check dam ( rock exposed)

DESIGN OF CHECK DAM ON EXPOSED ROCK

INPUT DATA

Maximum flood discharge Qa = 1022 m3/sec

Bed slope s 1 in 2000 = 0.0005

River width L = 130 m

Deepest bed level BLd = 8.35 m

Average bed level BLa = 9.171 m

Bottom level of weir/check dam BLW = 9.5 m

Maximum flood level MFL = 15.4 m

Height of check dam Ht = 2 m

Top width of weir a = 2 m

Density of concrete dc = 2.4 t/m3

Density of water dw = 1 t/m3

Afflux afx = 0.3 mRetrogression ret = 0.2 mAcceleration due to gravity g = 9.81 m2/sec

DESIGN

1 Discharge per unit width q = Qa/L = 7.862 m3/sec/m

2 Average depth of flow d = MFL-Bla = 6.229 m

3 Velocity of flow v = q/d = 1.262 m/sec

4 Velocity head hd = 0.081 m

5 Crest level Crl = BLW+Ht = 11.5 m

6 Downstream total energy line D/s TEL = MFL+hd = 15.481 m

7 Downstream water level DWL =D/s TEL-hd-ret = 15.2 m

8 Upstream total energy line U/s TEL = D/s TEL+afx = 15.781 m

9 Upstream water level UWL =U/s TEL-hd = 15.7 m

10 Depth of flow over weir H = UWL-Crl = 4.2 m

11 Drowning ratio DR =(DWL-Crl)/(UWL-Crl) = 0.881

= v2/2g

Display the value of DR to input the corresponding value of Cd

12 Coefficient of discharge Cd 1.56(Input taken from graph depending on the value of DR)

13 Calculated discharge Qth =Cd*L*H^(3/2) = 1745.589

Check whether Qth > Qa 1745.589 > 1022

Hence safe

Check for discharge for broad crested weir

No. of vents nV (L/10-1.1) = 11.90rounded 12

No. of end contractions n 2+2*nV = 26

Calculated discharge Qthb 1.705*(L-0.01*n*H)*H^1.5 = 1891.813 m3/sec

1891.81292959834 > 1022

Hence safe

Bottom width of weir

Case 1 When the u/s water is at crest level and there is no flow downsteam

Overturning moment Mo (dw*Ht^3)/6 = 1.333 tm

Resisting moment Mr (dw*Ht*dc*(b2+ab-a2))/6 =

0.8 (b2+ab-a2)

Equating Mo=Mr

Constant for calculation x a^2+(Ht^2)/dc = 5.667

Bottom width b (-a+sqrt(a2+4*x))/2 = 1.582 m

Case II When the weir is just submerged

Under maximum flow condition depth of flow over the weir H = 4.2

Depth of water in the d/s H2 H+Ht-afx-ret = 5.7

ie

Constant K H/H2 = 0.736842

Depth of flow over the crest whentail water is at crest level D K*Ht = 1.474

ie, when a depth of 1.474 m flows over the weir, the weir will be just submerged.If Ht is the height of weir and D is the difference in water level then

Max overturning moment Mo2 dw*Ht^2*D/2 = 2.948 tm

Moment of resistance Mr2 (Ht/6)*(dc-1)(b^2+ab-a^2)

Equating Mr2=Mo2

(Ht/6)*(dc-1)(b^2+ab-a^2) = 2.948

b^2+ab = a^2+Mo2*6/(Ht(dc-1))

Let x2= a^2+Mo2*6/(Ht(dc-1)) = 10.31714

b^2+ab-x2=0

Bottom width b = (-a+sqrt(a^2+4*x2))/2 = 2.364 m

Case III When the water is passing over the weir crest and the water is discharging withfree overfall (This case is to be done only if L>30*Ht)

