examples_chapter_3.pdf

HYDRAULIC STRUCTURES II: EXAMPLES ON DIVERSION HEADWORKS

1

1. Figure below shows the section of a vertical drop weir founded on sand. Determine:

a) The average hydraulic gradient

b) The uplift pressure at points A, B and C

c) The thickness of the floor at points A, B and C.

Use Bligh’s theory, specific gravity of floor material is equal to 2.24.

Solutions

Step 1: Determine creep Length

𝐿𝐵 = 2𝑑1 + 𝑏 + 2𝑑2 = 2(4𝑚) + 30𝑚 + 2(6𝑚) = 50𝑚

Step 2: Find average hydraulic gradient (𝑖)

𝑖 =𝐻

𝐿𝐵=

4𝑚

50𝑚=

1

12.5 𝑨𝒏𝒔.

Step3: Determine residual head (uplift pressure) at points A, B and C.

ℎ = 𝐻 (𝑆𝑒𝑒𝑝𝑎𝑔𝑒 ℎ𝑒𝑎𝑑) − 𝐻𝐿(𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠)

𝐻𝐿 = 𝐻𝑦𝑑𝑎𝑢𝑙𝑖𝑐 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 (𝑖) 𝑥 𝐶𝑟𝑒𝑒𝑝 𝐿𝑒𝑛𝑔𝑡ℎ (𝐿)

ℎ𝐴 = 𝐻 − (𝐻𝐿)𝐴 = ℎ − (𝑖 𝑥 𝐿𝐴) = 4𝑚 − (1

12.5 𝑥 18𝑚) = 2.56𝑚 𝑨𝒏𝒔.

ℎ𝐵 = 𝐻 − (𝐻𝐿)𝐵 = ℎ − (𝑖 𝑥 𝐿𝐵) = 4𝑚 − (1

12.5 𝑥 28𝑚) = 1.76𝑚 𝑨𝒏𝒔.

ℎ𝐶 = 𝐻 − (𝐻𝐿)𝐶 = ℎ − (𝑖 𝑥 𝐿𝐶) = 4𝑚 − (1

12.5 𝑥 38𝑚) = 0.96𝑚 𝑨𝒏𝒔.

Step 4: Determine the thickness of the floor at point A, B and C

ℎ =4

3 (

ℎ

𝐺 − 1)

ℎ𝐴 =4

3 (

ℎ𝐴

𝐺 − 1) =

4

3(

2.56𝑚

2.24 − 1) = 2.75𝑚 𝑨𝒏𝒔.


2

ℎ𝐵 =4

3 (

ℎ𝐵

𝐺 − 1) =

4

3(

1.76𝑚

2.24 − 1) = 1.89𝑚 𝑨𝒏𝒔.

ℎ𝐶 =4

3 (

ℎ𝐶

𝐺 − 1) =

4

3(

0.96𝑚

2.24 − 1) = 1.03𝑚 𝑨𝒏𝒔.

2. Determine the following for the sloping weir apron shown below using Khosla’s analytical

(mathematical) method and Khosla’s Graphical method.

a. The uplift pressures at key points

b. Whether the section provided is safe against uplift at A

c. Check against piping if the structure is founded on fine sand with a permissible

exit gradient of 1

6.

d. Draw sub-soil hydraulic gradient Line (H.G.L)

Use the thickness of the floor is 1m

Solutions

Determining uplift pressure at key points using the Khosla’s theory is a three step processes:

1. Decompose the composite profile into three simple profiles: u/s pile only, intermediate

pile only and d/s pile only.

2. Determine the uplift pressures at key points for each simple profiles either analytical or

Khosla’s curve method

3. Apply correction uplift pressure determined in step 2: Thickness, Interference and Slope

4. Determine uplift pressure at any points by interpolation


3

STEP1: DETERMINE UPLIFT PRESSURE AT KEY POINTS

1. FOR UPSTREAM SHEET PILE:

B. Graphical or Khosla’s Curve Method

Percentage of pressures at key points can also be determined by using Khosla’s

pressure curves (Figure 3-15 in your handout).

𝑏 = 57 𝑎𝑛𝑑 𝑑 = 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = (154𝑚 − 147𝑚 = 7𝑚

1

𝛼=

𝑑

𝑏=

7

57= 0.123

𝑈𝑠𝑖𝑛𝑔 1

𝛼 𝑟𝑒𝑎𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝜙𝐷 𝑎𝑛𝑑 𝜙𝐸 𝑓𝑟𝑜𝑚 𝑓𝑖𝑔𝑢𝑟𝑒 3

− 15 ( 𝐾ℎ𝑜𝑠𝑙𝑎′𝑠 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐶𝑢𝑟𝑣𝑒𝑠

𝜙𝐷 = 22% 𝑎𝑛𝑑 𝜙𝐸 = 31%

𝜙𝐷1 = 100 − 𝜙𝐷 = 100 − 22 = 78%

𝜙𝐶1 = 100 − 𝜙𝐸 = 100 − 31 = 69%

A. Analytical (Mathematical expression) method

𝑏 = 57 𝑎𝑛𝑑 𝑑 = 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = (154𝑚 − 147𝑚 = 7𝑚

𝛼 =𝑏

𝑑=

57

7= 8.143 𝑎𝑛𝑑 𝜆 =

1 + √1 + 𝛼2

2

=1 + √1 + 8.1432

2= 4.602

Φ𝐸1= 1 = 100%

Φ𝐷1=

1

𝜋cos−1 (

1 − 𝜆

𝜆) =

1

𝜋cos−1 (

1 − 4.602

4.602) = 0.7862 = 78.62%

Φ𝐶1 =1

𝜋cos−1 (

2 − 𝜆

𝜆) =

1

𝜋cos−1 (

2 − 4.602

4.602) = 0.6913 = 69.13%


4

Apply correction to Pressure calculated at key points by analytical method: Thickness and

Interference. No slope correction is needed since the point is not located at start and end of

sloping floor.

