Design of Shallow Foundation (Isolated Footings)

PROJECT:- Design of Shallow Foundation (Isolated Footings)

AIM:- Design a square footing of column size 400x400mm

INTRODUCTION: The spread footing behaves like an inverted cantilever with loads applied in the upward direc-tion. As a rule, a spread footing is a quite rigid element therefore, the applied soil stresses are almost linear and in case of a symmetric (with respect to the pedestal) footing, they are or-thogonal. These soil pressures constitutes the loads carried by the footing which behaves like a slab and is deformed.Soil pressures and deformation of an isolated spread footing. Footings and other foundation units transfer loads from the structure to the soil or rock supporting the structure. The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad.

The most common type of footings used are;

. Strip or wall footing. Spread footing . Stepped footing . Tapered footing . Pile cap . Combined footing . Mat or raft footing

QUESTION:- Design a square footing for a column size of 400x400. The compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN. The safe soil bearing capacity is 150 KN/m2 at a depth of 2 metres below E.G.L. Participants are free to go for either a tapered or stepped footing. Besides the total axial load, also account for the self-weight of the footing and soil above it. Assume M25 concrete grade. Do the following checks:

. If the gross bearing pressure under the footing is within safe bearing capacity

. Provide sufficient depth and longitudinal steel for resisting bending and shear

. Sufficient anchorage for column steel into the footing

. If the bearing pressure at column-footing joint is within 0.45 times the maximum concrete compressive strength Provide a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel.   Please do go through the solved examples in the document provided with the challenge.  

PROCEDURE:

GIVEN:- . Column Size = 400mm * 400mm

. Compression axial load = 2000 KN

. Safe soil bearing capacity = 150 KNm2">KNm2KNm2

. Compressive Strengthfck=M25">fck=M25fck=M25

. depth = 2 metres below E.G.L

. Soil depth = 20 KNm3">KNm3KNm3

. RCC density = 25 KNm3">KNm3KNm3

Steps involved in Design of  Footing:-

. Determine the size of footing

. Upward pressure

. Depth of footing on the basis of B.M

. Area of reinforcement

. Check for one way shear

. Check for two-way shear

SIZE OF FOOTING:-Assume self-weight of footing equal to 105 of superimposed load.

Wd=10%of2000">Wd=10%of2000Wd=10%of2000

=0.1⋅2000=200KN.">=0.1⋅2000=200KN.=0.1⋅2000=200KN.

Now, Total load, W=WL+Wd">W=WL+WdW=WL+Wd

=200+200">=200+200=200+200

W=2200KN.">W=2200KN.W=2200KN.

Now, Area of footing ; A=WSBC">A=WSBCA=WSBC

=2200150=14.67m³">=2200150=14.67m³=2200150=14.67m³

Now, let us take,

B= √AA = √14.67=3.82">B= √AA = √14.67=3.82B= √AA = √14.67=3.82

Hence, we provide the size of square footing is, =4m⋅4m">=4m⋅4m=4m⋅4m

Therefore, Area of footing is 16m².">16m².16m².

NET UPWARD PRESSURE (ρo)">(ρo)(ρo):- ρo =W⋅15sizeoffooting">ρo =W⋅15sizeoffootingρo =W⋅15sizeoffooting

ρo =2200⋅1.516">ρo =2200⋅1.516ρo =2200⋅1.516

ρo =206.25KNm³">ρo =206.25KNm3ρo =206.25KNm³

DEPTH OF FOOTING ON THE BASIS OF B.M:- The minimum B.M at the face of column and is given by,

M= ρ⋅B8(B-b)²">M= ρ⋅B8(B−b)2M= ρ⋅B8(B-b)²=206.25⋅48(4-0.4)2">=206.25⋅48(4−0.4)2=206.25⋅48(4-0.4)2

=1336.5KNm">=1336.5KNm=1336.5KNm

Now, Ultimate moment

Mu=1.5(M)">Mu=1.5(M)Mu=1.5(M)

=1.5⋅1336.5">=1.5⋅1336.5=1.5⋅1336.5

Mu=2004.75">Mu=2004.75Mu=2004.75

To get the ultimate moment we have calculated that,

Mu=1.5⋅FOS⋅Moment.">Mu=1.5⋅FOS⋅Moment.Mu=1.5⋅FOS⋅Moment.

Now, Depth of footing (D),

Mu=0.138fckBd2">Mu=0.138fckBd2Mu=0.138fckBd2

2004.75=0.138⋅25⋅4000⋅d2">2004.75=0.138⋅25⋅4000⋅d22004.75=0.138⋅25⋅4000⋅d2

d2=2004.75⋅1060.138⋅25⋅4000">d2=2004.75⋅1060.138⋅25⋅4000d2=2004.75⋅1060.138⋅25⋅4000

d2=145271.73">d2=145271.73d2=145271.73

d= √145271.73">d= √145271.73d= √145271.73

d=381.14">d=381.14d=381.14

By the thumb rule, we take 2.5 times higher value of for shear check.

d=381.14⋅2.5=952.85">d=381.14⋅2.5=952.85d=381.14⋅2.5=952.85

D=d+50">D=d+50D=d+50

D=952.85+50">D=952.85+50D=952.85+50

D=1003mm">D=1003mmD=1003mm

 

