Design of Spread Footing

Introduction to Spread Footing Design Flow Charts

STRUNETCONCRETE DESIGN AIDS

Strunet.com: Spread Footing Design V1.01 - Page 1

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 e<L/6 flexural equations are used, and for e>L/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.

As = area or reinforcement. b = column width dimension. bo = perimeter of critical shear section for footing. B = footing width dimension. d = distance from extreme compression fiber to centroid of tension

reinforcement. db = nominal diameter of bar. f’c = specified compressive strength of concrete. fy = specified tensile strength of reinforcement. h = overall member thickness. l = column length dimension. lava = available length for bar development. ld = development length of bar in tension. ls = compression lap splice length. ldb = basic development length of bar in compression. L = footing length dimension. Po = axial load, service. Pu = axial load, ultimate. qact = actual soil pressure based on service loads condition. qall = allowable soil bearing pressure. qs = factored actual soil pressure. Ru = coefficient of resistance. Vu = factored shear force at section considered. Vc = nominal shear strength of concrete. βc = ratio of long side to short side of column dimensions. ρ = ratio of tension reinforcement. ρb = ratio of tension reinforcement at balanced strain condition. ρmax = maximum ratio permissible by code. ρmin = minimum ratio permissible by code. ρreq’d = required ratio of tension reinforcement. ρprov’d = provided ratio of tension reinforcement. φ = strength reduction factor.

Notations for Spread Footing Design Flow Charts


Strunet.com: Spread Footing Design V1.01 - Page 2

Force transfer atcolumn/footing

for compressionforce only

Reinforcement

One Way Shear(Beam Action)

Two Way Shear(Slab Action)

Ultimate DesignForces Vu & Mu

RebarDevelopment

EquilibriumEquations


Ultimate SoilsPressure

FlexuralEquations


Ultimate SoilsPressure

Shear Check

Footing subjectedto vertical load only

Preliminary Size Preliminary Size

Main Input &Notation

Footing Subjectedto vertical loadand moment

Strunet.com: Spread Footing Design V1.01- Page 3


Spread Footing Analysis & DesignMain Chart

footing sizeis given?

= oF

all

PAq

square orrect. footing?

≅ = FB L A

Roundup B,L

FA =BL

us

F

Pq =A

YES

SquareRect.

NO

PreliminarySize

ultimate bearingpressure

oact all

F

Pq = < qA

l= column longer dimensionb= column shorter dimension

Proceed toultimate Design

forces

Footingsubjected to

vertical load only

proportion offooting w/ column

′ = 4a

′ = +2b (l b)

′ = − Fc lb A

′ ′ ′ ′+′ =′

2 42

-b b - a cka

= +2L l k'

FAB=L



Preliminary Size of Footing Subjected to Vertical Loads only.

one way shear(beam action)

( )u sV = q B 0.5L-0.5l - d

( )u sV = q L 0.5B-0.5b-d

two way shear(slab action)

( ) ( ) ob = 2 l +d + b+d use w/ Vccalculation

( ) ( ) u s FV = q A - l +d b+d

finding Mu

TransverseDirection

LongitudinalDirection

( )2u sM = 0.125q L B- b

( )2u sM = 0.125q B L- l

ShortDirection

long.Direction

L

l

bB

L

qs

d

d/2 Pud/2

l

qs

finding Vu

L

B

l+d

b+d

Note: the following footing forcescalculations are based on:

l= column dimension parallel to Lb= column dimension parallel to B



STRUNETCONCRETE DESIGN AIDS Ultimate Forces for Footing Subjected to Vertical Loads only

L>6e

o o2

all

P 6M1B = +q L L

max allq < q

o omin

F F

P Mq = -A S

oe

o

3ML =1.5L -P

o

all e

2PB =q L

B & L

omax

e

2Pq =BL

max allq > q

STOP.Increase B or L

Proceed to UltimateSoils BearingPressure, use

flexural equations

minq >0.0

Proceed to UltimateSoils BearingPressure, use

equilibrium equations

Lmax= maximumpermissible footing

length .

o

o

Me =P

L=Lmax

NO

YESYES

YES

NO NO

NO

Proceed to UltimateBearing Pressure, useequilibrium equations

L

qmax

Mo

Po

qmin

e

Soils Pressure distribution if

L

qmax

Mo

Poe

Soils Pressure distribution if

Le

PreliminarySize

YES


=FA BL

=2

6FBLS

= +o omax

F F

P MqA S

6Le <

6Le >


Preliminary Size of Footing Subjected to Vertical Load and Moment

ShortDirection

long.Direction

u umax

F F

P Mq = +A S

u umin

F F

P Mq = -A S

minq >0.0

mine

min max

q LL = L -q +q

STOP.go to equilibrium for

continuation

max minq = q -qδ

( )

