4.5.4. Design of vertical drop weir
The design of a vertical drop weir consists of (see Fig. 4.19): i. Hydraulic calculations to fix various elevations, ii. Design of weir wall, iii. Design of impervious floor (apron), iv. Design of protection works on upstream and downstream sides. The following data must be known for the design of the vertical drop weir: a) Maximum flood discharge, b) High flood level (H.F.L.) before construction of weir, c) Downstream bed level, d) Full supply level (F.S.L.) of canal taking off from the river, e) Allowable afflux, f) Lacey’s silt factor.
Figure 4.19 Vertical Drop Weir 4.5.4.1 Hydraulic Calculations
(i)
The length of the waterway, L is calculated from Lacey’s regime formula ( given in section4.5.2). section4.5.2). (ii) The discharge per unit length of the waterway, q is calculated as q = Q/L. (iii) The regime scour depth is calculated using Lacey’s formula 1
q 2 3 . R = 1.35 f (iv) The regime velocity and velocity head are calculated from Hydraulic Structures II – Lecture Note
Page 32
2
(v)
V = q/R, and velocity head = v /(2g) Water levels and total energy line (T.E.L.) on the downstream side and upstream side are calculated as 2
Level of d/s T.E.L. = (H.F.L. before construction) + v /(2g) Level of u/s T.E.L. = Level of d/s T.E.L. + Afflux 2
Level of u/s H.F.L. = Level of u/s T.E.L. – v /(2g) (vi)
Discharge over the crest of the weir is determined from,
q = 1.70 K 3 and
2
q K = 1.70
2 3
Therefore, crest level = u/s T.E.L. - K (vii)
Pond level = Level of top of gates = F.S.L. of canal + Head loss through regulator
Head loss through regulator may be taken as 0.5 to 1.0 m. Height of shutters = S = Level of top of gates – Crest level (viii) Protection against scour Level of bottom of u/s pile = u/s H.F.L. – 1.5 R Level of bottom of d/s pile = H.F.L. after retrogression – 2 R 4.5.4.2 Design of Weir Wall A weir wall is usually trapezoidal in cross-section with either both u/s and d/s faces inclined; or u/s face vertical and d/s face inclined. The design of the weir wall involves the determination of its top and bottom widths such that the section will be stable under the condition of maximum stress. In general the condition of maximum pressure on a hydraulic structure is that in which the head water is at the level of the crest of the weir or at the top of the crest of the shutters (if any) and no water is flowing over the weir so that there is no water on the d/s side. However, the condition of maximum stress on the weir may be different in certain cases. Hence, it is necessary to check the stability of the weir under the following three states:
State 1. When the u/s water or head water is at crest level or at the top of the crest shutter (if any) and there is no flow (Figure 4.20 a). State 2. When water is flowing over the weir crest and the weir is submerged (Fig. 4.20 b). State 3. When water is flowing over the weir crest and weir is discharging with a clear overfall (Fig. 4.20 c). Hydraulic Structures II – Lecture Note
Page 33
Figure 4.20 Calculations of flow over vertical drop weir Top width of weir wall
Top width of the weir wall is obtained using three methods: 1.
No tension criterion (for elementary profile)
The top width is determined on the consideration of no tension criterion for determining the width of an elementary (or triangular) profile of a dam or a weir, given by d B1 = Sm Where B1 = top width of the weir Hydraulic Structures II – Lecture Note
Page 34
d = maximum depth of water above the weir crest, which is equal to u/s H.F.L. – crest level, 2.
No sliding criterion (for elementary profile)
The top width of the weir is given by, d B1 = µ Sm Where m = coefficient of friction. Assuming m = 2/3 as a safe value, we get B1 3.
