Design Of Counter Fort Retaining Wall"Reference: Example 18.4, R.C.C. Design Vol. - I, B.C. Punamia"1 Design ConstantsHieght of cantilever wall from ground level = 11.5 mSaturated weight of Earth (?sat ) = 18 kN/m3Unit Weight of water (?w) = 10 kN/m3Angle of repose (?) = 30 SBC of Soil (qo) = 185 kN/m2Co-eff of friction () = 0.5Unit Weight of Concrete (fck) = 25 MPafy = 415 MPaCover = 40 mmFoundation Depth = 1.5 mWidth of counterfort = 0.5 m?cbc = 8.5 N/mm2?st = 230 N/mm2Neuteral axis constant (k) = 0.289Lever arm constant (j) = 0.904Moment of resistance constant = 1.109"Coefficient of earth pressure, Ka" = 0.3332 Dimension of Various PartsHeight of wall above base (H) = 11.5 + 1.5= 13 mThe ratio of length of slabe (DE) to base width b is given by eq.? = 1 - q02.2 y H= 1 - 1852.2 x 18 x 13.00? = 0.64Adopting ? = 0.40 . Eq (1)The width of base is given by Eq.b = 0.95 H x Ka(1-?) x (1+3?)b = 0.95 x 12.50 0.333( 1 - 0.40)x( 1 + 1.20 )= 5.97Normal practice is to provide b between 0.7 to 0.8 H A 0.6Taking maximum value of H = 0.7b = 9.10 mWidth of toe slab = ? x b ?= 3.64 mI1Provide toe slab = 3.60H1 ITaking the uniform thickness of stem = 0.6 mH 13.00Hence width of heel slab = 4.90 mLet thickness of base slab = 0.8 m11.5 GClear spacing of counter fort = 3.5 x H 1/4F3 G1?sat2.50= 3.23 mFProviding spacing of counterfort = 3.00 m3.60 1.50 F1 4.90 F2 D E B C3 Stability of Wall1.50Let w1 = Weight of rectangular portion of stemw2 = Weight of base slab9.10w3 = Weight of soil on heel slab.FIGURE 1Detailforce(kN) lever arm Moment about toe (KN-m)w1 1 x 0.60 x 12.20 x 25 =183 3.9 713.70w2 1 x 0.80 x 9.10 x 25 =182 4.55 828.10w3 1 x 4.90 x 12.20 x 18 =1076.04 6.65 7155.67 403.06 kN/m2-86.35 kN/m2Sw =1441.04 Total MR 8697Total resisting moment = 8697 kNmEarth pressure (p) = (Ka x ?' x H)+(?w x H)= 164.67 kNMoment = p*H2/6FIGURE 2= 5518.11 kNmF.O.S. against overturning = 1.58 > 1.5 Safe against overturningF.O.S. against sliding = Sw/p= 4.38 > 1.5 Safe against slidingPressure DistributionNet Moment (SM) = 3179 kNm"Distance x of the point of application of resultant, from toe is"x = SM/Sw= 2.21 me = b/2 - x= 2.34b/6 = 1.52Hence un safe as e > b/6Pressure p1 at toe = SW 1 + 6 eb b= 1441.04 x 1 + 6 x 2.349.10 9.10= 403.06 kN/m2Hence un safe as p1 > 125Pressure p2 at heel = SW 1 - 6 eb b1441.04 x 1 + 6 x 2.349.10 9.10= -86.35 kN/m2Hence un safe as p2 < 0"The Pressure intencity p1 under E,"p1 = 403.06 - 403.06 - -86.35 x 3.609.10= 209.45 kN/m2"The Pressure intencity p2 under B,"p2 = 403.06 - 403.06 - -86.35 x 4.209.10= 177.20 kN/m24 Design of Heel SlabClear spacing b/w counterforts = 3.00 mThe pressure distribution on the heel slab is shown in fig 2. Consider a strip 1 meter wide."Upward pressure intensity (near outer edge, C)" = -86.35kN/m2Down ward load due to weight of Earth. = 219.6 kN/m2Down ward weight of slab per unit area = 15 kN/m2Hence net pressure intensities will be P = 148.25 kN/m2M1 = Pl2/12= 111.19 kNmEffective depth required (d) = BMRxb= 316.71 mmProviding overall depth (D) = 500 mmEffective depth (d) = 460 mm"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 800 mm2"AstM=Ast ?st j d" = 1162.78 mm2Provide Ast1 = 1162.78 mm2Dia of bar = 20 mmSpacing of bar calculated = 270 mmSpacing of bar provided = 150 mmShear force (V) = PL/2= 222.38 kN100Ast/bd = 0.25?c = 0.26 N/mm2"Refernce: Pg. 84, Table 23, IS 456 : 2000"?v = V/(d x b)= 0.48 483.423913 0.48Hence shear reinforcement is required as ?v > ?cHence depth required from shear point of veiw (d) = V/(?c x b)= 855.29 mmProviding overall depth (D) = 1200 mmHence effective depth (d) = 1160 mm"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 1920 mm2"AstM=Ast ?st j d" = 461.10 mm2Provide Ast1 = 1920 