Loads

Dead loadsImposed loads floor roofDeterminingloadpermandm2Wind

Structurestransmitloadsfromoneplacetoanother

Wheredoloadscomefrom

Deadloads-permanentandstationary

StructureitselfPlantandequipment

Someroughfigures(notethatvaluesaresubjecttovariationdependingonspecifcmaterialtype)Alsonotevaluesareforcesperunitvolumenotmassperunitvolume.

UnitWeightsofbaiscconstructionmaterials kN/m3Aluminium 24Brick 22Concrete 24Steel 70Timber 6

Precastconcretebeamlength

1.Calculateweightofbeamperunitlength.2.Calculatetotalweightofbeaminitis10.5m.

Firstcrosssectionalareaofbeam=

(0.6x0.25)-(0.4x0.15)=0.09m2

Fromtableunitweightofconcrete=24kN/m3

Weightperunitlength=0.09x24=2.16kN/m

Totalweight=2.16x10.5=22.68kN

Oftenwearedealingwithsheetmaterialsorweknowalayerthicknessoffloororroofbuildup.

Figureshereareperunitarea

Againwhenusingthesetypeofchartssomecareisneededtoensureyouhavethecorrectfigure,orthatitcorrespondswithyourdesign.

Unitweightofbasicsheetmaterials kN/m2Asphalt(19mm) 0.45Aluminiumroofsheeting 0.04Glass(singleglazing) 0.1Plasterboardandskim 0.15Raftersbattensroofingfelt 0.14Sand/cementscreed(25mm) 0.6Slates 0.6Timberfloorboards 0.15Plasteronwallface 0.3

CalculatethedeadloadinkN/m2ofthefollowingfloorbuildup:

Timberfloorboards40mmsand/cementscreed125mmreinforcedconcreteslab

timberfloorboards =0.15screed =0.6x40/25=0.96concreteslab =0.125x24=3.00

dead load /m2 =4.11kN/m2

ifwearedealingwithawallactingonabeamweareinterestedinloadperlinearunitofthebeam

Inthisexamplecalculatetheloadpermetreonthebeam.Thebuildupisadoubleglazedwindowonacavitywallof102.5mmbrickouterfaceand100mmplasteredlightweightblock-workwhichis12kN/m3.

brickwork =1.2x0.1025x22 =2.71blockwork =1.2x0.1x12 =1.44plaster =1.2x0.3 =0.36doubleglazing =2x1.3x0.1 =0.26

loadonbeam =4.77kN/m

Imposedloads-orliveloads,movableloadsthatactonthestructurewhenitisinuse.

People,furniture,cars,computersandmachineryareallimposedloads.

NormallyweconsiderimposedloadsasfloorandroofloadsTypicalfloorloads kN/m2Artgalleries 4.0Bankinghalls 3.0bars 5.0Carparks 2.5Classrooms 3.0Churches 3.0Computerlabs 3.5Dancehalls 5.0Factoryworkshop 5.0Foundaries 20.0Hotelbedrooms 2.0Offices(general) 2.5Offices(filing) 5.0Privatehouses 1.5Shops 4.0Theatres(fixedseats) 4.0

Ifabarshouldbedesignedwithliveloadof5.0kN/m2andifanaveragepersonis80kghowmanypeopleareexpectedtobestandinginonesquaremetreoffloor?

Forceexertedbyoneperson =80x9.81 =785NNumberofpeopleperm2 =5000/785 =6.4people/m2

equivalentlyifyourhouseisdesignedwith1.5kN/m2andthetotalareawas22m2howmanypeoplecouldyouinvitetoaparty?

Forceexertedbyoneperson =785NNumberofpeopleperm2 =1500/785 =1.9people/m2Totalnumberofpeopleatparty =1.9x22 =42andabit.

certaintypesofdancingcancausedynamiceffectsthatincreasetheeffectofload.

Calculatingimposedroofloads.

Whatyouneedtoknow:

1.Isaccesstotheroofprovided?(aloadofadjacentfloorareaisrequired)2.Predominantloadissnow.• whichisdependanton• geographicallocation• heightabovesealevel• shapeofroof• windthatredistributessnowintodrifts

EstimatinggroundsnowloadsinCanada.InfofromCanadianCryosphericInformationNetworkFindworstcasedepthandmultiplybydensity(kg/m3)and9.81

TablesinNationalBuildingCodeprovidefurtherdetails

InUKsnowloadvariesfrom0.3kN/m2onsouthcoastto3.0kN/m2inScotland

CalculatingasnowloadinCanada.NationalBuildingCodePart44.1.7.