=
3d 2 Sm
Considering the height of the crest shutter
The top width of the weir is affected by height of crest shutter and it is given by, B1 = S + 1 (meters) Then, the largest of the three values is taken as the top width of the weir wall. Bottom width of weir wall
The bottom width of the weir wall, B, is determined by equating the overturning moments to the resisting moments taken at the outer middle third of the bottom width of the weir wall. In determining the bottom width all the three states discussed before are considered. State 1. Head water is at crest level or at the top of the crest shutter (if any) Considering the pressure diagram in Fig. 4.20(a), the overturning moment is given by M0
=
M r
=
γ (H + S)3
6 The resisting moment about the outer middle third point of the bottom is given by,
γ 1 2 2 + + + − − − { ( ) } ( ) S 1 . 5 H 2 . 5 S B B S H H S B B1 (H + 3S ) m 1 m 12 2
The above expression is valid when u/s and d/s faces have the same slope. If the u/s face is kept vertical, the resisting moment is given by M r
=
γHS m
(B
2
+ BB1 − B12 )
6 By equating the overturning moment to the resisting moment, the bottom width B is determined. State 2. Water is flowing over the weir crest and the weir is submerged (Fig. 4.20 b). Hydraulic Structures II – Lecture Note
Page 35
The overturning moment is given by,
=
M0
γhH 2
2 For maximum value of M0, h is taken corresponding to the case when the weir is just submerged. The resisting moment about the outer middle third of the bottom, with tail water at weir crest level, is given by, M r
γH(S m − 1)
=
12
(B
2
+ B1 B) - if both u/s and d/s faces have the same slope.
(B
2
+ B1 B − B12 )
If the u/s face is vertical, M r
γH(S m − 1)
=
6
Equating M0 and Mr , B can be determined. State 3. Water is flowing over the weir crest and weir is discharging with a clear over-fall (Fig. 4.20 c). M0
=
γ
(H
3
+ 3dH 2 − D 3 )
6 In this equation there are two unknowns, viz. d and D and the relation between the two must be known to find the maximum overturning moment. When the weir extends over the entire width of the river and the width of the river is of considerable width in comparison to its depth, d is roughly proportional to D, i.e. d = kD; where k is a constant and can be known if the river discharge for any one depth is known (i.e. gauged). Introducing this relation, M0 will be, M0
=
γ
(H
3
+ 3kDH 2 − D 3 )
6 The maximum value of M0 is obtained from, dM 0 dD
=
γ 6
[3kH
2
− 3D 2 ] = 0
Or D = H k Hence, the maximum value of M0 is given by M0
=
γH 3
[1 + 2k ] 3 2
6 The resisting moment may be considered to be approximately the same as that given for state 2. Like the previous cases, by equating M0 and Mr , B is determined. The greatest of the three values of B obtained above is adopted.
Hydraulic Structures II – Lecture Note
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4.5.4.3 Design of impervious floor (or apron)
For underseepage the worst condition occurs when the water on the u/s side is at the level of the weir crest or at the top of the crest shutters and there is no tailwater. Then, if the floor is designed on the basis of Bligh’s theory, the total creep length is given by L = CHs Where Hs is as shown in Figure 4.19. However, if the floor is to be designed on the basis of Khosla’s theory, the horizontal length b of the impervious floor is found by the consideration of the permissible exit gradient, GE, given by GE
=
HS
1
d
π λ
Knowing the permissible value of GE for the soil and values of Hs and d, calculated. From the exit gradient curve, for this value of
1
π λ
1
π λ
may be
, the corresponding value of α
may be found. Then from α = b/d, knowing α and d, the value of b can be determined. Out of the total impervious floor length b (or creep length L), the lengths L1 and L2 (Fig. 4.19) on the d/s and u/s of the weir wall, respectively are fixed on the basis of Bligh’s recommendation as (1) Downstream impervious floor length, L1: L1
= 2.21C
Hs
L1
= 2.21C
Hs
10
13
- for weirs without crest shutters
- for weirs with crest shutters
(2) Upstream impervious floor length, L2 L2 = L – L1 – (B + 2d1 + 2d2)
(according to Bligh’s theory)
L2 = b – L2 – B
(according to Khosla’s theory)
4.5.4.4 Design of protection works on upstream and downstream sides
(i) Upstream side Immediately at u/s end of the impervious floor, a block protection of length d1 is provided, where d1 is the depth of pile below the river bed or the impervious floor at the u/s end. U/s of the block protection a launching apron of length equal to 1.5d1 is provided. Hydraulic Structures II – Lecture Note
Page 37
(ii) Downstream side The total length of the impervious floor, inverted filter, and launching apron on the d/s side of the weir wall is also fixed on the basis of Bligh’s recommendations as L3
= 18C
Hs
L3
= 18C
Hs
10 13
x x
q 75 q 75
- for weirs without crest shutters - for weirs with crest shutters
The minimum length of the inverted and the launching apron is then equal to L3 – L1. However, an inverted filter of minimum length equal to 1.5d2 is to be provided immediately at the d/s end of the impervious floor, where d2 is depth of d/s sheet pile. After the inverted filter, a launching apron of length equal to 1.5 d2 and thickness of 1.5 m is provided. If the required length (L3 – L1) is more than 3d2 the lengths of the inverted filter and launching apron may be suitably increased.
