UT @ 23.85REV

1

PROJECT ALTARNATE CANAL TO NTR CANAL , SATHUPALLY

CLIENT GOVERNMENT OF ANDHRA PRADESH, IRRIGATION & CAD DEPT

EPC CONTRACTOR NAVAYUGA-TRANSTROY (JV), HYDERABAD

CONSULTANT23.85

SUBJECT DESIGN OF UNDER TUNNEL AT Km 23.850

DATE /10/2008 REVISION ONE

UT-H-

DESIGN CODE UT-H-23.850

Revision No. Date Remarks with respect to Revision

document.xls

Page 1 IN-RIMT,HYyderabad

DESIGN OF UNDER TUNNEL AT Km 23.850ON ALTARNATE CNAL TO NTR CANAL

CANAL DATA :

1) FULL SUPPLY DISCHARGE Required 12.040 Cumecs Design 13.670 Cumecs

2) FULL SUPPLY LEVEL U/S = 147.592 M. D/S = 147.592 M.

3) FULL SUPPLY DEPTH U/S = 1.700 M. D/S = 1.700 M.

4) BED LEVEL U/S = 145.892 M.D/S = 145.892 M.

5) BED WIDTH U/S = 9.000 M.D/S = 9.000 M.

6) BED SLOPE U/S = 1 In 5000D/S = 1 In 5000

7) FREE BOARD U/S = 0.750 M.D/S = 0.750 M.

8) VELOCITY U/S = 0.873 M/secD/S = 0.873 M/sec

9) VALUE OF "N" 0.0250

10) TYPE OF CANAL U/S = LinedD/S = Lined

11) SIDE SLOPE INNER 1.500 In 1OUTER 2.000 In 1

12) TOP WIDTH OF BANK U/S = Left/Right 7.00 4.00 MD/S = Left/Right 7.00 4.00 M

13) GROUND LEVEL 146.350 M

14) TOP BANK LEVEL U/S = 148.342 MD/S = 148.342 M

HYDRAULIC PARTICULARS OF DRAIN1) CATCHMENT AREA = 0.810 sq.km

2) DISCHARGE = 16.580 cumecs

3) BED LEVEL OF DRAIN = 146.350 M

4) MFL OF DRAIN = 146.808 M

document.xls


ALTARNATE CANAL TO NTR CANAL , SATHUPALLY

DESIGN OF UNDER TUNNEL AT Km 23.850

Calculation of Bed Fall

Level @ 100m U/S = 147.190

Level @ 0m D/S = 146.350

Bed fall =+ 147.190 - +146.350

= 0.0084100

Say 1 in 119.05

Level @ 0m U/S = 146.35

Level @ -100m D/S = 146.3

Bed fall =+ 146.350 - +146.300

= 0.0005100

Say 1 in 2000.0

Level @ 100m U/S = 147.19

Level @ -100m D/S = 146.3

Bed fall =+ 147.190 - +146.300

= 0.00445200

Say 1 in 224.7

Average bed Fall = 0.00445

Say 1 in 225

Maximum Flood Discharge Q =

Where ,

Q = Maximum Flood Discharge in Cumecs

M = Catchment Area in Sq. Km = 0.810

Catchment Area in Sq. Miles = 0.313 Sq.miles

C = Constant & value of C should be adopted as under

Value of C Catchment area in Sq. Miles

1400 Upto 1 Sq.miles

1200 1 to 30 Sq.miles

1060 More than Sq.miles

Q = =

= 1400 ( 0.313 ) = 585.5 Cusecs

= 16.579

= 16.58

C M3/4

KM2

C M3/43/4

M3/sec

document.xls


Section @100 m UP Stream

Maximun Flood Discharge 16.58

Maximum Flood Level 148.009

Bed Fall 224.72 0

Rugosity Coefficient 0.0225

S. No. Ordinate Distance Area Perimeter

1 147.82 0.189

2.500 0.459 2.500

2 147.83 0.179

2.500 0.484 2.500

3 147.800 0.209

2.500 0.572 2.500

4 147.760 0.249

2.500 0.647 2.500

5 147.740 0.269

2.500 0.622 2.500

6 147.780 0.229

2.500 0.559 2.500

7 147.790 0.219

2.500 0.559 2.500

8 147.780 0.229

2.500 0.584 2.500

9 147.770 0.239

2.500 0.572 2.500

10 147.790 0.219

2.500 0.540 2.500

11 147.795 0.214

2.500 0.465 2.501

12 147.850 0.159

2.500 0.309 2.501

13 147.920 0.089

2.500 0.159 2.500

14 147.970 0.039

2.500 0.247 2.503

15 147.850 0.159

2.500 1.222 2.586

16 147.190 0.819

2.500 1.897 2.503

17 147.310 0.699

2.500 1.322 2.523

18 147.650 0.359

2.500 0.809 2.501

19 147.720 0.289

2.500 0.709 2.500

20 147.730 0.279

2.500 0.709 2.500

21 147.720 0.289

Totals = 50.000 13.4 50.119

Hydraulic Radius = R=A

=13.44

P 50.12

= 0.268

Velocity = V = =n

= 1.233

Discharge = Q = A x V = 13.445 x 1.233

= 16.580 Cumecs

Reduced Level

R2/3x S1/2

document.xls


Section @ 0 m UP Stream



Bed Fall 224.72



1 146.870 0.000

2.500 0.000 0.000

2 146.870 0.000

2.500 0.022 2.500

3 146.790 0.018

2.500 0.257 2.506

4 146.620 0.188

2.500 0.595 2.502

5 146.520 0.288

2.500 0.757 2.500

6 146.490 0.318

2.500 0.820 2.500

7 146.470 0.338

2.500 0.820 2.500

8 146.490 0.318

2.500 0.932 2.502

9 146.380 0.428

2.500 1.082 2.500

10 146.370 0.438

2.500 1.107 2.500

11 146.360 0.448

2.500 1.132 2.500

12 146.350 0.458

2.500 1.107 2.500

13 146.380 0.428

2.500 1.057 2.500

14 146.390 0.418

2.500 1.020 2.500

15 146.410 0.398

2.500 0.857 2.502

16 146.520 0.288

2.500 0.595 2.502

17 146.620 0.188

2.500 0.332 2.502

18 146.730 0.078

2.500 0.097 2.501

19 146.890 0.000

2.500 0.000 0.000

20 146.890 0.000

2.500 0.000 0.000

21 146.880 0.000

Totals = 50.000 12.6 42.519


=12.59

P 42.52

= 0.296

Velocity = V = =n

= 1.317

Discharge = Q = A x V = 12.589 x 1.317

= 16.580 Cumecs

Reduced Level

R2/3x S1/2

document.xls


Section 100 m Down Stream



Bed Fall 224.72



1 146.590 0.112

2.500 0.317 2.500

2 146.560 0.142

2.500 0.367 2.500

3 146.550 0.152

2.500 0.417 2.500

4 146.520 0.182

2.500 0.504 2.500

5 146.480 0.222

2.500 0.604 2.500

6 146.440 0.262

2.500 0.692 2.500

7 146.410 0.292

2.500 0.754 2.500

8 146.390 0.312

2.500 0.829 2.500

9 146.350 0.352

2.500 0.917 2.500

10 146.320 0.382

2.500 0.979 2.500

11 146.300 0.402

2.500 0.992 2.500

12 146.310 0.392

2.500 0.942 2.500

13 146.340 0.362

2.500 0.879 2.500

14 146.360 0.342

2.500 0.817 2.500

15 146.390 0.312

2.500 0.742 2.500

16 146.420 0.282

2.500 0.667 2.500

17 146.450 0.252

2.500 0.592 2.500

18 146.480 0.222

2.500 0.554 2.500

19 146.480 0.222

2.500 0.492 2.500

20 146.530 0.172

2.500 0.379 2.500

21 146.570 0.132

Totals = 50.000 13.43 50.004


=13.43

P 50.00

= 0.269

Velocity = V = =n

= 1.234

Discharge = Q = A x V = 13.432 x 1.234

= 16.580 Cumecs

Reduced Level

R2/3x S1/2

document.xls


M.F.L. AT ENTRY

+148.009

+147.051

+146.808

20.247 +146.782

24.047 +146.702

100 U/S -100 D/S

0.00

M.F.L. @ 100m U/S = +148.009

M.F.L. @ 0m = +146.808

M.F.L. @ 100m D/S = +146.702

M.F.L. @ 20.25m U/S = +147.051

M.F.L. @ 24.0471007183371m D/S = +146.782

Vent Way Calculations

Discharge Q = 16.58 Cumecs

Assumed velocity in Barrel V = 3.00 m/sec

Area Required Q/V = = 5.53

Considering a Vent size of 1.80m x 1.20m

Height of Vent = 1.20 m

Width of Vent = 1.80 m

No. of vents required = 2.56 Nos.

