Footing design and analysis

8/18/2019 Footing design and analysis

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1

Footings

Footings

Definition

Footings are structural members used to support

columns and walls and to transmit and distribute

their loads to the soil in such a way that the load

bearing capacity of the soil is not exceeded,

excessive settlement, differential settlement,or

rotation are prevented and adequate safety

against overturning or sliding is maintained.



2

Types of Footings

Wall footings are used to

support structural walls that

carry loads for other floors

or to support nonstructural

walls.

Types of Footings Isolated or single footings

are used to support single

columns. This is one of the

most economical types of

footings and is used when

columns are spaced at

relatively long distances.





4

Types of Footings

Continuous footings

support a row of three or

more columns. They have

limited width and continue

under all columns.

Types of Footings Rafted or mat foundation

consists of one footing usually

placed under the entire building

area. They are used, when soil

bearing capacity is low, column

loads are heavy single footings

cannot be used, piles are not used

and differential settlement must

be reduced.



5

Types of Footings

Pile caps are thick slabs

used to tie a group of piles

together to support and

transmit column loads to the

piles.

Distribution of Soil Pressure

When the column load P is

applied on the centroid of the

footing, a uniform pressure is

assumed to develop on the soilsurface below the footing area.

However the actual distribution of the soil is not uniform,

but depends on may factors especially the composition of

the soil and degree of flexibility of the footing.



6

Distribution of Soil Pressure

Soil pressure distribution in

cohesionless soil.

Soil pressure distribution incohesive soil.

Design Considerations

Footings must be designed to carry the column loads

and transmit them to the soil safely while satisfying

code limitations.

The area of the footing based on the allowable bearing soil capacity

Two-way shear or punch out shear.

One-way bearing

Bending moment and steel reinforcement

required

1.

2.

3.

4.



7

Design Considerations

Footings must be designed to carry the column loads

and transmit them to the soil safely while satisfying

code limitations.

Bearing capacity of columns at their base

Dowel requirements

Development length of bars

Differential settlement

1.

2.

3.

4.

Size of Footings

The area of footing can be determined from the actual

external loads such that the allowable soil pressure is

not exceeded.

( ) pressure soil allowable

weight-self includingload Total footingof Area =

footingof area

uu

Pq =

Strength design requirements



8

Two-Way Shear (Punching Shear)

For two-way shear in slabs (& footings) Vc is smallest of

long side/short side of column concentrated

load or reaction area < 2

length of critical perimeter around the

column

where, βc =

b0 =

ACI 11-35d b f V 0c

c

c 4

2 ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ += β

When β > 2 the allowable Vc is reduced.

Design of two-way shear

Assume d.

Determine b0.

b0

= 4(c+d)

b0 = 2(c1+d) +2(c2+d)

1.

2.

for square columns

where one side = c

for rectangular

columns of sides c1

and c2.



9


The shear force Vu acts at a

section that has a length

b0 = 4(c+d) or 2(c1+d) +2(c2+d)

and a depth d; the section is

subjected to a vertical downward

load Pu and vertical upward

pressure q u.

3.

( )

( ) ( )

2

u u u

u u u 1 2

V P q c d

V P q c d c d

= − +

= − + +

for square columns

for rectangular columns


Allowable

Let Vu=φVc

4. d b f V 0cc 4φ φ =

0c

u

4 b f

V d

φ

=

If d is not close to the assumed d,

revise your assumptions



10

Design of one-way shear

For footings with bending

action in one direction the

critical section is located a

distance d from face of column

d b f V 0cc 2φ φ =


The ultimate shearing force at

section m-m can be calculated

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−−= d

c L

bqV 22 uu

If no shear reinforcement is to be

used, then d can be checked



11


b f

V d

2c

u

φ

=

If no shear reinforcement is

to be used, then d can be

checked, assuming Vu = φVc

Flexural Strength and Footingreinforcement

2y

us

⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

=a

d f

M A

φ

The bending moment in each

direction of the footing must be

checked and the appropriate

reinforcement must be provided.