Hydraulic Structures II – Lecture Note
Page 38
Example 4.3 Design a vertical drop weir on the basis of Bligh’s theory for the following site conditions: 3 (a) Maximum flood discharge = 2585 m /s (b) H.F.L. before construction = 255 m (c) Minimum water level = d/s bed level = 248 m (d) F.S.L. of canal = 254 m (e) Allowable afflux = 1 m (f) Coefficient of creep, C = 12 Assume any other data not given. If permissible exit gradient is 1/6, test the floor of the above weir by Khosla’s theory and make changes if found necessary. Solution 1) Hydraulic Calculations 1/2 1/2 i) L = 4.75 Q = 4.75 x (2585) = 241.50 m 3 q = Q/L = 2585/241.50 = 10.70 m /s/m ii) Regime scour depth, taking silt factor, f = 1 1
q 2 3 = 10.70 2 R = 1.35 1.35 f 1
iii)
iv) v)
vi)
1
3
= 6.56 m
Regime velocity, V = q/R = 10.70/6.56 = 1.63 m/s 2 2 Velocity head = V /2g = (1.63) /(2 x 9.81) = 0.14 m Level of d/s T.E.L. = H.F.L. before construction + velocity head = 255 + 0.14 = 255.14 m Afflux = 1 m Level of u/s T.E.L. = Level of d/s T.E.L. + Afflux = 255.14 + 1.0 = 256.14 m 2 Therefore, level of u/s H.F.L. = Level of u/s T.E.L. – V /2g = 256.14 – 0.14 = 256.0 m Actual d/s H.F.L. (allowing 0.5 m for retrogression) = 255.0 – 0.5 = 254.5 m q
3
= 1.70K 2 ⇒ K = (q/1.70)2/3 = (10.70/1.70)2/3 = 3.41 m
Crest level = u/s T.E.L. – K = 256.14 – 3.41 = 252.73 m Pond level = Level of top of gates = F.S.L. of canal + Head loss through regulator (taken as 0.5m) = 254 + 0.5 = 254.5 m Therefore, height of shutters, S = Level of top of gates – crest level = 254.5 – 252.73 = 1.77 m Bottom level of u/s pile (cutoff) = u/s H.F.L. – 1.5 R = 256.0 – 1.5 x 6.56 = 246.16 m
Hydraulic Structures II – Lecture Note
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The u/s pile may be taken up to a level of 246.0 m. Therefore, depth of u/s pile = 248.0 – 246.0 = 2.0 m Bottom level of d/s pile = d/s H.F.L. after retrogression – 2R = 254.5 – 2 x 6.56 = 241.38 m Hence, the d/s pile may be taken up to a level of 241.0 m. Depth of d/s pile = 248.0 – 241 = 7.0 m vii)
Head of water, Hs = Level of crest gates – bed level = 254.5 – 248.0 = 6.5 m Height of crest, H = Crest level – bed level = 252.73 – 248.0 = 4.73 m (Check: Hs = H + S = 4.73 + 1.77 = 6.5 m) 2) Design of Weir Wall viii) Top width B1 d = u/s H.F.L. – crest level = 256.0 – 252.73 = 3.27 m
-
Top width, B1
-
B1
-
ix)
=
3d 2S m
d
=
=
Sm
3x 3.27 2 x 2.24
=
3.27 2.24
= 2.18 m
= 2.19 m
From practical consideration, B1 = S + 1 = 1.77 + 1 = 2.77 m Hence, provide top width B1 = 3.0 m.
