Design of a Barrage

Hydraulic Structures

Design of Barrage

Design of Barrage Input Design Data Maximum Discharge, Q max Minimum Discharge, Qmin River Bed Level, RBL High Flood Level, HFL Lowest water level, LWL Numbers of canals on left side Numbers of canals on right side Maximum Discharge of one Canal Slope of river Lacey's Looseness Coefficient, LLC 1- Minimum Stable Wetted Perimeter Wetted perimeter, Pw = 2.67√ Qmax Width between abutment, Wa = LLC x Pw Number of bays Bay width Number of fish ladder Width of one fish ladder Number of divide walls Width of on divide wall width of one pier Total number of piers Total width of bays Total width of piers Width between abutment, Wa

500000 cusecs 12000 cusecs 582 ft 600 ft 587 ft 1 2 3500 cusecs 1 ft/mile 1.8

1888 ft 3398 ft 50 60 ft 1 26 ft 2 15 ft 7 ft 47 3000 ft 329 ft 3385 ft Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Discharge between abutments, qabt Discharge over weir, q weir 2- Calculation of Lacey's Silt Factor S = (1/1844) x f**5/3 / Q**1/6 Lacey's silt factor, f 3- Fixation of Crest Level Afflux Height of crest above river bed, P Scour depth, R = 0.9(qabt**2 / f)**1/3 Depth of water above crest, Ho = R- P Approach velocity, Vo = qabt / R Energy head, ho = Vo**2 / 2g Eo = Ho + ho Do = HFL - RBL E1 = Do + ho + Afflux Level of E1 = RBL + E1 Crest level = Level of E1 - Eo Maximum d/s water level h = d/s WL - Crest Level Using Gibson Curve h / Eo C' / C C C' = (C'/C) x C

Q = C' x W clear x Eo**3/2

147.71cusecs/ft 166.67cusecs/ft

2.04

3 ft 6 ft 19.82 ft 13.82 ft 7.45 ft/s 0.86 ft 14.69 ft 18.00 ft 21.86 ft 603.86 ft 589.18 ft 600 ft 10.82 ft 0.74 0.84 3.8 fps 3.19 538948 cusecs O.K

Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

4- Design of Undersluices

Difference between undersluices & main weir Number of undersluices (N1) Number of bays for one undersluices (N2) Flow through undesluices as % of main weir Crest level of undersluices b1 = N1 x Bay width qus = % flow flow x q weir Scour depth, R = 0.9(qus**2 / f)**1/3 Do, (may be Do = R) Approach velocity, Vo = qus / R Energy head, ho = Vo**2 / 2g Maximum U/S E.L = HFL + Afflux + ho Eo = U/S E.L - Crest Level h = (U/S E.L - Afflux) - Crest level h / Eo Using Gibson Curve C' / C C' = (C'/C) x C Q1 & Q3,

( Q = C' x W clear x Eo**3/2)

Q main weir = C' x (W clear(bays) - Wclear( us) )x Eo**3/2

Total Discharge = Q1 + Q3 + Q main weir %water through undersluices=(Q1+Q3)/Qmain undersluices=(Q1+Q3)/Qmain weir*100

Hence Crest Level of Undersluices Number of Bays on Each Side

3 ft 2 5 120 % 586.18 ft 300 ft 200 cusecs/ft cusecs/ft 24.26 ft 24.26 ft 8.24ft/sec 1.06 ft 604.06 ft 17.88 ft 14.88 ft 0.83 0.76 2.89 131000cusecs 431158cusecs 562158cusecs O.K 30.4%

1

586.18 ft 5



Design of Barrage

5- Determination of Water Levels and Energy Levels 5.1 Check for main weir

Q (cusecs)

DSWL (ft)

USWL (ft)

For normal state 600000 601.5 500000 600 250000 597 125000 592

D

ho

h

Ho

Eo

(USWL-RBL)

Vo qclear/D

h/Eo

(Vo**2/2g)

(DSWL-CL)

(USWL-CL)

(USWL+ho-CL)

(ft)

(ft/s)

(ft)

(ft)

(ft)

(ft)

C'/C

C'

qclear

Q

(cusecs/ft)

(cusecs)

(Gibson)

604.0 602.5 598.5 594.5

22.0 20.5 16.5 12.5

9.1 8.1 5.1 3.3

1.28 1.03 0.396 0.17

12.32 10.82 7.82 2.82

14.82 13.32 9.32 5.32

16.11 14.35 9.72 5.50

0.765 0.754 0.805 0.514

0.820 0.815 0.780 0.940

3.12 3.10 2.96 3.57

201.4 604304.51 168.4 50 5 05080.27 89.8 269469.2 46.0 1 13 38093.92

O.K

For retrogressed state 600000 595.5 602.0 500000 596 601.0 250000 591 597.0

20.0 19.0 15.0

10.0 8.8 5.6

1.55 1.19 0.48

6.32 6.82 1.82

12.82 11.82 7.82

14.38 13.02 8.30

0.440 0.524 0.220

0.960 0.930 0.970

3.65 3.53 3.69

198.9 596585.23 166.0 498022.7 88.2 2 26 64578.81

NOT O.K

For accreted state 600000 604 500000 602.5 250000 601.5

24.0 22.0 20.0

8.3 7.6 4.2

1.08 0.89 0.27

14.82 13.32 12.32

16.82 14.82 12.82

17.90 15.72 13.09

0.828 0.848 0.941

0.765 0.740 0.520

2.91 2.81 1.98

220.2 66 6 60591.93 175.2 525555.63 93.6 28 2 80866.24

O.K

606.0 604.0 602.0

O.K O.K O.K

NOT O.K O.K

O.K O.K

5.2 Check fo undersluices

Increase in flow Concentration of flow, Q

DSWL

20 % ### cusecs

USWL

D

Vo

ho

h

Ho

Eo

h/Eo

C'/C

C'

qclear

Q Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

(cusecs)

(ft)

(ft)

(USWL-RBL)

qclear/D

(Vo**2/2g)

(DSWL-CL)

(USWL-CL)

(USWL+ho-CL)

(ft)

(ft/s)

(ft)

(ft)

(ft)

(ft)

For normal state 157199 601.5 603.5 For Retrogressed state 157199 595.5 601.5 For accreted state 157199 604.0 605.5

(Gibson) (cusecs/ft)

(cusecs)

21.5

12.2

2.31

15.32

17.32

19.63

0.781

0.82

3.12

271.00 162602.65Err:508

19.5

13.4

2.80

9.32

15.32

18.13

0.514

0.94

3.57

275.68 165409.02Err:508

23.5

11.1

1.93

17.82

19.32

21.25

0.839

0.77

2.93

286.71 172025.03Err:508

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 (cusecs)

qclear (cusecs/ft)

USEL

DSEL

hL

E2

DSFL

(USWL+ho) (USWL+ho) (DSWL +ho) (USEL-DSEL)

(blench curve) (DSEL - E2)

(ft)

(ft)

(ft)

Normal state of river 600000 201.4 605.28 602.78 500000 168.4 603.53 601.03 250000 89.8 598.90 597.40 For Retrogressed state of river 600000 198.9 603.55 597.05 500000 166.0 602.19 597.19 250000 88.2 597.48 591.48 For accreted state of river 600000 220.2 607.08 605.08 500000 175.2 604.89 603.39 250000 93.6 602.27 601.77

(ft)

(ft)

2.50 2.50 1.50

19.1 583.68 17.3 583.73 11.2 586.20

6.50 5.00 6.00

21 576.05 18.3 578.89 13 578.48

2.00 1.50 0.50

19.6 585.48 16.6 586.79 10.4 591.37 Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Hence d/s Floor level

576.00 ft ft

6.2 Fixation of d/s floor levels for undersluices using blench curves Q (cusecs)

qclear (cusecs/ft)

USEL

DSEL

hL

E2

DSFL

(USWL+ho) (USWL+ho) (DSWL +ho) (USEL-DSEL)


(ft)

(ft)

(ft)

(ft)

Normal state of river 162603 271.00 605.81 603.81 For Retrogressed state of river 165409 275.68 604.30 598.30 For accreted state of river 172025 286.71 607.43 605.93 Hence d/s Floor level for underslu 573.00ft

(ft)