12


b f A f a

85.0 c

sy=

Another approach is to

calculated R u = Mu / bd 2 and

determine the steel percentage

required ρ . Determine As then

check if assumed a is close to

calculated a


The minimum steel percentage

required in flexural members is

200/f y with minimum area and

maximum spacing of steel barsin the direction of bending shall

be as required for shrinkage

temperature reinforcement.



13


The reinforcement in one-way footings

and two-way footings must be

distributed across the entire width of

the footing.

1

2

directionshortinentreinforcemTotal

width band inentReinforcem

+= β

footingof sideshort

footingof sidelong= β where

Bearing Capacity of Column atBase

The loads from the column act on the footing at the

base of the column, on an area equal to area of the

column cross-section. Compressive forces are

transferred to the footing directly by bearing on theconcrete. Tensile forces must be resisted by

reinforcement, neglecting any contribution by

concrete.



14


Force acting on the concrete at the base of the column

must not exceed the bearing strength of the concrete

( )1c1 85.0 A f N φ =

where φ = 0.65 and

A1 =bearing area of column


The value of the bearing strength may be multiplied by a

factor for bearing on footing when the

supporting surface is wider on all sides than the loaded

area.

0.2/ 12 ≤ A A

The modified bearingstrength

( )

( )1c2

121c2

85.02

/85.0

A f N

A A A f N

φ

φ

≤

≤



15

Dowels in Footings

A minimum steel ratio ρ = 0.005 of the column section

as compared to ρ = 0.01 as minimum reinforcement for

the column itself. The number of dowel bars needed is

four these may be placed at the four corners of the

column. The dowel bars are usually extended into the

footing, bent at the ends, and tied to the main footing

reinforcement. The dowel diameter shall not exceed

the diameter of the longitudinal bars in the column by

more than 0.15 in.

Development length of theReinforcing Bars

The development length for compression bars was given

but not less than

Dowel bars must be checked for proper development

length.

c byd /02.0 f d f l =

in.8003.0 by ≥d f



16

Example – Combined Loading

A 12-in. x 24 in. column of an unsymmetrical shed is

subjected to an axial load PD of 220 k and MD = 180 k-ft

due to dead load and an an axial load PL = 165 k and a

moment ML= 140 k-ft due to

live load. The base of the

footing is 5 ft below final

grade, and the allowable soil

bearing pressure is 5 k/ft2.Design the footing using

f c = 4 ksi and f y = 60 ksi


Find the combined actual loads, P0 and M0

ft-k 320ft-k 140ft-k 180

k 385k165k220

LLDL0

LLDL0

=+=+=

=+=+=

M M M

PPP

Determine the eccentricity of the footing

in.10Usein97.9

k 385

ft1

in12ft-k 320

0

0 ⇒=⎟⎟ ⎠

⎞⎜⎜⎝

⎛

==P

M e



17


Assume a depth of footing, 24 in. The weight of

concrete and the soil are:

23

c lb/ft300

in.12

ft.1*in.24*lb/ft150 === d W γ

23

sss lb/ft300

in.12

ft.1*in.24ft5*lb/ft100 =⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −== d W γ


The effective soil pressure is given as:

22

222

scseff

k/ft4.4lb/ft4400

lb/ft300lb/ft300lb/ft5000

⇒=

−−=

−−= W W qq





19


Calculate net upward pressure:

( ) ( )

( )( )2

n

ftk /5.87

ft01ft9

k528.0

pressureupward Net

k 528.0

k 1651.6k 2022.1

6.12.1LoadsActual

=

=

=

+=

+=

q

LL DL


Calculate the depth of the reinforcement use # 8 bars

with a crisscrossing layering.

( )

in.5.19

in0.15.1in3in.24

5.1cover b

=

−−=

−−=

d

d hd



20


( )

( )( ) k 4.169ft208.3ft9k/ft87.5

22

2

2nu

==

⎟ ⎠ ⎞⎜

⎝ ⎛ +−−= ed c LlqV

ft208.3

in12

ft1in10

in12

ft1in5.19

2

in12

ft1in42

2

ft10

22

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−=+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −− ed

c L

Vu =139.4 k in

short direction

The depth of the footing can be calculated by using the

one-way shear (long direction)

Example – Combined LoadingThe depth of the footing can be calculated by using

one-way shear design

in.53.16

ft1

in12ft9400020.75

k 1

lb1000

k 4.169

2 c

u =

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝ ⎛

==b f

V d

φ

The footing is 19.5 in. > 16.53 in. so it will work.



