23. دیزاین برای
Weir
شود آورده بدست ساحه از ذیل معلومات باید
:
•
سیالب اعظمی مقدار
(Q).
•
ساختمان اعمار از قبل سیالب اعظمی سطح
(H.F.L).
•
ع بستر سطح
بند قب
(Downstream bed level).
•
کانال در آب اعظمی سطح
(F.S.L).
•
مجازی جریان
(Allowable afflux).
•
لیسی ضریب
(Lacey’s silt factor).
1:- Hydraulic calculation
I. L= 4.75 Q1/2 ………..(1.1)
L = Length of water way in (m)
Q = discharge in (m3)
II. Q = Q/L ...…….(1.2)
q = the discharge per unit length of waterway (m3)
III. R = 1.35( q2/f )1/3 ………..(1.3)
R = scour depth in (m)
f = Lacey’s silt factor
IV. V = q / R ……….(1.4)
velocity head = v2/2g ……….(1.5)
g = 9.81
V. Water levels and total energy (T.E.L.) on the downstream side (d/s) and
upstream side (u/s) are calculated as follow:
Level of d/s T.E.L. = (H.F.L. before construction ) + v2/2g
Level of u/s T.E.L. = Level of d/s T.E.L. + Afflux
Level of u/s H.F.L. = Level of u/s T.E.L. – v2/ 2g
VI. Discharge over the Crest of the weir is given by:
24. q= 1.7k3/2 …………(1.6)
k= (q/1.7)2/3 …………(1.7)
٭
Crest level = u/s T.E.L. – k
VII. Pound level = Level of top of gates
= F.S.L. of canal + Head loss through regulator
Head loss through the regulator may be taken as ½ to 1 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. – 1.5R
Level of bottom of d/s pile = H.F.L. after retrogression – 2R.
2:- Design of Weir Wall:
دیوار
Weir
م شکل به معموآل
جناح هردو در یی ذوذنقه قطع
(u/s, d/s )
قسمت در که .میگردد اعمار
u/s
قسمت در و عمودی شکل به
d/s
مایل شکل به
.
مقابل در طوریکه میباشد دیوار پایینی قسمت و باالیی قسمت عرض نمودن تعین شامل ویر دیوار دیزاین
باشد پایدار اعظمی فشار
.
ب نظر باید بند استواری این بنابر
شود محاسبه ذیل حالت سه ه
:
:اول حالت
-
)نکند عبور بند باالی قسمت از (آب باشد بند تاج ارتفاع با مساوی آب ارتفاع صورتیکه در
.
:دوم حالت
-
نماید عبور بند باال قسمت از آب صورتیکه در
.
:سوم حالت
-
نماید عبور بند تاج باالی قسمت از آب درصورتنکه
.
:بند باالیی عرض
-
باال عرض
میگردد محاسبه ذیل های ازفرمول بند یی
:
B1= d/√G …….. (1.
فوق فرمول در
:
B1-
بند باالیی قسمت عرض
d-
به است مساوی که بند تاج باالی در آب اعظمی ارتفاع
:
u/s H.F.L. – Crest Level
26. Mr = {WH (G-1)/6} (B2+B1B-B12)
:سوم حالت
-
شونده واژگون مومنت حالت این در .نماید عبور بند تاج باالی قسمت از آب درصورتنکه
(Overturning Moment)
میگردد محاسبه ذیل فورمول از
.
Mo = WH3/6 + WdH2/2 – WD3/6
or
Mo = W/6 (H3+3dH2 – D3) ……. (1.16)
d-
به است مساوی که بند سطح از آب ارتفاع
:
d= kD
k-
است ثابت عدد یک
.
میگیرد خود به را ذیل شکل اعظمی مومنت فوق های فورمول درنظرداشت با
.
Mo = W/6 (H3 + 3k3/2H3 – k2/3H3)
Or
Mo = WH3/6 (1 + 2k3/2) ....... (1.17)
بند کف دیزاین
(Design of Apron):-
از عبارتند هریک که میگردد محاسبه ذیل های فورمول از بند کف دیزاین
:
• Downstream apron (L1). For weirs without crest shutters.