2.00

22.3 581.51

6.00

25.1 573.20

1.50

22.6 583.33

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

Q Maximum DSWL USWL USEL DSFL RBL Crest level Dpool (Max. DSWL - DSFL)

500000 cusecs 602.5 ft 604.0 ft 604.89 ft 576.00 ft 582 ft 589.18 ft 26.5 ft Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

d/s Velocity (Q/(Dpool x Wa)

5.57 ft/sec

d/s velocity head (V2/2g) DSEL (DSWL + velocity head) K (USEL - Crest Level ) L (USEL - DSEL ) q (Q / Total width of bays)

0.48ft 602.98 ft 15.71 ft 1.91 ft 166.67 cusecs/ft

Critical Depth, C, (q2 / g)1/3 L/C (K+F)/C ,(from crumps curve) F, [(K+F)/C x C - K] Level of intersection of jump with glacis = Crest level - F E2, ( DSEL - Level of intersection of jump)

9.52ft 0.20 1.92 2.56 586.61 ft

Submergency of jump, (Level of intersection of jump - DSFL ) Slope of d/s glacis Length of glacis d/s of jump, (slope x submergency) Length of stilling pool, (4.5 x E2) Length of d/s floor, (Length of stilling pool -Length of glacis d/s of jump ) Say

16.37ft 16.37ft 10.61ft 10.61ft 1: 3 31.84 f t 73.66ft 73.66ft 41.83ft 41.83ft 42ft

L K

y F

x



Design of Barrage

Fig:1 Various Parameters for using Crump's Curve

b)

Q Minimum DSWL USWL USEL DSFL RBL Crest level Dpool (Min. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa) d/s velocity head (V2/2g) DSEL, (DSWL + velocity head) K (USEL - Crest Level ) L (USEL - DSEL ) q (Q / Total width of bays) Critical Depth, C, (q2 / g)1/3 L/C (K+F)/C ,(from crumps curve) F, [(K+F)/C x C - K] Level of intersection of jump with glacis = Crest level - F E2, ( DSEL - Level of intersection of jump) Submergency of jump, (Level of intersection of jump - DSFL ) Slope of d/s glacis

500000 cusecs 596 ft 601 ft 602.19 ft 576 ft 582 ft 589.18 ft 20 ft 7.39 ft/sec 0.85ft 596.85 ft 13.01 ft 5.34 ft 166.67 cusecs/ft 9.52ft 0.56 2.48 10.59 ft 578.58 ft 18.27ft 18.27ft 2.58ft 1: 3 Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

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

7.74 f t

Say

82.19ft 82.19ft 74.45ft 74.45ft 75.00 ft ft

8 - Fixation of d/s floor length for undersluices a)

Q Maximum DSWL USWL USEL DSFL RBL Crest level Dpool (Max. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa)

### cusecs 604 ft 605.5 ft 607.43 ft 573 ft 582 ft 586.18 ft 31 ft 8.5 ft/sec




12.87ft 12.87ft 0.180 1.92 3.46 ft 582.72 ft 22.39ft 22.39ft Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage


9.72ft 1: 3 29.16 f t 101ft 71.60ft 71.60ft 72.00 ft

Q Minimum DSWL USWL USEL DSFL RBL Crest level Dpool (Min. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa)

### cusecs 595.5 ft 601.5 ft 604.3 ft 573 ft 582 ft 586.18 ft 22.5 ft 11.64 ft/sec



b)


12.87ft 12.87ft 0.520 2.32 11.73 ft 574.44 ft 23.16ft 23.16ft Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage


Hence we shall provide d/s floor length =

1.44ft 1: 3 4.32 f t 104ft 99.91ft 99.91ft 100 ft

100ft

9- Check for Adequacy for d/s floor levels using conjugate depth method. 9.1 For normal weir section

Ф Floor level of stilling pool Discharge in river, Q (cusecs) Discharge through main weir, Q1= 0.8Q (cusecs) USEL (ft) DSWL (ft) E = USEL - DSFL Intensity of flow on d/s floor, q = Q1/width of main weir

1.00 576.00 ft 500000

166.67 166.67

Depth in stilling pool, Dpool = DSWL - DSFL

26.50

f(z) = q/E3/2

1.073 0.145 0.635

z Conjugate depth coefficients z'

250000

400000 200000 max. min. max min 604.89 60 602.19 6 60 02.27 597.48 602.5 596.0 601.5 591 28.89 26.19 26.27 21.48 83.33

83.33

20.00

25.50

15.00

1.243 0.170 0.671

0.619 0.080 0.504

0.837 0.110 0.573 Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Conjugate depths

d1 = z x E d2 = z' x E

4.19

4.45

2.10

2.36

18.36

17.57

13.24

12.31

Jump submergency = Dpool - d2 Remarks

8.14 O.K

2.43 O.K

12.26 2.69 O.K O.K

9.2 For undersluices section

Ф Floor level of stilling pool Discharge in river, Q (cusecs) Disc Discha harg rge e thr throu ough gh U.S U.S wit with h 20% 20% conc concen entr trat atio ion, n, (1.2 (1.2 x (Q1 (Q1 + Q2) Q2))) USEL (ft) DSWL (ft) E = USEL - DSFL (ft) Intensity of flow on d/s floor, q = Q1/Total width of all U/S

1.00 573.00 ft 500000 1571 15 7199 99 Max. Min. 607.43 604.30 604.00 595.5 34.43 31.30 262

262

Depth in stilling pool, Dpool = DSWL - DSFL

31.00

22.50

f(z) = q/E3/2

1.30 0.180 0.684

1.50 0.217 0.695

6.20

6.79

23.54

21.75

7.46

0.75 O.K

z Conjugate depth coefficients z' d1 = z x E Conjugate depths d2 = z' x E Jump submergency = Dpool - d2 (ft) Remarks

O.K

10 - Scour Protection Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

For main weir Assume flow concentration q = Qmax /(total width of bays) x 1.2

20 % 200 cusecs/ft

R = 0.9 (q2/f)1/3

24.26ft 24.26ft

10.1 - d/s scour protection for main weir

Safety factor for d/s floor critical condition Depth, R' = safety factor x R Minimum DSWL for Q max

1.75 42.46 ft 596ft 576.00 ft 20.00 ft 0.50 ft 20.50 ft 21.96 ft 1: 3

d/s appron (floor) level, (DSFL) Depth of water on apron (Min DSWL - DSFL) Increase in depth due to concentration Depth of water with concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (12+32)x(R' Length of d/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

69.44ft 69.44ft 50ft MIN. W.L

D R'

2.5(RD)Level Bed

DSFL

1:3 t Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Deepest Possible Scour 3(R-D)

Fig:2 Scour Protection

10.2 U/s Scour Protection for main weir

Safety factor for u/s floor critical condition R' = Safety factor x R Minimum USWL for Qmax U/s apron level, (RBL) Depth of water on apron = USWL -RBL Increase in depth due to concentration Total depth with concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (12+32)x(R' Length of u/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

1.25 30.33 ft 601.00ft 582 ft 19.00 ft 0.50 ft 19.50 ft 10.83 ft 1: 3 34.24ft 34.24ft 25ft

For undersluices

Assume flow concentration q = (Q1+Q3)/Total width of undersluicesx 1.2 R = 0.9 (q2/f)1/3

20 % 262 cusecs/ft cusecs/ft 29.05ft 29.05ft

10.3 - d/s scour protection for undersluices Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Safety factor for d/s floor critical condition Depth, R' = safety factor x R Minimum DSWL for Q 1 + Q3 d/s appron (floor) level, (DSFL) Depth of water on apron (Min DSWL - DSFL) Increase in depth due to concentration Depth of water with concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (12+32)x(R' Length of d/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

1.75 50.83 ft 595.5ft 595 .5ft 573.00 ft 22.50 ft 0.5 ft 23.00 ft 27.83 ft 1: 3 88.02ft 88.02ft 63ft

10.4 U/s Scour Protection for undersluices

Safety factor for u/s floor critical condition R' = Safety factor x R Minimum USWL for Q1 + Q3 U/s apron level, (RBL) Depth of water on apron = USWL -RBL Increase in depth due to concentration Total depth with concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (12+32)x(R' Length of u/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

1.25 36.31 ft 601.5ft 601.5ft 582 ft 19.5 ft 0.5 ft 20 ft 16.31 ft 1: 3 51.58ft 51.58ft 37ft Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