21

Example – Combined LoadingCalculate perimeter for two-way shear or punch out

shear. The column is 12 in. x 24 in.

( ) ( )

( ) ( )

( )

( ) ft625.3

in12

ft1in.5.19in.24

ft625.2

in12

ft1in.5.19in.12

in.150in.5.19in.242in.5.19in.122

22

2

1

21o

=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=+

=⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ +=+

=+++=

+++=

d c

d c

d cd cb


Calculate the shear Vu

( )

( )( )k 2.472

ft625.3ft625.2k/ft5.87k0.5282

2

nuu

=

−=

+−= d cqPV

11.1

ft9

ft10== β

The shape parameter



22

Example – Combined LoadingCalculate d from the shear capacity according to

11.12.2.1 chose the largest value of d.

d b f V 0c

c

c 4

2 ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ += β

d b f V 0cc 4=

d b f b

d

V 0c

o

s

c 2⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+=

α αs is 40 for interior, 30 for edge

and 20 for corner column

Example – Combined LoadingThe depth of the footing can be calculated for the

two way shear

( )

in.84.11

in150400011.1

420.75

k 1

lb1000k 2.472

4

2 0c

u

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +

⎟ ⎠ ⎞⎜⎝ ⎛ =

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

=

b f

V d

β φ



23

Example – Combined LoadingThe third equation bo is dependent on d so use the

assumed values and you will find that d is smaller and

α = 40

( )( )( )

in.22.9

in15040002in150

in9.51400.75

k 1

lb1000k 2.472

240

0c

o

u

=⎟

⎠ ⎞

⎜⎝ ⎛ +

⎟

⎠

⎞⎜

⎝

⎛

=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

=

b f b

d

V d

φ

Example – Combined LoadingThe depth of the footing can be calculated by using

the two way shear

( )( )in.59.16

in150400040.75

k 1

lb1000k 72.24

4 0c

u

=

⎟ ⎠ ⎞⎜⎝ ⎛ ==

b f

V d

φ



24

Example – Combined LoadingCalculate the bending moment of the footing at the edge of

the column (long direction)

ft83.4

in12

ft1in10

2

in12

ft1in24

2

ft10

22

=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=+⎟⎟

⎠

⎞⎜⎜⎝

⎛ − e

c L

( )( )

( ) ft-k 2.616ft92

ft83.4ft83.4k/ft87.5

2

22

22nu

==

⎟ ⎠ ⎞⎜⎝ ⎛ +−⎟

⎠

⎞⎜⎝

⎛ +−= b

ec L

ec L

q M

Example – Combined LoadingCalculate R u for the footing to find ρ of the footing.

( )ksi1801.0

in5.19*ft1

in12ft9

ft1

in.12*ft-k 2.616

bd R

22

uu =

⎟ ⎠

⎞⎜⎝

⎛ ⎟

⎠ ⎞

⎜⎝ ⎛

⎟ ⎠ ⎞

⎜⎝ ⎛

== M



25

Example – Combined LoadingUse the R u for the footing to find ρ.

( )

( )( )

( )00344.0

ksi60

ksi405158.005158.0

05158.0

2

ksi49.0

ksi1801.07.147.17.1

07.1

7.159.01

c

y

2

c

u2

cu

==⇒=

=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

=

=+−⇒−=

ρ ρ

ω

φ ω ω ω ω

f

f

f

R f R

Example – Combined LoadingCompute the amount of steel needed

( ) 2

s in24.7in.5.19ft1

in.12ft900344.0 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ == bd A ρ

The minimum amount of steel for shrinkage is

( )( ) 2

s in67.4in.24in.1080018.00018.0 === bh A

The minimum amount of steel for flexure is

( )( ) 2

y

s in02.7in.9.51in.108

60000

200

200=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ == bd

f

A



26

Example – Combined LoadingUse As =8.36 in2 with #8 bars (0.79 in2). Compute

the number of bars need

bars11Use25.10

in79.0

in1.8

2

2

b

s ⇒=== A

An

Determine the spacing between bars

( )( ) in2.10

10

in32-in108

1

cover *2 ==−

−=n

Ls


Calculate the bending moment of the footing at the

edge of the column for short length

ft42

in12

ft1in12

2

ft9

22 =⎟⎟⎟

⎟⎟

⎠

⎞

⎜⎜⎜

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−

c L

( )( )

