425 footing design examples

Chp.12 Cont. Examples

to design Footings

Example

Design a square footing to support a 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 k and an axial live load of 200 k. The base of the footing is 4 ft. below final grade and allowable soil pressure is 5 k/ft2 Use fc = 3 ksi and fy = 60 ksi

Example 1

Assume a depth of footing. (2 ft or 24 in.) The weight of concrete and the soil are:

23c lb/ft 300

in. 12

ft. 1* in. 24*lb/ft 150 === dW

23sss lb/ft 200

in. 12

ft. 1* in. 24ft 4*lb/ft 100 =

== dW

Example 1

The effective soil pressure is given as:

22

222

scseff

k/ft 5.4lb/ft 4500

lb/ft 200lb/ft 300lb/ft 5000

=

=

= WWqq

Example 1

Calculate the size of the footing:

ft 10 Useft 94.9footing of Side

ft 98.9k/ft 5.4

k 445footing of Area

k 445 k 200 k 245Loads Actual

2

2

=

==

=+=+= LLDL

Example 1

Calculate net upward pressure:

( ) ( )

2

2n ftk / 6.83

ft 001

k 836 pressure upwardNet

k 683 k 2001.7 k 2454.1

7.14.1Loads Actual

==

=+=

+=

q

LLDL

Example 1

Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering.

( )in. 5.19

in 0.15.1in 3 in. 24

5.1cover b

=

=

=

d

dhd

Example 1Calculate perimeter for two-way shear or punch out shear. The column is 18 in. square.

( )( )

( ) ft 125.3in 12

ft 1in. 5.19 in. 18

in. 150in. 5.19 in. 184

4o

=

+=+

=+=

+=

dc

dcb

Example 1Calculate the shear Vu

( )

( ) k 616ft 125.3k/ft 6.83k 683 22

2nuu

==

+= dcqPV

1ft 10

ft 10==

The shape parameter

Example 1Calculate d value from the shear capacity according to 11.12.2.1 chose the largest value of d

dbfV 0cc

c 42

+=

dbfV 0cc 4=

dbfb

dV 0co

sc 2

+=

s is 40 for interior, 30 for edge and 20 for corner column

Example 1The depth of the footing can be calculated by using two way shear

( )( )in. 1.19

in 150400040.85k 1

lb 1000k 616

4 0c

u =

==bf

Vd

Example 1The second equation bo is dependent on d so use the assumed values and you will find that d is smaller and = 40

( ) ( )( )in. 6.10

in 15040002in 150

in 9.51400.85

k 1

lb 1000k 616

240 0co

u

=

+

=

+

=

bfb

d

Vd

Actual (d =14.02324 in.)

bo=128.93 in

Example 1The depth of the footing can be calculated by using one-way shear

( )

( )( ) k 3.179ft 625.2ft 10k/ft 83.622

2

2nu

==

= dcLlqV

ft 625.2

in 12

ft 1in 5.192

in 12

ft 1in 18

2

ft 10

22

=

=

dcL

Example 1The depth of the footing can be calculated by using one-way shear

in. 9.13

ft 1

in 12ft 10400020.85

k 1

lb 1000k 3.179

2 c

u =

==bf

Vd

The footing is 19.5 in. > 13.9 in. so it will work.

Example 1Calculate the bending moment of the footing at the edge of the column

ft 25.42

in 12

ft 1in 18

2

ft 10

22=

=

cL

( ) ( ) ( ) ft-k 8.616ft 102

ft 25.4ft 25.4k/ft 83.62

2222

nu ==

= b

cLcLqM

Example 1Calculate Ru for the footing to find of the footing.

( ) ( )ksi 1622.0

in 5.19*in 120

ft 1

in. 12*ft-k 8.616

bdR

22

uu =

==M

Example 1From Ru for the footing the value can be found.

( )

( )( )

( ) 0031.0ksi 60

ksi 404632.004632.0

04632.02

ksi 49.0

ksi 1622.07.147.17.1

07.17.159.01

c

y

2

c

u2cu

===

=

=

=+=

f

f

f

RfR

Example 1Compute the area of steel needed

( ) 2s in 23.7in. 5.19ft 1

in. 12ft 1000309.0 =

== bdA

The minimum amount of steel for shrinkage is

( )( ) 2s in 18.5in. 24in. 1200018.0 0018.0 === bhAThe minimum amount of steel for flexure is

( )( ) 2y

s in 8.7in. 9.51in. 12060000

200 200 =

== bd

fA Use

Example 1Use a #7 bar (0.60 in2) Compute the number of bars need

bars 13 Use13in 60.0

in 8.72

2

b

s ===A

An

Determine the spacing between bars

( )( ) in 5.9

12

in 32 -in 120

1

cover*2==

=

n

Ls

Example 1Check the bearing stress. The bearing strength N1, at the base of the column, 18 in x 18 in., = 0.7

( ) ( )( )( ) k 771in 18ksi 485.07.085.0 21c1 === AfN The bearing strength, N2, at the top of the footing is

1

1

212 2 N

A

ANN =

Example 1

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

( ) k 1542k 771222 6.67ft 25.2

ft 10012

2

2

1

2 ===>== NNA

A

( )

2

2

1

222

ft 25.2in. 12

ft 1in 18

ft 100ft 10

=

=

==

A

A

Example 1Pu =683 k < N1, bearing stress is adequate. The minimum area of dowels is required.

