Barrages

 Varshney, R.S., Gupta, S.C., Gupta, R.L., “Theory and Design of Irrigation Structure”. N.C. Jain, at Roorkee press, Roorkee, India, 1982.  Garg, S.K., “Irrigation Engineering and Hydraulic Structures”, Khanna Publishers, Delhi, 1976.  Abdulkadir, S. Z., Mero, S.K., Yaseen, D.S. “Design of Barrage on a Proposed River”, 4th Year Graduation Project, 2008.

Page  2

In addition to the weir proper, head works consist of the following components,  Undersluices.  Canal head regulator.

 Divide wall or groyne.  Fish ladder.  Piers and abutments.  Protection works.  River training wall.

Page  3

These are gates controlled openings in the weir with crest at low level. They are located on the same side as off-take canal. If two canal take off on either side of the river, it would be necessary to provide undersluices on either side.

Page  4

Functions of undersluices

 To preserve a clear and defined river channel approaching the canal regulator.  To scour silt deposited in front of canal regulator and control silt entry in the canal.  To facilitate working of weir crest shutters or gates. The flood can easily pass.

 To lower the highest flood level.

Page  5

Discharge Capacity of undersluices is provided of the following:

i.Qu = 2 (Qmax.)offtake ii.Qu = 20% (Qmax.)flood

Page  6

A canal head regulator is to serve the following functions:

Regulate the supply of water in the canal. Control the entry of silt in the canal.

Page  7

The head regulator is normally aligned between 90° - 120° in respect to the axis of the weir. The regulation done by means of gates, steel gates of spans ranging between (8 m-12 m) are used and operated by electric winches. Height of gates = pond level - crest level To check flood water entering the canal a breast wall between pond level and high flood level (H.F.L). Unless H.F.L - pond level is nominal, breast wall is usually more economical than high gates.

Page  8

If silt excluder is provided, it is necessary to further raise the crest of head regulator by minimum of 0.75 m. To calculate the discharge, the drowned weir formula can be used: 2 𝑄 = 𝐶1 𝐿 2𝑔 𝐻 + 𝑕𝑎 3

2

3

− 𝑕𝑎

2

where: C1=0.577, C2=0.8

H=difference of u.s and d.s water levels. L=clear length of waterway. d=depth of d.s water level above the crest.

ha=head due to velocity of approach.

Page  9

3

+ 𝐶2 𝐿𝑑 2𝑔 𝐻 + 𝑕𝑎

 In case of the waterway (L) worked out more than width of the canal, the crest level so adjusted as to keep (L) equal to the width of the canal.  In exceptional cases waterway more than canal width may be provided with a flared wall in the d.s of the regulator to join the canal width.  Note that, the most critical condition of uplift occurs when high flood level is passed down the weir and there is no flow in canal.

Page  10

It is a wall located between weir and undersluices extending a little u.s of canal regulator, and d.s up to end of loose protection of the undersluices. It is a concrete or masonry structure, with top width (t)=(l.5-3)m, and aligned at right angle to the weir axis.

Page  11

The functions of divide walls are

 To separate the floor of scouring sluices which is at lower level than the weir proper.  To isolated the pockets u.s of the canal head regulator to facilitate scouring operation.  To prevent formations of cross currents to a void their damaging effects. Additional divide walls are sometimes provided for this purpose.

Page  12

 The divide walls are costly structures.

 These walls are likely to be subjected to maximum differential pressure when the full discharge of the river is passing through the weir, (there will be difference in water level on the two sides ).  Also there may exit difference in silt pressure on the two side . The values of differential pressure are taken arbitrarily say 1.0 m for water heads and about 2.0 m for silt pressure (see Fig. (5)).

Page  13

Fish ladder or fish passes are generally provided to enable the fish to ascend the head waters of the river and thus reach their spawning grounds for propagation or to follow their migratory habits in search of food.

Page  14

The general requirements of a fish ladder are:

 The slope of the fish ladder should not be steeper than 1:10 (i.e velocity not exceeding 2 m/s in any portion of the fish-way).  The compartments of bays of the pass must be such dimensions that the fish do not risk collision with the sides and upper end of each bay when ascending.

Page  15

The general requirements of a fish ladder are (Continued)

 Plenty of light should be admitted in the fish-way.  The water supply should be ample at all times.  The top and sides of a fish-way should be above ordinary high water level.(see fig.(6)).

Page  16

In barrages piers are provided at an interval of 10 to 20 m. The piers support bridge decking, and working platform for the operation of gates. Cutwaters are usually simple in shape and the side face of piers is often vertical. Tapering if done, does not exceed 1/50 to 1/40. Piers should be provided with separate foundation (see Fig. (5)).

