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:
Wc  d  150 lb/ft * 24 in. *
3
1 ft.
 300 lb/ft 2
12 in.


1
ft.
  200 lb/ft 2
Ws   s d s  100 lb/ft *  4 ft  24 in. *


12 in. 

3
Example 1
The effective soil pressure is given as:
qeff  qs  Wc  Ws
 5000 lb/ft 2  300 lb/ft 2  200 lb/ft 2
 4500 lb/ft 2  4.5 k/ft 2
Example 1
Calculate the size of the footing:
Actual Loads  DL  LL  245 k  200 k  445 k
Area of footing 
445 k
 98.9 ft 2
4.5 k/ft 2
Side of footing  9.94 ft  Use 10 ft
Example 1
Calculate net upward pressure:
Actual Loads  1.4 DL  1.7 LL
 1.4245 k   1.7200 k   683 k
Net upward pressure qn 
683 k
100 ft 2
 6.83 k / ft 2
Example 1
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d  h  cover  1.5d b
d  24 in.  3 in  1.51.0 in 
 19.5 in.
Example 1
Calculate perimeter for two-way shear or
punch out shear. The column is 18 in.
square.
bo  4c  d 
 418 in.  19.5 in.   150 in.
 1 ft 
  3.125 ft
c  d  18 in.  19.5 in. 


 12 in 
Example 1
Calculate the shear Vu
Vu  Pu  qn c  d 
2
 683 k  6.83 k/ft 2 3.125 ft   616 k
2
The shape parameter

10 ft
10 ft
1
Example 1
Calculate d value from the shear capacity according to
11.12.2.1 chose the largest value of d


4
Vc   2   f c b0 d
c 

 d

Vc   s  2  f c b0 d
 bo

Vc  4 f c b0 d
s is 40 for interior, 30 for edge
and 20 for corner column
Example 1
The depth of the footing can be calculated by using
two way shear
d
Vu
 4 f c b0

 1000 lb 

616 k 


 1k 


0.85 4 4000 150 in 
 19.1 in.
Example 1
The second equation bo is dependent on d so use the
assumed values and you will find that d is smaller and
 = 40
d
Vu
Actual (d =14.02324 in.)
 40d

 
 2  f c b0

b
 o

bo=128.93 in
 1000 lb 

616 k 


 1k 

 10.6 in.
 4019.5 in  
0.85
 2  4000 150 in 


150
in




Example 1
The depth of the footing can be calculated
by using one-way shear


 1 ft 



18 in 


L c
  10 ft
 1 ft  
12
in


  d 


 19.5 in 

 


2
2
2 2

 12 in  




 2.625 ft
L c

Vu  qn l2    d 


2 2

 6.83 k/ft 2 10 ft 2.625 ft   179.3 k
Example 1
The depth of the footing can be calculated by using
one-way shear
 1000 lb 

179.3 k


Vu
 1k 
d

 13.9 in.





 2 fc b
12
in

0.85 2 4000 10 ft 




 1 ft   


The footing is 19.5 in. > 13.9 in. so it will work.
Example 1
Calculate the bending moment of the footing at the
edge of the column

 1 ft  


18 in 


 L c   10 ft
 12 in    4.25 ft
  




2
 2 2  2




L c
  
 L c   2 2 
4.25 ft  10 ft   616.8 k - ft
M u  qn   
b  6.83 k/ft 4.25 ft 


2
2
 2 2
Example 1
Calculate Ru for the footing to find r of the footing.
 12 in. 

616.8 k - ft * 


Mu
 1 ft   0.1622 ksi
Ru 

120 in * 19.5 in 2
bd 2
Example 1
From Ru for the footing the r value can be found.
Ru  f c 1  0.59     1.7 
2
1.7 Ru
f c
 

0
.
1622
ksi

1.7  1.7   41.7
  0.94 ksi   

 
0
2

 0.04632
2
r fy
fc
 0.04632  r 
0.046324 ksi 
60 ksi
 0.0031
Example 1
Compute the area of steel needed

 12 in.  
 19.5 in.   7.23 in 2
As  r bd  0.0030910 ft 



 1 ft  

The minimum amount of steel for shrinkage is
As  0.0018 bh  0.0018120 in. 24 in.   5.18 in 2
The minimum amount of steel for flexure is
 200 
200
120 in. 19.5 in.   7.8 in 2
As 
bd  


fy
 60000 
 Use
Example 1
Use a #7 bar (0.60 in2) Compute the number of bars
need
As
7.8 in 2
n

 13  Use 13 bars
Ab 0.60 in 2
Determine the spacing between bars
s
L  2 * cover
n  1

120 in - 23 in 
12
 9.5 in
Example 1
Check the bearing stress. The bearing strength N1, at
the base of the column, 18 in x 18 in.,   0.7


N1   0.85 f c A1   0.7 0.854 ksi 18 in   771 k
2
The bearing strength, N2, at the top of the footing is
N 2  N1
A2
A1
 2 N1
Example 1
A2  10 ft   100 ft 2
2
2

 1 ft  
   2.25 ft 2
A1  18 in 



 12 in.  

