double box
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SummarySUMMARY OF STRUCTURAL CALCULATION OF 2-BARREL BOX CULVERT(Batang Angkola Main Canal from BBA0 - BBAII)1Design Dimensions and Bar ArrangementsClass III Road (BM50)Type of box culvert2-B1.8xH2.0Clear width of 1-barrelm1.75Clear heightm2.00Height of filletm0.15ThicknessSide wallcm25.0Partition wallcm25.0Top slabcm25.0Bottom slabcm25.0Cover of reinforcement bar(between concrete surface and center of reinforcement bar)Side wallOutsidecm5.0Insidecm5.0Partition wallBoth sidescm5.0Top slabUppercm5.0Lowercm5.0Bottom slabLowercm5.0Uppercm5.0Bar arrangement(dia - spacing per unit length of 1.0 m)Side wallLower outsideTensile barmm12@250Distribution barmm12@250Middle insideTensile barmm12@125Distribution barmm12@250Upper outsideTensile barmm12@250Distribution barmm12@250PartitionBoth sidesTensile barmm12@250wallDistribution barmm12@250Top slabUpper edgeTensile barmm12@250Distribution barmm12@250Upper centerTensile barmm12@125Distribution barmm12@250Lower middleTensile barmm12@250Distribution barmm12@250Bottom slabLower edgeTensile barmm12@250Distribution barmm12@250Lower centerTensile barmm12@125Distribution barmm12@250Upper middleTensile barmm12@250Distribution barmm12@250FilletUpper edgeFillet barmm12@250Upper centerFillet barmm12@250Lower edgeFillet barmm12@250Lower centerFillet barmm12@2502Design ParametersUnit WeightReinforced Concretegc=2.4tf/m3Backfill soil(wet)gs=1.8tf/m3(submerged)gs'=1.0tf/m3Live LoadClass of roadClass III(BM50)P (t)Truck load at rear wheelP=5.0tfClass IBM10010.0Impact coefficient(for Class I to IV road)Ci=0.3(D4.0m)Class IIIBM505.0Pedestrian load(for Class V roads)0tf/m2Class IVBM353.5Class V0.0ConcreteDesign Strength scksck=175kgf/cm2(K175)Allowable Compressive Stress scasca=60kgf/cm2Allowable Shearing Stress tmtm=5.5kgf/cm2Reinforcement BarAllowable Tensile Stress ssassa=1,400kgf/cm2(U24, deformed bar)Yielding Point of Reinforcement Barssy=3,000kgf/cm2Young's Modulus Ration=24Coefficient of static earth pressureKa=0.33
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LoadSTRUCTURAL CALCULATION OF BOX CULVERTType: B1.75m x H2.00m x 2Class III RoadSoil Cover Depth:0.6 m1Dimensions and ParametersBasic ParametersBasic ConditionsKa:Coefficient of static earth pressure0.3333Classification of Live load by truckClass3ClassP (t)gw:Unit weight of water (t/m3)1.00t/m3PTM:Truck load of Middle Tire5.00t1Class IBM10010.0gd:Unit weight of soil (dry) (t/m3)1.80t/m3Ii:Impact coefficient (D4.0m:0, D FaFs=1.0792< 1.2checkwhere,Vd:Total dead weight (t/m)Vd=11.466tf/mU:Total uplift (t.m)U=BT*HT*gwU=10.625tf/mWs:Weight of covering soilWs =BT*{(D-Gwd)*(gs-gw)+Gwd*gd}=2.550tf/mWc:Self weight of box culvertWc =(HT*BT-2*H*B+4*Hf^2)*gc=8.916tf/mFa:Safety factor against upliftFa=1.23Load calculationCase 1: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L11)vertical load against top slabActing Load(tf/m2)Wtop= (t2*BT+2*Hf^2)*gc/BT0Wtop=0.6645Pvd=Gwd*gd+(D-Gwd)*gsPvd=1.2000Pvt1Pvt1=1.7012Pvt2Pvt2=0.0000Pv1=3.56572)horizontral load at top of side wallActing Load(tf/m2)Horizontal pressure by track tireP1=Ka*we1P1=0.0000we1=0.0000tf/m2P2=Ka*we2P2=0.0000we2=0.0000tf/m2P3=Ka*gd*GwdP3=0.0000P4=Ka*gs*(D1-Gwd)P4=0.4833P5=gw*(D1-Gwd)P5=0.7250Ph1=1.20833)horizontral load at bottom of side wallActing Load(tf/m2)P1=Ka*we1P1=0.0000P2=Ka*we2P2=0.0000P3=Ka*gd*GwdP3=0.0000P4=Ka*gs*(D1+H0-Gwd)P4=1.9831P5=gw*(D1+H0-Gwd)P5=2.9750Ph2=4.95814)self weight of side wall and partition wallActing Load(tf/m)Wsw=t1*H*gcWsw=1.2000Wpw=t4*H*gcWpw=1.20005)ground reactionActing Load(tf/m2)Wbot=(t3*BT+2*Hf^2)*gc/BT0Wbot=0.6645WtopWtop=0.6645Ws=(Wsw*2+Wpw)/BT0Ws=0.9000PvdPvd=1.2000Pvt1Pvt1=1.7012Pvt2Pvt2=0.0000Wiw=2*(hiw*B-2Hf^2)*gw/BT0Wiw=0.0000hiw: internal water depth0.00mUp=-U/BT0U=-2.6563Q=2.4739summary of resistance momentItemVHxyM(tf/m)(tf/m)(m)(m)(tf.m/m)acting point of resultant forceSelf weighttop slab2.6580-2.0000-5.3160X = SM/SV =2.000mside wall (left)1.2000-0.0000-0.0000e = BT0/2 - X =0.000mside wall (right)1.2000-4.0000-4.8000partition wall1.2000-2.0000-2.4000invert2.6580-2.0000-5.3160ground reactionload on top slabPvd4.8000-2.0000-9.6000q1 = SV/BT0 + 6SVe/BT0^2 =2.4739tf/m2Pvt16.8047-2.0000-13.6094q2 = SV/BT0 - 6SVe/BT0^2 =2.4739tf/m2Pvt20.0000-2.0000-0.0000soil pressureside wall (left)-6.9372-0.89706.2224side wall (right)--6.9372-0.8970-6.2224internal water0.0000-2.0000-0.0000uplift-10.6250-2.0000--21.2500total9.895719.79146)load against invertActing Load(tf/m2)Pvd1.2000Pvt11.7012Pvt20.0000Wtop0.6645Ws0.9000Pq=4.4657Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L21)vertical load against top slabActing Load(tf/m2)Wtop= (t2*BT+2*Hf^2)*gc/BT0Wtop=0.6645Pvd=Gwd*gd+(D-Gwd)*gsPvd=1.2000Pvt1Pvt1=3.9063Pvt2Pvt2=0.0000Pv1=5.77082)horizontral load at top of side wallActing Load(tf/m2)Horizontal pressure by track tireP1=Ka*we1P1=1.0416we1=3.1250tf/m2P2=Ka*we2P2=0.2604we2=0.7813tf/m2P3=Ka*gd*GwdP3=0.0000P4=Ka*gs*(D1-Gwd)P4=0.4833P5=gw*(D1-Gwd)P5=0.7250Ph1=2.51023)horizontral load at bottom of side wallActing Load(tf/m2)P1=Ka*we1P1=1.0416P2=Ka*we2P2=0.2604P3=Ka*gd*GwdP3=0.0000P4=Ka*gs*(D1+H0-Gwd)P4=1.9831P5=gw*(D1+H0-Gwd)P5=2.9750Ph2=6.26014)self weight of side wall and partition wallActing Load(tf/m)Wsw=t1*H*gcWsw=1.2000Wpw=t4*H*gcWpw=1.20005)ground reactionActing Load(tf/m2)Wbot=(t3*BT+2*Hf^2)*gc/BT0Wbot=0.6645WtopWtop=0.6645Ws=(Wsw*2+Wpw)/BT0Ws=0.9000PvdPvd=1.2000Pvt1Pvt1=3.9063Pvt2Pvt2=0.0000Wiw=2*(hiw*B-2Hf^2)*gw/BT0Wiw=0.0000hiw: internal water depth0.00mUp=-U/B0U=-2.6563Q=4.6790summary of resistance momentItemVHxyM(tf/m)(tf/m)(m)(m)(tf.m/m)acting point of resultant forceSelf weighttop slab2.6580-2.0000-5.3160X = SM/SV =2.0000mside wall (left)1.2000-0.0000-0.0000e = BT0/2 - X =0.0000mside wall (right)1.2000-4.0000-4.8000partition wall1.2000-2.0000-2.4000invert2.6580-2.0000-5.3160ground reactionload on top slabPvd4.8000-2.0000-9.6000q1 = SV/BT0 + 6SVe/BT0^2 =4.6790tf/m2Pvt115.6250-2.0000-31.2500q2 = SV/BT0 - 6SVe/BT0^2 =4.6790tf/m2Pvt20.0000-2.0000-0.0000soil pressureside wall (left)-9.8666-0.96479.5180side wall (right)--9.8666-0.9647-9.5180internal water0.0000-2.0000-0.0000uplift-10.6250-2.0000--21.2500total18.716037.43206)load against invertActing Load(tf/m2)Pvd1.2000Pvt13.9063Pvt20.0000Wtop0.6645Ws0.9000Pq=6.6708Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L11)vertical load against top slabActing Load(tf/m2)Wtop= (t2*BT+2*Hf^2)*gc/BT0Wtop=0.6645Pvd=D*gdPvd=1.0800Pvt1Pvt1=1.7012Pvt2Pvt2=0.0000Pv1=3.44572)horizontral load at top of side wallActing Load(tf/m2)Horizontal pressure by track tireP1=Ka*we1P1=0.0000we1=0.0000tf/m2P2=Ka*we2P2=0.0000we2=0.0000tf/m2P3=Ka*gd*D1P3=0.4350WP=-gw*0P4=0.0000Ph1=0.43503)horizontral load at bottom of side wallActing Load(tf/m2)P1=Ka*we1P1=0.0000P2=Ka*we2P2=0.0000P3=Ka*gd*(D1+H0)P3=1.7848WP=-gw*HP4=-2.0000Ph2=-0.21524)self weight of side wallActing Load(tf/m)Wsw=t1*H*gcWsw=1.2000Wpw=t4*H*gcWpw=1.20005)ground reactionActing Load(tf/m2)Wbot=(t3*BT+2*Hf^2)*gc/BT0Wbot=0.6645WtopWtop=0.6645Ws=(Wsw*2+Wpw)/BT0Ws=0.9000PvdPvd=1.0800Pvt1Pvt1=1.7012Pvt2Pvt2=0.0000Wiw=2*(hiw*B-2Hf^2)*gw/BT0Wiw=1.5200hiw: internal water depth1.75mUp=0U=0.0000Q=6.5302summary of resistance momentItemVHxyM(tf/m)(tf/m)(m)(m)(tf.m/m)acting point of resultant forceSelf weighttop slab2.6580-2.0000-5.3160X = SM/SV =2.0000mside wall (left)1.2000-0.0000-0.0000e = BT0/2 - X =0.0000mside wall (right)1.2000-4.0000-4.8000partition wall1.2000-2.0000-2.4000invert2.6580-2.0000-5.3160ground reactionload on top slabPvd4.3200-2.0000-8.6400q1 = SV/BT0 + 6SVe/BT0^2 =6.5302tf/m2Pvt16.8047-2.0000-13.6094q2 = SV/BT0 - 6SVe/BT0^2 =6.5302tf/m2Pvt20.0000-2.0000-0.0000soil pressureside wall (left)-0.2473-2.23430.5524side wall (right)--0.2473-2.2343-0.5524internal