Post on 03-Jan-2016
description
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