1. 1 Summary Sheet Session Number : Date : Subject Expert : 5 09.04.2007 Dr. M.C. Nataraja Professor Department of Civil Engineering, Sri Jayachamarajendra College of Engineering, Mysore – 570 006. Phone:0821-2343521, 9880447742 E-mail: nataraja96@yahoo.com

2. 2 Design and Detailing of Counterfort Retaining wall Dr. M.C. NATARAJA

3. 3 When H exceeds about 6m, Stem and heel thickness is more More bending and more steel Cantilever-T type-Uneconomical Counterforts-Trapezoidal section 1.5m -3m c/c Counterfort Retaining wall CRW CF Base Slab Stem

4. 4 Parts of CRW Same as that of Cantilever Retaining wall Plus Counterfort Stem Toe Heel Base slab Counterforts Cross sectionPlan

5. 5 The stem acts as a continuous slab Soil pressure acts as the load on the slab. Earth pressure varies linearly over the height The slab deflects away from the earth face between the counterforts The bending moment in the stem is maximum at the base and reduces towards top. But the thickness of the wall is kept constant and only the area of steel is reduced. Design of Stem BF p=K a γh

6. 6 Maximum Bending moments for stem Maximum +ve B.M= p l 2 /16 (occurring mid-way between counterforts) and Maximum -ve B.M= p l 2 /12 (occurring at inner face of counterforts) Where ‘ l ’ is the clear distance between the counterforts and ‘p’ is the intensity of soil pressure l p + -

7. 7 Design of Toe Slab The base width=b =0.6 H to 0.7 H The projection=1/3 to 1/4 of base width. The toe slab is subjected to an upward soil reaction and is designed as a cantilever slab fixed at the front face of the stem. Reinforcement is provided on earth face along the length of the toe slab. In case the toe slab projection is large i.e. > b/3, front counterforts are provided above the toe slab and the slab is designed as a continuous horizontal slab spanning between the front counterforts. b H

8. 8 The heel slab is designed as a continuous slab spanning over the counterforts and is subjected to downward forces due to weight of soil plus self weight of slab and an upward force due to soil reaction. Maximum +ve B.M= p l 2 /16 (mid-way between counterforts) And Maximum -ve B.M= p l 2 /12 (occurring at counterforts) Design of Heel Slab BF

9. 9 Design of Counterforts The counterforts are subjected to outward reaction from the stem. This produces tension along the outer sloping face of the counterforts. The inner face supporting the stem is in compression. Thus counterforts are designed as a T-beam of varying depth. The main steel provided along the sloping face shall be anchored properly at both ends. The depth of the counterfort is measured perpendicular to the sloping side. TC d

10. 10 Behaviour of Counterfort RW -M TOE COUNTERFORT +M STEM HEEL SLAB Important points Loads on Wall Deflected shape Nature of BMs Position of steel Counterfort details

11. 11 PROBLEM -Counterfort Retaining Wall A R.C.C. retaining wall with counterforts is required to support earth to a height of 7 m above the ground level. The top surface of the backfill is horizontal. The trial pit taken at the site indicates that soil of bearing capacity 220 kN/m 2 is available at a depth of 1.25 m below the ground level. The weight of earth is 18 kN/m 3 and angle of repose is 30°. The coefficient of friction between concrete and soil is 0.58. Use concrete M20 and steel grade Fe 415. Design the retaining wall.

12. 12 Draw the following: Cross section of wall near the counterfort Cross section of wall between the counterforts L/s of stem at the base cutting the counterforts Given: f ck = 20 N/mm 2, f y = 415N/mm 2, H = 7 m above G.L, Depth of footing below G.L. = 1.25 m, γ = 18 kN/m 3, μ = 0.58, f b =SBC= 220 kN/m 2

13. 13 a. Proportioning of Wall Components Coefficient of active pressure = k a = 1/3 Coefficient of passive pressure= k p = 3 The height of the wall above the base = H = 7 + 1.25 = 8.25 m. Base width = 0.6 H to 0.7 H (4.95 m to 5.78 m), Say b = 5.5 m Toe projection = b/4 = 5.5/4 = say 1.2 m Assume thickness of vertical wall = 250 mm Thickness of base slab = 450 mm H b=5.5 m 1.25 m h 1 = 7 m

14. 14 Spacing of counterforts l = 3.5 (H/γ) 0.25 = 3.5 (8.25/18) 0.25 = 2.88 m  c/c spacing = 2.88 + 0.40 = 3.28 m say 3 m  Provide counterforts at 3 m c/c. Assume width of counterfort = 400 mm  clear spacing provided = l = 3 - 0.4 = 2.6 m l