( )

ft-k 6.469

ft102

ft4ft4k/ft87.5

2

22

22nu

=

=⎟ ⎠ ⎞

⎜⎝ ⎛ −

⎟ ⎠

⎞⎜⎝

⎛ −= b

c L

c Lq M



27

Example – Combined LoadingCalculate R u for the footing to find ρ of the footing.

( )

ksi1235.0

in5.19*ft1

in12ft10

ft1

in.12*ft-k 69.64

bd R

22

uu

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟

⎠ ⎞

⎜⎝ ⎛

⎟ ⎠ ⎞

⎜⎝ ⎛

== M


Use R u for the footing to find ρ.

( )

( )

( )00234.0

ksi60

ksi403503.003503.0

03503.02

ksi4

ksi1235.07.147.17.1

07.1

7.159.01

c

y

2

c

u2

cu

==⇒=

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ −−

=

=+−⇒−=

ρ ρ

ω

φ

ω ω ω ω

f

f

f

R f R



28


Compute the amount of steel needed

( ) 2

s in46.5in.5.19ft1

in.12ft1000234.0 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ == bd A ρ


( )( ) 2

s in18.5in.24in.1200018.00018.0 === bh A


( )( ) 2

y

s in80.7in.9.51in.120

60000

200

200=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ == bd

f

A


Use As =9.36 in2 with #6 bar (0.44 in2) Compute the

number of bars need

bars18Use7.17

in44.0

in.807

2

2

b

s ⇒=== A

An

Calculate the reinforcement bandwidth

947.0

111.1

2

1

2

entreinforcemTotal

bandwidthinentReinforcem=

+=

+=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛

β



29


The number of bars in the 9 ft band is 0.947(18)=17 bars .

So place 17 bars in 9 ft section and 1 bars in each in(10ft - 9ft)/2 =0.5 ft of the band.

bars1 Use5.02

1718

2

bars band - bars#Total bar#outside

⇒=−

=

=


Determine the spacing between bars for the band of 9 ft

( )in75.6

16

in108

1

==−

=n

Ls

Determine the spacing between bars outside the band

in3

1

3in-in6cover ==

−=

n

Ls



30

Example – Combined LoadingCheck the bearing stress. The bearing strength N1, at

the base of the column, 12 in x 24 in., φ = 0.65

( ) ( )( )( )( ) k 5.636in42in12ksi485.065.085.0 1c1 === A f N φ

The bearing strength, N2, at the top of the footing is

1

1

212 2 N

A A N N ≤=



( ) k 1273k 636.5222 6.71ft2

ft90122

2

1

2 ===⇒>== N N A

A

( )

2

1

2

2

ft2

in.12

ft1in42

in.12

ft1in12

ft90ft10ft9

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ =

==

A

A



31

Example – Combined LoadingPu =628 k < N1, bearing stress is adequate. The

minimum area of dowels is required.

( )( ) 2

1 in44.1in42in12*005.0005.0 == A

Use minimum number of bars is 4, so use 4 # 8 bars

placed at the four corners of the column.


The development length of the dowels in compression

from ACI Code 12.3.2 for compression.

( )( )in19 Usein97.18

psi4000

psi60000in102.002.0

c

y b

d ⇒=== f

f d

l

The minimum ld , which has to be greater than 8 in., is

( )( ) in8in18 psi60000in10003.00003.0 y bd ≥=== f d l



32


Therefore, use 4#8 dowels in the corners of

the column extending 19 in. into the column

and the footing. Note that ld is less than the

given d = 19.5 in., which is sufficient

development length.


