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Footings Example 1Design of a square spread footing of a seven-story building

Design and detail a typical square spread footing of a six bay by five bay seven-story building,

founded on stiff soil, supporting a 24 in. square column. The building has a 10 ft high basement.

The bottom of the footing is 13 ft below finished grade. The building is assigned to Seismic

Design Category (SDC) B.

Given:

Column load

Service dead loadD = 541 kipService live loadL = 194 kip

Seismic load E= 18 kip

(Column force due to the building frame

resisting the seismic load)

Material propertiesConcrete compressive strengthf

c= 4 ksi

Steel yield strengthfy= 60 ksi

Normalweight concrete = 1Density of concrete = 150 lb/ft3

Allowable soil-bearing pressures

Donly: qall.D= 4000 p

sf

D +L: qall,D+L= 5600 psfD +L + E: qall.Lat= 6000 psf

Fig. 1.1Rectangular foundation plan.

ACI 318-14 Procedure Computation

Step 1: Foundation type

13.1.1 This bottom footing is 3 ft belowbasement slab. Therefore, it is

considered a shallow foundation.

Step 2: Material requirements

13.2.1.1 The mixture proportion mustsatisfy the durability requirements

of Chapter 19 (318) and structural

strength requirements. The

designer determines the durabilityclasses. Please see Chapter 2 of

SP-17 for an in-depth discussion

of the categories and classes.

ACI 301 is a referencespecification that is in sync with

ACI 318. ACI encourages

By specifying that the concrete mixture shallbe in accordance with ACI 301 and providing

the exposure classes, Chapter 19 requirements

are satisfied.

Based on durability and strengthrequirements, and experience with localmixtures, the compressive strength of concrete

is specified at 28 days to be at least 4000 psi.


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referencing 301 into job

specifications.

There are several mixture optionswithin ACI 301, such as

admixtures and pozzolans, which

the designer can require, permit,or review if suggested by the

contractor.

Example 1 provides a more

detailed breakdown ondetermining the concrete

compressive strength and

exposure categories and classes.

Step 3: Determine footing dimensions

13.3.1.1 To calculate the footing base area,

divide the service load by theallowable soil pressure.

Area of footing =

total service load ( )

allowable soil pressure, qa

P

Assuming a square footing.

The footing thickness is

calculated in Step 5, footing

design.

The unit weights of concrete and soil are 150

pcf and 120 pcf; close. Therefore, footingself-weight will be ignored:

2

.,

135 ft4

541 k

ksfall D

D

q (controls)

,

2541 k+ 194 k 131 ft5.6 ksf

all D L

D L

q

2

.,

D + L +E 541 k + 194 k + 18 k126 ft

6 ksfall Latq

2 135 ft 11.6 ftl

Therefore, provide 12 x 12 ft square footing.

Step 4: Soil pressure

Footing stability

Because there is no overturningmoment, overall footing stability

is assumed.

Calculate factored soil pressure

uPq

Areau


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5.3.1(a)

5.3.1(b)

5.3.1(d)

5.3.1(e)

Calculate the soil pressuresresulting from the applied

factored loads.

Load Case I: U= 1.4D

Load Case II: U= 1.2D+ 1.6L

Load Case IV: U= 1.2D+ E +L

Load Case IV: U= 0.9D+ E

The load combinations includesthe seismic uplift force. In this

example, uplift does not occur.

U = 1.4D = 1.4(541 kip) = 757 kip

2

757 kipq = 5.3 ksf

144 ftu

U = 1.2D + 1.6L = 1.2(541 kip) + 1.6(194 kip)

= 960kip

2

960 kipq = 6.7 ksf

144 ftu (controls)

U = 1.2D + 1.0E+ 1.0L

= 1.2(541 kip) + 18 kip + 1.0(194 kip) = 861 kip

2

861 kipq = 6.0 ksf

144 ftu

U = 0.9D + 1.0E= 0.9(541 kip)+ 18 kip = 505 kip

2

505 kipq = 3.5 ksf

144 ftu

13.3.2.1 Because the footing has equal dimension in plan, it will be designed in onedirection and symmetry is assumed.

Step 5: One-way shear design

Fig. 1.2One-way shear in longitudinal

direction.