Page  17

 In case, however, when raft is provided the piers may be constructed monolith with floor.  Abutments are usually gravity section and founded on well packed closely in either direction. Perhaps in case of higher abutments a better alternative is to provide a counter fort type structure on open foundation.

Page  18

The sediment transporting capacity of water depends mainly on the discharge, slope and grade of material. As a result of putting obstruction across the river in form of weir, the river regime will be affected in the following sequence:

Page  19

 A weir will pond up water leading to the flattening of water surface slope for some distance on the up-stream side.  River would drop a part of its sediment load resulting in the formation of shoals in the pond.  Clear water passes over the weir; this water scours the bed river to make up deficiency in its silt load and causes a progressive lowering or retrogression of downstream level.

Page  20

 The progressive silting a formation of shoals in the u.s increases the resistance to flow of water to recover this resistance increase head is required.  The overall effect of a weir is to take away the excess energy due to steep gradient by localizing it and ultimately dissipating it.

Page  21

The causes of failures may classify into:

 Failure due to seepage or subsurface flow.  Failure due to surface flow.

Page  22

Subsurface flow endangers the stability of a weir in the following two ways:  Piping or floatation.  Uplift pressure.

The surface flow endangers the weir in the following two ways:  Unbalanced head due to standing wave (i.e. due to formation of the jump very high unbalanced pressures are developed in the trough).  Scour in the upstream and downstream.(this occurs in the bed of alluvial rivers at flood).

Page  23

The design of weir and barrage like any hydraulic structure, consist of many phases. The two main phases are  The hydraulic design (evaluation hydraulic forces acting on the structure).  The structural design (dimensioning of the various part of the structure to enable it to resist safely all the forces acting on it).

Page  24

The problem involved in the hydraulic design of weir and barrage on permeable foundation may be treated under the following:  Sub soil flow, and  Surface flow.

Page  25

Page  26

a) Crest level of a barrage is fixed on the consolidation of existing river bed level, at the proposed site. b) The undersluices crest is usually kept as near the bed level in the deepest channel as is practically possible.  The barrage bay crest is kept slightly higher and at about the general bed level in the remaining portion of the river.  The undersluice crest is kept lower to attract a deep current in front of the regulators, so that the dry weather current may remain near the regulator.

Page  27

 Note that afflux and discharge per meter are related to crest levels. So lower crest levels, result in lesser afflux, but higher discharge per meter. Low set barrage with increase depth of water over crest may result in an increased height of gates, thickness of floor, and cost of superstructure above floor level.

Page  28

It's necessary to check that the maximum. Flood discharge passes down the works without exceeding the afflux. The following discharge formulas may be used for this purpose: a) For broad crested weir:

𝑄 = 1.705 𝐿 − 0.1 𝑛𝐻 𝐻3

2

b) For sharp crested weir:

𝑄 = 1.84 𝐿 − 0.1 𝑛𝐻 𝐻3 where: L= total clear waterway (m)

n= number of end contractions H= the head over crest in (m) Page  29

2

Afflux actually denotes loss of head and its magnitude is represented by the difference in total energy level on upstream and the downstream of the works. Afflux is generally limited to 1 meter, but may be kept higher if permissible. To adopt for waterway is given by the following formula representing Lacey's wetted perimeter, 𝑃 = 4.83 𝑄 where: P= Lacey's wetted perimeter, and Q= Maximum flood discharge.

Page  30

In boulder reaches of the river it would be economical to reduce the waterway to about (0.6 - 0.8) time Lacey's waterway. In plains where the silt factor is in the neighborhood of unity it is generally economical to keep the waterway (1.0 - 1.2) times the Lacey's waterway. Generally the shorter waterway is preferable.

Page  31

1. Maximum flood discharge "Q“ 2. Stage discharge curve of the river at barrage. 3. Minimum water level.

Page  32

4. Cross section of the river at barrage site the following have to be decided:  Lacey's silt factor (f) this is determined from the equation: 𝐹 = 1.76 𝑀𝑟

 Length of waterway, discharge per meter and afflux.  Safe exit gradient.  Depth of sheet piles: • Scour depth. • Exit gradient.

Page  33

5. Level and length of horizontal part of d.s. impervious floor in coordination with hydraulic jump. 6. Thickness of d.s. impervious floor. •

with reference to uplift pressure.

•

with reference to hydraulic jump or standing wave.

7. Length and thickness of protection works beyond pucca floor u.s. and d.s.

Page  34

Page  35

Determine head loss (HL) for different flow condition. HL = afflux (if there is no retrogression)* If allowance for retrogression is taken in d.s bed level then, HL = afflux + retrogression, usually, 0.5 m retrogression will be sufficient in most cases.