L1 = 2.21 C√Hs/10 ……. (1.18)
for weirs with crest shutters
L1 = 2.21 C√Hs/13 ……. (1.19)
• Upstream apron (L2). According to Bligh’s theory.
L2 = L – L1 – (B + 2d1 + 2d2) …….. (1.20)
According to the Khosla’s theory
L2 = b – L1 – B ……. (1.21)
• Total length of d/s apron
27. the total length L3 of the d/s impervious floor and the u/s impervious apron is
given by the following:
For weir with crest shutters.
L3 = 18C √ Hs/13 x q/75
for weir with no crest shutters.
L3 = 18C √ Hs/10 x q/75
:مثال
-
تیوری داشت درنظر با است ذیل مشخصات دارای که عمودی سربند یک
creep Bligh’s
دیزاین
نمایید؟
a) Maximum flood discharge: = 2800 M3/sec
b) H.F.L before construction: = 285 m
c) Minimum water level = d/s bed level = 278 m
d) F.S.L of canal = 284 m
e) Allowable afflux = 1 m
f) Coefficient of creep = 12
g) Permissible exit gradient = 1/6
دهید تغیر را فوق داتا که میتوانید ضرورت صورت در
.
(A)Hydraulic Calculations:
Step 1. Q=2800 m3/sec
L = 4.75 Q1/2 = 4.75 (2800)1/2 m
= 251 m
q= Q/L= 2800/251 = 11.2 M3/sec
Step 2.Regime scour depth: f=1
R= 1.35 (q2/f) 1/3 = 1.35 [(11.2)2/1]1/3
= 6.76 m
Regime Velocity = q/R = 11.2/6.76 = 1.66 m/sec
28. Velocity head = V2/2g = (1.66)2/2x9.81 = 0.14 m
Step 3. Level of d/s T.E.L. = H.F.L before cons. + V2/2g
=285 + 0.14 = 285.14 m
Afflux = 1 m
Level of u/s T.E.L. = d/s T.E.L. + Afflux
= 285.14 + 1 = 286.14 m
٭
u/s H.F.L. = u/s T.E.L. - V2/2g = 286.14 – 0.14
= 286
Actual d/s H.F.L. allowing 0.5 m for retrogression
= 285 – 0.5 = 284.5 m
Step 4. q= 1.7 k3/2
k = (q/1.7)2/3
= (11.2/1.7)2/3
= 3.56 m
٭
Crest level = u/s T.E.L. – k
= 286.14 – 3.56
= 282.58 m.
Step 5.Pond level= Level of top of gates
= F.S.L. in canal + Head loss through regulator
= 284 + 0.5
= 284.5
٭
Height of shutter = s = Level of top of gates – crest level
= 284.5 – 282.58
= 1.92 m.
29. Step 6. Level of bottom of u/s pile
= u/s H.F.L. – 1.5 R
= 286 – (6.75 x 1.5)
= 275.88 m.
u/s pile may take up to a level of 276 m.
Dept of u/s cut off = 278 – 276
= 2 m.
Hence provide concrete cutoff of 2 m depth below the bed of river at the u/s end
of the floor.
Level of bottom of d/s pile (d2)
= d/s H.F.L. after retrogression – 2R
= 284.5 – (6.75 x 2)
= 271 m.
d2= 278 – 271 = 7 m.
Step 7.
Head of water
Hs = Level of crest gates – Bed level
= 284.5 – 278
= 6.5 m.
Height of crest H = Crest level – Bed level
= 282.58 – 278
= 4.58 m.
(Check: Hs = H + s = 4.85 + 1.92 = 6.5 m.)
(B) Design of weir wall (B1)
Step 8.
30. Calculation of top width
d = u/s H.F.L. – Crest level
= 286 – 282.5
= 3.42 m.
٭
Top width B1 = d/√G
= 3.42/√2.24 = 2.3 m.
From sliding consideration,
a = 3d/2G = 3x3.42/ 2x2.4
= 2.3 m.
From practical considerations,
B1 = s + 1 = 1.92 + 1
= 2.92 m.
Hence provide top width
B1 = 3 m.