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21.2.1(b)

7.5.1.1

7.5.3.122.5.1.1

22.5.5.1

7.4.3.2

20.6.1.3.1

Shear reduction factor:

Vn Vu

n c sV V V

Therefore:

'2c c wV f b d

And satisfying: u cV V

The critical section for one-wayshear is at a distance dfrom the

face of the column (Fig. 1.2).

The engineer could either assume

a value for dthat satisfies thestrength Eq. (22.5.5.1) by iteration

or solve Eq. (7.5.1.1).

In this example, the first approach

is followed:

Assume that the footing is 30 in.thick.

The cover requirement is 3 in. to

bottom of reinforcement. Assume

that No. 8 bars are used in theboth directions and design for themore critical case (upper layer).

Therefore, the effective depth d:

d= 30 in.3 in.1 in.1 in./2

= 25.5 in.

( )2 2

n u ul c

V V d bq

shear= 0.75

Assume Vs= 0 (no shear reinforcement)

Vn= V

c

(12 ft) 24 in. 25.5 in.( )

in. in.22(12 ) 12

ft ft

(12 ft)(6.7 ksf) = 231 kip

uV

0.75(2) 4000 psi (12 ft)(25.5 in.)

(12 in./ft) = 348 kip

cV

Vc= 348 kip > Vu= 231 kip OK

Therefore, assumed depth is adequate:

h= 30 in.


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Step 6: Two-way shear design

22.6.1.2

22.6.1.422.6.4.1

22.6.2.1

22.6.5.1

22.6.5.2

The foundation will not be

reinforced with shear

reinforcement for two-way action.Therefore, the nominal shear

strength for two-way foundation

without shear reinforcement isequal to the concrete shear

strength:

vn= vc

Under punching shear theory,

inclined cracks are assumed to

originate and propagate at 45degrees away and down from the

column corners. The punch area is

calculated at an average distanceof d/2 from column face on all

sides (Fig. 1.3).

bo= 4(c+ d)

ACI 318 permits the engineer to

take the average of the effective

depth in the two orthogonal

directions when designing thefooting, but in this example the

smaller effective depth will beused.

The two-way shear strength

equations for nonprestressed

members must be satisfied and theleast calculated value of (a), (b),

and (c) controls:

'

'

4 (a)

4(2 ) (b)

c c

c c

v f

v f

where is ratio of the long side to

short side of column; = 1

'( 2) (c)sc co

dv f

b

Fig. 1.3Two-way shear.

bo= 4(24 + 25.5) = 198 in.

4(1.0)( 4000 psi) = 253 psicv

4

(2 )(1.0)( 4000 psi) = 379.5 psi1cv

(40)(25.5 in.)( 2)( 4000 psi) = 452 psi

198 in.cv

Equation (a) controls; vc= 253 psi


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22.6.5.3

21.2.1(b)

8.5.1.1

s= 40, considered interior

column

'4 b dc c o

V f

Use a reduction factor of 0.75:

'(0.75)4 b dc c oV f

2 2[(a) ( ) ]u uV q c d

Check if design strength exceeds

required strength:VcVu?

(253 psi)(198 in.)(25.5 in.)= 1277 kip

1000 lb/kipcV

= 0.75

Vc= 0.75(1277 kip) = 958 kip

224 in.+25.5 in.[(12 ft)(12 ft) ( ) ](6.7 ksf)

12 in./ft

851 kip

uV

Vc= 958 kip> Vu= 851 kip OK

Two-way shear strength is adequate.

Step 7: Flexure design

13.2.71 The critical section is permitted to

be at the face of the column (Fig.

1.4).

2( ) ( ) / 22

u ul c

M q b

Fig. 1.4Flexure in the longitudinal

direction.

2

24 in.12 ft

12 in./ft(6.7 ksf)( ) (12 ft) / 22

uM


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22.2.1.1

22.3.1.1

22.2.2.1

22.2.2.2

21.2.1(a)

8.5.1.1(a)

8.6.1.1

13.3.3.3(a)

21.2.1(a)

21.2.2

22.2.2.4.122.2.2.4.3

Set compression force equal to

tension force at the column face:

C= T

C = 0.85fcbaand T = Asfy

'0.85

s y

c

A fa

f b and

( )2

n s ya

M A f d

Substitute for ain the equation

above.

Use reduction factor from Table

21.2.1.

Setting Mn

Mu= 1005 ft-k andsolving for As:

Check the minimum

reinforcement ratio: l= 0.0018

Check if the tension controlled

assumption and the use of = 0.9 is correct.

