A five-storey office has the general layout shown in Figure 1. In addition to the self-weight of the structural members, a superimposed dead load of 1 kPa should be allowedfor. The live load rating of the office floor is 3 kPa. The roof should be designed for asuperimposeddeadloadof0.5kPaandliveloadratingof1kPa.Windloadisassumedtobe resisted by the shear walls located at the edges of the building and need not beconsidered.
Designforatypicalfloorslab,beamandcolumn.Youarerequiredtoprovidetwooptions:
Reinforcedconcreteslabsupportedbyreinforcedconcrete beamsandcolumns
Reinforcedconcreteslabsupportedbysteelbeamsand columns.
Note:
Ifyoudecidetochangethebeamlayoutforthesteeloptions,thereinforcedconcrete slab should normally be re-designed.You are not required to redesignthe slab for this assignment but you should take into account the change ofthickness(andhenceself-weight)inyourdesign.
Theslabcannotbeassumedtoprovidecontinuouslateralrestraintsforthesteelbeams.
The design should be supported by calculations to ensure that the slab, beams andcolumns and their connections (when applicable) are adequate in resisting the designload. The calculations for any horizontal bracings, should they be used in the designoption ii, are not required for this assignment. However, the locations and details oftheirconnectionshould beincluded inthefinaldrawings
Summary
This report is the part 1 of the continuous assignment and contains three individual task. Themain objective of this report is to understand the application of Space Gass for calculating themaximumbending moment and reaction forces. Furthermore,for task01 threeconceptualdeigns of the building are sketched. For each design its merits and demerits are also discussedandoneparticulardesignedisfinalized.Fortask2,modelisvalidatedbycomparingthebendingmomentandreactionforcevaluesfromthecomputeranalysiswithresultsfrommanualcalculation.Fortask3,twoloadcasesareappliedtotheselectedlayoutandsummaryofmaximumbending momentandshearforceiscalculated.
TASK01:
Conceptualdesignofthebuilding
Brainstorm with your partner and come up with three sketches of possible columns andbeams layout. Discuss the merits and limitations of each sketch. You are to choose onelayoutforfurtheranalysis.
Layout01:ConventionalSlabdesignwithflatslabsystemSlab area:7500mmx10000mm
⇨10000/7500=1.33<2(Therefore,Twowayslab)
Advantages:
Economicalstructuresincethenumberofcolumnsareless.
Canbeconstructedinlesstime,sincenocomplexstructureisinvolved.
CarpetareaisincreasedduetolessnumberofcolumnsDisadvantages:
Designwillbedifficultsincetheslabsaretwoway.
Extracarewhiledesigningsinceloadisdistributedonlessnumberofcolumnandbeam.
Layout02:OneWaySlab(ConventionalSlabdesignwithflatslabsystem)
Slabarea:5000mmx10000mm
⇨10000/5000=2=2(Therefore,onewayslab)
Advantages:
Designscalculationwillbeeasier,sinceit’saonewayslabdesign
LesscomplexityInvolved,sincestructureissimple.Disadvantages:
Foronewayslabs,duetotheirhugedifferenceinlengths,loadisnottransferredtotheshorterbeams
Layout03(hollowcoreslabdesign)Slab area:3000 mmx4000mm
4000/3000=1.33<2(Therefore,Twowayslab)
Advantages:
reducingthecostandself-weight
easyconstruction
excellentsoundandfireresistance
Disadvantages:
hardtostrengthenandrepair
hardtoliftandmove
easytobedamageduringtransport
expensiveforsmallspan
Reasonforchoosing
Since,thedesignhasonewayslab,calculationswillbemudistributionwillbealsouniformandthicknessofthecolumnLayout01.Layout02willbealsomoreeconomicalthanlayoutare farlesserthan layout03. In layout03,slab isdesignedaexpensivetoconstructascomparedtotheotherlayouts.