Correction for mutual interference: Interference of intermediate pile on upstream pile:

𝐶 = (+)19√𝐷

𝑏′ (

𝑑 + 𝐷

𝑏) 𝑖𝑓 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑖𝑛𝑔 𝑝𝑖𝑙𝑒

𝐶 = (−)19√𝐷

𝑏′ (

𝑑 + 𝐷

𝑏) 𝑖𝑓 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑖𝑛𝑔 𝑝𝑖𝑙𝑒

𝐷 = 𝐸𝑙𝐶1 − 𝐸𝐿𝐷2 = 153 − 147 = 6𝑚

𝑑 = 𝐸𝑙𝐶1 − 𝐸𝐿𝐷1 = 153 − 147 = 6𝑚

𝑏 = 57𝑚 𝑎𝑛𝑑 𝑏′ = 17𝑚

𝜙𝐶1𝑖 = (+)19√6

17 (

6 + 6

57) = 2.376% (+𝑣𝑒)

Thickness correction:

𝜙𝐶1 = 69.13% , 𝜙𝐷1 = 78.62%

𝜙𝑐1𝑡 =𝜙𝐷1−𝜙𝐶1

𝑑 𝑡 =

78.62−69.13

7𝑚 𝑥 1𝑚 = 1.355% (+ve)

Slope Correction: No slope correction is need since it is not located at start or end of

sloping floor

The corrected pressure at key point 𝐶1

𝜙𝐶1 = 69.13% + 2.37% + 1.355% = 72.86%

2. FOR INTERMEDIATE SHEET PILE:


𝑏1 = 17𝑚, 𝑏2 = 40 𝑚, 𝑑 = 7𝑚

𝛼1 =𝑏1

𝑑=

17𝑚

7𝑚= 2.429 𝑎𝑛𝑑 𝛼1 =

𝑏1

𝑑=

40𝑚

7𝑚= 5.714

𝜆1 =√1 + 𝛼1

2 − √1 + 𝛼2 2

2=

√1 + 2.4292 − √1 + 5.7142

2= −1.587

𝜆2 =√1 + 𝛼1

2 + √1 + 𝛼2 2

2=

√1 + 2.4292 + √1 + 5.7142

2= 4.214

Φ𝐸1= 1 = 100%

Φ𝐸2 =1

𝜋cos−1 (

𝜆1 − 1

𝜆2) =

1

𝜋cos−1 (

−1.587 − 1

4.214) = 0.7104 = 71.04%

Φ𝐷2 =1

𝜋cos−1 (

𝜆1

𝜆2) =

1

𝜋cos−1 (

−1.587

4.214) = 0.6229 = 62.29%

Φ𝐶2 =1

𝜋cos−1 (

𝜆1 + 1

𝜆2) =

1

𝜋cos−1 (

−1.587 + 1

4.214) = 0.5445 = 54.45%


5

Apply correction to pressure calculated at key points E2, D2, E2 by using analytical method:

Thickness, Interference, and slope. Correction for mutual interference: Interference of

intermediate pile on upstream pile:

o Correction for mutual interference: Interference of u/s pile on intermediate pile at point

E2 and Interference of d/s pile on intermediate pile at point C2

Mutual Interference Correction for 𝜙𝐸2: Interference of u/s pile on Intermediate

pile.

𝐷 = 𝐸𝑙𝐸2 − 𝐸𝐿𝐷1 = 153 − 147 = 6𝑚

𝑑 = 𝑡𝑟𝑢𝑒 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑠ℎ𝑒𝑒𝑡 𝑝𝑖𝑙𝑒 = 𝐸𝑙𝐸2 − 𝐸𝐿𝐷2 = 153 − 147 = 6𝑚

𝑏 = 57𝑚 𝑎𝑛𝑑 𝑏′ = 17𝑚

𝐶𝐸2 = (−)19√6

17 (

6 + 6

57) = −2.376% (−𝑣𝑒)

Mutual Interference Correction for 𝜙𝐶2 : Interference of d/s pile on Intermediate

pile.

𝐷 = 𝐸𝑙𝐶2 − 𝐸𝐿𝐷3 = 153 − 142 = 11𝑚

𝑑 = 𝑡𝑟𝑢𝑒 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑝𝑖𝑙𝑒 = 𝐸𝑙𝐶2 − 𝐸𝐿𝐷2 = 153 − 147 = 6𝑚


The value of 𝜙𝐶 can be read from the graph directly using 1

𝛼 and base ratio

(𝑏1

𝑏)

The value of 𝜙𝐷 can be read from the graph directly using 1

𝛼 and base ratio

(𝑏1

𝑏) for

𝑏1

𝑏 greater than 0.5.

To find 𝜙𝐷 for any value of base ratio𝑏1

𝑏, read 𝜙𝐷 for base ratio (1 −

𝑏1

𝑏) and

subtract from 100.

𝜙𝐷 = 100 − 𝜙𝐷(1 −𝑏1

𝑏)

To find 𝜙𝐸 for any value of 𝛼 and base ratio 𝑏1

𝑏, read 𝜙𝐶 for base ratio (1 −

𝑏1

𝑏) and subtract from 100.

𝜙𝐸 = 100 − 𝜙𝐶(1 −𝑏1

𝑏)

𝛼 =𝑏

𝑑= 8.14,

𝑏1

𝑏=

17

57= 0.3, (1 −

𝑏1

𝑏) = 0.7

From figure 3-15, 𝜙𝐶 = 54%, 𝜙𝐷 (1 −𝑏1

𝑏) = 38%, 𝜙𝐷 = 100 − 38 = 62%, 𝜙𝐶 (1 −

𝑏1

𝑏) = 29, 𝑎𝑛𝑑 𝜙𝐸 = 100 − 𝜙𝐶 (1 −

𝑏1

𝑏) = 100 − 29 = 71%


6

𝑏 = 57𝑚 𝑎𝑛𝑑 𝑏′ = 40𝑚

𝐶𝐶2 = (+)19√11

40 (

11 + 6

57) = 2.971% (+𝑣𝑒)

o Thickness correction:

Thickness correction for ϕE2

ϕE2 = 71.05% , ϕD2 = 62.29%

ϕc1t =ϕD2 −ϕE2

d t =

71.05−62.29

7m x 1m = −1.25% (-ve)

Thickness correction for ϕC2:

ϕC2 = 24.45% , ϕD2 = 62.29%

ϕc1t =ϕD2 −ϕE2

d t =

62.29−54.45

7m x 1m = 1.12% (-ve)

o Slope Correction: Slope Correction is need only at 𝐶2 since it is at the start of sloping

floor

𝐶 = ± (𝑏𝑠

𝑏1) 𝐶′ 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑓𝑜𝑟 𝑑𝑜𝑤𝑛𝑠𝑙𝑜𝑝𝑒 𝑎𝑛𝑑 𝑣𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑓𝑜𝑟 𝑢𝑝𝑠𝑙𝑜𝑝𝑒

𝐶′ = 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑠𝑙𝑜𝑝𝑒 3: 1 = 4.5 (𝑇𝑎𝑏𝑙𝑒 3 − 4)

𝑏𝑠 = 3𝑚 𝑎𝑛𝑑 𝑏1 = 40𝑚

𝐶 = − (𝑏𝑠

𝑏1) 𝐶′ = 𝐶 = − (

3𝑚

40𝑚) 𝑥 4.5% = −0.3375% (−𝑉𝑒)

The corrected pressure at key point 𝐸2 𝑎𝑛𝑑 𝐶2:

𝜙𝐸2 = 71.05% − 2.37% − 1.25% = 67.43%

𝜙𝐶2 = 54.45% + 2.972 + 1.12 − 0.34% = 58.2%

3. FOR DOWNSTREAM SHEET PILE:


𝑏 = 57 𝑎𝑛𝑑 𝑑 = 152𝑚 − 142𝑚 = 10𝑚

𝛼 =𝑏

𝑑=

57

10= 5.7 𝑎𝑛𝑑 𝜆 =

1 + √1 + 𝛼2

2=

1 + √1 + 5.72

2= 3.394

Φ𝐸3 =1

𝜋cos−1 (

𝜆 − 2

𝜆) =

1

𝜋cos−1 (

3.394 − 2

3.394) = 0.3653 = 36.53%

Φ𝐷3 =1

𝜋cos−1 (

𝜆 − 1

𝜆) =

1

𝜋cos−1 (

3.394 − 1

3.394) = 0.2508 = 25.08%

Φ𝐶3 = 0


7

Apply correction to uplift pressure calculated at key points: 𝐸3, 𝐷3 𝑎𝑛𝑑 𝐶3 by using analytical

method: Thickness and Interference. No slope correction is needed since the key points are

not located at start and end of sloping floor.

o Correction for mutual interference: Interference of intermediate pile on downstream

pile for key point 𝐸3.

𝐷 = 𝐸𝑙𝐶2 − 𝐸𝐿𝐷2 = 151 − 147 = 4𝑚

𝑑 = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = 𝐸𝑙𝐸3 − 𝐸𝐿𝐷3 = 151 − 142 = 9𝑚

𝑏 = 57𝑚 𝑎𝑛𝑑 𝑏′ = 40𝑚

𝐶𝐸3 = (−)19√4

40 (

4 + 9

57) = −1.37% (−𝑣𝑒)

Thickness correction for 𝐸3

𝑑 = 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = 𝐸𝑙𝐸3 − 𝐸𝐿𝐷3 = 152 − 142 = 10𝑚

𝜙𝐸3 = 36.53% , 𝜙𝐷3 = 25.08%

𝐶𝐸𝑡 = −𝜙𝐸3−𝜙𝐷3

𝑑 𝑡 = −

36.53−25.08

10𝑚 𝑥 1𝑚 = −1.145% (-ve)

Slope Correction: No slope correction is need since it is not located at start or end of

sloping floor

The corrected pressure at key point 𝐸3:

𝜙𝐸3 = 36.53% − 1.37% − 1.145% = 34.02%

SUMMARY OF PERCENTAGE OF PRESSURE AT ALL KEY POINTS


1

𝛼=

10

57= 0.18

From figure 3-15, for 𝛼 = 0.18,

𝜙𝐷 = 25% 𝑎𝑛𝑑 𝜙𝐸 = 37%


8

Key Points 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝜙 (%) Remarks

𝐸1 100.00

Upstream Pile 𝐷1 78.62

𝐶1 72.86

𝐸2 67.42

Intermediate Pile 𝐷2 62.29

𝐶2 58.21

𝐸3 34.02

Downstream Pile 𝐷3 25.08

𝐶3 0.00

STEP2: DETERMINE UPLIFT PRESSURE AT POINT A

Percentage of pressure at A can be obtained by interpolations. Point A is located

between𝐶2 𝑎𝑛𝑑 𝐸3.

𝐶2 𝐴 𝐸3

𝜙𝐶2 = 58.21 𝜙𝐴 =? 𝜙𝐸3 = 34.02

0𝑚 20𝑚 40𝑚

(58.21 − 𝜙𝐴)

(0 − 20)=

58.21 − 34.02)

(0 − 40)

𝜙𝐴 = 46.12%

ℎ𝐴 = 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 ℎ𝑒𝑎𝑑 = 0.4612 𝑥 𝐻 = 0.4612 𝑥 6𝑚 = 2.7672𝑚

Since the residual head (2.767m) is greater than thickness (t = 1m), the structure is not safe against

uplift.


9

SPP3: DETERMINE EXIT GRADIENT

𝑏 = 57 𝑎𝑛𝑑 𝑑 = 151𝑚 − 142𝑚 = 9𝑚

𝐻 =𝑢

𝑠𝑤𝑎𝑡𝑒𝑟 𝑙𝑒𝑣𝑒𝑟 −

𝑑

𝑠𝑤𝑎𝑡𝑒𝑟 𝑙𝑒𝑣𝑒𝑙 = 158𝑚 − 152𝑚 = 6𝑚

𝛼 =𝑏

𝑑=

57

10= 5.7 𝑎𝑛𝑑 𝜆 =

1 + √1 + 𝛼2

2=

1 + √1 + 5.72

2= 3.394

𝐺𝐸 =𝐻

𝑑

1

𝜋√𝜆=

6

9

1

𝜋√3.392=

1

8.68<

1

6⟹ 𝑆𝑎𝑓𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑝𝑖𝑝𝑖𝑛𝑔

STEP4: DRAW SUB-SOIL HYDRAULIC GRADIENT LINE (H.G.L)

Key points 𝜙 (%) Seepage

head (m)

Residual Head

𝒉 = 𝝓 𝒙 𝑯

Reference

Elevation(m)

H.G.L

(m)