10.5 - Thickness of Aprons

The following table gives the required valuves of "t" (Fig:2) for protection protection of various grades of sand and slope of rivers. Fall in inches/mile Sand Sand clas classi sifi fica cati tion on Very coarse Coarse Medium Fine Very fine

3 9 Thic Thickn knes ess s of ston stone e 16 19 22 25 28 31 34 37 40 43

12 18 pitc pitchi hing ng in inch inches es 22 25 28 31 34 37 40 43 45 49

24 28 34 40 46 52

Type of soil Medium sand Slope of river 12 in/mile Thickness if stone pitching, t 34 in Thickness of stone apron in horizontal position = 1.75xt/slo 5 ft Size of concrete blocks over filter 4 ft cube Summary Total Length of d/s stone apron 50 ft 4 fftt Thick bloke apron = 1/3 x total length 16 fftt (block= 4'x4'x4') 5 ft Thick stone apron 34 ft Total length of u/s apron 4 ft ft Thick bloke apron = 1/3 x total length 5 ft Thick stone apron

25 ft 8 ft ft (block= 4'x4'x4') 17 ft



Design of Barrage

11 - Inverted Filter Design Size of Concrete blocks Thickness of shingle (3' - 6") Thickness of coarse shingle (3/4" - 3") Thickness of fine shingle (3/16" - 3/4") Spacing b/w conc. Blocks filled with fine shingle

Spacing /Jhries (2")

4 ft cube 9 in 9 in 6 in 2 in in

Concrete Blocks (4'x4'x4')

9" Gravel 9" Coarse sand 6" Sand

Fig: 3 Inverted Filter

12- Design of guide banks



Design of Barrage

i)

Length Length of of each each guid guid bank bank measur measured ed in straig straight ht line line along along the barrage u/s , Lu/s = 1.5 x Wa

i i)

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

iii)

For the the nose nose of the the u/s guide bank and and the full length length of d/s d/s guide bank use Lacey's depth = 1.75 x R For rem remaini aining ng u/s u/s gui guide de bank bank lace lacey' y's s dep depth th = 1.2 1.25 5 xR Possible slope of scour Free board u/s Free board d/s

iv) v)

5078ft 677 ft

42.46 ft 30.3 30.33ft 3ft 1: 3 7 ft above HFL 6 ft above HFL

These free boards boards also include allowance allowance for accretion. accretion.

vi) To Top of guide bank vii) Si S ide slope of guide bank viii) Minimum apron thickness

10 ft 1: 3 4 ft

Length of barrage, Wa Length of u/s guide bank Length of d/s guide bank Radius of u/s curved part Radius of d/s curved part

3385 ft 5078 ft 677 ft 600 ft 400 ft

Maximum u/s angle protected

140o

Maximum d/s angle protected

57o - 80o

12.1 Determination of levels of guide banks Merrimen's backwater formula L =

d 1− d 2 S

 D

1

S

2

−

C

Φ

d 1 D

−Φ

d 2 D Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

L = length of back water curve

d2

d1

D

S

L Bresse Backwater Function

Chezy's Coefficient, C Bed slope of river, S RBL D/s HFL with accretion D = D/s HFL with accretion - RBL U/s HFL with accretion d1 = U/s HFL with accretion - RBL

71 ( max for earthen channels) 1/ 5000 582 ft 602.5 ft 20.5 ft 604.0 ft 22.0ft

Assume d2 (in between d1 and D)

21.78ft 21.78ft

d1/D

1.073

d2/D

1.062

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

0.7870

Ф (d2/D) L Length of guide bank Comments

0.83 5240 ft 5078 O.K Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

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

1.02 from barrage level 604.80 ft

i)Leve i)Levell at at the the nose nose of u/s u/s gui guide de bank bank = W/L W/L + fre free e boa board rd ii) Level at the barrage = HFL + free board iii)Water level d/s of barrage barrage D/s free board Level of guide bank d/s = W/L + Free board

611. 61 1.80ft 80ft 607 ft 602.5ft 6 ft 608.5 fftt

13 - Design of Guide Bank Apron W.L D

2.5 (R' - D') T=1.07 t

T R 1 :3

Deepest Possible Scour t

Working on same lines as in section 10, Length of of un unlanched ho horizontal ap apron = 2.5(R' - D')

27.07 f t Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Length of of la launched ap apron at at 1: 1:3 sl slope = 3.16(R ' - D' D') Thickness of stone apron, t (as calculated previously) say Volu Volume me of ston stones es in apro apron n = t x leng length th laun launch ched ed apro apron n Minimum thickness of unlaunched apron = 1.07t Mean Mean thic thickn knes ess s of of unl unlau aunc nche hed d apr apron on = vol volum ume/ e/ 2.5( 2.5(R' R' - D') D') Maximum thickness of unlaunched apron = 2tmean - tmin

34.22 f t 34 inches 3ft 102. 10 2.65 65ft3/unit width 3.2 ft ft 3.8 3.8 ft 5.0ft

14 - Design of Marginal Bunds i) Top width 20 ft ii) ii) Top Top leve levell abov above e esti estima mate ted d HFL HFL aft after er all allow owin ing g 1.5f 1.5ftt accr accret etio ion n 5ft iii) Front slope of marginal bunds (not pitched) 1: 3 iv) Back slope to be such as to provide minimum cover cover of 2 ft, over hydraulic gradient of 1:6 v) U/s water level at nose of guide bank 611.80 ft Free board of marginal bund 5 ft Hence level of marginal bund 616.80 ft Calculation of length of backwater curve:

Merrimen's equation can be used to calculate backwater length L =

d 1− d 2 S

 D



1

S

2

−

C g

[    ] Φ

d 1

D

−Φ

d 2

D



Design of Barrage

d2

d1

D


Maximum USWL at Qmax RBL Normal W.L without weir d1 = Maximum USWL - RBL D = Normal W.L - RBL Slope Table for length of backwater curve d1 d2 d 1− d 2 D S 1

2

18 18 18 18 18 18 18 18

22.0 21.5 21.0 20.5 20.0 19.5 19.0 18.5

3

4

21.5 21.0 20.5 20.0 19.5 19.0 18.5 18.1

2500 2500 2500 2500 2500 2500 2500 2000

604.0ft 604.0ft 582 ft 600 ft 22.0ft 18 ft 1: 5000

C 2 S g 1

−

5

4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4

T 1=

d 1 D

6

1.222 1.194 1.167 1.139 1.111 1.083 1.056 1.028

T 2 =

d 2 D

7

1.194 1.167 1.139 1.111 1.083 1.056 1.028 1.006

Φ1

    d 1

D

8

0.32 0.34 0.36 0.39 0.42 0.46 0.5 0.55

Φ2

d 2

D

9

0.34 0.36 0.39 0.42 0.46 0.5 0.55 0.62

Φ 2 − Φ1 10

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

12 12

0.02 2049 4549 0.02 2084 4584 0.03 2371 4871 0.03 2528 5028 0.04 3243 5743 0.04 3688 6188 0.05 4516 7016 0.07 6068 8068 Total 46047 ft 8.73miles Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Hence length of backwater cure = 8.73 miles

Part II Design of barrage profile for sub surface flow condition 15 - Fixing of Depth of Sheet Piles Scour depth, R Depth of u/s sheet pile from HFL = 1.5 R Max. USWL for Qmax

19.82 ft 30 ft

RL of bottomo fu/s sheet pile = Max. USWL - 1.5R Depth of d/s sheet pile below HFL = 2R RL of of bott bottom om of inte interm rmed edia iate te she sheet et pil pile e = Max Max.. USWL USWL - 2R Let RL of bottom of d/s sheet pile

604.0ft 604.0ft 574.0 ft ft 40 ft 564. 564.0 0 ft 550 ft

604.0ft 604 .0ft

589.18 589.18 ft

582.0

1: 4

A

K

1: 3

576.00 B

L

N

574.0

P

R

75.00 M

564.0 1.5H = 48

28.70

6

39.53

Q

550.0 Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

16 - Calculation of Exit Gradient

H 1 G E = x d π  λ λ H

b E

C

b1

d D

α = b/d

Parmeters of Khosla's Curve

Let Let the the wate waterr be be hea heade ded d up up to to Max. Max. accr accret eted ed leve levell u/s u/s and no flow d/s. Retogression DSFL Differential head causing seepeage, H = Max. u/s WL - (DSFL - Retrogressio Depth of d/s sheet pile, d = DSFL - RL of bottom of d/s sheet pi

