Beam Design 1

Sheet1ReferenceCalculationOutputBeam DesignAfter doing initial member sizing and frame analysis, final layoutdrawings were prepared. Design of First floor beam along grid 7 isillustrated here.Design of beam 7 (A-K)Calculation of loadingsBS8110Loadings of slab panels were transfered to adjacent beams usingT 3.16shear force co-efficients. When this coeffients are being used,distributionof load on beam should be limited to 0.75 times of beam span. Forsimplyfy the analysis this has been taken for full span length of the beam.Calculated shear force coeffients are shown in the following figure 7-1.Dead loads and Imposed loads:Specimen calculation for Grid 7- First floor level:Span K-J:Dead LoadWeight of slab=10.9kN/m(0.4x4x3.6)+(0.36x4x3.6)Weight of beam=2.7 kN/m(0.3x0.375x24)No wall load is present at this spanWeight of finishes=3.0 kN/m(0.4x4x1)+(0.36x4x1)Total dead load=16.6 kN/mProject - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page01 of 28Imposed loadImposed load for general office area=3.5 kN/m2Imposed load for span K-J=10.6kN/m(0.4x4x3.5)+ (0.36x4x3.5)Simillary for other spans also dead loads and imposed loads werecalculated. Specimen calculations for beam 7(K-A) of First floorare attached here with in figure 7-2.After calculating the loads for each span, moments were calculatedfor three critical load combinations.Load combination 1:All spans maximum loadedDesign load (W1)=(1.4x16.6 + 1.6x10.6)=40kN/mDesign load (W2)=(1.4x28.1+ 1.6x17)=66 kN/mDesign load (W3)=(1.4x16 + 1.6x10)=38 kN/mDesign load (W4)=(1.4x21+ 1.6x2)=33 kN/mDesign load (W5)=(1.4x21.6 + 1.6x4.76)=38 kN/mDesign load (P1)=(1.4x127 + 1.6x50)=258 kNDesign load (P2)=(1.4x77 + 1.6x15)=132 kNFrame is statically indeterminate. Hence moment distribution method isused to find bending moments and shear forces.Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page02 of 28Calculation of fixed end moments(FEM)BE=(FEM)EB=wl2=40x42=53 kNm1212(FEM)EH=(FEM)HE=wl2=66x7.52=309 kNm1212(FEM)HL=(FEM)LH=wl2=38x42=51 kNm1212(FEM)LO=(FEM)OL=wl2 + pl=33x4.52+258 x4.512101210=172kNm(FEM)OR=(FEM)RO=wl2 + pl=38x4.12+132 x4.112141214=92kNmCalculation of Distribution factorsFinding second moment of inertia (I):For beam 300x550 (I)=300x(550)3=4.16x109mm412For beam 300x400 (I)=300x(400)3=1.6x109mm412For column 500x500 (I)=500x(500)3=5.2x109mm412For column 300x300 (I)=300x(300)3=0.67x109mm412(DF)BE=KBEKBA+KBC+KBE=(4EI1/L1)(4EI2/L2) +(4EI2/L2)+(4EI1/L1)=(4.16x109/4000)(5.2x109/3500) +(5.2x109/3500) +(4.16x109/4000)=0.259Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page03 of 28(DF)EB=KEBKED+KEF+KEB+KEH=(4EI1/L1)(4EI2/L2) +(4EI2/L2)+(4EI1/L1)+(4EI3/L3)=(4.16x109/4000)((2x5.2/3500) +(4.16/4000) +(4.16/7500) )x109=0.227(DF)EH=(4.16x109/7500)((2x5.2/3500) +(4.16/4000) +(4.16/7500) )x109=0.121(DF)HE=(4.16x109/7500)((2x5.2/3500) +(4.16/4000) +(4.16/7500) )x109=0.121(DF)HL=(4.16x109/4000)((2x5.2/3500) +(4.16/4000) +(4.16/7500) )x109=0.227(DF)LH=(4.16x109/4000)((2x5.2/3500) +(4.16/4000) +(4.16/4500) )x109=0.21(DF)LO=(4.16x109/4500)((2x5.2/3500) +(4.16/4000) +(4.16/4500) )x109=0.187(DF)OL=(4.16x109/4500)((2x5.2/3500) +(4.16/4500) +(1.6/4100) )x109=0.215(DF)OR=(1.6x109/4100)((2x5.2/3500) +(4.16/4500)+(1.6/4100))=0.091Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page04 of 28(DF)RO=(1.6x109/4100)((2x0.67/3500) +(1.6/4100)=0.504Moment distribution table for support moments of load arrangement 1is as shown in the figure 7-3. To findout span moments , it is required todraw free bending moment diagrams for each span seperately.Consider span B-ER1=40x4=80 kN2Mx=R1.x - W1.x2/2=80x - 20x2x (m)M (kNm)0016028036040Consider span E-HR2=66x7.5=247.5 kN2Mx=R2.x - W2.x2/2=247.5x - 33x2Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page05 of 28x (m)M (kNm)001214.523633445.544625412.562977115.57.50Consider span H-LR3=38x4=76 kN2Mx=R3.x - W3.x2/2=76x - 19x2x (m)M (kNm)0015727635740Consider span L-OR4+R5=33x4.5+258=406.5kNR4=33x4.52 /2+258x2.5=217.6kN4.5x2Mx=R4.x - W4.x2/2-P1.(x-2)=(-40.4)x - 16.5x2+516Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page06 of 28x (m)M (kNm)001201.12369.23246.3490.44.50Consider span O-RR6+ R7=38x4.1+132=287.8 kNR6=38x4.12 /2+132x2.5=158.4kN4.1x1.6Mx=R6.x - W5.x2/2-P2.(x-1.6)=158.4x - 19x2-132x+211.2=26.4x -19x2+211.2x (m)M (kNm)001139.421883119.44.10Bending moment diagram for the beam 7(A-K) was obtained bycombineing support moments with free bending moments. Results areshown in the figure 7-6 and Table7-1.Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page07 of 28Load Combination II:Alternate span loadingDesign load (W1)=1.0x16.6=16.6 kN/mDesign load (W2)=(1.4x28.1+ 1.6x17)=66 kN/mDesign load (W3)=1.0x16=16 kN/mDesign load (W4)=(1.4x21+ 1.6x2)=33 kN/mDesign load (W5)=1.0x21.6=21.6 kN/mDesign load (P1)=(1.4x127 + 1.6x50)=258 kNDesign load (P2)=(1.0x77)=77 kNFrame is statically indeterminate. Hence moment distribution method isused to find bending moments and shear forces.Calculation of fixed end moments(FEM)BE=(FEM)EB=wl2=16.6x42=22 kNm1212(FEM)EH=(FEM)HE=wl2=66x7.52=309 kNm1212(FEM)HL=(FEM)LH=wl2=16x42=21 kNm1212(FEM)LO=(FEM)OL=wl2 + pl=33x4.52+258 x4.512101210=172kNm(FEM)OR=(FEM)RO=wl2 + pl=21.6x4.12+77 x4.112141214=53kNmProject - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page08 of 28Distribution factors are unchanged and could be used the valuescalculated for load combination 1. Moment distribution table is asshown in figure 7-4.Bending moment diagram for each span seperatly, were drawn to find outspan moments.Consider span B-ER1=16.6x4=33.2 kN2Mx=R1.x - W1.x2/2=33.2x - 8.3x2x (m)M (kNm)00124.9233.2324.940Consider span E-HR2=66x7.5=247.5 kN2Mx=R2.x - W2.x2/2=247.5x - 33x2Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page09 of 28x (m)M (kNm)001214.523633445.544625412.562977115.57.50Consider span H-LR3=16x4=32 kN2Mx=R3.x - W3.x2/2=32x -8x2x (m)M (kNm)0012423232440Consider span L-OR4+R5=33x4.5+258=406.5kNR4=33x4.52 /2+258x2.5=217.6kN4.5x2Mx=R4.x - W4.x2/2-P1.(x-2)=(-40.4)x - 16.5x2+516Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page10 of 28x (m)M (kNm)001201.12369.23246.3490.44.50Consider span O-RR6+ R7=21.6x4.1+77=165.56 kNR6=21.6x4.12 /2+77x2.5=91.2kN4.1x1.6Mx=R6.x - W5.x2/2-P2.(x-1.6)=91.2x -10.8x2-77x+123.2=14.2x -10.8x2+123.2x (m)M (kNm)00180.42108.4368.64.10Bending moment diagram for the beam 7(A-K) was obtained bycombining support moments with free bending moments. Results areshown in the figure 7-6 and Table 7-1.Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page11 of 28Load Combination III:Alternate span loadingDesign load (W1)=(1.4x16.6 + 1.6x10.6)=40 kN/mDesign load (W2)=1.0x28.1=28.1 kN/mDesign load (W3)=(1.4x16 + 1.6x10)=38 kN/mDesign load (W4)=1.0x21=21 kN/mDesign load (W5)=(1.4x21.6 + 1.6x4.76)=38 kN/mDesign load (P1)=1.0x127=127 kNDesign load (P2)=(1.4x77+1.6x15)=132 kNFrame is statically indeterminate. Hence moment distribution method isused to find bending moments and shear forces.Calculation of fixed end moments(FEM)BE=(FEM)EB=wl2=40x42=53 kNm1212(FEM)EH=(FEM)HE=wl2=28.1x7.52=132 kNm1212(FEM)HL=(FEM)LH=wl2=38x42=51 kNm1212(FEM)LO=(FEM)OL=wl2 + pl=21x4.52+127x4.512101210=93kNmProject - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page12 of 28(FEM)OR=(FEM)RO=wl2 + pl=38x4.12+132 x4.112141214=92kNmDistribution factors are unchanged and could be used the valuescalculated for load combination 1. Moment distribution table is asshown in figure 7-5.Consider span B-ER1=40x4=80 kN2Mx=R1.x - W1.x2/2=80x -20x2x (m)M (kNm)0016028036040Consider span E-HR2=28.1x7.5=105.38 kN2Mx=R2.x - W2.x2/2=105.38x - 14.05x2Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page13 of 28x (m)M (kNm)00191.332154.563189.694196.725175.656126.48749.217.50Consider span H-LR3=38x4=76 kN2Mx=R3.x - W3.x2/2=76x - 19x2x (m)M (kNm)0015727635740Consider span L-OR4+R5=21x4.5+127=221.5kNR4=21x4.52 /2+127x2.5=117.8kN4.5x2Mx=R4.x - W4.x2/2-P1.(x-2)=(-9.2)x -10.5x2+254Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page14 of 28x (m)M (kNm)001107.32193.63131.9449.24.50Consider span O-RR6+ R7=38x4.1+132=287.8 kNR6=38x4.12 /2+132x2.5=158.4kN4.1x1.6Mx=R6.x - W5.x2/2-P2.(x-1.6)=158.4x - 19x2-132x+211.2=26.4x -19x2+211.2x (m)M (kNm)001139.421883119.44.10Bending moment diagram for the beam 7(A-K) was obtained bycombining support moments with free bending moments. Results areshown in the figure 7-6 and Table 7-1.Project - DahampayaFeatureDesign of beamsDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodeD-676Page15 of 28ProjectFeatureDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodePageofProjectFeatureDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodePageofProjectFeatureDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodePageofProjectFeatureDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodePageofProjectFeatureDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodePageof

NF13.2770.785FHigh Flood Level

lood LevelExisting water LevelBed Rock LevelP2P1P3Pw18000 (approx.)4000 (approx.)4000 (approx.)18000/3 =6000(approx.)P2P1P3PwP3P4P2600 f600 f600 f600 f400x800400x800High Flood Level

lood LevelExisting water LevelPoint of virtual fixityW2W1CTba400x800=1.5x9.81x{42/(2x9.81)}=12.00 KN/m279 KNm340 KNmRiver bed Level400800Lateral load on the beam due to water presureNF13.2770.785FHigh Flood Level














































































































lood LevelExisting water LevelPoint of virtual fixityW2W1CTba400x800=1.5x9.81x{42/(2x9.81)}=12.00 KN/m279 KNm340 KNmRiver bed Level400800Lateral load on the beam due to water presure4m7.5m4m4.5m4.1m300x550300x550300x550300x550300x4000500x500500x500500x500500x500500x500300x300300x300500x500500x500500x500500x500500x5004.0mR1R1W1=40KN/mx7.5mR2W2=66kN/mxR24.0mR3R3W3=38kN/mxW1W2W3W4W5P14m7.5m4m4.5m4.1mADGKNQFJMPSCBEHLOR4.0mR1R1W1=16.6kN/mx7.5mR2W2=66kN/mxR24.0mR3R3W3=16kN/mx4.0mR1R1W1=40KN/mx7.5mR2W2=28.1kN/mxR24.0mR3R3W3=38kN/mx4m7.5m4m4.5m4.1mW1W2W3W4W5ADGKNQFJMPSCBEHLORP1P24m7.5m4m4.5m4.1mADGKNQFJMPSCBEHLORW1W2W3W4W5P1P24.5mR4R5W4=33kN/mxP1=258kN2.5m4.1mR6R7W5=38kN/mxP2=132kN2.5m4.5mR4R5W4=33kN/mxP1=258kN2.5m4.1mR6R7W5=21.6kN/mxP2=77kN2.5m4.5mR4R5W4=21kN/mxP1=127kN2.5m4.1mR6R7W5=38kN/mxP2=132kN2.5m

Sheet2ReferenceCalculationOutputsProjectFeatureDECDESIGN UNITDesignedDateEPC DIVISIONCheckedDateCENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)Job CodePageof

NF13.2770.785FHigh Flood Level
















































































































lood LevelExisting water LevelPoint of virtual fixityW2W1CTba400x800=1.5x9.81x{42/(2x9.81)}=12.00 KN/m279 KNm340 KNmRiver bed Level400800Lateral load on the beam due to water presure

Sheet3

Beam Design 1

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