15. 15 4.05m h=7.8 m θ d 250 mm 1.2 m b=5.5 m H=8.25 mh 1 =7 m 1.25m CF: 3m c/c, 400 mm T Details of wall

16. 16 Sr. No. Description of loads Loads in kN Dist. of e.g. from T in m Moment about T in kN-m 1 Weight of stem W 1 25x0.25x1x7.8 = 48.75 1.2 + 0.25/2 =1.325 64.59 2 Weight of base slab W 2 25x5.5x1x0.45 = 61.88 5.5/2 =2.75170.17 3 Weight of earth over heel slab W 3 18x4.05x1x7.8 = 568.62 1.45 +4.05/2 = 3.475 1975.95 TotalΣW = 679.25 ΣW =2210.71 b. Check Stability of Wall

17. 17 AB C D H 8250 250 mm 1200 mm4050 mm 450 D f = 1250 ΣW PAPA R e X b/2 T W3 W1 W2 PAPA Pressure distribution Cross section of wall-Stability analysis b/3 kaHkaH H/3 h 1 = 7000

18. 18 Stability of walls Horizontal earth pressure on full height of wall = P h = k a  H 2 /2 =18 x 8.25 2 /(3 x 2) = 204.19 kN Overturning moment = M 0 = P h x H/3 = 204.19 x 8.25/3 = 561.52 kN.m. Factor of safety against overturning = ∑ M / M 0 = 2210.71/561.52 = 3.94 > 1.55  safe.

19. 19 Check for sliding Total horizontal force tending to slide the wall = P h = 204.19 kN Resisting force = ∑µ.W = 0.58 x 679.25 = 393.97 kN  Factor of safety against sliding = ∑µ.W / P h = 393.97/204.19 = 1.93 > 1.55... safe.

20. 20 Check for pressure distribution at base Let x be the distance of R from toe (T),  x = ∑ M / ∑ W = 2210.71 -561.52 /679.25 = 2.43 m Eccentricity=e = b/2 - x = 5.5/2 - 2.43 = 0.32 < b/6 (0.91m)  Whole base is under compression. Maximum pressure at toe = p A = ∑W / b ( 1+6e/b) = 679.25/5.5 ( 1+ 6*0.32/5.5) = 166.61 kN/m 2 < f b (i.e. SBC= 220 kN/m 2 ) Minimum pressure at heel = p D = 80.39 kN/m 2 compression.

21. 21 Intensity of pressure at junction of stem with toe i.e. under B = p B = 80.39 + (166.61 - 80.39) x 4.3/5.5 = 147.8kN/m 2 Intensity of pressure at junction of stem with heel i.e. under C =P c = 80.39 + (166.61 - 80.39) x 4.05/5.5 = 143.9 kN/m 2

22. 22 80.39 kN/m 2 166.61 kN/m 2 143.9147.8153.9 AB C D H 8250 250 mm 1200 mm 5500 mm 4050 mm 450 1250 ΣW PAPA R e X b/2 T

23. 23 b) Design of Toe slab Max. BM B = psf x (moment due to soil pressure - moment due to wt. of slab TB] = 1.5 [147.8 x 1.2 2 /2 + (166.61 - 147.8) x 1.2 (2/3 x 1.2) -(25x 1.2 x 0.45 x 1.2/2) =174.57 kN-m. M u /bd 2 = 1.14 < 2.76, URS

24. 24 To find steel p t =0.34% <0.96%, A st =1326 mm 2, # 16 @150 However, provide # 16 @110 from shear considerations. Area provided =1827 mm 2, p t =0.47% Development length= 47 x 16=750 mm Distribution steel = 0.12 x 1000 x 450/100 = 540 mm 2 Provide #12 mm at 200 mm c/c. Area provided = 565 mm 2 b) Design of Toe slab- Contd.,

25. 25 Check for Shear Critical section for shear: At distance d (= 390 mm) from the face of the toe p E = 80.39 + (166.61 - 80.39) (4.3 + 0.39)/5.5 = 153.9kN/m 2 Net vertical shear = (166.61 + 153.9) x 0.81/2 - (25 x 0.45 x 0.81) =120.7 kN.  Net ultimate shear = V u.max = 1.5 x 120.7 =181.05 kN. ζ v = 181.05x 1000/1000x390 =0.46 MPa p t = 100 x 1827/ (1000 x 390) = 0.47 % ζ uc = 0.36 + (0.48 - 0.36) x 0.22/0.25 = 0.47N/mm 2 > ζ v safe d