Say 3.00 Nos.

Velocity in Barrel = 2.56 m/sec

Thickness of Pier = 0.90 m

Total Width of Vent Way including Piers = 7.20 m

Canal Bed Level = 145.892

Lining Thickness - 0.040 m

Bed filling - 0.000 m

Slab Thickness - 0.250 m

Height of Vent = 1.200 m

Barrel Floor Level 144.402

document.xls


Design of Tail Channel

Required discharge = 586 Cusecs = 16.58 CumecsSection adopted

0.60 Lacy's Bed width= 4.8*(Q)1/2

1.5 : 1 = 19.54490214864

2.38= 11.7269413

11.750~ 11.75 m

Where bed width 11.750Depth of Flow 2.380Top width 18.891

Area A =(11.75 + 18.89)

x 2.38 = (18.89 + 11.75)

x 2.38 = 36.467 m²2 2

Wetted Perimeter P = B + 2√(1+S.S²)d = 11.75 + 2(1 + 1.5x1.5) x 2.38 = 20.332 m

Hydraulic radious = A

=36.5

= 1.794 mP 20.332

Velocity (V) =

= (1/0.03)x1.794x sqrt(1/12000) = 0.449 m/sec

Discharge(Desi) = A x V =36.467 x 0.449 = 16.4 Cumecs

Hydraulic Particulars

1. Bed Width = 11.750 6. Velocity = 0.449

2. Depth of Flow = 2.380 7. Side Slopes = 1.5 : 1

3. Free board = 0.60 8. Value of 'n' = 0.03

4. Discharge(req) = 16.580 9. Bed fall = 12000

5. Discharge(Des) = 16.381

D/S MFL 146.782 Computed MFL as per Bed Slope

Barrel Floor Level 144.402

Scour Depth Calculations D/S

Maximum Flood Discharge Q = 16.580 Cumecs

Linear water way = 18.891 m

M F L in Tail Channel = 144.402 + 2.38 = 146.782 m

Discharge per meter q = 16.580= 0.878

18.891

Silt factor f = 1.50

Normal scour depth R = 1.35

q

f

=1.35

0.878

1.50

= 1.13 m

Maximum scour depth = 1.5 x R

= 1.5 x 1.126 = 1.689 m

Maximum scour level = M.F.L. - Max. scour depth

= 146.782 - 1.689 = +145.094 145.094

D/S Cutoff depth = 144.402 - 145.094 = -0.692 Provide 0 m = +144.402 mCut-Off may be provided from end to end of Return Walls

(1/n)R⅔S½ =

10. Bank Widhts L/R =

m3/sec/m

1/3.

1/3.

D43

Author:

document.xls


TRANSITIONS:

Upstrem

a) Let the length of drop wall = 11.75 M

b) Width at entry of barrel = 7.20 M

c) Side splay proposed = 1 in 2.0

d) Length of transition required = [(b-a)/2] x 2 = 4.55 M

Length of u/s transition provided = 5.80 M

( Drop bottom width+ solid apron)

U/S Return Level = U/S MFL + 0.50m = 148.150 M

Downstream

a) Width at the end of d/s transition = 11.75 M

b) Width at exit of barrel = 7.20 M

c) Side splay proposed = 1 in 3.0

d) Length of transition required = [(a-b)/2] x 3 = 6.825 M

Length of d/s transition provided = 6.90 M

D/S Return Level = D/S MFL + 030m = 147.100 M

Head Loss and Bed Levels at different sections along the length of vent:

Sections 7 5 3 1

8 6 4 2

11.75 m 7.20 11.75

rectangular trough

5.80 31.29 6.90

8 6 4 2

7 5 3 1

Bed level DOF MFL Vel TEL

At Section 1-1 144.402 2.380 146.782 0.455 146.793

At Section 2-2 144.402 2.377 146.779 0.594 146.797

At Section 3-3 144.402 2.357 146.759 0.977 146.807

At Section 4-4 144.402 1.200 145.602 2.559 147.278

At Section 5-5 144.402 1.200 145.602 2.559 147.278

At Section 6-6 144.402 2.904 147.306 0.793 147.338

At Section 7-7 146.402 1.216 147.618 1.208 147.687

At Section 8-8 146.402 1.242 147.644 0.335 147.693

147.592

7.00 0.750 4.00

147.644 1.70 146.779

145.892

146.402 9.00

1.200

144.402 + 144.402

document.xls


FLOW CONDITIONS IN DRAIN

FLOW CONDITION AT SECTION 1-1

COEFFICIENT OF RUGOSITY (n) 0.0300

DISCHARGE (Q) ( in Cumecs) 16.580

BED LEVEL (BL) 144.402

SIDE SLOPE (s) 1.500

BED WIDTH (B) ( in mtrs) 11.750

FLOW DEPTH (D) ( in mtrs) 2.380 MFL + 146.782

MFL 146.782 1.5 :1 2.380 m

AREA = (B+sD)D ( in Sqm) 36.467 BL + 144.402

VELOCITY (V ) = Q / A ( in m/s) 0.455 11.750

0.011

TEL @ 1-1 = MFL + VEL. HEAD 146.793

20.332

HYDRAULIC MEAN RADIUS (R) = A / P ( in m) 1.794

0.000085


LENGTH FROM PREVIOUS SECTION(L) ( in mtrs) 0.000





BED LEVEL 144.402


MFL 146.779 0.0 :1 2.377 m



0.018

16.504


0.000057

0.004

FRICTION LOSS (H ) ( in mtrs) 0.000

TEL wrt MFL = MFL + VEL HEAD 146.797 0.000

TEL wrt PREVIOUS TEL = PRE.TEL + EDDY LOSS +FRI.LOSS 146.797 0.000 <------ Difference in TELs

VELOCITY HEAD =V2//2g = ( m )

PERIMETER (P) = B + 2D(1 + s2 )½ (in m)

SURFACE SLOPE(S) =V2 n2/R4/3 =




EDDY LOSS(he) = 0.5(V1 - V2)/2g ( in mtrs)

m

m

document.xls








BED LEVEL 144.402


MFL 146.759 0.0 :1 2.357 m



0.049

11.913


0.000193

0.009

FRICTION LOSS(Hf) ( in mtrs) 0.001

TEL wrt MFL = MFL + VEL HEAD 146.807


Loss of Head Calculations Inside The Barrel (Using UNWIN'S formula)




NO OF VENTS 3



BED LEVEL 144.402 MFL + 145.602

FLOW DEPTH (D) ( in mtrs) 1.200 0.0 :1 1.200

MFL 145.602 BL + 144.402

AREA = (B+sD)D ( in Sqm) 6.480 1.800

VELOCITY (V ) = Q / A ( in m/s) 2.559

0.334

19.800


0.009405

f1 (coeffi. Of loss of head at entry) 0.08000

a (assuming that the walls inside the barrel are plastered) 0.00316

b 0.03050

f2 (at Entry) = a (1+ b/ R) 0.00345








m

m

m

B76

user: FOR BELL MOUTH ENTRY

document.xls


0.471

TEL@ 4 - 4 = PRE.TEL + h 147.278

FLOW CONDITION AT SECTION 6-6 <------ Difference in TELs





SIDE SLOPE (s) 0.000 MFL + 147.306

BED LEVEL 144.402

FLOW DEPTH (D) ( in mtrs) 2.904 BL + 144.402 0.0 :1 2.904

MFL 147.306 7.200

AREA = (B+sD)D ( in Sqm) 20.908

VELOCITY (V ) = Q / A ( in m/s) 0.793

0.032

13.008


0.000108

0.060

FRICTION LOSS(Hf) 0.000000


TEL wrt PREVIOUS TEL = PRE.TEL + EDDY LOSS +FRI.LOSS 147.338 0.000

<------ Difference in TELs

Loss of head in side the barrel (h) = (1+f1+f2 L/R )V2 / 2g





m

m

document.xls


DROP CALCULATIONS:

This drop is proposed at U/S transision of vent ie at section 7-7

INPUT:Design Discharge 16.580 CumecsWeir Length 11.750 mCrest Level 145.152 md/s MFL 147.306 md/s Bed level 144.402 m

MFL CALCULATIONS:

u/s MFL assumed 147.051 mflow depth 1.899 mCoefficient of discharge, C = 1.733 Discharge, Q = Coeff x Weir length x U/s Flow Depth ^ (3/2)

= 6.90 cumecsSince the calculated discharge is eqaul to design discharge, hence Safe.

