chp.12 cont. – examples to design footings
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Chp.12 Cont. – Examples to design Footings. Example. - PowerPoint PPT PresentationTRANSCRIPT
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Chp.12 Cont. – Examples
to design Footings
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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
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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
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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
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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
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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
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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
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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
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Example 1Example 1Calculate the shear Vu
k 616ft 125.3k/ft 6.83k 683 22
2nuu
dcqPV
1ft 10
ft 10
The shape parameter
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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.
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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
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Example 1 - Final DesignExample 1 - Final Design
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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.
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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
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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
![Page 30: Chp.12 Cont. – Examples to design Footings](https://reader035.vdocument.in/reader035/viewer/2022062721/56813665550346895d9df24b/html5/thumbnails/30.jpg)
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
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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
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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
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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)
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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.
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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
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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
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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
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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
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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
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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
![Page 41: Chp.12 Cont. – Examples to design Footings](https://reader035.vdocument.in/reader035/viewer/2022062721/56813665550346895d9df24b/html5/thumbnails/41.jpg)
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
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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
![Page 43: Chp.12 Cont. – Examples to design Footings](https://reader035.vdocument.in/reader035/viewer/2022062721/56813665550346895d9df24b/html5/thumbnails/43.jpg)
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
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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
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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
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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
![Page 47: Chp.12 Cont. – Examples to design Footings](https://reader035.vdocument.in/reader035/viewer/2022062721/56813665550346895d9df24b/html5/thumbnails/47.jpg)
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
![Page 48: Chp.12 Cont. – Examples to design Footings](https://reader035.vdocument.in/reader035/viewer/2022062721/56813665550346895d9df24b/html5/thumbnails/48.jpg)
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
![Page 49: Chp.12 Cont. – Examples to design Footings](https://reader035.vdocument.in/reader035/viewer/2022062721/56813665550346895d9df24b/html5/thumbnails/49.jpg)
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
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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
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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
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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
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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
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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
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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.
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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
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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.
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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
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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
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Example 2 - Final designExample 2 - Final design
12 #823 #6