mm2Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mmDistribution Steel = 1920 mm2Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm"Let us check this reinforcement for development length at point of contraflexure which is situated at distance of 0.211 x L. In this case, the slab is continuous, but we will assume the same position of contraflexure."Hence point of contraflexure is at = 0.633 mmShear force at this point is given = P(L/2 - x)= 128.53 kN= 128532.75 N"Assuming that all the bars will avilable at point of contraflexure,"M = Ast ?st j d= 462979286.1 NmmLo = "12? or d, whichever is more"= 1160 mmLd = 45?= 900 mmM/V + Lo = 4762.03 mmHence safe as M/V+Lo > LdCotinue these bars by a distance Lo = d = 1160 "mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length. ""At the point of curtailment, length of each bar available" =1793 mmHence safe These bars will be provide at the top face of heel slab. Maximum passive B.M. = PL2/16= 3/4 x M1Area of bottom steel (Ast2) = 3/4 x Ast1= 1440 mm2"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 1920 mm2Provide Ast2 = 1920 mm2Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm"Let us check this reinforcement for development length crierion at point of contraflexur,"M = Ast ?st j d= 462979286.1 NmmLo = "12? or d, whichever is more"= 1160 mmLd = 45?= 900 mmM/V + Lo = 4762.03 mmHence safe as M/V+Lo > LdCotinue these bars by a distance Lo = d = 1160 mm from the point of contraflexure.i.e. upto distance = -527 mm from the centre of support." At this point half bars can be discontinued. Since this distance is quite small, it is better to continue these bars upto center of counterfors."5 Design of Toe SlabProviding counterfort over toe slab upto ground level.Assuming total depth of toe slab = 500 mmTotal weight of toe slab = 12.5 kN/m2Net upward intensity at D = 403.06-12.5 kN/m2= 390.56 kN/m2Net upward intensity at E = 209.45-12.5 kN/m2= 196.95 kN/m2Cosidering strip of unit width at D. Max. negative B.M. (M1) = wL2/12= 292.92 kNmEffective depth required (d) = BMRxb= 514.05 mmProviding overall depth (D) = 500 mmEffective depth (d) = 460 mm"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 800 mm2"AstM=Ast ?st j d" = 3063.30 mm2Provide Ast1 = 3063.30 mm2Dia of bar = 20 mmSpacing of bar calculated = 102 mmSpacing of bar provided = 100 mmShear force (V) = PL/2= 585.84 kN100Ast/bd = 0.67?c = 0.32 N/mm2"Refernce: Pg. 84, Table 23, IS 456 : 2000"?v = V/(d x b)= 1.27Hence shear reinforcement is required as ?v > ?cHence depth required from shear point of veiw (d) = V/(?c x b)= 1830.75 mmProviding overall depth (D) = 1500 mmHence effective depth (d) = 1460 mm"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 2400 mm2"AstM=Ast ?st j d" = 965.15 mm2Provide Ast1 = 2400 mm2Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mmDistribution Steel = 2400 mm2Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mmLet us check this reinforcement for development length crierion at point of contraflexur.Point of contraflexure is at = 0.633 mmShear force at this point is given = P(L/2 - x)= 338.62 kN= 338615.52 NM = Ast ?st j d= 728394135.5 NmmLo = "12? or d, whichever is more"= 1460 mmLd = 45?= 900 mmM/V + Lo = 3611.09Hence safe as M/V+Lo > LdCotinue these bars by a distance Lo = d = 1460 "mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length. ""At the point of curtailment, length of each bar available" =2093 mmHence safe These bars will be provide at the top face of toe slab. Maximum passive B.M. = PL2/16= 3/4 x M1Area of bottom steel (Ast2) = 3/4 x Ast1= 1800.00 mm2"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 2400 mm2Provide Ast2 = 2400 mm2Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm"Let us check this reinforcement for development length crierion at point of contraflexur,"M = Ast ?st j d= 728394135.5 NmmLo = "12? or d, whichever is more"= 1460 mmLd = 45?