Retaining wall

TheoryRCC design B.C.PunmiaRETAINING WALLA retaining wall or retaining structure is used for maintaining the ground surfgaces at defrent elevations on either side of it. Whenever embankments are involed in construction ,retaining wall are usually necessary. In the construction of buildins having basements, retaining walls are mandatory. Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining walls , to resist earth pressure along with superimposed loads. The material retained or supported by a retaining wall is called backfill lying above the horizontal plane at the elevation of the top of a wall is called the surcharge, and its inclination to horizontal is called the surcharge angle bIn the design of retaining walls or other retaining structures, it is necessary to compute the lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical experiment work has been done in this field and many theory and hypothesis heve benn proposed.18.2 TYPE OF RETAINING WALLSRetaining walls may be classified according to their mode of resisting the earth pressure,and according to their shape. Following are some of commen types of retaining walls (Fig)1Gravity walls2Cantilever retaining wallsa. T- shapedb. L- shaped3Counterfort retainig walls.4Buttresssed walls.A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any where, and the resultant of forces remain withen the middle third of the base.The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal

Data sheetBased on example 18.1 (RCC design by B.C.Punmia)DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill00mm F0mm F0@ c/cHieght of cantilever wall from ground level=7.00m0@ c/c70000.00Unit weight of Earth=18KN/m300mm FAngle of repose=30Degree0mm F0@ c/cSafe Bearing capacity of soil=180KN/m30@ c/c0Coffiecent of friction=0.500.000mm FConcreteM-200mm F0@ c/cwt. of concrete=25000N/m30@ c/cSteelfe415N/mm2sst230N/mm200mm Fm13.33scbc7N/mm2000@ c/cNominal cover=30mm00mm FFoundation depth=1.00m0ToeHeel0@ c/cCounter forts width=0.50m000DESIGN SUMMARYStem thicknessAt footing0mmAt top0mm0mm F2000Heel width2000mmToe width1700mm0@ c/c1700400Footing width4100mm Key0x0mm4100Reinforcement Summary17004002000STEM:-0mm F0@ c/c0mm FMain (from top of Retaining wall)0@ c/c100% Reinforcement upto m0.0010mm F@90mm c/c0F10mm F50% Reinforcement upto m0.0010mm F@0mm c/c00.000@ c/c25% Reinforcement upto mTop10mm F@0mm c/c0.00Distribution8mm F@0mm c/c0Tamprecture8mm F@0mm c/c10mm FTOE:-0@ c/cMain10mm F@100mm c/cDistribution8mm F@0mm c/c10mm FHEEL:-0@ c/cMain10mm F@0mm c/cDistribution8mm F@180mm c/cOut sideEarth sidePut the value in BLU cell till value of BROWN cell =o, zero or near zeroh =2.830.00h =2.190.00

&Lpk_nandwana@yahoo.co.inPut value for h in this cell Till the value of cell R 227 ZeroPut value for h in this cell Till the value of cell R 211 Zero

Design 151152045334418402040204018401634163426Design of Catilever retaing wall with horizontal back fill 18.7 (Design example 18.4)DESIGN OF COUNTERFORT RETAINING WALLHieght of cantilever wall from ground level=7.00mUnit weight of Earthg=18kN/m3=18000N/m2Angle of repose=30DegreeSafe Bearing capacity of soilq0=180kN/m3Coffiecent of frictionm=0.5Concrete=M20Steelfe=415Nominal cover=30mmeffective cover40mmFoundation depth=1.00mCounter forts width assume=0.50m500mm1Design Constants:-For HYSD BarsCocrete M=20sst ==230N/mm2wt. of concrete=25000N/mm2=25kN/mm2scbc ==7N/mm3m=13.33k=m*c=13.33x7=0.289m*c+sst13.33x7+230j=1-k/3=1-0.289/3=0.904R=1/2xc x j x k=0.5x7x0.904x0.289=0.913For soil, F=30\Ka =1-sin F=1-0.5=0.33Kp=1=31+sinF1+0.5Ka2Diamension of various parts:-Hence height of wall above base H=7.00+1.00=8.00mThe ratio of length of slabe (DE) to base width b is given by eq.Value of sin anglea=1-q0=1-180=0.432DegreeValue2.2 y H2.2x18x8.00250.422Keep a=0.43.Eq (1)300.50The width of base is given by Eq.350.573b=0.95 HxKa400.643(1- a)x(1+3 a)450.707500.766b=0.95x8.000.333=3.84m550.819(1-0.43)x(1+1.29)600.866650.906The base width from the considration of sliding is given by