To answer the question, the

calculated tensile strain inreinforcement is compared to the

values in Table 21.2.2. The strain

in reinforcement is calculatedfrom similar triangles (refer to

Fig. 1.5):

c= ( )

c

t d c

where:1

a=

c

and a= 0.28As

Mu= 1005 ft-kip

ss

A (60 ksi)0.1A

0.85(5 ksi)(12 ft)a

= 0.9

s(0.1)A

(0.9) (60 ksi)(25.5 in. )2n sM A

As= 8.91 in.2

As,min= 0.0018(12 ft)(12 in./ft)(30 in.)

= 7.8 in.2 0.005


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Section is tension controlled and = 0.9

Fig. 1.5Strain distribution across footing.

Step 8: Transfer of column forces to the base

16.3.1.1

22.8.3.2

22.8.3.2(a)

22.8.3.2(b)

Factored forces are transferred tothe foundation at the base of the

column by bearing on concrete

and the reinforcement dowels.

The foundation is wider on allsides than the loaded area.

Therefore, the nominal bearing

strength,Bn, is the smaller of the

two equations.

'21

1

(0.85 )n cA

B f AA

and

'12(0.85 )n cB f A

Check if 2

1

2.0A

A where

A1is the bearing area of thecolumn andA2is the area of thepart of the supporting footing that

is geometrically similar to and

concentric with the loaded area.

22

21

[(12 ft)(12 in./ft)]6 2

(24 in.)

A

A

Therefore, Eq. (22.8.3.2(b)) controls.


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21.2.1(d)

16.3.4.1

16.3.5.4

25.4.9.2

25.3.1

The reduction factor for bearing is

0.65:

Column factored forces aretransferred to the foundation by

bearing and throughreinforcement dowels. Providedowel reinforcement area of at

least 0.005Agand at least four

bars.

Bars are in compression for all

load combinations. Therefore, the

bars must extend into the footinga compression development

length, ldc, the larger of the two:

'50

(0.0003 )

y rb

dc c

y r b

fd

l f

f d

The footing depth must satisfy the

following inequality so that thevertical reinforcement can be

developed within the provided

depth:

, ,2 3 in.dc b dwl b barsh l r d d

where

r= radius of No. 6 bent = 6db

bearing= 0.65

2nB (0.65)(2)(0.85)(4000 psi)(24 in.)

Bn= 2546 kip > 960 kip (Step 4) OK

As,dowel= 0.005(24 in.)2= 2.88 in.2

Use eight No. 6 bars

0.02(60,000 psi)(0.75 in.) 14.3 in.

4000 psidcl

0.0003(60,000 psi)(0.75 in.) 13.5 in.dcl

ldc= 14.3 in. (controls)

hreqd= 14.3 in. + 6(0.75 in.) + 0.75 in.

+ 2(0.75 in.) + 3 in. = 24.05 in.

hreqd= 24.1 in. < hprov.= 30 in. OK

Step 9: Footing details

13.2.8.3

13.2.7.1

Development length

Reinforcement development is

calculated at the maximumfactored moment, which occurs atthe column face. Bars must extend


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25.4.2.2

25.4.2.4

a tension development length

beyond the critical section.

'

3{ }

40

y t e sd b

trc

b

fl d

c Kf

d

where

tbar location; not more than12 in. of fresh concrete belowhorizontal reinforcement

e coating factor; uncoated

sbar size factor; No. 8 andlarger

cb= spacing or cover dimension

to center of bar, whichever is

smaller

tr= transverse reinforcementindex

It is permitted to useKtr= 0.

But the expression:b tr

b

c K

d

must

not be taken greater than 2.5.

3 60, 000 (1.0)(1.0)(1.0){ }

40 2.5(1.0) 4000

= 28.5d

d b

b

psil d

psi

No. 6:b tr

b

c K

d

=

3.5 in. 03.5 2.5

1.0 in.

No.8 bars: 28.5(1.0 in.) = 28.5 in. > 12 in. OK

ldin the longitudinal direction:

ld,prov. = ((12 ft)(12 in./ft)24 in.)/23 in.

ld,prov. = 57 in. > ld,reqd= 28.5 in. OK

use straight No. 8 bars in both directions.


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Step 10: Detailing

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