water6.0800-2.0000-12.1600uplift0.0000-2.0000-0.0000total26.120752.24146)load against invertActing Load(tf/m2)Pvd1.0800Pvt11.7012Pvt20.0000Wtop0.6645Ws0.9000Pq=4.3457Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L21)vertical load against top slabActing Load(tf/m2)Wtop= (t2*BT+2*Hf^2)*gc/BT0Wtop=0.6645Pvd=D*gdPvd=1.0800Pvt1Pvt1=3.9063Pvt2Pvt2=0.0000Pv1=5.65082)horizontral load at top of side wallActing Load(tf/m2)Horizontal pressure by track tireP1=Ka*we1P1=1.0416we1=3.1250tf/m2P2=Ka*we2P2=0.2604we2=0.7813tf/m2P3=Ka*gd*D1P3=0.4350WP=-gw*0P4=0.0000Ph1=1.73693)horizontral load at bottom of side wallActing Load(tf/m2)P1=Ka*we1P1=1.0416P2=Ka*we2P2=0.2604P3=Ka*gd*(D1+H0)P3=1.7848WP=-gw*HP4=-2.0000Ph2=1.08684)self weight of side wallActing Load(tf/m)Wsw=t1*H*gcWsw=1.2000Wpw=t4*H*gcWpw=1.20005)ground reactionActing Load(tf/m2)Wbot=(t3*BT+2*Hf^2)*gc/BT0Wbot=0.6645WtopWtop=0.6645Ws=(Wsw*2+Wpw)/BT0Ws=0.9000PvdPvd=1.0800Pvt1Pvt1=3.9063Pvt2Pvt2=0.0000Wiw=(hiw*B-2Hf^2)*gw/BT0Wiw=1.5200hiw: internal water depth1.75mUp=0U=0.0000Q=8.7353summary of resistance momentItemVHxyM(tf/m)(tf/m)(m)(m)(tf.m/m)acting point of resultant forceSelf weighttop slab2.6580-2.0000-5.3160X = SM/SV =2.0000mside wall (left)1.2000-0.0000-0.0000e = BT0/2 - X =0.0000mside wall (right)1.2000-4.0000-4.8000partition wall1.2000-2.0000-2.4000invert2.6580-2.0000-5.3160ground reactionload on top slabPvd4.3200-2.0000-8.6400q1 = SV/BT0 + 6SVe/BT0^2 =8.7353tf/m2Pvt115.6250-2.0000-31.2500q2 = SV/BT0 - 6SVe/BT0^2 =8.7353tf/m2Pvt20.0000-2.0000-0.0000soil pressureside wall (left)-3.1766-1.21133.8480side wall (right)--3.1766-1.2113-3.8480internal water6.0800-2.0000-12.1600uplift0.0000-2.0000-0.0000total34.941069.88206)load against invertActing Load(tf/m2)Pvd1.0800Pvt13.9063Pvt20.0000Wtop0.6645Ws0.9000Pq=6.5508Summary of Load CalculationItemPv1Ph1Ph2PqWswWpwq1Case(tf/m2)(tf/m2)(tf/m2)(tf/m2)(tf/m)(tf/m)(tf/m2)Case.13.56571.20834.95814.46571.20001.20002.4739Case.25.77082.51026.26016.67081.20001.20004.6790Case.33.44570.4350-0.21524.34571.20001.20006.5302Case.45.65081.73691.08686.55081.20001.20008.7353
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MSN4Analysis of Plane FrameCase 1: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L11) Calculation of Load TermPh1Horizontal Pressure at top of side wall1.208tf/m2Ph2Horizontal Pressure at bottom of side wall4.958tf/m2Ph3Horizontal Pressure at top of side wall1.208tf/m2Ph4Horizontal Pressure at bottom of side wall4.958tf/m2Pv1Vertical Pressure(1) on top slab3.566tf/m2Q1Reaction to bottom slab4.466tf/m2H0Height of plane frame2.250mB0Width of plane frame2.000mt1Thickness of side wall0.250mt2Thickness of top slab0.250mt3Thickness of bottom slab0.250mt4Thickness of partition wall0.250mcUnit weight of reinforcement concrete2.400tf/m3L1Height of side wall2.000mCAB = CFE = (2Ph3+3Ph4)H02/60=1.45893tfmCBA = CEF = (3Ph3+2Ph4)H02/60=1.14253tfmCBC = CCB = CCE = CEC = Pv1B02/12=1.18856tfmCDA = CAD = CFD = CDF = Q1B02/12=1.48856tfmPv1=3.57 tf/m2t2=0.25mt2=0.25mPh1=1.21 tf/m2Ph3=1.21 tf/m2H0=2.25mt1=0.25mt4=0.25mt1=0.25mt3=0.25mt3=0.25mPh2=4.96 tf/m2Ph4=4.96 tf/m2Q1=4.47 tf/m2B0 =2.00mB0 =2.00m2) Calculation of Bending Moment at jointk1=1.0k2 = H0t23/(B0t13)=1.12500k3 = H0t33/(B0t13)=1.12500k4 = H0t43/(H0t13)=1.00000As load has bilateral symmetry, the equation shown below is formed.qA = -qFqB = -qEqC = qD = 0R =02(k1+k3)k1-3k3qACAB - CADPreparatory caculation for qA,qB,Rk12(k1+k2)-3k2qB=CBC - CBAk3k2-2(k2+k3)RC4.250001.00000-3.37500D1.000004.25000-3.37500=-52.10156251.125001.12500-4.50000RBC = B0Pv1/2=3.5657tf-0.029631.00000-3.37500P = t1L1c=1.2000tf/mD10.046024.25000-3.37500=2.68030313960.200001.12500-4.50000C = [CBC - CCB + CDA - CAD + (RBC - B0Q1/2 + P)B0]/3=0.2000tfm4.25000-0.02963-3.375004.250001.00000-3.37500qA-0.02963D21.000000.04602-3.37500=1.46744055181.000004.25000-3.37500qB=0.046021.125000.20000-4.500001.125001.12500-4.50000R0.200004.250001.00000-0.02963By solving above equation, the result is led as shown below;D31.000004.250000.04602=3.35256092291.125001.125000.20000qA=-0.05144R=-0.064347qB=-0.02817qA=D1/DqB=D2/DR=D3/DMAB = -MFE = k1(2qA +qB) - CAB=-1.5900tfmMBA = -MEF = k1(2qB+qA)+CBA=1.0348tfmMBC = -MEC = k2(2qB+qC - 3R) - CBC=-1.0348tfmMCB = -MCE = k2(2qC+qB - 3R)+CCB=1.3740tfmMDA = -MDF = k3(2qD+qA - 3R) - CDA=-1.3293tfmMAD = -MFD = k3 (2qA+ qD - 3R)+CAD=1.5900tfmSMA = MAB + MAD=0.0000tfmo.k.SMB = MBA + MBC=0.0000tfmo.k.MCD = k4(2qC+qD)=0.0000tfmMDC = k4(2qD+qC)=0.0000tfm2) Calculation of Design Force2-1) Side Walla)Shearing Force at jointMBA=1.03 tf.mw1Load at end A4.958tf/m2w2=1.21 tf/m2w2Load at end B1.208tf/m2MABBending moment at end A-1.5900tfmMBABending moment at end B1.0348tfmL=2.25 mLLength of member2.250mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mw1=4.96 tf/m2SAB = (2w1+w2)L/6 - (MAB+MBA)/L=4.4185tfMAB=-1.59 tf.mSBA = SAB - L(w1+w2)/2=-2.5187tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SAB - w1x - (w2 - w1)x2/(2L)(i) In case of x1 =0.4mSx1=2.5685tf(ii) In case of x2 =1.85mSx2=-1.9021tfc)Bending MomentMA = MAB=-1.5900tfmMB = -MBA=-1.0348tfmThe maximum bending moment occurs at the point of that shearing force equal to zero.Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)=4.4184725877-4.9581x+0.8333x^2,x=4.85871.0913Check of SxBending moment at x =1.0913m is;Sx =SAB - w1x - (w2 - w1)x2/(2L)Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB=0.6405tfm=0.000tf2-2) Top Slaba)Shearing Force at jointw1Uniform load on top slab3.566tf/m2w1=3.57 tf/m2MBCBending moment at end B-1.0348tfmMCBBending moment at end C1.3740tfmLLength of member2.000mchProtective covering height0.050mMBC =tThickness of member (height)0.250m-1.03 tf.mMCB =dEffective height of member0.200m1.37 tf.mSBC = w1L/2-(MBC+MCB)/L=3.3960tfL=2.00 mSCB = SBC -w1L=-3.7353tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SBC- w1x(i) In case of x1 =0.400mSx1=1.9698tf(ii) In case of x2 =1.6mSx2=-2.3090tfc)Bending MomentMB = MBC=-1.0348tfmMC = -MCB=-1.3740tfmSx = SBC - w1x=3.39603-3.566x=0,x=0.9524Check of SxBending moment at x =0.9524m is;Sx =SBC- w1xMmax = SBCx - w1x2/2 + MBC=0.58247tfm=0.000tf2-3) Invert (Bottom Slab)a)Shearing Force at jointw1Reaction to bottom slab4.466tf/m2L=2.00 mMDABending moment at end D-1.3293tfmMADBending moment at end A1.5900tfmLLength of member2.000mchProtective covering height0.050mtThickness of member (height)0.250mMAD =MDA =dEffective height of member0.200m1.59 tf.m-1.33 tf.mw1=4.47 tf/m2SDA = w1L/2 - (MDA+MAD)/L=4.3353tfSAD = SDA - Lw1=-4.5960tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SDA- w1x(i) In case of x1 =0.400mSx1=2.5490tf(ii) In case of x2 =1.600mSx2=-2.80976tfc)Bending MomentMD = MDA=-1.3293tfmMA = -MAD=-1.5900tfmSx = SDA - w1x=4.33531-4.466x=0,x=0.9708Check of SxBending moment at x =0.9708mis;Sx =SDA- w1xMmax = SDAx - w1x2/2 + MDA=0.7751tfm=0.000tf2-4) Partition Walla)Shearing Force at jointw1Load at end C0.000tf/m2w2Load at end D0.000tf/m2MCDBending moment at end C0.000tfmMDCBending moment at end D0.000tfmLLength of member2.250mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mSCD = w1L/2 - (MCD+MDC)/L=0.000tfSDC = SCD - Lw1=0.000tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SCD- w1x(i) In case of x1 =0.400mSx1=0tf(ii) In case of x2 =1.850mSx2=0tfc)Bending MomentMC = MCD=0.0000tfmMD = -MDC=0.0000tfmSx = SCD - w1x=0.000-0.000x=0,x=-Check of SxBending moment at x =1.1250mis;Sx =SCD- w1xMmax = SCDx - w1x2/2 + MCD=0.000tfm=0.000tfCase 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L21) Calculation of Load TermPh1Horizontal Pressure at top of side wall2.510tf/m2Ph2Horizontal Pressure at bottom of side wall6.260tf/m2Ph3Horizontal Pressure at top of side wall2.510tf/m2Ph4Horizontal Pressure at bottom of side wall6.260tf/m2Pv1Vertical Pressure(1) on top slab5.771tf/m2Q1Reaction