RCC Combined Footing design

DESIGN OF COMBINED FOOTINGS(6,7) I) Design Parametres:Footing designation = Concrete mix Steel Cover to Reinforcement Safe Bearing Capacity(From soil Report)= Unit weight of RCC = W idth of Column C6 = Depth of Column C6 = W idth of Column C7 = Depth of Column C7 = Cha Characte cteristi istic c co compressiv ssive e str stre ength of co concr ncrete = Yield strength of steel = Distance between columns =

F1 M25 Fe415 50mm 73.50KN/sqm 25.0KN/cum 0.23m 0.30m 0.23m 0.30m 25.00 .00N/sq /sqmm 415.00N/sqmm 2.06m

II) Load calculations:Factored Load fr from co column C6 C6 fr from analysis = ----do----- from column C7 from analysis =

424.91KN 354.23KN TOTAL TOTAL 779.1 779.14KN 4KN  Add self weight weight of the footing&Weight 116.871KN 116.871KN of back fill (15%) Total Load =

896.01KN

Ultimate Bearing Capacity of the soil qu = 1.5xSBC =

110.25KN/sqm

Size of the footing required: Area of the footing required required =

8.13Sqm

It is necess necessary ary that that the the resulta resultant nt of the the loads loads of two two columns columns and and the centr centroid oid of the the foo coincide coincide so so that a unifor uniform m distribu distribution tion of of soil pressu pressure re is obtain obtained. ed. Thus, Thus, y, y, the distan distance ce of the the cent footing from column C6 can be obtained as given below: Taking Taking mom moment ents s of the the colum column n loads loads & resul resultan tantt wrt wrt C7,th C7,then en y =

0.94m 0.94m

Rectang Rectangular ular footi footing ng is to to be desig designed ned with with a total total length length of 2.24m 2.24m,wh ,whose ose centro centroid id coinc coinc resultant of the column loads. However provide footing with 3.31m length.The plan of the footing is as given below:-

3310

2060 600

535

        0         0         7

        2

The size of the footing is 2.70mx3.31m,the area comes to Width = Length =

8.94Sqm

2.70m 3.31m

The net ultimate bearing pressure acting on the footing due to direct load = Design moment about tranverse axis =

100.26KN/sqm

2.86KN-m 2

(1/6)xbd  =

Section Modulus for the above section =

4.93cum

Soil pressure due to moment = M/Z =

0.58KN/sqm

Max.Soil pressure =

pu + mu =

100.84KN/sqm Hence O.K

Min.Soil pressure =

pu - mu =

99.68KN/sqm Hence O.K

Hence,the design soil pressure =

100.84KN/sqm

The UDL on the longitudinal beam of unit width =

272.27KN/m

The base pressure and loading on the footing is as given below:-

C6

C7

424.91

354.23

715

535

2060

Loads &Soil pressure

        4         2   .         0         3         2

        7         2   .         2         7         2

        7         6   .         6         4         1

Shear force dia ram

        6         6         2         5         9         2

        1   .         0         9

Bending Moment diagram

        6   .         9         6

        7         9   .         8         3

The maximum sagging bending moment(Considering 1m width of footing) =

The maximum hogging bending moment(Considering 1m width of footing) = III)Depth of the footing required:i)Bending moment criteria:The depth of section required for Max.bending moment as shown in the figure.Hence,t Maximum factored bending moment = ML = 90.10KN-m (Considering 1m width of footing) 2

 Adopting Limit state method of design Mu = 0.138 f ckbd The effective depth of footing required = d =

0.5

[Mu/(0.138f ckb)]

 = 161.60mm Over all depth required assuming 12mm dia bars = However assume

500.00mm

 = 217.60mm

overall depth,then the effective depth comes to

Longitudinal direction:Calculation of reinforcement at top :The actual depth of neutral axis =

23.05mm

 Area of steel required =

574.58sqmm

 Assuming 12mm dia bars,the spacing comes to Provide 12mm bars at a spacing of

197mm 150mm

along width

Then the area of reinforcement provided =

753.60Sqm

Percentage of reinforcement provided =

0.17

ii)Check for one way shear:The critical section of one way shear is at a distance of 'd' from the face of the column Hence,the factored design shear force V Fd =

Nominal shear stress T v =

113.81KN (Calculated from Analysis)

0.256N/sqmm 0.256 Hence,no shear reinforcement is required. Calculation of reinforcement at bottom :Maximum factored bending moment = MB = (Considering 1m width of footing)