The bearing strength, N2, at the top of the footing is
A2
A1

100 ft 2
2.25 ft 2
 6.67  2  N 2  2 N1  2771 k   1542 k
Example 1
Pu =683 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1  0.005 * 18 in   1.62 in 2
2
Use minimum number of bars is 4, so use 4 # 8 bars
placed at the four corners of the column.
Example 1
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld 
0.02d b f y
fc

0.021 in 60000 psi 
 18.97 in  Use 19 in
4000 psi
The minimum ld , which has to be greater than 8 in., is
ld  0.0003d b f y  0.00031 in 60000 psi   18 in  8 in
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 1
The development length, ld for the #7 bars for the
reinforcement of the footing.
fy
f ydb

ld
60000 psi 0.875 in 

 ld 

 41.5 in
d b 20 f c
20 f c
20 4000 psi
There is adequate development length provided.
ld 
L
2
 cover 
c
2

120 in
2
 3 in 
18 in
2
 48 in
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:
Wc  d  150 lb/ft * 24 in. *
3
1 ft.
 300 lb/ft 2
12 in.


1
ft.
  200 lb/ft 2
Ws   s d s  100 lb/ft *  4 ft  24 in. *


12 in. 

3
Example 2
The effective soil pressure is given as:
qeff  qs  Wc  Ws
 5000 lb/ft 2  300 lb/ft 2  200 lb/ft 2
 4500 lb/ft 2  4.5 k/ft 2
Example 2
Calculate the size of the footing:
Actual Loads  DL  LL  245 k  200 k  445 k
Area of footing 
445 k
 98.9 ft 2
4.5 k/ft 2
Side of footing 
98.9 ft 2
8.5 ft
 11.64 ft  Use 12 ft
Example 2
Calculate net upward pressure:
Actual Loads  1.4 DL  1.7 LL
 1.4245 k   1.7200 k   683 k
Net upward pressure qn 
683 k
8.5 ft 12 ft 
 6.70 k / ft 2
Example 2
Calculate the depth of the reinforcement use # 8 bars
with a crisscrossing layering.
d  h  cover  1.5d b
d  24 in.  3 in  1.51.0 in 
 19.5 in.
Example 2
The depth of the footing can be calculated
by using the one-way shear (long direction)


 1 ft 



18 in 


L c
  12 ft
 1 ft  
12
in


  d 


 19.5 in 

 


2
2
2 2

 12 in  




 3.625 ft
L c

Vu  qn l2    d 


2 2

Vu =150.7 k in
short direction
 6.7 k/ft 2 8.5 ft 3.625 ft   206.4 k
Example 2
The depth of the footing can be calculated by using
one-way shear design
 1000 lb 

206.4 k


Vu
 1k 
d

 18.8 in.





 2 fc b
12
in

0.85 2 4000  8.5 ft 




 1 ft   


The footing is 19.5 in. > 18.8 in. so it will work.
Example 2
Calculate perimeter for two-way shear or
punch out shear. The column is 18 in.
square.
bo  4c  d 
 418 in.  19.5 in.   150 in.
 1 ft 
  3.125 ft
c  d  18 in.  19.5 in. 


 12 in 
Example 2
Calculate the shear Vu
Vu  Pu  qn c  d 
2
 683 k  6.7 k/ft 2 3.125 ft   617.6 k
2
The shape parameter

12 ft
8.5 ft
 1.41
Example 2
Calculate d from the shear capacity according to
11.12.2.1 chose the largest value of d.


4
Vc   2   f c b0 d
c 

 d

Vc   s  2  f c b0 d
 bo

Vc  4 f c b0 d
s is 40 for interior, 30 for edge
and 20 for corner column
Example 2
The depth of the footing can be calculated for the
two way shear
 1000 lb 

617.6 k 


Vu
 1k 
d

 15.8 in.





