Strunet Spread Footing Design Flow

7/30/2019 Strunet Spread Footing Design Flow

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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 soilpressure for concentric load, and a triangular or trapezoidal soil

pressure for eccentric loading (combined axial and bending)

• Establish preliminary size under service loads, and proportionrectangular 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 nominalconcrete shear strength with restricted foot ing depth.

• Calculate required flexural reinforcement ratio and compared with the

minimum and maximum permitted by code, and provide requiredtensile 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 requiredinto 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 tensionreinforcement.

d b = nominal diameter of bar.

f’ c = specified compressive strength of concrete.

f y = specified tensile strength of reinforcement.

h = overall member thickness.

l = column length dimension.

l ava = available length for bar development.

l d = development length of bar in tension.

l s = compression lap splice length.

l db = basic development length of bar in compression.

L = footing length dimension.

P o = axial load, service.

P u = axial load, ultimate.

qact = actual soil pressure based on service loads condition.

qall = allowable soil bearing pressure.

qs = factored actual soil pressure.

R u = coefficient of resistance.

V u = factored shear force at section considered.

V c = nominal shear strength of concrete.

β c = ratio of long side to short side of column dimensions.

ρ = ratio of tension reinforcement.

ρ b = ratio of tension reinforcement at balanced strain condition.

ρ max = maximum ratio permissible by code.

ρ min = minimum ratio permissible by code.

ρ req’d = required ratio of tension reinforcement.

ρ prov’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 V u & M u

Rebar Development

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?

=o

F

all

P A

q

square or

rect. footing?

≅ = F B L A

Roundup B,L

F

A = BL

us

F

P q =

A

YES

SquareRect.

NO

PreliminarySize

ultimate bearingpressure

oact all

F

P q = < q

A

l= column longer dimensionb= column shorter dimension

Proceed toultimate Design

forces

Footingsubjected to

vertical load only

proportion of footing w/ column

′ = 4a

′ = +2b (l b)

′ = − F c lb A

′ ′ ′ ′+′ =

′

2 4

2

-b b - a c k

a

= +2L l k'

F AB =

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/ V c

calculation

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

finding M u

TransverseDirection

LongitudinalDirection

( )2

u sM = 0.125q L B - b

( )2

u sM = 0.125q B L - l

ShortDirection

long.Direction

L

l

b B

L

qs

d

d/2 Pud/2

l

qs

finding V u

L

B

l+d

b + d

e: the following footing forcesculations are based on:column dimension parallel to L

= column dimension parallel to B

Ultimate Design

Forces V u & M u


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



L>6e

o o

2

all

P 6M 1B = +

q L L

max all q < q

o omin

F F

P M q = -

A S

oe

o

3M L =1.5L-

P

o

all e

2P B =

q L

B & L

omax

e

2P q =

BL

max all q > q

STOP.Increase B or L

Proceed to UltimateSoils BearingPressure, use

flexural equations

minq >0.0

Proceed to UltimateSoils BearingPressure, use

equilibrium equations

Lmax = maximum

permissible footinglength .

o

o

M

e = P

L=Lmax

NO

YESYES

YES

NO NO

NO

oceed to Ultimate

ring Pressure, useuilibrium equations

L

qmax

Mo

Po

qmin

e

Soils Pressure distribution if

L

qmax

Mo

Poe

Soils Pressure distribution if

Le

PreliminarySize

YES


=F A BL

=2

6

F

BLS

= +o omax

F F

P M q

A S

6

Le <

6

Le >


Preliminary Size of Footing Subjected to Vertical Load and Moment



ShortDirection

long.Direction

u umax

F F

P M q = +

A S

u umin

F F

P M q = - 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 - d q = q + q

L0.5 L- l

q = q + qL

0.5 L+l q = q + q

L

0.5 L+l+d q = q + q

L

0.5 L+ l + d q = 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 5 V =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 V u

finding M u

( )( )2

u 3 max M =0.0625B q +q L- l

( )( )2

u min max M =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.ncrease L

YESNO


TransverseDirection


finding M u

NO


2 u

max e

P

q BL=

31 5 u

e

u

M L . L

P = −

( )

( )

( )

( )

( )

1

2

3

4

5

0 52

0 52

0 52

0 5 2

0 52 2

max e

e

max e

e

max e

e

max e

e

max e

e

. qq L L l d

L

. qq L L l

L

. qq L L l

L

. qq L L l d L

. qq L L l d

L

= − − −

= − −

= − +

= − + +

= − + +

( ) ( )50 5 0 5 0 5u max V . 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 > +

( )2

0 0625u e max M . L q B b= −

( )( )2

30 0625u max M . B q q L l = + −

YES


L

qmax

q5

d

d/2

Mu

Pu d/2

q2q1

l

q3q4

Le

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



is f ct given?

use ACI11.3.2.1

repeat check

one wayshear o.k.

NO

YESNO

NO

ACI 15.5.1 ACI 11.12

ACI 11.3.1.1

'

100c f psi≤

ACI 11.1.2

ACI 11.2.1.2 ACI 11.2.1.1

YESNO

see V ucalculations

ACI 9.3.2.3

As = provided

flexural reinf.

YES

can increase f c '

or footing depth?