26. 26 -M TOE COUNTERFORT +M STEM HEEL SLAB Counterfort RW

27. 27 Continuous slab.  Consider 1 m wide strip near the outer edge D The forces acting near the edge are Downward wt. of soil=18x7.8xl= 140.4 kN/m Downward wt. of heel slab = 25 x 0.45 x 1= 11.25 kN/m Upward soil pressure 80.39 kN/m 2 = 80.39 x 1= 80.39 kN/m  Net down force at D= 140.4 + 11.25 - 80.39 = 71.26 kN/m Also net down force at C = 140.4 + 11.25 - 143.9 = 7.75 kN/m Negative Bending Moment for heel at junction of counterfort M u = (psf) pl 2 /12 = 1.5 x 71.26 x 2.6 2 /12 = 60.2 kN-m (At the junction of CF) (c) Design of Heel Slab

28. 28 71.26 kN/m 7.75 kN/m D C Forces on heel slab 80.39 kN/m 2 166.61 kN/m 2 143.9147.8153.9 5500 mm

29. 29 To find steel M u /bd 2 =60.2x10 6 /(1000x390 2)= 0.39 < 2.76, URS To find steel p t =0.114% <0.12%GA (Min. steel), <0.96%(p t,lim. ) Provide 0.12% of GA A st = 0.12x1000x450/100 = 540 mm 2  Provide # 12 mm @ 200 mm c/c,  Area provided = 565 mm 2 p t = 100 x 565/ (1000 x 390) = 0.14 %

30. 30 Maximum shear = V u,max = 1.5 x 71.26 x 2.6/2 = 139 kN For P t, = 0.14 % and M20 concrete, ζ uc = 0.28 N/mm 2 ζ v = V umax /bd =0.36 N/mm 2, ζ uc < ζ v, Unsafe, Hence shear steel is needed Using #8 mm 2-legged stirrups, Spacing=0.87x415x100/[(0.36-0.28)x1000] = 452 mm < (0.75 x 390 = 290 mm or 300 mm )  Provide #8 mm 2-legged stirrups at 290 mm c/c.  Provide for 1m x 1m area as shown in figure Check for shear (Heel slab)

31. 31 AB C 1200 mm 4050 mm 450 1250 R e X b/2 2600 3000 4050 mm CD SFD Net down force dia. TOE HEEL 139 71.28 kN/m 7.75 kN/m x1 y1 Shear analysis and Zone of shear steel Area for stirrups

32. 32 Area of steel for +ve moment (Heel slab) Maximum +ve ultimate moment = psf x pl 2 /16 = 3/4 M u = 0.75 x 60.2= 45.15 kN-m. M u /bd 2 =Very small and hence provide minimum steel. A st,min = 540 mm 2  Provide # 12 mm bars at 200 mm c/c.  Area provided = 565 mm 2 > 540 mm 2

33. 33 Check the force at junction of heel slab with stem The intensity of downward force decreases due to increases in upward soil reaction. Consider m width of the slab at C Net downward force= 18 x 7.8 +25 x 0.45 - 143.9 = 7.75 kN/m.  Provide only minimum reinforcement. Distribution steel A st = 0.12 x 1000 x 450/100 = 540 mm 2 Using # 12 mm bars, spacing = 1000 x 113/468 = 241 mm. Provide # 12 mm at 200 mm c/c. Area provided = 565 mm 2

34. 34 (d) Design of Stem (Vertical Slab). Continuous slab spanning between the counterforts and subjected to earth pressure. The intensity of earth pressure = p h = k a γh =18 x 7.8/3=46.8 kN/m 2 Area of steel on earth side near counterforts : Maximum -ve ultimate moment, M u = 1.5 x p h l 2 /12 = 1.5 x 46.8 x 2.6 2 /12 = 39.54 kN.m. Required d = √ (39.54 x 106/(2.76 x 1000)) = 119 mm However provide total depth = 250 mm M u /bd 2 = 39.54x10 6 /1000x390 2 =1.1 < 2.76, URS

35. 35 To find steel: P t =0.34% <0.96%, A st =646 mm 2, #12 mm @ 170 mm c/c, However provide #12 mm @ 110 mm c/c, Area provided = 1027.27 mm 2, P t = 0.54 %. As the earth pressure decreases towards the top, the spacing of the bars is increased with decrease in height. Max.ult. shear = V umax = 1.5 x 46.8 x 2.6/2 = 91.26 kN For P t, = 0.54 % and M20 concrete ζ uc = 0.5 N/mm 2 ζ v = V umax /bd =91.28 x1000/(100X190)=0.48 N/mm 2, Shear steel is not needed and hence safe.