30.Ht = 60 < L

Under maximum flow condition depth of flow over the weir H = 4.2

Depth of water in the d/s H2 H+Ht-afx-ret = 5.7

Constant K H/H2 = 0.736842

Overturning moment Mo3 dw*Ht^3(1+2K^(3/2))/6 = 3.020004

Resisting moment Mr3 dwHt(dc-1)(b^2+ab-a^2)/6

dw.Ht.(dc-1)/6 = 0.466667

Mr3 = Mo3

0.46666667 (b^2+ab-a^2) = 3.020004

(b^2+ab-a^2) = 6.471438

constant x3 = a^2 + 6.47143791094897 = 10.47144

b^2+ab-x3=0

Bottom width b = (-a+sqrt(a^2+4*x3))/2 = 2.387 m

OUTPUTCase Bottom width

1 1.5822 2.3643 2.387

Enter the value of Bottom width b = 2.6 m(maximum rounded)

Check for stress at the foundation level of weir

Case 1 Water upto crest level and no tail water

Offset of foundation c = 0.15 m

Depth of foundation df = 0.3 m

Height of weir from foundn level Hf Ht+df = 2.3 m

Base width at foundation level B b+2c = 2.9

Let bma b-a = 0.6

ANALYSIS

Body wall W1 bma*Ht*dc/2 = 1.44la1 c+bma*2/3 = 0.55

W2 a*Ht*dc = 9.6la2 b-a/2+c = 1.75

Foundation W3 B*df*dc = 2.088la3 B/2 = 1.45

Wt of water W4 c*Ht*dw = 0.3la4 B-c/2 = 2.825

Water pressure H (1/2)*Hf^2 = 2.645la Hf/3 = 0.766667

Vertical force V W1+W2+W3+W4 = 13.428 t

Horizontal force H = 2.645 t

Resisting moment MR W1*la1+W2*la2+ = 21.4671

W3*la3+W4*la4

Overturning moment MO H*la = 2.027833

constant x' (MR-MO)/V = 1.447667

Eccentricity e B/2-x' = 0.002333

Maximum pressure Pmax (V/B)*(1+6e/B) = 4.652699 t/m2

Minimum pressure Pmin (V/B)*(1-6e/B) = 4.60799 t/m2

Check for stability at bed level

Body wall W1 bma*Ht*dc/2 = 1.44la1 bma*2/3 = 0.4

W2 a*Ht*dc = 9.6la2 b-a/2 = 1.6

Water pressure H (1/2)*Ht^2 = 2la Ht/3 = 0.666667

Vertical force V W1+W2 = 11.04 t

Horizontal force H = 2 t

Resisting moment MR W1*la1+W2*la2 = 15.936



Eccentricity e b/2-x' = 0.022705

Maximum pressure Pmax (V/b)*(1+6e/b) = 4.469 t/m2

Minimum pressure Pmin (V/b)*(1-6e/b) = 4.024 t/m2

DESIGN OF LEFT ABUTMENT

MFL after construction MFLac MFL+afx = 15.7 m

Bottom level of weir BLW = 9.5 m

Left Bank Level LBL = 49.83 m

Top level of abutment TLA MFLac+0.1 = 15.8 m

Hard Rock Level HRL = 46.93 m

Height of abutment Hta(cal) TLA-HRL = -31.13 m

Hta Say 3.2 m

Top width of abutment tw = 0.6 m

Bottom width of abutment bwa = 1.92 m

bw Say 1.9 m

ka = 0.333333

r = 2.1 t/m3

ct1 ka.r.tw = 0.42

ct2 ka.r.Hta+ct1 = 2.66

constant btw bw-tw = 1.3

Case 1 Backfill saturated and no water in the river

Earth pressure EP (ct1+ct2)Hta/2 = 4.928 t/m2

Lever Arm LA ((ct2+2ct1)/(ct2+ct1))(Hta/3) 1.212121 m

ANALYSIS

Vertical Force W1 tw.Hta.dc = 4.608 tla1 tw/2 = 0.3 m

