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 central

beam for supporting two columns 400x400 mm in size to carry a load of 1000kN each. Center to center distance between the columns is 3.5m. The projection of the footing on either side of the column with respect to center is 1m. Safe bearing capacity of the soil can be taken as 190kN/m2. Use M20 concrete and Fe-415 steel.

Draw to a suitable scale the following

1. The longitudinal sectional elevation

2. Transverse section at the left face of the heavier column

3. Plan of the footing

Marks 60

Solution: Data fck = 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 with central beam joining the two columns.

Ultimate loads PuA= 1.5 x 1000 = 1500 kN, PuB = 1.5 x 1000 = 1500 kN

Proportioning of base size

Working load carried by column A = PA = 1000 kN Working load carried by column B = PB = 1000 kN Self weight of footing 10 % x (PA + PB) = 200 kN Total working load = 2200 kN Required area of footing = Af = Total load /SBC=2200/190 = 11.57 m2

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

For uniform pressure distribution the C.G. of the footing should coincide with the C.G. of column loads. As the footing and columns loads are symmetrical, this condition is satisfied.

1000 kN 1000 kN

3.5 m 1 m 1 m

x R p=173.16 kN/m2

(safe) w=363.64 kN/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/m Cantilever projection of the slab =1050 - 400/2 = 850mm Maximum ultimate moment = 173.16 x 0.8502/2 = 62.55 kN-m. (Working condition)

For M20 and Fe 415, Q u max = 2.76 N/mm2 Required effective depth = √ (62.15 x1.5 x 106/(2.76 x 1000)) = 184.28 mm Using 20 mm bars, effective cover =20/2 +50 say 75 mm Required total depth = 184.28 + 75 = 259.4 mm. However provide 300 mm from shear consideration as well. Provided effective depth = d = 300-75 = 225 mm

p=173.16 kN/m2

1m

850mm

2.1m

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 shear

considerations

Design shear force=Vu=1.5x173.16x(0.850-0.225)= 162.33 kN Nominal shear stress = τv

= Vu/bd =162330/(1000x225) =0.72 MPa Permissible shear stress pt = 100 x 1365 /(1000 x 225 ) = 0.607 %, τuc = 0.51 N/mm2 Value of k for 300 mm thick slab =1 Permissible shear stress = 1 x 0.51= 0.51 N/mm2

τuc < τv and hence unsafe.

p=173.16 kN/m2

1m

850mm

2.1m

Critical s/n for shear d

The depth may be increased to 400 mm so that d = 325mm Mu/bd2=1.5 x 62.15 x106/1000x3252 = 0.883< 2.76, URS pt=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 shear

considerations

Design shear force=Vu=1.5x173.16 x (0.850-0.325) = 136.36 kN Nominal shear stress = τv = Vu/bd =136360/(1000x325) = 0.42 MPa Permissible shear stress pt = 100 x 874/(1000 x 325 ) = 0.269 %, τuc = 0.38 N/mm2 Value of k for 400 mm thick slab =1 Permissible shear stress = 1 x 0.38= 0.38 N/mm2

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

Check for development length

Ldt= 47 times diameter = 47x16=768 mm Available length of bar = 850 - 25 =825mm > 768 mm and hence safe.

Transverse reinforcement

Required A st = 0.12 bD / 100 = 0.12 x 1000 x 400/100 = 480mm2

Using 10 mm bars, s= 1000 x 79 / 480 = 164.58 mm Provide distribution steel of #10 mm at 160 mm c/c

p=173.16 kN/m2

1m

850mm

2.1m

Design of Longitudinal Beam

The net upward soil pressure per meter length of the beam under service. = w = 173.16 x 2.1 = 363.64 kN/m Shear Force and Bending Moment at service condition

VAC=363.64 x 1= 363.14 kN, VA =1000-363.14 = 636.36 kN VBD = 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 E ME = 363.64 x 2.752 / 2 - 1000 (2.75 - 1) = -375 kN.m

MC=375 kN-m

MA=181.82 kN-m MB=181.82 kN-m

V2=636.36 kN

V3=636.36 kN

BMD at

working

+

-

V1=363.63 kN

V4=363.63 kN

SFD

+ _ +

-

1000 kN 1000 kN

363.64 kN/m

1m 1m 3.5 m

1.3 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) = 0 Therefore 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.

MC=375 kN-m

MA=181.82 kN-m MB=181.82 kN-m

V2=636.36 kN

V3=636.36 kN

BMD at working

+

-

V1=363.63 kN

V4=363.63 kN

SFD

+

_ +

-

1000 kN 1000 kN

363.64 kN/m

1.3m

4.2m

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

4000 x 400 400 x 400 400

5500 mm

2100

mm

+

0.85m

1.75m 1.75m 0.8m

A B

a=1 m b=1

m

3.5m

400

2100 B

D

D+db

D+ds

db/2

400

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 Ks

where, 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 Reinforcement

Cantilever portion BD and AC Length 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-m Therefore Section is singly reinforced. Mu/bd2 =177.53x106/(400x7252) =0.844 <2.76, URS pt=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 of cantilever for both columns. If required two bottom bars of 2-16mm will be curtailed at a distance d (= 725 mm) from the point of contra flexure in the portion 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 mm b= actual width of flange = 2100 mm, b w = 400 mm bf= [2900/(2900/2100) + 4] + 400 =938.9mm < 2100mm Df = 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.

MC=375 kN-m

MA=181.82 kN-m MB=181.82 kN-m BMD at working

+

-

1000 kN 1000 kN

363.64 kN/m

1.3m

4.2m

Design of shear reinforcement

Portion between column i.e. AB

Max. shear force at A or B=Vumax=1.5 x 636.36 =954.54 kN SF at the point of contra flexure = 954.54-1.5x 363.64x0.3 = 790.9 kN τv=790900/(400x725) =2.73 MPa < τc,max.(2.8 MPa) Area of steel available = 2354 mm2, 0.805 % τc=0.59MPa, τv > τc

Design 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/c Zone of shear reinforcements is between τv to τc

Cantilever portion BD and AC

Vumax = 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 kN τv=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

400x400 400x400

#12@300, 2L Stp

#12@190, 4L Stp

#12@300, 2L Stp

#12@190, 4L Stp

#12@300, 2L Stp

4-#25 + 2-#16 2-#16

4-#16

1 m 3.5 m 1 m

2-#16 Side face

2-#12

2-#16

4-25 2-16

C/S at Centre C/S at the junction (Right of B)

400 400

800 800

400

2100

2-16

4-16

2-φ16

5500 mm

2.1

m

Plan of footing slab

Φ20@130

Φ8@160