Vertical Drop Weir

VERTICAL DROP WEIR
Vertical drop weir consists of a masonry wall with a vertical tot Vertical nearly downstream face and a
horizontal concrete floor. The shutters are provided at the crest, which are dropped during flood so as
to reduce afflux. The water is pounded up to the shutters during the rest of the period.
The Weir floor immediately designed as gravity section at the upstream end of the floor block protection
and at the downstream end a graded inverted filter provided. Launching aprons are provided at the uls
and dis ends. the floor to safeguard against scouring action. This type of Wheir is suitable for hard clay
foundation as consolidated gravel foundation, and where the al well drop is small. Such weirs are
absolute now a days.
Design of Vertical drop Weirs:

The Complete design of a vertical drop weir consists of the design calculations for the following:
1. Hydraulic calculations for fixing various elevations

2. Design of Weir wall.
3. Design of impervious apron.
4. Design of inverted and d/s talus.
Design example of a Vertical drop Weir
a) Design a Vertical drop weir on Blights theory for the following site conditions: as Maximum
flood discharge = 2800 cumecs
b) H.F.L before construction = 285.0 mts
c) Minimum bed level = d/s bed level = 278.0 mts
d) F.S.L of canal = 284 mts
e) Allowable afflux = 1 mts
f) Coefficient of creep= 12
g) Permissible exit gradient = 1/6
Assume other data not given any
Solutions:
1. Hydraulic Calculations:
Given Q = 2800 cumecs
* Length of water way (L) = 4.75 Q1/2
=4-75 (2800)1/2
= 250 mts
*Discharge per unit width of river (q) = Q/L=2800/250=11.2 cumecs/m
* Regime scour depth (R) = 1·35 (q²/f) 1/3 6.75 mts
~Take f=1
* Regime Velocity (v) = q/R=11.2/6.75=1.66m/sec
*Velocity head =V 2/2g= (1.66)2/2(9.81) = 0.14 mts
*Level of d/s T.E·L= H.F.L before construction + V2/2g
= 285.00+ 0.14=285.14 mts
*level of u/s T·E·L= d/s T·E·L + afflux= 286.14 mts
* u/s high flood level (H·F·L) = u/s T·E·L- V2/2g=286.14-0.14=286 mts
*Actual d/s H.F.L allowing 0.5m for retrogation = 285-0.5=284.5 mts
* Discharge Over the crest of the weir
Q = 1.7(k)3/2
*K = (11.2/1.7)2/3
k= 3.56 mts
*Crest level = u/s T.E.L - K
= 286-14 - 3.56
= 282.58 mts
* Pond level = FSL of canal + Head loss through regulator
= 284 +0.5 (say)
= 284.5 mts
* weight of shutter (s) = Level of top of gates - Crest level
=284.15 – 282.58
=1.92 mts
*RL of bottom of u/s pile =u/s HFL-1.5R
=286-1.5 (6.75) = 275.88 mts=276 mts
* Depth of uls cut off = 278-0-276-0 = 2 mts
* R.L of bottom of d/s pile

=d/s H.F.L. after retrogation -2R
= 284-50-2 (6.75)
= 271.0 mts
*Depth of d/s cutoff = 278 -271 = 7 mts
* Seepage head of water (Hs) = R.L. of gate crest- R.L. of bed=284.5-278= 6.50 mts
* Height of crest (H) = R.L. of Crest - R.L. of bed
=282-58- 278
= 4.58 mts
* {check Hs = H+S = 4·58 +1-92 = 6·50 m}
2) Design of Weir wall:
* d = u/s H.F.L. -crest level
=286-282.58
= 3.42 mts
* Top Width, a=d/sqrt(p) = 2.3 mts
From Sliding consideration, a=3d/2f=3(3.42)/2(2.24) =2.29m
From practical considerations,
a = s+1 = 1.92 +1 = 2.92 mts
Hence, provide a top width a = 3 mts
*Bottom Width,
→ Considering state 1, when the water is to the best gates, with no tail water, the overturning moment
is given by
M0 =W(H+S)3/6= 449 KN-m
The Moment of resistance is given by
Mr=w/12(((p+1.5) H+2.5S) b2 + a(pH-H-S) b-1/2a2 (H+3S))
B= 4.96m =5 mts
Consider state II, when the water is flowing over the weir and the weir is submerged.
Considering the tail water just at the rest of the weir, the overturning moment is given by
M0= whH2/2
M0= 360.1 KN-m
Minimum length of inverted filter =1.5 d₂
= 1·5 (7) = 10.5 m
Minimum horizontal length of launching apron
= 2.5 d₂ = 2.5 (7) = 17.5 mts
length of both = 10-5+17-5 = 28 mts
Total minimum length of both= 10.5+17.5 = 28 mts
But we have to provide length of both = 40 m
Hence, provide the filter of 20m length launching apron of 20m length
4) Design of launching aprons, block protection & invented filter:
*U/s slide,
Launching apron = 2d₁ = 2(2) = 4 mts
Block protection = d₁ = 2m
* D/s Side,
Launching apron = 2.5 d₂ = 2.5 (7) = 17.5 mts
Inverted filter = 1.5 d₂ = 1·5 (7) = 10.5 mts

Vertical Drop Weir

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VERTICAL DROP WEIR

Design of Vertical drop Weirs:

1. Hydraulic calculations for fixing various elevations

* R.L of bottom of d/s pile

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