Total length of concrete floor = b α = b/d 1

π  λ λ

form α ~

1

curve

604. 604.0ft 0ft 4 ft 576.0 ft 32.0ft 26.0ft 197.23 ft 7.59 0.15

π  λ λ



Design of Barrage

GE

0.187 SAFE

17 - Calculation of Uplift Pressure After Applying Correction 17.1 U/s pile line:

Length of concrete floor upto u/s u/s sheet pile, b1 Total length of concrete floor, b Depth of u/s sheet pile,d Assume tf u/s floor thickness

48ft 197.23 ft 8.0 ft 2.5ft 0.0406 24.65

1/α = d/b α = b/d b1/b

0.243

1 - b1/b Φ B = Φ D From khosla's curve Φ A = Φ E From khosla's curve

0.757 67 % 69 % 64 %

100 - 33 100 - 31

Φ K = Φc From khosla's curve

i) Correction for floor thickness Correction in Correction in

Φ K =

t f

Φ A=

d

0.938 %

 Φ B−Φ K 

d t f

-ve

 Φ A−Φ B 

0.63 %

ii) Correction for interface of sheet pile Correction in

Φ K

[ ]

due to second pile =19 d  D b

D b ' Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Depth of u/s sheet pile,d D = RBL - RL of bottom of second pile Total length of concrete floor, b Distance between two piles, b' Correction in Φ K

8.0 ft 18.0 ft 197.23 ft 74.23 ft +ve 1.23 %

iii) Slope correction for Φ K Correction for Φ =− F K S

b s b1

For 1:4 slope, Fs (f ( from slope correction curve) 3.3 Distance between two piles, b1 74.23ft 74.23ft Horizontal projection of u/s glacis, bs = (crest level - RBL) x 1/slop 28.70ft 28.70ft Correction for Φ K -1.28 Hence Correted Φ A 68.38 % Corrected Corrected Φ B 67 % Corrected Φ 64.895 % K

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

Assume floor thickness DSFL RL of intermediate sheet pile d = DSFL - RL of Intermediate sheet pile Total length of concrete floor, b Length of concrete floor up to sheet pile, b1 b1/b

10 ft 576.00 ft 564.00 ft 12.00 ft 197.23 ft 122.23ft 0.620 Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

1 - b1/b = b/d Φ L = Φ E Φ M = Φ D

From khosla's curve

Φ N =Φ

From khosla's curve

α

C

100 - 55 100 - 58

From khosla's curve

0.380 16.44 45 % 42 % 36.5%

i) Correction due to floor thickness Correction in Φ L =

t f

Correction in Φ N =

 Φ L−Φ M 

-2.5 %

 Φ M −Φ N 

4.58 %

d t f d

ii) Correction due to interference of pile

[ ]

Correction in Φ L due to u/s sheet pile19 d  D b

D b '

Depth of Intermediate sheet pile,d D = RBL - RL of bottom of u/s sheet pile Total length of concrete floor, b Distance between two piles, b' Correction in Φ L

[ ]

= 19 d  D b

D b '

12.0 ft 18.00 ft 197.23 ft 74.23 ft -1.42 %

[ ]

Correction in Φ N due to d/s sheet pile19 = Depth of Intermediate sheet pile,d D = DSFL - RL of bottom of d/s sheet pile Total length of concrete floor, b Distance between two piles, b'

d  D b

D b '

12.00 ft 26.00 ft 197.23 ft 75.00 ft



Design of Barrage

Correction in Φ N = 19

b

2.16 %

b '

Φ L iii) Slope correction for for '1:3 slope, Fs

4.5

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

39.53ft 39.53ft

Distance between two piles, b1

74.23ft 74.23ft

Correction in Φ = F × L S

b s

2.40 %

b1

Hence Corrected Φ L Corrected Φ M Corrected Φ N

43.47 % 42 % 43.24 %

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

Assume floor thickness Depth of d/s sheet pile, d Total length of concrete floor, b 1/α = d/b Φ P = Φ E From khosla's curve ΦQ = Φ D From khosla's curve Φ R= ΦC From khosla's curve

100 - 68 100 - 78

7 ft 26.00 ft 197.23 ft 0.132 32 % 22 % 0%

i) Correction due to floor thickness Correction in Φ P

-2.69 % Muhammad Azhar Saleem 2003/II-MS-C-STRU-01


Design of Barrage

Correction in Φ R

5.92 %

ii) Correction due to interface of piles Correction in Φ P Depth of d/s sheet pile,d D = DSFL - RL of bottom of intermediate sheet pile Total length of concrete floor, b Distance between two piles, b' Correction in Φ P Hence Corrected Corrected Φ P Corrected Corrected ΦQ Corrected Corrected Φ R

[ ] 

=19 d  D b

D b '

26.00 ft 12.00 f t 197.23 ft 75.00 ft -1.46 %

27.84 % 22 % 5.92 %

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

Symbol used in Khosla cueve Φ E Φ D ΦC

u/s Pile line Φ A = 68.38% 67% Φ B = = 6 4 . 89% Φ K

Intermediate Line Φ L = 43.47% 42% Φ M = 43.24% Φ N =

d/s Pile Line Φ P = 27.84% ΦQ = 22% 5.92% Φ R =

18 - Calculation For Floor Thickness:



Design of Barrage

t f =

100

 G− 1

× H

where tf = Thickness of floor in ft Φ = % Uplift pressure

H = Maximum differential head causing seepage G = Specific gravity of concrete

2.4

a) Thickness of floor at A Assumend thickness H Thickness from uplift pressure say

2.5 ft 32 ft 15.63 ft 16.00 ft ft

say

10 ft 9.94 ft 10.00 ft ft

say

10 ft 9.88 ft 10.00 ft ft

say

7 ft 6.36 ft 7.00 ft

b) Thickness of floor at L Assumend thickness Thickness from uplift pressure c) Thickness of floor at N Assumend thickness Thickness from uplift pressure d) Thickness of floor at P Assumend thickness Thickness from uplift pressure e) Thickness of floor at crest



Design of Barrage

K

x

L

Φ L

Φ K

Pressure at crest y

Φ −Φ L Uplift pressure at crest= Φ L  K  x  y Hence Thickness of floor at crest d/s of gate

54.88 %

say

12.54 ft 13 ft


Design of Barrage Input Design Data Maximum Discharge, Q max Minimum Discharge, Qmin River Bed Level, RBL High Flood Level, HFL Lowest water level, LWL Numbers of canals on left side Numbers of canals on right side Maximum Discharge of one Canal Slope of river Lacey's Looseness Coefficient, LLC 1- Minimum Stable Wetted Perimeter Wetted perimeter, Pw = 2.67√ Qmax Width between abutment, Wa = LLC x Pw Number of bays Bay width Number of fish ladder Width of one fish ladder Number of divide walls Width of one divide wall width of one pier Total number of piers piers Total width of bays Total width of piers piers Width between abutment, Wa Discharge between abutments, q abt Discharge over weir, q weir 2- Calculation of Lacey's Silt Factor S = (1/1844) x f**5/3 / Q**1/6 Lacey's silt factor, f 3- Fixation of Crest Level Afflux Height of crest above river bed, P Scour depth, R = 0.9(qabt**2 / f)**1/3 Depth of water above crest, Ho = R- P Approach velocity, Vo = qabt / R Energy head, ho = Vo**2 / 2g Eo = Ho + ho

540000 cusecs 12000 cusecs 582 ft 600 ft 587 ft 1 2 3500 cusecs 1 ft/mile 1.8

1962 ft 3532 ft 55 60 ft 1 25 ft 2 15 ft 7 ft 52 3300 ft 364 ft 3719 ft 145.20 145.20 cusecs/f cusecs/f 163.64 163.64 cusecs/f cusecs/f

2.06

3 ft 6 ft 19.55 ft 13.55 ft 7.43 ft/s 0.86 ft 14.41 ft

Do = HFL - RBL E1 = Do + ho + Afflux Level of E1 = RBL + E1 Crest level = Level of E1 - Eo Maximum d/s water level h = d/s WL - Crest Level Using Gibson Curve h / Eo C' / C C C' = (C'/C) x C