S=Ss (CbxCwxCsxCa)+Sr

Snowloadperm2

groundsnowloadinkPa(kN/m2)

roofsnowloadfactor=0.8???

windexposurefactor

slopefactor

accumulationfactor

associatedrainload

NationalBuildingCodeofCanadaappendixcfortablesofclimaticinformation

windexposurefactor

is1.0butcanbereducedto0.75orinexposedareasnorthoftreelineto0.5 if buildingisanexposedlocationandexposedonallsides noobstructionsaroundbuilding noobstructionsonroofsuchasparapet snowcannotdriftontorooffromadjacentsurfaces

slopefactorbasedonroofangleaandsurfacetype. is1.0ifa<=30⁰ is(70⁰ -a)/40⁰whena>30⁰ is0ifa>70⁰

ifroofisaslipperysurface(wheresnowandiceslideoff)

slopefactor is1.0ifa<=15⁰ is(60⁰ -a)/40⁰whena>15⁰ is0ifa>60⁰

NationalBuildingCodeofCanadaappendixcfortablesofclimaticinformation

accumulationfactor

is1.0 exceptwhen forlargeflatroofswhen 1.2x[1-(30/l)2]butnotlessthan1.0forroofswithwindfactor=1.0 1.6x[1-(120/l)2]butnotlessthan1.0forroofswithwindfactor=0.75or0.5

w=smallerplandimension L=largerplandimension and lis2xw-(w2/L)inmetres

canbeassignedothervalueswhen: roofshapesarearched,curvedordomes snowloadsinvalleys snowdriftsfromanotherroof projectionsonadjacentroofs snowslidingordrainagefromadjacentroofs

Theresmore:

inrealityfullandpartialloadinghastobeconsidered

Inadditiontotheloadcalculationaboveroofsofslopelessthan15⁰andarchedorcurvedroofsmustbedesignedwithaccumulationfactor1.0ononeportionwhilehalfthatloadisappliedtotheremainder.

Calculatesnowloadonthisroofstructure

Whatisthesnowloadpermetrelengthoftruss?

Whatisthetotalsnowloadononerooftruss?

Whatistheloadpermetreonthesupportingwall?Assumethatloadsfromtrussesareevenlydistributed

CalculatesnowloadforHalifaxS

groundsnowHalifax=1.7

snowloadfactor=0.8

windexposurefactor=1.0

slopefactor=(70-40)/40=0.75

accumulationfactor=1.0

rainloadHalifax=0.5

S=1.7x(0.8x1x0.75x1)+0.5

S=1.52kN/m2

Snowloadperm2

groundsnowloadinkPa(kN/m2)

roofsnowloadfactor=0.8???

windexposurefactor

slopefactor

accumulationfactor

associatedrainload

S=Ss (CbxCwxCsxCa)+Sr

Trussesareat0.6mcentres

Sosnowloadpermetrelengthoftrussis:

0.6x1.52=0.9kN/m

Noteloadisverticalso1mdimen-sionismeasuredhorzontally

For7mtrussloadis

7x0.9=6.4kN

Loadpermonwall=1.52x3.5=5.32kN/m

Windloadsactnormal(orperpendicular)tobuildingsurfaces

windscancausepressureorsuction.

Forthisreasonbuildingstructuresmustresisthorizontalforcesaswellasverticalforces.

Inadditionsomelightweightstructurescanbesubjecttoupliftforcesfromthewindsoneedtobead-equatelyhelddown.

Windloadslikesnowloadsvarydependingon:

geographiclocation degreeofexposure buildingheightandsize buildingshape winddirectioninrelationstostructure positiveornegativepressuresinthebuilding

Fastermovingaircreateslowerpressure(bernoullieffect)asinplanewings.

Thesameprinciplecausesforcestoactonbuildingsurfaces.

Structureforresistingwindloads

Theseprinciplesshouldbewellunderstoodbynowifnot

Lookat: FrancisChing.BuildingConstructionIllustrated

EdwardAllen.Architect’sStudioCompanion

Forstructuraldesignitisoftennecessarytoconsiderseveralloadcasesduetothewindblowingfromdifferentdirections.

DesigningabuildinginHalifaxcalculatingwindloads.NationalBuildingCodeofCanadaPart44.1.8.

p=qxCexCgxCp

externalpressureactingstaticallyandnormaltosurface

referencevelocitypressure

exposurefactor

gusteffectfactor

externalpressurecoefficient

NationalBuildingCodeofCanadaappendixcfortablesofclimaticinformation

netpressureonasurfaceisthedifferencebetweeninternalandexternal

similartoexternalpressureinternalpressureiscalculatedaccordingtotheNBC

p=qxCexCgxCp

internalpressureactingstaticallyandnormaltosurface

referencevelocitypressure

exposurefactor

gusteffectfactor

internalpressurecoefficient

referenceveloctypressurethreeareshownintable1in10,1in30,1in100

theseareprobabilitiesofpressureoccuring

so1in10isusedforcladdingandstucturaldesignforvibrationanddeflection

1in30forstructuralstrength

post-disasterbuldingsusethe1in100pressurevalues.

exposurefactor

exposureincreasewithheightheightm exposurefactor>0and<=6 0.9>6and<=12 1.0>12and<=20 1.1>20and<=30 1.2>30and<=44 1.3>44and<=64 1.4andsoonmoreheightsgiveninnbc

gustfactor

1.0or2.0forinternalpressurestobefoundsomewhereinthe500pagesofappendixA!! we’lluse1.0fornow.

