Retaing Wall With Surcharge

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DESIGN OF R.C.C. CANTILEVER RETAINING WALL DATA Height ofembankment above G.L. = 4 m = 16 = 160 Coeffcient offriction µ = 0.5 Angle of Reo!e "Angle of !hearing re!i!tance #$= %0 &eg '(rcharge )re!!(re*! = 40 +,(ivalent height of!(rcharge h! = -.5 m Gra&e ofConcrete = -0 Gra&e of'teel = 415 4 m +arth re!!(re coeifficient! h = 5.-5m Ca = 0.%%% C = % PRELIMINARY PROPORTIONS inim(m &ethoffo(n&ation = 1.1 m 1.-5m )rovi&e &eth offo(n&ation = 1.-5 m /verall &ethof*all h = 5.-5 m hickne!! of2a!e 'lab = 6-0 mm 6-0 )rovi&e hickne!! of2a!e 'lab = 6-0 mm mm 'tem hickne!! "A!!(me$ = 650 mm 'tem hickne!! ato = -00 mm Re,(ire& *i&th of2a!e 'lab * = %.3 5 m Re,(ire& )ro ection ' X ' = -.53 m )rovi&e Re,(ire& *i&th of2a!e 'lab * = %. m )rovi&e )ro ection ' X ' = -.6 m STABILITY AGAINST OVERTURNING 7orce &(e to Active )re!!(re )a = 160 k8 )a1 = 0 k8 )a- = 4 k8 /vert(rning oment o = %1-.4 k89m "er m length of:all$ 7orce "k8$ ;i!tance from Heel oment "k89m$ *1 = - %.3 1.- = %- *- = -%.15 -.50 = 53 *% = .4 -.-5 = -1 *4 = 60.45 1. 5 = 113 *i = %6 : = 5-5 ;i!tance of vertical force from Heel <: = : *i = 1.4%- m 'tabili!ing oment at oe r = *i " *9<:$ = 06 k89m "er m length of :all$ ;en!it> of !oil ? e k8 m % '2C of !oil , a k8 m - k8 m - 8 mm - 8 mm -