The development length, ld for the #8 bars

( )( )in4.47

psi400020

in0.1 psi60000

2020 c

by

d

c

y

b

d ===⇒=

f

d f l

f

f

d

l

There is adequate development length provided.

in60

2

in18in3

2

in144

2

cover

2d =−−=−−=

c Ll



33


The development length, ld for the #6 bars

( )( )in5.28

psi400025

in75.0 psi60000

2525 c

by

d

c

y

b

d ===⇒= f

d f l

f

f

d

l

There is adequate development length provided.

in39

2

in18in3

2

in102

2

cover

2d =−−=−−=

c Ll

Example – Multi-Column Footing

Design a rectangular footing to support two square

columns. The exterior column (I) has a section 16 x

16 in., which carries DL of 180 k and a LL of 120 k.

The interior column (II) has a section of 20 x 20 in.,which carries a DL of 250 k

and a LL of 140 k. The base of

the footing is 5 ft. below final

grade and allowable soil

pressure is 5 k/ft2 Use f c = 4 ksi

and f y = 60 ksi The external

column is located 2 ft from the

property line.



34

Example – Multi-Column FootingDetermine the location of an equivalent point and

its location select the datum at column I

( ) ( )

( ) ( )

ft.9 Useft.04.9

k 120k 180k 140k 250

k 120k 180ft0k 140k 250ft16

i

ii

⇒=

+++

+++==

∑∑

F

F x x

Extend the footing up to the property line, so the length

is l = 9 ft + 2 ft. = 11 ft. So the length of the footing is

2(11 ft.) = 22 ft.


Assume a depth of footing. (36 in.) The weight

of concrete and the soil are:

23

c lb/ft450

in.12

ft.1*in.36*lb/ft150 === d W γ

23

sss lb/ft200

in.12

ft.1*in.36ft5*lb/ft100 =⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −== d W γ



35


The effective soil pressure is given as:

22

222

scseff

k/ft35.4lb/ft4350

lb/ft200lb/ft450lb/ft5000

⇒=

−−=

−−= W W qq


Calculate the size of the footing:

ft7.5Useft21.7

ft22

ft8.651footingof Side

ft8.651

k/ft35.4

k906footingof Area

k 690k300k390LoadsTotal

k 300k120k180LoadsActual

k 390k140k250LoadsActual

2

2

2

21

⇒==

==

=+=+=

=+=+=

=+=+=

AL AL

LL DL

LL DL



36

Example – Multi-Column FootingCalculate net upward pressure:

( ) ( )

( ) ( )

( )( )

2

n ftk /6.33

ft7.5ft22

k1044 pressureupward Net

k 1044

k588k456

k 1401.7k 2504.1

k 1201.7k 1804.1

7.14.1LoadsActual

==

=

+=

++

+=

+=

q

LL DL


Calculate the depth of the reinforcement use # 8 bars

with a crisscrossing layering.

( )

in.5.31

in0.15.1in3in.36

5.1cover b

=

−−=

−−=

d

d hd



37

Example – Multi-Column FootingCompute the shear and bending moment diagrams.

Shear Forces

-400

-300

-200

-100

0

100

200

300

400

0 2 4 6 8 10 12 14 16 18 20 22

location (ft)

F o r c e (

k i p s )

63.3 k

-329.5 k

358.7 k

-150.3 k

The columns are

considered point loads

but shear values are

taken at each side of the

column.

( ) ( )

( )

( )

26.32 k/ft 7.5 ft

47.454 k/ft

V x qb x w

x w

x w

= −

= −

= −


The location of the maximum moment is

( ) ft9.6ft5.14

k 7.358k 329.5

k 329.5x

ft5.14in12ft1in10

in12ft1in8ft16

=+

=

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −⎟⎟

⎠

⎞⎜⎜⎝

⎛ −



38


Compute the shear and bending moment diagrams.