* Retrogression is lowering d.s bed caused by water scour

Page  36

For known values of q and HL find Ef2 from blench curve (from Blench curves (Next slide)). With known values of Ef2 read corresponding values of D2. Cistern Level = D.S.T.E.L – Ef2

Ef2 = down stream specific energy. D2 = post jump corresponding to Ef2 Ef1 = Upstream specific energy. D1 = per jump depth corresponding to Ef1

Page  37

Page  38

Ef1 = Ef2 + HL Knowing Ef1 , Ef2 & q read values of D1 and D2 from Fig. 10 (next slide), energy of flow curves. Provide minimum cistern length = 5(D2–D1). = 6(D2–D1).

Page  39

Page  40

Determine scour depth from the formula

𝑅 = 1.35

1 3 2 𝑞

𝑓

Depth of u.s. sheet pile =(1–1.25)R Depth of d.s. sheet pile =(1.25–1.5)R

Page  41

Work out the value 1 𝜋 𝜆

GE =h/d.

1 𝜋 𝜆

=

from equation:

𝑑 G.E. 𝑕

1 𝜋 𝜆

∝ =[(2𝛌–1)²–1]½ b=∝ 𝑑

Page  42

(for given GE.)

Provide total length of floor = ∝ 𝑑

Disposition of total floor length may be as follows 1. Cistern length = 5(D2–D1) to 6(D2–D1) 2. Glacis length = 3 to 5 times (crest level – cistern level) For 3:1 to 5:1 slope of glacis 3. Upstream floor = the balance.

Page  43

Find u.s pile and d.s pile % pressure; the pressure distribution assumed to be linear; used Khosla’s method; correct due to floor thickness, interference of sheet pile and slope.

Page  44

Page  45

Plotting water profile before and after the jump formation (Fr=

𝑞 𝑞𝐷1³

)

The intercept between the profile of hydraulic jump and the gradient gives the unbalance dynamic head. The floor thickness is, however, designed for ⅔ the max. unbalance head in jump through.

Floor thickness = H/(G–1) where: H = max static head or ⅔ max. unbalance head. (use the greater head of the two)

Page  46

The protection works are now designed in respect to the scour depth.

Page  47

Page  48

A barrage is to be constructed on Tigris River having high flood discharge 8500 cumecs. The relevant data are as follows: Average bed level of river

100 m

High flood level

105.75 m

Normal pond level

103.5 m

Permissible afflux

1m

Lacey silt factor

1m

Safe exit gradient for bed material

1/6

Concentration

20٪

Bed retrogression

0.5

Pier contraction coefficient

0.1 m

Page  49

Design a suitable regulator with reference of the barrage with the following data: 1. Full supply discharge of off-take canal

=220 cumecs

2. full supply level of canal

=102 m

3. water depth in canal

=3.5m

4. bed level of canal

=98.5 m

5. angel of off take canal

=115˚

6. safe exit gradient

=1/6

7. side slopes

=2H:1V

Page  50

-Crest of under sluice is to be provided at bed level of the river = 100 m

-Crest level of other barrage is to be (1.25)m higher than under sluice crest = 100+1.25=101.25 m -Minimum water way = 4.83 𝑄. from Lacey new edition using factor (4.75) instead of (4.83)‚ but the factor (4.83) is used for more safe. = 4.83 𝑄 = 4.83 8500 = 445.3 𝑚

Page  51

-Assume water way is as follows:

Under sluice water way: -

5 bays (13 m) each

=65m

-

4 piers(2.5m) each

=10m

Total under sluice

=75m

Other barrage water way -

30 bays (10m) each

=300m

-

29 piers(2m) each

=58m

-

Total other barrage

=358m

-

Divide wall thickness

=3m

Total water way Page  52

= 75+358+3=436m

Page  53

D.S.H.F.L =105.75 m

U.S.H.F.L =D.S.H.F.L + Afflux =105.75+1=106.75 m q =Q/B =8500/436=19.5 cumec /m 1 3

𝑅 = 1.35

𝑞2 𝑓

𝑞 V= 𝑅

1.99 m/sec

19.5 = = 9.78

= 1.35

9/52 1

1 3

= 9.78 𝑚

Velocity head =V²/2g =1.99²/2(9.81) =0.2 m U.S.T.E.L =106.75+0.2 = 106.95 m

Page  54

Head over under sluice crest =106.95-100 = 6.95 m

Head over other barrage crest =106.95-101.25=5.7 m Qunder sluice = 1.705(L-0.1nH) H3/2 =1.705(65-0.1×8×6.95)(6.95)3/2 =1856.87 m³/sec Qother barrage =1.84(L-0.1 nH) H3/2 =1.84(300-0.1×58×5.7) ×5.73/2 =6684.1 m³/sec Qtotal 1856.87 + 6684.1 = 8540.98 m³/sec >8500 m³/sec

