Footing design

Learning Outcomes:
2
• After this students will be able design and detail
combined footings through drawing and bar bending
schedule.

3
The function of a footing or a foundation is to transmit the load
form the structure to the underlying soil.
The choice of suitable type of footing depends on the depth at
which the bearing strata lies, the soil condition and the type of
superstructure.

4
Whenever two or more columns in a straight line are carried on a
single spread footing, it is called a combined footing. Isolated
footings for each column are generally the economical.
Combined footings are provided only when it is absolutely
necessary, as
1. When two columns are close together, causing overlap of
adjacent isolated footings
2. Where soil bearing capacity is low, causing overlap of
adjacent isolated footings
3. Proximity of building line or existing building or sewer,
adjacent to a building column.

• The combined footing may be rectangular, trapezoidal or Tee-shaped
in plan.
The geometric proportions and shape are so fixed that the centeroid
of the footing area coincides with the resultant of the column loads.
This results in uniform pressure below the entire area of footing.
• Trapezoidal footing is provided when one column load is much
more than the other. As a result, the both projections of footing
beyond the faces of the columns will be restricted.
• Rectangular footing is provided when one of the projections of the
footing is restricted or the width of the footing is restricted.

•
Longitudinally, the footing acts as an upward loaded beam
spanning between columns and cantilevering beyond. Using statics,
the shear force and bending moment diagrams in the longitudinal
direction are drawn. Moment is checked at the faces of the column.
Shear force is critical at distance ‘d’ from the faces of columns or
at the point of contra flexure. Two-way shear is checked under the
heavier column.
The footing is also subjected to transverse bending and this
bending is spread over a transverse strip near the column.

1
Locate the point of application of the column loads on the footing.
Proportion the footing such that the resultant of loads passes through
the center of footing.
Compute the area of footing such that the allowable soil pressure is
not exceeded.
Calculate the shear forces and bending moments at the salient points
and hence draw SFD and BMD. Fix the depth of footing from the
maximum bending moment.
Calculate the transverse bending moment and design the transverse
section for depth and reinforcement.
Check for anchorage and shear.

1
Check the footing for longitudinal shear and hence
design the longitudinal steel
Design the reinforcement for the longitudinal moment
and place them in the appropriate positions.
Check the development length for longitudinal steel
Curtail the longitudinal bars for economy
Draw and detail the reinforcement Prepare the bar
bending schedule

1
Detailing of steel (both longitudinal and transverse) in a combined footing is similar
to that of conventional beam- SP-34
Detailing requirements of beams and slabs should be followed as appropriate-SP-34

1
Two interior columns A and B carry 700 kN and 1000 kN loads
respectively. Column A is 350 mm x 350 mm and column B is 400 mm
X 400 mm in section. The centre to centre spacing between columns is
4.6 m. The soil on which the footing rests is capable of providing
resistance of 130 kN/m2. Design a combined footing by providing a
central beam joining the two columns. Use concrete grade M25 and mild
steel reinforcement.
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

Solution:
Data fck= 25 Nlmm2,
fy= 250 N/mm2,
fb= l30 kN/m2(SBC),
Column A = 350 mm x 350 mm, Column B = 400 mm x 400
mm, c/c spacing of columns =4.6 m, PA= 700 kN and PB=
1000 kN
Required: To design combined footing with central beam
joining the two columns.
Ultimate loads
PuA=1.5 x 700 = 1050 kN, PuB= 1.5 x 1000 = 1500kN
1

1
Working load carried by column A = PA
Working load carried by column B = PB
Self weight of footing 10 % x (PA + PB)
= 700 kN
= 1000 kN
= 170 kN
Total working load = 1870 kN
Required area of footing = Af = Total load /SBC
= 1870/130 = 14.38 m2
Let the width of the footing = Bf = 2m
Required length of footing = Lf = Af /Bf = 14.38/2 = 7.19m
Provide footing of size 7.2m X 2m,Af = 7.2 x 2 = 14.4 m2

