Footing Design 2

Chp.12 Cont. – Examplesto design Footings

ExampleExampleDesign 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 1Example 1

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

23c lb/ft 300

in. 12ft. 1* in. 24*lb/ft 150 dW

23sss lb/ft 200

in. 12ft. 1* in. 24ft 4*lb/ft 100

dW

Example 1Example 1

The effective soil pressure is given as:

22

222

scseff

k/ft 5.4lb/ft 4500lb/ft 200lb/ft 300lb/ft 5000

WWqq

Example 1Example 1Calculate the size of the footing:

ft 10 Useft 94.9footing of Side

ft 98.9k/ft 5.4k 445footing of Area

k 445 k 200 k 245Loads Actual2

2

LLDL

Example 1Example 1Calculate net upward pressure:

2

2n ftk / 6.83

ft 001k 836 pressure upwardNet

k 683 k 2001.7 k 2454.17.14.1Loads Actual

q

LLDL

Example 1Example 1Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering.

in. 5.19

in 0.15.1in 3 in. 245.1cover b

ddhd

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

ft 125.3in 12ft 1in. 5.19 in. 18

in. 150in. 5.19 in. 1844o

dc

dcb

Example 1Example 1Calculate the shear Vu

k 616ft 125.3k/ft 6.83k 683 22

2nuu

dcqPV

1ft 10ft 10

The shape parameter

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

dbfV 0c

c

c 42

dbfV 0cc 4

dbfbdV 0c

o

sc 2

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

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

in. 1.19

in 150400040.85k 1lb 1000k 616

4 0c

u

bf

Vd

Example 1Example 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 150in 9.51400.85

k 1lb 1000k 616

2400c

o

u

bfbd

Vd

Actual (d =14.02324 in.)

bo=128.93 in

Example 1Example 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 12ft 1in 5.19

2in 12ft 1in 18

2ft 10

22

dcL

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

in. 9.13

ft 1in 12ft 10400020.85

k 1lb 1000k 3.179

2 c

u

bf

Vd

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

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

ft 25.42

in 12ft 1in 18

2ft 10

22

cL

ft-k 8.616ft 102ft 25.4ft 25.4k/ft 83.6

222

22nu

b

cLcLqM

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

ksi 1622.0

in 5.19*in 120ft 1in. 12*ft-k 8.616

bdR

22u

u

M

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

0031.0ksi 60

ksi 404632.004632.0

04632.02

ksi 49.0ksi 1622.07.147.17.1

07.17.159.01

c

y

2

c

u2cu

f

f

fRfR

Example 1Example 1Compute the area of steel needed

2s in 23.7in. 5.19

ft 1in. 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

2

y

s in 8.7in. 9.51in. 12060000200 200

bd

fA Use

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

bars 13 Use13in 60.0in 8.7

2

2

b

s AAn

Determine the spacing between bars

in 5.9

12in 32 -in 120

1cover*2

nLs

Example 1Example 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 2

1c1 AfN

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

11

212 2 N

AANN

Example 1Example 1


k 1542k 771222 6.67ft 25.2ft 100

122

2

1

2 NNAA

22

1

222

ft 25.2in. 12ft 1in 18

ft 100ft 10

A

A

Example 1Example 1Pu =683 k < N1, bearing stress is adequate. The minimum area of dowels is required. 22

1 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 1Example 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 1Example 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 1Example 1The development length, ld for the #7 bars for the reinforcement of the footing. in 5.41

psi 400020in 875.0psi 60000

2020 c

byd

c

y

b

d f

dfl

f

f

dl

There is adequate development length provided.

in 482in 18in 3

2in 120

2cover

2d

cLl

Example 1 - Final DesignExample 1 - Final Design

Example 2Example 2Design 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 2Example 2

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

23c lb/ft 300

in. 12ft. 1* in. 24*lb/ft 150 dW

23sss lb/ft 200

in. 12ft. 1* in. 24ft 4*lb/ft 100

dW

Example 2Example 2

The effective soil pressure is given as:

22

222

scseff

k/ft 5.4lb/ft 4500lb/ft 200lb/ft 300lb/ft 5000

WWqq

Example 2Example 2Calculate the size of the footing:

ft 12 Useft 64.11ft 5.8ft 98.9footing of Side

ft 98.9k/ft 5.4k 445footing of Area

k 445 k 200 k 245Loads Actual

2

22

LLDL

Example 2Example 2Calculate net upward pressure:

2

n ftk / 6.70ft 21ft .58

k 836 pressure upwardNet

k 683 k 2001.7 k 2454.17.14.1Loads Actual

q

LLDL

Example 2Example 2Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering.