to bottom slab6.671tf/m2H0Height of plane frame2.250mB0Width of plane frame2.000mt1Thickness of side wall0.250mt2Thickness of top slab0.250mt3Thickness of bottom slab0.250mt4Thickness of partition wall0.250mcUnit weight of reinforcement concrete2.400tf/m3L1Height of side wall2.000mCAB = CFE = (2Ph3+3Ph4)H02/60=2.0082tfmCBA = CEF = (3Ph3+2Ph4)H02/60=1.6918tfmCBC = CCB = CCE = CEC = Pv1B02/12=1.9236tfmCDA = CAD = CFD = CDF = Q1B02/12=2.2236tfmPv1=5.77 tf/m2t2=0.25mt2=0.25mPh1=2.51 tf/m2Ph3 =2.51 tf/m2H0=2.25mt1=0.25mt4=0.25mt1=0.25mt3=0.25mt3=0.25mPh2 =6.26 tf/m2Ph4 =6.26 tf/m2Q1=6.67 tf/m2B0 =2.00mB0 =2.00m2) Calculation of Bending Moment at jointk1=1.0k2 = H0t23/(B0t13)=1.12500k3 = H0t33/(B0t13)=1.12500k4 = H0t43/(H0t13)=1.00000As load has bilateral symmetry, the equation shown below is formed.qA = -qFqB = -qEqC = qD = 0R =02(k1+k3)k1-3k3qACAB - CADPreparatory caculation for qA,qB,Rk12(k1+k2)-3k2qB=CBC - CBAk3k2-2(k2+k3)RC4.250001.00000-3.37500D1.000004.25000-3.37500=-52.10156251.125001.12500-4.50000RBC = B0Pv1/2=5.7708tf-0.215401.00000-3.37500P = t1L1c=1.2000tf/mD10.231794.25000-3.37500=5.65834135530.200001.12500-4.50000C = [CBC - CCB + CDA - CAD + (RBC - B0Q1/2 + P)B0]/3=0.2000tfm4.25000-0.21540-3.375004.250001.00000-3.37500qA-0.2154D21.000000.23179-3.37500=-1.51059766391.000004.25000-3.37500qB=0.23181.125000.20000-4.500001.125001.12500-4.50000R0.20004.250001.00000-0.21540By solving above equation, the result is led as shown below;D31.000004.250000.23179=3.35256092291.125001.125000.20000qA=-0.1086R=-0.0643qB=0.0290qA=D1/DqB=D2/DR=D3/DMAB = -MFE = k1(2qA +qB) - CAB=-2.1964tfmMBA = -MEF = k1(2qB+qA)+CBA=1.6412tfmMBC = -MEC = k2(2qB+qC - 3R) - CBC=-1.6412tfmMCB = -MCE = k2(2qC+qB - 3R)+CCB=2.1734tfmMDA = -MDF = k3(2qD+qA - 3R) - CDA=-2.1286tfmMAD = -MFD = k3 (2qA+ qD - 3R)+CAD=2.1964tfmSMA = MAB + MAD=0.0000tfmo.k.SMB = MBA + MBC=0.0000tfmo.k.MCD = k4(2qC+qD)=0.0000tfmMDC = k4(2qD+qC)=0.0000tfm2) Calculation of Design Force2-1) Side Walla)Shearing Force at jointMBA=1.64 tf.mw1Load at end A6.260tf/m2w2=2.51 tf/m2w2Load at end B2.510tf/m2MABBending moment at end A-2.1964tfmMBABending moment at end B1.6412tfmL=2.25 mLLength of member2.250mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mw1=6.26 tf/m2SAB = (2w1+w2)L/6 - (MAB+MBA)/L=5.8832tfMAB=-2.20 tf.mSBA = SAB - L(w1+w2)/2=-3.9834tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SAB - w1x - (w2 - w1)x2/(2L)(i) In case of x1 =0.4mSx1=3.5125tf(ii) In case of x2 =1.85mSx2=-2.8460tfc)Bending MomentMA = MAB=-2.1964tfmMB = -MBA=-1.6412tfmThe maximum bending moment occurs at the point of that shearing force equal to zero.Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)=5.8831698533-6.2601x+0.83330x^2,x=6.41121.1012Check of SxBending moment at x =1.1012m is;Sx =SAB - w1x - (w2 - w1)x2/(2L)Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB=0.8574tfm=0.000tf2-2) Top Slaba)Shearing Force at jointw1Uniform load on top slab5.771tf/m2w1=5.77 tf/m2MBCBending moment at end B-1.6412tfmMCBBending moment at end C2.1734tfmLLength of member2.000mchProtective covering height0.050mMBC =tThickness of member (height)0.250m-1.64 tf.mMCB =dEffective height of member0.200m2.17 tf.mSBC = w1L/2-(MBC+MCB)/L=5.5047tfL=2.00 mSCB = SBC -w1L=-6.0368tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SBC- w1x(i) In case of x1 =0.400mSx1=3.1964tf(ii) In case of x2 =1.6mSx2=-3.7285tfc)Bending MomentMB = MBC=-1.6412tfmMC = -MCB=-2.1734tfmSx = SBC - w1x=5.50465-5.771x=0,x=0.9539Check of SxBending moment at x =0.9539m is;Sx =SBC- w1xMmax = SBCx - w1x2/2 + MBC=0.9842tfm=0.000tf2-3) Invert (Bottom Slab)a)Shearing Force at jointw1Reaction to bottom slab6.671tf/m2L=2.00 mMDABending moment at end D-2.1286tfmMADBending moment at end A2.1964tfmLLength of member2.000mchProtective covering height0.050mtThickness of member (height)0.250mMAD =MDA =dEffective height of member0.200m2.20 tf.m-2.13 tf.mw1=6.67 tf/m2SDA = w1L/2 - (MDA+MAD)/L=6.6368tfSAD = SDA - Lw1=-6.7047tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SDA- w1x(i) In case of x1 =0.400mSx1=3.9685tf(ii) In case of x2 =1.600mSx2=-4.03635tfc)Bending MomentMD = MDA=-2.1286tfmMA = -MAD=-2.1964tfmSx = SDA - w1x=6.6368-6.671x=0,x=0.9949Check of SxBending moment at x =0.9949mis;Sx =SDA- w1xMmax = SDAx - w1x2/2 + MDA=1.1730tfm=0.000tf2-4) Partition Walla)Shearing Force at jointw1Load at end C0.000tf/m2w2Load at end D0.000tf/m2MCDBending moment at end C0.000tfmMDCBending moment at end D0.000tfmLLength of member2.250mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mSCD = w1L/2 - (MCD+MDC)/L=0tfSDC = SCD - Lw1=0tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SCD- w1x(i) In case of x1 =0.400mSx1=0.000tf(ii) In case of x2 =1.850mSx2=0.000tfc)Bending MomentMC = MCD=0.00000tfmMD = -MDC=0.00000tfmSx = SCD - w1x=0-0.000x=0,x=-Check of SxBending moment at x =1.1250mis;Sx =SCD- w1xMmax = SCDx - w1x2/2 + MCD=0.000tfm=0.000tfCase 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L11) Calculation of Load TermPh1Horizontal Pressure at top of side wall0.435tf/m2Ph2Horizontal Pressure at bottom of side wall-0.215tf/m2Ph3Horizontal Pressure at top of side wall0.435tf/m2Ph4Horizontal Pressure at bottom of side wall-0.215tf/m2Pv1Vertical Pressure(1) on top slab3.446tf/m2Q1Reaction to bottom slab4.346tf/m2H0Height of plane frame2.250mB0Width of plane frame2.000mt1Thickness of side wall0.250mt2Thickness of top slab0.250mt3Thickness of bottom slab0.250mt4Thickness of partition wall0.250mcUnit weight of reinforcement concrete2.400tf/m3L1Height of side wall2.000mCAB = CFE = (2Ph3+3Ph4)H02/60=0.01893tfmCBA = CEF = (3Ph3+2Ph4)H02/60=0.07379tfmCBC = CCB = CCE = CEC = Pv1B02/12=1.14856tfmCDA = CAD = CFD = CDF = Q1B02/12=1.44856tfmPv1=3.45 tf/m2t2=0.25mt2=0.25mPh1=0.43 tf/m2Ph3 =0.43 tf/m2H0=2.25mt1=0.25mt4=0.25mt1=0.25mt3=0.25mt3=0.25mPh2 =-0.22 tf/m2Ph4 =-0.22 tf/m2Q1=4.35 tf/m2B0 =2.00mB0 =2.00m2) Calculation of Bending Moment at jointk1=1.0k2 = H0t23/(B0t13)=1.12500k3 = H0t33/(B0t13)=1.12500k4 = H0t43/(H0t13)=1.00000As load has bilateral symmetry, the equation shown below is formed.qA = -qFqB = -qEqC = qD = 0R =02(k1+k3)k1-3k3qACAB - CADPreparatory caculation for qA,qB,Rk12(k1+k2)-3k2qB=CBC - CBAk3k2-2(k2+k3)RC4.250001.00000-3.37500D1.000004.25000-3.37500=-52.10156251.125001.12500-4.50000RBC = B0Pv1/2=3.445671875tf-1.429631.00000-3.37500P = t1L1c=1.20000tf/mD11.074774.25000-3.37500=24.86292531590.200001.12500-4.50000C = [CBC - CCB + CDA - CAD + (RBC - B0Q1/2 + P)B0]/3=0.2tfm4.25000-1.42963-3.375004.250001.00000-3.37500qA-1.42963D21.000001.07477-3.37500=-15.28566888431.000004.25000-3.37500qB=1.074771.125000.20000-4.500001.125001.12500-4.50000R0.200004.250001.00000-1.42963By solving above equation, the result is led as shown below;D31.000004.250001.07477=4.70993910791.125001.125000.20000qA=-0.47720R=-0.09040qB=0.29338qA=D1/DqB=D2/DR=D3/DMAB = -MFE = k1(2qA +qB) - CAB=-0.6800tfmMBA = -MEF = k1(2qB+qA)+CBA=0.1834tfmMBC = -MEC = k2(2qB+qC - 3R) - CBC=-0.1834tfmMCB = -MCE = k2(2qC+qB - 3R)+CCB=1.7837tfmMDA = -MDF = k3(2qD+qA - 3R) - CDA=-1.6803tfmMAD = -MFD = k3 (2qA+ qD - 3R)+CAD=0.6800tfmSMA = MAB + MAD=0.0000tfmo.k.SMB = MBA + MBC=0.0000tfmo.k.MCD = k4(2qC+qD)=0.0000tfmMDC = k4(2qD+qC)=0.0000tfm2) Calculation of Design Force2-1) Side Walla)Shearing Force at jointMBA=0.18 tf.mw1Load at end A-0.215tf/m2w2=0.43 tf/m2w2Load at end B0.435tf/m2MABBending moment at end A-0.6800tfmMBABending moment at end B0.1834tfmLLength of member2.250mL=2.25 mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mw1=-0.22 tf/m2SAB = (2w1+w2)L/6 - (MAB+MBA)/L=0.2224tfMAB=-0.68 tf.mSBA = SAB - L(w1+w2)/2=-0.0248tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SAB - w1x - (w2 - w1)x2/(2L)(i) In case of x1 =0.4mSx1=0.2854tf(ii) In case of x2 =1.85mSx2=0.1261tfc)Bending MomentMA = MAB=-0.6800tfmMB = -MBA=-0.1834tfmThe maximum bending moment occurs at the point of that shearing force equal to zero.Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)=0.22243673250.2152x+-0.14447x^2,x=-0.7022.192Check of SxBending moment at x =2.1918m is;Sx =SAB - w1x - (w2 - w1)x2/(2L)Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB=-0.1826tfm=0.000tf2-2) Top Slaba)Shearing Force at jointw1Uniform load on top slab3.446tf/m2w1=3.45 tf/m2MBCBending moment at end B-0.1834tfmMCBBending moment at end C1.7837tfmLLength of member2.000mchProtective covering height0.050mMBC =tThickness of member (height)0.250m-0.18 tf.mMCB =dEffective height of member0.200m1.78 tf.mSBC = w1L/2-(MBC+MCB)/L=2.6455tfL=2.00 mSCB = SBC -w1L=-4.2459tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SBC- w1x(i) In case of x1 =0.400mSx1=1.2672tf(ii) In case of x2 =1.6mSx2=-2.8676tfc)Bending MomentMB = MBC=-0.1834tfmMC = -MCB=-1.7837tfmSx = SBC - w1x=2.6455-3.446x=0,x=0.7678Check of SxBending moment at x =0.7678m is;Sx =SBC- w1xMmax = SBCx - w1x2/2 + MBC=0.8322tfm=0.000tf2-3) Invert (Bottom Slab)a)Shearing Force at jointw1Reaction to bottom slab4.346tf/m2L=2.00 mMDABending moment at end B-1.6803tfmMADBending moment at end C0.6800tfmLLength of member2.000mchProtective covering height0.050mtThickness of member (height)0.250mMAD =MDA =dEffective height of member0.200m0.68 tf.m-1.68 tf.mw1=4.35 tf/m2SDA = w1L/2 - (MDA+MAD)/L=4.8459tfSAD = SDA - Lw1=-3.8455tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SDA- w1x(i) In case of x1 =0.400mSx1=3.1076tf(ii) In case of x2 =1.600mSx2=-2.1072tfc)Bending MomentMD = MDA=-1.6803tfmMA = -MAD=-0.6800tfmSx = SDA - w1x=4.84585-4.346x=0,x=1.1151Check of SxBending moment at x =1.1151mis;Sx =SDA- w1xMmax = SDAx - w1x2/2 + MDA=1.0215tfm=0.000tf2-4) Partition Walla)Shearing Force at jointw1Load at end C0.000tf/m2w2Load at end D0.000tf/m2MCDBending moment at end C0.000tfmMDCBending moment at end D0.000tfmLLength of member2.250mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mSCD = w1L/2 - (MCD+MDC)/L=0tfSDC = SCD - Lw1=0tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SCD- w1x(i) In case of x1 =0.400mSx1=0tf(ii) In case of x2 =1.850mSx2=0tfc)Bending MomentMC = MCD=0.00000tfmMD = -MDC=0.00000tfmSx = SCD - w1x=0-0.000x=0,x=-Check of SxBending moment at x =1.1250mis;Sx =SCD- w1xMmax = SCDx - w1x2/2 + MCD=0tfm=0.000tfCase 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L21) Calculation of Load TermPh1Horizontal Pressure at top of side wall1.737tf/m2Ph2Horizontal Pressure at bottom of side wall1.087tf/m2Ph3Horizontal Pressure at top of side wall1.737tf/m2Ph4Horizontal Pressure at bottom of side wall1.087tf/m2Pv1Vertical Pressure(1) on top slab5.651tf/m2Q1Reaction to bottom slab6.551tf/m2H0Height of plane frame2.250mB0Width of plane frame2.000mt1Thickness of side wall0.250mt2Thickness of top slab0.250mt3Thickness of bottom slab0.250mt4Thickness of partition wall0.250mcUnit weight of reinforcement concrete2.400tf/m3L1Height of side wall2.000mCAB = CFE = (2Ph3+3Ph4)H02/60=0.56819tfmCBA = CEF = (3Ph3+2Ph4)H02/60=0.62305tfmCBC = CCB = CCE = CEC = Pv1B02/12=1.88358tfmCDA = CAD = CFD = CDF = Q1B02/12=2.18358tfmPv1=5.65 tf/m2t2=0.25mt2=0.25mPh1=1.74 tf/m2Ph3 =1.74 tf/m2H0=2.25mt1=0.25mt4=0.25mt1=0.25mt3=0.25mt3=0.25mPh2 =1.09 tf/m2Ph4 =1.09 tf/m2Q1=6.55 tf/m2B0 =2.00mB0 =2.00m2) Calculation of Bending Moment at jointk1=1.0k2 = H0t23/(B0t13)=1.12500k3 = H0t33/(B0t13)=1.12500k4 = H0t43/(H0t13)=1.00000As load has bilateral symmetry, the equation shown below is formed.qA = -qFqB = -qEqC = qD = 0R =02(k1+k3)k1-3k3qACAB - CADPreparatory caculation for A,B,Rk12(k1+k2)-3k2qB=CBC - CBAk3k2-2(k2+k3)RC4.250001.00000-3.37500D1.000004.25000-3.37500=-52.10156251.125001.12500-4.50000RBC = B0Pv1/2=5.6508tf-1.615391.00000-3.37500P = t1L1c=1.2000tf/mD11.260534.25000-3.37500=27.84096353160.200001.12500-4.50000C = [CBC - CCB + CDA - CAD + (RBC - B0Q1/2 + P)B0]/3=0.2tfm4.25000-1.61539-3.375004.250001.00000-3.37500qA-1.61539D21.000001.26053-3.37500=-18.26370709991.000004.25000-3.37500qB=1.260531.125000.20000-4.500001.125001.12500-4.50000R0.200004.250001.00000-1.61539By solving above equation, the result is led as shown below;D31.000004.250001.26053=4.70993910791.125001.125000.20000qA=-0.53436R=-0.09040qB=0.35054qA=D1/DqB=D2/DR=D3/DMAB = -MFE = k1(2qA +qB) - CAB=-1.2864tfmMBA = -MEF = k1(2qB+qA)+CBA=0.7898tfmMBC = -MEC = k2(2qB+qC - 3R) - CBC=-0.7898tfmMCB = -MCE = k2(2qC+qB - 3R)+CCB=2.5830tfmMDA = -MDF = k3(2qD+qA - 3R) - CDA=-2.4796tfmMAD = -MFD = k3 (2qA+ qD - 3R)+CAD=1.2864tfmSMA = MAB + MAD=0.0000tfmo.k.SMB = MBA + MBC=0.0000tfmo.k.MCD = k4(2qC+qD)=0.0000tfmMDC = k4(2qD+qC)=0.0000tfm2) Calculation of Design Force2-1) Side Walla)Shearing Force at jointMBA=0.79 tf.mw1Load at end A1.087tf/m2w2=1.74 tf/m2w2Load at end B1.737tf/m2MABBending moment at end A-1.2864tfmMBABending moment at end B0.7898tfmLLength of member2.250mL=2.25 mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mw1=1.09 tf/m2SAB = (2w1+w2)L/6 - (MAB+MBA)/L=1.6871tfMAB=-1.29 tf.mSBA = SAB - L(w1+w2)/2=-1.4895tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SAB - w1x - (w2 - w1)x2/(2L)(i) In case of x1 =0.4mSx1=1.2293tf(ii) In case of x2 =1.85mSx2=-0.8179tfc)Bending MomentMA = MAB=-1.28637tfmMB = -MBA=-0.78977tfmThe maximum bending moment occurs at the point of that shearing force equal to zero.Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)=1.6871339981-1.0868x+-0.14447x^2,x=-8.8431.321Check of SxBending moment at x =1.3206m is;Sx =SAB - w1x - (w2 - w1)x2/(2L)Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB=-0.1169tfm=0.000tf2-2) Top Slaba)Shearing Force at jointw1Uniform load on top slab5.651tf/m2w1=5.65 tf/m2MBCBending moment at end B-0.7898tfmMCBBending moment at end C2.5830tfmLLength of member2.000mchProtective covering height0.050mMBC =tThickness of member (height)0.250m-0.79 tf.mMCB =dEffective height of member0.200m2.58 tf.mSBC = w1L/2-(MBC+MCB)/L=4.7541tfL=2.00 mSCB = SBC -w1L=-6.5474tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SBC- w1x(i) In case of x1 =0.400mSx1=2.4938tf(ii) In case of x2 =1.6mSx2=-4.2871tfc)Bending MomentMB = MBC=-0.7898tfmMC = -MCB=-2.5830tfmSx = SBC - w1x=4.7541-5.651x=0,x=0.8413Check of SxBending moment at x =0.8413m is;Sx =SBC- w1xMmax = SBCx - w1x2/2 + MBC=1.2101tfm=0.000tf2-3) Invert (Bottom Slab)a)Shearing Force at jointw1Reaction to bottom slab6.551tf/m2L=2.00 mMDABending moment at end B-2.4796tfmMADBending moment at end C1.2864tfmLLength of member2.000mchProtective covering height0.050mtThickness of member (height)0.250mMAD =MDA =dEffective height of member0.200m1.29 tf.m-2.48 tf.mw1=6.55 tf/m2SDA = w1L/2 - (MDA+MAD)/L=7.1474tfSAD = SDA - Lw1=-5.9541tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SDA- w1x(i) In case of x1 =0.400mSx1=4.5271tf(ii) In case of x2 =1.600mSx2=-3.3338tfc)Bending MomentMD = MDA=-2.4796tfmMA = -MAD=-1.2864tfmSx = SDA - w1x=7.14738-6.551x=0,x=1.0911Check of SxBending moment at x =1.0911mis;Sx =SDA- w1xMmax = SDAx - w1x2/2 + MDA=1.41954tfm=0.000tf2-4) Partition Walla)Shearing Force at jointw1Load at end C0.000tf/m2w2Load at end D0.000tf/m2MCDBending moment at end C0.000tfmMDCBending moment at end D0.000tfmLLength of member2.250mchProtective covering height0.050mtThickness of member (height)0.250mdEffective height of member0.200mSCD = w1L/2 - (MCD+MDC)/L=0tfSDC = SCD - Lw1=0tfb)Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.Sx = SCD- w1x(i) In case of x1 =0.400mSx1=0tf(ii) In case of x2 =1.850mSx2=0tfc)Bending MomentMC = MCD=0.00000tfmMD = -MDC=0.00000tfmSx = SCD - w1x=0-0.000x=0,x=-Check of SxBending moment at x =1.1250mis;Sx =SCD- w1xMmax = SCDx - w1x2/2 + MCD=0tfm=0.000tf