18.00 ft 21.86 ft 603.86 ft 589.45 ft 600 ft 10.55 ft 0.73 0.86 3.8 fps 3.27

Q = C' x W clear x Eo**3/2

589630 cusecs O.K

4- Design of Undersluices

Difference between undersluices & main weir Number of undersluices (N1) Number of bays for one undersluices (N2) Flow through undesluices as % of main weir Crest level of undersluices b1 = N1 x Bay width qus = % flow x q weir Scour depth, R = 0.9(qus**2 / f)**1/3 Do, (may be Do = R) Approach velocity, Vo = qus / R Energy head, ho = Vo**2 / 2g Maximum U/S E.L = HFL + Afflux + ho Eo = U/S E.L - Crest Level h = (U/S E.L - Afflux) - Crest level h / Eo Using Gibson Curve C' / C C' = (C'/C) x C

3 ft 2 5 120 % 586.45 ft 300 ft 196.36 196.36 cusecs/f cusecs/f 23.91 ft 23.91 ft 8.21 8.21 ft/sec ft/sec 1.05 ft 604.05 ft 17.60 ft 14.60 ft 0.83

Q1 & Q3,

129586 cusecs 129586 482425 4824 25 cusecs 612010 6120 10 cusecs 26.9%

0.77 2.93

( Q = C' x W clear x Eo**3/2)

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

Total Discharge = Q1 + Q3 + Q main weir %water through undersluices=(Q1+Q3)/Qmain weir*100

O.K 1

Hence Crest Level of Undersluices Number of Bays on Each Side 5- Determination of Water Levels and Energy Levels

586.45 ft 5

5.1 Check for main weir

Q (cusecs)

DSWL (ft)

For normal state 648000 602 540000 600.5 270000 597.5 135000 593

USWL D (ft)

ho

h

Ho

(USWL-RBL)

Vo qclear/D

(Vo**2/2g)

(DSWL-CL)

(USWL-CL)

(ft)

(ft/s)

(ft)

(ft)

(ft)

604.5 603.0 598.5 594.5

22.5 21.0 16.5 12.5

8.7 7.8 5.0 3.3

1.18 0.94 0.382 0.17

12.55 11.05 8.05 3.55

15.05 13.55 9.05 5.05

For retrogressed state 648000 596 602.5 540000 596.5 601.5 270000 592 597.0

20.5 19.5 15.0

9.6 8.4 5.5

1.42 1.09 0.46

6.55 7.05 2.55

13.05 12.05 7.55

For accreted state 648000 604.5 540000 603 270000 602

23.5 22.0 20.5

8.4 7.4 4.0

1.08 0.86 0.25

15.05 13.55 12.55

16.05 14.55 13.05

605.5 604.0 602.5

5.2 Check for undersluices

Increase in flow Concentra Concentration tion of flow Q (cusecs)

DSWL (ft)

20 % ### cusecs USWL D (ft)

For normal state 155503 594 601.5 For Retrogressed state 155503 590 601 For accreted state 155503 598.0 602.5

ho

h

Ho

(USWL-RBL)

Vo qclear/D

(Vo**2/2g)

(DSWL-CL)

(USWL-CL)

(ft)

(ft/s)

(ft)

(ft)

(ft)

19.5

13.3

2.74

7.55

15.05

19

13.6

2.89

3.55

14.55

20.5

12.6

2.48

11.55

16.05

6- Fixation of d/s Floor Levels and Length Length of d/s Glacis and d/s Floor Floor 6.1 Fixation of d/s floor levels for normal weir section using blench c Q

qclear

USEL

DSEL

(USWL+ho)(DSWL +ho)

hL

E2

(USEL-DSEL)


DSFL

(cusecs)

(cusecs/ft)

(ft)

(ft)

(ft)

Normal state of river 648000 201.3 605.68 603.18 540000 170.8 603.94 601.44 270000 83.6 59 5 98.88 597.88 For Retrogressed state of river 648000 198.8 603.92 597.42 540000 168.4 602.59 597.59 270000 84.4 59 5 97.46 592.46 For accreted state of river 648000 196.7 606.58 605.58 540000 167.8 604.86 603.86 270000 86.6 602.75 602.25 Hence d/s Floor level ## # ft

(ft)

(ft)

2.50 2.50 1.00

18.6 584.58 17.6 583.84 10 587.88

6.50 5.00 5.00

20.8 576.62 17.3 580.29 12 580.46

1.00 1.00 0.50

17 588.58 15.4 588.46 10.5 591.75

6.2 Fixation of d/s floor levels for undersluices using blench curves Q (cusecs)

qclear (cusecs/ft)

hL

E2

(USWL+ho)(DSWL +ho)

(USEL-DSEL)


(ft)

(ft)

(ft)

USEL

DSEL (ft)

Normal state of river 157414 262.36 604.24 596.74 For Retrogressed state of river 156064 260.11 603.89 592.89 For accreted state of river 165503 275.84 604.98 600.48 Hence 566.00 566 .00 ft d/s Floor level for undersluices

DSFL (ft)

7.50

27.5 569.24

11.00

26.5 566.39

4.50

28.4 572.08

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

Q Maximum DSWL USWL USEL DSFL RBL Crest level Dpool (Max. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa) d/s velocity head (V2/2g)

540000 cusecs 603.00 ft 604.00 ft 604.86 ft 577.00 ft 582 ft 589.45 ft 26 ft 5.58 ft/sec 0.48 0.48 ft

DSEL (DSWL + velocity head) K (USEL - Crest Level ) L (USEL - DSEL ) q (Q / Total width of bays)

603.48 ft 15.41 ft 1.38 ft 163.64 cusecs/f

Critical Depth, C, (q 2 / g)1/3 L/C (K+F)/C ,(from crumps curve) F, [(K+F)/C x C - K] Level of intersection of jump with glacis = Crest level - F E2, ( DSEL - Level of intersection of jump)

9.40 9.40 ft 0.15 1.92 2.65 586.80 ft

Submergency of jump, (Level of intersection of jump - DSFL ) Slope of d/s glacis Length of glacis d/s of jump, (slope x submergency) Length of stilling pool, (4.5 x E 2) Length of d/s floor,

(Length of stilling pool -Length of glacis d/s of jump)

Say

16.68 16.68 ft 9.80 9.80 ft 1: 3 29.41 f t 75.06 75.06 ft 45.64 45.64 ft 46ft

L K

y F

x

Fig:1 Various Parameters for using Crump's Curve

b)

Q Minimum DSWL USWL USEL DSFL RBL Crest level Dpool (Min. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa) d/s velocity head (V2/2g) DSEL, (DSWL + velocity head) K (USEL - Crest Level )

540000 cusecs 596.5 ft 601.5 ft 602.59 ft 577.00 ft 582 ft 589.45 ft 19.5 ft 7.45 ft/sec 0.86 0.86 ft 597.36 ft 13.14 ft

L (USEL - DSEL ) q (Q / Total width of bays)

5.23 ft 163.64 cusecs/f

Critical Depth, C, (q 2 / g)1/3 L/C (K+F)/C ,(from crumps curve) F, [(K+F)/C x C - K] Level of intersection of jump with glacis = Crest level - F E2, ( DSEL - Level of intersection of jump)

9.40 9.40 ft 0.56 2.48 10.18 ft 579.27 ft 18.09 18.09 ft 2.27 2.27 ft 1: 3 6.81 ft ft

Submergency of jump, (Level of intersection of jump - DSFL ) Slope of d/s glacis Length of glacis d/s of jump, (slope x submergency) Length of stilling pool, (4.5 x E 2) Length of d/s floor,

(Length of stilling pool -Length of glacis d/s of jump)

Say

81.42 81.42 ft 74.61 74.61 ft 75.00 ft ft

8 - Fixation of d/s floor length for undersluices a)

Q Maximum DSWL USWL USEL DSFL RBL Crest level Dpool (Max. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa)

### cusecs 598 ft 602.5 ft 604.98 ft 566.00 ft 582 ft 586.45 ft 32 ft 8.1 ft/sec


1.02 1.02 ft 599.02 ft 18.53 ft 5.96 ft 259.2 cusecs/f

Critical Depth, C, (q 2 / g)1/3 L/C (K+F)/C ,(from crumps curve) F, [(K+F)/C x C - K] Level of intersection of jump with glacis = Crest level - F E2, ( DSEL - Level of intersection of jump) Submergency of jump, (Level of intersection of jump - DSFL ) Slope of d/s glacis Length of glacis d/s of jump, (slope x submergency)

