TOCClient : Kuwait Oil CompanyProject : Facility Upgrade and Relocation of Under Ground ProcessJob No :JI-180Doc No :JI-180-000-ECV-CAL-050Subject :Design of RCC Bund Wall ForRev No :0Wet Crude Tank 21-TK-001.Prep. By :DipakCheckd. By :GK4.5 Design of Concrete Bund Wall forWet Crude Tank 21-TK-001.SR NOTABLE OF CONTENTSPAGE NO1.0Design Data2.0Design Philosophy1psi7.5psi3.0Design Of Wall1.5754.0Design Of Base SlabAPPENDIX - ITypical details of Cantilever Wall

Wall DesignClient : Kuwait Oil CompanyProject : Facility Upgrade and Relocation of Under Ground ProcessJob No :JI-180Doc No :JI-180-000-ECV-CAL-050Subject :Design of RCC Bund Wall ForRev No :0Wet Crude Tank 21-TK-001.Prep. By :DipakCheckd. By :GK1.0 DESIGN DATA :1.1 Geometrical Data :Height of soil filling, ( H2 )=1.6mDepth of base, ( Wb )=0.5mB1Depth of retaining wall below finished ground level, ( H )=2.1mFFree Board ( F )=0mBcH1Total Height of retaining wall, ( H1 )=5.25mDepth of soil, ( D )=2.1mTop of raft from GL of tank ( D1)=1.6mHWidth of heel slab ( Bh )=2.7mH2Thickness of wall at bottom, ( B )=0.5mWidth of toe slab ( Bt )=0.6mThickness of wall at top, ( B1 )=0.3mD1Thickness of wall at H2 from raft top(Bc)=0.433mDBClear distance between face of shear key & face of heel slab ( Dsh )=2.7mBhBtHsWbWidth of shear key, ( Bs )=0.5mDshDstClear distance between face of shear key & face of toe slab ( Dst )=0.6mBsHeight of shear key, ( Hs )=0.2m1.2 Soil Data : ( As Per JI-180-000-ECV-SPE-001 )Unit weight of soil,( g )=18kN/m3Coefficient of soil pressure, ( Ko )=0.5( considered at rest condition as per geotechnical report)Net Allowable Bearing capacity of soil=125KN/m2Angle of internal friction, ( f )=32Coefficient of active earth pre. ( Ka )=( 1-SINf ) / ( 1+SINf ) =0.31Coefficient of passive earth pre. ( KP )=( 1+SINf ) / ( 1-SINf ) =3.25Factor of safety against sliding=1.75Factor of safety against overturning=1.751.3 Material Data: ( As Per JI-180-000-ECV-SPE-001 )Grade of Concrete ( Fck )=30N/mm2Yield Strength of reinforcement ( Fy )=414N/mm2Dia of Reinforcement in wall ( dwall )=20mmDia of Reinforcement in base ( dbase )=20mmClear Cover to Reinforcement ( c )=75mmUnit Weight of Concrete ( gc )=24KN/m3Density of retained liquid, ( gw )=8.77KN/m32.0 DESIGN PHILOSOPHY :Here, Bund wall has been desinged as a cantilever retaining wall for 1 m length and for that following criticalcases has been consideredCase 1 ) Empty on tank side & soil pressure and wind on other side of retaining wall.In this case cantilever wall has been analyzed for the active earth pressure from one side only,while checking for the stability, wt of earth from both the side has been considered.Case 2 ) Hydrostatic pressure due to stored liquid during spillages or tank burst conditions.In this case cantilever wall has been analyzed for the submerged liquid pressure from tank sideonly passive earth pressure on the other side of retaining wall to the possible extent of 2/3height of overburden soil. While checking for the stability, wt of earth from both the side & wt ofliquid form tank side has been considered.The stability and base pressure check for the retaining wall have been carried out to decide thesize and other details of the assumed retaining wall. The structural calculations are carried out later.3.0 DESIGN OF WALL :3.1 BM Calculation For Case 1 :a )Active earth pressure ( Pa )=g*Ka*H2Qw=8.870KN/m2Shear at base of stem due to active=Pa * H2 / 2earth pressure ( Va )=7.096kN/mMoment at base of stem due active=Pa * H2 / 2 * H2 / 3earth pressure ( Ma )=3.785kNm/mb )PaWind pressure ( Qw )=0.76KN/m2( Refer Civil & Structural designphilosophy JI-180-000-ECV-SPE-001)Shear at base of stem due to wind=Qw * (H1-H)pressure ( Vw )2.394kN/mMoment at base of stem due to Wind=Qw * (H1-H) * ((H1-H)/2+H2)pressure ( Mw )=7.601kNm/mc )Passive earth pressure, ( Pp )=g*Kp*D1( Passive pressure for moment=0.000KN/m2calculation has been ignored soShear at base of stem due to passive=Pp * D1 / 2as to be on conservative side )earth pressure ( Vp )=0.000kN/mMoment at base of stem due to=Pp * D1 / 2 * D1 / 3passive earth pressure ( Mp )=0.000kNm/md )Additional Shear ( Vadd )=0.00kN/m( Due to Walkway at top )Additional Moment ( Madd )=0.53kNm/m( Due to Walkway at top )Total shear at base ( V1 )=Va + Vw + Vadd - Vp=9.49kN/mTotal moment at base ( M1 )=Ma + Mw + Madd - Mp=11.92kNm/m3.2 BM Calculation For Case 2 :a )Contained Liquid pressure ( Pw )=gw*(H1-Wb-F)=41.658KN/m2Shear at base of stem due to liquid=Pw * (H1-Wb-F)/2pressure ( Vl )=98.937kN/mMoment at base of stem due to liquid=Pw * (H1-Wb-F)2/6Qwpressure ( Ml )=156.650kNm/mb )Active earth pressure ( Pa )=Ka * (g-gw) * D1=4.549KN/m2Shear at base of stem due to active=Ka*(g-gw)*D12/2earth pressure ( Va )=3.639kN/mPaPwPpMoment at base of stem due to active=Ka*(g-gw)*D13/6earth pressure ( Ma )=1.941kNm/mc )Passive earth pressure ( Pp )=g*Kp*H2( Passive pressure for moment=0.000KN/m2calculation has been ignored soShear at base of stem due to passive=Pp * H2 / 2as to be on conservative side )earth pressure ( Vp )=0.000kN/mMoment at base of stem due to=Pp * H2 / 2 * H2 / 3passive earth pressure ( Mp )=0.000kNm/md )Wind pressure ( Qw )=0.76KN/m2Shear at base of stem due to wind=Qw * Fpressure ( Vw )=0kN/mMoment at base of stem due to wind=Qw * F * (F/2 + (H1-F-Wb))pressure ( Mw )=0kNm/me )Additional Shear ( Vadd )=0.00kN/m( Due to Walkway at top )Additional Moment ( Madd )=0.53kNm/m( Due to Walkway at top )Total shear at base ( V2 )=Vl + Va + Vw + Vadd - Vp=102.58kN/mTotal moment at base ( M2 )=Ml + Ma + Mw + Madd - Mp=159.12kNm/m3.3 Rebar Calculation :Design factored bending=1.4 * ( Maximum of M1 & M2 )moment ( Mu )=222.77kNm( Load factor for soil pressure is1.4, as per BS:8110-Part I )Effective depth ( d )=( B * 1000 ) - (dwall/2) - c=415mmNow as per clause 3.4.4.4 of BS 8110 ( Part - I )k=Mu / fcu bd2=222.77x 10630x 1000x 415x 415=0.043 0.25%, Steel is sufficient for crack control3.4 Distribution Steel :Provide Min Reinforcement As Dist. Steel, As per table 3.25 of BS : 8110 ( Part - I )Distribution steel, Adist=0.13 % of Area of concrete=0.13x 1000x 500/ 100=650mm2/mDia of distribution rebar=16mmProvide 16 mm bar at a spacing of250mm c/cArea of steel provided Astprov1=804mm23.5 Check For Shear :Maximum design shear at face ( Vf )=Maximum Of V1 & V2=102.58kNShear V at face of the support ( Vu1 )=1.4 * Vf( Load factor for soil pressure is=143.61kN1.4, as per BS:8110-Part I )Shear stress at face of support, ( v1 )=Vu1/bd=0.35N/mmMaximum allowed shear stress ( vmax )=4.38N/mm2or 5 N/mm2Max( 0.8fcu , 5 )v1 < vmax, O.K. (Clause 3.7.7.2)100Ast / bd=0.50%From BS 8110, Part 1 Table 3.8, ( vc )=0.43N/mmFor fck =25 N/mm2Revise value of ( vc )=vc*( fck /25 )1/3For fck =30 N/mm2=0.46N/mmSince vc > v1, Hence section is SAFE in shearShear V at d distance from the=1.4 * Vf *( H1-Wb-d)/(H1-Wb)support ( Vu2 )=131.06kN ( Load factor for soil pressure is 1.4)Shear V at 1.5d distance from the=1.4 * Vf *( H1-Wb-1.5d)/(H1-Wb)support ( Vu3 )=124.79kN ( Load factor for soil pressure is 1.4)Actual shear stress at d dist.=Vu2/bdfrom support ( v2 )=0.32N/mmActual shear stress at 1.5d dist.=Vu3/bdfrom support ( v3 )=0.30N/mm