Eq.b=0.7HKa=0.7x8.00x0.3333333333=6.55m(1-a) m(1-0.43)x0.5This width is excessive. Normal practice is to provide b between 0.5 to 0.6 H .Taking maximum value of H =0.5b=0.50x8.00=4.00mHence Provided b=4.00mfor design purposeThe wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .Width of toe slab=a x b=0.43x4.00=1.72mProvided toe slab =1.70mtaking the uniform thickness of stem =300mm=0.30mfor design purposeHence width of heel slab=4.00-0.30-1.70=2.00mLet the thickness of base slab=300mm=0.3mClear spacing of counter fort =3.5 xH1/4=3.50x8.001/4=2.86my18keep them at3.00m apart.Let us provide counterfort over toe slab, upto ground level at3.00m clear distance.0.33Stability of wall:-Full dimension wall is shown in fig 1aLet w1=weight of rectangular portion of stemw2=weight of base slabw3=weight of soil on heel slab.The calculation are arrenged in Table7.50Detailforce(kN)lever armMoment about toe (KN-m)7.00w11x0.30x7.70x25=57.751.85106.8375w21x0.30x4.00x25=30260w31x2.00x7.70x18=277.23831.60Sw=364.95total MR998.440.3Total resisting moment=998.44kN-m..(1)1.702.00DEBC1.00Earth pressure p=Ka x y x H2=0.33x18x(8.00)2=192kN..(2)1.802.000.50224.10Over turningOver turning moment Mo=192x8=512kN-m4.003\F.S. against over turning=998.44=1.95>2Hence safe512F.S. against sliding =mSw0.50x364.95=0.950.667Hence un safe22Pressure p1 at toe=SW1 +6 e=364.95x1 +6x0.67=183>180Hence un safebb4.004.00kN -m2In order to make it safe, increase the length of toe slab DE to1.80m, So that total wiodth is =4.10The revised computations are arranged in tableDetailforce(kN)lever armMoment about toe (KN-m)w11x0.30x7.70x25=581.95113w21x0.30x4.10x25=312.0563w31x2.00x7.70x18=2773.1859Sw=365.70total MR1035.00net moment SM =1035-512.00=523.00kN-m\Distance x of the point of application of resultant, from toe isx=SM=523.00=1.43mb=4.10=0.6833333333Sw365.7066Eccenticity e=b-x=4.10-1.43=0.62mLd=540mmHence safeThese bars will be provide at the top face of heel slab. Maximum Passive B.M. =PL2=3M1164\Area of Bottom steel Ast2=3xAst1=3x1130=847mm244using12mm barsA=3.14xdia2=3.14x12x12=113mm244\SpacingA x1000 / Ast=113x1000/847=133mmHence Provided12mm F bar, @130mm c/cActual Ast=1000x113=870mm2130Let us check this reinforcement for development length crierion at point of contraflexur,Inherent in criterion :M+Lo>LdWhere V = Shear at point of contraflexure=119886NVDistance from face of supports=0.63mAssuming that all bars are available at point of contraflexure,M = sst x Ast x j x d =230x870x0.904x440=79531705N-mmLo=12 For d,witch ever is more=440mmLd=45x F=45x12=540mmas before\M+Lo>Ld=79531705+440=1103.3944330447>540Hence safeV119886Thus continue all bottom bars to a point distance Lo=440mm from the point of contraflexure,i.e. upto a distance =630-440=190mm from the center of sports.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.Reinforcement near B :-The c/c spacing of reinforcement near B may be increased, because P decreasedue to increase in upward soil reaction. Consider a strip 1 m wide near BUpward soil reaction at B is =87.2kN/m2As found earlier.Net downward load p' =weight of earth + weight of counterforts - upward soil reaction\Net downward load p'=138.60+(0.50x25) -87.2=63.9kN/m2This is about=63.90/137.8=0.464of load intencity at CHence spacing of steel bars=100/0.464=200mm c/c at the top face, near supportsSpacing of steel bars at the bottom face, at mid span=133/0.464=300mm c/cDistribution steel=0.12x1000x500=600mm21000.3Using12mm F bars, Area=P D2=3.14x (12)'2=113mm244\Spacing =1000x113=188mm say=180mm c/c600Shear reinforcement.shear stress at C = tv =0.471.802.00% of steel provided =100x1130=0.26%1000x440Permissible shear stress for0.26%steel provided tc=0.21N/mm2(See Table 3.1)Safe if tv< tcHere0.47>0.21Hence shear reinforcement requiredVc = tc b x d=0.21x1000x440=92400Nor92.4kN3.00Consider a section distance x1 from face of counterfort, where shear force is=92.4kNThe position is given by=92.4=1.50-x1or x1=1.50-0.70=0.80m2071.50Hence shear stirrups are required upto distance=0.80m on either side of each counterforts.D1C1The