OK Page  55

2.2.1 High flood condition

I. For high flood without concentration and retrogression q =C H3/2 =1.705(6.95)3/2=31.24 cumec/m D.S.H.F.L= 105.75 m U.S.H.F.L= 106.75 m U.S.T.E.L= 106.95 m D.S.T.E.L=105.95 m HL = U.S.T.E.L-D.S.T.E.L =106.95- 105.95 = 1 m

Page  56

II. For high flood with 20% concentration and 0.5 m bed retrogression q =1.2×31.24 = 37.488 cumec/m q =C H3/2 𝑞 𝐶

H=( )2/3 =(

37.488 2/3 ) 1.705

=7.848 m

U.S.T.E.L= 100+7.848 = 107 .848 m

D.S.H.F.L with 0.5 retrogression = 105.25 m D.S.T.E.L=105.25+0.2 = 105.45 m HL= 107.848- 105.45 = 2.398 m

Page  57

2.2.1.2 Flow at pond level

Ι. without concentration and retrogression Head over under sluice crest = 103.5-100 =3.5 m Head over other barrage crest = 103.5-101.25 =2.25 m Q =1.705(65-0.1×8×3.5×3.53/2 +1.84(300-0.1×58×2.25) ×2.253/2 = 694.41 + 1781.959 = 2476.369 m³/sec 𝑄 𝐵

q= =

Page  58

2476.369 436

=5.68 m³/sec/m

Page  59

𝑞² 1/3 R=1.35( ) 𝑓

= 4.297 m

𝑞 𝑅

V= =1.32 m/sec 𝑉² hv = 2𝑔

=0.089 m

U.S.T.E.L =103.5+ 0.089 = 103.59 m D.S water level when flood discharge of 2476.369 m³/sec is passing = 102.7 m

Page  60

From stage discharge curve Fig .8

D.S.T.E.L =102.7 + 0.089 = 102.789 m q between piers = C H3/2

= 1.705(3.5) 3/2= 11.164 m³/sec/m HL = 103.59- 102.789 = 0.801 m Page  61

II. with 20% concentration and 0.5 m retrogression

q = 1.2×11.164 = 13.3968 cumec/m head for this intensity =(

13.3968 2/3 ) 1.705

=3.9523 m

U.S.T.E.L =100+3.9523 = 103.9523 m D.S.W.L with 0.5 m retrogression = 102.7-0.5 =102.289 m

D.S.T.E.L = 102.2 + 0.089 =102.289 m HL = 103.9523-102.289 = 1.663 m

Page  62

2.2.2 Hydraulic jump calculations High Flood No.

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

Item

Discharge q D.S.W.L U.S.W.L D.S.T.E.L U.S.T.E.L HL Ef2 Ef1=Ef2+HL Level at which jump occurs D1 D2 Length of floor 5(D2-D1) 𝐪 √𝐠𝐃𝟏³

13  63 Page F=

Pond Flow

31.24 105.75 106.75 105.95 106.95 1 8 9

With retrogression and concentration 37.488 105.25 106.75 105.45 107.85 2.40 9.75 12.15

Without retrogression and concentration 11.16 102.7 103.5 102.79 103.59 0.80 4.20 5.00

With retrogression and concentration 13.39 102.2 103.5 102.29 103.95 1.663 5.07 6.733

97.95

95.70

98.59

97.22

2.81 6.85

2.70 8.63

1.25 3.71

1.25 4.51

20.2

29.68

12.3

16.3

2.117

2.70

2.55

3.06

Without retrogression and concentration

2.2.2 Hydraulic jump calculations

Provide the D.S floor at elevation 95.65 m with horizontal length of 32 m

Page  64

Table (2):the water surface profile u.s of the jump

Distance from u.s end of crest

U.S.T.E.L = 107.848 m q= 37.488 cumec/m R .L of glacis

High flood

Ef1=U.S.T.E.LR.L of glacis 3.00 4.23 6.00 6.15 8.34 9.00 12.00 12.90 Page  65

99.00 98.59 98.00 97.95 97.22 97.00 96.00 95.70

8.848 9.259 9.848 9.898 10.629 10.848 11.848 12.148

U.S.T.E.L=103.952 m q=13.396 cumec/m Pond flow

D1

3.76 3.50 3.31 3.25 3.10 3.00 2.76 2.60

Ef1=U.S.T.E.LR.L of glacis 4.952 5.363 5.952 6.002 6.733 6.952 7.952 8.252

D1

1.75 1.70 1.45 1.40 1.35 1.30 1.15 1.10

Page  66

Table (3): the water surface profile d.s the jump SL No .