( ) 221 in 62.1in 18*005.0005.0 ==A

Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column.

Example 1The development length of the dowels in compression from ACI Code 12.3.2 for compression.

( )( ) in 19 Usein 97.18psi 4000

psi 60000in 102.002.0

c

ybd ===

f

fdl

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

( )( ) in 8 in 18psi 60000in 10003.00003.0 ybd === fdl

Example 1

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.

Example 1The development length, ld for the #7 bars for the reinforcement of the footing.

( )( ) in 5.41psi 400020

in 875.0psi 60000

2020 c

byd

c

y

b

d ====f

dfl

f

f

d

l

There is adequate development length provided.

in 482

in 18in 32

in 120

2cover

2d ===

cLl

Example 1 - Final Design

Example 2

Design a footing to support a 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 k and an axial live load of 200 k. The base of the footing is 4 ft. below final grade and allowable soil pressure is 5 k/ft2 Use fc = 3 ksi and fy = 60 ksi. Limit one side of the footing to 8.5 ft.

Example 2

Assume a depth of footing. (2 ft or 24 in.) The weight of concrete and the soil are:

23c lb/ft 300

in. 12

ft. 1* in. 24*lb/ft 150 === dW

23sss lb/ft 200

in. 12

ft. 1* in. 24ft 4*lb/ft 100 =

== dW

Example 2

The effective soil pressure is given as:

22

222

scseff

k/ft 5.4lb/ft 4500

lb/ft 200lb/ft 300lb/ft 5000

=

=

= WWqq

Example 2

Calculate the size of the footing:

ft 12 Useft 64.11ft 5.8

ft 98.9footing of Side

ft 98.9k/ft 5.4

k 445footing of Area

k 445 k 200 k 245Loads Actual

2

2

2

==

==

=+=+= LLDL

Example 2

Calculate net upward pressure:

( ) ( )

( )( )2

n ftk / 6.70ft 21ft .58

k 836 pressure upwardNet

k 683 k 2001.7 k 2454.1

7.14.1Loads Actual

==

=+=

+=

q

LLDL

Example 2

Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering.

( )in. 5.19

in 0.15.1in 3 in. 24

5.1cover b

=

=

=

d

dhd

Example 2

( )

( )( ) k 4.206ft 625.3ft .58k/ft 7.622

2

2nu

==

= dcLlqV

ft 625.3

in 12

ft 1in 5.192

in 12

ft 1in 18

2

ft 12

22

=

=

dcL

Vu =150.7 k in short direction

The depth of the footing can be calculated by using the one-way shear (long direction)

Example 2The depth of the footing can be calculated by using one-way shear design

in. 8.18

ft 1

in 12ft .58400020.85

k 1

lb 1000k 4.206

2 c

u =

==bf

Vd

The footing is 19.5 in. > 18.8 in. so it will work.

Example 2Calculate perimeter for two-way shear or punch out shear. The column is 18 in. square.

( )( )

( ) ft 125.3in 12

ft 1in. 5.19 in. 18

in. 150in. 5.19 in. 184

4o

=

+=+

=+=

+=

dc

dcb

Example 2Calculate the shear Vu

( )

( ) k 6.617ft 125.3k/ft 6.7k 683 22

2nuu

==

+= dcqPV

41.1ft 8.5

ft 12==

The shape parameter

Example 2Calculate d from the shear capacity according to 11.12.2.1 chose the largest value of d.

dbfV 0cc

c 42

+=

dbfV 0cc 4=

dbfb

dV 0co

sc 2

+=

s is 40 for interior, 30 for edge and 20 for corner column

Example 2The depth of the footing can be calculated for the two way shear

( )in. 8.15

in 150400041.1

420.85

k 1

lb 1000k 6.617

42 0c

u =

+

=

+

=

bf

Vd

Example 2The third equation bo is dependent on d so use the assumed values and you will find that d is smaller and = 40

( ) ( )( )in. 64.10

in 15040002in 150

in 9.51400.85

k 1

lb 1000k 6.617

240 0co

u

=

+

=

+

=

bfb

d

Vd

Actual (d =14.032 in.)

bo=128.173 in

Example 2The depth of the footing can be calculated by using the two way shear

( )( )in. 14.19

in 150400040.85k 1

lb 1000k 6.617

4 0c

u =

==bf

Vd

Example 2Calculate the bending moment of the footing at the edge of the column (long direction)

ft 25.52

in 12

ft 1in 18

2

ft 12

22=

=

cL

( ) ( ) ( ) ft-k 8.784ft .582

ft 25.5ft 25.5k/ft 7.62

2222

nu ==

= b

cLcLqM


( )ksi 2428.0

in 5.19*ft 1

in 12ft 8.5

ft 1

in. 12*ft-k 8.784

bdR

22

uu =

==M

Example 2Use the Ru for the footing to find .

