Design Of Spread Footing

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Introduction to Spread Footing Design Flow Charts

Spread Footing Charts in Bullets: • • • • • •

• • •

All code provisions are listed, where applicable, on the charts for quick reference. Analysis assume rigid footing condition, resulting in a uniform soil pressure for concentric load, and a triangular or trapezoidal soil pressure for eccentric loading (combined axial and bending) Establish preliminary size under service loads, and proportion rectangular footing dimensions, if required, around a rectangular column. Calculate in one single equation one -way shear, two -way shear, and design moment, under factored loads, respectively. Deal separately with two eccentricity conditions, while eL/6 equilibrium equations are used. Drive the nominal shear strength of the concrete for bo th beam shear (one way) and punching shear (two way, or slab shear). Alternatively, provide reference to the code provisions where shear reinforcement may be used in case of factored shear force exceeded nominal concrete shear strength with restricted foot ing depth. Calculate required flexural reinforcement ratio and compared with the minimum and maximum permitted by code, and provide required tensile reinforcement, and calculate rebar development length. Address axial force transfer at the column base (f or compression only), and fully detailing the dowels design and development length required into footing and column. Include sketches illustrating the subject under investigation.

Include notations sheet explaining in details all symbols used in the char s.

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Notations for Spread Footing Design Flow Charts

As b bo B d

= = = = =

db f’c fy h l lava ld ls ldb L Po Pu qact qall qs Ru

= = = = = = = = = = = = = = = =

area or reinforcement. column width dimension. perimeter of critical shear section for footing. footing width dimension. distance from extreme compression fiber to centroid of tension reinforcement. nominal diameter of bar. specified compressive strength of concrete. specified tensile strength of reinforcement. overall member thickness. column length dimension. available length for bar development. development length of bar in tension. compression lap splice length. basic development length of bar in compression. footing length dimension. axial load, service. axial load, ultimate. actual soil pressure based on service loads condition. allowable soil bearing pressure. factored actual soil pressure. coefficient of resistance.

Vu

=

factored shear force at section considered.

Vc

= = = = = = = = =

nominal shear strength of concrete. ratio of long side to short side of column dimensions. ratio of tension reinforcement. ratio of tension reinforcement at balanced strain condition. maximum ratio permissible by code. minimum ratio permissible by code. required ratio of tension reinforcement. provided ratio of tension reinforcement. strength reduction factor.

βc ρ ρb ρmax ρmin ρreq’d ρprov’d φ

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Spread Footing Analysis & Design Main Chart Main Input & Notation

Footing subjected to vertical load only

Footing Subjected to vertical load and moment

Preliminary Size

Preliminary Size

Ultimate Design Forces Vu & Mu

Flexural Equations

Equilibrium Equations

Ultimate Soils Pressure

Ultimate Soils Pressure

Ultimate Design Forces Vu & Mu

Ultimate Design Forces Vu & Mu

Shear Check

One Way Shear (Beam Action)

Reinforcement

Rebar Development

Force transfer at column/footing

for compression force only

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Two Way Shear (Slab Action)

Preliminary Size of Footing Subjected to Vertical Loads only.

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Footing subjected to vertical load only

Preliminary Size

footing size is given?

NO

AF = l= column longer dimension b= column shorter dimension Rect.

Po qall

square or rect. footing?

proportion of footing w/ column

Square

B ≅ L = AF

a′ = 4

Roundup B,L

b′ = 2(l + b)

AF = BL

c′ = lb − AF

k′ =

YES

-b′ + b′2 - 4a′c′ 2a′

L = l + 2k'

B=

AF L

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qact =

ultimate bearing pressure

Po < qall AF

qs =

Pu AF

Proceed to ultimate Design forces

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Ultimate Forces for Footing Subjected to Vertical Loads only

L Ultimate Design Forces Vu & Mu

b

B

l

l finding Vu Note: the following footing forces calculations are based on: l= column dimension parallel to L b= column dimension parallel to B

