DesignDESIGN OF SLAB CULVERT1.0DATAThickness of slabD=500.00mmCarriageway widthCw=7.50mmFootpath widthFp=1.00mmClear spanL=6.00mThickness of wearing courseWc=80.00mmWidth of end support/bearing=0.40mConcrete Grade=M25Steel Grade=Fe415Permissible stress bc=8.3N/sq.mmPermissible stress st=200N/sq.mmModular ratiom=10j=0.90q=1.10Live load dimensiona1=3.60mLive load dimensionb1=0.85mLive load dimensionb2=1.20mTotal live loadW1=700kNClear cover to reinforcement=30mmUnit weight of concreteYc=24N/cumUnit weight of wearing courseYwc=22N/cum2.0CALCULATING EFFECTIVE SPANOverall thickness of slabD=500.00mmEffective thickness of slabd=460.00mmEffective span is lesser than of i) Clear span + Effective depth=6.46mii) C/C of End Supports=6.40mSo, Effective Spanl=6.40m

3.0BENDING MOMENT BY PERMANENT LOADSWeight of slabW1=(D/1000)*Yc=12kN/sq.mmWeight of wearing courseW2=(Wc/1000)*Ywc=1.76kN/sq.mmTotal loadW=13.76kN/sq.mmSay=14.00kN/sq.mmBending moment for permanent loads=w*l2/8M1=71.68kN-m

4.0BENDING MOMENT BY VEHICLE LOAD / LIVE LOADFor 5m span impact factor=25%For 9m span impact factor=10%So, for 6.40 m span Impact Factor=19.75%Length of loada1=3.60mLength of load including 450 dispersal=a1+2*((D+Wc)/1000)ld=4.76mEffective width of slab perpendicular to spanbe=K * X * (1-(X/L)) + bwPlacing the load symmetrically on the spanDistance from centre of end X=l/2support to centre of load =3.20mWidth of slab B=Cw + (2 * Fp) =9.50mB/l=1.484mbw=b1+(2*(Wc/1000))=1.01mFrom Table of IRC 21:2000For B/l = 1.484, for simply supported slab, K =2.84So, Effective Width of Load be=K * X * (1-(X/L)) + bw=5.55mWidth of load with 450 dispersalWd=(2 * (be/2)) + (2 * (b1/2)) + b2=7.604mTotal Live Load including impactTLL=w1 * ((fact/100)+1)=838.25kNLive Load per Unit AreaLLUA=TLL/(ld * Wd)=23.16kN/sq.mmBending Moment for Live LoadM2=((LLUA * ld)/2) * (l/2) - ((LLUA * ld)/2) * (ld/4)=110.79kN-mDesign Bending MomentM=M1+M2=182.47kN-m5.0SHEAR FORCE BY LIVE LOADEffective spanl=6.40mLength of load incl. 450 dispersal ld=4.76mTo get maximum Shear Force at support let us place the load coinciding thestart point of the above lengths.X=ld/2=2.38mB/l=1.48From IRC 21:2000,K=2.84bw=1.01mEffective width of loadbe=K * X * (1-(X/l)) + bw=5.26mWidth of load with 450 dispersalwd=((2*be)/2) + ((2*b1)/2) + b2=7.31mLive Load per Unit AreaLLUA=TLL/(ld*wd)=24.11mShear force by live loadV1=[LLUA*ld*2*(b1+2*(Wc/1000)+2*(D/1000)]/l=72.07kNShear force by dead loadV2=W * l/2=44.80kNTotal Design ShearV=V1 + V2=116.87kN6.0STRUCTURAL DESIGN OF SLABRequired effective depthd=sqrt((M * 10^6)/(Q*b))=407.29mmEffective depth provideddeff=460mmRequired steel reinforcementAst=(M*10^6)/(st*j*d)=2488.9641919116sq.mmUsing 20mm dia bars, Spacing=((*d^2)/4)/Ast)*1000=126.22mmProvide 20 mm dia bars @ 120 mm c/c as main steel.Bending Moment for Distribution Steel=0.2*M1+0.3*M2=47.57kN-m=48.00kN-mEffective depth=444mmRequired steel=600.6006006006sq.mmUsing 12mm dia bars, Spacing=188.31mmProvide 12 mm dia bars @ 150 mm c/c as distribution steel.7.0SHEAR REINFORCEMENTSDesign ShearT=(V*10^3)/(b*deff)=0.25N/sq.mmK1=1.14 - 0.7 * (deff/1000)=0.818> 0.50o.k.Bending up alternate bars, provide 20 mm bars @ 240 mm c/c.Ast provided=1308.9969389958sq.mm%Ast=0.28%K2=0.5+0.25*%Ast=0.571< 1.0Not o.kPermissible shear stress=K1*K2*T_C0=0.33> 0.25 N/Sq.mm, o.k

Table1

Concrete GradeM15M20M25M30M35M40T_Co0.280.340.400.450.500.50