For uniform pressure distribution the C.G. of the footing should
coincide with the C.G. of column loads. Let x be the distance of C.G.
from the centre line of column A
Then x = (PB x 4.6)/(PA + PB) = (1000 x 4.6)/(1000 +700)
= 2.7 m from column A.
If the cantilever projection of footing beyond column A is ‘a’ then, a +
2.7 = Lf /2 = 7.2/2, Therefore a = 0.9 m
Similarly if the cantilever projection of footing beyond B is 'b' then, b
+ (4.6-2.7) = Lf /2 = 3.6 m,
Therefore b = 3.6 - 1.9 = 1.7 m The details are shown in Figure
1

2
Total ultimate load from columns = Pu= 1.5(700 + 1000) = 2550 kN. Upward
intensity of soil pressure wu= P/Af= 2550/14.4 = 177 kN/m2
Design of slab
Intensity of Upward pressure = wu=177 kN/m2
Consider one meter width of the slab (b=1m)
Load per m run of slab at ultimate = 177 x 1 = 177 kN/m Cantilever
projection of the slab (For smaller column)
=1000 - 350/2 = 825 mm
Maximum ultimate moment = 177 x 0.8252/2 = 60.2 kN-m.

0.825 m
1m
pu=177 kN/m2
For M25 and Fe 250, Q u max = 3.71 N/mm2
Required effective depth = √ (60.2 x 106/(3.71 x 1000)) = 128 mm
Since the slab is in contact with the soil clear cover of 50 mm is
assumed.
Using 20 mm diameter bars
Required total depth = 128 + 20/2 + 50 =188 mm say 200
mmProvided effective depth = d = 200-50-20/2 = 140 mm
1m
0.35m

Area provided =1000 x 314 / 130 = 2415 mm2
22
Mu/bd2 =3.07 3.73, URS
Mu=0.87 fy Ast[d-fyAst/(fckb)]
pt=1.7%
Ast = 2380 mm2
Use Φ20 mm diameter bar at spacing= 1000 x 314 / 2380
= 131.93 say 130 mm c/c

25
• Design shear force=Vu=177x(0.825-0.140)=121kN
• Nominal shear stress=τv=Vu/bd=121000/ (1000x140)=0.866MPa
• Permissible shear stress
• Pt = 100 x 2415 /(1000 x 140 ) = 1.7 %, τuc = 0.772N/mm2
• Value of k for 200 mm thick slab =1.2
• Permissible shear stress = 1.2 x 0.772 = 0.926 N/mm2
τuc > τv and hence safe
The depth may be reduced uniformly to 150 mm at the edges.

Ldt= [0.87 x 250 / (4 x 1.4)]Ф =39 Ф
= 39 x 20 = 780 mm
Available length of bar=825 - 25 = 800mm
> 780 mm and hence safe.
Transverse reinforcement
Required Ast=0.15bD/100
=0.15x1000 x 200/100 = 300mm2
Using Ф8 mm bars, Spacing=1000x50/300
= 160 mm
Provide distribution steel of Ф8 mm at 160 mm
c/c,<300, <5d
2

2
Load from the slab will be transferred to the beam. As the
width of the footing is 2 m, the net upward soil pressure per
meter length of the beam
= wu = 177 x 2 = 354 kN/m
Shear Force and Bending Moment
VAC= 354 x 0.9 =318.6 kN, VAB = 1050-318.6 =731.4 kN
VBD= 354 x 1.7 = 601.8kN, VBA = 1500-601.8 = 898.2 kN
Point of zero shear from left end C X1 = 1050/354 = 2.97m
from C or X2 = 7.2-2.97 = 4.23 m from D

Maximum B.M. occurs at a distance of 4.23 m from D MuE = 354 x 4.232 / 2 -
1500 (4.23 - 1.7) = -628 kN.m
Bending moment under column A= MuA=354x0.92 /2 =
143.37 kN.m
Bending moment under column B = MuB = 354 x 1.72
= 511.5 kN-m
Let the point of contra flexure be at a distance x from the centre of column A
Then, Mx= I050x - 354 (x + 0.9 )2/ 2 = 0
Therefore x = 0.206 m and 3.92 m from column A
i.e. 0.68 m from B.
2