12.78 12.78 ft 0.467 2.2 9.58 ft 576.87 ft 22.15 22.15 ft 10.87 10.87 ft 1: 3 32.61 f t

Length of stilling pool, (4.5 x E 2) Length of d/s floor, (Length of stilling pool -Length of glacis d/s of jump) Say

100ft 67.06 67.06 ft 68.00 ft

Q Minimum DSWL USWL USEL DSFL RBL Crest level Dpool (Min. DSWL - DSFL) d/s Velocity (Q/(Dpool x Wa)

### cusecs 590 ft 601 ft 603.89 ft 566.00 ft 582 ft 586.45 ft 24 ft 10.80 ft/sec


1.81 1.81 ft 591.81 ft 17.44 ft 12.08 ft 259.2 cusecs/f

b)

Critical Depth, C, (q 2 / g)1/3 L/C (K+F)/C ,(from crumps curve) F, [(K+F)/C x C - K] Level of intersection of jump with glacis = Crest level - F E2, ( DSEL - Level of intersection of jump) Submergency of jump, (Level of intersection of jump - DSFL ) Slope of d/s glacis Length of glacis d/s of jump, (slope x submergency) Length of stilling pool, (4.5 x E 2) Length of d/s floor, (Length of stilling pool -Length of glacis d/s of jump) Say

Hence we shall provide d/s floor length =

12.78 12.78 ft 0.945 2.9 19.62 ft 566.84 ft 24.97 24.97 ft 0.84 0.84 ft 1: 3 2.51 ft ft 112ft 109.88 109 .88 ft 110 ft

110ft

9- Check for Adequacy for d/s floor levels using conjugate depth me 9.1 For normal weir section

Ф Floor level of stilling pool Discharge in river, Q (cusecs)

1.00 577.00 ft 540000

Discharge through main weir, Q 1= 0.8Q (cusecs) USEL (ft) DSWL (ft) E = USEL - DSFL Intensity of flow on d/s floor, q = Q1/width of main weir

432000 max. min. 604.86 602.59 603 596.5 27.86 25.59 160.00 160.00

Depth in stilling pool, D pool = DSWL - DSFL

26.00

19.50

f(z) = q/E3/2

1.088 0.150 0.643

1.236 0.170 0.671 0.

4.18

4.35

17.92

17.17

8.08

2.33 O.K

Conjugate depth coefficients Conjugate depths

z z' d1 = z x E d2 = z' x E

Jump submergency = Dpool - d2 Remarks

O.K

9.2 For undersluices section

Ф

1.00 566.00

Floor level of stilling pool Discharge in river, Q (cusecs) Disch scharge through U.S with 20% concentrati ation, (1.2 x (Q1 + Q2)) USEL (ft) DSWL (ft) E = USEL - DSFL (ft) Intensity of flow on d/s floor, q = Q1/Total width of all U/S

540 155 Max. 604.98 598.00 38.98 259

Depth in stilling pool, D pool = DSWL - DSFL

32.00

f(z) = q/E3/2

1.06 0.145 0.635

Conjugate depth coefficients Conjugate depths

z z' d1 = z x E

5.65

d2 = z' x E

Jump submergency = Dpool - d2 (ft) Remarks

24.75 7.25 O.K

10 - Scour Protection For main weir Assume flow concentration q = Qmax /(total width of bays) x 1.2

R = 0.9 (q2/f)1/3

20 % 196.36 cusecs/f 23.91 23.91 ft

10.1 - d/s scour protection for main weir

Safety factor for d/s floor critical condition Depth, R' = safety factor x R Minimum DSWL for Q max

1.75 41.84 ft 596.5 ft 596.5 577.00 ft 19.50 ft 0.50 ft 20.00 ft 21.84 ft 1: 3

d/s appron (floor) level, (DSFL) Depth of water on apron (Min DSWL - DSFL) Increase in depth due to concentration Depth of water with concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (1 2+32)x(R Length of d/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

69.05 69.05 ft 50ft MIN. W.L

D R'

2.5(RBedD)Level

DSFL

1:3 t

Deepest Possible Scour

3(R-D)

Fig:2 Scour Protection

10.2 U/s Scour Protection for main weir

Safety factor for u/s floor critical condition R' = Safety factor x R Minimum USWL for Q max U/s apron level, (RBL) Depth of water on apron = USWL -RBL Increase in depth due to concentration Total depth with concentration, concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (1 2+32)x(R Length of u/s stone apron in horizontal position

1.25 29.88 ft 601.50 ft 601.50 582 ft 19.50 ft 0.50 ft 20.00 ft 9.88 ft 1: 3 31.25 31.25 ft

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

23ft

For undersluices

Assume flow concentration q = (Q1+Q3)/Total width of undersluicesx 1.2 R = 0.9 (q2/f)1/3

20 % 259 cusecs/f cusecs/f 28.76 28.76 ft

10.3 - d/s scour protection for undersluices

Safety factor for d/s floor critical condition Depth, R' = safety factor x R Minimum DSWL for Q 1 + Q3 d/s appron (floor) level, (DSFL) Depth of water on apron (Min DSWL - DSFL) Increase in depth due to concentration Depth of water with concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (1 2+32)x(R Length of d/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

1.75 50.34 ft 590ft 566.00 ft 24.00 ft 0.5 ft 24.50 ft 25.84 ft 1: 3 81.71 81.71 ft 59ft

10.4 U/s Scour Protection for undersluices

Safety factor for u/s floor critical condition R' = Safety factor x R Minimum USWL for Q 1 + Q3 U/s apron level, (RBL) Depth of water on apron = USWL -RBL Increase in depth due to concentration Total depth with concentration, concentration, D' Depth of scour below apron = R' - D' Slope of protection Length of apron to cover surface of scour = Sqrt (1 2+32)x(R Length of u/s stone apron in horizontal position = length of apron x (1.25t/1.75t)

1.25 35.96 ft 601ft 582 ft 19 ft 0.5 ft 19.5 ft 16.46 ft 1: 3 52.04 52.04 ft 38ft

10.5 - Thickness of Aprons

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

Fall in inches/mile Sand classification Very coarse Coarse Medium Fine Very fine

3 9 Thickness of stone 16 19 22 25 28 31 34 37 40 43

12 18 pitching in inches 22 25 28 31 34 37 40 43 45 49

24 28 34 40 46 52

Type of soil Medium sand Slope of river 12 in/mile Thickness if stone pitching, t 34 in Thickness of stone apron in horizontal horizontal position = 1.75xt/slo 1.75xt/slo 5 ft Size of concrete blocks over filter 4 ft cube Summary Total Length of d/s d/s stone apron 50 ft 4 ft ft Thick bloke apron = 1/3 x total length 16 ft ft (block 5 ft Thick stone apron 34 ft Total length of u/s apron apron 4 ft ft Thick bloke apron = 1/3 x total length 5 ft Thick stone apron

23 ft 7 fftt (block 16 ft

11 - Inverted Filter Design Size of Concrete blocks Thickness of shingle (3' - 6") Thickness of coarse shingle (3/4" - 3") 3") Thickness of fine shingle (3/16" - 3/4") 3/4") Spacing b/w conc. Blocks filled with fine shingle

Spacing /Jhries (2")

4 ft cube 9 in 9 in 6 in 2 in in

Co

9"

Fig: 3 Inverted Filter

12- Design of guide banks i) Length of each guid bank measured measured in straight line along the barrage u/s , L u/s = 1.5 x Wa

5579 55 79 ft

ii) Length of each guid bank d/s of barrage, L d/s = 0.2 x Wa

743.8 ft

iii) For the nose of the u/s guide guide bank and the full length of of d/s guide bank use Lacey's depth = 1.75 x R For remaining u/s gu guide bank lacey's depth = 1.25 x R iv) Possible slope of scour v) Free board u/s Free board d/s

41.84 ft 29.88 ft 1: 3 7 ft above 6 ft above

These free boards also include allowance allowance for accretion.

vi) Top of guide bank vii) Side slope of guide bank viii) Minimum apron thickness

10 ft 1: 3 4 ft

Length of barrage, Wa Length of u/s guide bank Length of d/s guide bank Radius of u/s curved part Radius of d/s curved part

