Cantilever Retaining Wall Analysis
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The details of a cantilever retaining wall are shown below, is the design of t Calculation Reference Craig Soil Mechanics. R.F. Craig Calculation Validation Check against a worked example in the reference above. Calculation Unit Surcharge Pressure Q = 10000 N/m² Thickness of the Stem 0.3 m Thickness of the Base 0.4 m Length of Backfill 1.75 m Height of Wall from the Base 5.4 m Width of the Base 3m Characteristics of Backfill Drained (effective stress) Shear Strength Parameter c' 0 N/m² Unit Weight of the backfill γ = 17000 N/m³ Drained (effective stress) Shear Strength Parameter 36 ° Characteristics of Concrete Unit Weight of Concrete 23500 N/m³ Angle of friction between the base and the foundation soil is δ = 27 ° Purpose of calculation : t1 = t2 = w1 = h1 = b1 = ϕ' = γ concrete = Q t₁ h₁ w₁ t₂ 1 2 b₁
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Transcript of Cantilever Retaining Wall Analysis
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RetWallPurpose of calculation:Table of compatible unitsThe details of a cantilever retaining wall are shown below, is the design of the wall satisfactory?LengthAreaVolumeTimeAngleMassVelocityAccelerationForceUnit WeightMomentLDForceWorkEnergyPowerDensityStressFundamentalmmmskgm/sm/skg.m/sN/mN.mkg/skg.m/s.mkg.m/s.mkg.m/skg/mkg/m.sCalculation ReferenceSImmmskgm/sm/sNN/mN.mN/mJJWkg/mN/mCraig Soil Mechanics. R.F. CraigSI (mm)mmmmmmsgmm/smm/sNN/mmN.mmN/mmmJmJmWg/mmMPaUS Unit (ft)ftftftsslugft/sft/slbflbf/ftlbf.ftlbf/ftft.lbft.lbft.lb/sslug/ftlbf/ftCalculation ValidationUS Unit (in)inininslbf.s/inin/sin/slbflbf/inlbf.inlbf/inin.lbfin.lbfin.lbf/slbf.s/inpsiCheck against a worked example in the reference above.Calculation Units2Surcharge PressureQ =10000N/mThickness of the Stemt1 =0.3mThickness of the Baset2 =0.4mLength of Backfillw1 =1.75mHeight of Wall from the Baseh1 =5.4mWidth of the Baseb1 =3mCharacteristics of BackfillDrained (effective stress) Shear Strength Parameterc'=0N/mUnit Weight of the backfill =17000N/mDrained (effective stress) Shear Strength Parameter' =36Characteristics of ConcreteUnit Weight of Concrete concrete =23500N/mAngle of friction between the base and the foundation soil is =27Active earth pressure coefficientKa =0.25961618370Horizontal ForcesFH1 =14019.273918855N0FH2 =64348.4672875444N0Sum of Horizontal ForcesFH =78367.7412063994N0Vertical ForcesFstem =35250N0Fbase =28200N0Fsoil =148750N0Sum of Vertical ForcesFV =212200N0LeverLH1 =2.7m0LH2 =1.8m0Lstem =1.1m0Lbase =1.5m0Lsoil =2.125m0MomentsMH1 =37852.0395809085N.m0MH2 =115827.2411175799N.m0MH =153679.2806984884N.m0Mstem =38775N.m0Mbase =42300N.m0Msoil =316093.75N.m0MV =397168.75N.m0M =243489.4693015116N.m0Lever arm of base resultantL =1.14745273m0Note, the resultant acts within the middle third of the baseFoS against overturningFoS =2.58440011040Eccentricity of base reactione =0.35254727m0Maximum and minimum base pressurespmax =120607.0204656589N/m0pmin =20859.6462010078N/m0FoS against slidingFoS =1.37966590230The design of the wall is satisfactory
Qthbwt12b=1 - sin[{p'} / 180]1 + sin[{p'} / 180][Eqn. 1]= KaQh1 [Eqn. 2]= 0.5Kah12 [Eqn. 3]= FH1 + FH2 [Eqn. 4]= (h1 - t2)t1 concrete [Eqn. 5]= t2b1 concrete [Eqn. 6]= (h1 - t2)w1 [Eqn. 7]= Fsoil + Fbase + Fstem [Eqn. 8]=h12[Eqn. 9]=h13[Eqn. 10]= b1 - w1 -t12[Eqn. 11]=b12[Eqn. 12]=w12+ (b1 - w1) [Eqn. 13]= FH1LH1 [Eqn. 14]= FstemLstem [Eqn. 17]= FbaseLbase [Eqn. 18]= FsoilLsoil [Eqn. 19]= Mstem + Mbase + Msoil [Eqn. 20]= MV - MH [Eqn. 21]=MFV[Eqn. 22]=MVMH[Eqn. 23]=b12- L [Eqn. 24]=FVb11 +6eb1[Eqn. 25]=FVb11 -6eb1[Eqn. 26]=FVtan[{p} / 180]FH[Eqn. 27]