requirement is there form a strip of unit width paassing through C,x2x1such that shear force at the counterforts isd=92.4kNNet down ward pressure at C =137.80kN/m2,Net down ward pressure at B =63.90kN/m2Dy2E1 EBB1 y1 CLt net down ward pressure at B1=w1 x 3/2=1.5w1This is equal to=92.4kNx2x1w=92.40=61.6kN/m2However at Y1 from C,1.50w1=137.80-137.80-64y1=137.8-36.95Y1=61.62.00137.80\y1 =2.10m63.90Hence shear reinforcement is required in triangular portion on the other side of counterforts shown hatched in fig .y2However, we will provide shear strirrups in reangular portion x1 x y1=0.80x2.10=1.68m on157.6086.57w1either side of counterforts.Let us provide4legged stirrups of8mm F wireUsing8mm F bars, Area=P D2=3.14x (8)'2=201mm2w2FIG 244Sv=Asv.ssv.d=201x230x440=178mmV - Vc207-92.40)x1000Hence provided these8mm F4lgd strirrups @170mm c/c either side of each counterforts.5Design of toe slab :-Since the toe slab is also large, provide counterforts over it, upto ground level at3.00m clear distance face to face. The toe slab will thus bend like a contious slab.Assume total depth of toe slab=500mm or0.5mTotal weight of toe slab=0.50x1x1x25=12.50kN//m2Net upward pressure intencity at D=170.10-12.50=157.60kN//m2Similarly Net upward pressure intencity at E=99.07-12.50=86.57kN//m2Cosidering strip of unit width at D.Max. negative B.M.wl2=157.60x3.002=118.2x10'6kN/m21212Effective depth required =BM=118.20x10 6=359.8mmRxb0.913x1000Shear force V=157.6x3.00=236.4kN2Taking a permissible stress tc=0.30N/mm2assuming % steel0.5%table 3.1The depth of slab required from shear point of view is given by d= V / (b x tc)d=236x1000=788mm1000x0.300This is excessive ,However we will keep the same depth as that of heeland provide shear strirrups to take up excessive shearing stress.However keep =500mm providing effective cover =60mm d=500-60=440mmArea of steel at supports, at bottom is Ast=118.20x10'6=1292mm230x0.904x440using12mm barsA=3.14xdia2=3.14x12x12=113mm244\SpacingA x1000 / Ast=113x1000/1292=87.5mmHence Provided12mm F bar, @80mm c/cActual Ast=1000x113=1413mm280Let us check this reinforcement for development length crierion at point of contraflexur,Inherent in criterion :M+Lo>LdWhere the point of contraflexure occure at distance x rom supports=0.63mVHence shear force at the point of contraflexure is V =V =w L-x=157.6x3.00-0.63=137112.0N22M =230x1413x0.904x440=129239020NmmLo=12 For d,witch ever is more=440mmLd=45x F=45x12=540mm\M+Lo>Ld=129239020.21+440=1382.5799361814>540Hence safeV137112Hence satisfied , continue these bars, at the bottom of toe slab, beyond the point of contraflexureby a distance of Lo=440.0mm i.e. by a distance of630+440.0=1070mmfrom the face of counterfortsAgain, positive B.M.3x M1=3x118.20x10'6=88.65x10 6N-m44\Area of Bottom steel Ast2=3xAst1=3x1292=969mm244using12mm barsA=3.14xdia2=3.14x12x12=113mm244\Spacing =A x1000 / Ast=113x1000/969=117mmHence Provided12mm F bar, @110mm c/cActual Ast=1000x113=1028mm2110Let us check this reinforcement for development length crierion at point of contraflexur,Inherent in criterion :M+Lo>LdWhere V = Shear at point of contraflexure=137112NVAssuming that all bars provided at top face,are available at point of contraflexure,M = sst x Ast x j x d =230x1028x0.904x440=93992015N-mmLo=12 For d,witch ever is more=440mmLd=45x F=45x12=540mmas before\M+Lo>Ld=93992015+440=1125.512683062>540Hence safeV137112Thus continue all bottom bars to a point distance Lo=440mm from the point of contraflexure,i.e. upto a distance =630-440=190mm from the center of sports.