&C&"Times New Roman,"&9&P/&N&R&"Times New Roman,"&9(2)&F, &AABw1w1xw1w1xBCxDABACDEFBACDEFBACDEFBACDEFxw1w1xDABACDEFBACDEFABw1w1xw1w1BACDEFBACDEFABw1w1xw1w1xDAABw1w1xw1w1xBCxDAxBCxDAxDAxBCxDA
Sum MSNSummary of Internal forcesMemberCaseCheck PointMNS (tf)(tfm)(tf)at jointat 2dCase.1Case.2Case.3Case.4Side wallA-1.5904.5964.4182.569joint4.4184735.8831700.2224371.687134(left)Case.1Middle0.6404.0140.000-2d2.5685473.5124630.2853921.229308B-1.0353.396-2.519-1.902joint2.5187503.9834470.0248141.489511A-2.1966.7055.8833.5122d1.9021082.8460240.1260530.817863Case.2Middle0.8576.1170.000-B-1.6415.505-3.983-2.846A-0.6803.8450.2220.285Case.3Middle-0.1832.6770.000-B-0.1832.645-0.0250.126A-1.2865.9541.6871.229Case.4Middle-0.1175.2500.000-B-0.7904.754-1.490-0.818Top slabB-1.0352.5193.3961.970joint3.3960305.5046542.6454924.754116(left)Case.1Middle0.5822.5190.000-2d1.9697623.1963541.2672232.493816C-1.3742.519-3.735-2.309joint3.7353136.0368464.2458526.547384B-1.6413.9835.5053.1962d2.3090453.7285462.8675834.287084Case.2Middle0.9843.9830.000-C-2.1733.983-6.037-3.729B-0.1830.0252.6451.267Case.3Middle0.8320.0250.000-C-1.7840.025-4.246-2.868B-0.7900.0254.7542.494Case.4Middle1.2100.0250.000-C-2.5830.025-6.547-4.287PartitionC0.0007.4710.0000.000joint0.0000000.0000000.0000000.000000wallCase.1Middle0.0008.0710.000-2d0.0000000.0000000.0000000.000000D0.0008.6710.0000.000joint0.0000000.0000000.0000000.000000C0.00012.0740.0000.0002d0.0000000.0000000.0000000.000000Case.2Middle0.00012.6740.000-D0.00013.2740.0000.000C0.0008.4920.0000.000Case.3Middle0.0009.0920.000-D0.0009.6920.0000.000C0.00013.0950.0000.000Case.4Middle0.00013.6950.000-D0.00014.2950.0000.000InvertD-1.3294.4184.3352.549joint4.3353136.6368464.8458527.147384(left)Case.1Middle0.7754.4180.000-2d2.5490453.9685463.1075834.527084A-1.5904.418-4.596-2.810joint4.5960306.7046543.8454925.954116D-2.1295.8836.6373.9692d2.8097624.0363542.1072233.333816Case.2Middle1.1735.8830.000-A-2.1965.883-6.705-4.036D-1.6800.2224.8463.108Case.3Middle1.0210.2220.000-A-0.6800.222-3.845-2.107D-2.4801.6877.1474.527Case.4Middle1.4201.6870.000-A-1.2861.687-5.954-3.334MemberCaseCheck PointMNS (tf)(tfm)(tf)at jointat 2dTop slabC-1.3742.5193.7352.309joint3.7353136.0368464.2458526.547384(right)Case.1Middle0.5822.5190.000-2d2.3090453.7285462.8675834.287084E-1.0352.519-3.396-1.970joint3.3960305.5046542.6454924.754116C-2.1733.9836.0373.7292d1.9697623.1963541.2672232.493816Case.2Middle0.9843.9830.000-E-1.6413.983-5.505-3.196C-1.7840.0254.2462.868Case.3Middle0.8320.0250.000-E-0.1830.025-2.645-1.267C-2.5830.0256.5474.287Case.4Middle1.2100.0250.000-E-0.7900.025-4.754-2.494Side wallE-1.0353.3962.5191.902joint2.5187503.9834470.0248141.489511(right)Case.1Middle0.6404.0140.000-2d1.9021082.8460240.1260530.817863F-1.5904.596-4.418-2.569joint4.4184735.8831700.2224371.687134E-1.6415.5053.9832.8462d2.5685473.5124630.2853921.229308Case.2Middle0.8576.117-0.000-F-2.1966.705-5.883-3.512E-0.1832.6450.025-0.126Case.3Middle-0.1832.6770.000-F-0.6803.845-0.222-0.285E-0.7904.7541.4900.818Case.4Middle-0.1175.250-0.000-F-1.2865.954-1.687-1.229InvertF-1.5904.4184.5962.810joint4.5960306.7046543.8454925.954116(right)Case.1Middle0.7754.4180.000-2d2.8097624.0363542.1072233.333816D-1.3294.418-4.335-2.549joint4.3353136.6368464.8458527.147384F-2.1965.8836.7054.0362d2.5490453.9685463.1075834.527084Case.2Middle1.1735.8830.000-D-2.1295.883-6.637-3.969F-0.6800.2223.8452.107Case.3Middle1.0210.2220.000-D-1.6800.222-4.846-3.108F-1.2861.6875.9543.334Case.4Middle1.4201.6870.000-D-2.4801.687-7.147-4.527
&C&"Times New Roman,"&9&P/&N&R&"Times New Roman,"&9(3)&F, &A
R-bar req5Calculation of Required Reinforcement Bar5-1Calculation of Required Reinforcement Bar1)At Joint "A"of side wallCase.1M=1.5900tfmsca =60kgf/m2h =25cm (height of member)N=4.5960tfssa =1400kgf/m2d =20cm (effective height of member)S0=4.4185tfn =24d' =5cm (protective covering depth)S2d=2.5685tfc =7.50cm (distance from neutral axis)b =100cme = M/N =34.5946cmSolving the fomula shown below, sc =31.3325kgf/cm2(0.0000003029kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+72.99sc^2-1692.85sc-49374.67s = nsc/(nsc+ssa) =0.3494Asreq = (sc*s/2 - N/(bd))bd/ssa =4.5376cm2Case.2M=2.1964tfmsca =60kgf/m2h =25cm (height of member)N=6.7047tfssa =1400kgf/m2d =20cm (effective height of member)S0=5.8832tfn =24d' =5cm (protective covering depth)S2d=3.5125tfc =7.50cm (distance from neutral axis)b =100cme = M/N =32.7593cmSolving the fomula shown below, sc =38.9162kgf/cm2(0.0000454666kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+67.26sc^2-2361.84sc-68887.05s = nsc/(nsc+ssa) =0.4002Asreq = (sc*s/2 - N/(bd))bd/ssa =6.3346cm2Case.3M=0.6800tfmsca =60kgf/m2h =25cm (height of member)N=3.8455tfssa =1400kgf/m2d =20cm (effective height of member)S0=0.2224tfn =24d' =5cm (protective covering depth)S2d=0.2854tfc =7.50cm (distance from neutral axis)b =100cme = M/N =17.6818cmSolving the fomula shown below, sc =20.4316kgf/cm2(0.0000002274kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+80.24sc^2-847.32sc-24713.45s = nsc/(nsc+ssa) =0.2594Asreq = (sc*s/2 - N/(bd))bd/ssa =1.0389cm2Case.4M=1.2864tfmsca =60kgf/m2h =25cm (height of member)N=5.9541tfssa =1400kgf/m2d =20cm (effective height of member)S0=1.6871tfn =24d' =5cm (protective covering depth)S2d=1.2293tfc =7.50cm (distance from neutral axis)b =100cme = M/N =21.6047cmSolving the fomula shown below, sc =29.2154kgf/cm2(0.0000026242kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+74.5sc^2-1516.31sc-44225.83s = nsc/(nsc+ssa) =0.3337Asreq = (sc*s/2 - N/(bd))bd/ssa =2.7109cm2The maximum requirement of reinforcement bar is6.3346cm2 in Case.2from above calculation.Case.1234Requirement4.53766.33461.03892.7109(cm2)2)At Joint "B"of side wallCase.1M=1.0348tfmsca =60kgf/m2h =25cm (height of member)N=3.3960tfssa =1400kgf/m2d =20cm (effective height of member)S0=2.5187tfn =24d' =5cm (protective covering depth)S2d=1.9021tfc =7.50cm (distance from neutral axis)b =100cme = M/N =30.4697cmSolving the fomula shown below, sc =24.3042kgf/cm2(0.0002507842kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+77.83sc^2-1128.28sc-32908.12s = nsc/(nsc+ssa) =0.2941Asreq = (sc*s/2 - N/(bd))bd/ssa =2.6800cm2Case.2M=1.6412tfmsca =60kgf/m2h =25cm (height of member)N=5.5047tfssa =1400kgf/m2d =20cm (effective height of member)S0=3.9834tfn =24d' =5cm (protective covering depth)S2d=2.8460tfc =7.50cm (distance from neutral axis)b =100cme = M/N =29.8144cmSolving the fomula shown below, sc =32.5591kgf/cm2(0.0000000816kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+72.09sc^2-1797.27sc-52420.49s = nsc/(nsc+ssa) =0.3582Asreq = (sc*s/2 - N/(bd))bd/ssa =4.3989cm2Case.3M=0.1834tfmsca =60kgf/m2h =25cm (height of member)N=2.6455tfssa =1400kgf/m2d =20cm (effective height of member)S0=0.0248tfn =24d' =5cm (protective covering depth)S2d=0.1261tfc =7.50cm (distance from neutral axis)b =100cme = M/N =6.9307cmSolving the fomula shown below, sc =11.9223kgf/cm2(0.0000006127kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+84.64sc^2-334.04sc-9742.89s = nsc/(nsc+ssa) =0.1697Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2Case.4M=0.7898tfmsca =60kgf/m2h =25cm (height of member)N=4.7541tfssa =1400kgf/m2d =20cm (effective height of member)S0=1.4895tfn =24d' =5cm (protective covering depth)S2d=0.8179tfc =7.50cm (distance from neutral axis)b =100cme = M/N =16.6123cmSolving the fomula shown below, sc =22.6200kgf/cm2(0.0000000114kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+78.9sc^2-1003.04sc-29255.26s = nsc/(nsc+ssa) =0.2794Asreq = (sc*s/2 - N/(bd))bd/ssa =1.1188cm2The maximum requirement of reinforcement bar is4.3989cm2 in Case.2from above calculation.Case.1234Requirement2.68004.39890.00001.1188(cm2)3)At Joint "B"of top slabCase.1M=1.0348tfmsca =60kgf/m2h =25cm (height of member)N=2.5187tfssa =1400kgf/m2d =20cm (effective height of member)S0=3.3960tfn =24d' =5cm (protective covering