36. 36 At any section at any depth h below the top, the total horizontal earth pressure acting on the counterfort = 1/2 k a y h 2 x c/c distance between counterfort = 18 x h 2 x 3 x 1/6 = 9 h 2  B.M. at any depth h = 9h 2 xh/3 = 3h 3 B.M. at the base at C= 3 x 7.8 3 = 1423.7 kN.m. Ultimate moment = M u = 1.5 x 1423.7 = 2135.60 kN.m. Counterfort acts as a T-beam. Even assuming rectangular section, d =√(2135.6 x 10 6 (2.76 x 400)) = 1390 mm (e) Design of Counterfort

37. 37 The effective depth is taken at right angle to the reinforcement. tan θ = 7.8/4.05 =1.93, θ = 62.5°,  d = 4050 sin θ - eff. cover = 3535 mm > > 1390 mm M u /bd 2 =2135.6x10 6 /(400x3535 2 ) =0.427, p t =0.12%, A st =1696mm 2 Check for minimum steel 4.05m h =7.8 m θ d

38. 38 A st.min = 0.85 bd/f y = 0.85 x 400 x 3535/415 = 2896 mm 2  Provided 4- # 22 mm + 4 - # 22 mm,  Area provided = 3041 mm 2 p t = 100 x 3041/(400 x 3535) = 0.21 % The height h where half of the reinforcement can be curtailed is approximately equal to √H= √7.8=2.79 m Curtail 4 bars at 2.79-L dt from top i.e, 2.79-1.03 =1.77m from top.

39. 39 Design of Horizontal Ties The direct pull by the wall on counterfort for 1 m height at base = k a γh x c/c distance =1/3x18 x 7.8 x 3 = 140.4 kN Area of steel required to resist the direct pull = 1.5 x 140.4 x 10 3 /(0.87 x 415) = 583 mm 2 per m height. Using # 8 mm 2-legged stirrups, A st = 100 mm 2 spacing = 1000 x 100/583 = 170 mm c/c.  Provide # 8 at 170 mm c/c. Since the horizontal pressure decreases with h, the spacing of stirrups can be increased from 170 mm c/c to 450 mm c/c towards the top.

40. 40 Design of Vertical Ties The maximum pull will be exerted at the end of heel slab where the net downward force = 71.26 kN/m. Total downward force at D = 71.26 x c/c distance bet. CFs = 71.28 x 3 = 213.78 kN. Required A st = 1.5 x 213.78 x 10 3 /(0.87 x 415) = 888 mm 2 Using # 8 mm 2-legged stirrups, A st = 100 mm 2 spacing = 1000 x 100/888 = 110 mm c/c.  Provide # 8 mm 2-legged stirrups at 110 mm c/c. Increase the spacing of vertical stirrups from 110 mm c/c to 450 mm c/c towards the end C

41. 41 DRAWING AND DETAILING COUNTERFORT RETAINING WALL

42. 42 250 mm 1200 mm 4050 mm 450 1250 #16@120#12@200 8250 mm 7000 Cross section between counterforts 0-200mm STEMCOUNTERFORT TOEHEEL

43. 43 Cross section through counterforts

44. 44 Section through stem at the junction of Base slab. Backfill With straight bars 0.3l 0.25 l Backfill With cranked bars STEM STRAIGHT BARS

45. 45 Cross section of heel slab Backfill

46. 46 Examination Problems July 2006 Single bay Fixed Portal Frame Combined footing (Beam and slab type) December 2006 T-shaped Cantilever Retaining wall Combined footing (Type not mentioned)

47. 47 Exam Problem (Dec. 2006) Design a T shaped cantilever retaining wall to retain earth embankment 3.2 m high above the ground level. The unit weight of the earth is 18 kN/m2 and its angle of repose is 30 degrees. The embankment is horizontal at it top. The SBC of soil is 120 kN/m 2 and the co- efficient of friction between soil and concrete is 0.5. Use M20 concrete and Fe 415 steel. Draw the following to a suitable scale: 1.Section of the retaining wall 2.Reinforcement details at the inner face of the stem. 60 Marks Data: h 1 =3.2 m, µ=0.5, γ=18 kN/m 2, ө=30º, SBC= 120 kN/m 2, M20 Concrete and Fe 415 steel Find H= h 1 + D f

48. 48