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.Reinforcement at E :-At a section distance 1 meter from E,upward soil pressure=170.1-170.1-8.30x0.80=138.50kN/m24.10\Net upward pressure=138.5-12.50=126.0kN/m2This is about=126.0/157.60=0.80of w at D\Spacing of bottom steel=87.5/0.80=109mmsay=100mmSpacing of top steel=117/0.80=146mmsay=140mmDistribution steel=0.12x1000x500=600mm2100Using12mm F bars, Area=P D2=3.14x (12)'2=113mm244\Spacing =1000x113=188mm say=180mm c/c600Shear reinforcementshear force at D=236.4kNShear stress tv=shear force=236.40x1000=0.54N/mm2Beam Ht.x beam wt.1000x440% of steel provided =100x1413=0.32%\tc=0.24N/mm2table1000x4403.1Permissible shear stress tc for0.32%steel provided tc=0.24N/mm2(See Table 3.1)Safe if tv< tcHere0.54>0.24Shear reinforcement requiredVc = tc b x d=0.24x1000x440=105600Nor105.6kNConsider a section distance x1 from face of counterfort, where shear force is=105.6kNThe position is given by=105.6=1.50-x2or x2=1.50-0.70=0.80m236.401.50Hence shear stirrups are required upto distance=0.80m on either side of each counterforts.The requirement is there form a strip of unit width paassing through D, Let us consider a strip through E1,distance y2 from D, such that shear force at the counterforts is105.6kN. To find the position of Y2consider the net pressure distribution below the toe.Self weight of toe slab=12.50Hence net pressure intencity below D an dE arebelow D=170.1-12.50=157.6kN/m2, and below E99.1-12.50=86.6kN/m2Let the net pressure intencity at E1 = w2 x 3/2 = 1.5 w2 kN/m2\ Shear force at the counterforts at E1 = w2 x 3/2 =1.5m w2 kN/M2This is equal to=105.6kNw2=105.60=70.4kN/m2.(1)1.50However at Y2 from D,w2=157.60-157.60-86.6y2=157.6-39.4634146341Y2(2)1.80Equating the two we get,=157.60-39.46Y2=70.4\y2=2.20mThis is>than DEDE=1.80mHence shear force at E is more than=105.6kN/m2Actual shear forceat E =1.50x86.57=129.8kN/m2Considered a section distence Z from the face of dounterforts (Point E), where S.F. is105.6kNThe position of Z is given =105.60=1.50-Z=or Z=1.50-1.22=0.30m129.851.50Hence shear stirrups are to be procided for a region DEE2D1, where EE2=0.30m only.However, we will provide shear strirrups for whole of rectangular area (shown dotted),for width DD1,= x2 =0.80mand length DE=0.30mLet us provide8legged stirrups of8mm F wireUsing8mm F bars, Area=P D2=3.14x (8)'2=402mm244Sv=Asv.ssv.d=402x230x440=311mmV - Vc236.4-105.60)x1000Hence provided these8mm F8lgd strirrups @300mm c/c either side of each counterforts.6Design of stem (vertical slab)The stem acts as a continuous slab. Considred 1 m strip at B .The intencity of earth pressure is given by.ph= KayH1=0.33x18x7.50=45.0kN/m2Hence revised H1=8.00-0.50=7.50mB.M.=phxL2=45.00x(3)2=33750000N-mm1212Effective depth required =BM=33750000=192mmRxb0.913x1000Providing effective cover =60mm, so total depth=192+60=252mmHowever provide total depth d =300mm and effective thickness=300-60=240mmthis increased thickness will keep the shear stress within limit so that additional shearreinforcement not required.Shear force V=45.0x3=67.5kN\tc=67.50x1000=0.28N21000x240mm2this is less than=tc=0.3N/mm2 at 0.5% reinfocement (see Table 3.1)Area of steel near conuterforts is=33750000=677mm2230x0.904x240Reinforcement corresponding to p=0.50%is = pbd/100=0.50x1000x240=1200mm2100Using12mm F bars, Area=P D2=3.14x (12)'2=113mm244\Spacing =1000x113=94mm say=90mm c/c1200Actual AS provided=1000x113=1256mm2 and100As=100x1256=0.5233333333%90bd1000x240Let us check these bars for devlopment length, near points of contraflexure, so as to satisfy the criterion=M+ Lo > LdFor fixed beam or slab carrying U.D.L. , the point of contraflexure is at a distance of 0.211 LVcontraflexure point =0.211x3.00=0.63mfrom the face of counterforts ,shear force at this point given by V= pL/2(l/2-x)+(L/2)V=pLl- x+L=p xl- x=67.5x3-0.6322222=58.7kNAssuming that all the bars will be available at the point of contrflexure,M=sst x Ast x jc x d=230x1256x0.904x240=62661343N-mmLo=12 F or D , whichever is more=240mmLd=45 F=45x12=540mmM+Lo=62661343+240=1307mm>540Hence safeV58.70x1000It is thus essencial to continue all the bars upto a point distance=240mm beyondpoint of contraflexure, i.e. upto a point240+630=870mm say=900mm from theface of counterforts. These bars are to be provided at the inner face of stem slab.Maximum positive B.M.