comb_foot-slab-mcn

7/29/2019 Comb_Foot-Slab-MCN

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1000 kN 1000 kN

3.5 m 1 m1 m

x

Rp=173.16 kN/m

2

w=363.64 kN/m

Design of combined footingSlab type-3

Design a rectangular combined footingslab type [with out a central beam] for supporting twocolumns 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.

Solution: Data

fck= 20Nlmm2,

fy= 415mm2,

fb = 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 loadsP uA= 1.5 x 1000 = 1500 kN, P uB = 1.5 x 1000 = 1500 kN

Proportioning of base sizeWorking 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.


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The details are shown in Figure

Total load from columns = P = (1000 + 1000) = 2000 kN.

Upward intensity of soil pressure=Column loads, P/Af= 2000/5.5 x2.1 =173.16 kN/m2< SBC

Design of slab for longitudinal bendingAs the width ofthe footing is 2.1m, 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 conditionVAC = 363.64 x 1 = 363.14 kN, VAB = 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, i.e., at E

Maximum B.M. occurs at E

ME = 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= 363.63x2/21000(x-1) = 0

Therefore x = 1.30 m and 4.2m from C

Depth of footing from B.M. Considerations:

d = (375 x 1.5 x 106/ (2.76 x 2100)) = 311.52 mmD=d + 75 =400 mm, d=325 mm

Longitudinal Bending Transverse Bending

T

TTT

Pa Pb

SLAB TYPE COMBINED FOOTING


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Check the depth for Two-way Shear:

MC=375 kN-m

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

V2=636.36 kN

V3=636.36 kN

BMD

+_

-

V1=363.63 kN

V4=363.63 kN

SFD

+

_+

-

1000 kN 1000 kN

363.64 kN/m


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Load on heavy column = Pu =1500 kN.

Load causing the punching shear = Design shear=Pud= column loadW u x area at critical section

Pud = 1500- 173.15x1.5x (0.4 + 0.325)2

= 1363.48 kN

v=Pud/bod =1363.48 x 1000 /(725x4x325) = 1.45 MPa

Shear stress resisted by concrete = uc = uc x K swhere, uc = 0.25 fck= 0.25 20 = 1.11 N/mm

2

K s = 0.5 + d / D = 0.5 + 400/400 = 1.5 1 Hence K s = 1uc = 1 x 1.11 = 1.11 N/mm

2, Therefore unsafe

Increase the depth to 500 mm with d =425 mm

Pud = 1500- 173.15x1.5x (0.4 + 0.425)2

= 1285.73 kN

v=Pud/bod =1285.73 x 1000 /(725x4x325) = 0.92 Mpa. Therefore ok.

Area of Reinforcement

Region AB between points of contra flexuresMu = 1.5 x 375=562.5 kN-m

Mu/bd2

= 562.5 x 106/2100 x 425

2=1.48, pt= 0.456, Area = 4070 mm

2

Provide 13 bars of 20 mm, Area provided = 4084 mm2

400

2100BD

D+ds

D+ds

ds/2

400

4000 x 400 400 x 400

5500 mm

2100

mm

0.85m

1.75m1.75m

0.8m

A B

a=1 m b=1m3.5m


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pt = 0.46 %

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. Extend the remaining bars upto the end offooting.

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 0.8

2/ 2 = 177.53 kN-m

Therefore Section is singly reinforced.

Mu/bd2

=177.53x106/(2100x425

2) =0.47 2.76, URS

Pt=0.135%, A st =0.135x2100x425/100 = 1205 mm2

Minimum steel = 0.12x2100x500/100 =1260 mm2

Provide 13 - 12 mm at bottom face, Area provided = 1470 mm2, pt=0.16%

Ld = 47x12 = 564 mm

Curtailment: All bottom bars will be continued up to the end of cantilever for both columns. Alternate

bars are to be curtailed at a distance d (= 725 mm) from the point of contra flexure as shown.

Design of shear reinforcement

Portion between column i.e. ABIn this case the crack due to diagonal tension will occur at the point of contra flexure because the

distance of the point of contra flexure from the column is less than the effective depth d(= 725mm)

(i) Maximum shear force at A or B = Vumax = 1.5 x 636.36 =954.54 kN

Shear at the point of contra flexure = 954.54-1.5x 363.64x0.3 = 790.9 kN

v=790900/(2100x425) =0.886 MPa c,max.(2.8 MPa)Area of steel available = 4084 mm

2, 0.46 %

c=0.47MPa, v > cDesign shear reinforcement is required.

Using 12 mm diameter 4 - legged stirrups,

Spacing = 0.87 x 415 x (4 x 113) /(0.886-0.47)x2100 =186 mm say 180 mm c/c

Zone of shear reinforcements between v to c= m from support B towards A

Cantilever portion BD and ACSimilarly design for shear at other critical sections can be made. However minimum steel is

required for the two projections of the footing. Provide 12mm, 4 legged at 200 mm c/c.

Design of slab:Intensity of Upward pressure = w=173.16 kN/m

2

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

Cantilever projection of the slab (For smaller column) =1050 - 400/2 = 850mm

Maximum ultimate moment = 173.16 x 0.8502/2 = 62.55 kN-m. (Working condition)


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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, which is more than theprovided 425 mm.

To find steel

Mu/bd2

=1.5 x 62.15 x106/1000 x 425

2= 0.516 2.76, URS

Pt=0.15 %Ast = 0.15 x 1000 x 425/100 = 637 mm

2

Use # 12 mm diameter bar at spacing = 1000 x113 / 637 = 177mm say170 mm c/c

This is required at the central band whose length is 400 mm + 2x425 =1250 mm under both columns.

For the remaining length minimum steel can be provided, which 0.12 % of grass area.

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

Using 12 mm bars, spacing = 1000 x 113 / 480 = 235.4 mm

Provide distributionsteel of#12 mm at 230 mm c/c

Check for development lengthLdt= 47 times diameter = 47x12 mm

Available length of bar = 850 - 25 =825mm and hence safe.

Transverse reinforcementRequired A st = 0.12 bD / 100 = 0.12 x 1000 x 400/100 = 480mm

2

Using 12 mm bars, spacing = 1000 x 113 / 480 = 164.58 mm

Provide distributionsteel of#12 mm at 230 mm c/c

Details can be seen in drawings. Provide all dimensions and the details of the reinforcements.

End

Dr. M. C. Nataraja

Professor, SJCE, Mysore-570 006

p=173.16 kN/m2

1m


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1000 kN 1000 kN

3500 10001000

comb_foot-slab-mcn

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