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retaining wall design

Transcript of Retaing Wall With Surcharge

DataDESIGN OF R.C.C. CANTILEVER RETAINING WALLDATAHeight of embankment above G.L.=4mDensity of soil e=16kN/m3SBC of soil qa=160kN/m2200 mmhs =2.50Coeffcient of friction =0.5mAngle of Repose (Angle of shearing resistance )=30degSurcharge Pressure Ws=40kN/m2Equivalent height of surcharge hs=2.5mGrade of Concrete=20N/mm2Grade of Steel=415N/mm24 mPa1h' =Earth pressure coeifficientsh = 5.25 m7.75Ca=0.333Pa2mCp=3h/2PRELIMINARY PROPORTIONSh/3Minimum depth of foundation=1.1m1.25 m650 mmProvide depth of foundation=1.25mOverall depth of Wall 'h'=5.25mThickness of Base Slab=620mm620Provide Thickness of Base Slab=620mm0mm1.3X = 2.61.25Stem Thickness ( Assume)=650mmmmCa x e x hStem Thickness at Top=200mmCa x e x hs = Ca x WsRequired Width of Base Slab 'W'=3.875m03.9 m620Required Projection' X '=2.58mProvide Required Width of Base Slab 'W'=3.9mProvide Projection' X '=2.6mSTABILITY AGAINST OVERTURNING(a) Forces on Wall (With preliminary proportions)Force due to Active Pressure Pa=160kNPa1=70kNPa2=74kNOverturning Moment Mo=312.4kN-m(per m length of wall)Force (kN)Distance from HeelMoment (kN-m)W1 =273.81.2=329W2 =23.152.50=58W3 =9.42.25=21W4 =60.451.95=118Wi =367Mw =525Distance of vertical force from Heel Xw = Mw/Wi=1.432mStabilising Moment at Toe Mr = Wi ( W-Xw)=906kN-m(per m length of wall)(FS) overturning = 0.9 Mr/Mo=2.6Overturning Check StatisfiedMo =312.375 kN-m/mSOIL PRESSURE AT FOOTING BASE1.432mResultant vertical reaction = R = Wi=367kNDistance from heel Lr = (Mw + Mo)/R=2.283mEccentricity e = Lr - W/2=0.350mOKW =367 kN/mPressure at BasePmax=145kN/m2< than SBC hence okPmin=43kN/m214543kN/m2kN/m2STABILITY AGAINST SLIDING1.95 mSliding Force Pa = Pa1 +Pa2=144kNR=WResisting Force F = R ( Ignoring passive pressure=184kNXr = 2.2875 mon toe side)(b) Pressure distribution diagram(FS)Sliding=0.9F/Pa cos =1.150Shear Key Has to be Designed0.3 m neglect1.3 mSize of Shear KeyHeight=300mmBreadth=400mmh10.95Distance from Toe=1600mmh2 =2.171.25mPp = Cp x e x (h22-h12)/2=92kNh1Therefore,h2(FS)Sliding=1.73Stabilty Check satisfiedPp30deg0.3m0.4 mToe1.6HeelProcecdure to hide.m3.9 mSub Hide()Dim Z As Long(c)Design of Shear KeyDim x As RangeZ = Range("A" & Rows.Count).End(xlUp).RowSet x = Range("A2:A" & Z)Application.ScreenUpdating = FalseFor Each Cell In Range("A18:A" & Z)Cell.EntireRow.Hidden = Cell.Value = TrueNext CellEnd Sub'Procecdure to unhide.Sub UnHide()Dim Z As LongDim x As RangeZ = Range("A" & Rows.Count).End(xlUp).RowSet x = Range("A2:A" & Z)Application.ScreenUpdating = FalseFor Each Cell In Range("A18:A" & Z)Cell.EntireRow.Hidden = FalseNext CellEnd Sub