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Page  67 16

𝑋 𝐷1 1 2 4 6 8 10 11.88 12 14 16 18 20 22 24 25.6 26.4

High Flood Condition Fr=2.698, Fr²=7.279, D1=2.7 m 𝒀 𝑫𝟏 1.3 1.70 2.40 2.7 2.9 3.09 3.21

X

Y

2.7 5.4 10.8 16.2 21.6 27 32

3.51 4.6 6.48 7.28 7.83 8.34 8.66

Pond Flow Condition Fr=3.061, Fr²=9.37, D1=1.25 m 𝒀 𝑫𝟏 1.3 1.70 2.40 2.75 3.10 3.40 3.6 3.65 3.85 3.90 3.95 4.00 4.00 4.00 4.00 4.00

X 1.25 2.5 5.0 7.5 10.0 12.5 14.8 15.0 17.5 20.0 22.5 25 27.5 30 32 33

Y 1.625 2.125 3.00 3.44 3.875 4.25 4.50 4.56 4.81 4.88 4.94 5.00 5.00 5.00 5.00 5.00

Page  68

2.2.3 Depth of sheet pile lines from scour condition

Ι. Depth of scour Total discharge passing through under sluice =1856.87 cumec Under sluice over all water way = 75 m q =average discharge =

1856.87 75

=24.758 m²/sec

1/3 = 11.468 m R (depth of scour) = 1.35(𝑞² ) 𝑓

Page  69

II. U.S sheet pile

On the U.S side allow

1.1R = 1.1 ×11.468 = 12.615 m

R.L of bottom of scour hole =106.75-12.615 = 94.135 m Provide sheet pile line at elevation 94 m Depth of U.S sheet pile = 100-94 = 6 m

Page  70

III. D.S sheet pile

On D.S sheet pile allow 1.25 R = 1.25×11.468 = 14.34 m R.L of bottom of scour hole =105.25-14.34 =90.91 m Provide D.S sheet pile line at elevation 90 m D.S sheet pile depth = 95.65 - 90 = 5.65 m

Page  71

2.2.4. Total floor length and exit gradient:

Safe exit gradient =

1 6

Maximum static head =103.5 – 95.65 =7.85 m Depth of D.S sheet pile = 95.65 – 90.0 = 5.65 m 𝐻 1 1 7.85 1 𝐺𝐸 = , = 𝑑𝜋 𝜆

6

5.65 𝜋 𝜆

λ=7.704, 𝛼 = [ 2𝜆 − 1

Page  72

2

− 1]½=13.0437

b = 𝛼𝑑

b = 13.0437 ×5.65 = 73.697 m provide D.S sheet pile = 6 m λ = 6.243, 𝛼 = 11.44 𝑚 b = 6.243 × 11.44 = 68.64 m provide total floor length = 69 m provide D.S sheet pile line at elevation 89.65 m

Page  73

Floor length should be distributed as following:

D.S horizontal floor =32 m D.S glacis length =3(100- 95.65) =13.05 m The balance should be provided as U.S floor = 23.95 m Total length = 69 m

Page  74

Assume U.S floor thickness 1.0 m and D.S floor thickness 1.5 m

I. Upstream pile line: d =100 – 94 = 6 m 1 𝛼

=

𝑑 𝑏

=

6 69

=0.084

∅𝐷1 = 100% − ∅% , ∅𝐷 = 18% ∅𝐷1 = 82% ∅𝐶1 = 100% − ∅𝐸% , ∅𝐸 = 27% ∅𝐶1 = 73%

Page  75

Correction of Thickness: 𝑡 𝑑

Ct = (∅𝐷1 − ∅𝐶1 ) 1 6

= (82%-73%) = 1.5%

Page  76

Correction of Interference of Piles:

D =(100 – 1) – 89.65 = 9.35 m d =(100 - 1) - 94 =5 m b'=69 – 0.5 – 0.5 =68 m 𝐷 𝐷:𝑑 ) 𝑏′ 𝑏

cp =19√ (

9.35 9.35:5 =19√ ( ) 68 69

= 1.465٪

Corrected ∅𝐶1 =73% +1.5% +1.465% =75.965% Page  77

II. D.S pile line

Depth of d.s sheet pile = 6 m 1 ∝

𝑑 𝑏

= =

6 69

=0.08695~ 0.087

∅𝐸 = 27٪ , ∅𝐷 = 18٪ Correction of thickness: 𝑡 𝑑

ct = - (∅𝐸 − ∅𝐷) =-

Page  78

1.5 ( 6

27 – 18 ) = - 2.25٪

Correction of interference:

d =(95.65 – 1.5 ) – 89.65 =4.5 m D =(95.65 – 1.5 ) – 94 =0.15 m 0.15 0.15:4.5 ( )= 68 69

Cp = - 19 √

- 0.06 ٪

Correction of ∅𝐸 = 27 − 2.25 – 0.06 =24.69٪

Page  79

condition D.S water level datum (m) 1 No flow maximu m static head High Flood

Flow at pond level

Page  80

U.S water level datum (m)

H

2

3

4

95.65

103.5

7.85

105.25

106.75

102.289 103.5

𝑕𝑒𝑖𝑔𝑕𝑡 Of sub Soil H. 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 U.S Pile line

G line

above

datum

D.S Pile line

∅𝐸 100٪ 5 7.85

∅𝐷1 82٪ 6 6.437

∅𝐶1 ∅𝐸 75.965٪ 24.69٪ 7 8 5.963 1.938

∅𝐷 18٪ 9 1.413

∅𝐶 0 10 0

103.5 1.5

102.087 1.23

101.613 97.582 1.14 0.3704

97.063 0.27

95.65 0

106.75 1.211

106.48 0.9930

106.39 0.9199

105.52 0.2180

105.25 0

103.5

103.282

103.210 102.588

1.5 105.620 0.2990

1.211 102.507 102.289

1. down stream floor thickness : Distance from D.S end

Page  81

Static head (m)

3

2.0860

6 9 12 15 18 21 24 27 30 32

2.2636 2.4411 2.6187 2.7963 2.9738 3.1514 3.3290 3.5066 3.6841 3.8025

Dynamic head (m)

Minimum thickness from calculation 2 1.68 1.6* =1.067 3 1.33 1.83 1.467 1.97 1.87 2.11 2 2.26 2.267 2.40 2.67 2.54 3.267 2.68 4.0 3.22 4.867 3.92 5.13 4.13

Provided thickness

1.70 1.90 2.00 2.2 2.40 2.5 2.70 2.80 3.3 4.0 4.2

 Provide 4.0 m of floor thickness extend by 3 m inside beyond the toe of glacis. 2.U.S floor thickness: The floor thickness is required however minimum thickness of shall be provided.

Page  82

1m

Ι. Upstream protection

a) Block protection R = 11.467 m Provide 1.5 R = 1.5×11.467 =17.20 m R.L of bottom of scour hole =106.75 – 17.2 =89.55 m Scour depth below U.S floor = 100 – 89.55 = 10.45 m say 11 m Volume of block = D m³/m = 11m³/m Provide 1.6m × 1.6m ×1.0m c/c block over 0.4 m thick graded filter.

Length Page  83

11 = 1:0.4

= 7.857 m

Provide 5 row of c/c block in a length of 8 m

Ι. Upstream protection (continued)

b) Launching apron provide 2.25 D m³/m = 2.25 * 11 =24.75 m³/m Length =

24.75 1.4

= 17.67857 m

Provide launching apron of a length of 18 m and 1.4 m thickness.

Page  84

II. Down stream protection

Anticipated scour depth = 2 R = 2 ×11.467 = 22.934 m R.L of bottom scour hole = 105.25 – 22.934 = 82.316 m Depth of scour hole below D.S floor = 95.65 – 82.316 = 13.334 m Say 13.4 m

Page  85

II. Down stream protection (continued)

a) Block protection Length = 1.0 D = 1.0 × 13.4 = 13.4 m Provide 1.6 × 1.6 × 1.0 c/c block with 10 cm gaps filled with bajri over 1.0 m thick graded filter . No. of rows required =13.4/(1+1) = 6.7 m Provide 7 rows of blocks in a length of 11.9 m

Page  86

II. Down stream protection (continued)

b) Launching apron Thickness of launching apron = 2.0 m Length of launching apron =

2.25×13.4 2

=15.0075 m

Provide 20 m length of launching apron in the downstream.