W2 btw.Hta.dc/2 = 4.992la2 tw+btw/3 = 1.033333

W3 btw.Hta.r/2 = 4.368la3 tw+btw*2/3 = 1.466667

Total V W1+W2+W3 = 13.968

Resisting Moment MR W1*la1+W2*la2+W3*la3 = 12.9472

Horizontal Force H EP = 4.928

Overturning Moment MO EP*LA = 5.973333

Let xa (MR-MO)/V = 0.499275

ea bw/2-xa = 0.450725

Maximum pressure Pmax (V/bw)*(1+6*ea/bw) = 17.8154

Minimum pressure Pmin (V/bw)*(1-6*ea/bw) = -3.112244

Factor of safety (overturning) FSo MR/MO = 2.1675 >2

Factor of safety (sliding) FSsl 0.6*V/H = 1.700649 >1.5

Case II Water upto MFL and backfill submerged

ct22 ka(r-1).Hta+ct1 = 1.593

Earth pressure EP (ct1+ct22)Hta/2 = 3.2208

Lever Arm LA ((ct22+2ct1)/(ct22+ct1))* 1.28922 (Hta/3)

ANALYSIS

Vertical Force W1 tw.Hta.dc = 4.608la1 tw/2 = 0.3



Buoyancy W4 (tw+bw)Hta*1/2 = -4la4 (tw^2/2+btw(tw+btw/3)/2/ 0.681333

(tw+btw/2)Total V W1+W2+W3+W4 = 9.968

Resisting Moment MR W1*la1+W2*la2+W3*la3 = 10.22187 +W4*la4



Let xa (MR-MO)/V = 0.608903

ea bw/2-xa = 0.341097



Factor of safety (overturning) Fso MR/MO = 2.461724 >2


Foundation for Left Abutment

Backfill saturated and no water in river

INPUT

Offset ca = 0.2 m

Depth dfa = 0.5 m

Height from foundn level Hfa Hta+dfa = 3.7

Bottom width Ba bw+2*ca = 2.3

Let ct2a ka.r.Hfa+ct1 = 3.01

Earth pressure EP (ct1+ct2a)Hfa/2 = 6.3455 t/m2

Lever Arm LA ((ct2a+2ct1)/(ct2a+ct1)) 1.384354 m*Hfa/3

ANALYSIS

Vertical Force W1 tw.Hta.dc = 4.608 tla1 tw/2+ca = 0.5 m

W2 btw.Hta.dc/2 = 4.992la2 tw+btw/3+ca = 1.233333

Wt of soil W3 btw.Hta.r/2 = 4.368la3 tw+btw*2/3+ca = 1.666667

Foundation W4 Ba*dfa*dc = 2.76 la4 Ba/2 = 1.15

Wt of soil W5 ca*Hta*r = 1.344la5 Ba-ca/2 = 2.2

Resisting Moment MR W1*la1+W2*la2+W3*la3 = 21.8716 +W4*la4+W5*la5



Vertical Force V W1+W2+W3+W4+W5 = 18.072

Let xa (MR-MO)/V = 0.724169

ea Ba/2-xa = 0.425831

Maximum pressure Pmax (V/Ba)*(1+6*ea/Ba) = 16.58588

Minimum pressure Pmin (V/Ba)*(1-6*ea/Ba) = -0.871096



16.5858790170132

x0.871096 2.3 -x

x Pmin*Ba/(Pmax+Pmin) = 0.127493

(x/Ba)*100 = 5.543166

< 20%

Check for tan a

tan a dfa/ca = 2.5

0.9*sqrt(qo/fck+1) 0.9*sqrt(Pmax/15+1) = 1.306001

< 2.5

Hence safe

DESIGN OF RIGHT ABUTMENT (with surcharge)

MFL after construction MFLac MFL+afx = 15.7 m

Bottom level of weir BLW = 9.5 m

Right Bank Level RBL = 52.8 m

Top level of abutment TLA MFLac+0.1 = 15.8 m

Hard Rock Level HRL = 46.93 m

Height of abutment Hta(cal) TLA-BLW = 6.3 m

Let Hta 3.2 m

Surcharge height SHt RBL-TLA = 37 m

Bank height at the end of abutment after pitching at a slope of 1:1is taken as surcharge