Step 9.Calculation of bottom width (B)
Consideration state I when the water is to top of the crest gates, with no tail
water, the over turning moment is given by:
Mo = W (H + s)3/ 6 = WHs3/ 6
= 1 x (6.5)3/6 = 45.77 ≈ 46 t-m.
The moment of resistance is given by equation (1.12) as
Mr = W/12 [ { (G+1.5) H+2.5S} B2+B1 (GH-H-S) B-1/2B12 (H+3S) ]
H = 4.58; s = 1.92; B1 = 3 m; G = 2.24.
Introducing these values in the above equation and equating it to the overturning
moment, we get
46 = 1/12 [ { (2.24 + 1.5) 4.58 + 2.5 x 1.92} B2 + 3 (2.24 x 4.58 – 4.58
31. – 1.92) B – 1/2 x 32 (4.58 + 3 x 1.92) ]
قیمت دو درجه مجهوله یک معادله حل از بعد
B
به میشود مساوی
:
B = 4.97 ≈ 5 m.
Consideration II.
When the water is flowing over the weir and the weir is submerged. Considering
the tail water just at the crest of the weir, the overturning moment is given by:
Mo = wh H2/2
When the water is at the crest, d and h will be equal. For this case, the value of d
is given by
d = [q2/ (2/3C)2 x 2g]1/3
(Neglecting the velocity of approach) where C = 0.58 is the coefficient of
discharge.
Substituting q= 11.2 M3/sec/m, we get
d = h = [(11.2)2/ (2/3 x 0.58)2 x 2 x 9.81]1/3
= 3.5 m.
Mo = wh H2/2
= 1 x 3.5 (4.58)2/ 2
= 36.6 t-m.
The moment of resistance is given by:
Mr = {wH(G-1)/12} (B2 + B1B)
= 1 x 4.85 (2.24 – 1) / 12} (B2 + 3B)
= 0.473 (B2 + 3B)
Equating the two, we get
0. 473 (B12 + 3B) = 36.6
From which B = 7.4 m.
32. Provide B = 8 m.
Thus the weir wall has top of 3 m and bottom width equal to 8 m, with equal
slopes of u/s and d/s faces.
(C) Design of impervious and pervious aprons:
Step 10.Assuming C = 12, total creep length is
L = CHs = 12 x 6.5
= 78 m.
Step 11.
d/s impervious apron (L1):
L1= 2.21C √ Hs/13
= 2.21 x 12 √ 6.5/13
= 18.8 ≈ 19 m.
Step 12.u/s impervious apron (L2):
L2 = L – L1 – (B + 2d1 + 2d2)
d1= 278 – 276 = 2 m.
d2= 278 – 271 = 7 m.
B = 8 m.
L2 = 78 – 19 – (8 + 2 x 2 + 2 x 7)
= 33 m.
Step 13.
Total length of d/s apron
L3 = 18C √ (Hs/13) (q/75)
L3 = 18 x 12 √ (6.5/13) (11.2/75)
L3 = 53.2 m. ≈ 54 m.
Step 14.
33. Length of filter + launching apron
= L3 – L1 = 54 – 19 = 35 m.
Minimum length of inverted filter
= 1.5 d2 = 1.5 x 7 = 10.5 m.
Minimum horizontal length of launching apron
= 2.5 d2 = 2.5 x 7 = 17.5 m.
Total minimum length of both
= 10.5 + 17.5 = 28 m.
But we have to provide 35 m length of both.
Step 15.
u/s block protection and launching apron:
d1 = 2 m
Length of u/s block protection = d1 = 2 m.
Length of u/s Talus = 2d1 = 2x 2 = 4 m.
Step 16. Thickness of impervious floor
Provide a nominal thickness of 1 m to the u/s of the weir wall and 1.5 m below
the weir wall
Residual pressure at point A of figure just at the d/s of weir wall
Hr = Hs – (Hs/L) x
= 6.5 – 6.5/78 (2 x 2 + 33 +
Hr = 2.6 m.
٭
Thickness (t) = 4/3 (Hr/G – 1)
= 4/3 (2.6/2.24 – 1) = 2.8 m.
Hence provide a thickness of 2.8 m from the d/s of weir wall to a point 6 m from
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