ISOLATED FOOTING DESIGN EXAMPLEAND EXCEL SHEET

Isolatedfootingdesignexamplewithstepbystepprocedureandisolatedfootingdesignexcelsheet(spreadsheet)isalsoprovidedforeasyandfastcalculation.

Learningdesignwithexamplesisalwaysthebestmethodoflearning.Stepbystepprocedureforstructuraldesignofisolatedfootingisdiscussedbelow:

IsolatedFootingDesignExample:

LetusconsideranisolatedfootingforanRCCcolumnofsize450mmx450mm.Loadsfromthiscolumntothefoundationare:

VerticalLoad:1000kN

UniaxialMoment:100kNm

Thesafebearingcapacity(SBC)ofsoilis300kN/m2.ThegradeofconcretetobeusedisM30andgradeofsteelisFe415.

StepbyStepProcedureofFootingDesign:

Step1:Determiningsizeoffooting:

Loadsonfootingconsistsofloadfromcolumn,selfweightoffootingandweightofsoilabovefooting.Forsimplicity,selfweightoffootingandweightofsoilonfootingisconsideredas10to15%oftheverticalload.

Loadoncolumn=1000kN

Extraloadat10%ofloadduetoselfweightofsoil=1000x10%=100kN

Therefore,totalloadP=1100kN.

Sizeoffootingtobedesignedcanbesquare,rectangularorcircularinplan.Herewewillconsidersquareisolatedfooting.

Therefore,lengthoffooting(L)=Widthoffooting(B)

Thereforeareaoffootingrequired=

=1100/300=3.67m

ProvideLengthandwidthoffooting=2m

Areaoffooting=2x2=4m

Nowthepressureonisolatedfootingiscalculatedas

Whencalculated,pmax=325kN/m

pmin=175kN/m

ButpmaxisgreaterthanSBCofsoil,soweneedtorevisethesizeoffootingsothatPmaxisbelow300kN/m .

Considerwidthandlengthoffooting=L=B=2.25m

Now,pmax=250.21kN/m2(OK)

andpmin=144.86kN/m2>0(OK)

Hence,factoredupwardpressureofsoil=pumax=375.315kN/m2

pumin=217.29kN/m2

Further,averagepressureatthecenterofthefootingisgivenbyPu,avg=296.3

2

2

2

2

2

kN/m2

and,factoredload,Pu=1500kN,factoreduniaxialmoment,Mu=150kNm.

Step2:Twowayshear

Assumeanuniformoverallthicknessoffooting,D=500mm

Assuming16mmdiameterbarsformainsteel,effectivedepthoffootingdis

d=500508=452mm

Thecriticalsectionforthetwowayshearorpunchingshearoccursatadistanceofd/2fromthefaceofthecolumn(Fig.1),whereaandbarethedimensionsofthecolumn.

Fig1:CriticalsectionforTwoWayShear(PunchingShear)

Hence,punchingareaoffooting=(a+d) =(0.45+0.442) =0.796m

wherea=b=sideofcolumn

2 2 2

Punchingshearforce=Factoredload(Factoredaveragepressurexpunchingareaoffooting)

=1500(296.3x0.0.796)

=1264.245kN

Perimeteralongthecriticalsection=4(a+d)=4(450+442)=3568mm

Therefore,nominalshearstressinpunchingorpunchingshearstress iscalculatedasbelow:

=1264.245x1000/(3568442)=0.802N/mm2

Allowableshearstress=

where =1.369N/mm2

= =1

therefore,allowableshearstress=11.369=1.369N/mm

Sincethepunchingshearstress(0.802N/mm2)islessthantheallowableshearstress(1.369N/mm2),theassumedthicknessissufficienttoresistthepunchingshearforce.Hence,theassumedthicknessoffootingD=500mmissufficient.Pleasenote,thereismuchdifferencebetweenallowableandactualvaluesofshearstress,sodepthoffootingcanberevisedandreduced.Forourexample,wewillcontinuetouseD=500mm.

2

Step3:Designforflexure:

Thecriticalsectionforflexureoccursatthefaceofthecolumn(Fig.2).

Fig.2Criticalsectionforflexure

Theprojectionoffootingbeyondthecolumnfaceistreatedasacantileverslabsubjectedtofactoredupwardpressureofsoil.

Factoredmaximumupwardpressureofsoil,pu,max=375.315kN/m2

Factoredupwardpressureofsoilatcriticalsection,pu=312.1kN/m2

Projectionoffootingbeyondthecolumnface,l=(2250450)/2=900mm

Bendingmomentatthecriticalsectioninthefootingisgivenby:

Mu=TotalforceXDistancefromthecriticalsection

Consideringuniformsoilpressureof375.315,Mu=180kN/m

0.92

fromSP16,percentageofreinforcementcanbefoundforM30concrete,fe415steel

2

forabove pt=0.265%

Ast=ptxbxd

considering1mwidefooting,Astrequired=1171.1mm /mwidth

Provide16diabar@140mmc/c

Repeatthisexerciseforotherdirectionaswell.Since,uniformbasepressureisassumed,anditisasquarefooting,MuandAstforotherdirectionwillbesame.

Step4:CheckforOneWayShear:

Thecriticalsectionforonewayshearoccursatadistanceofdfromthefaceofthecolumn.

Factoredmaximumupwardpressureofsoil,pu,max=375.315kN/m2

Factoredupwardpressureofsoilatcriticalsection,pu=375.315kN/m2

Forthecantileverslab,totalShearForcealongcriticalsectionconsideringtheentirewidthBis

Vu=TotalForceX(ld)XB

=375.315X(0.90.442)X2=343.8kN

Nominalshearstress=Vu/(Bxd)=0.346N/mm

For,pt=0.265,andM30,allowableshearforcefromTable19,IS456isgreaterthan0.346N/mm

Therefore,thefoundationissafeinonewayshear.

Step5:Checkfordevelopmentlength

2

2

2

Sufficientdevelopmentlengthshouldbeavailableforthereinforcementfromthecriticalsection.

Here,thecriticalsectionconsideredforLdisthatofflexure.

Thedevelopmentlengthfor16mmdiameterbarsisgivenby

Ld=47xdiameterofbar=47x16=752mm.

Providing60mmsidecover,thetotallengthavailablefromthecriticalsectionis

0.5x(La)60=0.5x(2250450)60=840>Ld,HenceO.K.

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