L

one way shear (beam action)

d d/2

Short Direction

long. Direction

Vu = qsL ( 0.5B - 0.5b - d )

qs

qs

two way shear (slab action)

b+d

B

L l+d

Vu = qsB (0.5L - 0.5l - d )

bo = 2 ( l + d ) + ( b+ d ) Vu = qs  AF - ( l + d )( b+ d )

finding M u

Longitudinal Direction

Mu = 0.125qsB (L - l )

Transverse Direction

Mu = 0.125qsL (B - b )

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2

2

Pu d/2

use w/ Vc calculation

Preliminary Size of Footing Subjected to Vertical Load and Moment

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Preliminary Size

e=

L e Po

Mo Po

Mo

qmin

qmax

Lmax= maximum permissible footing length .

L=Lmax Soils Pressure distribution if

NO

e<

L 6

YES

L>6e

L e Po

Le =1.5L -

B=

3Mo Po

B=

2Po qall Le

qmax =

Proceed to Ultimate Bearing Pressure, use equilibrium equations

qmax Le

Soils Pressure distribution if

e>

L 6

BL2 SF = 6

2Po BLe

qmax > qall

Mo

AF = BL

B&L

NO

1  Po 6Mo   + 2  qall  L L 

qmax =

YES

NO

Po Mo + AF SF

qmax < qall

YES

qmin =

Po Mo AF SF

STOP. Increase B or L NO

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Proceed to Ultimate Soils Bearing Pressure, use equilibrium equations

qmin > 0.0

YES

Proceed to Ultimate Soils Bearing Pressure, use flexural equations

l

Ultimate Forces with Flexural Equations for Footing Subjected to Vertical Load and Moment

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L

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P M qmax = u + u AF SF qmin =

Le = L -

d/2

Pu Mu AF SF

qmin > 0.0

NO

d

qmin L qmin + qmax

Mu qmin qmax

YES

δ q = qmax - qmin

q2 q1

 0.5 ( L - l )  q2 = qmin + δ q   L  

continuation

Ultimate Design Forces Vu & Mu

 0.5 ( L + l )  q3 = qmin + δ q   L    0.5 ( L + l + d )  q4 = qmin + δ q   L  

finding Vu

 0.5 ( L + l ) + d  q5 = qmin + δ q   L  

one way shear (Beam Action)

long. Direction

q q5 q4 3

 0.5 ( L - l - d )  q1 = qmin + δ q   L  

STOP. go to equilibrium for

Short Direction

Pu d/2

Vu = 0.5B (qmax + q5 )(0.5L - 0.5l - d )

Vu = 0.5L (qmax + qmin )(0.5B - 0.5b - d ) two way shear (Slab Action)

Vu = 0.25B (q min + q1 )(L - l - d ) + 0.5 (q1 + q 4 )(B - b - d )(l + d ) + 0.25B (q 4 + q max )(L - l - d ) Strunet.com: Spread Footing Design V1.01- Page 7

finding M u

Longitudinal Direction

Transverse Direction

Mu = 0.0625B (q3 + qmax )(L - l )

2

Mu = 0.0625L (qmin + qmax )(B - b )

2

Ultimate Forces with Equilibrium Equations for Footing Subjected to Vertical Load and Moment

Ultimate Bearing Pressure using Equilibrium Equations

l

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Le = 1.5L −

L d d/2

Pu

qmax =

d/2

Mu qmax q5

q4

q3

q2

Ultimate Design Forces Vu & Mu

one way shear (Beam Action)

Short Direction

long. Direction

Vu = 0.5B ( q5 + qmax )( 0.5L − 0.5l − d ) Vu = 0.5qmaxLe ( 0.5B − 0.5b − d )

