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2. SALIENT FEATURES OF ARCH BUTTRESSES CHECK DAM 1)Horizontal thrust of water is resisted by the arch action of the arches which uniform transfer the load to the buttresses. 2)The buttresses are supported on mat or raft foundation, in soft soil (where the hard foundation is available at deeper strata). In case of hard/rocky foundations the buttresses are supported on a spread footing. 3)To prevent the flow / seepage underneath the check dam, a cutoff wall with minimum reinforcement is provided at the upstream edge of the foundation. The cutoff wall also to prevent the horizontal sliding of the check dam (it acts as key wall). 2

3. SALIENT FEATURES OF ARCH BUTTRESS CHECK DAM 4)Minimum reinforcement is provided in the arch portion of the dam because all points in the arch portion are in compression. The permissible compressive strength in concrete is taken as fc=40kg/sqcm or 400 tonne /m 2. 5)The buttresses designed as cantilever fixed at the base as well as at the junction of arch and buttresses. Suitable reinforcement is worked out for the same. 6) Pressure relief pipes of 75mm dia GI material are to be provided in staggered manner at a spacing of 3.0 to 4.0mtrs on down stream apron, with a bend on the top facing in the flow direction to counter act uplift pressure in sandy, black cotton soils. 3

4. SALIENT FEATURES OF ARCH BUTTRESS CHECK DAM 7)The check dam is checked for (a) stability against overturning. The factor of safety against overturning shall be > 1.5 (b) Factor of safety against sliding shall be > 1 8)The angle (Ø) subtended by the arch at the centre varies from 100 0 to 140 0. However this angle Ø is taken as 133 0 34’ ( Ø = 2.331 radians) for which the section of the arch will be the least and it is economical. 4

5. HYDRAULIC PARTICULARS TO BE COLLECTED FOR PREPARATION OF ESTIMATE 1)Lowest nala bed level. 2)Nala bank levels on either sides. 3)Observed H.F.L. of nala. 4)Normal / maximum scour level of the foundation. 5)Linear waterway of the nala. 6)Catchment area of the nala should be arrived by the Topo sheet. 7)The discharge shall be calculated either by Ryve’s formula or Area velocity method. 5

6. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM By Ryve’s formula Discharge Q = C (A) 2/3 Where C = varying constant factor depends on catchment area A = area in Sqkm Q = discharge in Cumec The normal scour depth D = 0.473(Q) 1 / 3 f 1 / 3 Where f = Silt factor 1.0 In case of loose foundation the maximum scour depth D1=1.5XD should be adopted. The scour level of the foundation should be works out from H.F.L. -- Scour Depth. 6

7. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM Consider span of nalla as 23.0m. Let us take spacing of buttress c/c as 4.25m. Provide 5 arch bays i.e. 5 x 4 2.5 = 21.25m and for remaining (23.0-21.25) = 1.75m provide 1.0m RCC wall on either side of bank which inturn acts as thrust walls to take care of horizontal thrust at ends of extreme arches at nalla banks. 7

8. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM 8

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11. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM thickness of arch required ‘t’= 1.0 x 3.0 x 2.31 400 = 0.0173mts or 1.73 Cums which is very-very less and is difficult for execution at site for such heights. however provide 15cm thick for workability and ease fabrication & casting concrete, with a minimum reinforcement of 0.15% as per IS 456. Area of steel required Ast= 0.15 x 1.0 x 0.15 100 = 2.25 x 10 -4 = 0.000225 m 2 or= 225.00 mm 2 Area of 8mm= π x 82 = 50.27mm 2 4 11

12. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM Spacing of 8mm φ = 50.27 x 100 225.0 = 22.34cm c/c say 20.0 cm c/c  Provide for both vertical and circular arch using 8mm φ. HYSD bard @ spacing of 20cm c/c Design of Buttresses: The buttresses are designed as cantilever having fixed ends both at the junction at Arch and at the base at mat. 12

13. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM 13

14. DESIGN DETAILS OF MULTIPLE ARCH BUTTRESS DAM 14

15. STABILITY CHECK FOR ONE SPAN OF ARCH 15

16. STABILITY CHECK FOR ONE SPAN OF ARCH CG of arch from: ӯ c = 4/3 R Sin 3 0/2 = 4/3 – 2.31 x 0.919 3 (Ø-SinØ) (2.33-0.724) ʘ re of segment ӯ c = 1.48m ӯ t = r - ӯ c ӯ t = 2.31 – 1.48 = 0.83m Lever arm @ B= 2.09 + 1.475 – 0.075 – 0.83 = 2.66m 16