DESIGN OF DROP:

The depth of drop = 145.152 - 144.402 = 0.750 m

Top width of drop =Where 'd' is depth of water over sill 0.486 mTop width of drop = 0.395 mProvide top width 0.400 m

Bottom wdth = (H + d )/ p1/2Where 'H' is height of wall = 0.750 m 'p' is specific gravity of wall material 2.4 mBottom wdth 0.798 mProvide Bottom wIdth 0.800 m

Length of Apron L = 2 d + 2( d h )1/2Where 'h' is the diff.of u/s and d/s water levels 1.668 mLength of Apron 2.772 mProvide Length of Apron 2.800 m

Thickness of Apron t = (( d + h )^.5 )1/2Thickness of Apron 0.349 mProvide thickness of Apron 0.350 m

Depth of water cushion dw = 0.905 d h^1/2 - tDepth of water cushion -0.350 mHowever, provide depth of water cushion of 0.000 m

= C.L.h^3/2

0.5 d + 0.1524 to 0.5 d + 0.3048

document.xls


FLOW CONDITION AT SECTION 7-7 (ie @ DROP)


BED WIDTH (B) (in mtrs) 28.250


CREST LEVEL OF DROP 145.152

COEFICIENT OF DISCHARGE ( C ) 1.733

FLOW DEPTH (D) = (Q/C B)^2/3 0.486 MFL + 145.638

MFL 145.638 0.0 :1 0.486



0.074

29.222



145.638

145.152

147.306

144.402








BED LEVEL 145.152


MFL 146.767 1.5 :1 1.615



0.006

34.073


0.000083

0.021








7

2

m

m

+

+

+

+

m

m

7

NOTE:IF THIS VALUE IS STRIKED OFF, IT IS AN INDICATION THAT THE WEIR IS OF SUBMERGED. HENCE GO TO THE OTHER SHEET NAMED "6-6 & 7-7 SECTIONS FOR SUBMERGED".

D19

NOTE: IF THIS VALUE IS STRIKED OFF, IT IS AN INDICATION THAT THE WEIR IS OF SUBMERGED. HENCE GO TO THE OTHER SHEET NAMED "6-6 & 7-7 SECTIONS FOR SUBMERGED".

document.xls



TEL wrt PREVIOUS TEL = PRE.TEL + EDDY LOSS +FRI.LOSS 145.733 -1.040 <------ Difference in TELs

document.xls


145.152

1.615

146.767

0.335

146.773

145.152

0.486

145.638

1.208

145.712

144.402

2.904

147.306

0.793

147.338

144.402

1.200

145.602

2.559

147.278

BE

D L

EV

EL

DE

PT

H O

F F

LO

W

M F

L

VE

LO

CIT

Y

T E

L

144.402

1.200

145.602

2.559

147.278

144.402

2.357

146.759

0.977

146.807

144.402

2.377

146.779

0.594

146.797

144.402

2.380

146.782

0.455

146.793

1

2

6

7 7

6

2

1

8 8

555

4 4

document.xls


DROP CALCULATIONS:

This drop is proposed at U/S transision of vent ie at section 7-7

INPUT:Design Discharge 16.580 CumecsWeir Length 11.750 mCrest Level 146.402 md/s MFL 147.306 md/s Bed level 144.402 m

MFL CALCULATIONS:

u/s MFL assumed 147.051 mflow depth 0.649 mDrowning Ratio = (D/S MFL - Crest Level)x100 / (U/S MFL - Crest Level)

147.306 - 146.402 x 100147.051 - 146.402

139.29 %Reading from Malikapur graph (in MKS units),Coefficient of discharge, C = 1.052 Discharge, Q = Coeff x Weir length x U/s Flow Depth ^ (3/2)

= 16.58 cumecsSince the calculated discharge is eqaul to design discharge, hence Safe.

DESIGN OF DROP:

The depth of drop = 146.402 - 144.402 = 2.000 m

Top width of drop =Where 'd' is depth of water over sill 1.216 mTop width of drop = 0.760 mProvide top width 0.800 m

Bottom wdth = (H + d )/ p1/2Where 'H' is height of wall = 2.000 m 'p' is specific gravity of wall material 2.4 mBottom wdth 2.076 mProvide Bottom wIdth 2.100 m

Length of Apron L = 2 d + 2( d h )1/2Where 'h' is the diff.of u/s and d/s water levels 0.312 mLength of Apron 3.664 mProvide Length of Apron 3.700 m

Thickness of Apron t = (( d + h )^.5 )1/2Thickness of Apron 0.618 mProvide thickness of Apron 0.650 m

Depth of water cushion dw = 0.905 d h^1/2 - tDepth of water cushion -0.035 mHowever, provide depth of water cushion of 0.000 m

= C.L.h^3/2

0.5 d + 0.1524 to 0.5 d + 0.3048

document.xls




BED WIDTH (B) (in mtrs) 11.750


CREST LEVEL OF DROP 146.402

COEFICIENT OF DISCHARGE ( C ) 1.052

FLOW DEPTH (D) (i.e. Arrived from Malikapur graph) 1.216 MFL 147.618

MFL 147.618 0.0 :11.216

AREA = (B+sD)D ( in Sqm) 14.288 BL 146.402


0.069

14.182



147.618

147.306

146.402

144.402


FLOW CONDITION AT SECTION 8-8 (APPROACH CHANNEL)






BED LEVEL 146.402

FLOW DEPTH (D) ( in mtrs) 1.242 MFL 147.644

MFL 147.644 1.5 :11.242

AREA = (B+sD)D ( in Sqm) 16.901 BL 146.402

VELOCITY (V ) = Q / A ( in m/s 0.981 11.75

0.049

16.227


0.001117

0.006










7

7

document.xls


146.402

1.242

147.644

0.981

147.693

146.402

1.216

147.618

1.160

147.687

144.402

2.904

147.306

0.793

147.338

144.402

1.200

145.602

2.559

147.278

BE

D L

EV

EL

DE

PT

H O

F F

LO

W

M F

L

VE

LO

CIT

Y

T E

L

144.402

1.200

145.602

2.559

147.278

144.402

2.357

146.759

0.977

146.807

144.402

2.377

146.779

0.594

146.797

144.402

2.380

146.782

0.455

146.793

1

2

6

7 7

6

2

1

8 8

555

4 4

document.xls


Design of Approach ChannelRequired discharge = 586 Cusecs = 16.58 CumecsSection adopted

0.60 Lacy's Bed width= 4.8*(Q)1/21.5:1

= 19.544902149

1.756= 11.73

11.750Where bed width 11.750 ~ 11.75Depth of Flow 1.24156Top width 15.475

Area A =(11.75 + 15.47)

x 1.24 =(15.47 + 11.75)

x 1.24 = 16.901 m²2 2

Wetted Perimeter P = B + 2√(1+S.S²)d = 11.75 + 2(1 + 1.5) x 1.24 = 16.227 m

Hydraulic radious = A

=16.901

= 1.042 mP 16.227

Velocity (V) =

= (1/0.03)x1.042x sqrt(1/1000) = 1.083 m/sec

Discharge(Desi) = A x V 16.901 x 1.083 = 18.305 Cumecs

Hydraulic Particulars

1. Bed Width = 11.750 6. Velocity = 1.203

2. Depth of Flow = 1.242 7. Side Slopes = 1.5:1

3. Free board = 0.60 8. Value of 'n' = 0.03

4. Discharge(req) = 16.580 9. Bed fall = 1000

5. Discharge(Des) = 18.305 10. Bank Widths L/R =

drain bed level = 146.402M F L = 147.051

Depth of flow as per MFLs = 0.649Afflux = 0.593

Scour Depth Calculations U/S Maximum Flood Discharge Q = 16.58 Cumecs Linear water way = 15.47 m Maximum Flood Level = 147.644 m Discharge per meter q = 16.58

= 1.07115.47

Silt factor f = 1.50 Normal scour depth R =

1.346 f

=1.346

1.0711.50

= 1.231 m Maximum scour depth =

= 1.5 x 1.231 = 1.847 m Maximum scour level = M.F.L. - Max. scour depth

= 147.644 - 1.847 = +145.797 Cutoff depth = 146.402 - 145.797 = 0.605 Provide 0.75 m = +145.65 mCut-Off may be provided from end to end of Return Walls

(1/n)R⅔S½ =

m3/sec/m

q2 1/3.

1/3.