2
The width of beam is kept equal to the maximum width of the column i.e. 400
mm. Determine the
depth of the beam where T- beam action is not available. The beam acts as a
rectangular section in the cantilever portion, where the maximum positive
moment = 511.5 kN/m.
d =√ (511.5 x 106/ (3.73 x 400)) = 586 mm
Provide total depth of 750 mm. Assuming two rows of bars with effective
cover of 70 mm.
Effective depth provided = d= 750-70 = 680 mm (Less than 750mm and hence no
side face steel is needed.

2
The heaver 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. 680/2 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, db=680 mm, ds=140 mm
Area resisting two - way shear
= 2(b x db + ds x d s) + 2 (D + db)ds
= 2 (400 x 680+ 140 x 140) + 2(400+680) 140= 885600 mm2
Design shear=Pud= column load – W u x area at critical section
= 1500 - 177 x(b + ds) x (D + db)
=1500-177 x (0.400+0.140) x (0.400+ 0.680)
=1377.65kN
τv=Pud/bod= 1377.65x1000/885600=1.56 MPa
Shear stress resisted by concrete = τuc = τuc x Ks
where, τuc = 0.25 √ f ck= 0.25√ 25 = 1.25 N/mm2
K s = 0.5 + d / D = 0.5 + 400/400 = 1.5≤ 1 Hence K s =1
τuc = 1 x 1.25 = 1.25 N/mm2 . Therefore Unsafe
3

3
Length of cantilever from the face of column
=1.7- 0.4/2 = 1.5 m.
Ultimate moment at the face of column
=354x1.52/2=398.25 kN-m
Mumax = 3.71 x 400 x 6802 x 10 -6 = 686 kN-m > 398.25 kN-m
Therefore Section is singly reinforced.
Mu/bd2 =398.25x106/(400x6802)=2.15 3.73, URS
Pt=1.114%
A st =3030 mm2, Provide 3-Φ32 mm + 4-Φ16 mm at bottom face, Area
provided = 3217 mm2
Ldt = 39 x 32 =1248 mm

Curtailment
All bottom bars will be continued up to the end of cantilever. The
bottom bars of 3 - Ф 32 will be curtailed at a distance d (= 680
mm) from the
point of contra flexure (λ = 680 mm) in the portion BE with its
distance from the centre of support equal to 1360 mm from B.
Cantilever portion AC
Length of cantilever from the face of column =900-350/2 = 725 mm
Ultimate moment = 354 x 0.7252 /2 = 93 kN-m
Mu/bd2 =93x106/(400x6802) =0.52 <3.73, URS
Pt=0.245% (Greater than minimum steel) Ast =660 mm2
Provide 4 - Ф 16 mm at bottom face, Area provided = 804 mm2
Continue all 4 bars of 16 mm diameter through out at bottom.
35

The beam acts as an isolated T- beam. bf = [Lo/( Lo / b +4)]+ bw, where,
Lo = 4.6 - 0.206 - 0.68 = 3.714 m = 3714 mm
b= actual width of flange = 2000 mm, b w = 400mm
bf = [3714/(3714/2000+4) + 400] =1034mm <2000mm
Df =200 mm, Mu= 628 kN-m
Moment of resistance Muf of a beam for x u =Dfis : Muf= [0.36 x 25 x1034
x 200(680 - 0.42x200)]x10-6
= 1109 kN.m > Mu ( = 628 kN-m)
3

Therefore Xu <Df
Mu=0.87fyAst(d - fyAst/fckbf) Ast= 4542 mm2
Provide 5 bars of Ф 32 mm and 3 bars of Ф 16 mm, Area
provided= 4021 + 603 = 4624 mm2 >4542 mm2 pt= 100 x
4624/(400x680) = 1.7 %
3