The columns are

considered point loads

but moments are takenat each side of the

column. It will not

balance because center

is at 9.04 ft

Bending Moment

-1400

-1200

-1000

-800

-600

-400

-200

0

200

400

0 2 4 6 8 10 12 14 16 18 20 22

Location (ft)

B e n

d i n g M o m e n t ( k - f t )

-1278.9 k-ft @ 9.61 ft

42.2 k-ft

249.9 k-ft

( ) ( )

( ) ( )

( )

2

i i

22

i i

2

i i

2

6.32 k/ft 7.5 ft2

47.454 k/ft2

x M x qb w x x

xw x x

xw x x

= − −

= − −

= − −


The maximum shear force occurs at the edge of

the 20 in. column. So maximum shear is measured

at distance d from the column.

( )

k 1.234

in12ft1in31.5k/ft454.47k7.358max

=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=− d qV



39


The depth of the footing can be calculated by using

one-way shear

in.2.24

ft1

in12ft.57400020.85

k 1

lb1000k 34.12

2 c

u =

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜⎝

⎛

==b f

V d

φ

The footing is 31.5 in. > 24.2 in. so it will work.


Calculate perimeter for two-way shear or

punch out shear. The column is 20 in.

square.

( )( )

( ) ft292.4

in12

ft1in.5.31in.02

in.206in.5.31in.024

4o

=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=+

=+=+=

d c

d cb



40


Calculate the shear Vu

( )

( )

k 6.464

ft292.4k/ft6.70k58822

2

nuu

=

−=

+−= d cqPV

The other column will not be critical,

Pu = 456 k for the 16 in. column


The depth of the footing can be calculated by using

two way shear

( )( )in.5.10

in206400040.85

k 1

lb1000k 64.64

4 0c

u =⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

==b f

V d

φ



41


Calculate R u for the footing to find ρ of the footing.

( ) ( )

ksi1719.0

in5.31*in90

ft1

in.12*ft-k 278.91

bd

R 22

uu =

⎟⎟

⎠

⎞⎜⎜⎝

⎛

== M


From R u for the footing the ρ value can be found.

( )

( )( )

( )00328.0

ksi60

ksi404917.004917.0

04917.0

2

ksi49.0

ksi1719.07.147.17.1

07.1

7.159.01

c

y

2

c

u2

cu

==⇒=

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

=

=+−⇒−=

ρ ρ

ω

φ

ω ω ω ω

f

f

f

R f R



42


Compute the area of steel needed

( ) 2

s in29.9in.5.31

ft1

in.12ft5.700277.0 =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ == bd A ρ


( )( ) 2

s in38.5in.63in.900018.00018.0 === bh A


( )( ) 2

y

s in45.9in.1.53in.90

60000

200

200=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ == bd

f

A Use⇐


Use a #9 bar (1.00 in2) Compute the number of bars

needed bars10Use45.9

in0.1

in45.9

2

2

b

s ⇒=== A

An


( )

( )in33.9

9

in32-in90

1

cover *2==

−

−=

n

Ls



43


The minimum amount is steel is going to be due to the

flexural restrictions. So below the columns with

positive moment, the reinforcement will be 10 # 9 bars

running longitudinally. The development length will

have to be calculated.


The development length, ld for the #7 bars for the

reinforcement of the footing.

( )( )in5.53

psi400020

in.1281 psi60000

2020 c

by

d

c

y

b

d ===⇒=

f

d f l

f

f

d

l

The bars have more than 12-in. of concrete below

them, therefore ld = 1.3 ld .

( ) in.70Usein6.69in3.553.1d ⇒==l



44


To determine the reinforcement in the short direction.

The bandwidth of the two columns must be determined

for the 16 in. column.

ft5.5 Useft3.5

in.12

ft1in5.31

in.12

ft1

2

in16 ft2in16Band ⇒=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ +

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+=

Compute the moment at the edge

k/ft8.60

ft7.5

k 456net ==q ft08.3

in12

ft1in8

2

ft5.7=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −= L


The bending moment will be

Compute the R u

( )( )

ft-k 0.289

2

ft3.08k/ft8.60

2

22

netu === l

q M

( )

ksi053.0

in5.31*

ft1

in12ft5.5

ft1

in.12*ft-k 289

bd

R

22

uu =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟

⎠

⎞⎜⎜⎝

⎛

== M



45



( )

( )( )