E41

Author:

document.xls


Rivetment & Bed pitching:

Depth of Flow in Approach Channel = 1.242 m

provide length of rivetment on u/s = 3 x depth of water in drain

= 3 x 1.24

3.72 M.

Provide length of U/S rivetment = 3.75 M.

length of bed pitching on U/S = 1/2 length of rivetment on U/S

= 0.5 x 3.75

= 1.88 m.

Provide length of bed pitching on U/S = 2.00 m

Depth of water in Tail Channel = 2.380 m

provide length of rivetment on D/S 4 x depth of water in drain

= 4 x 2.38

9.52 M.

Provide length of U/S rivetment = 9.75 M.

length of bed pitching on D/S = 1/2 length of rivetment on D/S

= 0.5 x 4.88

= 4.88 m.

Provide length of bed pitching on U/S = 5.00 m

Live Load Calculations

2.490 m

Clear Span of Slab = 1.80 m

Maximum Axle Load = 11.4 tonnes

Load on each wheel = 11.4 / 2 = 5.7 tonnes

Ground Contact Area of each Wheels 250 x 500

Spacing of wheels in Longitudinal direction = 1.2 m

Spacing of wheels in each Axle = 1.8 m

Load Dispersion on Longitudinal direction = 250 + 2 H Tan 45 + 1.2 Clause 403.3.1 45

6.43 m

Load Dispersion on Transverse direction = 500 + 2 H Tan 45 + 1.8

7.28 m

Load Dispersion each wheel = ( 250 + 2 H Tan 45 ) x ( 500 + 2 H Tan 45 )

6.43 x 7.28

= 28.660

Live load on slab = (5.7) / (28.66) = 0.199

0.199 2.615

0.199 2.740

Total Live Load on slab = 0.597

H Height of Banking =

m2

t/m2

Live Load on slab due to Over lapping longitudinal direction =

Live Load on slab due to Over lapping in transverse direction =

t/m2

t/m2

DESIGN OF U/S HEAD WALL

GIVEN DATA: DENSITY:

Head Wall Top Level + 147.606 Concrete 2.4 TBL 148.342 min. Ht head wall 1

Slab Top Level + 145.85 Earth 2.1Slab Bottom Level + 145.602

TBL 2.0 :10.500

DERIVED DATA: 0.50 147.606Slab Thickness = 0.25 m Load on SlabHead Wall Height = 1.754 m 7.299

1.754 Average Stress6.310

WIDTHS: 9.086 0.759Top-Width = 0.50 m 145.852Rear - Batter = 1.00 m 0.25 1.00 145.602

1.50STRESSES IN CONCRETE:By taking moments about 'A'Base Width = 1.5 m

SL.NO. FORCE PARTICULARS MAGNIT L.A MOMENT`m' `m' `t/m2' constant `t' `m' `tm'

1 W1 0.5 1.75395 2.4 2.105 1.25 2.6312 W2 1 1.75395 2.4 0.5 2.105 0.667 1.4033 W3 1 2.254 2.1 0.5 2.367 0.333 0.7894 PV 4.830 2.1 0.0796 0.807

SUM-V= 7.3845 PH 4.830 2.1 0.3046 3.090 0.737 2.276

SUM-M= 7.099

=SUM-M/SUM-V= 7.099 7.384 = 0.961 mEccentricity , e=X-b/2 = 0.961 1.5 0.211 m

<b/6= 0.250 mAs `e' is <b/6 ,Eccentricity is acceptable .SUM-V/b 4.9226*e/b = 0.846STRESSES : Maximum stress,fmax = SUM-V/b (1+6e/b)

9.086 t/m2Minimum stress,fmin = SUM-V/b (1-6e/b)

0.759 t/m2

t/m3

t/m3

W1

`

A

W2

W3

/- /2 =

DESIGN OF D/S HEAD WALL


Head Wall Top Level + 147.606 Concrete 2.4 TBL 148.342

Slab Top Level + 145.85 Earth 2.1Slab Bottom Level + 145.602

TBL 2.0 :10.500

DERIVED DATA: 0.50 147.606Slab Thickness = 0.25 mHead Wall Height = 1.754 m

1.754

WIDTHS:Top-Width = 0.50 m 145.852Rear - Batter = 1.00 m 0.25 1.00 145.602

1.50STRESSES IN CONCRETE:By taking moments about 'A'Base Width = 1.5 m

SL.NO. FORCE PARTICULARS MAGNIT L.A MOMENT`m' `m' `t/m2' constant `t' `m' `tm'

1 W1 0.5 1.75395 2.4 2.105 1.25 2.6312 W2 1 1.75395 2.4 0.5 2.105 0.667 1.4033 W3 1 2.254 2.1 0.5 2.367 0.333 0.7894 PV 4.830 2.1 0.0796 0.807

SUM-V= 7.3845 PH 4.830 2.1 0.3046 3.090 0.737 2.276

SUM-M= 7.099

=SUM-M/SUM-V= 7.099 7.384 = 0.961 mEccentricity , e=X-b/2 = 0.961 1.5 0.211 m

<b/6= 0.250 mAs `e' is <b/6 ,Eccentricity is acceptable .SUM-V/b 4.9226*e/b = 0.846STRESSES : Maximum stress,fmax = SUM-V/b (1+6e/b)


0.759 t/m2

t/m3

t/m3

W1

`

A

W2

W3

/- /2 =

min. Ht head wall 1

Load on Slab7.299

Average Stress6.3109.086 0.759

U/S WING WALLSGIVEN DATA: DENSITY:

Wing Wall Top Level = + 147.606 Concrete 2.4

Earth Top Level = + 147.606 Earth 2.1Foundation Top Level = + 144.327Foundation Bottom Level= + 143.727Concrete Offsets = 0.3 mDERIVED DATA: 0.50 147.606Footing Thickness = 0.6 mWing Height & Earth height = 3.27895 mEarth Height From Bottom 3.279Of Foundation = 3.87895 mWIDTHS:Top-Width = 0.50 m 0.3 0.3 144.327Rear - Batter = 1.80 m 0.6 143.727

w5STRESSES IN CONCRETE: 2.90By taking moments about 'A'Base Width,b = 2.3 m

SL.NO. FORCE PARTICULARS MAGNITU L.A MOMENT

`m' `m' constant `t' `m' `tm'1 W1 0.5 3.27895 2.4 3.935 2.05 8.0662 W2 1.8 3.27895 2.4 0.5 7.083 1.200 8.4993 W3 1.8 3.27895 2.1 0.5 6.197 0.600 3.7184 PV 10.752 2.1 0.0395 0.892

SUM-V= 18.1065 PH 10.752 2.1 0.158 3.567 1.377 4.913

SUM-M= 25.196

X=SUM-M/SUM-V= 25.196 18.106 = 1.392 mEccentricity , e=X-b/2 = 1.392 2.3 0.242 m

<b/6= 0.383 mAs `e' is <b/6 ,Eccentricity is acceptable .SUM-V/b= 7.8726*e/b = 0.630STRESSES : Maximum stress,fmax = SUM-V/b (1+6e/b)


2.911 t/m2

SRESSES ON SOIL:Base Width ,b = 2.90 m


`m' `m' constant `t' `m' `tm'1 W1 0.5 3.27895 2.4 3.935 2.35 9.2472 W2 1.8 3.27895 2.4 0.5 7.083 1.500 10.624

t/m3

t/m3

`t/m3'

`t/m3'

W1

`

A

B

W2

W3W4

- /2 =

4 W3 1.8 3.27895 2.1 0.5 6.197 0.900 5.5775 W4 0.300 3.27895 2.1 2.066 0.150 0.3106 W5 2.900 0.6 2.4 4.176 1.450 6.0557 PV 15.046 2.1 0.0395 1.248

SUM-V= 24.7046 PH 15.046 2.1 0.158 4.992 1.629 8.133

SUM-M= 39.946




5.576 t/m2

//- /2 =

Head @ TBL 148.342

Head @ MFL 147.60595Wing Wall Top Level 147.60595

12.8 11.462.9 5.58

144.327 abutment Foundation Level = +145.652Foundation @ U/S Scour Level

144.327 Foundation Level

MOMENT

`MOMENT

D/S WING WALLSGIVEN DATA: DENSITY:



w5STRESSES IN CONCRETE: 3.50By taking moments about 'A'Base Width,b = 2.9 m



SUM-V= 24.2055 PH 12.336 2.1 0.158 4.093 1.475 6.038

SUM-M= 40.473




4.511 t/m2

SRESSES ON SOIL:Base Width ,b = 3.50 m


`m' `m' constant `t' `m' `tm'1 W1 0.5 3.51233 2.4 4.215 2.95 12.4342 W2 2.4 3.51233 2.4 0.5 10.116 1.900 19.219

t/m3

t/m3

`t/m3'