depth)S2d=1.9698tfc =7.50cm (distance from neutral axis)b =100cme = M/N =41.0822cmSolving the fomula shown below, sc =23.5376kgf/cm2(0.0004860488kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+78.32sc^2-1070.71sc-31228.95s = nsc/(nsc+ssa) =0.2875Asreq = (sc*s/2 - N/(bd))bd/ssa =3.0344cm2Case.2M=1.6412tfmsca =60kgf/m2h =25cm (height of member)N=3.9834tfssa =1400kgf/m2d =20cm (effective height of member)S0=5.5047tfn =24d' =5cm (protective covering depth)S2d=3.1964tfc =7.50cm (distance from neutral axis)b =100cme = M/N =41.2000cmSolving the fomula shown below, sc =31.3869kgf/cm2(0.0000002861kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+72.95sc^2-1697.44sc-49508.81s = nsc/(nsc+ssa) =0.3498Asreq = (sc*s/2 - N/(bd))bd/ssa =4.9976cm2Case.3M=0.1834tfmsca =60kgf/m2h =25cm (height of member)N=0.0248tfssa =1400kgf/m2d =20cm (effective height of member)S0=2.6455tfn =24d' =5cm (protective covering depth)S2d=1.2672tfc =7.50cm (distance from neutral axis)b =100cme = M/N =738.91cmSolving the fomula shown below, sc =8.0001kgf/cm2(0.0009516341kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+86.11sc^2-162.06sc-4726.74s = nsc/(nsc+ssa) =0.1206Asreq = (sc*s/2 - N/(bd))bd/ssa =0.6715cm2Case.4M=0.7898tfmsca =60kgf/m2h =25cm (height of member)N=0.0248tfssa =1400kgf/m2d =20cm (effective height of member)S0=4.7541tfn =24d' =5cm (protective covering depth)S2d=2.4938tfc =7.50cm (distance from neutral axis)b =100cme = M/N =3182.82cmSolving the fomula shown below, sc =18.1280kgf/cm2(0.000004604kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+81.56sc^2-692.68sc-20203.08s = nsc/(nsc+ssa) =0.2371Asreq = (sc*s/2 - N/(bd))bd/ssa =3.0522cm2The maximum requirement of reinforcement bar is4.9976cm2 in Case.2from above calculation.Case.1234Requirement3.03444.99760.67153.0522(cm2)4)At Joint "A"of invertCase.1M=1.5900tfmsca =60kgf/m2h =25cm (height of member)N=4.4185tfssa =1400kgf/m2d =20cm (effective height of member)S0=4.5960tfn =24d' =5cm (protective covering depth)S2d=2.8098tfc =7.50cm (distance from neutral axis)b =100cme = M/N =35.9848cmSolving the fomula shown below, sc =31.1945kgf/cm2(0.0000003501kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+73.09sc^2-1681.19sc-49034.81s = nsc/(nsc+ssa) =0.3484Asreq = (sc*s/2 - N/(bd))bd/ssa =4.6077cm2Case.2M=2.1964tfmsca =60kgf/m2h =25cm (height of member)N=5.8832tfssa =1400kgf/m2d =20cm (effective height of member)S0=6.7047tfn =24d' =5cm (protective covering depth)S2d=4.0364tfc =7.50cm (distance from neutral axis)b =100cme = M/N =37.3336cmSolving the fomula shown below, sc =38.3259kgf/cm2(0.0000728003kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+67.72sc^2-2307.93sc-67314.67s = nsc/(nsc+ssa) =0.3965Asreq = (sc*s/2 - N/(bd))bd/ssa =6.6523cm2Case.3M=0.6800tfmsca =60kgf/m2h =25cm (height of member)N=0.2224tfssa =1400kgf/m2d =20cm (effective height of member)S0=3.8455tfn =24d' =5cm (protective covering depth)S2d=2.1072tfc =7.50cm (distance from neutral axis)b =100cme = M/N =305.6833cmSolving the fomula shown below, sc =16.8183kgf/cm2(0.0000239535kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+82.28sc^2-609.56sc-17778.70s = nsc/(nsc+ssa) =0.2238Asreq = (sc*s/2 - N/(bd))bd/ssa =2.5295cm2Case.4M=1.2864tfmsca =60kgf/m2h =25cm (height of member)N=1.6871tfssa =1400kgf/m2d =20cm (effective height of member)S0=5.9541tfn =24d' =5cm (protective covering depth)S2d=3.3338tfc =7.50cm (distance from neutral axis)b =100cme = M/N =76.2460cmSolving the fomula shown below, sc =25.7138kgf/cm2(0.0000718588kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+76.9sc^2-1236.29sc-36058.56s = nsc/(nsc+ssa) =0.3059Asreq = (sc*s/2 - N/(bd))bd/ssa =4.4142cm2The maximum requirement of reinforcement bar is6.6523cm2 in Case.2from above calculation.Case.1234Requirement4.60776.65232.52954.4142(cm2)5)At Middle of side wall (between A and B)Case.1M=0.6405tfmsca =60kgf/m2h =25cm (height of member)N=4.0140tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =15.9566cmSolving the fomula shown below, sc =20.0915kgf/cm2(0.0000003577kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+80.44sc^2-823.85sc-24029.00s = nsc/(nsc+ssa) =0.2562Asreq = (sc*s/2 - N/(bd))bd/ssa =0.8094cm2Tensile is on inside of memberCase.2M=0.8574tfmsca =60kgf/m2h =25cm (height of member)N=6.1173tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =14.0166cmSolving the fomula shown below, sc =24.6130kgf/cm2(0.0001914167kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+77.63sc^2-1151.72sc-33591.70s = nsc/(nsc+ssa) =0.2967Asreq = (sc*s/2 - N/(bd))bd/ssa =0.8473cm2Tensile is on inside of memberCase.3M=0.1826tfmsca =60kgf/m2h =25cm (height of member)N=2.6765tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =6.8235cmSolving the fomula shown below, sc =11.9510kgf/cm2(0.00000058kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+84.62sc^2-335.45sc-9783.97s = nsc/(nsc+ssa) =0.1700Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2Tensile is on outside of memberCase.4M=0.1169tfmsca =60kgf/m2h =25cm (height of member)N=5.2498tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =2.2270cmSolving the fomula shown below, sc =14.0580kgf/cm2(0.0006792107kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+83.67sc^2-446.82sc-13032.22s = nsc/(nsc+ssa) =0.1942Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2Tensile is on outside of memberThe maximum requirement of reinforcement bar is0.8473cm2 in Case.2from above calculation.Case.1234Requirement0.80940.8473--(cm2)Sideinsideinsideoutsideoutside6)At Middle of top slab (between B and C)Case.1M=0.5825tfmsca =60kgf/m2h =25cm (height of member)N=2.5187tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =23.1254cmSolving the fomula shown below, sc =17.8538kgf/cm2(0.0000065267kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+81.71sc^2-674.96sc-19686.20s = nsc/(nsc+ssa) =0.2343Asreq = (sc*s/2 - N/(bd))bd/ssa =1.1894cm2Tensile is on inside of memberCase.2M=0.9842tfmsca =60kgf/m2h =25cm (height of member)N=3.9834tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =24.7081cmSolving the fomula shown below, sc =24.2291kgf/cm2(0.0002677035kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+77.88sc^2-1122.62sc-32743.07s = nsc/(nsc+ssa) =0.2935Asreq = (sc*s/2 - N/(bd))bd/ssa =2.2335cm2Tensile is on inside of memberCase.3M=0.8322tfmsca =60kgf/m2h =25cm (height of member)N=0.0248tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =3353.8895cmSolving the fomula shown below, sc =18.6958kgf/cm2(0.0000022224kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+81.24sc^2-729.82sc-21286.39s = nsc/(nsc+ssa) =0.2427Asreq = (sc*s/2 - N/(bd))bd/ssa =3.2235cm2Tensile is on inside of memberCase.4M=1.2101tfmsca =60kgf/m2h =25cm (height of member)N=0.0248tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =4876.8077cmSolving the fomula shown below, sc =23.3996kgf/cm2(0.0005467204kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+78.41sc^2-1060.47sc-30930.45s = nsc/(nsc+ssa) =0.2863Asreq = (sc*s/2 - N/(bd))bd/ssa =4.7674cm2Tensile is on inside of memberThe maximum requirement of reinforcement bar is4.7674cm2 in Case.4from above calculation.Case.1234Requirement1.18942.23353.22354.7674(cm2)Sideinsideinsideinsideinside7)At Middle of invert (between D and A)Case.1M=0.7751tfmsca =60kgf/m2h =25cm (height of member)N=4.4185tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =17.5427cmSolving the fomula shown below, sc =22.1400kgf/cm2(0.0000000222kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+79.2sc^2-968.19sc-28238.90s = nsc/(nsc+ssa) =0.2751Asreq = (sc*s/2 - N/(bd))bd/ssa =1.1948cm2Tensile is on inside of memberCase.2M=1.1730tfmsca =60kgf/m2h =25cm (height of member)N=5.8832tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =19.9377cmSolving the fomula shown below, sc =27.9391kgf/cm2(0.0000090881kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+75.39sc^2-1412.43sc-41195.84s = nsc/(nsc+ssa) =0.3238Asreq = (sc*s/2 - N/(bd))bd/ssa =2.2606cm2Tensile is on inside of memberCase.3M=1.0215tfmsca =60kgf/m2h =25cm (height of member)N=0.2224tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =459.2269cmSolving the fomula shown below, sc =21.3043kgf/cm2(0.0000000701kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+79.71sc^2-908.4sc-26495.01s = nsc/(nsc+ssa) =0.2675Asreq = (sc*s/2 - N/(bd))bd/ssa =3.9120cm2Tensile is on inside of memberCase.4M=1.4195tfmsca =60kgf/m2h =25cm (height of member)N=1.6871tfssa =1400kgf/m2d =20cm (effective height of member)n =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =84.1391cmSolving the fomula shown below, sc =27.1953kgf/cm2(0.0000183862kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+75.9sc^2-1352.82sc-39457.10s = nsc/(nsc+ssa) =0.3180Asreq = (sc*s/2 - N/(bd))bd/ssa =4.9715cm2Tensile is on inside of memberThe maximum requirement of reinforcement bar is4.9715cm2 in Case.4from above calculation.Case.1234Requirement1.19482.26063.91204.9715(cm2)Sideinsideinsideinsideinside8)At Top of partition wall (joint C)Case.1M=1.3740tfmsca =60kgf/m2h =25cm (height of member)N=2.5187tfssa =1400kgf/m2d =20cm (effective height of member)S0=3.7353tfn =24d' =5cm (protective covering depth)S2d=2.3090tfc =7.50cm (distance from neutral axis)b =100cme = M/N =54.5525cmSolving the fomula shown below, sc =27.3793kgf/cm2(0.0000154593kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+75.78sc^2-1367.58sc-39887.73s = nsc/(nsc+ssa) =0.3194Asreq = (sc*s/2 - N/(bd))bd/ssa =4.4479cm2Case.2M=2.1734tfmsca =60kgf/m2h =25cm (height of member)N=3.9834tfssa =1400kgf/m2d =20cm (effective height of member)S0=6.0368tfn =24d' =5cm (protective covering depth)S2d=3.7285tfc =7.50cm (distance from neutral axis)b =100cme = M/N =54.5600cmSolving the fomula shown below, sc =36.7237kgf/cm2(0.0002489747kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+68.96sc^2-2163.11sc-63090.80s = nsc/(nsc+ssa) =0.3863Asreq = (sc*s/2 - N/(bd))bd/ssa =7.2887cm2Case.3M=1.7837tfmsca =60kgf/m2h =25cm (height of member)N=0.0248tfssa =1400kgf/m2d =20cm (effective height of member)S0=4.2459tfn =24d' =5cm (protective covering depth)S2d=2.8676tfc =7.50cm (distance from neutral axis)b =100cme = M/N =7188.4588202266cmSolving the fomula shown below, sc =29.7731kgf/cm2(0.000001504kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+74.11sc^2-1562.37sc-45569.25s = nsc/(nsc+ssa) =0.3379Asreq = (sc*s/2 - N/(bd))bd/ssa =7.1687cm2Case.4M=2.5830tfmsca =60kgf/m2h =25cm (height of member)N=0.0248tfssa =1400kgf/m2d =20cm (effective height of member)S0=6.5474tfn =24d' =5cm (protective covering depth)S2d=4.2871tfc =7.50cm (distance from neutral axis)b =100cme = M/N =10409.8044421029cmSolving the fomula shown below, sc =37.8189kgf/cm2(0.0001082406kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+68.11sc^2-2261.79sc-65968.79s = nsc/(nsc+ssa) =0.3933Asreq = (sc*s/2 - N/(bd))bd/ssa =10.6073cm2The maximum requirement of reinforcement bar is10.6073cm2 in Case.4from above calculation.Case.1234Requirement4.44797.28877.168710.6073(cm2)9)At Bottom of partition wall (joint D)Case.1M=1.3293tfmsca =60kgf/m2h =25cm (height of member)N=4.4185tfssa =1400kgf/m2d =20cm (effective height of member)S0=4.3353tfn =24d' =5cm (protective covering depth)S2d=2.5490tfc =7.50cm (distance from neutral axis)b =100cme = M/N =30.0842cmSolving the fomula shown below, sc =28.4400kgf/cm2(0.0000056076kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+75.05sc^2-1453.07sc-42381.10s = nsc/(nsc+ssa) =0.3278Asreq = (sc*s/2 - N/(bd))bd/ssa =3.5020cm2Case.2M=2.1286tfmsca =60kgf/m2h =25cm (height of member)N=5.8832tfssa =1400kgf/m2d =20cm (effective height of member)S0=6.6368tfn =24d' =5cm (protective covering depth)S2d=3.9685tfc =7.50cm (distance from neutral axis)b =100cme = M/N =36.1810cmSolving the fomula shown below, sc =37.6719kgf/cm2(0.000121245kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+68.23sc^2-2248.6sc-65584.17s = nsc/(nsc+ssa) =0.3924Asreq = (sc*s/2 - N/(bd))bd/ssa =6.3565cm2Case.3M=1.6803tfmsca =60kgf/m2h =25cm (height of member)N=0.2224tfssa =1400kgf/m2d =20cm (effective height of member)S0=4.8459tfn =24d' =5cm (protective covering depth)S2d=3.1076tfc =7.50cm (distance from neutral axis)b =100cme = M/N =755.4109cmSolving the fomula shown below, sc =28.8316kgf/cm2(0.0000038299kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+74.77sc^2-1484.87sc-43308.70s = nsc/(nsc+ssa) =0.3308Asreq = (sc*s/2 - N/(bd))bd/ssa =6.6530cm2Case.4M=2.4796tfmsca =60kgf/m2h =25cm (height of member)N=1.6871tfssa =1400kgf/m2d =20cm (effective height of member)S0=7.1474tfn =24d' =5cm (protective covering depth)S2d=4.5271tfc =7.50cm (distance from neutral axis)b =100cme = M/N =146.9735cmSolving the fomula shown below, sc =38.0243kgf/cm2(0.0000922636kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+67.95sc^2-2280.4sc-66511.77s = nsc/(nsc+ssa) =0.3946Asreq = (sc*s/2 - N/(bd))bd/ssa =9.5128cm2The maximum requirement of reinforcement bar is9.5128cm2 in Case.4from above calculation.Case.1234Requirement3.50206.35656.65309.5128(cm2)10)At Middle of partition wallCase.1M=0.0000tfmsca =60kgf/m2h =25cm (height of member)N=8.0706tfssa =1400kgf/m2d =20cm (effective height of member)S0=0.0000tfn =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =0.0000cmSolving the fomula shown below, sc =15.5007kgf/cm2(0.0001208493kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+82.96sc^2-529.63sc-15447.68s = nsc/(nsc+ssa) =0.2099Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2Case.2M=0.0000tfmsca =60kgf/m2h =25cm (height of member)N=12.6737tfssa =1400kgf/m2d =20cm (effective height of member)S0=0.0000tfn =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =0.0000cmSolving the fomula shown below, sc =20.2061kgf/cm2(0.0000003073kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+80.37sc^2-831.71sc-24258.24s = nsc/(nsc+ssa) =0.2573Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2Case.3M=0.0000tfmsca =60kgf/m2h =25cm (height of member)N=9.0917tfssa =1400kgf/m2d =20cm (effective height of member)S0=0.0000tfn =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =0.0000cmSolving the fomula shown below, sc =16.6097kgf/cm2(0.0000310341kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+82.39sc^2-596.64sc-17402.09s = nsc/(nsc+ssa) =0.2216Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2Case.4M=0.0000tfmsca =60kgf/m2h =25cm (height of member)N=13.6948tfssa =1400kgf/m2d =20cm (effective height of member)S0=0.0000tfn =24d' =5cm (protective covering depth)c =7.50cm (distance from neutral axis)b =100cme = M/N =0.0000cmSolving the fomula shown below, sc =21.1666kgf/cm2(0.0000000845kgf/cm2)o.k.sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 00 = sc^3+79.8sc^2-898.72sc-26212.64s = nsc/(nsc+ssa) =0.2662Asreq = (sc*s/2 - N/(bd))bd/ssa =0.0000cm2The maximum requirement of reinforcement bar is0.0000cm2 in Case.1from above calculation.Case.1234Requirement0.00000.00000.00000.0000(cm2)11)Summary of required reinforcementRequired reinforcement for design is the maximum required reinforcment calculated above in 1) - 4).ItemSide wallSide wallTop slabInvertSide wallTop slabInvertTop slabInvertPar. WallPointbotomtopendendmiddlemiddlemiddlejoint Cjoind DmiddleSideoutsideoutsideoutsideoutsideinsideinsideinsideoutsideoutsidebothCalculation1)2)3)4)5)6)7)8)9)10)Requirment6.33464.39894.99766.65230.84734.76744.971510.60739.51280.0000(cm2)