= 900 mmM/V + Lo = 3611.09 mmHence safe as M/V+Lo > LdCotinue these bars by a distance Lo = d = 1460 mm from the point of contraflexure.i.e. upto distance = -827 mm from the centre of support.6 Design of StemThe stem acts as a continuous slab. Considred 1 m strip at B . The intencity of earth pressure is given by (ph) = Ka?H1= 70.80 kN/m2Hence revised H1 = 11.80 m"Negative B.M. in slab, (M1)" = phL2/12= 53.1 kNmEffective depth required (d) = BMRxb= 218.87 mmProviding overall depth (D) = 600 mmEffective depth (d) = 560 mm"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 960 mm2"AstM=Ast ?st j d" = 456.15 mm2Provide Ast1 = 960.00 mm2Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mmDistribution Steel = 960 mm2Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mmShear force (V) = PL/2= 106.20 kN100Ast/bd = 0.17?c = 0.20 N/mm2"Refernce: Pg. 84, Table 23, IS 456 : 2000"?v = V/(d x b)= 0.19 N/mm2Hence no shear reinforcement is required as ?v < ?cLet us check this reinforcement for development length crierion at point of contraflexur.Point of contraflexure is at = 0.633 mmShear force at this point is given (V) = V(L/2 - x)= 92.08 kN= 92075.40 NM = Ast ?st j d= 111753620.8 NmmLo = "12? or d, whichever is more"= 560 mmLd = 45?= 720 mmM/V + Lo = 1773.72Hence safe as M/V+Lo > LdCotinue these bars by a distance Lo = d = 560 "mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length. ""At the point of curtailment, length of each bar available" =1193 mmHence safe These bars are to be provided at the inner face of the stem. Maximum passive B.M. = PL2/16= 3/4 x M1Area of steel (Ast2) = 3/4 x Ast1= 720.00 mm2"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 960 mm2Provide Ast2 = 960 mm2Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm"Let us check this reinforcement for development length crierion at point of contraflexur,"M = Ast ?st j d= 111753620.8 NmmLo = "12? or d, whichever is more"= 560 mmLd = 45?= 720 mmM/V + Lo = 1773.72 mmHence safe as M/V+Lo > Ld7 Design of Main CounterfortLet us assuming thickness of counterforts is = 500 mmSpacing of counterforts = 350 cm c/c"At any section at depth h below the top A, the earth pressure acting on each counter forts will be " = Ka x ?sat x h x L= 21 h kN/m"Similarly, net down ward pressure on heel at C is" = (11.8 x 18) + (1200/1000) x 25 - -86.35= 328.75 kN/m2"Similarly, net down ward pressure on heel at B is" = (11.8 x 18) + (1200/1000) x 25 - 177.2= 65.2 kN/m2"Hence reaction transferrred to each counterfort will be,"At C = 328.75 x (350/100)= 1150.625 kN/mAt B = 65.2 x (350/100)= 228.2 kN/mPressure intencity at h = 11.5 m= 241.5 kN/mShear force at F (Q) = 0.5 x 241.5 x 11.5= 1388.63 kNB.M. = h/3 x 1388.625= 5323.06 kNm= 5323062500 NmmEffective depth required (d) = BMRxb= 3099.03 mmProviding total depth (D) = 3150 mmEffective depth (d) = 3110 mm"Angle ? of the face AC is given by," =tan ? = 4.9/11.8tan ? = 0.415254237? = 22.55 DegreesSin ? = 0.3835Cos ? = 0.9235F1G1 = AF1 x sin?= 4.41 m= 4410.29 mmFG = 5011.00 mmAsssuming that the steel reinforcement is provided in two layers and providing a nominal cover of 40 mm and 20 mm dia bar.Effective depth (d) = 5011 - (40 + 20 + 10)= 4941.00 mm"Area of steel at supports, at bottom (Ast ) M=Ast ?st j d" = 5182.56 mm2Using dia. of bar = 20 mmA? = 314.16 mm2No. of bars = 17 barsProvide the bars in two layers.Effective shear force = Q-(M/d')tan?d' = d/cos?