( )001.0

ksi60

ksi404917.001484.0

01484.0

2

ksi49.0

ksi053.07.147.17.1

07.1

7.159.01

c

y

2

c

u2

cu

==⇒=

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

=

=+−⇒−=

ρ ρ

ω

φ

ω ω ω ω

f

f

f

R f R



( ) 2

s in08.2in.5.31

ft1

in.12ft5.5001.0 =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ == bd A ρ


( )( ) 2

s in28.4in.63in.660018.00018.0 === bh A


( )( ) 2

y

s in93.6in.1.53in.66

60000

200

200=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ == bd

f

A Use⇐



46



needed bars7Use93.6

in0.1

in93.6

2

2

b

s ⇒=== A

An


( )

( )in5.10

6

in3-in66

1

cover ==

−

−=

n

Ls


To determine the reinforcement in the short direction.

The 20-in. interior column extends beyond 4 ft from the

center therefore the band is 7.5 ft x 7.5 ft. Compute the

moment at the edge

k/ft4.78

ft7.5

k 885net ==q ft92.2

in12

ft1in10

2

ft5.7=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ −= L



47


The bending moment will be

Compute the R u

( )( )

ft-k 3.334

2

ft2.92k/ft4.78

2

22

netu === l

q M

( )ksi045.0

in5.31*

ft1

in12ft7.5

ft1

in.12*ft-k 334.3

bd R 2

2

u

u =⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟

⎠

⎞⎜⎜⎝

⎛

==

M



( )

( )( )

( )00084.0

ksi60

ksi401257.001257.0

01257.0

2

ksi49.0

ksi045.07.147.17.1

07.1

7.159.01

c

y

2

c

u2

cu

==⇒=

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

=

=+−⇒−=

ρ ρ

ω

φ

ω ω ω ω

f

f

f

R f R



48



( ) 2

s in38.2in.5.31

ft1

in.12ft5.700084.0 =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ == bd A ρ


( )( ) 2

s in83.5in.63in.090018.00018.0 === bh A


( )( ) 2

y

s in45.9in.1.53in.09

60000

200

200=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ == bd

f

A Use⇐


Check the bearing stress. The bearing strength N1, at


( ) ( )( )( ) k 609in16ksi485.07.085.02

1c1 === A f N φ


1

1

212 2 N

A

A N N ≤=



49



( ) k 1218k 096222 4.125ft78.1ft30.25 12

2

2

1

2 ===⇒>== N N A A

( )

2

2

1

22

2

ft78.1

in.12

ft1in16

ft25.30ft.55

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

==

A

A


Pu =456 k < N1, bearing stress is adequate. The


( ) 22

1 in28.1in16*005.0005.0 == A



Note if the Pu > N1 the area of steel will be

( )

y

1us

f

N P A

−=

As long as the area of

steel is greater than the

minimum amount.



50




( )( )in17 Usein6.16

psi4000

psi60000in.875002.002.0

c

y b

d ⇒=== f

f d l


( )( ) in8in75.15 psi60000in.87500003.00003.0 y bd ≥=== f d l




and the footing. Note that ld is less than thegiven d = 31.5 in., which is sufficient

development length.



51



need bars10Use45.9

in0.1

in45.9

2

2

b

s ⇒=== A

An


( )

( )in67.9

9

in3-in90

1

cover ==

−

−=

n

Ls


Check the bearing stress. The bearing strength N1, at


( ) ( )( )( ) k 952in02ksi485.07.085.02

1c1 === A f N φ


1

1

212 2 N

A

A N N ≤=



52



( ) k 1904k 952222 4.5ft78.2ft56.25 12

2

2

1

2 ===⇒>== N N A A

( )

2

2

1

22

2

ft78.2

in.12

ft1in02

ft25.56ft.57

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

==

A

A


Pu =588 k < N1, bearing stress is adequate. The


( ) 22

1 in0.2in02*005.0005.0 == A








( )( )in19 Usein7.18

psi4000

psi60000in102.002.0

c

y b

d ⇒=== f

f d l


( )( ) in8in18 psi60000in10003.00003.0 y bd ≥=== f d l




and the footing. Note that ld is less than thegiven d = 31.5 in., which is sufficient

development length.

Documents

Footing design and analysis