69.60KN-m

Over all depth provided =

500.00mm

Effective depth assuming 10mm dia bars The actual depth of neutral axis =

17.67mm

 Area of steel required =

440.54sqmm

 Assuming 12mm dia bars,the spacing comes to

257mm

Provide 12mm bars at a spacing of

150mm

Then the area of reinforcement provided = Percentage of reinforcement provided =

along width 523.33Sqm

0.118

iii)Two way shear criteria:(a) Around column C6:The critical section for two-way shear is along the perphery of the square at a distance d/2 from the face of the column Hence perimetre of the preriphery b 0 =

2836mm

Hence,the factored shear force V Fd = qu(B -AB )= Nominal shear stress Tv = V Fd/b0d =

850.64KN

0.68 N/sqmm

Permissible shear stress in concrete for two-way shear for M20 grade concrete Tc' =ks . Tc ks = (0.5+l/b)> 1 Hence ks = 1 Tc = 0.25(f ck)

=

1.25 N/sqmm

Hence,Tc' =

1.25 N/sqmm 1.25 >0.680

Hence,the section provided is safe from two-way shear point of view (b) Around column C7:The critical section for two-way shear is along the perphery of the square at a distance d/2 from the face of the column Hence perimetre of the preriphery b 0 = Hence,the factored shear force V Fd = qu(B -AB )= Nominal shear stress Tv = V Fd/b0d =

2836mm 850.64KN

0.68 N/sqmm

Permissible shear stress in concrete for two-way shear for M20 grade concrete Tc' =ks . Tc ks = (0.5+l/b)> 1 Hence ks = 1 Tc = 0.25(f ck)

=

Hence,Tc' =

1.25 N/sqmm 1.25 N/sqmm 1.25 >0.680

Hence,the section provided is safe from two-way shear point of view IV) Transverse direction:The column strips are to be designed as transverse cantilever beams having a width c 2 times effective depth,ie, one effective depth on each side.

a)Column strip for C6 :Width of column strip = Length of column strip =

1118mm 1200mm

The effective soil pressure under column strip =

161.88KN/sqm

Maximum bending moment =

130.31KN-m

Over all depth provided =

500.00mm

Effective depth assuming 12mm dia bars The actual depth of neutral axis =

33.68mm

 Area of steel required =

839.63sqmm

 Assuming 12mm dia bars,the spacing comes to

135mm

Provide 12mm bars at a spacing of

125mm

Then the area of reinforcement provided =

within column strip 904.32Sqm

Percentage of reinforcement provided =

0.204

Check for one way shear:The critical section of one way shear is at a distance of 'd' from the face of the column Hence,the factored design shear force V Fd = Nominal shear stress T v =

136.82KN

0.308N/sqmm 0.308 Hence,no shear reinforcement is required. b)Column strip for C7 :Width of column strip = Length of column strip = The effective soil pressure under column strip =

1118mm 1200mm 134.95KN/sqm

Maximum bending moment =

108.63KN-m

Over all depth provided =

500.00mm

Effective depth assuming 12mm dia bars The actual depth of neutral axis =

27.92mm

 Area of steel required =

696.02sqmm

 Assuming 12mm dia bars,the spacing comes to

162mm

Provide 12mm bars at a spacing of

150mm

Then the area of reinforcement provided =

within column strip 753.60Sqm

Percentage of reinforcement provided =

0.17

Check for one way shear:The critical section of one way shear is at a distance of 'd' from the face of the column Hence,the factored design shear force V Fd = Nominal shear stress T v =

114.06KN

0.257N/sqmm 0.257 Hence,no shear reinforcement is required. CHECK FOR TRANSFER OF BEARING STRESSES:(a) For column C6:Pu =

424.91KN 1/2

Compressive bearing resistance = 0.45 f ck(A1/A2) . For the column face A 1/A2 = 1 and for the other face A1/A2 > 2 but should be taken as 2. In any case, the column face governs.Force transferred to the base through column at the interface =

776.25KN >424.91KN

Hence O.K There is no need to design separate dowel bars to tranfer the load to the base of the fo (b) For column C7:Pu =

354.23KN

Compressive bearing resistance = 0.45 f ck(A1/A2) . For the column face A 1/A2 = 1 and for the other face A1/A2 > 2 but should be taken as 2. In any case, the column face governs.Force transferred to the base through column at the interface =

776.25KN >354.23KN Hence O.K

There is no need to design separate dowel bars to tranfer the load to the base of the fo CALCULATION OF DISTRIBUTION STEEL:Provide temperature re inforcement @ 0.12%  Area required = Taking 10mm dia HYSD bars,the spacing comes to Hence,provide 10mm dia bars @ 125mm c/c

600.00Sqm 131mm

 

ting roid of the

ides with the

0

90.10KN-m (From Analysis) 69.60KN-m (From Analysis)

e

444.00mm

445.00mm

lumn width+

444.00mm

444.00mm

 

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