cheasierthanothertwo.Loadwillnotbethatmuchhighunlike03,sincethenumberofcolumn
shollowslab,whichwillbevery
ChosenLayout:Layout03
TASK02:
Maximum+vemomentlocations18.75–10x=10
x=1.875metersand8.125meters(symmetrical)Maximum-vemoment at5meters
Maximum+vemoment=0.5×1.875 x18.75=17.58kN
ii.ValidationofmodelusingSpaceGas
Fig.1:Reactionforcesforthecontinuesbeam
Fig.2:ShearForceDiagramfor thecontinuousbeam
Fig.3:BendingMomentdiagramfor thecontinuous beam
Fig.4:Positionvsbendingmomentfrom0-5m
Fig.5:Positionvsbendingmomentfrom5-10m
TASK03:
DesignLoadCombinations:
DensityforReinforcedConcreteGrade32=2450kg/m2
Tributaryareaof5mactingalongthebeamandthicknessofslabis180mm=0.18mWeknowthat,
LiveLoadRatingforOfficeflooris3kPaSuperimposedDeadloadis1kPa
ForRoof,Deadload=0.5kPaandLiveLoad=1kPa
AssumingfloorsprovidesthemaximumbendingmomentandShearforceSelf-weightofslab=25.5kN/m3x0.18mx 1m=4.41kNm
Now,WSDL=1kPax1m=1kNm
Therefore,totalDeadload,G=4.41+1=5.51kNmWLL=3kPax1=3kNm
Therefore,Q=3kN/m
Asperassignmenttherearetwoloadcases:
Case1:1.2G+1.5Q=1.2x(5.51)+1.5 x(3)=10.992kN/m
Case2:0.8G=0.8 x(5.41)=4.328kN/m
Assumptions:
Designcalculationsforroofareexemptedforthepreliminarycalculations,sincetheloadson the roof aresmaller ascomparedtothefloors.
Thedesignlayouthasonewayslab,thereforeanalysisisdoneonlyonthecriticaldirectioni.e. (15m)of theslabs.
Formax.bendingmomentandshearforcecalculations,slabistreatedasabeamwith1meterwidth15meterlength and0.18mdepth.
Fordesigncalculations,externalsupportistakenaspinnedandinternalsupportasroller supports.
Forthisreport,Windloadisexemptedsincewindloadisassumedtoberesistedbytheshearwallslocatedattheedges ofthebuilding.
SPACEGASSANALYSIS:
Theconsideredbeamhas3spanstherefore,itwillhavetotal8combinationforapplicationofloadsonitsdifferentspans. Followingtableshowsthedifferentloadcombinations:
Figure6.Maximum&minimumbendingmomentfromallloadcombinations
Figure7.Maximum&minimumshearforcefromallloadcombinations
Maximum bending moment and shear force from Space Gass (For appendix for individualloadcase BMandSF diagrams)
MaximumBendingMoment
-
Combinations
MaximumPositive(kNm)
MaximumNegative(kNm)
1
21.95
-27.48
2
20.86
–30.26
3
25.42
-19.15
4
20.86
-30.26
5
24.26
-21.93
6
15.2
-19.15
7
24.26
-21.93
8
8.64
-10.82
Maximumcriticalnegativemomentsis-30.26kNmat5.0m&10mfromtheleftfromCombinations2andcombination4.
Maximumcriticalpositivemomentsis25.42kNmat2.08m&12.92mfromtheleftfromCombinations3.
MaximumShearForce
-
Combinations
MaximumPositive(kN)
MaximumNegative(kN)
1
32.98
-32.98
2
33.53
-30.26
3
31.31
-31.31
4
30.26
-33.53
5
23.09
-31.87
6
27.48
-27.48
7
31.87
-23.09
-
8
12.98
-12.98
MaximumcriticalnegativeShearforceis-33.53kNat5.0mfromtheleftfromCombinations4.
Maximumcriticalpositivemomentsisalso33.53kNat10mfromtheleftfromCombinations2.
SIMPLIFIEDCALCULATIONS:
UsingAS3600(2018)Cl.6.10.2.