`t/m3'

W1

`

A

B

W2

W3W4

- /2 =

4 W3 2.4 3.51233 2.1 0.5 8.851 1.100 9.7365 W4 0.300 3.51233 2.1 2.213 0.150 0.3326 W5 3.500 0.6 2.4 5.040 1.750 8.8207 PV 16.911 2.1 0.0395 1.403

SUM-V= 31.8376 PH 16.911 2.1 0.158 5.611 1.727 9.691

SUM-M= 60.233




6.883 t/m2

//- /2 =

Head @ TBL 148.342

Head @ MFL 147.60595Wing Wall Top Level 147.60595

on Concret on soil12.18 11.31

4.51 6.88

144.327 abutment Foundation Level144.094 Foundation @ D/S Scour Level144.094 Foundation Level

MOMENT

`MOMENT

U/S RETURN WALL GIVEN DATA: DENSITY:



w5STRESSES IN CONCRETE: 3.60By taking moments about 'A'Base Width,b = 3 m



SUM-V= 25.7795 PH 13.090 2.1 0.158 4.343 1.520 6.600

SUM-M= 44.544

X=SUM-M/SUM-V= 44.544 25.779 = 1.728 mEccentricity , e=X-b/2 = 1.728 3 0.228 m



4.676 t/m2SRESSES ON SOIL:

Base Width ,b = 3.60 m


`m' `m' constant `t' `m' `tm'1 W1 0.5 3.618 2.4 4.342 3.05 13.2422 W2 2.5 3.618 2.4 0.5 10.854 1.967 21.3464 W3 2.5 3.618 2.1 0.5 9.497 1.133 10.764

t/m3

t/m3

`t/m3'

`t/m3'

W1

`

A

B

W2

W3W4

- /2 =

5 W4 0.300 3.618 2.1 2.279 0.150 0.3426 W5 3.600 0.6 2.4 5.184 1.800 9.3317 PV 17.792 2.1 0.0395 1.476

SUM-V= 33.6326 PH 17.792 2.1 0.158 5.903 1.772 10.458

SUM-M= 65.483




7.053 t/m2

//- /2 =

12.510 11.632

4.676 7.053

MOMENT

MOMENT

D/S RETURN WALL


Wing Wall Top Level 147.100 Concrete 2.4


. w5STRESSES IN CONCRETE: 2.60By taking moments about 'A'Base Width,b = 2 m

SL.NO. FORCEPARTICULARS MAGNITU L.A MOMENT


SUM-V= 14.5045 PH 9.038 2.1 0.158 2.999 1.263 3.787

SUM-M= 17.879

X=SUM-M/SUM-V= 17.879 14.504 = 1.233 mEccentricity , e=X-b/2 = 1.233 2 0.233 m



2.189 t/m2SRESSES ON SOIL:Base Width ,b = 2.60 m

SL.NO. FORCEPARTICULARS MAGNITU L.A MOMENT

`m' `m' constant `t' `m' `tm'1 W1 0.5 3.006381 2.4 3.608 2.05 7.3962 W2 1.5 3.006381 2.4 0.5 5.411 1.300 7.0354 W3 1.5 3.006381 2.1 0.5 4.735 0.800 3.788

t/m3

t/m3

`t/m3'

`t/m3'

W1

`

A

B

W2

W3W4

- /2 =

5 W4 0.300 3.006381 2.1 1.894 0.150 0.2846 W5 2.600 0.6 2.4 3.744 1.300 4.8677 PV 13.006 2.1 0.0395 1.079

SUM-V= 20.4716 PH 13.006 2.1 0.158 4.315 1.515 6.536

SUM-M= 29.906




4.950 t/m2

//- /2 =

12.315 10.797

2.189 4.950

MOMENT

MOMENT

C1.734128192 0.451.733427419 0.51.725229557 0.55

1.7195039 0.61.713145932 0.651.699994483 0.71.676021257 0.751.639692713 0.81.587504317 0.851.504687149 0.91.351086878 0.951.042215089 1

1.052 1.000

Drowning Ratio

1 2 3 4 5 6 7 8 9 10 11 121

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

Malikapur Graph

Drowning Ratio

Co

ff.o

f D

isc.

'C'

1 2 3 4 5 6 7 8 9 10 11 121

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

Malikapur Graph

Drowning Ratio

Co

ff.o

f D

isc.

'C'

document.xls


Design OF SLAB UNDER CANAL BED:The slab is designed as a water retaining structure with HYSD bars with a mix M20 with respect to IS-456

Stress in steel to HYSD bars upto 20 mm diastress in steel (t) = 1500stress in concrete( c) = 70shear stress(s) = 17Local bond stress = 11.20Modular ratio (m) = 13

n= (mc / (mc+t))n= (13.000 X 70 ) / ( (13.000 X 70 + 1500 )n= 0.378

j = 1- n/3 = 0.874Q = 11.55Moment of Resistance

Effective depth (d) = √(M.R./Qb)

Clear span = 1.80 MF.S.D. = 1.70 M

Max depth water considered in the design1.70 + 0.750 = 2.45 M

Unit Wt. Of water = 1000Unit Wt. of RCC = 2500Unit Wt. of concrete = 2400Assume overall depth of slab = 0.25 mWearing Coat thickness = 0.04 mClear corver = 40 MM 40 mmMain reinforcement = 12 mm Effective depth = 25.00 - 4.00 - 0.600

= 20.40 cmEffective span of the slab = clear span + eff. Depth

= 1.80 + 0.204= 2.004 M

considering width of slab, total loads on the slab are

wt. Of water 2.45 x 1000 2450.00

wt. Of w.c. 0.04 m x 2400 96.00

wt. Of R.C.C. 0.25 m x 2500 625.00

D.L. 721.00

Total Load = 3171.00

Max. D.L. B M = = (721 X 2.004 X 2.004 )12 12.00

= 241.30 Kg-M

Max. L. L. B M = = (2450 X 2.004 X 2.004 )10 10.00

= 983.92 Kg-MMax. Mid Span B M = 241.30 + 983.92 = 1225.22

Max. D.L. B M = = (721 X 2.004 X 2.004 )10 10.00

= 289.55 Kg-MMax. L. L. B M = = (2450 X 2.004 X 2.004 )

9 9.00= 1093.25 Kg-M

Max. Support B M = 289.55 + 1093.25 = 1382.80

Mix cc(1:11/2:3) in M20

kg/cm2

kg/cm2

kg/cm2

kg/cm2

=Qbd2

Kg/m3

Kg/m3

Kg/m3

kg/m2

kg/m2

kg/m2

kg/m3

kg/m2

WL 2

WL 2

-WL 2

-WL 2

document.xls


Q = 11.552 1500 j= 0.874

Effective Depth required = √ M/(QXb)

= Sqrt( 1382.80 X 100 ) / ( 11.552 X 100 ) 10.941 Cm

= 10.941 < 20.40 CMHence ok

So Provide thickness of slab = 0.25 m M

Area of steel required = Mбst j d

= 1225.220 / ( 1500 x ### 20.400

= 4.581 Sq.cmSapcing of 12 mm dia. Bars

Spacing = П /4 xArea of Steel Required

= P /4 x 12 X 12 = 24.691 Cm4.581

3 times effective depth = 3 x 20.4 = 61.2cmMaximum =300mm Clause 26.3.3.b.1 of IS 456-2000

So Provide 12mm Dia bars @ 210 mm C/C

Area of steel required at Support = Mбst j d

= 1382.804 / ( 1500 x ### 20.400

= 5.170 Sq.cmSapcing of 12 mm dia. Bars


= P /4 x 12 X 12 = 21.877 Cm5.170

3 times effective depth = 3 x = 0cmMaximum =300mm Clause 26.3.3.b.1 of IS 456-2000

So Provide 12mm Dia bars @ 215 mm C/C

Alternate bar can be cranked to take of care of shear resistance.At top provide 12 mm dia bar 420 mm

area of 2.693

Area of steel Provided = П /4 xSpacing

= П /4 x 12 X 12 21

= 5.386 > 4.581DISTRIBUTION STEEL:( As per IS 3370 (Part-II)- 1965)

0.30%450 0.20% 100

For 250 mm= 0.30% - 150 x 0.10%

350= 0.26%

sst =

D2 x 100

D2 x 100

cm2 in addition to crank bars to take care of partial fixity at support.

