Page 1

DESIGN OF A WEIR The purpose of this exercise is to acquaint with the basic scope of a small weir calculation. It must cover the following topics: Calculation part (calculations should be illustrated with sketches):

- designation of spillway width, - selecting the shape of the spillway (trapezoidal), - calculation of stilling basin (length and depht), - calculation of the seepage in weir’s subsoil, - checking the stability of weir’s body and stilling basin

Drawing part: - horizontal projection and cross section of the weir (scale 1:50 or 1:100)

Term: 15. January. 2013 SDF = Spillway design flood [m3/s] Max storage level = maximum storage elevation in the reservoir

No. Name River SDF [m3/s]

Aver. flow rate

[m3/s]

Max storage

level [m.a.s.l.]

Subsoil Class of hydro-

structure

decline of

water surface

i [-]

1 Alvaro Correa Andres Alcazar Pagona 28,00 6,00 93,50 Gravel IV 0,00050

2 Adan Gomez Roberto Lopez Pagona 26,00 5,50 93,20 Soft clay III 0,00053

3 Ana Lopez Val Irene Rivera Pagona 24,00 5,00 93,20

Sand-gravel

mix II 0,00055

4 Piotr Prus Marcin Mosiejko Pagona 22,00 4,50 92,90 Fine

sand II 0,00058

5 Krzysztof

Jednachowski Sergiusz Szreder

Pagona 20,00 4,00 93,20 Gravel IV 0,00060

6 Borys Sajko Irmina Byzdra Pagona 19,00 4,00 92,70

Sand-gravel

mix IV 0,00065

7 Samuel Rodriguez Inigo De la Calle Prosna 100,00 20,00 149,00 Gravel II 0,00045

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No. Name River SDF [m3/s]

Aver. flow rate

[m3/s]

Max storage

level [m.a.s.l.]

Subsoil Class of hydro-

structure

decline of

water surface

i [-]

8 Aleksandra Latiszewska

Beata Mikołajczuk Prosna 90,00 18,00 149,00 Fine

sand IV 0,00048

9 Mateusz Samociuk

Krzysztof Skrzydłowski

Prosna 85,00 19,00 148,80 Sand-gravel

mix III 0,00050

10 Tomasz Zybała Antoni Żabniak Prosna 80,00 16,00 148,70 Soft clay II 0,00052

11 Piotr Stolarski Maciej Serwin Prosna 75,00 15,00 148,50 Medium

sand III 0,00056

12 Anna Jaczewska Timothy Apiyo Prosna 70,00 14,00 148,40 Coarse

sand IV 0,00045

13 Paweł Stanisławski

Maksymilian Mokrzycki

Prosna 95,00 14,00 148,40 Coarse sand IV 0,00055

14 Karolina Zamiar Wojciech Podleś Pagona 15,00 2,00 92,00 Coarse

sand IV 0,00061

15 Krzysztof Tomczak Maciej Zemfler Pagona 16,00 2,50 92,40

Sand-gravel

mix III 0,0004

16 Przemysław

Wróbel Tomasz Szymczak

Pagona 17,00 3,00 92,80 Fine sand III 0,00045

17 Łukasz Jagalski Remi Mordome Pagona 18,00 3,50 93,00 Gravel IV 0,00050

18 Mohammad Ramadan

Marta Gosz Pagona 19,00 4,00 93,40 Fine

sand III 0,00063

19 Daniel Wrzosek

Marcin Mosakowski

Prosna 100,00 18,00 149,0 Sand- gravel

mix IV 0,0007

20 Jędrzej Sznajder Ilga Navitski Prosna 70,00 20,00 150,0 Fine

sand IV 0,0006

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PROSNA river

144.00

146.00

148.00

150.00

0 5 10 15 20 25 30

Valley profile

144.5

145.5

146.5

147.5

148.5

0 50 100

Flow-rate curie (vertical axis: elevation [m a.s.l.]; horizontal axis: flow rate [m3/s])

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PAGONA river

89.00

91.00

93.00

0 2 4 6 8 10 12 14 16

Valley profile

89.089.590.090.591.091.592.092.5

0 5 10 15 20 25 30 35

Flow-rate curie (vertical axis: elevation [m a.s.l.]; horizontal axis: flow rate [m3/s])

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DESIGN OF A WEIR 1. The minimum width of the spillway may be determined depending on the allowable maximum unit discharge::

maxmin q

Qb

35max

21max

1 tin

q

where:: Q – discharge of spillway [[m3/s],, -- coefficient of river bed erosion resistance, qmax - maximum flow found in naturally flowing river ((stream) prior to its construction [m3/sm]] 2. Discharge capacity of spillway

2/32

2gvhbMQ o

where: c - width of the crest [m], M – discharge coefficient, b – is the width of the spillway [m], h - is the head from the water level in the upper pool to the dam crest [m], - St Venant ratio = 1,1, v0 - velocity of the water which flows to the spillway (velocity of approach;) [m/s], g - acceleration due to gravity [m/s2], - side throttling ratio, - submerging coefficient of the spillway

Soil in river bed λ

Rocks 1,80

Stones 1,60

Gravel 1,40

Sand–gravel mixing 1,30

Coarse sand 1,20

Medium sand 1,15

Fine sand 1,10

Loamy sand 1,08

Organic soils 1,05

Medium clay 1,10

Heavy clay 1,15

Clay 1,20

inclination factor m

H / c > 2,0 H / c = 1,0 – 2,0

H / c = 0,5 – 1,0

1 1,86 1,78 1,64-1,70

2 1,78 1,70 1,55-1,60

3 1,73 1,64 *

5 1,67 1,55 *

10 1,55 * *

C 1 : m

H.

v202g

Page 6

2

1h dh

...23

21

oh

q

bhn2,01

where: n - number of bays of the spillway [-], b - width of the spillway [m], h - depth of water above the crest of the spillway[m], ξ - coefficient depending on the shape of pillars and abutments [-].