Letthewidthofthebeambe300mm,
Fdistheuniformlydistributeddesignloadperunitlength=10.992(FromCase1,1.2G.+0.8Q)Therefore,
Ln1= 5000-300-300/2= 4550 mmLn2=5000-300/2-300/2=4700mmLn3=5000-300-300/2=4550mm
NegativeBendingMoments:(UsingAS3600(2018)Cl.6.10.2.2)
0mfromleft,
Atinteriorfacesofexteriorsupportsformembersbuiltintegrallywiththeirsupports
Forslabsandbeamswherethesupportisabeam
• =FdLn2/24
• =10.992x(4.55)2/24
• =-9.48kNm
5mfromleft,
Atthefirstinteriorsupport
morethantwospans
• =FdLn2/10
• =10.992x(4.77)2/10
• =-25.01kNm
10mfromleft,
Atthefirstinteriorsupport
morethantwospans
• =FdLn2/10
• =10.992x(4.77)2/10
• =-25.01kNm
15mfromleft,
Atinteriorfacesofexteriorsupportsformembersbuiltintegrallywiththeirsupports
Forslabsandbeamswherethesupportisabeam
• =FdLn2/24
• =10.992x(4.55)2/24
• =-9.48kNm
PositiveBendingMoments:(UsingAS3600(2018)Cl.6.10.2.3)
Inbothendspans,midspan1and midspan2
o=FdLn2/11
o=10.992 x(4.55)2/11
o=20.68kNm
IninteriorspansforDuctilityClassN
o=FdLn2/16
o=10.992 x(4.70)2/16
o=15.07kNm
ShearForceCalculations:(UsingAS3600(2018)Cl.6.10.2.4)
University of MelbourneCVEN90049StructuralTheoryandDesign2
DesignProject:Assignment01
Inanendspan,VA
Atthefaceoftheendsupport
VA=FdLn/2=10.992x(4.55)/2=25.01kN
Inanendspan,VB’
Atthefaceoftheinteriorsupport
VB’=1.15FdLn/2= 1.15 x10.992x (4.55)/2= 28.76kN
Ininteriorspans,VB
Atthefaceofsupports
VB=FdLn/2=10.992x(4.70)/2=25.83kN
Ininteriorspans,VC
Atthefaceofsupports
VC=FdLn/2=10.992x(4.70)/2=25.83kN
Ininteriorspans,VC’
Atthefaceofsupports
VC’==1.15FdLn/2=1.15×10.992x (4.55)/2=28.76kN
Inanendspan,VD
Atthefaceoftheendsupport
VD=FdLn/2=10.992x(4.55)/2=25.01kN
Inanendspan,VABmid-span
Atmid-span
VD=FdLn/7=10.992x(4.55)/7=7.14kN
Ininteriorspans,VBCmid-span
Atmid-span
VD=FdLn/8=10.992x(4.70)/8=6.45kN
COMPARISONOFSPACEGASSANDMANUALCALCULATION:
|
Category |
Position |
CalculatedValues |
Maximumvalues inSpaceGass(kNm) |
Scenario |
Criticallocation |
Difference |
Percentage |
|
Negativedesignmoment |
5 m and 10mfromleft(interiorsupport) |
-25.01kNm |
-30.26kNm |
Combi.2and Combi.04 |
5 m and 10mfromleft |
-5.25 |
17.3% |
|
Positivedesignmoment |
In both endspans, midspan1and midspan2 |
20.68kNm |
25.42kNm |
Combi.3 |
2.08mand 12.92 mfromleft |
4.74 |
18.6% |
|
Shearforce |
Va=Vd |
25.01kN |
23.65kN |
Combi.3 |
0mfromleft |
-1.36 |
5.4% |
University of MelbourneCVEN90049StructuralTheoryandDesign2
DesignProject:Assignment01
|
Vb=Vc |
25.83kN |
30.26kN |
Combi.2& Combi.4 |
5 m and 10mfromleft |
4.43 |
14.6% |
|
|
Vb’=Vc’ |
28.76kN |
33.53kN |
Combi.2& Combi.4 |
5 m and 10mfromleft |
4.77 |
14.2% |
Above table shows the discrepancies between the values obtained from space gass andManual Calculation. It is to be noted that percentage deviation between both is not morethan 20 percentage. Reason for this might be that the Space gass have considered all thepossible load combinations which can be made from case 1 and case 2. While manualcalculation depends upon the formulations given in code and it wold not have taken allthecombinationintoaccount.
Anotherreasonmightbe