=3 x M1=3x33750000=25312500N-mm44Area of steel=3x Ast=3=1256=942mm244Using12mm F bars, Area=P D2=3.14x (12)'2=113mm244\Spacing =1000x113=120mm say=120mm c/c942Actual AS provided=1000x113=942mm2 and100As=100x942=0.1%120bd1000x942Let us check these bars for devlopment length, near points of contraflexure, so as to satisfy the criterion=M+ Lo > LdAssuming that all reinforcement is extended upto poin of contraflexure.VM=sst x Ast x jc x d=230x942x0.904x240=46996007N-mmLo=12 F or D , whichever is more=240mmLd=45 F=45x12=540mmV=58.7As before\ M+Lo=46996007+240=1041mm>540Hence safeV58.70x1000The spacing of reinforcement at B, found above can be increased with height .The pressure ph and hence the bending moment decreaases linearly with height.\AsthHence the spacing of bars can be increase gradually to say300mm c/c near top.\Distribution reinforcement=0.12x1000x300=360mm2100Using10mm F bars, Area=P D2=3.14x (10)'2=79mm244\Spacing =1000x79=218mm say=200mm c/c3607Design of main counterfort.Let us assuming thickness of counterforts is =500mm. The counterfort will thusbe spaced @300x50=350cm c/c. They will thus receive earth pressure froma width of3.5m and down ward reaction from heel slab for width of3.5mAt any section at depth h below the top A, the eerth pressure acting on each counter forts will be=1x18.00xhx3.5=21hkN/m3similarly, net down ward pressure on heel at c is=7.50x18+0.50x25-8.30=139.2kN/m2Aand that at B=7.50x18+0.50x25-87.2=60.3kN/m2Hence reaction transferrred to each counterfort are will beAt C,=139.2x3.50=487.20kN/mhAt B,=60.3x3.50=211.05kN/mThe variations of horizontal and vertical forces on counterfort are shown in fig.21h kn/mThe critical section for the counterfort will be F, since below this, enormous depth will be available to resist bending.7.008.00Pressure intencity at h =7.00m is=21x7.00=147kN/m7.50Shear force at F=1x147x7.00=514.5kN2B.M.=514.5x7/3=1200.5kN-mor1200500000N-mmGConterforts act as a T beam. However, even as a reactangular beam, depth requiredFG1147.00kN/mEffective depth required =BM=1200500000=1622mmF1F2Rxb0.913x5001.00EBProviding effective cover =60mm, so total depth=1622+60=1682mmD0.5CHowever provide total depth d =1700mm and effective thickness=1700-60=1640mm2.00168.00kN/mAngle F of face AC is given byTan F=2.00/7.50=0.3\F=15\sin F=0.259andcos F=0.966211.05487.2Depth F1C1=AF1 sin F =7x0.259=1.82mor1820mmkN/m\ Depth FG=1820+300=2120Asssuming that the steel reinforcement is provided in2layer with20mm spaceFIG. 3between them and providing a nominal cover30mm and main bars of20mm F diathe effective depth will be=2120-(30+20+12+10=2048mmArea of steel at supports, at bottom is Ast=1200500000=2820mm230x0.904x2048using20mm barsA=3.14xdia2=3.14x20x20=314mm244\No. of bars2820/314=9No. provode these in2layersEffective shear force=Q-Mtan Fwhere d'=d=2048=2120mmd'cos F0.966\ Effective shear force=514500-1200500000x0.27=361606.132075472N2120\tv=361606.132075472=0.3411378604N/mm2500x2120area of steel100xAs=100x9x314=0.3%\ tc=0.21N/mm2b xd500x2120thus the shear stress tv is more than permisssible shear stress tc. 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 will be equal to HH=8.00=2.8mbelow A, i.e. at point H. To locate the position of point of curtailmenton AC, draw Hl parallel to FG.Thus half bars can be curtailed at l. However these should be extent by a distance 12 F =240mmbeyond l, i.e. extented upto l1. The location of H corresponding to l1 can be locate by drawing line l1H1parallel FG. It should be noted that l1G should not less than 45 F =900mm similarly, other bars can becurtailed, if desired,Design of Horizontal ties:-The vertical stem has a tendency to saprate outfrom the counterforts, and henceshould be tie to it by horizontalties. At any depth h below the top, force causing sepration1x18h x3.00=18hkN/m3hereh=7\force=18x7=126kN/m\ steel required=126x1000=548mm230using10mm f2legged ties, As=2x3.14x(10)2=157mm24spacing=1000x157=286mm548however provide200mm c/c at bottom and gradualy increase to300mm at topDesign of vertical ties:-Similar to the stem slab, heel slab has also tendency to seprate out from counterforts,due to net down ward force, unless tied properly by vertical ties.The down wars force ar C will be487.20x3.00417.6kN/m3.50The