17. TAKING MOMENTS @ POINT (B) Component MeasurementWtLevel armMoment W1 5.38 x 0.15 x 3.0 x 2.45.81 3.4920.277 W2 ½ x 3.0 x 3.0 x 0.15 x 2.41.62 1.393 2.257 W3 0.6 x 0.15 x 4.25 x 2.40.918 3.49 3.203 W4: Rect. 0.3 x 4.10 x 3.0 x 2.48.856 1.5 13.284 Arch 0.3 x 5.3823 x 2.43.87 2.66 10.294 W5: Rect. 0.15 x 4.10 x 2.0 x 30.44.428 1.5 6.642 Arch 0.15 x 5.3823 x 2.4 1.9376 2.66 5.154 Positive moments ∑ w = 27.439T ∑ m=61.111T-M P = ½ x 3.0 x 3.0 x 4.25 19.125 3/319.125 Uplift: ½ x 3.0 x 3.565 5.347 2.3712.709 Negative moments ∑ w =24.472T ∑ m=31.834T-M 17

18. TAKING MOMENTS @ POINT (B) ∑ M +ve = 61.111T-M ∑ M -ve = 31.83T-M Factor of safety against overturning = ∑ M +ve = 64.111 = 1.919 > 1.92 hence safe ∑ M -ve 31.83 X = ( ∑ M +ve - M -ve ) = (61.111 – 31.834) ∑ W 27.439 = 1.067mt.  Eccentricity = e = b/2 – X = 3.565/2 – 1.067 = 0.715mt 18

19. STRESSES DIAGRAM Stress = P = ∑ w x (1 ± 6 e ) b b P@ toe= 27.439 x (1 + 6x0.715) 3.565 3.565 = 7.696 x (1 + 1.203) = 16.95 t/m 2 19

20. STRESSES DIAGRAM P@ heel = 27.439 x (1 – 6 x 0.715) 3.565 3.565 = 7.696 x (1 - 1.203) = - 1.56 t/m 2 20

21. DESIGN OF FOOTING Max uplift pressure: Pmax = 16.95 t/m2 which is > than SBC of soil and crushing strength of concrete. Moment = (0.15 x 1 ) x (16.95) x 0.15 2 = 0.19t-m = 0.19 x 1000 x 100 M= Qbd 2 = 0.19 x 10 5 kg-cm 21

22. DESIGN OF FOOTING For M15Q = 8.75 a = 0.87  depth of footing required = √ M √ Qb = √ 0.19 x 10 5 √ 8.75 x 100 = √ 21.71 = 4.65 cm. However provided in 15cm thick. Area of steel required= Ast = M ta = 0.19 x 10 5 1900 x 0.87 x 7 = 1.642 Sqcm = 164.20Sqmm 22

23. DESIGN OF FOOTING Area of one bar of 8mm= 50.26mm 2  Spacing = 1000 x 50.26 164.20 = 306.09mm Say 31.00cm However provided in 20cm. 23

24. DESIGN OF BUTTRESS Average structure of Buttress: = (21.52 – 6.12) = 7.70 t/m2 2 Uplift force= 7.70 x 0.45 x 3.0 = 10.395 t/m 24 21.52 t/m 2 6.12 t/m 2

25. DESIGN OF BUTTRESS Moment= Load x lever arm = 10.395 x 3/2 = 15.59 t-m Depth required= d = √ M/Qb = √ 15.59 x 1000 x 100 8.5 x 100 = √ 1834.41 = 42.83cm However provided is 300.00cm. Area of Steel: Ast = M/ta = 15.59 x 1000 x 100 1900 x 0.87 x 300 = 3.144 cm 2 Say 314.4mm 2 Nos of 8mm = 314.4/50.26 Required. = 6.25 Say 8 Nos 25

26. DESIGN OF BUTTRESS V= 2 R Sin Ø/2 = 2 x whr x Sin Ø/2 = 2 x 1.0 x 3.0 x 2.31 x 0.919 = 12.737 tonne /mt higher Moment= ½ x 12.737 x (2/3 x 3) = 12.737 t-m = 12.737 x 10 5 kg-cm 26

27. DESIGN OF BUTTRESS Depth required= √ 12.737 x 105 8.75 x 100 = √ 1455.696 = 38.153cm However provided is Area of steel required= M/txa (Ast) = 12.737 x 10 5 1900 x 0.87 x 300 = 2.568cm 2 Say 256.8mm 2 Nos of 8mm bar = 256.80 = 5.11 50.26 Say 6 Nos of 8mm 27

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