AREA OF REINFORCEMENT:- Ast=0.5fckf(y)[1-√1−4.6MufckBd²]Bd">Ast=0.5fckf(y)[1−√1−4.6MufckBd²]BdAst=0.5fckf(y)[1-√1−4.6MufckBd²]Bd

=0.5⋅25415[1-√1−(4.6⋅2004.75⋅106fckBd²)]⋅4000⋅953">=0.5⋅25415[1−√1−(4.6⋅2004.75⋅106fckBd²)]⋅4000⋅953=0.5⋅25415[1-√1−(4.6⋅2004.75⋅106fckBd²)]⋅4000⋅953

Ast=5985.3mm2">Ast=5985.3mm2Ast=5985.3mm2

Now, we have to find the % of the reinforcement

P= AstBd⋅100">P= AstBd⋅100P= AstBd⋅100

P= 5985.34000⋅953100">P= 5985.34000⋅953100P= 5985.34000⋅953100

P=0.157%">P=0.157%P=0.157%

No of bars required for the reinforcement, we assume that 20mm dia bars,

Ast=5985.3mm2">Ast=5985.3mm2Ast=5985.3mm2

π4202=314.5">π4202=314.5π4202=314.5

No:of bars = 5985.3314.5 =20">= 5985.3314.5 =20= 5985.3314.5 =20 bars.

Therefore, Spacing b/w the bars

= BNo.ofBars">= BNo.ofBars= BNo.ofBars

= 400020">= 400020= 400020

=200mmcc">=200mmcc=200mmcc

CHECK FOR ONE WAY SHEAR:-

For, One-way shear, critical section is located at a distance ‘d’ from the face of the column.

Now, shear force V= ρo⋅B{12(B-b)-d}">V= ρo⋅B{12(B−b)−d}V= ρo⋅B{12(B-b)-d}=206.25⋅4{12(4-0.4)–0.952}">=206.25⋅4{12(4−0.4)–0.952}=206.25⋅4{12(4-0.4)–0.952}

=699.6KN">=699.6KN=699.6KN

Ultimate shear force Vu=1.5⋅V">Vu=1.5⋅VVu=1.5⋅V

=1.5⋅699.6=1049.4KN">=1.5⋅699.6=1049.4KN=1.5⋅699.6=1049.4KN

Nominal shear stress = VuBd">= VuBd= VuBdas per IS 456-2000

=1049.4⋅10³4000⋅952 =0.27Nmm2">=1049.4⋅10³4000⋅952 =0.27Nmm2=1049.4⋅10³4000⋅952 =0.27Nmm2

Now, design shear strength of concrete τ">ττc

τc = τc⋅K">τc = τc⋅Kτc = τc⋅K

K=1">K=1K=1, because depth > mm

τc=0.28Nmm²">τc=0.28Nmm²τc=0.28Nmm²

Then,τv< τc">τv< τcτv< τc................Hence OK.

Now,τv< τc">τv< τcτv< τc

VuBd=K⋅τc">VuBd=K⋅τcVuBd=K⋅τc

d=VuB⋅K⋅τc">d=VuB⋅K⋅τcd=VuB⋅K⋅τc

d=1049.4⋅104000⋅1⋅0.28">d=1049.4⋅104000⋅1⋅0.28d=1049.4⋅104000⋅1⋅0.28

d=937mm">d=937mmd=937mm

d provided =953>937mm">=953>937mm=953>937mm

Therefore, OK for One-way shear.

 

CHECK FOR TWO WAY SHEAR:- For, two-way shear, critical section is located at a distance ‘d/2’ from the face of the column all around.

The width bo=b+d=400+953=1353mm">bo=b+d=400+953=1353mmbo=b+d=400+953=1353mm

Now the Net shear force acting in the perimeter.

F = ρo [B²-bo2]">F = ρo [B²−b2o]F = ρo [B²-bo2]

= ρo [B²-(b+d)²]">= ρo [B²−(b+d)²]= ρo [B²-(b+d)²]

=206.25[4–(1.35)²]">=206.25[4–(1.35)²]=206.25[4–(1.35)²]

=449.10KN">=449.10KN=449.10KN

Now ultimate shear force, Fu=1.5⋅F">Fu=1.5⋅FFu=1.5⋅F

=1.5⋅449.10=673.65KN">=1.5⋅449.10=673.65KN=1.5⋅449.10=673.65KN

Now, from IS 456-2000

τv= Fu4bd= =0.13Nmm²">τv= Fu4bd= =0.13Nmm²τv= Fu4bd= =0.13Nmm²

Now, permissible shear stress(ks τc)">(ks τc)(ks τc)

Where,Ks=1">Ks=1Ks=1

τc =0.25 √fck=0.25 √25=1.25">τc =0.25 √fck=0.25 √25=1.25τc =0.25 √fck=0.25 √25=1.25

ks ⋅τc =1⋅1.25=1.25">ks ⋅τc =1⋅1.25=1.25ks ⋅τc =1⋅1.25=1.25

Therefore, τv< τc">τv< τcτv< τc  .........Hence, safe in the two-way shear

 

RESULT:- Hence the given 400*400 footing has been designed successfully.