Tank Farm area

Check For Crack WidthClient : Kuwait Oil CompanyProject : Facility Upgrade and Relocation of Under Ground ProcessJob No :JI-180Doc No :JI-180-000-ECV-CAL-050Subject :Design of RCC Bund Wall ForRev No :0Wet Crude Tank 21-TK-001.Prep. By :DipakCheckd. By :GK3.6 Calculation Of Crack WidthMaximum allowable crack width=0.3mm (Per BS 8110-2 : 1985 clause 3.2.4)1 ) Crack width for drying shrinkage / thermal movement :fcu=Characteristic strength of reinforced concrete=30N/mm2fy=Characteristic strength of reinforcing steel as per table 3.1 of BS 8110=414N/mm2( As per design philosophy 0.9fy )Thermal strain er=0.8*Dt*a*R( Refer equation 14 of clause 3.8.4.2 of BS 8110-2)R=0.6( Per Table 3.3 of BS 8110-2)a=Coefficient of thermal expansion of mature concrete=0.000012Table 7.3 of BS 8110 ( Part 2 )DT=Fall in temperatue between hydration peak and ambient=20( per Table 3.2 of BS 8110-2)Thermal strain er=0.0001152Design surface crack width, W1=3*acr*er/(1+2*((acr-cmin)/(h-x)))Where,acr=Dist from point considered to the surface of the nearest long bar=Sqrt( S/22+ (c+f/2)2 ) - (f/2)=113.31mmf=Size of each reinforcing bar=20D=Depth of wall = B=500S=Spacing of reinforcement=150As=Area of steel=2093.3333333333W1=0.034mm