Chp.12 Cont. – Examples

to design Footings

ExampleExample

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 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. 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 1Example 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 1Example 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 1Example 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 1Example 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 1Example 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 1Example 1Calculate the shear Vu

k 616ft 125.3k/ft 6.83k 683 22

2nuu

dcqPV

1ft 10

ft 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 4

2

dbfV 0cc 4

dbfb

dV 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 150400040.85

k 1

lb 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 150

in 9.51400.85

k 1

lb 1000k 616

240

0c

o

u

bfb

d

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 83.6

222

2nu

dcL

lqV

ft 625.2

in 12

ft 1in 5.19

2

in 12

ft 1in 18

2

ft 10

22

dcL

Example 1Example 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 1Example 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 83.6

2

22

22nu

b

cL

cLqM

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

0031.0

ksi 60

ksi 404632.004632.0

04632.02

ksi 49.0

ksi 1622.07.147.17.1

07.1

7.159.01

c

y

2

c

u2cu

f

f

f

RfR

Example 1Example 1Compute the area of steel needed

2s in 23.7in. 5.19

ft 1

in. 12ft 1000309.0

bdA

The minimum amount of steel for shrinkage is

2s in 18.5in. 24in. 1200018.0 0018.0 bhA

The minimum amount of steel for flexure is

2

y

s in 8.7in. 9.51in. 12060000

200

200

bd

fA Use

Example 1Example 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 5.912

in 32 -in 120

1

cover*2

n

Ls

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 21c1 AfN

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

1

1

212 2 NA

ANN

Example 1Example 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 1Example 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 1Example 1

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

in 19 Usein 97.18

psi 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 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 3

2

in 120

2cover

2d

cLl

Example 1 - Final DesignExample 1 - Final Design

Example 2Example 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 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. 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 2Example 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 2Example 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 2Example 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 2Example 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 2Example 2

k 4.206ft 625.3ft .58k/ft 7.6

222

2nu

dcL

lqV

ft 625.3

in 12

ft 1in 5.19

2

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

dbfV 0c

c

c 4

2

dbfV 0cc 4

dbfb

dV 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 150400041.1

420.85

k 1

lb 1000k 6.617

4

2 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 150

in 9.51400.85

k 1

lb 1000k 6.617

240

0c

o

u

bfb

d

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 150400040.85

k 1

lb 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 12

ft 1in 18

2

ft 12

22

cL

ft-k 8.784ft .582

ft 25.5ft 25.5k/ft 7.6

2

22

22nu

b

cL

cLqM

Example 2Example 2Calculate Ru for the footing to find of the footing.

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

00469.0

ksi 60

ksi 407036.007036.0

07036.02

ksi 49.0

ksi 2428.07.147.17.1

07.1

7.159.01

c

y

2

c

u2cu

f

f

f

RfR

Example 2Example 2Compute the amount of steel needed

2s in 33.9in. 5.19

ft 1

in. 12ft 5.800469.0

bdA

The minimum amount of steel for shrinkage is

2s in 41.4in. 24in. 1020018.0 0018.0 bhA

The minimum amount of steel for flexure is

2

y

s in 63.6in. 9.51in. 10260000

200

200

bd

fA

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

bars 12 Use8.11in 79.0

in 33.92

2

b

s A

An

Determine the spacing between bars

in 73.811

in 32 -in 102

1

cover*2

n

Ls

Example 2Example 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 7.6

2

22

22nu

b

cL

cLqM

Example 2Example 2Calculate Ru for the footing to find of the footing.

ksi 1079.0

in 5.19*ft 1

in 12ft 12

ft 1

in. 12*ft-k 92.54

bdR

22

uu

M

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

00203.0

ksi 60

ksi 40305.00305.0

0305.02

ksi 49.0

ksi 1079.07.147.17.1

07.1

7.159.01

c

y

2

c

u2cu

f

f

f

RfR

Example 2Example 2Compute the amount of steel needed

2s in 72.5in. 5.19

ft 1

in. 12ft 1200203.0

bdA

The minimum amount of steel for shrinkage is

2s in 22.6in. 24in. 1440018.0 0018.0 bhA

The minimum amount of steel for flexure is

2

y

s in 36.9in. 9.51in. 14460000

200

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 44.0

in 36.92

2

b

s A

An

Calculate the reinforcement bandwidth

83.0141.1

2

1

2

entreinforcem Total

bandwidthin ent Reinforcem

Example 2Example 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 2Example 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 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 21c1 AfN

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

1

1

212 2 NA

ANN

Example 2Example 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 2Example 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 2Example 2

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

in 19 Usein 97.18

psi 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 2

The development length, ld for the #8 bars

in 4.47

psi 400020

in 0.1psi 60000

2020 c

byd

c

y

b

d f

dfl

f

f

d

l

There is adequate development length provided.

in 602

in 18in 3

2

in 144

2cover

2d

cLl

Example 2Example 2

The development length, ld for the #6 bars

in 5.28

psi 400025

in 75.0psi 60000

2525 c

byd

c

y

b

d f

dfl

f

f

d

l

There is adequate development length provided.

in 392

in 18in 3

2

in 102

2cover

2d

cLl

Example 2 - Final designExample 2 - Final design

12 #823 #6