D2 x 100

CM2 CM2

document.xls


As HYSD bars of Fe-415 grade steel is used , reduce it by 20%= 0.8 x 0.26% = 0.00205714286= 0.206 x 100 x 5.386

100

= 1.108As thickness of slab is more than 225 mm Provide steel on both the faces.

( As per Clause 7.1 I.S. 3370 Part-II )

Hence area of steel on each face = 0.55 = 0.554

Spacing of 8 mm Dia. Bars


= 8 X 8 0.554

= 90.741 CM5 times effective depth = 5 x 20.4 = 102cmMaximum =450mm Clause 26.3.3.b.2 of IS 456-2000

So Provide 8 mm MM Dia Bars at 450 MM C/C @ both CHECK FOR SHEAR : top and bottom

Max SF = WL = 3171 x 2.0042 2

= 3177.34 Kg

Shear Force @ "d" from support 0.204= 3177 ( 1.002 - 0.204 )

1.002 2.004= 2530.5 Kg

Nominal shear stress = Vbd

= 253020.40 x 100

= 1.240

Shear Stress taken by Steel: 50 % of bars are cranked up and top bar provided 12 mm dia @ 420 mm c/c


= 12 X 120 42.0

= 2.693

The area of steel available = 0.5 X 5.386 + 2.693

= 5.386

As per Clause 47.2.1 & 47.2.1.1 of I.S. 456-2000Permissible shear stress :

100Ast = 5.386 X 100bd 100 X 20.40

= 0.264 %For M-20 Permissible shear stress for 0.2640 % of Tensile reinforcement

0.25 % 0.220.5 % 0.3

For 0.264 %

= 0.22 + ( 0.08 X 0.014 ) / 0.250= 0.224= 2.245 > 1.240

Hence ok

cm2

cm2

D2 x 100

P /4 x

tv =

Kg/cm2

D2 x 100

P /4 x

CM2

cm2

N/mm2

N/mm2

N/mm2

Kg/cm2 Kg/cm2

document.xls


DESIGN OF THE SLAB UNDER CANAL BANKThe mix proposed for the RCC slab unless earth bank is M-20

Mix adopted is M-20 Grade

stress in steel for HySD bars upto 20 mm j (if) = 1500

stress on concrete ( c) = 70Modular ratio (m) = 13Neutial axis (m) = 0.378 dLevel (a) = 0.874 d

MR = 11.552Unit we. Of soil = 2100 kg/m2Weight . Of earth filling over the slab Assumed thickness of slab = 0.250 MBottom level of slab below canal bed = C.B.L. -Th. of W.C. - Th.of slab under canal bed

= + 145.892 - 0.040 - 0.25= 145.602 M

Top of slab = Bottom of slab + th of slab under canal bank= 145.602 + 0.250 = 145.852 M

Height of earth fill = 148.342 - 145.852= 2.49 M

LOAD CALCULATIONS:Wt. Of earth = 2.49 x 2100 = 5229Wt. Of slab = 0.25 x 2500 = 625

5854Live load due to class A Loading = 437Thickness of slab = 0.250 mClear corver = 25 mmMain reinforcement = 16 mm

= 25.00 - 2.50 - 0.8= 21.70 cm

clear span + eff. Depth= 1.80 + 0.217 = 2.017 m

Max. D. L. B M = = ### 2.017 X 2.017 )12 12

= 1984.6 Kg-M

Max. L.L. B M = = ### 2.017 X 2.017 )10 10

= 177.9 Kg-MMax. Mid Span B M = 1984.65 + 177.88 = 2162.53

Max. D. L. B M = = ### 2.017 X 2.017 )

10 10= 2381.6 Kg-M

Max. L.L. B M = = ### 2.017 X 2.017 )9 9

= 197.6 Kg-MMax. Support B M = 2381.58 + 197.65 = 2579.22

Q = 11.55 st = 1500 j= 0.874

Effective Depth required = MQ x b

= 2579.223 x 10011.55 x 100

= 14.942 < 21.7 CMHence ok

kg/cm2

kg/cm2

bd2

Kg/m2

Kg/m2

Kg/m2

Kg/m2

WL 2

WL 2

WL 2

WL 2

document.xls


So Provide thickness of slab = 0.25 M

Area of steel required = Mst j d

= 2162.529 / ( 1500 X 0.874 X 21.700= 7.60 Sq.cm

Sapcing of 16 mm dia. Bars

Spacing of Steel Required= П /4 x 1.6 X 1.60

7.60= 26.454 cm

3 times effective depth = 3 x 21.7 = 65.1cmMaximum =300mm Clause 26.3.3.b.1 of IS 456-2000

So Provide 16 mm Dia Bars at 230 MM C/C


= П /4 x 1.6 X 1.623

= 8.742 > 7.600

Area of steel required = Mst j d

= 2579.223 / ( 1500 X 0.874 X 21.700= 9.06 Sq.cm

Sapcing of 16 mm dia. Bars

Spacing = П /4 xSpacing of Steel Required

= П /4 x 1.6 X 1.609.06

= 22.180 cm3 times effective depth = 3 x 21.7 = 65.1cmMaximum =300mm Clause 26.3.3.b.1 of IS 456-2000

Alternate bar be cranked to take of care of shear resistance.So Provide 16 mm Dia Bars at 220 MM C/C

DISTRIBUTION STEEL:( As per IS 3370 (Part-II)- 1965)

0.30%0.20% 100

For 250 MM 450= 0.20% - 100 x 0.02%

300= 0.26%

As HYSD bars of Fe-415 grade steel is used , reduce it by 20%= 0.8 x 0.26% = 0.00205714286= 0.206 x 100 x 8.742

100= 1.798

As thickness of slab is more than 225 mm Provide steel on both the faces. ( As per Clause 7.1 I.S. 3370 Part-II )Hence area of steel on each face = 1.798 = 1.798

D2 x 100

CM2 CM2

D2 x 100

cm2

cm2

document.xls


Sapcing of 8 mm Dia Bars

Spacing = P /4 xArea of Steel Required

= П /4 x 8 X 81.798

= 27.951 CM5 times effective depth = 5 x 21.7 = 108.5cmMaximum =450mm Clause 26.3.3.b.2 of IS 456-2000

So Provide 8 mm Dia Bars at 270 MM C/C @ both at top and bottom

CHECK FOR SHEAR : Max SF = WL = 6291 x 2.017

2 2= 6344.7 Kg

Shear Force @ "d" from support = 6345 ( 1.009 - 0.217 )

1.009 0.217= 4979.516 Kg

2.017 Nominal shear stress = V

bd= 4980

2.017 X 100= 2.469

Shear Stress taken by Steel: 50 % of bars are cranked up andAlso top bar provided 16 mm dia brs 220 mm c/c


= 16 X 1622

= 9.139

The area of steel available = 0.5 X 8.742 + 9.139

= 13.510

As per Clause 47.2.1 & 47.2.1.1 of I.S. 456-2000Permissible shear stress :

100Ast = 13.510 X 100bd 21.700 X 100

= 0.623 %For M-20 Permissible shear stress for 0.623 % of Tensile reinforcement

0.25 % 0.22

0.5 % 0.3For 0.623 %

= 0.22 + ( 0.08 X ### 0.250

= 0.339

= 3.392 > 2.469Hence ok

D2 x 100

tv =

Kg/cm2

D2 x 100

П /4 x

CM2

cm2

N/mm2

N/mm2

N/mm2

Kg/cm2 Kg/cm2

document.xls

Page 24IN-RIMT,HYyderabad

DESIGN OF ABUTMENT (UNDER EARTH BANK)

INPUT:

Barrel floor level (ie Drain Bed Level) = + 144.402 m Concrete Density = 2.4

Level at bottom of Abutment = + 144.327 m Earth Density = 2.1

Canal Bed Level (CBL) ------> = + 145.892 m

Bottom of Bund level (ie Slab top under E' bank) = + 145.852 m

Top of Bank Level ------> = + 148.342 m

Width of Bed block ------> = 0.450 m

Height of Bed block ------> = 0.300 m

Width of rear batter ------> = 3.000 m

Width of ( w3 ) ------> = 0.450 m

Foundation offset ------> = 0.300 m

Depth of foundation ------> = 0.600 m

Total span of slab ------> = 2.700 m

Thickness of slab under Earth Bank = 0.250 m

Thickness of Sealing Coat on Canal Bed = 0.040 m

Thickness of Wearing Coat on Barrel Floor = 0.075 m

T B L = + 148.342

2.490 W6 W9 2.490

R

+ 145.852

0.250 145.602

0.3 145.302 W1

W5

W8 h

1.52 0.975 W3 1.525

W2 W4 2.125 Pv

0.30 0.30 Ph

+ 144.327 0.450 0.450 3.000

0.600 + 143.727 3.900 W7 A B

4.500

+

document.xls


DESIGN DATA:

Height of Earth Bank (Height of Surcharge) = 2.490 m

Height of rear batter ------> = 1.525 m

Base width of Abutment ------> = 3.900 m

Reaction on abutment under earth bank = Wt of slab + Wt of Earth on span + Live Load

= ( 2.700 /2 x 0.250 ) x 2.500

+ ( 2.700 /2 x 2.490 ) x 2.100

+ 2.700 /2 x 0.437

= 8.493 t

Total Height of soil on foundation concrete = Ht. of Surcharge+ Ht. of Abutment

= 2.490 + 1.525 = 4.015 m

Total Height of soil on foundation soil = Ht. ot Surcharge + Ht. of Abutment + Foundation Depth

= 2.490 + 1.525 + 0.600 = 4.615 m

Taking moments about A(Stresses on concrete)LOAD DESCRIPTION MAGNITUDE LEVER MOMENT

ARM

in Tons in mtrs. in t-m

R Reaction as calculated 8.493 3.675 31.212

W1 1.000 x 0.450 x 0.300 x 2.500 = 0.338 3.675 1.242

W2 1.000 x 0.450 x 0.975 x 2.400 = 1.053 3.675 3.870

W3 1.000 x 0.450 x 1.525 x 2.400 = 1.647 3.225 5.312

W4 0.500 x 3.000 x 1.525 x 2.400 = 5.490 2.000 10.980

W5 0.500 x 3.000 x 1.525 x 2.100 = 4.804 1.000 4.804

W6 1.000 x 3.450 x 2.490 x 2.100 = 18.040 1.725 31.119

Pv 0.040 x ( 4.015 2.490 2.100 = 0.833

Total load W = 40.698

Ph 0.158 x ( 4.015 2.490 2.100 = 3.291 0.640 2.108

Total moment M = 90.647

Bottom width of abutment = 3.900 m

L.A. = M / W---> 90.647 / 40.698 = 2.227 m

e = b/2-L.A---> 3.900 / 2 - 2.227 = 0.277 m

ep = b/6---> 3.900 / 6 = 0.650 m

e < b/6 = 0.277 < 0.650 Hence NoTension.

Max. Stress = W/b(1+6*e/b)---> = 14.882 T/Sq.m

Min. Stress = W/b(1-6*e/b)---> = 5.988 T/Sq.m

Taking moments about B(Stresses on soil)

- ) x

- ) x

2 2

2 2

-

document.xls



NO. DESCRIPTION MAGNITUDE LEVER MOMENT

ARM


R Reaction as calculated 8.493 3.975 33.760

W1 1.000 x 0.450 x 0.300 x 2.500 = 0.338 3.975 1.344

W2 1.000 x 0.450 x 0.975 x 2.400 = 1.053 3.975 4.186

W3 1.000 x 0.450 x 1.525 x 2.400 = 1.647 3.525 5.806

W4 0.500 x 3.000 x 1.525 x 2.400 = 5.490 2.300 12.627

W5 0.500 x 3.000 x 1.525 x 2.100 = 4.804 1.300 6.245

W6 1.000 x 3.450 x 2.490 x 2.100 = 18.040 2.025 36.531

W7 1.000 x 4.500 x 0.600 x 2.400 = 6.480 2.25 14.580

W8 1.000 x 0.300 x 1.525 x 2.100 = 0.961 0.150 0.144

W9 1.000 x 0.300 x 2.490 x 2.100 = 1.569 0.150 0.235

Pv 0.040 x ( 4.615 2.490 2.100 = 1.268


Ph 0.158 x ( 4.615 2.490 ) x 2.100 = 5.010 0.892 4.471

TOTAL MOMENT M = 119.929

b = Width of foundation concrete = 4.500 m

L.A. = M / W ---> 119.929 / 50.143 = 2.392 m

e = b/2-L.A ---> 4.500 / 2 - 2.392 = 0.142 m

ep = b / 6 ---> 4.500 / 6 = 0.750 m

e < b/6 = 0.142 < 0.750 Hence No Tension.

Max. Stress=W/b(1+6e/b)---> = 13.253 t/sq.m

Min. Stress=W/b(1-6e/b)---> = 9.033 t/sq.m

- ) x

-

2 2

2 2

-

document.xls


DESIGN OF PIER UNDER EARTH BANK :

T.B.L + 148.342

2.490

+ 145.852

0.25 + 145.602 Concrete Offset 300

900

1.275

+ 144.402

0.60 144.327

1500 + 143.727

Load coming on the top of Pier :

( a ) Weight of Earth = 1.80 0.45 2.49 2.1 = 11.765 t

( b ) Weight of Slab = 1.80 0.45 0.25 2.5 = 1.406 t

Total Load = 13.171 t

1 Stress in Concrete @ top of Pier :

= 13.171 = 14.635 t/m

0.90 X 1

2 Stress in Concrete @ top of Foundation Concrete :

Total Weight on top of Pier = 13.171 t

Self Weight of Pier = 0.90 1.275 2.40 = 2.754 t

Total Weight = 15.925 t

Stress at bottom of pier = 15.925 = 17.695 t/m

0.9 X 1

3 Stress on Soil :

Total weight including Self wt. of pier at bottom of pier = 15.925 t

Weight of foundation concrete = 1.50 0.60 2.40 = 2.160 t

(Considering 2/3:1 dispersion)

Weight of wearing coat = 0.075 2.40 = 0.180 t

Total Weight = 18.265

Stress on Soil = 18.265 = 12.2 t/m

1.50 x 1.0

W.C. 0.075

2

2

2

+

+

(

(

)

)

X

X

X X

XX X

X

X

X

document.xls


DESIGN OF ABUTMENT (UNDER CANAL TROUGH) ( CONSIDERING 50 % HYDROSTATIC PRESSURE)

( CANAL FULL & DRAIN EMPTY )

INPUT:

Barrel floor level (ie Drain Bed Level) = + 144.402 m Concrete Density = 2.4

Level at bottom of Abutment = + 144.327 m Earth Density = 2.1

Canal Bed Level (CBL) ------> = + 145.892 m

Full Supply Level of canal (FSL) ------> = + 147.592 m

Top of Bund Level of canal (TBL) ------> = + 148.342 m

Width of Bed block ------> = 0.450 m

Height of Bed block ------> = 0.300 m

Width of rear batter ------> = 1.850 m

Width of ( w3 ) ------> = 0.450 m

Foundation offset ------> = 0.300 m

Depth of foundation ------> = 0.600 m

Total span of slab ------> = 2.100 m

Thickness of slab under Canal trough = 0.250 m

Thickness of Sealing Coat on Canal Bed = 0.040 m

Thickness of Wearing Coat on Barrel Floor = 0.075 m

TBL + 148.342

FSL+0.3 = + 147.892

W9

2.000 W6 2.000

R 2.000

0.040 CBL + 145.892

0.250 145.602

0.3 W1

145.302 W5

W8

1.565 0.975 W3 1.565

W2 W4 2.165 P1

0.3 0.3 P2

144.327 0.45 0.45 1.85 1.565

0.600 143.727 2.75 W7 A 2.165

3.35 B

2.000

+

+

t/cu.m

t/cu.m

document.xls


DESIGN DATA:

Full Supply Depth in Canal (FSD) ------> = 1.700 m

FSD+Extra depth considered for design (=0.3) = 2.000 m

Height of rear batter ------> = 1.565 m

Base width of Abutment ------> = 2.750 m

Width of Foundation ------> = 3.350 m.