Longitudinal Direction

Transverse Direction

NO

STOP. increase L

Le > 0.5L + 0.5l

0.5qmax ( 2Le − L − l − d ) Le

q2 =

0.5qmax ( 2Le − L − l ) Le

q3 =

0.5qmax ( 2Le − L + l ) Le

q4 =

0.5qmax ( 2Le − L + l + d ) Le

q5 =

0.5qmax ( 2Le − L + l + 2d ) Le

finding M u

two way shear (Slab Action)

NO

Le > 0.5 ( L + l + d )

2Pu BLe

q1 = q1

Le

3Mu Pu

Mu = 0.0625B ( q3 + qmax )( L − l )

Mu = 0.0625Leqmax ( B − b )

2

YES

YES

Vu = 0.25B ( q4 + qmax )( L − l − d ) + 0.25q4 (2Le − L + l + d )(B − b − d )

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Vu = 0.25q1B ( 2Le − L − l − d ) + 0 .5 ( q1 + q4 )( B − b − d )( l + d ) + 0 .25B ( q4 + qmax )( L − l − d )

2

One-Way Shear Check for Spread Footing

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ACI 11.12.1.1

ACI 15.5.1 ACI 11.12

One way Shear

bw = L , footing length long direction

Normal or Light Wt Concrete

LIGHT

ACI 11.2.1.2

is fct given?

(

all-Light wt : Vc = 0.75 2 fc′bw d

(

f c' ≤ 100 psi

ACI 11.2.1

d = h - 3.5 , h = footing depth

NO

ACI 11.1.2

finding Vc

bw = B , footing width short direction

)

Sand Light wt : Vc = 0.85 2 fc′bwd

NORMAL

YES ACI 11.2.1.1

ACI 11.3.1.1

 f  Vc = 2  ct  bw d  6 .7   fct  '   ≤ fc  6 .7 

)

Vc see Vu calculations

ACI 9.3.2.3

φ = 0.85

Vu

NO

Vu > φVc

one way shear o.k.

Vc = 2 fc' bw d

YES

NO

can increase fc' or footing depth?

YES

use ACI 11.3.2.1

As = provided flexural reinf.

ρw =

Req'd increase Vu =φVc find d or f'c

As bw d

 Vd Vc = 1.9 fc' + 2500 ρw u  bw d ≤ 3.5 fc' bwd Mu  

NO

Vu > φVc

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YES

repeat check

Two-Way Shear Check for Spread Footing

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ACI 11.12.1.2 Two way Shear

ACI 11.1.2

bo = 2 ( b + d ) + ( l + d ) 

Vc

l b

 4  ' Vc =  2 +  fc bo d βc    α d Vc =  2 + s  fc' bod bo  

b+d

b & l are column width and lenght

L l+d B

βc =

fc' ≤ 100psi

Vc = 4 fc' bod ACI 9.3.2.3

Vu

φ = 0.85 Vu > φVc

NO

NO N.G. increase footing depth d or f'c

YES

option to use shear reinforcement?

ACI 11.12.3.2 NO

Repeat Check NO

Vu > 6 fc' bod

ACI 11.12.3 YES

Vc > 2 fc' bod YES

ACI 11.12.3.1 YES N.G., increase footing depth d or f'c

φVs =Vu - φVc ACI 11.5.6.2

Vs =

two way shear is o.k.

Av fy d s

Proceed to reinforcement

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Repeat Check

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Area of Reinforcement for Spread Footing

ACI 7.12.2

finding ρmin

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see M u calculations

Mu

ACI 9.3.2.1

φ = 0.9

Ru =

ρreq' d =

NO

Mu φ bd 2

fy > 60 ksi ?

fy > 60 ksi ?