𝑬𝟏 100.00

H = 6m

6m

152

158.00

𝑫𝟏 78.62 4.7m 156.70

𝑪𝟏 72.86 4.37m 156.37

𝑬𝟐 67.42 4.05m 156.05

𝑫𝟐 62.29 3.74m 155.74

𝑪𝟐 58.21 3.49m 155.49

𝑬𝟑 34.02 2.04m 154.04

𝑫𝟑 25.08 1.50m 153.50

𝑪𝟑 0.00 0 152.00


10

3. Design the weir sections of a diversion headwork’s with the following data:

Design flood discharge = 3000m3/s

Maximum dry (winter) flood discharge = 300 m3/s

HFL before construction of Weir = 255.5 m

Deepest bed level of the river = 249.5 m

Pond level = 254.5 m

Lacey’s Silt Factor (f) = 0.9

Permissible afflux = 1 m

Off-taking canal discharge = 200 m3/s

Looseness factor = 1.1

Concentration factor = 1.2

Bed retrogression = 0.5 m

Permissible exit gradient (𝐺𝐸) = 1/7

Pier contraction coefficient (K) = 0.1

The stage discharge curve of the river is shown below:


11

SOLUTIONS

STEP 1: FIX DISCHARGE OVER WEIR BAY AND UNDERSLUICE SECTION

o Undersluice Section: The Undersluice bay carry maximum of the following three

discharges:

1. Twice the discharge of the off-taking canal = 2 x 200 = 400 m3/s

2. 20 % of the design flood discharge = 0.2 x 3000 = 600 m3/s

3. Maximum winter (dry) flood discharge = 300 m3/s

Thus, the discharge over the Undersluice section 𝑄𝑢𝑛𝑑𝑒𝑟𝑠𝑙𝑢𝑖𝑐𝑒 = 600𝑚3

𝑠

o Weir Bay Section:

𝑄𝑤𝑒𝑖𝑟 = 𝑄𝑓𝑙𝑜𝑜𝑑 − 𝑄𝑢𝑛𝑑𝑒𝑟𝑠𝑙𝑢𝑖𝑐𝑒 = 3000 − 600 = 2400𝑚3

𝑠

STEP 2: DETERMINE LEVEL OF WEIR CREST AND UNDERSLUICE

o Lacey’s water way: 𝑃 = 4.75√𝑄 = 4.75√3000 = 260.17𝑚

o Actual width of waterway (overall waterway):

𝐿 = 𝐿𝑜𝑜𝑠𝑒𝑛𝑒𝑠𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑥 𝑃 = 1.1 𝑥 260.17 = 286.2𝑚

o Lacey’s Regime approaching Velocity: va = (Qf2

140)

1

6= (

3000 x0.92

140)

1

6= 1.61

m

s

o Approaching velocity head: ha =va

2

2g=

1.612

19.62= 0.13m

o D/s TEL:

o 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 = 𝐻𝐹𝐿 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 + ℎ𝑎 = 255.5𝑚 + 0.13𝑚 = 255.63𝑚

o 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 = 255.63𝑚 − 𝑅𝑒𝑡𝑟𝑜𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 = 255.13𝑚 𝑎𝑓𝑡𝑒𝑟 𝑟𝑒𝑔𝑟𝑜𝑔𝑟𝑎𝑡𝑖𝑜𝑛

o U/s TEL:

o 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 = 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 + 𝑎𝑓𝑓𝑙𝑢𝑥 = 255.63𝑚 + 1𝑚 = 256.63𝑚

o 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 = 256.63𝑚 − 0.5𝑚 = 256.13𝑚 𝑎𝑓𝑡𝑒𝑟 𝑟𝑒𝑔𝑟𝑜𝑔𝑟𝑎𝑡𝑖𝑜𝑛

a) Undersluice Section: Assuming broad crested (Coefficient of discharge = 1.71)

Fix Undersluice crest level at the deepest river bed level = 249.5m

Total head over the crest level of the Undersluice during flooding condition

𝐻𝑒 = 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 − 𝐵𝑒𝑑 𝐿𝑒𝑣𝑒𝑙 = 256.63 𝑚 − 249.5𝑚 = 7.13𝑚

Discharge intensity (𝑞𝑢): 𝑞𝑢 = 1.71 (7.13)3

2 = 32.56𝑚3

𝑠

𝑚

Width of Undersluice (𝐿𝑢): 𝐿𝑢 =𝑄𝑢

𝑞𝑢=

600

32.56= 18.43𝑚

Let us provide two bays of 10m each: 𝑳𝒖 = 𝟐𝟎𝒎

Since the width provided (20m) is greater than calculated value (18.43m), we have to adjust

the discharge over the crest of Undersluice using discharge formula:


12

𝑄 = 𝐶𝑑(𝐿 − 𝑘𝑛𝐻𝑒)𝐻𝑒

32

𝑛 = 2𝑁, 𝑤ℎ𝑒𝑟𝑒 𝑁 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑖𝑒𝑟 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑦𝑠 − 1

𝑄𝑢 = 1.71(20 − 0.1 𝑥2 𝑥 1𝑥 7.13)(7.13)32 = 604

𝑚3

𝑠

Height of gate over Undersluice crest level : 𝑯𝒈𝒂𝒕𝒆 = 𝟐𝟓𝟒. 𝟓𝒎 – 𝟐𝟒𝟗. 𝟓𝒎 = 𝟓𝒎

b) Weir Bay Section: Assume sharp crest weir (coefficient of discharge = 1.84)

Actual width of waterway of Weir Bay section:

𝐿𝑊 = 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑤𝑎𝑡𝑒𝑟𝑤𝑎𝑦 − 𝐿𝑢 = 286.2𝑚 − 20𝑚 = 266.2𝑚

𝐿𝑊 = 270𝑚 (𝑆𝑎𝑦)

Adjust discharge over the weir bay section: 𝑄𝑤 = 3000 − 604 = 2396𝑚3

𝑠

Discharge intensity (𝑞𝑢): 𝑞𝑢 =𝑄𝑤

𝐿𝑤=

2396

270= 8.87

𝑚3

𝑠

𝑚

Height of TEL over weir crest (𝐻𝑒): 𝑞 = 𝐶(𝐻𝑒)3

2 for sharp crested weir 𝐶 = 1.84

𝐻𝑒 = (𝑞

1.84)