Req'd increase

V u=φ V c

find d or f' c

One way Shear

Normal or LightWt Concrete

finding V c

LIGHT

ACI 11.12.1.1

ACI 11.2.1

YES

NORMAL

,

,

w

w

b = B

b = Ld = h-3.5 , h=

footing width short direction

footing length long directionfooting 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

ct c w

' ct c

f V b d

.

f f .

= 2 '

c c w V f b d

c V

uV φ = 0 85.

φ >u c V V

ρ = sw

w

A

b d

ρ

= + ≤ 1 9 2500 3 5' ' u

c c w w c w

u

V d V . f b d . f b d

M

φ >u c V V


One-Way Shear Check for Spread Footing




Two wayShear

option to useshear

reinforcement?

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

Repeat

Check

two wayshear is o.k.

Proceed toreinforcement

NO YES

YES

ACI 11.12.1.2

( ) ( )2 +o

b b d l d = + +

b & l are columnwidth and lenght

ACI 9.3.2.3

ACI 11.12.3

ACI 11.12.3.1 ACI 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'

c

f psi

c V

β =c

l

b

β

α

= +

= +

=

42

2

4

'

c c o

c

' sc c o

o

'

c c o

V f b d

d V f b d

b

V f b d

uV

φ = 0 85.

φ >u c V V

> 2 '

c c oV f b d

= v y

s

A f d V

s

s u c V =V - V φ φ

6 '

u c oV f b d >


Two-Way Shear Check for Spread Footing



YES

use deeper section or higher

strength

NO

NO ACI 10.5.2

YES

YES

NO

NO YES

NO YES

ACI 10.2.7.3

ACI 8.4.3

ACI 10.3.3

YES

proceed to rebar development

ACI 7.12.2

ACI 10.3.3

ACI 10.2.7.3

see M ucalculations

ACI 9.3.2.1

NO


uM

φ =

2u

Mu

R bd φ = 0 9.

ρ ′

= − − ′

0 85 21 1

0 85c u

req' d

y c

. f R

f . 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 .e Min ρ ρ =

ρ ρ = ≥ =mins s min A bd A bh

ρ finding min

> 60 ksi ?y f

> 60 ksi ?y f 60,000

0.0018min

y f ρ

=

0.002min ρ =0.0018min ρ =

ρ min

MAX b=0.75 ρ ρ

1

0 85 87 000

87 000

c b

y y

. f ,

f , f

ρ β ′

= +

′ ≤ 4000c f psi

ρ finding max

β =1 0 85.c 1

f -4000 = 0.85 - 0.05 0.65

1000 β

′ ≥


Area of Reinforcementfor Spread Footing




c=one-half bar spacing , or

center of bar to the nearestconcrete surface, which is smaller

ACI 12.2.4

ACI 12.2.3

ACI 318-9512.2.3

ACI 12.2.3

Rebar Development

k tr =0.0 for footing

+≤ 2 5tr

b

c k .

d

αβ ≤1 7.

′ ≤ 100c f psi

αβγλ =

+′

3

40

y

d btr c

b

f l d

c k f d

γ

γ

=

=

0 8 for bar size 6 or smaller.

1 0 for bar size 7 or larger.

.

.

=

=

′= ≥

1 0 , normal weight concrete.1 3 , light weight concrete, if is not s pecified.

6 71 0 , light weight concrete, if is speci fied

ct

c ct

ct

.

. f

.f . f

f

β β

β

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

= 1.5 uncoated

α

α

=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

for compression

force only

Proceed todowels

development

ACI 10.17.1

the least

ACI 10.17.1

φ

=

=

1

0 7 ACI 9.3.2.4

A bl

.

′ ′> 2cc cf f f

′ ′=

′ ′=

footing

column

cf c

cc c

f f

f f

[ ]φ φ ′=2

1

1

0 85nb cf

AP ( . f A )

A

φ φ ′= 10 85nb cc P ( . f A )

φ ′= 11 19nb cf P . f Aφ ′= 10 595nb cc P . f A

φ nbP

φ

φ

−=

u nbs

y

P P A

f = 10 005

mins A . A

prov' d s A

=req' d

prov' d

s

r

s

Ak

A

≤2

12 0

A. A

( )= maxreq' d mins s s A A ,A

u nbP P φ >


Forces Transfer at Column/Footing Interface




col. bars are #14or #18 and in

compression only?

dowel comp. lapsplice l s

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

STOP. dowels arefully developed.

development isthe largest of

RepeatCheck

ACI 15.8.2.3

col. bar (d b 14 &

18) develop. length

ACI 15.8.2.3

ACI 12.3.2

ACI 12.16.1

CI 12.3.2

ACI 12.3.3.1

dowelsdevelopment

into footingl 1

into columnl 2

NOYES

NO YES

NO YES

the largest

NOYES

l 2

l 1

= column rebar

& dowels

yc y f f

= ≥′

0 02

0 0003b yc

db b yc

cc

. d f

l . d f f

= ≥′

0 020 0003

b yc

db b yc

cf

. d f . d f

f ≤ 60y f ksi

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

= ≥′

0 020 0003

b yc

db b yc

cc

. d f l . d f

f

′ < 3000c f psi

= 1 33s sl . l =s sl l

sl

>d aval l

= − 6aval h

( )2 max s dbl l ,l =


Column Dowels Development

Documents

Strunet Spread Footing Design Flow