Design and Detailing of steel in Combined Footings

Dr. M.C. NATARAJA

Design of combined footing –Slab and Beam type- Problem 2

2 Design a rectangular combined footing with a centralbeam for supporting two columns 400x400 mm in size tocarry a load of 1000kN each. Center to center distancebetween the columns is 3.5m. The projection of thefooting on either side of the column with respect tocenter is 1m. Safe bearing capacity of the soil can betaken as 190kN/m2. Use M20 concrete and Fe-415steel.

Draw to a suitable scale the following1. The longitudinal sectional elevation2. Transverse section at the left face of the heavier

column3. Plan of the footing

Marks 60

Solution: Datafck = 20Nlmm2,fy= 415mm2,f b (SBC)= l90 kN/m2,Column A = 400 mm x 400 mm,Column B = 400 mm x 400 mm,c/c spacing of columns = 3.5,PA = 1000 kN and PB = 1000 kN

Required: To design combined footing withcentral beam joining the two columns.

Ultimate loadsPuA= 1.5 x 1000 = 1500 kN, PuB = 1.5 x 1000 = 1500 kN

Proportioning of base sizeWorking load carried by column A = PA = 1000 kNWorking load carried by column B = PB = 1000 kNSelf weight of footing 10 % x (PA + PB) = 200 kNTotal working load = 2200 kNRequired area of footing= Af = Total load /SBC=2200/190 = 11.57 m2

Length of the footing Lf = 3.5 +1 +1 =5.5mRequired width of footing = b= Af /Lf = 11.57/5.5 = 2.1mProvide footing of size 5.5 x 2.1 m

Total load from columns = P = (1000 + 1000) = 2000 kN.Upward intensity of soil pressure=Column loads/A provided= P/Af= 2000/5.5 x2.1 =173.16 kN/m2< SBC

Design of slab:Intensity of Upward pressure = p =173.16 kN/m2

Consider one meter width of the slab (b=1m)Load per m run of slab at service = 173.16 x 1 = 173.16 kN/mCantilever projection of the slab

=1050 - 400/2 = 850mmMaximum ultimate moment = 173.16 x 0.8502/2 = 62.55 kN-m.(Working condition)

To find steel

Mu/bd2 =1.5 x 62.15 x106/1000 x 2252=1.84< 2.76, URS

Pt=0.584 %Ast = 1314 mm2

Use #20 mm diameter bar at spacing= 1000 x 314 / 1314 =238.96 mm say 230 mm c/c

#20@230

Area provided =1000 x 314 / 230 = 1365 mm2. Hence safe.This steel is required for the entire length of the footing.

Check the depth from one-way shearconsiderations

Design shear force=Vu=1.5x173.16x(0.850-0.225)= 162.33 kNNominal shear stress = τv= Vu/bd =162330/(1000x225) =0.72 MPa

Permissible shear stresspt = 100 x 1365 /(1000 x 225 ) = 0.607 %, τuc = 0.51 N/mm2

Value of k for 300 mm thick slab =1Permissible shear stress = 1 x 0.51= 0.51 N/mm2

τuc < τv and hence unsafe.

The depth may be increased to 400 mm so thatd = 325mm

Mu/bd2=1.5 x 62.15 x106/1000x3252 = 0.883< 2.76, URSpt=0.26 %, Ast = 845 mm2

Use #16 mm diameter bar at spacing = 1000 x 201 / 845=237.8 mm, say 230 mm c/c

Area provided =1000 x 201 / 230 = 874 mm2.

#16@230

Check the depth from one - way shearconsiderations

Design shear force=Vu=1.5x173.16 x (0.850-0.325)= 136.36 kNNominal shear stress = τv = Vu/bd=136360/(1000x325) = 0.42 MPaPermissible shear stresspt = 100 x 874/(1000 x 325 ) = 0.269 %, τuc = 0.38 N/mm2

Value of k for 400 mm thick slab =1Permissible shear stress = 1 x 0.38= 0.38 N/mm2

Again τuc < τv and hence slightly unsafe.However provide steel at closure spacing, #16 @150Ast=201 x 1000/150 =1340 mm2 and pt =0.41% and henceτuc=0.45 MPa and safe.

Design of Longitudinal BeamThe net upward soil pressure per meter length of the beamunder service.= w = 173.16 x 2.1 = 363.64 kN/m

Shear Force and Bending Moment at service conditionVAC=363.64 x 1= 363.14 kN, VA =1000-363.14 = 636.36 kNVBD = 363.14 kN, VBA= 636.36 kN

Point of zero shear is at the center of footing at L/2,Maximum B.M. occurs at EME = 363.64 x 2.752 / 2 - 1000 (2.75 - 1) = -375 kN.m

Bending moment under column A= MA = 363.64 x 12 / 2 = 181.82 kN.m

Similarly bending moment under column B= MB = 181.82 kN-m

Let the point of contra flexure be at a distance x from C

Then, Mx= 363.63x2/2 – 1000(x-1) = 0Therefore x = 1.30 m and 4.2m from C

Depth of beam from B.M. Considerations:

The width of beam=400 mm. Assume the beam as rectangular at the center of span where the moment is maximum, we have,

d =√ (375 x 1.5 x 106/ (2.76 x 400)) = 713.8 mm

Provide total depth of 800 mm. Assuming two rows of bars at an effective cover of 75 mm, the effective depth provided = d= 800-75 = 725 mm.