Consider that 2 - Ф 32 mm are to be curtailed No. of bars to be
continued = 3 - Ф16 + 3 - Ф 32 giving area = Ast =3016 mm2
3
Moment of resistance of continuing bars
Mur= (0.87 x 250 x 3016 (680 – ((250 x 3016) / (25 x 400) x 10-6
= 396.6 kN-m
Let the theoretical point of curtailment be at a distance x from
the free end C,
Then, Muc= Mur Therefore -354 x2 / 2 + 1050 (x- 0.9) = 396.6 x2-
5.93x + 7.58 =0, Therefore x = 4.06m or 1.86m from C

Actual point of curtailment = 4.06 + 0.68 = 4.74 m from C or 1.86 - 0.68 =
1.18 m from C
Terminate 2 - Φ 32 mm bars at a distance of 280 mm (= 1180 - 900) from
the column A and 760mm (= 5500 - 4740) from column B. Remaining
bars 3 - Φ 32 shall be continued beyond the point of inflection for a
distance of 680 mm i.e. 460 mm from column A and up to the outer face
of column
B. Remaining bars of 3 - Φ 16 continued in the cantilever portion.
3

3
In 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(= 680mm)
(i) Maximum shear force at B = Vumax = 898.2 kN
Shear at the point of contra flexure
= VuD - 898.2-354 x 0.68 = 657.48 kN
τv=657000/(400x680) =2.42 MPa < τc,max.

Area of steel available 3 - Φ 16 + 3 - Φ 32 ,
Ast = 3016 mm2
pt = 100 x 3016 / (400 x 680) = 1.1% τc=0.664MPa
τv > τc
Design shear reinforcement is required. Using 12 mm diameter 4
- legged stirrups,
Spacing= [0.87 x 250x(4x113)] /(2.42-0.664)x400 =139 mm
say 120 mm c/c
Zone of shear reinforcements between τv to τc
= m from support B towards A
3

42
(ii) Maximum shear force at A
= Vu max= 731.4 kN.
Shear at the point contra flexure = VuD = 731.4 - 0.206x 354 =
658.5 kN
τv=658500/(400x680) =2.42MPa < τc,max.
Area of steel available = 4624 mm2, pt= 100 x 4624 / (400 *
680) = 1.7 %
τuc = 0.772 N/ mm2,
τv > τc

Design shear reinforcement is required.
Using 12 mm diameter 4 - legged stirrups,
Spacing = 0.87 x 250 x (4 x 113) /(2.42-0.774)x400
=149 mm say 140 mm c/c
Zone of shear reinforcement.
From A to B for a distance as shown in figure
For the remaining central portion of 1.88 m provide minimum
shear reinforcement using 12 mm diameter 2 - legged stirrups
at
Spacing , s = 0.87 x 250 x (2 x 113)/(0.4 x 400)=307.2
mm, Say 300 mm c/c< 0.75d
4

4
Vumax = 601.8kN,
VuD=601.8-354(0.400/2 + 0.680) = 290.28kN.
τv=290280/(400x680) =1.067MPa < τc,max.
Ast = 3217 mm2 and pt = 100 x 3217/(400 x 680) = 1.18% τc
=0.683N/mm2 (Table IS:456-2000)
τv > τc and τv - τc <0.4. Provide minimum steel.
Using 12 mm diameter 2- legged stirrups,
Spacing = 0.87 x 250 x (2 x 113) /(0.4x400) =307.2 mm
say 300 mm c/c

4
Minimum shear reinforcement of Ф 12 mm
diameters 2 - legged stirrups at 300mm c/c will be
sufficient in the cantilever portions of the beam as
the shear is very less.

Thank you

Mr. VIKAS MEHTA
School of Mechanical and civil engineering
Shoolini University
Village Bajhol, Solan (H.P)
vikasmehta@shooliniuniversity.com
+91 9459268898

Footing design

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