23

= (8.87

1.84)

23

= 2.85𝑚

Required level of the weir crest: 256.63𝑚 − 2.85𝑚 = 253.78𝑚

Thus Let us provide weir crest level at 253.7m

Required height of weir gate: 𝑯𝒈𝒂𝒕𝒆 = 𝑷𝒐𝒏𝒅 𝒍𝒆𝒗𝒆𝒍 − 𝒄𝒓𝒆𝒔𝒕 𝒍𝒆𝒗𝒆𝒍

𝑯𝒈𝒂𝒕𝒆 = 𝟐𝟓𝟒. 𝟓𝒎 − 𝟐𝟓𝟑. 𝟕𝟎𝒎 = 𝟎. 𝟖𝒎

STEP3: FIX THE LEVEL OF THE DOWNSTREAM FLOOR AND THE LENGTH OF

THE DOWNSTREAM HORIZONTAL LENGTH OF FLOOR FOR WEIR BAY SECTION

Two flow conditions exist: Pond level (PL) and high flood level conditions (HFL).

1. At Pond level conditions

o The total discharge through weir and Undersluice section:

𝑄 = 𝑄𝑤𝑒𝑖𝑟 + 𝑄𝑢𝑛𝑑𝑒𝑟𝑠𝑙𝑢𝑖𝑐𝑒

𝐻𝑒 = 254.5𝑚 − 253.7𝑚 = 0.8 𝑓𝑜𝑟 𝑤𝑒𝑖𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛

𝐻𝑒 = 254.5𝑚 − 249.5 = 5𝑚 𝑓𝑜𝑟 𝑈𝑛𝑑𝑒𝑟𝑙𝑠𝑢𝑖𝑐𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛

𝑄𝑃𝐿 = 1.71(270 − 0.1 𝑥 2 𝑥 17 𝑥 0.8)0.832 + 1.84(20 − 0.1 𝑥 2 𝑥1 𝑥5)5

32 = 715

𝑚3

𝑠

The corresponding river stage when 𝑸 = 𝟕𝟏𝟓𝒎𝟑

𝒔 from stage discharge curve = 251.75m


13

o Downstream water level after retrogression = 251.75𝑚 − 0.5𝑚 = 𝟐𝟓𝟏. 𝟐𝟓𝒎

o Lacey’s regime velocity at pond level: 𝑣𝑎 = (𝑄𝑓2

140)

1

6= (

715 𝑥 0.92

140)

1

6= 1.267

𝑚

𝑠

o Approaching velocity head at pond level : ℎ𝑎 =𝑣𝑎

2

2𝑔=

1.2672

19.62= 0.08𝑚

o Downstream TEL: 𝑫𝒐𝒘𝒏𝒔𝒕𝒓𝒆𝒂𝒎 𝑷𝑳 + 𝒉𝒂 = 𝟐𝟓𝟏. 𝟐𝟓𝒎 + 𝟎. 𝟎𝟖 = 𝟐𝟓𝟏. 𝟑𝟑𝒎

o Upstream TEL: 𝑼𝒑𝒔𝒕𝒓𝒆𝒂𝒎 𝒑𝒐𝒏𝒅 𝒍𝒆𝒗𝒆𝒍 + 𝒉𝒂 = 𝟐𝟓𝟒. 𝟓𝒎 + 𝟎. 𝟎𝟖𝒎 = 𝟐𝟓𝟒. 𝟓𝟖𝒎

2. At High Flood Level Conditions

o Downstream TEL after retrogression:

𝐻𝐹𝐿 𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 + ℎ𝑎 = 255.5𝑚 + 0.13𝑚 − 0.5𝑚 = 255.13𝑚

o Head over crest taking into account concentration factor of 1.2

𝐻𝑒 = (𝑞

1.84)

23

= (11.07

1.84)

23

= 3.31𝑚

o 𝑫𝒐𝒘𝒏𝒔𝒕𝒓𝒆𝒂𝒎 𝑭𝒍𝒐𝒐𝒅 𝑳𝒆𝒗𝒆𝒍 = 𝟐𝟓𝟓. 𝟏𝟑 − 𝒉𝒂 = 𝟐𝟓𝟓. 𝟏𝟑 − 𝟎. 𝟏𝟑 = 𝟐𝟓𝟓𝒎

o Upstream TEL: 𝑾𝒆𝒊𝒕 𝒄𝒓𝒆𝒔𝒕 𝑳𝒆𝒗𝒆𝒍 + 𝑯𝒆 = 𝟐𝟓𝟑. 𝟕 + 𝟑. 𝟑𝟏𝒎 = 𝟐𝟓𝟕. 𝟎𝟏𝒎

3. Discharge intensity taking into account concentration factor of 1.2

𝑞 = (1.84𝐻𝑒

32) 𝑥 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

𝐻𝑒 = 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 𝑎𝑡 𝑃𝑜𝑛𝑑 𝐿𝑒𝑣𝑒𝑙 − 𝑊𝑒𝑖𝑟 𝐶𝑟𝑒𝑠𝑡 𝐿𝑒𝑣𝑒𝑙

o At pond level: 𝐻𝑒 = 254.58𝑚 − 253.7𝑚 = 0.88𝑚

𝑞𝑃𝐿 = 1.84 𝑥0.81.5 𝑥 1.2 = 𝟏. 𝟖𝟐𝒎𝟑

𝒔/𝒎

o At HFL: 𝐻𝑒 = 256.63𝑚 − 253.7𝑚 = 2.93𝑚

𝑞𝑃𝐿 = 1.84 𝑥 2.931.5 𝑥 1.2 = 𝟏𝟏. 𝟎𝟕𝒎𝟑

𝒔/𝒎

o Critical depth of flow (𝑦𝑐)

𝑦𝑐 = (𝑞2

𝑔)

13

(𝑦𝑐)𝑃𝐿 = (𝑞2

𝑔)

13

= (11.072

9.81)

13

= 𝟐. 𝟑𝟐𝒎 𝑎𝑛𝑑 (𝑦𝑐)𝑃𝐿 = (𝑞2

𝑔)

13

= (1.822

9.81)

13

= 𝟎. 𝟕𝟎𝒎

Note. Downstream floor level is fixed based on the location of hydraulic jump while downstream

floor length is fixed based on the height of hydraulic jump (𝑦2 − 𝑦1)


