Example-1 (Design of Glacis Type Weir)

i) Design Data

Particulars U/S D/S Units

Discharge 1800.000 m3/s

H.F.L. 300.000 m

River bed level 293.000 293.000 m

Bed Width m

F.S.L. 299.000 m

Afflux 1.000 m

Permissible exist gradient 0.167

FSL D/S of canal regulator 296.000 m

Canal bed level D/S of regulator m/m

Canal bed level D/S of regulator 1.70

ii) Weir length and RL of weir crest

Regime width,B ;

B= 200 mU/S Max water depth= 7.000 mVelocity of approach,V= 1.3 m/sVelocity head,hv= 0.09 m

U/S total energy level=HFL+hv= 300.090

Effective width of weir=

n ( No of spans) 10 spans= 19Kp( weir contraction coefficient)= 0.01Ka(abutment coeff. ) 45 degree wing w 0.1Pier thickness= 1.5

B-n*pier thickness= 171.5beffective= 169.55H= (Q/bclear*C)^(2/3) 3.37Actual energy head, He=(Q/beff*C)^(2/3) 3.39RL of weir Crest= HFL-He= 296.610D/S HFL= U/S HFL-Afflux= 299.000

iii) U/S and D/S cutoff depths

Check if flow is free or submerged

H1(U/S head above weir crest)= 3.390 mH2(D/S head above weir crest)= 2.390 mH2/H1 (submergence ratio)= 0.71

Hence flow is; flow is free; Q not affected

Correction factor ,f= 1Q (corrected)=Q.f= 1800 m3/s

Regime scoure depth= 5.78 m

Lecey's silt factor,f= 1mean diammeter of bed mater,d(mm)= 0.35 mm

U/S scour depth=1.75*Rs= 10.12 mRL of bottom of U/S cutoff= U/S HFL-Rs= 289.88 mU/S cutoff depth=RBL-RL U/Sbottom cutoff= 3.00 m

D/S scour depth=2*Rs= 11.56 mRL of bottom of D/S cutoff=D/S HFL-Rs 287.440 mD/S cutoff depth=RBL-RL D/S bottom cutoff= 6.000 m

Pond level (FSL of regulator) 296.000 mHence Max. seppage head=PL-RBL= 6.000 m(assuming tail water depth=0)

iv) Total Weir length and proposed weir dimentions

Exit gradient (Ge);

Accuracy level b is solved; -1.279149E-11Total floor length,b= 37.82 m

D/S floor length (length of stilling basin)= 5(Y2-Y1) where Y1 & Y2 are conjugate depths of a hydraulic Jump.

Bernouli @ U/S and Toe of D/S weir; Continuity EquationQ=A1V1=Y1*B*V1

Z0 +vo^2/2g+Hd =Z1+Y1+V1^2/2g V1=Q/(Y1*B)V1= 6.15 m/sY1= 1.46 m/s

-4.6540549E-13Froud No at point 1, F1= V1/(g*Y1)^0.5= 1.62Y2=y1 ((1+8F1^2)^0.5-1)/2= 2.71 m

For stilling basin Type IV ; L= 4*Y2= 11.0 m

Tail water depth, Y3= 6.0 m

Hence the jump occurs; on sloping surface of the weirY3/Y2= 2.22

0.25L/Y2 (read from chart where L= length of jump 5.75L,length of jump= 15.6 mTherefore, provide;

i) U/S horizontal apron,L1= 5 mii) H:V 2:1 sloped U/S glacis,L2= 7.2 miii) Crest width , L3= 1 miv)H:V 4:1 sloped D/S glasis,L4= 14.4 mV) D/S horizontal apron,,L5= 16.0 m

TOTAL PROPOSED FLOOR LENGTH,L= 44 m OK !

bclear

tanα where α is the angle the D/S sloping glacis made with the horizontal

Design Weir Dimentions

A B

C D

E F

v) Uplift Pressure Using Khosla's Theory and floor thickness of weir

Calculation of Uplift pressure :-Let the floor thickness at U/S end = 0.40 m

Let the floor thickness at D/S end = 2.05 m

For u/s cutoff b= 44 md= 3 m

14.55 m 0.068713

7.79

0.74

0.87

23.32 %

16.30 %

76.68 %

83.70 %

(i) Correction for floor thickness :-Correction = 0.94 (+) %

(ii) Correction due to the effect of d/s cutoff on u/s cutoff :-d = 2.60D= 3.95b = 44b' (assuming the piles placed @ 0.5m from the end) 43

0.87 (+) %

So, correction due to effect of d/s cutoff on u/s cutoff = 0.87

78.48 %

For d/s cutoff b= 44 md= 6.00 m

7.28 m

4.17

0.52

0.76

32.57 %

22.50 %

(i) Correction for floor thickness :-Correction = 3.43 (-) %

(ii) Correction due to the effect of u/s cutoff on d/s cutoff :-d = 3.95D= 2.60b = 44b' = 43

0.70 (-) %

28.43 %

5 7.2 15.4 16.0

2.81 1.713.87

4.37Pressure 4.71Total head between U/S and Downstream= FSL-D/S river bed level= 6.000 m

Hence Pressure above weir floor at point C= 0.26 mE= 2.81 mF= 1.71 m

Weir floor thickness at point C= 0.31 mE= 3.37 mF= 2.05 m

0.00

Vi) Protection works

i) For D/S concrete apron block with open joints filled with spawls;

Length= 1.5*d2= 1.5*6= 9 mHence use cast-insitu concrete blocks of 1m * 1.5m *1m deep laid open jointed over a graded filter construction.

ii) For D/S flexible apron of brocken stone

length= 2.5 d2 =2.5*6= 15 mVolume content per meter length=2.63*d2= 15.78 m^3/mThickness of apron=Volume/length= 1.05 mMinimum stone size= 0.10Mean Velocity= 1800/200*6= 1.5 m/s

1.65iii) For U/S impervious concrete block apron

length=1.5d1= 4.5Use cast insitu concrete block 1m*1.5m*0.6m deep

Vi) For U/S flexible apron

length=2*d1= 6 mVolume=2.25*d1= 6.75 m^3/mThickness of apron= 1.13 m

a= (b/d) =

l=(1+(1+a2)0.5)/2 =((l-2)/l) =((l-1)/l) =

fE =(1/p) * cos-1(l-2/l) =

fD =(1/p) * cos-1(l-1/l) =

fC1 = 100-fE =

fD1 = 100-fD =

Correction for fC1

Correction = 19*(D/b')0.5*((d+D)/b) =

fC1 Corrected =

a= (b/d) =

l=(1+(1+a2)0.5)/2 =((l-2)/l) =((l-1)/l) =

fE2 =(1/p) * cos-1(l-2/l) =

fD2 =(1/p) * cos-1(l-1/l) =

Correction for fE2

Correction = 19*(D/b')0.5*((d+D)/b) =

fE2 Corrected =

∆=

A B C E F