( )

( )

( )

( )

1 min

2 min

3 min

4 min

5 min

0.5 L - l - dq = q + q

L

0.5 L - lq = q + q

L

0.5 L+ lq = q + q

L

0.5 L+ l + dq = q + q

L

0.5 L+ l + dq = q + q

L

δ

δ

δ

δ

δ

L

qmax

q5

d

d/2

Mu

Pud/2

q2q1

l

qmin

q3q4


one way shear(Beam Action)

( )( )u max 5V = 0.5B q +q 0.5L-0.5l - d

( )( )u max minV = 0.5L q +q 0.5B-0.5b-d

two way shear(Slab Action)

finding Vu

finding Mu

( ) ( )2u 3 maxM = 0.0625B q +q L- l

( ) ( )2u min maxM = 0.0625L q +q B- bTransverseDirection


YESNO

( )( )( )( ) ( )

( )( )

u min 1

1 4

4 max

V = 0.25B q +q L - l - d

+0.5 q +q B - b - d l +d

+0.25B q +q L - l - d



Ultimate Forces with Flexural Equations for FootingSubjected to Vertical Load and Moment

Ultimate BearingPressure using

Equilibrium Equations

one way shear(Beam Action)

ShortDirection

long.Direction

two way shear(Slab Action)

STOP. increase L

YESNO


TransverseDirection


finding Mu

NO


2 umax

e

PqBL

=

31 5 ue

u

ML . LP

= −

( )

( )

( )

( )

( )

1

2

3

4

5

0 5 2

0 5 2

0 5 2

0 5 2

0 5 2 2

maxe

e

maxe

e

maxe

e

maxe

e

maxe

e

. qq L L l dL

. qq L L lL

. qq L L lL

. qq L L l dL

. qq L L l dL

= − − −

= − −

= − +

= − + +

= − + +

( )( )50 5 0 5 0 5u maxV . B q q . L . l d= + − −

( )0 5 0 5 0 5u max eV . q L . B . b d= − −

( )0 5eL . L l d> + +

( )( )( )( )

4

4

0 25 0 25 2

u max

e

V . B q q L l d

. q L L l d B b d

= + − −

+ − + + − −

( )( ) ( ) ( )

( ) ( )

1

1 4

4

0 25 2 0 5 0 25

u e

max

V . q B L L l d

. q q B b d l d

. B q q L l d

= − − −

+ + − − +

+ + − −

0 5 0 5eL . L . l> +

( )20 0625u e maxM . L q B b= −

( )( )230 0625u maxM . B q q L l= + −

YES


L

qmax

q5

d

d/2

Mu

Pud/2

q2q1

l

q3q4

Le

Ultimate Forces with Equilibrium Equations forFooting Subjected to Vertical Load and Moment

is fct given?

use ACI11.3.2.1

repeat check

one wayshear o.k.

NO

YESNO

NO

ACI 15.5.1ACI 11.12

ACI 11.3.1.1

' 100cf psi≤

ACI 11.1.2

ACI 11.2.1.2 ACI 11.2.1.1

YESNO

see Vucalculations

ACI 9.3.2.3

As = providedflexural reinf.

YES

can increase fc'

or footing depth?

Req'd increaseVu=φVcfind d or f'c

One way Shear

Normal or LightWt Concrete

finding Vc

LIGHT

ACI 11.12.1.1

ACI 11.2.1

YES

NORMAL

, ,

w

w

b = Bb = L d = h- 3.5 , h=

footing width short directionfooting length long direction

footing depth


( )′=all-Light wt 0 75 2c c w:V . f b d

( )′=Sand Light wt 0 85 2c c w:V . f b d

=

≤

26 7

6 7

ctc w

'ctc

fV b d.

f f.

= 2 'c c wV f b d

cV

uV φ =0 85.

φ>u cV V

ρ = sw

w

Ab d

ρ

= + ≤ 1 9 2500 3 5' 'u

c c w w c wu

V dV . f b d . f b dM

φ>u cV V


One-Way Shear Check for Spread Footing


Two wayShear

option to useshear

reinforcement?

N.G., increase footingdepth d or f'c

RepeatCheck

two wayshear is o.k.

Proceed toreinforcement

NO YES

YES

ACI 11.12.1.2

( ) ( )2 +ob b d l d= + +

b & l are columnwidth and lenght

ACI 9.3.2.3

ACI 11.12.3

ACI 11.12.3.1ACI 11.12.3.2

YESNO

ACI 11.5.6.2

ACI 11.1.2

N.G. increase footingdepth d or f'c

YESNORepeatCheck

NO

L

B

l+d

b+d

≤100'cf psi

cV

β =clb

β

α

= +

= +

=

42

2

4

'c c o

c

'sc c o

o

'c c o

V f b d

dV f b db

V f b d

uVφ =0 85.