Calculation of bottom width, B - Considering State 1, the overturning moment is given by M0
=
γ (H + S)3 6
=
γH 3s 6
3
M0
=
M r
=
9.81 x (6.5)
= 449 kN − m 6 Resisting moment is given by γ 1 2 2 + + + − − − + { ( ) } ( ) ( ) S 1 . 5 H 2 . 5 S B B S H H S B B H 3 S m 1 m 1 12 2
For H = 4.73 m, S = 1.77 m, B1 = 3 m, and Sm = 2.24 M r
=
9.81
1
2
12
2 {(2.24 + 1.5)x 4.73 + 2.5x1.77}B + 3 x (2.24x 4.73 − 4.73 − 1.77 )B −
2
Therefore, equating M0 and Mr , 2 22.115B + 12.286 B – 594.42 = 0, and solving the quadratic equation, we get the value of the bottom width, B = 4.914 m Hydraulic Structures II – Lecture Note
x3 (4.73 + 3x1.77 )
Page 40
- Considering state 2, the overturning moment is calculated as follows: When the tailwater is just at the crest, d and h will be equal. For this condition, d (head over the crest) is given by d
q = 2 Cdx 3
2g
2
3
Neglecting the velocity of approach, and taking Cd = 0.58,
10.70 d= 2 x 0.58 x 2x 9.81 3
2
3
= 3.39 m
Since d = h = 3.39 m M0
=
M r
=
γhH 2
=
9.81x 3.39x 4.73
2 The resisting moment is
γH(S m − 1)
2
= 372.02 kN − m
2
(B
2
+ B1 B) =
9.81x 4.73(2.24 − 1)
(B
2
+ 3B) = 4.795 (B2 + 3B)
12 12 Equating Mo and Mr, 2 372.02 = 4.795 (B + 3B) From which, B = 7.44 m Adopting the greater of the two, we get B = 7.44 m. Provide B = 7.50 m 3) Design of Impervious and pervious aprons x) Total creep length L = CHs = 12 x 6.5 = 78.0 m
xi)
Downstream impervious apron L1 = 2.21C
xii)
Hs 13
= 2.21x12
6.5 13
= 18.75m , provide L1 = 19 m
Upstream impervious apron, L2 L2 = L – L1 – (B + 2d1 + 2d2) = 78 – 19 – (7.5 + 2 x 2 + 2 x7) = 33.5 m
xiii)
Total length of d/s apron L3
xiv)
= 18C
Hs 13
x
q 75
= 18x12
6.5 10.7 x 13 75
= 57.69m , provide L3 = 57.7 m
Length of inverted filter and launching apron on d/s side = L3 – L1 = 57.7 – 19 = 38.70 m
Hydraulic Structures II – Lecture Note
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d2 = 7.0 m Min. length of inverted filter = 1.5 d 2 = 1.5 x 7 = 10.5 m Min. length of launching apron = 1.5 d2 = 1.5 x 7 = 10.5 m Total length of downstream pervious apron = 2 x 10.5 = 21 m But the total length to be provided for both is 38.7 m. Hence, provide inverted filter of length 17.7 m with 1 m thick stone or concrete blocks laid on 0.5 m thick graded filter; and launching apron of 21 m length and 1.5 m thickness. xv)
U/s block protection and launching apron d1 = 2 m,
Hence, provide block protection of length = d1 = 2 m consisting of 1 m thick stone or concrete blocks laid on 0.5 m thick loosely packed stone. Minimum length of u/s launching apron = 1.5 d1 = 1.5 x 2 = 3 m Provide 3 m long and 1.5 m thick launching apron. xvi)
Thickness of impervious floor
Provide a nominal thickness of 1 m for the portion u/s of the weir wall and 1.5 m for the portion below the weir wall. At point A, just at the d/s toe of the weir wall, the residual seepage head H r = H s