in. 5.19

in 0.15.1in 3 in. 245.1cover b

ddhd

Example 2Example 2

k 4.206ft 625.3ft .58k/ft 7.622

2

2nu

dcLlqV

ft 625.3

in 12ft 1in 5.19

2in 12ft 1in 18

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

in. 8.18

ft 1in 12ft .58400020.85

k 1lb 1000k 4.206

2 c

u

bf

Vd

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

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

ft 125.3in 12ft 1in. 5.19 in. 18

in. 150in. 5.19 in. 1844o

dc

dcb

Example 2Example 2Calculate the shear Vu

k 6.617ft 125.3k/ft 6.7k 683 22

2nuu

dcqPV

41.1ft 8.5ft 12

The shape parameter

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

dbfV 0c

c

c 42

dbfV 0cc 4

dbfbdV 0c

o

sc 2

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

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

in. 8.15

in 150400041.1420.85

k 1lb 1000k 6.617

42 0c

u

bf

Vd

Example 2Example 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 150in 9.51400.85

k 1lb 1000k 6.617

2400c

o

u

bfbd

Vd

Actual (d =14.032 in.)

bo=128.173 in

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

in. 14.19

in 150400040.85k 1lb 1000k 6.617

4 0c

u

bf

Vd

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

ft 25.52

in 12ft 1in 18

2ft 12

22

cL

ft-k 8.784ft .582ft 25.5ft 25.5k/ft 7.6

222

22nu

b

cLcLqM


ksi 2428.0

in 5.19*ft 1in 12ft 8.5

ft 1in. 12*ft-k 8.784

bdR

22u

u

M

Example 2Example 2Use the Ru for the footing to find .

00469.0ksi 60

ksi 407036.007036.0

07036.02

ksi 49.0ksi 2428.07.147.17.1

07.17.159.01

c

y

2

c

u2cu

f

f

fRfR

Example 2Example 2Compute the amount of steel needed

2s in 33.9in. 5.19

ft 1in. 12ft 5.800469.0

bdA


2

y

s in 63.6in. 9.51in. 10260000200 200

bd

fA

Example 2Example 2Use As =8.36 in2 with #8 bars (0.79 in2). Compute the number of bars need bars 12 Use8.11

in 79.0in 33.9

2

2

b

s AAn

Determine the spacing between bars

in 73.8

11in 32 -in 102

1cover*2

nLs

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

ft 5.32

in 12ft 1in 18

2ft .58

22

cL

ft-k 5.492ft 122ft 5.3ft 5.3k/ft 7.6

222

22nu

b

cLcLqM


ksi 1079.0

in 5.19*ft 1in 12ft 12

ft 1in. 12*ft-k 92.54

bdR

22u

u

M

Example 2Example 2Use Ru for the footing to find .

00203.0ksi 60ksi 40305.00305.0

0305.02

ksi 49.0ksi 1079.07.147.17.1

07.17.159.01

c

y

2

c

u2cu

f

f

fRfR

Example 2Example 2Compute the amount of steel needed

2s in 72.5in. 5.19

ft 1in. 12ft 1200203.0

bdA


2

y

s in 36.9in. 9.51in. 14460000200 200

bd

fA

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

bars 22 Use3.21in 44.0in 36.9

2

2

b

s AAn

Calculate the reinforcement bandwidth

83.0141.1

21

2entreinforcem Total

bandwidthin ent Reinforcem

Example 2Example 2The 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 Use5.121922

2bars band - bars # Totalbar # outside

Example 2Example 2

Determine the spacing between bars for the band of 8.5 ft

in 67.5

18in 102

1

nLs

Determine the spacing between bars outside the band

in 923in-in 12cover

n

Ls

Example 2Example 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 2

1c1 AfN


11

212 2 N

AANN

Example 2Example 2


k 1542k 771222 6.67ft 25.2ft 100

122

2

1

2 NNAA

22

1

222

ft 25.2in. 12ft 1in 18

ft 100ft 10

A

A

Example 2Example 2Pu =683 k < N1, bearing stress is adequate. The minimum area of dowels is required. 22

1 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 2Example 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 2Example 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 2Example 2The development length, ld for the #8 bars

in 4.47psi 400020in 0.1psi 60000

2020 c

byd

c

y

b

d f

dfl

f

f

dl


in 602in 18in 3

2in 144

2cover

2d

cLl

Example 2Example 2

The development length, ld for the #6 bars

in 5.28psi 400025

in 75.0psi 600002525 c

byd

c

y

b

d f

dfl

f

f

dl


in 392in 18in 3

2in 102

2cover

2d

cLl

Example 2 - Final designExample 2 - Final design

12 #823 #6

Footing Design 2

Documents