down wars force ar B will be211.05x3.00180.9kN/msee fig.3.50Near end C, the heel slab is tied to counterforts with the help of main reinforcement of counterforts.steel required at C=417.6x1000=1816mm230using12mm f2legged ties, As=2x3.14x(12)2=226mm24\ spacing of ties=1000x226=124mmsay120mm1816steel required at B=180.9x1000=787mm230using12mm f2legged ties, As=2x3.14x(12)2=226mm24\ spacing of ties=1000x226=287mmsay280mm787Thus the spacing of vertical tie can be increase gradually from120mm c / c at C to280mm8Design of front counterforts :-Refer fig 1 The upward pressure intencity varies from170.1kN/m2 at D, to99.07kN/m2 at E.Down ward weight of500mm thick toe slab=0.5x25=12.5kN/m2hence net w at D=170.1-12.5=157.6kN/m2F1and at E=99.07-12.5=86.6kN/m20.40The center to center spacing of counterforts,500mm wide is3.50m.Hence upward force transmittedG. LevelFto counterforts at D157.6x3.50=551.6kN/m and at E86.6x3.50=302.9804878049kN/m1.001.40The counterforts bent up as cantilever about face FE. Hence DF will be in compression while D1E1 will beDFEin tension, and main reinforcement will be provide at bottom face D1E10.5Total upward force(551.6+303.0)x1.80=769kN/m2.0D1E1acting at x=303+2x551.6x1.80=0.99m from E303+551.63303\B.M.=769x0.99=759KN-mOR759000000N-m551.6\d=759000000=1289mm500x0.913FIG 4Providing effective cover =80mm, so total depth=1289+80=1369mmHowever provide total depth d =1400mm and effective thickness=1400-80=1320mmThus project the counterforts400mm above groud level,to point F1 as shown in fig 4Area of steel near conuterforts is=759000000=2767mm2230x0.904x1320Using25mm F bars, Area=P D2=3.14x (25)'2=491mm244\No of bars2767=6Noprovide these in one layer and continue by a distance of 45 F beyond E491Effective shear force=Q-Mtan FFrom fig 4 tanF =0.90d'1.80V=769-759x0.90=481.5kN481500N1.321.80and tv=482x1000=0.730N/mm2500x1320area of steel100xAs=100x6x491=0.45%\ tc=0.280N/mm2table 3.1b xd500x1320sincetv>tcshear reinforcement is requiredusing12mm f2legged ties, As=2x3.14x(12)2=226mm24Vc=tcxbxd=0.280x500x1320=184800V1=V-Vc=481500-184800=296700N\ sv=ssv.Asv.d=230x226x1320=231mmsubject to a maxi. 300 mmVs296700However providethese @230mm c/c provide 2 x12mm f bars on top for holding.9Fixing effect in stem, toe and heel :-At the junction of stem, toe and heel slab fixing moment are included,whichare at right angles to their normal direction of bending. These moment are not determine , butnormal reinforcement given below may be provided.(I)In stem@ 0.8x0.3 =0.24% of cross section, to be provided at inner face,in vertical direction,for a length 45 F\Ast=0.24x1000x300=720mm2100Using10mm F bars, Area=P D2=3.14x (10)'2=79mm244\Spacing =1000x79=109mm say=100mm c/c720Length embedment in stem, above heel slab =45x10=450mm(II)In toe slab @ 0.12% to be provided at the lowae face\Ast=0.12x1000x500=600mm2100Using10mm F bars, Area=P D2=3.14x (10)'2=79mm244\Spacing =1000x79=131mm say=130mm c/c600Length embedment in stem, above toe slab =45x10=450mm(III)In heel slab @ 0.12% to be provided in upper face\Ast=0.12x1000x500=600mm210010mm FUsing10mm F bars, Area=P D2=3.14x (10)'2=79mm2100mm c/c45044\Spacing =1000x79=131mm say=130mm c/c600Length embedment in stem, above heel slab =45x10=450mm45010mm FEach of above reinforcement should anchored properly in adjoining slab, as shown in fig 5130mm c/cFIG. 510. Design of shear key:-The wall is in unsafe in sliding, and hence shear key will have to be provided, as shown in fig. 61.803002.00Let the depth of key =a intensity of passive pressure Pp devloped in front of key depend upon the soil pressure P in front of the keyDEBC4.10Pp=KpP=3.00x99.07=297.20kN/m2\ total passive pressure Pp =Pp x a=297.20aaSliding force at level D1C1=0.33x18x(8+a)2D1C12.00Ppor PH=3.00x(8+a)2.(2)170.1099.07Weight of the soil between bottom of the base and D1C1=4.10ax18=73.80a\SW=364.95+73.80aRefer force calculation tableHence equilibrium of wall, permitting F.S.=1.5 against sliding we haveFIG. 