−
Hs L
(2d 1 + L 2 + B) = = 6.5 −
6.5 78
(2x 2 + 33.5 + 7.5) = 2.75m
Therefore, required thickness of floor, t
=
4 H r 3 Sm
−1
=
4
2.75
3 2.24 − 1
= 2.96m ,
Provide 3 m a thickness of 3 m from the d/s toe of the weir wall up to a point 6 m from it. For the rest of the d/s portion of the floor same procedure can be followed and thickness calculated (e.g. for portion d/s of the 6 m). 4) Check by Khosla’s Theory
i)
Check for exit gradient Total length of impervious floor = 33.5 + 7.5 + 19 = 60 m d2 = 7 m (d/s pile) α = b/d = 60/7 = 8.57 = 8.6
Hydraulic Structures II – Lecture Note
Page 42
1
From Khosla’s curve for exit gradient, for α = 8.6,
Therefore, G E
=
Hs
1
d
π λ
=
6.5 7
x 0.145
=
1 7.4
π λ
<
1 6
= 0. 145
(permissible exit gradient)
Hence, SAFE ii)
Check for the floor thickness
Pressures at key points C1 and D1 of the u/s pile 1
α
=
d1 b
=
2 60
= 0.033
From Khosla’s curves φD1 = 89%; φC1 = 82% Correction for floor thickness for φ C1 =
φ D1 − φ C1 d
xt
=
89 − 82 2
x1.0
= 3.5% (+ ve)
x 2.0
= 2.6% (− ve)
Correction for interference of d/s pile for φ C1 C = 19
D d + D
b ′ b
b = b’ = 60 m; D = 7 m; d = 2 m
2 + 7 = 0.97% ≈ 1% (+ ve) 60 60 7
C = 19
Hence, corrected value φC1 = 82 + 3.5 + 1 = 86.5% Residual head at C1 = Hs x φC1 = 6.5 x 0.865 = 5.62 m Pressure at key points E2 and D2 of the d/s pile 1
=
d
α b
=
7 60
= 0.117
From Khosla’s curves φD2 = 21%; φE2 = 30% Correction for floor thickness for φ E2 =
φE2 − φD2 d
xt
=
30 − 21 7
(Thickness of floor near pile = 2 m) Correction for interference of u/s pile for φ E2 C = 19
D d + D
b ′ b
b = b’ = 60 m; D = 2 m; d = 7 m Hydraulic Structures II – Lecture Note
Page 43
C = 19
7 + 2 = 0.5%(− ve) 60 60 2
Hence, corrected value φE2 = 30 – 2.6 – 0.5 = 26.9% Residual head at E2 = Hs x φE2 = 6.5 x 0.269 = 1.75 m Assuming linear variation of pressure for intermediate points, the pressures and floor thickness for points A, B and C are as follows: Point A just at the d/s toe of the weir wall PA
5.62 − 1.75 = 5.62 − (33.5 + 7.5) = 2.98m 60
tA
=
2.98 2.24 − 1
= 2.40m
Actual thickness provided according to Bligh’s theory = 3 m Hence, safe Point B, 6 m from d/s toe of the weir wall PB
5.62 − 1.75 = 5.62 − (33.5 + 7.5 + 6 ) = 2.59m 60
tB
=
2.59 2.24 − 1
= 2.04m
Actual thickness provided according to Bligh’s theory = 2.5 m Hence, safe Point C, 12 m from d/s toe of the weir wall PC
5.62 − 1.75 = 5.62 − (33.5 + 7.5 + 12) = 1.77 m 60
tC
=
2.2 2.24 − 1
= 1.77 m
Actual thickness provided according to Bligh’s theory = 2 m Hence, safe
Hydraulic Structures II – Lecture Note
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Figure E-4.3 Section of the example weir
Hydraulic Structures II – Lecture Note
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