( )

( )( )

( ) 00469.0ksi 60

ksi 407036.007036.0

07036.02

ksi 49.0

ksi 2428.07.147.17.1

07.17.159.01

c

y

2

c

u2cu

===

=

=

=+=

f

f

f

RfR

Example 2Compute the amount of steel needed

( ) 2s in 33.9in. 5.19ft 1

in. 12ft 5.800469.0 =

== bdA



( )( ) 2y

s in 63.6in. 9.51in. 10260000

200 200 =

== bd

fA

Example 2Use As =8.36 in2 with #8 bars (0.79 in2). Compute the number of bars need

bars 12 Use8.11in 79.0

in 33.92

2

b

s ===A

An

Determine the spacing between bars

( )( ) in 73.8

11

in 32 -in 102

1

cover*2==

=

n

Ls

Example 2Calculate the bending moment of the footing at the edge of the column for short length

ft 5.32

in 12

ft 1in 18

2

ft .58

22=

=

cL

( ) ( ) ( ) ft-k 5.492ft 122

ft 5.3ft 5.3k/ft 7.62

2222

nu ==

= b

cLcLqM


( )ksi 1079.0

in 5.19*ft 1

in 12ft 12

ft 1

in. 12*ft-k 92.54

bdR

22

uu =

==M

Example 2Use Ru for the footing to find .

( )

( )( )

( ) 00203.0ksi 60

ksi 40305.00305.0

0305.02

ksi 49.0

ksi 1079.07.147.17.1

07.17.159.01

c

y

2

c

u2cu

===

=

=

=+=

f

f

f

RfR

Example 2Compute the amount of steel needed

( ) 2s in 72.5in. 5.19ft 1

in. 12ft 1200203.0 =

== bdA



( )( ) 2y

s in 36.9in. 9.51in. 14460000

200 200 =

== bd

fA

Example 2Use As =9.36 in2 with #6 bar (0.44 in2) Compute the number of bars need

bars 22 Use3.21in 44.0

in 36.92

2

b

s ===A

An

Calculate the reinforcement bandwidth

83.0141.1

2

1

2

entreinforcem Total

bandwidthin ent Reinforcem=

+=

+=

Example 2

The number of bars in the 8.5 ft band is 0.83(22)=19 bars .

So place 19 bars in 8.5 ft section and 2 bars in each in (12ft -8.5ft)/2 =1.75 ft of the band.

bars 2 Use5.12

1922

2

bars band - bars # Totalbar # outside ===

Example 2

Determine the spacing between bars for the band of 8.5 ft

( )in 67.5

18

in 102

1==

=

n

Ls

Determine the spacing between bars outside the band

in 92

3in-in 12cover==

=

n

Ls

Example 2Check the bearing stress. The bearing strength N1, at the base of the column, 18 in x 18 in., = 0.7

( ) ( )( )( ) k 771in 18ksi 485.07.085.0 21c1 === AfN The bearing strength, N2, at the top of the footing is

1

1

212 2 N

A

ANN =

Example 2

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

( ) k 1542k 771222 6.67ft 25.2

ft 10012

2

2

1

2 ===>== NNA

A

( )

2

2

1

222

ft 25.2in. 12

ft 1in 18

ft 100ft 10

=

=

==

A

A

Example 2Pu =683 k < N1, bearing stress is adequate. The minimum area of dowels is required.

( ) 221 in 62.1in 18*005.0005.0 ==A

Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column.

Example 2The development length of the dowels in compression from ACI Code 12.3.2 for compression.

( )( ) in 19 Usein 97.18psi 4000

psi 60000in 102.002.0

c

ybd ===

f

fdl

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

( )( ) in 8 in 18psi 60000in 10003.00003.0 ybd === fdl

Example 2

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.

Example 2

The development length, ld for the #8 bars

( )( ) in 4.47psi 400020

in 0.1psi 60000

2020 c

byd

c

y

b

d ====f

dfl

f

f

d

l


in 602

in 18in 32

in 144

2cover

2d ===

cLl

Example 2

The development length, ld for the #6 bars

( )( ) in 5.28psi 400025

in 75.0psi 60000

2525 c

byd

c

y

b

d ====f

dfl

f

f

d

l


in 392

in 18in 32

in 102

2cover

2d ===

cLl

Example 2 - Final design

12 #823 #6

Chp.12 Cont. Examplesto design FootingsExampleExample 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1Example 1 - Final DesignExample 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2Example 2 - Final design

425 footing design examples

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