3719 ft 5579 ft 743.8 ft 600 ft 400 ft

Maximum u/s angle protected

140o

Maximum d/s angle protected

57o - 80o

12.1 Determination of levels of guide banks Merrimen's backwater formula L =

d 1− d 2 S



2

C  D − S g 1

L = length of back water curve

[     ] Φ

d 1

D

−Φ

d2

d 2

D

D

1

S


Chezy's Coefficient, C Bed slope of river, S RBL D/s HFL with accretion D = D/s HFL with accretion - RBL U/s HFL with accretion d1 = U/s HFL with accretion - RBL

71 ( max for ear 1/ 5000 582 ft 603 ft 21 ft 604.0 ft 22.0 22 .0 ft

Assume d2 (in between d1 and D)

21.8 21 .8 ft

d1/D

1.048

d2/D

1.038

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

0.8999

Ф (d2/D) L Length of guide bank Comments

0.96 6909 ft 5579 O.K

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

1.12 from barrage 604.92 ft

i) Level at at th the no nose of of u/s gu guide ba bank = W/L + free board ard ii) Level at the barrage = HFL + free board iii) Water level d/s of barrage barrage D/s free board Level of guide bank d/s = W/L + Free board

611.92 ft 607.5 ft 603.0 ft 6 ft 609.0 ft

13 - Design of Guide Bank Apron W.L D

2.5 (R' - D') T=1.07 t

T R 1 :3

Deepest Possible Scour t

Working on same lines as in section 10, Length of unlanched horizontal apron = 2.5(R' - D') Length of launched apron at 1:3 slope = 3.16(R' - D') Thickness of stone apron, t (as calculated previously) say

24.71 ft ft 31.23 ft 34 inches inches 3 ft

Volume of stones in apron = t x length launched apron Minimum thickness of unlaunched apron = 1.07t Mean Mean thick hickne ness ss of un unla laun unch ched ed apr apron = vol volu ume/ me/ 2.5 2.5(R' (R' - D') D') Maximum thickness of unlaunched apron = 2tmean - tmin

93.69 ft3/unit 3.2 ft ft 3.8ft 5.0ft

14 - Design of Marginal Bunds i) Top width width ii) Top level ab above es estimated HF HFL af after allowing 1.5ft accretio iii) Front slope of marginal bunds (not pitched) iv) Back slope to be such as to provide provide minimum cover of 2 ft, ft, over hydraulic gradient of 1:6 v) U/s water level at nose of guide bank Free board of marginal bund Hence level of marginal bund

20 ft 5 ft 1: 3

611.92 ft 5 ft 616.92 ft

Calculation of length of backwater curve:

Merrimen's equation can be used to calculate backwater length L =

d 1− d 2 S



2

C  D − S g 1

[     ] Φ

d 1

D

−Φ

d 2

D

d2

d1

L Bresse Backwater

D

Maximum USWL at Qmax RBL Normal W.L without weir d1 = Maximum USWL - RBL D = Normal W.L - RBL Slope Table for length of backwater curve d1 d2 d 1 − d 2 1 D 1

2

18 18 18 18 18 18 18 18

22.0 21.5 21.0 20.5 20.0 19.5 19.0 18.5

604.0 ft 604.0 582 ft 600 ft 22.0 ft 22.0 18 ft 1: 5000

S

C 2 − S g

3

4

5

21.5 21.0 20.5 20.0 19.5 19.0 18.5 18.1

2500 2500 2500 2500 2500 2500 2500 2000

4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4 4843.4

T 1 =

d 1 D

6

1.222 1.194 1.167 1.139 1.111 1.083 1.056 1.028

T 2 =

d 2 D

Φ1

7

1.194 1.167 1.139 1.111 1.083 1.056 1.028 1.006

  d 1

D

8

0.46 0.48 0.53 0.63 0.65 0.8 0.86 1.16

Hence length of backwater cure = 22.4 22.4 miles

Part II Design of barrage profile for sub surface flow condition 15 - Fixing of Depth of Sheet Piles Pi les Scour depth, R Depth of u/s sheet pile from HFL = 1.5 R Max. USWL for Qmax RL of bottomo fu/s sheet pile = Max. USWL - 1.5R Depth of d/s sheet pile below HFL = 2R RL of bott bottom om of inte interm rmed edia iate te shee sheett pil pile e = Max. Max. USWL USWL - 2R 2R Let RL of bottom of d/s sheet pile

604.0 604 .0 ft

589.45 589 .45 ft

582.0

1: 4

19.55 ft 30 ft 604.0 ft 604.0 574.0 ft ft 40 ft 564. 56 4.0 0 ft 548 ft

A

K

1: 3

577.00 B

L

N

574.0

P

R

75.00 M

564.0 29.81

1.5H 1.5H = 46.5 46.5

6

Q

37.35

548.0

16 - Calculation of Exit Gradient 1 H G E = x d π  λ λ H

b E

C

b1

d D

α = b/d

Parmeters Parmeters of Khosla's Curve

Let the water be headed up to Max. accreted level u/s and no flow d/s. Retogression DSFL

604.0 ft

4 ft 577.0 ft 31.0 31 .0 ft Differential head causing seepeage, H = Max. u/s WL - (DSFL - Retrogressi Depth of d/s sheet pile, d = DSFL - RL of bottom of d/s sheet p 29.0 29 .0 ft Total length of concrete concrete floor = b 194.66 ft α = b/d 1

π  λ λ

form α ~

1

curve

6.71 0.15

π  λ λ

GE

0.164 SAFE

17 - Calculation of Uplift Pressure After Applying Correctio 17.1 U/s pile line:

Length of concrete floor floor upto u/s sheet pile, b 1 Total length of concrete concrete floor, b

46.5 ft 46.5 194.66 ft

Depth of u/s sheet pile,d Assume tf u/s floor thickness 1/α = d/b α = b/d b1/b 1 - b1/b Φ B = Φ D From khosla's curve Φ A = Φ E From khosla's curve

8.0 ft 2.5ft 0.0411 24.33 0.239 0.761 66 % 67 % 61 %

100 - 34 100 - 33

Φ K = Φc From khosla's curve

i) Correction for floor thickness t f

Correction in

Φ K =

Correction in

Φ A=

 Φ B−Φ K 

1.563 %

d t f

-ve

 Φ −Φ B  d A

0.31 %

ii) Correction for interface of sheet pile Correction in

Φ K

[ ]

due to second pile =19 d  D b

Depth of u/s sheet pile,d D = RBL - RL of bottom of second pile Total length of concrete concrete floor, b Distance between two piles, b' Correction in Φ K

D b '

8.0 ft 18.0 ft 194.66 ft 73.16 ft +ve 1.26 %

iii) Slope correction for Φ K Correction for Φ =− F K S

b s b1

For 1:4 slope, Fs (from slope correction curve) 3.3 Distance between two piles, b1 73.16 73.16 ft Horizontal projection of u/s glacis, bs = (crest level - RBL) x 1/slo 29.81 29.81 ft -1.34 Correction for Φ K Hence Correted Φ A 66.69 % Corrected Corrected Φ B 66 % Corrected Corrected Φ 62.477 62.4 77 % K

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

Assume floor thickness DSFL

10 ft 577.00 ft

RL of intermediate sheet pile d = DSFL - RL of Intermediate sheet pile Total length of concrete concrete floor, b Length of concrete floor up to sheet pile, b1

564.00 ft 13.00 ft 194.66 ft 119.66 119 .66 ft

b1/b

0.615

1 - b1/b α = b/d Φ L = Φ E Φ M = Φ D

0.385 14.97 52 % 54 % 36.5%

Φ N =Φ

C

100 - 48 100 - 46

From khosla's curve From khosla's curve From khosla's curve

i) Correction due to floor thickness Correction Correction in Φ L =

t f

1.54 %

t f

13.46 %

Correction Correction in Φ N =

 Φ −Φ M  d L  Φ −Φ N  d M

ii) Correction due to interference of pile

[ ]

d  D Correction in Φ L due to u/s sheet pile 19 b

D b '

Depth of Intermediate sheet pile,d D = RBL - RL of bottom of u/s sheet pile Total length of concrete concrete floor, b Distance between two piles, b' Correction in Φ L

[ ]

= 1 9 d  D b

D b '