YES

NO

0.85fc′  2Ru  1− 1−  0.85fc′  fy 

YES

 60,000   f y  

ρmin = 0.0018 

ρmin = 0.002

ρmin = 0.0018

ACI 10.3.3

ρmin

finding ρmax ACI 10.2.7.3

fc′ ≤ 4000psi

NO

YES

ACI 10.2.7.3

 fc′ - 4000   ≥ 0.65  1000 

β1 = 0.85 - 0.05  NO

ρreq' d ≥ ρmin

YES ACI 8.4.3

ρreq' d ≤ ρmax ACI 10.5.2

NO

ρ = 1.33ρreq' d

NO

use ρ = ρMin

ρ < ρMin

β1 = 0.85

use deeper section or higher strength YES

ρ = 1.33ρreq' d

As = ρbd ≥ Asmin = ρminbh proceed to rebar development Strunet.com: Spread Footing Design V1.01- Page 11

ρb = YES

0.85fc′  87,000 β1  fy  87,000 + fy

ACI 10.3.3

ρMAX =0.75 ρb

  

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ACI 318-95 12.2.3 Rebar Development

ACI 12.2.4

c=one-half bar spacing , or center of bar to the nearest concrete surface, which is smaller

k tr=0.0 for footing ACI 12.2.3

c + ktr ≤ 2.5 db

α =1.3 fresh concrete below bars is more t han 12" α =1.0 fresh concrete below bars is 12" or less β = 1.5 Epoxy coated w/ cover < 3db and cl ear spacing < 6db β = 1.5 all other epoxy coated β = 1.5 uncoated

αβ ≤ 1.7

fc′ ≤ 100psi ACI 12.2.3

γ = 0.8 for bar size 6 or smaller. γ = 1.0 for bar size 7 or larger.

λ = 1.0 , normal weight concrete. λ = 1.3 , light weight concrete, if fct is not s pecified. λ=

6.7 fc′ ≥ 1.0 , light weight concrete, if fct is speci fied fct

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ld =

3 fy αβγλ db 40 fc′ c + ktr db

Forces Transfer at Column/Footing Interface

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ACI 15.8

fcf′ = fc′ footing fcc′ = fc′ column

for compression force only

A1 = bl

φ = 0.7 ACI 9.3.2.4

fcc′ > 2fcf′ Bearing strength of column

Bearing strength of footing

ACI 10.17.1

φPnb = φ( 0.85fcc′ A1 )

φPnb =

A2 [φ( 0.85fcf′ A1 )] A1

A2 ≤ 2.0 A1

φPnb = 0.595fcc′ A1

φPnb = 1.19fcf′ A1

ACI 15.8.1.1the least

φ Pnb

ACI 15.8.2.1

NO

Pu > φPnb

YES ACI 15.8.1.2

Asmin = 0.005A1

As =

Pu − φPnb φ fy

the largest

Asreq' d = max ( As ,Asmin ) select dowels reinforcement

ACI 10.17.1

Asprov ' d

kr =

Asreq' d Asprov ' d

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Proceed to dowels development

Column Dowels Development

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l1

dowels development

into footing l1

into column l2

fyc = fy column rebar & dowels

ACI 15.8.2.3 NO

col. bars are #14 or #18 and in compression only?

YES ACI 15.8.2.3 development is the largest of

ACI 12.3.2

ldb =

0.02dbfyc fcc′

≥ 0.0003dbfyc

ACI 12.3.2

ldb =

0.02dbfyc fcf′

col. bar (db 14 & 18) develop. length

dowel comp. lap splice ls

ACI 12.16.1

≥ 0.0003dbfyc

fy ≤ 60 ksi

NO

YES

ldb =

0.02dbfyc fcc′

≥ 0.0003dbfyc

ACI 12.3.3.1

ld = kr ldb

ls = ( 0.0009fyc − 24) db > 12"

the largest

lava = h − 6 NO

NO

ld > lava

STOP. dowels are fully developed.

ls = 0.0005dbfyc > 12"

fc′ < 3000 psi

ls = 1.33ls

ls = ls

YES

use larger number of smaller size dowels, or increase footing depth

Repeat Check

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YES

ls

l2 = max ( ls ,ldb )