14

S.No Variables Symbol Unit

Flow Conditions

High Flood Condition

Pond Level Condition

1 Discharge intensity 𝑞 m3/s/m 11.07 1.82

2 Critical depth of flow 𝑦𝑐 𝑚 2.32 0.70

4 Downstream Total Energy Level d/s TEL 𝑚 255.13 251.33

5 Upstream Total Energy Level u/s TEL 𝑚 257.01 254.58

6 Head Loss HL 𝑚 1.88 3.25

7 𝐻𝐿/𝑦𝑐 𝑍 - 0.81 4.64

8 Specific Energy after Jump (From table) 𝐸𝑓2 𝑚 4.66 1.88

9 Specific Energy before Jump 𝐸𝑓1 𝑚 6.54 4.93

10 Depth of flow Before jump 𝑦1 𝑚 1.08 0.19

11 Depth of flow after jump 𝑦2 𝑚 4.31 1.83

12 Height of Jump 𝑦2 − 𝑦1 𝑚 3.25 1.64

13 Length of d/s floor 5(𝑦2 − 𝑦1 𝑚 16.15 8.2

14 Location of Hydraulic jump 𝑃 𝑚 250.5 249.5

𝐿𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑗𝑢𝑚𝑝 = 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑇𝐸𝐿 − 𝐸𝑓2

𝑷𝑯𝑭𝑳 = 𝟐𝟓𝟓. 𝟏𝟑 − 𝟒. 𝟔𝟔 = 𝟐𝟓𝟎. 𝟒𝟕𝒎 𝒂𝒏𝒅 𝑷𝑷𝑳 = 𝟐𝟓𝟏. 𝟑𝟑 − 𝟏. 𝟖𝟖 = 𝟐𝟒𝟗. 𝟒𝟓𝒎

Flow conditions 𝑍 𝑋 𝑌

𝐸1

𝑌𝑐

𝐸2

𝑦𝑐

HFL

0.50 0.52 1.73 2.40 1.90

0.81 0.47 1.86 2.81 2.00

1.00 0.44 1.94 3.07 2.07

𝑦1 =𝑦1

𝑦𝑐 𝑥 𝑦𝑐 𝑦2 =

𝑦2

𝑦𝑐 𝑥 𝑦𝑐 𝐸1 =

𝐸1


𝐸1

𝑦𝑐 𝑥 𝑦𝑐

𝑦𝑐 = 2.32𝑚 1.08 4.31 6.54 4.66

PL

4.50 0.27 2.59 7.17 2.67

4.64 0.27 2.61 7.04 2.68

5.00 0.26 2.65 6.73 2.73

𝑦1 =𝑦1

𝑦𝑐 𝑥 𝑦𝑐 𝑦2 =

𝑦2


𝐸1


𝐸1

𝑦𝑐 𝑥 𝑦𝑐

𝑦𝑐 = 0.7𝑚 0.19 1.83 4.93 1.88

Note. 𝒁 =𝑯𝑳

𝒚𝒄, 𝑿 =

𝒚𝟏

𝒚𝒄, 𝒀 =

𝒚𝟐

𝒚𝒄

The level of the downstream floor should be at or lower than 249.6 m and the length of the

downstream horizontal floor should be equal to or more than 16.25 m. Therefore, provide

downstream horizontal floor of length 17 m at 249.5 m.


15

STEP4: DETERMINE UPSTREAM AND DOWNSTREAM CUTOFF WALLS (SHEET PILES)

OF WEIR BAY SECTION

For the determination of cutoff walls (piles) a concentration factor is accounted for in the

computations of scour depth. The scour depth below HFL can be computed as follow:

𝑅 = 1.35 (𝑞2

𝑓)

13

= 1.35 (11.072

0.9)

13

= 6.95𝑚

o Reduced level of bottom of the scour depth on upstream side:

= 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝐻𝐹𝐿 − 𝑋𝑅 = 256.5 − 1.25 𝑥 6.95 = 247.81𝑚

𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝒑𝒓𝒐𝒗𝒊𝒅𝒆 𝒖𝒑𝒔𝒕𝒓𝒆𝒂𝒎 𝒔𝒉𝒆𝒆𝒕 𝒖𝒑 𝒂𝒕 𝒆𝒍𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝟐𝟒𝟕. 𝟖𝒎

o Reduced level of bottom of the scour depth on downstream side:

= 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝐻𝐹𝐿 − 𝑋𝑅 = 255.5 − 1.5 𝑥 6.95 = 245.08𝑚

𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝒑𝒓𝒐𝒗𝒊𝒅𝒆 𝒅𝒐𝒘𝒏𝒔𝒕𝒓𝒆𝒂𝒎 𝒔𝒉𝒆𝒆𝒕 𝒖𝒑 𝒂𝒕 𝒆𝒍𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝟐𝟒𝟓𝒎

STEP5: DETERMINE LENGTH OF CONCRETE IMPERVIOUS FLOOR OF WEIR BAY

SECTION

Given that safe exit gradient of 1/7.

𝐺𝐸 =𝐻

𝜋

1

𝑑√𝜆= 1/7

𝐻 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑎𝑡𝑖𝑐 ℎ𝑒𝑎𝑑 = 𝑃𝑜𝑛𝑑 𝑙𝑒𝑣𝑒𝑙 − 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑓𝑙𝑜𝑜𝑟 𝑙𝑒𝑣𝑒𝑙

𝐻 = 254.5𝑚 − 249.5𝑚 = 5𝑚

𝑑 = 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑓𝑙𝑜𝑜𝑟 𝑙𝑒𝑣𝑒𝑙 − 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑙𝑒𝑣𝑒𝑙 𝑜𝑓 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑐𝑜𝑢𝑟 𝑑𝑒𝑝𝑡ℎ

𝑑 = 249.5𝑚 − 245 = 4.5𝑚

5𝑚

4.5𝑚

1

𝜋√𝜆=

1

7⇒ 𝜆 = 6.13

𝜆 =1 + √1 + 𝛼2

2= 6.13 ⇒ 𝛼 = 11.21 =

𝑏

𝑑⇒ 𝑏 = 50.46𝑚 = 𝟓𝟏𝒎(𝒔𝒂𝒚)

Let the crest width be 3m, upstream slope with 1𝑉: 2𝐻 and downstream glacis with1𝑉: 3𝐻.