Page  87

Page  88

Page  89

I) High flood level

i) Without concentration and retrogression U.S.W.L = 105.75 + 1 = 106.75 m U.S.T.E.L = 106.75 + 0.2 = 106.95 m Head over other barrage crest = 106.95 – 101.25 = 5.7 m q =C.H3/2 = 1.84 (5.7)3/2 =25.04 m D.S.W.L = 105.75 m D.S.T.E.L =105.75 + 0.2 = 105.95 m HL = 106.95 – 105.95 = 1 m

Page  90

I) High flood level

ii) with 20٪ concentration and 0.5 m retrogression q =1.2 × 25.04 = 𝑞 𝑐

H =( )2/3 =(

30.048 2/3 ) = 1.84

30.048 cumecs/m 6.43673 m

U.S.T.E.L

101.25 + 6.437 =107.687 m

D.S.W.L

105.75 – 0.5 = 105.25 m

D.S.T.E.L

105.25 +0.2 = 105.45 m

HL = 107.687 – 105.45 =2.237 m

Page  91

II. Pond flow condition

i)With out concentration and retrogression: head over under sluice crest = 103.5 – 100 = 3.5 m head over other barrage crest = 103.5 – 101.25 = 2.25 m Q =1.705 (65 – 0.1 × 8 × 3.5 ) × 3.5 + 1.84 (300 – 0.1 × 58 × 2.25) × 2.253/2 = 694.41 + 1781.9595 =2476.369 cumecs q

Page  92

2476.369 = 436

= 5.679 m²/sec ~ 5.68

i)With out concentration and retrogression (continued):

R= 1.35 V

𝑞 = 𝑅

hv

Page  93

5.68² 1/3 ( ) 1

5.68 = 4.297

𝑉² = 2𝑔

= 4.297

=1.32 m/sec

=0.089 m

i)With out concentration and retrogression (continued):

U.S.T.E.L =103.5 +0.089 = 103.589 m D.S water level when flood discharge of 2476.369 cumecs is passing =102.7 m ,for stage discharge curve

D.S.T.E.L = 102.7 + 0.089 =102.789 m HL =103.589 – 102.789 = 0.80 m q =1.84 (2.25)^1.5 = 6.21 cumecs/m

Page  94

ii) with 20٪ concentration and 0.5 retrogression

q = 1.2 * 6.21 = 7.452 cumecs/sec H=(

7.452 2 )^ 1.84 3

=2.54 m

U. S.T.E.L =101.25 + 2.54 = 103.79 m D.S.W.L = 102.7 – 0.5 = 102.2 m

D.S.T.E.L =102.2 +0.089 = 102.289 m HL = 103.79 – 102.289 = 1.501 m

Page  95

High No.

1 2 3 4 5 6 7 8 9

Discharge q D.S.W.L U.S.W.L D.S.T.E.L U.S.T.E.L HL Ef2 Ef1 =Ef2+HL

10 11 12

D1 D2 Length of floor 5(D2 - D1) 𝑞 F=

13

Page  96

Item

Level at which jump occurs

√𝑔𝐷1³

Flood

Pond

Flow

Without concentration and retrogression

With concentration and retrogression

Without concentration and retrogression

With concentration and retrogression

25.04 105.75 106.75 105.95 106.95 1 7.0 8.0 98.95

30.048 105.25 106.75 105.45 107.687 2.237 8.43 10.667 97.02

6.21 102.7 103.5 102.789 103.589 0.80 3.0 3.8 99.778

7.54 102.2 103.5 102.289 103.79 1.501 3.506 5.007 98.78

2.25 6.06 19.05

2.28 7.45 25.85

0.81 2.66 9.237

0.81 3.18 11.85

2.369

2.786

2.72

3.302

Provide the downstream floor at elevation 96.8 m with horizontal length of 27 m .

Distance from R.L of U.S end glacis of crest

0.75 3 3.75 6.0 6.75 9 9.75 12 12.69

Page  97

100.25 100 99.25 99 98.25 98 97.25 97.02

U.S.T.E.L=107.687 q =30.048 U.S.T.E.L= 103.79 cumecs/m q =7.54 cumecs / m High flood Pond flow Ef1 =U.S.T.E.L – Ef1=U.S.T.E.L R.L of glacis – R.L of glacis D1 D1 7.437 7.687 8.437 8.687 9.437 9.687 10.437 10.667

3.4375 3.218 2.844 2.75 2.560 2.50 2.344 2.281

2.79 3.54 3.79 4.54 4.79 5.54

1.5625 1.125 1.06 0.906 0.84375 0.78125

Page  98

Sl No.