61.5=m Sw+Pp=0.5x (364.95+73.80a)+297.20aPH3x(8.00+a2)1.5x(8+a)2=0.5x (364.95+73.80a)+297.20a3x1.5(8+a)2=182.48+36.9a+297.20a4.5(8+a)2=40.55+74.243902439a64+16a +a2=74.24a-16a +40.55-64a2=58.24a-23.45or a=a2-58.24a+23.45a =b+b2-4*a*cor a=0.4054395436msay=410mm2*aHence keep depth of key=400mm. and width of key400mmNow size of key=400x400mmPH=3.00x(8.00+a)2=3.00x(8.00+0.40)2PH=211.68kN=297.20a=297.20x0.40=118.88kNHenceSW=364.95+73.80a=364.95+73.80x0.40=394.5kNActual force to be resisted by the key at F.S. 1.5 is=1.5PH - mSW=1.5x211.68-0.5x394.5=120.29kN\ shear stress=120.29x1000=0.30N/mm2400x1000Bending stress=120.29x200x10001/6x1000x(400)2=0.90N/mm2Hence safeThe details reinforcement shown in fig 7

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DrawingDESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill0.3`0.31.802.007.507.003.000.31.702.00D1C1DEBC1.00x2x11.802.000.50E1 EBB1 C1C14.10Dy2y24.00x2x1170.199.0787.208.30137.8063.90y2157.6086.57w1w2FIG. 1FIG. 2DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill0.30AF1h0.40G.LevelF21h kn/m0.501.407.008.00DFE7.500.51.00GD1E1G1F147.00kN/mF1F2551.61.00EBkN/m0.5CFIG. 42.00168.00kN/m211.05487.2kN/m1.803002.00DECFIG. 30.504.1010mm Fa100mm c/c450D1C1Pp170.1099.0745010mm F130mm c/cFIG. 5FIG. 6DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill30030012mm F10mm F20mm F300mm c/c300mm c/c5Nos.12mm F110mm F240mm c/c265mm c/c20mm F4Nos.12mm F10mm F12mm F180mm c/c230mm c/c2 lgd vertical ties8000120to0mm c/c12mm F10mm F120mm c/c200mm c/c10mm F12mm F0mm F200mm c/c110mm c/c12mm F0mm c/c3900180mm c/c2x12mm FHolding bars12mm F22100mm c/c500200012mm F2 lgd1800230mm c/c340012mm F12mm F40012mm F12mm F130mm c/c80mm c/c180mm c/c130mm c/c400125mm F4000mm F6No.Bars0mm F0mm c/c0mm c/cCross -section mid way between counterfoortsCross -section mid way between counterfoorts(e) Sectional elevation oftoe slaband front counterforts at section 3 - 3mmFront counterfortsmmmm

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IS-Table64646464646464646464110646464646464646464Table 1.15. PERMISSIBLE DIRECT TENSILE STRESSTable 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000)100AsPermissible shear stress in concrete tc N/mm2Grade of concreteM-10M-15M-20M-25M-30M-35M-40bdM-15M-20M-25M-30M-35M-40Tensile stress N/mm21.22.02.83.23.64.04.4< 0.150.180.180.190.200.200.200.250.220.220.230.230.230.230.500.290.300.310.310.310.32Table 1.16.. Permissible stress in concrete (IS : 456-2000)0.750.340.350.360.370.370.38Grade of concretePermission stress in compression (N/mm2)Permissible stress in bond (Average) for plain bars in tention (N/mm2)1.000.370.390.400.410.420.42Bending acbcDirect (acc)1.250.400.420.440.450.450.46(N/mm2)Kg/m2(N/mm2)Kg/m2(N/mm2)in kg/m21.500.420.450.460.480.490.49M 103.03002.5250----1.750.440.470.490.500.520.52M 155.05004.04000.6602.000.440.490.510.530.540.55M 207.07005.05000.8802.250.440.510.530.550.560.57M 258.58506.06000.9902.500.440.510.550.570.580.60M 3010.010008.08001.01002.750.440.510.560.580.600.62M 3511.511509.09001.11103.00 and above0.440.510.570.60.620.63M 4013.0130010.010001.2120M 4514.5145011.011001.3130Table 3.2. Facor kM 5016.0160012.012001.4140Over all depth of slab300 or more275250225200175150 or lessTable 1.18. MODULAR RATIOk1.001.051.101.151.201.251.30Grade of concreteM-10M-15M-20M-25M-30M-35M-40Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)Modular ratio m31 (31.11)19 (18.67)13 (13.33)11 (10.98)9 (9.33)8 (8.11)7 (7.18)Grade of concreteM-15M-20M-25M-30M-35M-40tc.max1.61.81.92.22.32.5Table 2.1. VALUES OF DESIGN CONSTANTSGrade of concreteM-15M-20M-25M-30M-35M-40Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000)Modular Ratio18.6713.3310.989.338.117.18Grade of concreteM-10M-15M-20M-25M-30M-35M-40M-45M-50scbc N/mm2578.51011.513tbd (N / mm2)--0.60.80.911.11.21.31.4m scbc93.3393.3393.3393.3393.3393.33(a) sst = 140 N/mm2 (Fe 250)kc0.40.40.40.40.40.4Table 3.5. Development Length in tensionjc0.8670.8670.8670.8670.8670.867Rc0.8671.2141.4741.7341.9942.254Grade of concretePlain M.S. BarsH.Y.S.D. BarsPc (%)0.71411.2141.4291.6431.857tbd (N / mm2)kd = Ld Ftbd (N / mm2)kd = Ld F(b) sst = 190 N/mm2kc0.3290.3290.3290.3290.3290.329M 150.6580.9660jc0.890.890.890.890.890.89M 200.8441.2845Rc0.7321.0251.2441.4641.6841.903M 250.9391.4440Pc (%)0.4330.6060.7360.8660.9971.127M 301351.636(c ) sst = 230 N/mm2 (Fe 415)kc0.2890.2890.2890.2890.2890.289M 351.1321.7633jc0.9040.9040.9040.9040.9040.904M 401.2291.9230Rc0.6530.9141.111.3061.5021.698M 451.3272.0828Pc (%)0.3140.440.5340.6280.7220.816M 501.4252.2426(d) sst = 275 N/mm2 (Fe 500)kc0.2530.2530.2530.2530.2530.253jc0.9160.9160.9160.9140.9160.916Rc0.5790.8110.9851.1591.3321.506Pc (%)0.230.3220.3910.460.530.599Shear stress tcReiforcement %100AsM-20M-20100Asbdbd0.150.180.180.150.160.180.190.180.170.180.20.210.180.190.210.240.190.190.220.270.20.190.230.30.210.20.240.320.220.20.250.350.230.20.260.380.240.210.270.410.250.210.280.440.260.210.290.470.270.220.300.50.280.220.310.550.290.220.320.60.30.230.330.650.310.230.340.70.320.240.350.750.330.240.360.820.340.240.370.880.350.250.380.940.360.250.391.000.370.250.41.080.380.260.411.160.390.260.421.250.40.260.431.330.410.270.441.410.420.270.451.500.430.270.461.630.440.280.461.640.450.280.471.750.460.280.481.880.470.290.492.000.480.290.502.130.490.290.512.250.50.300.510.300.520.300.530.300.540.300.550.310.560.310.570.310.580.310.590.310.60.320.610.320.620.320.630.320.640.320.650.330.660.330.670.330.680.330.690.330.70.340.710.340.720.340.730.340.740.340.750.350.760.350.770.350.780.350.790.350.80.350.810.350.820.360.830.360.840.360.850.360.860.360.870.360.880.370.890.370.90.370.910.370.920.370.930.370.940.380.950.380.960.380.970.380.980.380.990.381.000.391.010.391.020.391.030.391.040.391.050.391.060.391.070.391.080.41.090.41.100.41.110.41.120.41.130.41.140.41.150.41.160.411.170.411.180.411.190.411.200.411.210.411.220.411.230.411.240.411.250.421.260.421.270.421.280.421.290.421.300.421.310.421.320.421.330.431.340.431.350.431.360.431.370.431.380.431.390.431.400.431.410.441.420.441.430.441.440.441.450.441.460.441.470.441.480.441.490.441.500.451.510.451.520.451.530.451.540.451.550.451.560.451.570.451.580.451.590.451.600.451.610.451.620.451.630.461.640.461.650.461.660.461.670.461.680.461.690.461.700.461.710.461.720.461.730.461.740.461.750.471.760.471.770.471.780.471.790.471.800.471.810.471.820.471.830.471.840.471.850.471.860.471.870.471.880.481.890.481.900.481.910.481.920.481.930.481.940.481.950.481.960.481.970.481.980.481.990.482.000.492.010.492.020.492.030.492.040.492.050.492.060.492.070.492.080.492.090.492.100.492.110.492.120.492.130.502.140.502.150.502.160.502.170.502.180.502.190.502.200.502.210.502.220.502.230.502.240.502.250.512.260.512.270.512.280.512.290.512.300.512.310.512.320.512.330.512.340.512.350.512.360.512.370.512.380.512.390.512.400.512.410.512.420.512.430.512.440.512.450.512.460.512.470.512.480.512.490.512.500.512.510.512.520.512.530.512.540.512.550.512.560.512.570.512.580.512.590.512.600.512.610.512.620.512.630.512.640.512.650.512.660.512.670.512.680.512.690.512.700.512.710.512.720.512.730.512.740.512.750.512.760.512.770.512.780.512.790.512.800.512.810.512.820.512.830.512.840.512.850.512.860.512.870.512.880.512.890.512.900.512.910.512.920.512.930.512.940.512.950.512.960.512.970.512.980.512.990.513.000.513.010.513.020.513.030.513.040.513.050.513.060.513.070.513.080.513.090.513.100.513.110.513.120.513.130.513.140.513.150.51

degreeValue of angleValue of angleDegreesincostantanDegreesincos100.170.980.180.18100.170.98110.190.980.190.19110.190.98120.210.980.210.21120.210.98130.230.970.230.23130.230.97140.240.970.250.25140.240.97150.260.970.270.27150.260.97160.280.960.290.29160.280.96170.290.960.310.31170.290.96180.310.950.320.32180.310.95190.330.950.340.34190.330.95200.340.940.360.36200.340.94210.360.930.380.38210.360.93220.370.930.400.40220.370.93230.390.920.420.42230.390.92240.410.920.450.45240.410.92250.420.910.470.47250.420.91300.500.870.580.58300.500.87350.570.820.700.70350.570.82400.640.770.840.84400.640.77450.710.711.001.00450.710.71500.770.641.191.19500.770.64550.820.571.431.43550.820.57600.870.501.731.73600.870.50650.910.422.142.14650.910.42