2.0forthebuildingasawholeandmainstructuralmembers

2.5forsmallelements

externalandinternalpressurecoefficients

againappendixAwe’lluse1.0fornow.

Forcesduetowindonsimplebuilding

externalpressurep=qxCexCgxCp

1in30yearPressureHalifax=0.52kPa(kN/m2)

Wallsbelow6msoexposurefactoris0.9

Gustfactor=2.0externaland1.0internal

Pressurecoefficient=1.0

externalp=0.52x0.9x2.0x1.0=0.936kN/m2

internalp=0.52x0.9x1.0x1.0=0.468kN/m2

so0.936-0.468=0.468kN/m2actingnormaltoverticalsurfaceswindward

and0.936+0.468=1.4kN/m2leeward

Forcesduetowindonsimplebuilding

externalpressurep=qxCexCgxCp

1in30yearPressureHalifax=0.52kPa(kN/m2)

Roofabove6msoexposurefactoris1.0

Gustfactor=2.0externaland1.0internal

Pressurecoefficient=1.0

externalp=0.52x1.0x2.0x1.0=1.04kN/m2

internalp=0.52x1.0x1.0x1.0=0.52kN/m2

1.04kN/m2actingnormaltoverticalsurfacesatrooflevel

normaltoroof1.04xSin(40)=0.67 windward=0.67-0.52=0.15kN/m2 leeward=0.67+0.52=1.19(suction)

Nextup:Acoupleofotherloadtypes(toknowabout)UniformandpointloadsSafetyfactorsCalculatingloadonbeamsLoadpaths

PinJointedstructures

A couple of other load types (to know about)Uniform and point loadsSafety factors Calculating load on beamsLoad paths

Hydrostatic pressure loads from soils and liquids

Increases linearly with depth.

Application of safety factors to loads

Loads discussed are realistic estimates of loads or characteristic loads

when checking ultimate strength characteristic loads are increased by multiplying by a safety factor.

The result is the design load.Safety factors

load combination dead imposed winddead and imposed 1.4 or 1.0 1.6 -dead and wind 1.4 or 1.0 - 1.4dead, imposed and wind 1.2 1.2 1.2

For example

to obtain the maximum compressive design in the support at Btwo load combinations should be checked and the larger value used

1.4 x dead + 1.6 imposed

or

1.2 x dead + 1.2 x imposed + 1.2 wind

to obtain maximum tensile design load in the support at A

we need to minimise the effect of the dead and imposed loads by using

1.0 x dead + 1.4 x wind

A B

wind

imposed roof

imposed floor

dead load

Point load (kN)

Uniformly distributed load (kN/m)

Calculating loads on beams

Example

Building type - officeFloor construction = 4.11kN/m2Perimeter wall construction = 4.77kN/m2self weight of beams = 0.6 kN/m

safety factors are 1.4 for dead load and 1.6 for live load to find design load

design load kN/m2floor 1.4 x 4.11 5.75wall 1.4 x 4.77 6.68beams 1.4 x 0.6 0.84imposed 1.6 x 2.5 4.00

total design floor load = 5.75 + 4.00 = 9.75kN/m2

B6

Beam B1 (8m span)supports a total width of 6m

load from floor = 9.75 x 6 = 58.50kN/mself weight of beam = 0.84 kN/m

design UDL = 59.34 kN/m

symmetry indicates reactions will be equal

reaction = (59.34 x 8) / 2 = 237.4 kN

beam B2 is the same as B1

B6

Beam B3 suports a 3m width of floor plus the perimeter wall

Load from floor = 3 x 9.75 = 29.25 kN/mload from wall = 6.68kN/mself weight of beam = 0.86kN/m

total design UDL = 36.79kN/m

symmetry indicates reactions will be equal

reaction = (36.9 x 8) /2 = 147.6kN

Beam B4 has same UDL as B1 and B2 but span is 6m

B6

Beam B6

Supports perimeter wall and a point load from the reaction of B2.

Load from wall = 6.68kN/mself weight of beam = 0.86kN/m

design UDL = 7.54kN/m

design point load = 237.4kN

Reaction from symmetry

= ((7.54 x 12) + 237.4) / 2 = 163.9kN

B6

Now work out reactions for B5

Load paths

Working through load path for a simple sign.