φ>u cV V

> 2 'c c oV f b d

= v ys

A f dV

s

s u cV =V - Vφ φ

6 'u c oV f b d>


Two-Way Shear Check for Spread Footing

YES

use deepersection or higher

strength

NO

NOACI 10.5.2YES

YES

NO

NO YES

NO YES

ACI 10.2.7.3

ACI 8.4.3

ACI 10.3.3

YES

proceed to rebardevelopment

ACI 7.12.2

ACI 10.3.3

ACI 10.2.7.3

see Mucalculations

ACI 9.3.2.1

NO


uM

φ= 2uMuRbd

φ = 0 9.

ρ ′

= − − ′

0 85 21 10 85

c ureq' d

y c

. f Rf . f

ρ ρ≥req' d min

ρ ρ≤req' d max

ρ ρ=1 33 req' d.

ρ ρ< Min

ρ ρ=1 33 req' d.use Minρ ρ=

ρ ρ= ≥ =mins s minA bd A bh

ρfinding min

> 60 ksi ?yf

> 60 ksi ?yf 60,0000.0018minyf

ρ

=

0.002minρ =0.0018minρ =

ρmin

MAX b=0.75ρ ρ

10 85 87 000

87 000c

by y

. f ,f , f

ρ β ′

= +

′ ≤ 4000cf psi

ρfinding max

β =1 0 85.c1

f - 4000=0.85 - 0.05 0.651000

β′ ≥


Area of Reinforcementfor Spread Footing


c=one-half bar spacing , orcenter of bar to the nearestconcrete surface, which is smaller

ACI 12.2.4

ACI 12.2.3

ACI 318-95 12.2.3

ACI 12.2.3

RebarDevelopment

ktr=0.0 for footing

+≤ 2 5tr

b

c k .d

αβ ≤1 7.

′ ≤100cf psi

αβγλ=

+′340

yd b

trc

b

fl dc kf

d

γγ==

0 8 for bar size 6 or smaller.1 0 for bar size 7 or larger...

λλ

λ

==

′= ≥

1 0 , normal weight concrete.1 3 , light weight concrete, if is not s pecified.6 7 1 0 , light weight concrete, if is speci fied

ct

c ctct

.

. f. f . ff

βββ

= 1.5 Epoxy coated w/ cover < 3db and cl ear spacing < 6db= 1.5 all other epoxy coated= 1.5 uncoated

αα

=1.3 fresh concrete below bars is more t han 12"=1.0 fresh concrete below bars is 12" or less


Rebar Development


ACI 15.8

Bearing strengthof column

Bearing strengthof footing

select dowelsreinforcement

the largest

ACI 15.8.1.1

ACI 15.8.2.1 ACI 15.8.1.2YESNO

forcompression

force only

Proceed todowels

development

ACI 10.17.1

the least

ACI 10.17.1

φ==

1

0 7 ACI 9.3.2.4A bl

.′ ′> 2cc cff f

′ ′=′ ′=

footing column

cf c

cc c

f ff f

[ ]φ φ ′= 21

1

0 85nb cfAP ( . f A )A

φ φ ′= 10 85nb ccP ( . f A )

φ ′= 11 19nb cfP . f Aφ ′= 10 595nb ccP . f A

φ nbP

φφ−

= u nbs

y

P PAf

= 10 005minsA . A

prov ' dsA

= req' d

prov ' d

sr

s

Ak

A

≤2

1

2 0A .A

( )=max req' d mins s sA A ,A

u nbP Pφ>


Forces Transfer at Column/Footing Interface


col. bars are #14or #18 and in

compression only?

dowel comp. lapsplice ls

use larger number ofsmaller size dowels, orincrease footing depth

STOP. dowels arefully developed.

development isthe largest of

RepeatCheck

ACI 15.8.2.3

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

ACI 15.8.2.3

ACI 12.3.2

ACI 12.16.1

ACI 12.3.2

ACI 12.3.3.1

dowelsdevelopment

into footingl1

into columnl2

NOYES

NO YES

NO YES

the largest

NO YES

l 2l 1

= column rebar & dowelsyc yf f

= ≥′

0 020 0003b yc

db b yccc

. d fl . d f

f

= ≥′

0 020 0003b yc

db b yccf

. d fl . d f

f ≤ 60 yf ksi

= >0 0005 12s b ycl . d f "( )= − >0 0009 24 12s yc bl . f d "=d r dbl k l

= ≥′

0 020 0003b yc

db b yccc

. d fl . d f

f

′ <3000 cf psi

=1 33s sl . l=s sl l

sl

>d aval l

= −6aval h

( )2 max s dbl l ,l=


Column Dowels Development

Documents

Design of Spread Footing