Submerging coefficient

9,0010

4

2

0

Hadla

Ha or 0,19,0181,0

0

16

0

Hadla

Ha

where: a - the difference between the downstream water level and the level of crest of spillway. If the downstream water level is arranged below the crest of the spillway, then it has no impact on its discharge (spillway is unsubmerged) and = 1.

Condition to meet: Q > SDF

3. Stilling basin

Condition to meet: The first step is to calculate the

fraction where: q=Q/b , =1,1 The second step is to choose the

shape of spillway inlet, and to determine the velocity index i.e.=0,92 for rounded shape of inlet)

The minimum depth of stilling basin: dmin = 0,30 m. The length of stilling basin: l = 5(h2 - h1)

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Page 8

4. Seepage

To avoid soil piping and ensure the stability of dam the outline of the underground part of the foundation should be so developed to provide a sufficiently long filter path to ensure the proper seepage velocity reduction. In an approximate manner the required (minimum) length of filtration path is determined by the formula

hClp C - factor depending on the type of soil and method of calculation h - the maximum difference in water levels on the upstream and downstream (max head) Bligh method

ii hll

li – length of individual horizontal sections hi – length of individual vertical sections

Lane method

ii hll 31

li – length of individual horizontal sections hi – length of individual vertical sections

Condition to meet: pll

With insufficient length of filtration path under the building, we have to increase its length by forcing water to flow around more developed contour. Here we can use a deck (a waterproof element made of clay or geomembrane) located on the reservoir’s bottom and / or sheet pile wall at the front of the dam’s body. In addition, we can also put the sheet pile wall on the end of the stilling basin (at point 7). This solution will reduce the velocity of seepage, but also will increase the uplift pressure acting on the slab (stilling basin).

Subsoil of dam Bligh CB Lane CL

Rocks - 2,5

Stones - 3,0

Gravel 7 3,5

Sand–gravel mixing 9 4,0

Coarse sand 12 5,0

Medium sand 13 6,0

Fine sand 15 7,0

Loamy sand 18 8,5

Organic soils - -

Medium clay 8 3,0

Heavy clay 6 2,0

Clay - 1,6

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5. Uplift pressure

Max storage level=149,00

145,80 (for average flow rate Q=18 m3/s)

300 480

143,7

144,50 144,50 (the bottom level)

143,70

1 3

7

8

6

4

5

2

139,70

144,2030

1 2 3 4 5 6 7 8

h= 3.20

1.30

1.60

1.60

6.10

2.10 1.30

2.10 2.10

1.40 0.90

250

An example of a calculation the uplift pressure acting on the body of the weir.

144,20

V =1,4+0,9

2 *3,0*10 = 34,5 kN/md

V = 2,1*3,0*10 = 63,0 kN/ms V = 34,5 + 63,0 = 97,5 kN/mtot

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6. Loads

Max storage level=149,00

143,7

144,501

3

6

4

5

2

139,70

30

An example of a calculation the water load and self-weight loadacting on the body of the weir.

144,20

148,50w*h1

h1h2

h3

w*(h1+h2)

w*(h1+h2)

P G

centroid ofprofile area

Vd+Vs

Vd+VsVtot

v

g p

Water load

]/_[13636100

8,05,40,42

5,45,02 3212

211

mkN

hhhhhhh

P www

www

Self-weight load

]/_[8,197236,8 mkNAG cb

where: Ab – is the area of weir’s cross section [m2],

c – is the unit weight of concrete = 23 [kN/m3] Uplift pressure load (as in p.5)

]/_[5,97 mkNVtot

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7. Stability of the body of dam Overturning stability - relation of the moments of the hypothetical point of rotation (point

n.6 – see the drawing in p.6)

oM

M mVvPp

Ggmo

u

mo Class of hydro structure

I II III IV

The basic layout of loads 1,20 1,15 1,10 1,05

Sliding stability - relation of forces in the foundation

oF

F nP

VGfnp

u

f = tan- friction coefficient between soil and concrete - internal friction angle of the soil no Class of hydro structure

I II III IV

The basic layout of loads 1,20 1,15 1,10 1,05

To improve the sliding stability of the dam the inclined (towards the reservoir) basis of foundation can be used, teeth or spurs at the front (and rear) wall.

In order to improve the stability of a shift, dam foundation basis shall be inclined. Then the stability of the shift is checked by formula:

onGtgP

PtgVGf

GPVPGfn

cos

sincossincos

– angle of inclination

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8. Stability of stilling basin The minimum thickness of stilling basin must meet the condition:

mdhd

w

cn 3,0min__,

1min

where: c – is the unit weight of concrete = 23 [kN/m3] w – is the unit weight of water = 10 [kN/m3] n – safety factor [-] h – maximum value of hydrodynamical pressure acting on the bottom of stilling basin [m]

n Class of hydro structure

I II III IV

The basic layout of loads 1,20 1,15 1,10 1,05