Check for two way shear

The column B can punch through the footing only if it shears against the depth of the beam along its two opposite edges, and along the depth of the slab on the remaining two edges. The critical section for two-way shear is taken at distance d/2 (i.e. 725/2 mm) from the face of the column. Therefore, the critical section will be taken at a distance half the effective depth of the slab (ds/2) on the other side as shown in Fig

In this case b=D= 400 mm, d b =725 mm, ds= 325 mm

Area resisting two - way shear= 2(b x db + ds x ds) + 2 (D + db)ds= 2 (400 x 725+ 325 x 325) + 2(400+725) 325 = 1522500 mm2

Design shear=Pud= column load – Wu x area at critical section= 1500 – 173.16 x1.5 x(b + ds) x (D + db)=1500-173.16 x 1.5 x (0.400+0.325) x (0.400+ 0.725)=1288.14 kN

τv=Pud/bod= 1288.14x1000/1522500=0.845 MPa

Shear stress resisted by concrete = τuc=τucx Kswhere, τuc = 0.25 √ f ck= 0.25√ 20 = 1.11 N/mm2

Ks = 0.5 + d / D = 0.5 + 400/400 = 1.5 ≠ 1 Hence Ks = 1

τuc = 1 x 1.11 = 1.11 N/mm2

Therefore safe

Area of ReinforcementCantilever portion BD and ACLength of cantilever from the face of column = 0.8 m.Ultimate moment at the face of column= 363.64x1.5 x 0.82 / 2 = 177.53 kN-m

Mumax =2.76 x400 x7252x10 -6= 580.29 kN.m >177.53 kN-mTherefore Section is singly reinforced.Mu/bd2 =177.53x106/(400x7252) =0.844 <2.76, URSpt=0.248%, A st =719.2 mm2

Provide 4 - 16 mm at bottom face, Area provided = 804 mm2,pt=0.278%Ld = 47x16 =752 mm

Curtailment:

All bottom bars will be continued up to the end ofcantilever for both columns. If required two bottombars of 2-16mm will be curtailed at a distance d (=725 mm) from the point of contra flexure in theportion BE as shown in figure.

Region AB between points of contra flexures

The beam acts as an isolated T- beam.bf = [ L o /( L o /b + 4 )] + b w, where,

L o = 4.2-1.3=2.9 m = 2900 mmb= actual width of flange = 2100 mm, b w = 400 mmbf= [2900/(2900/2100) + 4] + 400 =938.9mm < 2100mmDf = 400 mm,Mu= 1.5 x 375=562.5 kN-m

Moment of resistance Muf of a beam for x u=D f is :

(Muf)= [0.36 x 20 x 938.9 x 400 (725 - 0.42x400)] x10-6

= 1506 kN.m > Mu ( = 562.5 kN-m)

Therefore Xu<Df

Mu=0.87fyAst(d - fyAst/fckbf), Ast= 2334 mm2

Provide 4 bars of 25 mm and 2 bars of 16 mm,

Area provided = 2354 mm2 >2334 mm2

pt = 100 x 2334/(400x725) = 0.805 %

Curtailment:

Curtailment can be done as explained in the previous problem. However extend all bars up to a distance ‘d’ from the point of contra flexure i.e up to 225 mm from the outer faces of the columns. Extend 2-16mm only up to the end of the footing.

Design of shear reinforcementPortion between column i.e. ABMax. shear force at A or B=Vumax=1.5 x 636.36 =954.54 kNSF at the point of contra flexure

= 954.54-1.5x 363.64x0.3 = 790.9 kNτv=790900/(400x725) =2.73 MPa < τc,max.(2.8 MPa)Area of steel available = 2354 mm2, 0.805 %τc=0.59MPa, τv > τcDesign shear reinforcement is required.Using 12 mm diameter 4 - legged stirrups,Spacing = 0.87 x 415 x (4 x 113) /(2.73-0.59)x400

=190.6 mm say 190 mm c/cZone of shear reinforcements is between τv to τc

Cantilever portion BD and ACVumax = 363.64 x 1.5 = 545.45 kN,Shear from face at distance d= VuD = 545.45-363.64 x1.5(0.400 / 2 + 0.725) = 40.90 kNτv=40900/(400x 725) =0.14 MPa < τc,max. This is very small

Steel at this section is 4 – 16 mm, 804 mm2, pt=0.278%τc =0.38N/mm2 (IS:456-2000). No shear steel is needed.Provide minimum steel.Using 12 mm diameter 2- legged stirrups,Spacing = 0.87 x 415 x (2 x 113) /(0.4x400)=509.9 mm say 300 mm c/c