Surcharge Sr bw-tw = 1.6 m

Top width of abutment tw = 0.6 m

Bottom width of abutment bw(cal) = 1.92 m

bw INPUT 2.2 m

ka = 0.333333

r = 2.1 t/m3

ct1 ka.r.Sr = 1.12

ct2 ka.r.Hta+ct1 = 3.36

constant btw bw-tw = 1.6

Case 1 Backfill saturated and no water in the river

Earth pressure EP (ct1+ct2)Hta/2 = 7.168 t/m2

Lever Arm LA ((ct2+2ct1)/(ct2+ct1))(Hta/3) 1.333333 m

ANALYSIS

Vertical Force W1 tw.Hta.dc = 4.608 tla1 tw/2 = 0.3 m



W4 Sr.Sr.r/2 = 2.688la4 Sr*2/3+tw = 1.666667

Total V W1+W2+W3+W4 = 18.816

Resisting Moment MR W1*la1+W2*la2+W3*la3 = 21.7856 +W4*la4



Let xa (MR-MO)/V = 0.649887

ea bw/2-xa = 0.450113





Foundation for Right Abutment

Backfill saturated and no water in river

Offset ca = 0.2 m

Depth dfa = 0.5 m

Height from foundn level Hfa Hta+dfa = 3.7

Bottom width Ba bw+2*ca = 2.6

Let ct2a ka.r.Hfa+ct1 = 3.71

Earth pressure EP (ct1+ct2a)Hfa/2 = 8.9355 t/m2

Lever Arm LA ((ct2a+2ct1)/(ct2a+ct1)) 1.519324 m*Hfa/3

ANALYSIS

Vertical Force W1 tw.Hta.dc = 4.608 tla1 tw/2+ca = 0.5 m

W2 btw.Hta.dc/2 = 6.144la2 tw+btw/3+ca = 1.333333

Wt of soil W3 btw.Hta.r/2 = 5.376la3 tw+btw*2/3+ca = 1.866667

Foundation W4 Ba*dfa*dc = 3.12 la4 Ba/2 = 1.3

Wt of soil W5 ca*(Hta+Sr)*r = 2.016la5 Ba-ca/2 = 2.5

W6 Sr.Sr.r/2 = 2.688

la6 Sr*2/3+tw+ca = 1.866667

Resisting Moment MR W1*la1+W2*la2+W3*la3 = 34.6448+W4*la4+W5*la5+W6*la6



Vertical Force V W1+W2+W3+W4+W5 = 23.952 +W6

Let xa (MR-MO)/V = 0.879629

ea Ba/2-xa = 0.420371

Maximum pressure Pmax (V/Ba)*(1+6*ea/Ba) = 18.14904

Minimum pressure Pmin (V/Ba)*(1-6*ea/Ba) = 0.275577



18.1490384615385

x0.275577 2.6 -x

x Pmin*Ba/(Pmax+Pmin) = 0.038888

(x/Ba)*100 = 1.4957

< 20%

Check for tan a

tan a dfa/ca = 2.5

0.9*sqrt(qo/fck+1) 0.9*sqrt(Pmax/15+1) = 1.337927

< 2.5

Hence safe

Check for stress at deepest foundation level

Bottom level of weir/check dam BLW = 47 m

Deepest Ground Level DGL = 46.085 m

Allowing 30cm embedment into FL DGL-.3 = 45.785 Mrock, founding Level of Check Dam

Offset of foundation c = 0.15 m

Depth of foundation dfd BLW-FL = 1.215 m

Height of weir from foundn level Hfd Ht+dfd = 3.215 m

Base width at foundation level B b+2c = 2.9

Let bma b-a = 0.6

ANALYSIS

Body wall W1 bma*Ht*dc/2 = 1.44la1 c+bma*2/3 = 0.55

W2 a*Ht*dc = 9.6la2 b-a/2+c = 1.75

Foundation W3 B*dfd*dc = 8.4564la3 B/2 = 1.45

Wt of water W4 c*Ht*dw = 0.3la4 B-c/2 = 2.825

Water pressure H (1/2)*Hfd^2 = 5.168112la Hfd/3 = 1.071667

Vertical force V W1+W2+W3+W4 = 19.7964 t

Horizontal force H = 5.168112 t

Resisting moment MR W1*la1+W2*la2+ = 30.70128W3*la3+W4*la4



Eccentricity e B/2-x' = 0.178921

Maximum pressure Pmax (V/B)*(1+6e/B) = 9.353332 t/m2

Minimum pressure Pmin (V/B)*(1-6e/B) = 4.299358 t/m2



Design of Apron

Discharge per unit width q = Qa/L = 7.861538 m3/sec/m

Afflux afx = 0.3 m

From Blench curvesEf2 = 1.7 m

Ef1 Ef2+afx = 2 m

D1 = 0.7

D2 = 1.5

Length of cistern 5(D2-D1) = 4 m

Provide a nominal apron of 5m length d/s and 3m length u/s with a nominal thickness of 0.3 m