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑠𝑙𝑜𝑝𝑖𝑛𝑔 𝑓𝑙𝑜𝑜𝑟 = (253.7𝑚 − 249.5)𝑥 2 = 𝟖. 𝟒𝒎


16

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑠𝑙𝑜𝑝𝑖𝑛𝑔 𝑓𝑙𝑜𝑜𝑟 = (253.7𝑚 − 249.5)𝑥 3 = 𝟏𝟐. 𝟔𝒎

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑖𝑚𝑝𝑒𝑟𝑣𝑖𝑜𝑢𝑠 𝑓𝑙𝑜𝑜𝑟 = 51𝑚 − 8.4𝑚 − 12.6𝑚 − 3𝑚 − 17𝑚 = 𝟏𝟎𝒎

STEP6: DESIGN OF PROTECTION WORKS

Concentration factor is not be accounted for in computation of scour depth for design of protection

works.

𝑞𝐻𝐹𝐿 = 1.84(2.93)32 = 9.23

𝑚3

𝑠/𝑚

𝑅 = 1.35 (𝑞2

𝑓)

13

= 1.35 (9.232

0.9)

13

= 6.15𝑚

o Reduced level of bottom of the scour depth on upstream side:

= 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝐻𝐹𝐿 − 𝑋𝑅 = 256.5 − 1.5 𝑥 6.15 = 247.3𝑚

𝐷1 = 249.5𝑚 − 247.3𝑚 = 2.22𝑚

Length of u/s block protection: = 1.5𝐷1 = 1.5 𝑥 2.22 = 3.33𝑚

Length of u/s Launching apron: = 2 𝑥 𝐷1 = 2 𝑥2.22 = 6.66𝑚

o Reduced level of bottom of the scour depth on downstream side:

= 𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝐻𝐹𝐿 − 𝑋𝑅 = 255.5 − 2 𝑥 6.15 = 243.2𝑚

𝐷2 = 249.5𝑚 − 246.28𝑚 = 6.3𝑚

Length of d/s filter and block protection: = 2𝐷2 = 2 𝑥 6.3 = 12.6𝑚

Length of d/s Launching apron: = 2 𝑥 𝐷2 = 2 𝑥 6.3 = 12.6𝑚𝑚


17

STEP6: DESIGN OF FLOOR THICKNESS

A. DETERMINE SUB-SURFACE PERCENTAGE OF UPLIFT PRESSURE AT KEY

POINTS

Let assume a thickness of 1m at upstream and 1.5m at downstream end.

1. FOR UPSTREAM SHEET PILE:

Apply correction to pressure calculated at key points: Thickness and Interference. No slope

correction is needed since the point is not located at start and end of sloping floor.

Correction for mutual interference: Interference of intermediate pile on upstream pile:

𝐶 = (+)19√𝐷

𝑏′ (

𝑑 + 𝐷

𝑏)

𝐷 = 𝐸𝑙𝐶1 − 𝐸𝐿𝐷2 = 248.5𝑚 − 245𝑚 = 3.5𝑚

𝑑 = 𝐸𝑙𝐶1 − 𝐸𝐿𝐷1 = 248.5𝑚 − 147.8𝑚 = 0.7𝑚

𝑏 = 57𝑚 𝑎𝑛𝑑 𝑏′ = 51𝑚

Analytical method

𝑏 = 51 𝑎𝑛𝑑 𝑑 = 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = (249.5𝑚 − 147.8𝑚 = 1.7𝑚

𝛼 =𝑏

𝑑=

51

1.7= 30 𝑎𝑛𝑑 𝜆 =

1 + √1 + 𝛼2

2=

1 + √1 + 302

2= 15.51

Φ𝐸1= 1 = 100%

Φ𝐷1=

1

𝜋cos−1 (

1 − 𝜆

𝜆) =

1

𝜋cos−1 (

1 − 15.51

15.51) = 0.8851 = 88.51%

Φ𝐶1 =1

𝜋cos−1 (

2 − 𝜆

𝜆) =

1

𝜋cos−1 (

2 − 15.51

15.51) = 0.8366 = 83.66%


18

𝜙𝐶1𝑖 = (+)19√6

51 (

0.7 + 3.5

51) = 0.3668% (+𝑣𝑒)

Thickness correction:

𝜙𝐶1 = 83.66% , 𝜙𝐷1 = 88.51%

𝜙𝑐1𝑡 =𝜙𝐷1−𝜙𝐶1

𝑑 𝑡 =

88.51−83.66

1.7𝑚 𝑥 1𝑚 = 2.85% (+ve)

The corrected pressure at key point 𝐶1

𝜙𝐶1 = 83.66% + 0.3668% + 2.85% = 86.88%

2. FOR DOWNSTREAM SHEET PILE:

Apply correction to uplift pressure calculated at key points: 𝐸2, 𝐷2 𝑎𝑛𝑑 𝐶2: Thickness and

Interference. No slope correction is needed since the key points are not located at start and

end of sloping floor.

o Correction for mutual interference: Interference of intermediate pile on downstream

pile for key point 𝐸2.

𝐷 = 𝐸𝑙𝐸2 − 𝐸𝐿𝐷1 = 248.5 − 247.8 = 0.7𝑚

𝑑 = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = 𝐸𝑙𝐸2 − 𝐸𝐿𝐷2 = 248.5 − 145 = 3.5𝑚

𝑏 = 51𝑚 𝑎𝑛𝑑 𝑏′ = 51𝑚

𝐶𝐸2 = (−)19√0.7

51 (

0.7 + 3.5

51) = −0.1833% (−𝑣𝑒)

Thickness correction for 𝐸2

𝑑 = 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = 𝐸𝑙𝐸2 − 𝐸𝐿𝐷2 = 249.5 − 145 = 4.5𝑚

𝜙𝐸2 = 26.33% , 𝜙𝐷2 = 18.35%

𝐶𝐸𝑡 = −𝜙𝐸2−𝜙𝐷2

𝑑 𝑡 = −

26.33−18.35

4.5𝑚 𝑥 1.5𝑚 = −2.66% (-ve)

The corrected pressure at key point 𝐸2:

Analytical (Mathematical expression) method

𝑏 = 51 𝑎𝑛𝑑 𝑑 = 249.5𝑚 − 145𝑚 = 4.5𝑚

𝛼 =𝑏

𝑑=

51

4.5= 11.33 𝑎𝑛𝑑 𝜆 =

1 + √1 + 𝛼2

2=

1 + √1 + 11.332

2= 6.89

Φ𝐸2 =1

𝜋cos−1 (

𝜆 − 2

𝜆) =

1

𝜋cos−1 (

6.19 − 2

6.19) = 0.2633 = 26.33%

Φ𝐷2 =1

𝜋cos−1 (

𝜆 − 1

𝜆) =

1

𝜋cos−1 (

6.19 − 1

6.19) = 0.1835 = 18.35%

Φ𝐶2 = 0


19

𝝓𝑬𝟐 = 𝟐𝟔. 𝟑𝟑% − 𝟎. 𝟏𝟖% − 𝟐. 𝟔𝟔% = 𝟐𝟑. 𝟒𝟗%

B. DETERMINE AND DRAW SUB-SOIL HYDRAULIC GRADIENT LINE (H.G.L)

Flow

Conditions

Key points 𝜙 (-) Seepage head (m) Residual Head

𝒉 = 𝝓 𝒙 𝑯 (m)

RE

(m)

H.G.L

(m)

No flow

𝐸1 1.0000

H = 254.5 − 249.5

H = 5m

5.00

249.5

254.50

𝐷1 0.8851 4.43 253.93

𝐶1 0.8688 4.34 253.84

𝐸2 0.2349 1.17 250.67

𝐷2 0.1835 0.92 250.42

𝐶2 0.0000 0.00 249.50

Pond Level

𝐸1 1.0000

H

= 254.5 − 251.25

H = 3.25m

3.25

251.25

254.50

𝐷1 0.8851 2.88 254.13

𝐶1 0.8688 2.82 254.07

𝐸2 0.2349 0.76 252.01

𝐷2 0.1835 0.60 251.85

𝐶2 0.0000 0.00 251.25

HFL

𝐸1 1.0000

H = 256.5 − 255

H = 1.5m

1.50

255

256.50

𝐷1 0.8851 1.33 256.33

𝐶1 0.8688 1.30 256.30

𝐸2 0.2349 0.35 255.35

𝐷2 0.1835 0.28 255.28

𝐶2 0.0000 0.00 255.00


20

C. DETERMINE DYNAMIC PRESSURES AND MAXIMUM STATIC PRESSURES

For high flood condition and pond level condition, the suction pressure is computed from hydraulic

jumps.

R.L of Glacis (m)

HFL PL

𝑞 = 11.07 𝑚^3/𝑠/𝑚, 𝑈/𝑠 𝑇𝐸𝐿 = 257.01𝑚

𝑞 = 1.82 𝑚^3/𝑠/𝑚, 𝑈/𝑠 𝑇𝐸𝐿 = 254.58𝑚

𝐸𝑓(𝑚) 𝑦(𝑚) 𝑊𝐿(𝑚) 𝐸𝑓(𝑚) 𝑦(𝑚) 𝑊𝐿(𝑚)

253.7

252.9 4.11 1.5 254.4 1.68 0.37 253.27

252.1 4.91 1.4 253.5 2.48 0.25 252.35

251.3 5.71 1.2 252.5 3.28 0.19 251.49

250.5* 6.51 1 251.5 4.08 0.15 250.65

249.6** 4.98 0.14 249.74

Note. 𝑬𝒇 = specific energy, 𝒚 =depth of flow, 𝑾𝑳 = Water Level, 𝑹. 𝑳 = Reduced Level,

𝑯𝑭𝑳 = High Flood Level, 𝑷𝑳 = Pond Level. *Location of hydraulic (P) jump at HFL and **

Location of Hydraulic jump (P) at pond level.

a). Maximum suction pressure (𝑼𝒅)

Determine the level of hydraulic gradient line at the location of hydraulic jump (P) by using linear

integration.

HFL PL

Points H.G.L (m) Distance from point C1 (m) Points H.G.L (m)

Distance from point C1 (m)

C1 256.3 0 C1 254.07 0

P 255.72 31.15 P 252.71 33.7

E2 255.35 51 E2 252.01 51

Note. P = location of Hydraulic jump.

(𝒖𝒅)𝑯𝑭𝑳 = (𝑯. 𝑮. 𝑳)𝑷 − (𝑾𝑳)𝑷 = 𝟐𝟓𝟓. 𝟕𝟐𝒎 − 𝟐𝟓𝟏. 𝟓𝒎 = 𝟒. 𝟐𝟐𝒎

(𝒖𝒅)𝑷𝑳 = (𝑯. 𝑮. 𝑳)𝑷 − (𝑾𝑳)𝑷 = 𝟐𝟓𝟐. 𝟕𝟏𝒎 − 𝟐𝟒𝟗. 𝟕𝟒𝒎 = 𝟐. 𝟗𝟕𝒎

b). Maximum static pressure for no overflow (𝒖)

No overflow

Points H.G.L (m) Distance from point C1 (m)

C1 253.84 0

P 251.73 34

E2 250.67 51


21

𝑢 = 251.73 − 249.5 = 2.23𝑚

Floor thickness is given as

𝑡 =ℎ

𝑆𝑐 − 1

Where ℎ = uplift pressure and 𝑆𝑐 = specific gravity of floor material. Of course, h is equal to the

grater of:

ℎ = 𝑀𝑎𝑥𝑖𝑚𝑎𝑚 {2

3𝑢𝑑 , 𝑢} = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 {

2

3(4.22𝑚), 2.23𝑚} = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚{2.81𝑚, 2.23𝑚}

𝒉 = 𝟐. 𝟖𝟏𝒎

Where 𝑢𝑑 = maximum ordinate between HGL and water surface profile, which usually occur near

the location of hydraulic jump. 𝑢 = the uplift pressure at any point for no flow conditions.

Therefore, the thickness of the floor at downstream end of sloping (glacis) for concrete of specific

gravity 2.24 is:

𝒕 =𝒉

𝑺𝒄 − 𝟏=

𝟐. 𝟖𝟏𝒎

𝟐. 𝟐𝟒 − 𝟏= 𝟐. 𝟐𝟕𝒎

examples_chapter_3.pdf

Documents