𝑋 𝐷1

High Fr=2.78 𝑦 𝐷1 1 1 1.3 2 2 1.67 3 4 2.286 4 6 2.71 5 8 3.0 6 10 3.19 7 12 3.286 8 12.28 3.3 9 14 10 16 11 18 12 20

Page  99

flood Condition Pond Fr²=7.76 D1=2.28m Fr=3.302 𝑌 X Y 𝐷1 2.28 2.964 1.3 4.56 3.8070 2.714 9.12 5.212 2.38 13.68 6.179 2.86 18.24 6.84 3.143 22.8 7.27 3.524 27.36 7.49 3.62 28 7.524 3.64 3.86 4.0 4.095 4.19

Flow Condition Fr²=10.9 D1=0.81m X Y 0.81 1.62 3.24 4.86 6.48 8.1 9.72 9.947 11.34 12.96 14.58 16.2

1.053 1.39 1.93 2.316 2.546 2.85 2.93 2.95 3.126 3.24 3.317 3.4

Page  100

Ι. Depth of scour

Total discharge passing through other barrage =6684.1 cumecs Other barrage over all water way =358 m q=

6684.1 358

= 18.6706 cumecs/m

𝑞² 𝑓

R =1.35( )1/3 = 9.5 m

Page  101

II. U.S sheet pile

Allow 1.1 R = 1.1 × 9.5 = 10.45 m R.L of the bottom of scour hole = 106.75 – 10.45 = 96.3 m Provide sheet pile lines at elevation 96 m Depth of U.S sheet pile = 100 – 96 = 4 m

Page  102

III. D.S sheet pile :

Allow 1.25 R = 1.25 × 9.5 = 11.675 m R.L of bottom of scour = 105.25 – 11.875 = 93.375 m Provide sheet pile lines at elevation 92 m

Page  103

1 6

Safe exit gradient = max. static head = 103.5 – 96.8 = 6.8 m Depth of D.S cut off = 96.8 – 92 = 4.8 m GE = 1 6

=

𝐻 1 𝑑 𝜋 𝜆

6.8 1 4.8 𝜋 𝜆

𝛼 =

, λ = 7.32

2𝜆 − 1

2

− 1 ½ = 13.6

b = 𝛼𝑑 = 13.6 × 4.8 =65.28 m say 66 m provide floor length = 66 m

Page  104

The floor length is distributed as follows:

down stream horizontal floor =27 m down stream glacis length = 3(101.25 – 96.8) = 13.35 m width of crest = 2 m Upstream glacis length = 22.4 m Total floor length = 66 m

Page  105

Assume up stream floor thickness 1.0 m and down stream floor thickness 1.5 m Ι. Upstream pile line d = 100 – 96 = 4 m 1 ∝

𝑑 𝑏

= =

4 66

= 0.0606

∅𝐷1 = 100 − ∅𝐷 ,

∅𝐷 = 15٪

= 100 – 15 = 85٪ ∅𝐶1 = 100 − ∅𝐸 , ∅𝐸 = 22 ٪ = 100 – 22 = 78 ٪

Page  106

Correction of thickness : 𝑡 𝑑

Ct = (∅𝐷1 − ∅𝐶1) 1 4

= (85 – 78 ) = 1.75 ٪

Correction of interference : Cp =

𝐷 𝑑:𝐷 19√ ( ) 𝑏` 𝑏

= 19√

7 3:7 ( ) 65 66

= 0.945 ٪

Corrected ∅𝐶 = 78 + 1.75 + 0.945 = 80.695 ٪

Page  107

II. Down stream sheet pile lines 1 ∝

𝑑 𝑏

= =

4.8 66

= 0.0727

∅𝐷 = 16٪

,

∅𝐸 = 24 ٪

Correction of thickness : 𝑡 𝑑

Ct = - (∅𝐸 − ∅𝐷) =-

1.5 4.8

(24 – 16) = - 2.5 ٪

Corrected ∅𝐸 = 24 − 2.5 = 21.5٪ Correction of interference: There is no correction of pile interference . Page  108

Condition

1 No flow max. static head High Flood Flow at pond level

Page  109

Height/Elevation of sub soil H.G. line above datum

D.S water level datum (m)

U.S water level datum (m)

H

2

3

4

96.8

103.5

6.7

U.S Pile Line

D.S Pile Line

∅𝑬𝟏 100

∅𝑫𝟏 85

∅𝑪𝟏 80.695

∅𝑬 21.5

∅𝑫 16

∅𝑪 0

5

6

7

8

9

10

5.695

5.4065

1.4405

1.072

0

102.495

102.2065

98.2405

97.872

96.8

1.275

1.21

0.3225

0.24

0

106.525 1.105

106.46 1.049

105.5725 0.2795

105.49 0.208

105.25 0

103.305

103.249

102.4795

102.408

102.2

6.7

105.25

106.75

1.5

1.5

102.2

103.5

1.3

1.3