-1.50 %

[ ]

d  D Correction in Φ N due to d/s sheet pile19= b Depth of Intermediate sheet pile,d D = DSFL - RL of bottom of d/s sheet pile Total length of concrete concrete floor, b Distance between two piles, b'

[ ]

Correction Correction in Φ N = 1 9

d  D b

D b '

13.0 ft 18.00 ft 194.66 ft 73.16 ft

D b '

13.00 ft 29.00 ft 194.66 ft 75.00 ft 2.55 %

iii) Slope correction for Φ L for '1:3 slope, Fs

4.5

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

37.35 37.35 ft

Distance between two piles, b1

73.16 73.16 ft

Correction in Φ = F ×

b s

2 30 %

1

Hence Corrected Φ L Corrected Φ M Corrected Φ N

54.34 % 54 % 52.51 %

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

Assume floor thickness Depth of d/s sheet pile, d Total length of concrete concrete floor, b 1/α = d/b Φ P = Φ E From khosla's curve ΦQ = Φ D From khosla's curve Φ R= ΦC From khosla's curve

100 - 68 100 - 78

7 ft 29.00 ft 194.66 ft 0.149 32 % 22 % 0%

i) Correction due to floor thickness Correction Correction in Φ P Correction Correction in Φ R

-2.41 % 5.31 %

ii) Correction due to interface of piles Correction in Φ P Depth of d/s sheet pile,d D = DSFL - RL of bottom of intermediate sheet pile Total length of concrete concrete floor, b Distance between two piles, b'

[ ]

Correction in Φ P =1 9 d  D b Hence Corrected Corrected Φ P Corrected Corrected ΦQ Corrected Corrected Φ R

D b '

29.00 ft 13.00 ft ft 194.66 ft 75.00 ft -1.71 %

27.88 27. 88 % 22 % 5.31 % 5.31

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

Symbol used in Khosla cueve Φ E Φ D

u/s Pile line Φ A = 66.69% 66% Φ B =

Intermediate Line Φ L = 54.34% 54% Φ M =

d/s Pile Line Φ P = 27.88% ΦQ = 22%

Φ C

Φ K =

62.48%

Φ N

=

52.51%

Φ R

=

18 - Calculation For Floor Thickness: Φ × H 10 0  G − 1 

t f =

where tf = Thickness of floor in ft Φ = % Uplift pressure

H = Maximum differential head causing seepage G = Specific gravity of concrete

2.4

a) Thickness of floor at A Assumend thickness H Thickness from uplift pressure say

2.5 ft 27.00 ft 12.86 ft 13.00 ft ft

say

10 ft 10.48 ft 11.00 ft ft

say

10 ft 10.13 ft 11.00 ft ft

say

7 ft 5.38 ft 6.00 ft

b) Thickness of floor at L Assumend thickness Thickness from uplift pressure c) Thickness of floor at N Assumend thickness Thickness from uplift pressure d) Thickness of floor at P Assumend thickness Thickness from uplift pressure e) Thickness of floor at crest

K

x

Φ K

L

Φ L Pressure at

5.31%

y

Φ −Φ Uplift pressure at crest=Φ L  K L  x  y Hence Thickness of floor at crest d/s of of gate

58.49 %

say

11.28 ft 12 ft

Eo

h/Eo

C'/C

C'

qclear

Q

(cusecs/ft)

(cusecs)

(Gibson)

(USWL+ho-CL)

(ft)

16.23 14.49 9.43 5.21

0.77 0.76 0.85 0.68

0.810 0.815 0.760 0.920

3.08 3.10 2.89 3.50

201.3 170.8 83.6 41.6

664221.95 563797.82 275999.79 137391.44

O.K

14.47 13.14 8.01

0.45 0.54 0.32

0.950 0.930 0.980

3.61 3.53 3.72

198.8 655955.61 168.4 555619.13 84.4 278626.51

O.K

17.13 15.41 13.30

0.88 0.88 0.94

0.730 0.730 0.470

2.77 2.77 1.79

196.7 649176.95 167.8 553639.38 86.6 285743

O.K

Eo

h/Eo

C'/C

C'

qclear

Q

(cusecs/ft)

(cusecs)

O.K O.K O.K

O.K O.K

O.K O.K

(Gibson)

(USWL+ho-CL)

(ft)

17.79

0.42

0.92

3.50

262.36 157414.34 Err:508

17.44

0.20

0.94

3.57

260.11 156063.65 Err:508

18.53

0.62

0.91

3.46

275.84 165503.38 Err:508

urves

hod.

270000

216000 max min 602. 60 2.7 75 597 97.4 .46 6 602 597 25.75 20.46 80.00

80.00

25.00

20.00

0.612 0.080 0.504

0.864 0.115 0.585

2.06

2.35

12.98

11.97

12.02 8.03 O.K O.K

ft 000 503 Min. 603.89 590.00 37.89 259 24.00 1.11 0.155 0.630 5.87 23.87 0.13 O.K

4'x4'x4')

4'x4'x4')

ncrete Blocks (4'x4'x4')

9" Gravel Coarse sand

6" Sand

HFL HFL

then channels)

level

idth

Φ2

  d 2

D

9

0.48 0.53 0.63 0.65 0.8 0.86 1.16 1.59

Φ 2− Φ1 10

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

12

0.02 2145 4645 0.05 4760 7260 0.09 7916 10416 0.02 1744 4244 0.15 13103 15603 0.06 5301 7801 0.3 26033 28533 0.43 37759 39759 Total 118260 ft 22.40 22.40 miles miles

n

Table F Conjugate depth d 2 for different discharges under gated and ungated flows. Basic Data: b EL 700000 842000 950000

Clear width of the weir section of the barrage Barrage Crest Level Ungated Discharge Ungated Discharge Ungated Discharge

2520 678 692.74 694.55 695.84

H = Z − d / 2

Theoretical velocity for gated control flow

H = Z − h / 2 − d / 2

For ungated flow

Trial 1 q=Q/b (cfs/ft)

Q (cfs)

ft ft ft ft ft

H (ft)

Vth ft/sec

Vact ft/sec

d1 (ft)

H

Trial 2 Vth Vact d1 FR1 (ft) ft/sec ft/sec ft

d2 / d1

d2

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 300000

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

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 700000

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

208.33

15.37

31.46

26.74

7.79

11.47

27.18

23.11

9.02

1 .36

1.48

13.36

683.36

250.00

16.28

32.37

27.52

9.08

11.73

27.49

23.36

10.70

1.26

1.35

14.43

684.43

282.74

16.92

33.01

28.06

10.08

11.88

27.66

23.51

12.02

1.19

1.26

15.18

685.18

840000 950000

EL694

EL 678 d

Z

EL673 1:3 EL 670

h EL 678 d

Z

EL673 1:3 EL 670

Part I Design of barrage for overflow condition 1 2 3 4 5

6

7 8 9

10

11 12 13 14

Minimum stable wetted perimeter Calculation of Lacey's silt factor Fixation of crest level Design of undersluices Determination of water levels and energy levels 5.1 Check for main weir 5.2 Check fo undersluices undersluices Fixation of d/s floor levels and length of d/s glacis and d/s 6.1 Fixation of d/s floor levels levels for normal weir section using using bl 6.2 Fixation of floor floor levels for undersluices undersluices Fixation of d/s floor level for normal barrage section using Crump's method and determination of floor length Fixation of d/s floor length for undersluices Check for the adequacy for d/s fl floor levels using conjugate depth method 9.1 For normal normal weir section section 9.2 For undersluice undersluice section section Scour protection 10.1 d/s scour protection protection 10.2 u/s scour protection protection 10.3 Thickness of of aprons Inverted filter design Design of guide banks 12.1 Determination Determination of levels of guide banks banks Design of guide bank aprons Design of marginal bunds

Part II Design of barrage profile for sub surface flow condition 15 16 17

18

Fixation of depth of sheet piles Calculation of exit gradients Calculation of uplift pressure after applying correction 17.1 u/s pile pile length length 17.2 Intermediate Intermediate sheet pile at toe of d/s glacis 17.3 d/s sheet pile at the end of impervious impervious floor Calculation of floor thickness

days Done

0

floor nch curves

Done

0 Done

0 Done

0

Done

0 Done Done

0 0

Done

0

Done

total

0

0

days days left left

Design of a Barrage

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