= 5350.07 mmEffective shear force = 975466.99 N?v = V/(d x b)= 0.36 N/mm2100Ast/bd = 0.20?c = 0.21 N/mm2Hence shear reinforcement is required as ?v > ?c"However, the vertical and horizontal ties provided in counterforts will bear the excess shear stress."The height h where half of the reinforcement can curtailed (H) =h m= 3.61 "m below A, i.e. at point H.""To locate the position of point of curtailmenton AC, drawing Hl parallel to FG."Thus half bars can be curtailed at l. However these should be extent by a distance 12? = 240 "mm beyond I, i.e. extented upto I1"The location of H corresponding to I1 can be locate by drawing line I1H1 parallel FGIt should be noted that I1G should not less than 45? = 900mmDesign of Horizontal Ties"At any depth h below the top, force causing sepration" = Ka x ?sat x h x L= 207 kN/mAst required = Force/Stress= 900 mm2Using 2 legged ties of dia. = 10 mmA? = 2?/4 x D2= 157.08 mm2Spacing required = 174.53 mmSpacing provided = 150 mmDesign of Vertical TiesThe downward force at C = (1150.625 x 3)/(350/100)= 986.25 kN/mThe downward force at B = (228.2 x 3)/(350/100)= 195.6 kN/mAst required at C = Force/Stress= 4288.04 mm2Using 2 legged ties of dia. = 20 mmA? = 2?/4 x D2= 628.32 mm2Spacing required = 146.53 mmSpacing provided = 150 mmAst required at B = Force/Stress= 850.43 mm2Using 2 legged ties of dia. = 10 mmA? = 2?/4 x D2= 157.08 mm2Spacing required = 184.71 mmSpacing provided = 150 mm8 Design of Front CounterfortThe upward pressure intensity varies from 403.06 kN/m2 at D to 209.45 kN/m2 at E. Downward weight of toe slab = 37.5 kN/m2Net weight at D = 365.56 kN/m2Net weight at E = 171.95 kN/m2The center to center spacing of counterforts = 3.5 mHence upward force transmitted to counterforts at D = 365.56 x 3.5 kN/m= 1279.46 kN/mUpward force transmitted to counterforts at E = 171.95 x 3.5kN/m= 601.825 kN/mTotal upward force = 1/2 x (1279.46 + 601.825) x 3.6= 3386.31 kNForce will be acting at x = ((601.825 + 2 x 1279.46)/(1279.46 + 601.825)) x (3.6/3)= 2.02 m from EB.M. = 3386.31 x 2.02 kNm= 6840.35 kNmEffective depth required (d) = BMRxb= 3513.05Providing total depth (D) = 4000 mmEffective depth (d) = 3960 mmThus height of counterfort above ground level = 2500 mmi.e. upto point F3."Area of steel at supports, at bottom (Ast ) M=Ast ?st j d" = 8309.61 mm2"Minimum steel required 0.2% of b D ? further 20% [IS :3370,7.1]"= 6400 mm2Provide Ast1 = 8309.61 mm2Dia of bar = 32 mmA? = 804.25 mm2No. of bars = 11 barsEffective shear force (V) = Q-(M/d)tan?tan ? = 1.11Effective shear force (V) = 1467020.94 N?v = V/(d x b)= 0.74 N/mm2100Ast/bd = 0.15?c = 0.19 N/mm2Hence shear reinforcement is required as ?v > ?cUsing 2 legged ties of dia. = 10 mmA? = 2?/4 x D2= 157.08 mm2Vc = ?c x b x d= 376200 NVs = V - Vc= 1090820.94 NSpacing required (sv) = (?st x A? x d)/Vs= 131.16Spacing required (sv) = 150 mm c/c Provide 2 x 10 mm holding bar at top9 "Fixing Effect in Stem, Toe and Heel""At the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided."(i) "In stem @0.3% of cross section, to be provided at inner face, in vertical direction,for a length 58?"Ast = (0.3/100) x 1000 x 600= 1800 mm2Dia. of bar = 16 mmA? = 201.06 mm2Spacing required = (201.06 x 1000)/1800= 111.70 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 928 mm(ii) In toe slab @0.15% to be provided at the lowar faceAst = (0.15/100) x 1000 x 1500= 2250 mm2Dia. of bar = 20 mmA? = 314.16 mm2Spacing required = (314.16 x 1000)/2250= 139.63 mmSpacing provided = 100 mm16 mm ?Length of embedment in slab above heel slab = 1160 mm0.928 100 mm c/c0.928(iii) In heel slab @ 0.15% to be provided in upper faceAst = (0.15/100) x 1000 x 1200= 1800 mm220 mm ? 20 mm ?Dia. of bar = 16 mm100 mm c/c 100 mm c/cA? = 201.06 mm21.16Spacing required = (201.06 x 1000)/1800= 111.70 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 928 mm