W1W4W3W2surcharge

DesignDESIGN OF TOE SLABThe net pressures, acting upward, are obtained by reducing the uniformly distributedself weight of the toe slab from the gross pressures at the base.0.65Self weight loading=15.5kN/m21.3m1.95mmEffective depth 'd'=535mmTc, N/mm2Vu = 1.5 x (129+95)/2 x (1.3-0.535)=129.0kN/m15.5129.58pt=(100*0.5*(fck/fy))*(1-sqrt(1-(4.6*k/fck)))Grade of ConcreteMu = 1.5 x [(95x1.3x/2)+(129-95)x0.5x1.3x1.3x2/3]=149.6kN-m/mkN/m2kN/m2100 As/bd1520253035403440.150.280.280.290.290.290.3k = Mu/bd2=0.523N/mm218500.250.350.360.360.370.370.38pt=0.149%21078.50.50.460.480.490.50.50.51Ast=800mm2/m433121130.750.540.560.570.590.590.6Dia of Bar=113mm145kN/m241620110.60.620.640.660.670.68Spacing=141mmkN/m25203141.250.640.670.70.710.730.74Provided spacing=100mm6254911.50.680.720.740.760.780.79Check for shear111947328041.750.710.750.780.80.820.84Tv=0.24N/mm2kN/m2kN/m212161620.710.790.820.840.860.88Tc=0.329N/mm22.250.710.810.850.880.90.92Provide Nominal Shear Reinforcement8686Toe2.50.710.820.880.910.930.95kN/m2100As/bd0.21121495332.750.710.820.90.940.960.98DEVELOPMENT LENGTH129950.250.1530.710.820.920.960.991.01Ld = ( x s) / (4 x Tbd)1290.280.280.290.290.290.3kN/m235350.350.360.360.370.370.38 = Dia of bar=12mm95kN/m2Heelfor interpolation valuess = 0.87 x fy=361.05kN/m2100As/bd0.1878504673Tc1 =0.28Tbd = Design bond stress N/mm2=1.92Tc2 =0.36Therefore,Net soil pressures acting along base slabLd =564mmvertical stem100As/bd0.1703389831DESIGN OF HEEL SLABValues of Tbd Design bond stress N/mm2The distributed loading acting downward on the heel slab is given byGrade of concrete2504151) Overburden + Surcharge @16 x(7.75 - 0.62)=114.08kN/m2201.21.922) Heel slab @ 25 x 0.62=15.5kN/m2251.42.24Vu = 1.5 x (86+35)/2 x 1.95=177.0kN/m301.52.4Mu = 1.5 x [(35x1.95x1.95/2+(86-35)x0.5 x 1.95x1.95x2/3]164.5kN-m/m351.72.72401.93.04k = Mu/bd2=0.575N/mm2pt=0.165%01.92Ast=882mm2/m00Dia of Bar=201mm00Spacing=228mm00Provided Spacing=200mm00Check for shearTv=0.33N/mm201.92Tc=0.31N/mm2Provide Shear ReinforcementLd=752mmDESIGN OF VERTICAL STEMHeight of cantilever above base 'h'=4.63mEffective depth=590mmBending Moment Mu = 1.5[Ca x Ws x h2/2 + Ca x e x h3/6]=346.7kN-m/mk = Mu/bd2=1.00pt=0.294%Ast=1734mm2/mDia of Bar=201mmSpacing=116mmProvided Spacing=100mmCheck for shear at BaseCritical section is at d = 0.59 m above base i.e at Zs = 4.63 - 0.59m=4.04mbelow top edgeShear force at critical section = 1.5 x[ Ca x Ws x Zs + Ca x s x Zs2/2]=146.1kN/mTv=0.248N/mm2Tc=0.296N/mm2Provide Nominal Shear ReinforcementTEMPERATURE AND SHRINKAGE REINFORCEMENTProvide two-thirds of (Horizontal) bars near the front face remaining one-third near the rear face.For the lowermost one-third height of the stem above base,Ast = (0.0012 x 1000 x 650) x 2/3=520mm2/mBar Dia=8mmSpacing Required=97mmProvided Spacing=100mmProvide8 dia @100C/C near front face andIn the lowermost one-third height of the wall8 dia @200C/C near rear face8 dia @200C/C near front face andIn the midlle one-third height of the wall8 dia @400C/C near front face and8 dia @300C/C near front face andIn the top one-third height of the wall8 dia @600C/C near front face and

SummarySUMMARY450mm200 mmFck=20N/mm2Fy=415N/mm2Depth of foundation=1.25mY8 @ 300 mm C/CY8 @ 600 mm C/C1500Thickness of Base Slab=620mmmmStem Thickness Bottom=650mmY16 @ 300 mm C/CStem Thickness at Top=200mmFS against overturning=2.6OK4 mPmax=145OKPmin=43Y8 @ 200 mm C/CY8 @ 400 mm C/C1500FS against sliding=1.73OKmmY16 @ 200 mm C/CY8 @ 100 mm C/CY8 @ 200 mm C/CY10 @ 300 mm C/CY16 @ 100 mm C/C1.25 mY16 @ 200 mm C/C620mmY10Y12 @ 100 mm C/C300mm1600400mmmm3900mm

Notes1)LegendValues to be entered by userResults shown by program2)All advice or information from the program is intended for those who will evaluate the significanceand limitations of its contents and take responsibility for its use and applicationNo liability (including that for negligence shall be taken for its use and application.3)Figures are indicative only4)Reference:"Reinforced Concrete Design", by S Unnikrishnan Pillai Devdas Menon, Second Edition,2004, Tata McGraw-Hill Publications,5)DateVersionPublicationSize17.04.091.0Public Release114 KB6)For Comments and Suggestions Please [email protected]

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