INPUT

INPUT

INPUT

INPUT

INPUT

INPUT

INPUT

INPUT

INPUT

INPUT Increase by 10cm until calculated discharge Qth/Qthbis more than the max flood discharge Qa

(Input taken from graph depending on the value of DR)

If calculated discharge is less than the actual discharge (Qth < Qa), increase the afflux(afx) by 0.05m.If afflux is more than 1 Display the message "CHECK DAM IS NOT FEASIBLE AT THIS SITE" and exit.Otherwise recalculate the steps from 8 to 11 to get the new value of DR. Input the value of Cd corresponding to the new value of DR and calculate the new Qth and repeat the process until Qth>=Qa.

If calculated discharge is less than the actual discharge (Qth < Qa), increase the afflux(afx) by 0.05m.If afflux is more than 1 Display the message "CHECK DAM IS NOT FEASIBLE AT THIS SITE" and exit. Otherwise recalculate the steps from 8 to 10 to get the new value of H. Calculate the new Qthb and repeat the process until Qthb>=Qa

a Crl

HtBLW

b

UWL

D H

Ht

wHt wD

H

DHt

wHt wH wD

INPUT

a

HtHf

c bdf

B O

Taking moments about O

Sl.No Particulars Vertical Horizontal LeverForce Force Arm

V H 1 Body wall

W1 1/2x0.6x2x2.4 1.44 0.55

W2 2x2x2.4 9.6 1.75

2 FoundationW3 2.9x0.3x2.4 2.088 1.45

3 Wt of waterW4 0.15x2x1 0.3 2.825

4 Water pressureH 1/2x2.3x2.3 2.645 0.766667

13.428 2.645

a

Ht

w.Ht b O



V H 1 Body wall

W1 1/2x0.6x2x2.4 1.44 0.4

W2 2x2x2.4 9.6 1.6

2 Water pressureH 1/2x2x2 2 0.666667

11.040 2.000

INPUT

ct1 tw

INPUT

W3

W1Hta

INPUT

W2

ct2 bw



V H 1 Body wall

W1 0.6x3.2x2.4 4.608 0.3

W2 1.3x3.2x2.4/2 4.992 1.0333332 Wt of soil

W3 1.3x3.2x2.1/2 4.368 1.466667

3 Eath pr (0.420.42+2.66)x3.2/2 4.928 1.212121

13.968 4.928


V H

1 Body wallW1 0.6x3.2x2.4 4.608 0.3

W2 1.3x3.2x2.4/2 4.992 1.033333

W3 1.3x3.2x2.1/2 4.368 1.4666672 Buoyancy

W4 (1/2)x(0.6+1.9)x3.2 -4 0.681333

3 Earth pressure H (0.42+1.593)x3.2/2 3.2208 1.28922

9.968 3.221


V H

1 Body wallW1 0.6x3.2x2.4 4.608 0.5

W2 1.3x3.2x2.4/2 4.992 1.233333

2 Soil W3 1.3x3.2x2.1/2 4.368 1.666667

3 Foundn W4 2.3x0.5x2.4 2.76 1.15

4 Soil W5 0.2x3.2x2.1 1.344 2.2

5 H (0.42+3.01)x3.7/2 6.3455 1.384354

18.072 6.345

INPUT

INPUT

Try different values starting from bw(cal)


