Design and Detailing of steel in Combined Footings

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Design and Detailing of steel in Combined Footings

• Dr. M.C. NATARAJA

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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.

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• Draw to a suitable scale the following
• The longitudinal sectional elevation
• Transverse section at the left face of the
  heavier column
• Plan of the footing
• Marks 60

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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.
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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 /SBC2200/190
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
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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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Total load from columns P (1000 1000)
2000 kN. Upward intensity of soil pressureColumn
loads/A provided P/Af 2000/5.5 x2.1 173.16
kN/m2lt SBC Design of slab Intensity of Upward
pressure p 173.16 kN/m2 Consider one meter
width of the slab (b1m) 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)
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For M20 and Fe 415, Q u max 2.76 N/mm2
Required effective depth v (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
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To find steel Mu/bd2 1.5 x 62.15 x106/1000 x
22521.84? 2.76, URS Pt0.584 Ast 1314 mm2
Use 20 mm diameter bar at spacing 1000 x 314
/ 1314 238.96 mm say 230 mm c/c 20_at_230 Area
provided 1000 x 314 / 230 1365 mm2. Hence
safe. This steel is required for the entire
length of the footing.
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Check the depth from one-way shear
considerations Design shear forceVu1.5x173.16
x(0.850-0.225) 162.33 kN Nominal shear stress
tv Vu/bd 162330/(1000x225) 0.72
MPa Permissible shear stress pt 100 x 1365
/(1000 x 225 ) 0.607 , tuc 0.51 N/mm2
Value of k for 300 mm thick slab 1 Permissible
shear stress 1 x 0.51 0.51 N/mm2 tuc lt tv
and hence unsafe.
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The depth may be increased to 400 mm so that d
325mm Mu/bd21.5 x 62.15 x106/1000x3252
0.883? 2.76, URS pt0.26 , 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_at_230
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Check the depth from one - way shear
considerations Design shear forceVu1.5x173.16
x (0.850-0.325) 136.36 kN Nominal shear stress
tv Vu/bd 136360/(1000x325) 0.42
MPa Permissible shear stress pt 100 x 874/(1000
x 325 ) 0.269 , tuc 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 tuc lt tv and hence
slightly unsafe. However provide steel at
closure spacing, 16 _at_150 Ast201 x 1000/150
1340 mm2 and pt 0.41 and hence tuc0.45 MPa
and safe.
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Check for development length Ldt 47 times
diameter 47x16768 mm Available length of bar
850 - 25 825mm gt 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
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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 VAC363.64 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
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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
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Depth of beam from B.M. Considerations The
width of beam400 mm. Assume the beam as
rectangular at the center of span where the
moment is maximum, we have, d v (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.
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Check the depth for Two-way Shear
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
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In this case bD 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(400725) 325 1522500 mm2 Design
shearPud 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 (0.4000.325) x
(0.400 0.725) 1288.14 kN
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tvPud/bod 1288.14x1000/15225000.845 MPa Shear
stress resisted by concrete tuctucx Ks where,
tuc 0.25 v f ck 0.25v 20 1.11 N/mm2 Ks
0.5 d / D 0.5 400/400 1.5 ? 1 Hence Ks
1 tuc 1 x 1.11 1.11 N/mm2 Therefore safe
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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 gt177.53
kN-m Therefore Section is singly
reinforced. Mu/bd2 177.53x106/(400x7252) 0.844
?2.76, URS pt0.248, A st 719.2 mm2 Provide
4 - 16 mm at bottom face, Area provided 804
mm2, pt0.278 Ld 47x16 752 mm
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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.
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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
4.2-1.32.9 m 2900 mm b actual width of
flange 2100 mm, b w 400 mm bf
2900/(2900/2100) 4 400 938.9mm lt 2100mm Df
400 mm, Mu 1.5 x 375562.5 kN-m
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Moment of resistance Muf of a beam for x uD f
is (Muf) 0.36 x 20 x 938.9 x 400 (725 -
0.42x400) x10-6 1506 kN.m gt Mu ( 562.5
kN-m) Therefore XultDf Mu0.87fyAst(d -
fyAst/fckbf), Ast 2334 mm2 Provide 4 bars of
25 mm and 2 bars of 16 mm, Area provided
2354 mm2 gt2334 mm2 pt 100 x 2334/(400x725)
0.805
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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.
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Design of shear reinforcement Portion between
column i.e. AB Max. shear force at A or
BVumax1.5 x 636.36 954.54 kN SF at the point
of contra flexure 954.54-1.5x 363.64x0.3
790.9 kN tv790900/(400x725) 2.73 MPa ?
tc,max.(2.8 MPa) Area of steel available 2354
mm2, 0.805 tc0.59MPa, tv gt tc 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 tv to
tc
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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 tv40900/(400x 725) 0.14 MPa ? tc,max.
This is very small Steel at this section is 4
16 mm, 804 mm2, pt0.278 tc 0.38N/mm2
(IS456-2000). 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
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