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R-bar stress6 Bar Arrangement and Calculation of StressType: B1.75m x H2.00m x 2Data of Reinforcement BarSide wallTop slabfSectionalPerimeterArrengmentAreaPerimeterbottommiddletopendmiddlejoint CAreaoutsideinsideoutsideoutsideinsideoutside(mm)(cm2)(cm)(cm2)(cm)Bending momentMkgfcm219,64085,745164,118164,118121,011258,[email protected] force (joint)Skgf5,88303,9835,50506,[email protected] force (2d)S2dkgf3,512-2,8463,196-4,[email protected] forceNkgf6,7056,1175,5052,[email protected] of [email protected] depthd'[email protected] [email protected] [email protected] areabdcm2200020002000200020002000192.8355.96919@12522.6847.75Young's modulus [email protected]@25011.3423.88Required R-barAsreqcm26.3350.8474.3994.9984.76710.60719@[email protected] arrangement12@25012@12512@25012@25012@25012@[email protected]@25015.2127.65ReinforcementAscm24.529.054.524.524.529.0522@30012.6723.04Permeter of R-barU15.0830.1615.0815.0815.0830.16254.9097.85425@12539.2762.83M/Necm32.75914.01729.81465.1584876.80810409.80425@15032.7252.36Dist. from neutral [email protected]'[email protected]'262.0280.4242.9472.931791.0135758.3328.04210.05332@12564.3480.42c'-5240.8-5608.1-4857.4-9458.4-635819.6-2715166.232@15053.6267.02x7.0711.837.246.285.607.4032@25032.1740.210.0000.0000.0000.0000.0000.00032@30026.8133.51Calculation Check(check)okokokokokok12@250 + 16@25012,[email protected],[email protected] stresssckgf/cm243.2913.8632.2732.5323.8739.8312,[email protected] stressscakgf/cm260.0060.0060.0060.0060.0060.0012,[email protected],[email protected] stresssskgf/cm21901.24229.821365.971704.021473.191626.4016,[email protected] stressssakgf/cm21400.001400.001400.001400.001400.001400.0016,[email protected],[email protected] stress at jointtkgf/cm22.940.001.992.750.003.2716,[email protected] stresstakgf/cm211.0011.0011.0011.0011.0011.0019,[email protected],[email protected] stress at 2dt2dkgf/cm21.76-1.421.60-2.1419,[email protected] stresst2dakgf/cm25.50-5.505.50-5.5022,[email protected],[email protected] stresstokgf/cm221.510.0014.5620.130.0012.3825,[email protected] stresstoakgf/cm26.506.506.506.506.506.5012@300 + 16@30012,[email protected],[email protected],[email protected] = As/(bd)0.002260.004530.002260.002260.002260.0045312,[email protected] = (2np+(np)^2)^0.5-np0.279560.369930.279560.279560.279560.3699312,[email protected] = 1-k/30.906810.876690.906810.906810.906810.8766916,[email protected],[email protected] MomentMrkgfcm278,512380,599251,930190,960145,123318,46416,[email protected] for compressionMrckgfcm303,090380,599302,794302,983304,193389,13816,[email protected] for Mrccm6.6208.2396.4225.9565.5957.40219,[email protected] for Mrckgf/cm22910.52055.43044.63395.73707.82450.919,[email protected] for tensileMrskgfcm278,512473,674251,930190,960145,123318,46419,[email protected] for Mrscm9.45311.3509.0087.7896.6009.33022,[email protected] for Mrskgf/cm252.376.547.837.228.751.022,[email protected],[email protected] bar (>As/6 and >Asmin)12@25012@25012@25012@25012@25012@250ReinforcementAscm24.524.524.524.524.524.52okokokokokokReinforcement bar for fillet12@25012@[email protected] requirement of reinforcement barAs min =4.5cm2InvertPartition wallendmiddlejoint Dmiddleoutsideinsideoutsideboth sidesBending momentMkgfcm219,640141,954247,9640Shearing force (joint)Skgf6,70507,1470Shearing force (2d)S2dkgf4,036-4,527-Axial forceNkgf5,8831,6871,6878,071Height of memberhcm25252525Covering depthd'cm5555Effective heightdcm20202020Effective widthbcm100100100100Effective areabdcm22000200020002000Young's modulus ration-24242424Required R-barAsreqcm26.6524.9719.5130.000R-bar arrangement12@25012@25012@[email protected] of R-barU15.0815.0830.1615.08M/Necm37.33484.139146.9740.000Dist. from neutral axisccm7.507.507.507.50a'74.5214.9403.4-37.5b'291.8596.52013.148.8c'-5836.3-11929.2-40262.0-976.3x6.866.127.750.000.0000.0000.0000.000(check)okokok0.00000Compressive stresssckgf/cm243.3928.1338.633.23Allowable stressscakgf/cm260.0060.0060.0060.00okokokokTensile stresssskgf/cm21993.051531.261466.890.00Allowable stressssakgf/cm21400.001400.001400.001400.00checkcheckcheck0Shearing stress at jointtkgf/cm23.350.003.570.00Allowable stresstakgf/cm211.0011.0011.0011.00okokokokShearing stress at 2dt2dkgf/cm22.02-2.26-Allowable stresst2dakgf/cm25.50-5.50-ok-ok-Adhesive stresstokgf/cm224.510.0013.520.00Allowable stresstoakgf/cm26.506.506.506.50checkokcheckokp = As/(bd)0.002260.002260.004530.00226k = (2np+(np)^2)^0.5-np0.279560.279560.369930.27956j = 1-k/30.906810.906810.876690.90681Resisting MomentMrkgfcm260,178175,188356,945303,673Mr for compressionMrckgfcm302,865303,278386,481303,673x for Mrccm6.4845.8327.6206.853ss for Mrckgf/cm23001.83497.92339.52762.7Mr for tensileMrskgfcm260,178175,188356,945310,374x for Mrscm9.1517.4149.9229.936sc for Mrskgf/cm249.234.457.457.6Distribution bar (>As/6 and >Asmin)12@25012@25012@[email protected] bar for [email protected] requirement of reinforcement barAs min =4.5cm2
&C&"Times New Roman,"&9&P/&N&R&"Times New Roman,"&9(5)&F, &A12@25012@12512@25012@125
Plot R-barDATA PLOT PEMBESIANSection of Culvert1. Design Datab1=0.250mb2=1.750mb3=0.250mb4=1.750mh1=0.250mh2=2.000mh3=0.250mfillet =0.150mBb=4.250mHa=2.500md=0.05m2. Data PembesianBottom slab :TumpuanLapanganTulangan bagi :As1 (cm2):O12@250O12@250O12@250As2 (cm2):O12@250O12@250As1 (cm2):O12@125(Tumpuan Tengah)O12@250(Tumpuan Tepi)Side wall :TumpuanLapanganTulangan bagi :As1 (cm2):O12@250O12@125O12@250As2 (cm2):O12@125O12@250As1 (cm2):O12@250(Top wall)Top slab :TumpuanLapanganTulangan bagi :As1 (cm2):O12@250O12@250O12@250As2 (cm2):O12@250O12@250As1 (cm2):O12@125(Tumpuan Tengah)3.Tulangan Miring (fillet) :Partition wall :Bottom slab:O12@250O12@250Top slab:[email protected] Bangunan:Double Box-Culvert Bba.20.a (2.4 m x 1.5 m)Lokasi:. Irrigation Sub-Project
b4b3b1b1b2h3h2h1BbHab4b3b1b1b2h3h2h1BbHaPlot Re-Bar
PlotMomentDATA PLOT MOMENT AND SHEAR FORCE DIAGRAMSection of Culvert1. Design Datab1=0.250mb2=1.750mb3=0.250mb4=1.750mh1=0.250mh2=2.000mh3=0.250mBb=4.000mHa=2.250m2. Result CalculationSide wall (left)MaMmaxMbSaSbNaNbMABMBACase 1-1.5900.640-1.0354.418-2.5194.5963.396-1.590-1.035Case 3-0.680-0.183-0.1830.222-0.0253.8452.645-0.680-0.183Side wall (right)MeMmaxMfSeSfNeNfMEFMFECase 1-1.0350.640-1.5902.519-4.4183.3964.596-1.035-1.590Case 3-0.183-0.183-0.6800.025-0.2222.6453.845-0.183-0.680Top slab (left)MbMmaxMcSbScNbNcMBCMCBCase 1-1.0350.582-1.3743.396-3.7352.5192.519-1.035-1.374Case 3-0.1830.832-1.7842.645-4.2460.0250.025-0.183-1.784Bottom slab (left)MaMmaxMdSaSdNaNdMADMDACase 1-1.5900.775-1.329-4.5964.3354.4184.418-1.590-1.329Case 3-0.6801.021-1.680-3.8454.8460.2220.222-0.680-1.680Partition wallMcMmaxMdScSdNcNdMCDMDCCase 10.0000.0000.0000.0000.0007.4718.6710.0000.000Case 30.0000.0000.0000.0000.0008.4929.6920.0000.000Top slab (right)McMmaxMeScSeNcNeMCEMECCase 1-1.3740.582-1.0353.735-3.3962.5192.519-1.374-1.035Case 3-1.7840.832-0.1834.246-2.6450.0250.025-1.784-0.183Bottom slab (right)MdMmaxMfSdSfNdNfMDFMFDCase 1-1.3290.775-1.590-4.3354.5964.4184.418-1.329-1.590Case 3-1.6801.021-0.680-4.8463.8450.2220.222-1.680-0.6803.Nama Bangunan:Double Box-Culvert Bba.20.a (2.4 m x 1.5 m)Lokasi:. Irrigation Sub-ProjectCase 1Mmaxxx1x2x3x4x5x6x7q1q2Mab =1.091-0.540.160.540.640.490.14-0.384.9581.208Mef =1.159-0.540.160.540.640.490.14-0.381.2084.958Mmaxy1y2y3y4y5y6y7q1Mbc =0.952-0.300.220.510.580.420.05-0.553.566Mad =1.029-0.580.150.600.770.670.28-0.384.466Ha =2.2500.2810.5630.8441.1251.4061.6881.969Bb =2.0000.2500.5000.7501.0001.2501.5001.750xx1x2x3x4x5x6x7Sab2.2503.091.890.83-0.10-0.91-1.58-2.11Case 2Mmaxxx1x2x3x4x5x6x7q1q2Mab =2.192-0.61-0.53-0.44-0.36-0.29-0.23-0.19-0.2150.435Mef =0.058-0.61-0.53-0.44-0.36-0.29-0.23-0.190.435-0.215Mmaxy1y2y3y4y5y6y7q1Mbc =0.7680.370.710.830.740.43-0.09-0.833.446Mad =0.8850.150.700.980.990.730.20-0.604.346Ha =2.2500.2810.5630.8441.1251.4061.6881.969Bb =2.0000.2500.5000.7501.0001.2501.5001.750xx1x2x3x4x5x6x7Sab2.2500.270.300.300.280.240.170.09
b4b3b1b1b2h3h2h1BbHaPlot Moment