V H

1 W1 0.6x3.2x2.4 4.608 0.3

2 W2 1.6x3.2x2.4/2 6.144 1.133333

3 W3 1.6x3.2x2.1/2 5.376 1.666667

4 W4 1.6x1.6x2.1/2 2.688 1.666667

5 H (1.12+3.36)x3.2/2 7.168 1.333333

18.816 7.168

INPUT

INPUT


V H

1 Body wall W1 0.6x3.2x2.4 4.608 0.5

W2 1.6x3.2x2.4/2 6.144 1.333333

2 Soil W3 1.6x3.2x2.1/2 5.376 1.866667

3 Foundn W4 2.6x0.5x2.4 3.12 1.3

4 Soil W5 0.2x(3.2+1.6)x2.1 2.016 2.5

W6 1.6x1.6x2.1/2 2.688 1.866667

5 H (1.12+3.71)x3.7/2 8.9355 1.519324

23.952 8.936

INPUTSl.No Particulars Vertical Horizontal Lever

Force Force ArmV H

1 Body wall W1 0.6x2x2.4/2 1.44 0.55

W2 2x2x2.4 9.6 1.75

2 Foundn W3 2.9x1.215x2.4 8.4564 1.45

3 Wt of W4 0.15x2x1 0.3 2.825water

4 Water pr H 3.215x3.215/2 5.168112 1.071667

19.796 5.168

Input from blench curves

Input from graph

Input from graph

If calculated discharge is less than the actual discharge (Qth < Qa), increase the afflux(afx) by 0.05m.If afflux is more than 1 Display the message "CHECK DAM IS NOT FEASIBLE AT THIS SITE" and exit.Otherwise recalculate the steps from 8 to 11 to get the new value of DR. Input the value of Cd corresponding to the new value of DR and calculate the new Qth and repeat the

If calculated discharge is less than the actual discharge (Qth < Qa), increase the afflux(afx) by 0.05m.If afflux is more than 1 Display the message "CHECK DAM IS NOT FEASIBLE AT THIS

+ve -ve moment moment

MR MO

0.792

16.8

3.0276

0.8475

2.027833

21.467 2.028


MR MO

0.576

15.36

1.333333

15.936 1.333

O


MR MO

1.3824

5.1584

6.4064

5.973333

12.947 5.973


MR MO

1.3824

5.1584

6.4064

-2.725333

4.15232

10.222 4.152


MR MO

2.304

6.1568

7.28

3.174

2.9568

8.784417

21.872 8.784


MR MO

1.3824

6.9632

8.96

4.48

9.557333

21.786 9.557


MR MO

2.304

8.192

10.0352

4.056

5.04

5.0176

13.57592

34.645 13.576


MR MO

0.792

16.8 12.26178

0.8475

5.538494

30.701 5.538

DATA FOR DESIGN OF CHECK DAM ON ROCK

Maximum flood discharge Qa = m3/sec

Bed slope s = 1 in

River width L = m

Deepest bed level BLd = m

Average bed level BLa = m

Bottom level of weir/check dam BLW = m

Maximum flood level MFL = m

Height of check dam Ht = m

Top width of weir a = m

Afflux afx = m

Coefficient of discharge Cd =( Taken from graph depending on the value of DR)

Bottom width b = m( Rounded from the maximum of the values calculated in three cases)

Offset of foundation of weir c = m

Depth of foundation of weir df = m

Deepest Ground Level DGL = m(for checking stress at deepest foundation level of weir)

LEFT ABUTMENT

Left Bank Level LBL = m

Hard Rock Level HRL = m

Height of abutment Hta m(Rouded from the calculated value)

Top width of abutment tw = m

Bottom width of abutment bw = m(Rouded from the calculated value)

Foundation

Offset ca = m

Depth dfa = m

RIGHT ABUTMENT

Right Bank Level RBL = m

Hard Rock Level HRL = m

Height of abutment Hta m(Rouded from the calculated value)

Top width of abutment tw = m

Bottom width of abutment bw = m(Rouded from the calculated value)

Foundation

Offset ca = m

Depth dfa = m

Body wall

Foundation

Wt of water

Water pressure

check dam ( rock exposed)

Documents

dc2 twbtw3ca

dc2 twbtw3

dc tw2ca btw

dc tw2 btw

ve moment

soil foundn

particulars

htdc