Reaction on abutment under canal trough = Wt of slab + Wt of Water on span + Wt of Wearing Coat

= ( 2.100 /2 0.250 ) x 2.500

+ ( 2.100 /2 2.000 ) x 1.000

+ ( 2.100 /2 0.040 ) x 2.400

= 2.857 t

Eq. Height of surcharge = Ht of water x ( Unit Wt of Water / Unit Wt of Earth)

= 2.000 x 1.0 / 2.1 = 0.952 m

Total Height of soil on foundation concrete = Ht. ot Surcharge + Ht. of Abutment

= 0.952 + 1.565 = 2.517 m

Total Height of soil on foundation soil

= Ht. of Surcharge + Ht. of Abutment + Foundation Depth

= 0.952 + 1.565 + 0.600 = 3.117 m

Taking moments about A(Stresses on concrete)LOAD DESCRIPTION MAGNITUDE LEVER MOMENT

ARM


R Reaction as calculated = 2.857 2.525 7.214

W1 1.000 x 0.450 x 0.300 x 2.500 = 0.338 2.525 0.853

W2 1.000 x 0.450 x 0.975 x 2.400 = 1.053 2.525 2.659

W3 1.000 x 0.450 x 1.565 x 2.400 = 1.690 2.075 3.507

W4 0.500 x 1.850 x 1.565 x 2.400 = 3.474 1.233 4.283

W5 0.500 x 1.850 x 1.565 x 2.100 = 3.040 0.617 1.876

W6 1.000 x 2.300 x 2.000 x 1.000 = 4.600 1.150 5.290

Pv 0.040 x ( 2.517 0.952 2.100 = 0.450


Ph 0.158 x ( 2.517 0.952 2.100 = 1.801 0.657 1.184

P1(50%) 0.5 x 1.0 x 2.000 x 1.565 x 1.000 = 1.565 0.782 1.224

P2(50%) 0.5 x 0.5 x 1.565 x 1.565 x 1.000 = 0.612 0.522 0.319

Total moment = 28.409

Bottom width of Abutment = 2.750 m

L.A. = M / W---> 28.409 / 17.502 = 1.623 m

e = b/2-L.A---> = 0.248 m

ep = b/6---> = 0.458 m


Max. Stress = W/b(1+6*e/b)---> = 9.808 T/Sq.m

2 - 2 ) x

2 - 2 ) x

X

X

X

document.xls


Min. Stress = W/b(1-6*e/b)---> = 2.921 T/Sq.m

document.xls



NO. DESCRIPTION MAGNITUDE LEVER MOMENT

ARM


R Reaction as calculated = 2.857 2.825 8.071

W1 1.000 x 0.450 x 0.300 x 2.500 = 0.338 2.825 0.955

W2 1.000 x 0.450 x 0.975 x 2.400 = 1.053 2.825 2.975

W3 1.000 x 0.450 x 1.565 x 2.400 = 1.690 2.375 4.014

W4 0.500 x 1.850 x 1.565 x 2.400 = 3.474 1.533 5.326

W5 0.500 x 1.850 x 1.565 x 2.100 = 3.040 0.917 2.788

W6 1.000 x 2.300 x 2.000 x 1.000 = 4.600 1.450 6.670

W7 1.000 x 3.350 x 0.600 x 2.400 = 4.824 1.675 8.080

W8 1.000 x 0.300 x 1.565 x 2.100 = 0.986 0.150 0.148

W9 1.000 x 0.300 x 2.000 x 1.000 = 0.600 0.150 0.090

Pv 0.040 x ( 3.117 0.952 2.100 = 0.740

Total load = 24.202

Ph 0.158 x ( 3.117 0.952 2.100 = 2.930 0.909 2.664

P1(50%) 0.5 x 1.0 x 2.000 x 2.165 x 1.000 = 2.165 1.082 2.343

P2(50%) 0.5 x 0.5 x 2.165 x 2.165 x 1.000 = 1.172 0.722 0.846

TOTAL MOMENT M = 44.970

b = Width of foundation concrete = 3.350 m

L.A. = M / W ---> 44.970 / 24.202 = 1.858 m

e = b/2-L.A ---> = 0.183 m

ep = b / 6 ---> = 0.558 m


Max. Stress=W/b(1+6e/b)---> = 9.592 t/sq.m

Min. Stress=W/b(1-6e/b)---> = 4.857 t/sq.m

2 - 2 ) x

2 - 2 ) x

document.xls


DESIGN OF PIER UNDER CANAL TROUGH :

+ 147.892

2.00

+ 145.892

0.25 + 145.602

900

1.27

+ 144.402

0.60 144.327

1500 + 143.727

Loads coming on the top of Pier :

( a ) Weight of Water = 1.8 0.45 2.00 1.00 = 4.500 t

( b ) Weight of Slab = 1.8 0.45 0.29 2.5 = 1.631 t

Total Load = 6.131 t

1 Stress in Concrete @ top of Pier :

= 6.131 = 6.812 t/m

1.0 x 0.9

2 Stress in Concrete @ top of Foundation Concrete :

Total Weight on top of Pier = 6.131 t

Self Weight of Pier = 0.9 1.275 2.40 = 2.754 t


Stress at bottom of pier = 8.885 = 9.872 t/m

1.0 x 0.9

3 Stress on Soil :

Total weight including Self wt. of pier at bottom of pier = 8.885 t

Weight of foundation concrete = 1.5+0.6 0.60 2.40 = 1.512 t

(Considering 2/3:1 dispersion) 2

Weight of wearing coat = 0.075 2.40 = 0.180 t


Stress on Soil = 10.577 = 7.05 t/m

1.50 X 1

W.C. 0.075

2

2

+

+

(

(

)

)

X

X

X X

XX X

X

sealing coat 0.04

X

X

2

D18

Srikanth: Clear span in metres.

document.xls


DESIGN OF HOLDING DOWN BOLTS

HEAD OF WATER FROM = M.F.L @ ENTRY OF BARREL - BOTTOM LEVEL OF SLABBOTTOM OF SLAB

= 147.3059 - 145.602

= 1.704 M.

UPLIFT HEAD ON THE SLAB = 1.704 x 1.0

= 1.704

DEAD WEIGHT OF SLAB = 0.25 x 2.5 + 0.040 x 2.4+ WT. OF SEALING COAT( SLAB UNDER TROUGH) = 0.721

BALANCE UPWARD THRUST = UPLIFT ON SLAB - TOTAL WT. OF SLAB

= 1.704 - 0.721

= 0.983

MAXIMUM UPWARD THRUST = 1.5 x 0.983(FACTOR OF SAFETY= 1.5 )

= 1.474

LENGTH OF THE SLAB = 9.0 + 1.0 + 1.0UNDER TROUGH

= 11.0 M.

TOTAL UPWARD THRUST = 1.474 x 11.0 x 0.25 ON SLAB

= 4.055 T.

UPWARD THRUST ON = 4.055 / 2EACH ABUTMENT

= 2.03 T.

PROVIDE 20 mm. DIA. M.S. BOLTS

TENSILE STRENGTH OF M.S. = 1150( IS: 3370, PAGE-8,TABLE-2)

TENSILE STRENGTH OF = 3.142 x 1150EACH BOLT 1000

= 3.613 T.

NO. OF BOLTS REQUIRED = 2.03 / 3.613

= 0.561

HOWEVER PROVIDE 1 NOs. OF 20 mm. DIA. BOLTS @ 11 m C/CON EACH SIDE OF THE SLAB FOR A HEIGHT OF 0.9 m. IN EACH ABUTMENT

T/M2

T/M2

T/M2

T/M2

KG/M2

document.xls


UPLIFT PRESSURE:Uplift pressure under BarrelFSL of canal = 147.592 mBottom level of Barrel Floor = 144.027 mUplift head = 3.565 mW.C. thickness = 0.075 mAssume Floor thicness = 0.300 m

Uplift head resisted by weight of the floor = 0.375 x 2.400

= 0.900 mRemaining uplift head = 2.665 mUplift Pressure on barrel Floor

145.892145.602

1.565

2.750 144.402144.327143.727

Creep lost in percolation = 1.565 + 0.300 + 0.600

+ 3.350 + -0.30= 5.515 m

Assuming creep gradient of 1 in 4Head lost in percolation = 1.379 m

Head resisted by barrel floor = 0.900

Total uplift head resisted = 1.379 + 0.900= 2.279 m

Residual head to be resisted = 3.565 - 2.279= 1.286 m

Uplift head resisted by the arch action of the floor:

αSpan of the floor barrel, L = 1.80

radius of the arch = R R(2R-y)* y = (L/2)2

y = 0.19 0.900R = 2.2538 0.375

1.800 mCentral angle subtended by arch = 2α

cotα = (R-y)/(L/2)= 2.296 m

ρ x t/2 = μ x L/2 x cotαConcrete mix used is M10

ρ = 0.70 x fcwith FS=3, ρ = 0.7 x 100/3

= 4.041= 40.4 t/m2

t = 0.375ρ x t/2 = μ x L/2 x cotα

μ = 7.335 > 1.286Hence safe for uplift

However Provide Barrel floor thickness of 375 mm

When there is no water in the barrel and the canal is flowing full for worst situation. In this condition, water is likely to flow down, seeping by the side of the abutment, percolate under the foundations and then exert pressure with the residual head.

If 'μ' is the maximum uplift head that can be resisted by the arch action and 'p' is the mean pressure at the crown of the arch, then

kg/cm2

UT @ 23.85REV

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Transcript of UT @ 23.85REV