=ΦE

ΦQ=ΦDΦR=ΦC

ΦPΦR

ΦP

19[ d+Db ]√ D

b'ΦP

ΦP

ΦR

ΦQ

ΦEΦD

ΦC

ΦA

ΦB

ΦK

ΦL

ΦM

ΦN

ΦP

ΦQ

ΦR

= 62.48% = 52.51% = 5.31%

18 - Calculation For Floor Thickness:

where

H = Maximum differential head causing seepageG = Specific gravity of concrete 2.4

a) Thickness of floor at A

Assumend thickness 2.5 ftH 27.00 ftThickness from uplift pressure 12.86 ft

say 13.00 ftb) Thickness of floor at L


say 11.00 ftc) Thickness of floor at N


say 11.00 ftd) Thickness of floor at P


say 6.00 fte) Thickness of floor at crest

tf = Thickness of floor in ft = % Uplift pressure

K

L

Pressure at crest

x

ΦD

ΦC

ΦA

ΦB

ΦK

ΦL

ΦM

ΦN

ΦP

ΦQ

ΦR

t f=Φ

100 (G−1 )×H

Φ

ΦK

ΦL

Uplift pressure at crest 58.4922 %HenceThickness of floor at crest d/s of gate 11.2806 ft

say 12 ft

Pressure at crest

y

=Φ L+ΦK−ΦL

y( x )

cusecs/ftcusecs/ft

cusecs/ft

Eo h/Eo C'/C C' Q(USWL+ho-CL) (Gibson)

(ft) (cusecs/ft) (cusecs)

16.23 0.77 0.810 3.08 201.3 664221.95 O.K

14.49 0.76 0.815 3.10 170.8 563797.82 O.K

9.43 0.85 0.760 2.89 83.6 275999.79 O.K

5.21 0.68 0.920 3.50 41.6 137391.44 O.K

14.47 0.45 0.950 3.61 198.8 655955.61 O.K

13.14 0.54 0.930 3.53 168.4 555619.13 O.K

8.01 0.32 0.980 3.72 84.4 278626.51 O.K

17.13 0.88 0.730 2.77 196.7 649177 O.K

15.41 0.88 0.730 2.77 167.8 553639.38 O.K

13.30 0.94 0.470 1.79 86.6 285743 O.K

Eo h/Eo C'/C C' Q(USWL+ho-CL) (Gibson)

(ft) (cusecs/ft) (cusecs)

17.79 0.42 0.92 3.50 262.36 157414.34 O.K

17.44 0.20 0.94 3.57 260.11 156063.65 O.K

18.53 0.62 0.91 3.46 275.84 165503.38 O.K



qclear

qclear

cusecs/ft

cusecs/ft

cusecs/ft

cusecs/ft


270000

216000max min

602.75 597.46602 597

25.75 20.46

80.00 80.00

25.00 20.00

0.612 0.8640.080 0.1150.504 0.585

2.06 2.35

12.98 11.97

12.02 8.03O.K O.K

ft

540000155503

Min.603.89590.0037.89

259

24.00

1.110.1550.630

5.87

23.87

0.13O.K

cusecs/ft

cusecs/ft

ft (block= 4'x4'x4')

ft (block= 4'x4'x4')


9" Gravel

9" Coarse sand

6" Sand

ft above HFLft above HFL


9" Gravel

9" Coarse sand

6" Sand

( max for earthen channels)

from barrage level

ft3/unit width

(5)x(10) Lx(1) =(11)+(4)

9 10 11 12

0.4799 0.0246 2145 46450.5345 0.0546 4760 72600.6253 0.0908 7916 104160.6453 0.02 1744 42440.7956 0.1503 13103 156030.8564 0.0608 5301 78011.155 0.2986 26033 28533

1.5881 0.4331 37759 39759Total 118260 ft

22.40 miles

Φ2( d2

D )Φ2−Φ1


Basic Data:Clear width of the weir section of the barrage b 2520 ftBarrage Crest Level EL 678 ftUngated Discharge 700000 692.74 ftUngated Discharge 842000 694.55 ftUngated Discharge 950000 695.84 ft

For ungated flow

Q (cfs)

Trial 1 Trial 2

H (ft) D/S WL

50000 14.88 24.00 39.31 33.42 0.45 23.78 39.13 29.35 0.51 7.26 9.78 4.96 674.96

100000 29.76 24.00 39.31 33.42 0.89 23.55 38.95 29.21 1.02 5.10 6.73 6.86 676.86

200000 59.52 24.00 39.31 33.42 1.78 23.11 38.58 28.93 2.06 3.55 4.55 9.37 679.37

300000 89.29 24.00 39.31 33.42 2.67 22.66 38.20 32.47 2.75 3.45 4.41 12.12 682.12

400000 119.05 24.00 39.31 33.42 3.56 22.22 37.83 32.15 3.70 2.94 3.69 13.68 683.68

500000 148.81 24.00 39.31 33.42 4.45 21.77 37.45 31.83 4.68 2.59 3.20 14.97 684.97

700000700000208.33 15.37 31.46 26.74 7.79 11.47 27.18 23.11 9.02 1.36 1.48 13.36 683.36

840000840000250.00 16.27 32.37 27.52 9.08 11.73 27.49 23.36 10.70 1.26 1.35 14.43 684.43

950000950000282.74 16.92 33.01 28.06 10.08 11.88 27.66 23.51 12.02 1.19 1.26 15.18 685.18

Table F Conjugate depth d2 for different discharges under gated and ungated flows.

Theoretical velocity for gated control flow

q=Q/b (cfs/ft)

H (ft)

Vth ft/sec

Vact

ft/secd1

(ft)Vth ft/sec

Vact ft/sec

d1 ft FR1 d2/ d1 d2 ft

H=Z−d /2H=Z−h/2−d /2

EL694

EL 678d Z

EL673

EL 670

h

EL 678d Z

EL673

EL 670

1:3

1:3

Part IDesign of barrage for overflow condition

1 Minimum stable wetted perimeter2 Calculation of Lacey's silt factor3 Fixation of crest level4 Design of undersluices5 Determination of water levels and energy levels

5.1 Check for main weir5.2 Check fo undersluices

6 Fixation of d/s floor levels and length of d/s glacis and d/s floor6.1 Fixation of d/s floor levels for normal weir section using blench curves6.2 Fixation of floor levels for undersluices

7 Fixation of d/s floor level for normal barrage section usingCrump's method and determination of floor length

8 Fixation of d/s floor length for undersluices9 Check for the adequacy for d/s floor levels using conjugate

depth method9.1 For normal weir section9.2 For undersluice section

10 Scour protection10.1 d/s scour protection10.2 u/s scour protection10.3 Thickness of aprons

11 Inverted filter design12 Design of guide banks

12.1 Determination of levels of guide banks13 Design of guide bank aprons14 Design of marginal bunds


15 Fixation of depth of sheet piles16 Calculation of exit gradients17 Calculation of uplift pressure after applying correction

17.1 u/s pile length17.2 Intermediate sheet pile at toe of d/s glacis17.3 d/s sheet pile at the end of impervious floor

18 Calculation of floor thickness

days

Done

0

Done

0

Fixation of d/s floor levels and length of d/s glacis and d/s floorFixation of d/s floor levels for normal weir section using blench curves

Done0

Check for the adequacy for d/s floor levels using conjugate

Done

0

Done

0

Done 0

Done 0

Done

0

0Done

total 0 days left

44025203 Design of a Barrage MS

Documents

angle protected

muhammad azhar

divide walls

angle protected

fish ladder

soil medium

inverted filter

hflthese free