4
4
 4000 150 in 
  2   f c b0 0.85  2 




1
.
41






Example 2
The third equation bo is dependent on d so use the assumed
values and you will find that d is smaller and  = 40
d
Vu
Actual (d =14.032 in.)
 40d



 2  f c b0

b
 o

bo=128.173 in
 1000 lb 

617.6 k 


1
k



 10.64 in.
 4019.5 in  
0.85
 2  4000 150 in 


 150 in



Example 2
The depth of the footing can be calculated by using
the two way shear
d
Vu
 4 f c b0

 1000 lb 

617.6 k


 1k 


0.85 4 4000 150 in 
 19.14 in.
Example 2
Calculate the bending moment of the footing at the edge of
the column (long direction)

 1 ft  


18 in 


 L c   12 ft
 12 in    5.25 ft
  




2
 2 2  2




L c
  
 L c   2 2 
5.25 ft  8.5 ft   784.8 k - ft
M u  qn   
b  6.7 k/ft 5.25 ft 


2
2
 2 2
Example 2
Calculate Ru for the footing to find r of the footing.
 12 in. 

784.8 k - ft * 


Mu
 1 ft   0.2428 ksi
Ru 

 12 in  
bd 2 
  * 19.5 in 2
8.5 ft 



 1 ft  

Example 2
Use the Ru for the footing to find r.
Ru  f c 1  0.59     1.7 
2
1.7 Ru
f c
 

0
.
2428
ksi

1.7  1.7   41.7
  0.94 ksi   

 
0
2

 0.07036
2
r fy
fc
 0.07036  r 
0.070364 ksi 
60 ksi
 0.00469
Example 2
Compute the amount of steel needed

 12 in.  
 19.5 in.   9.33 in 2
As  r bd  0.00469 8.5 ft 



 1 ft  

The minimum amount of steel for shrinkage is
As  0.0018 bh  0.0018102 in. 24 in.   4.41 in 2
The minimum amount of steel for flexure is
 200 
200
102 in. 19.5 in.   6.63 in 2
As 
bd  


fy
 60000 
Example 2
Use As =8.36 in2 with #8 bars (0.79 in2). Compute
the number of bars need
As 9.33 in 2
n

 11.8  Use 12 bars
Ab 0.79 in 2
Determine the spacing between bars
s
L  2 * cover
n  1

102 in - 23 in 
11
 8.73 in
Example 2
Calculate the bending moment of the footing at the edge of
the column for short length

 1 ft  


18 in 


 L c   8.5 ft
 12 in    3.5 ft
  




2
 2 2  2




L c
  
 L c   2 2 
3.5 ft  12 ft   492.5 k - ft
M u  qn   
b  6.7 k/ft 3.5 ft 


2
2
 2 2
Example 2
Calculate Ru for the footing to find r of the footing.
 12 in. 

492.5 k - ft * 


Mu
 1 ft   0.1079 ksi
Ru 

 12 in  
bd 2 
  * 19.5 in 2
12 ft 



 1 ft  

Example 2
Use Ru for the footing to find r.
Ru  f c 1  0.59     1.7 
2
1.7 Ru
f c
 

0
.
1079
ksi

1.7  1.7   41.7
  0.94 ksi   

 
0
2

 0.0305
2
r fy
fc
 0.0305  r 
0.03054 ksi 
60 ksi
 0.00203
Example 2
Compute the amount of steel needed

 12 in.  
 19.5 in.   5.72 in 2
As  r bd  0.0020312 ft 



 1 ft  

The minimum amount of steel for shrinkage is
As  0.0018 bh  0.0018144 in. 24 in.   6.22 in 2
The minimum amount of steel for flexure is
 200 
200
144 in. 19.5 in.   9.36 in 2
As 
bd  


fy
 60000 
Example 2
Use As =9.36 in2 with #6 bar (0.44 in2) Compute the
number of bars need
As 9.36 in 2
n

 21.3  Use 22 bars
Ab 0.44 in 2
Calculate the reinforcement bandwidth
 Reinforcem ent in bandwidth


Total reinforcem ent


  2  2  0.83

   1 1.41  1
Example 2
The number of bars in the 8.5 ft band is 0.83(22)=19 bars .
outside # bar 
Total # bars - band bars
2

22  19
 1.5  Use 2 bars
2
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.
Example 2
Determine the spacing between bars for the band of 8.5 ft
s
L
n  1

102 in
 5.67 in
18
Determine the spacing between bars outside the band
s
L  cover
n

21 in - 3in
2
 9 in
Example 2
Check the bearing stress. The bearing strength N1, at
the base of the column, 18 in x 18 in.,   0.7


N1   0.85 f c A1   0.7 0.854 ksi 18 in   771 k
2
The bearing strength, N2, at the top of the footing is
N 2  N1
A2
A1
 2 N1
Example 2
A2  10 ft   100 ft 2
2
2

 1 ft  
   2.25 ft 2
A1  18 in 



 12 in.  

The bearing strength, N2, at the top of the footing is
A2
A1

100 ft 2
2.25 ft 2
 6.67  2  N 2  2 N1  2771 k   1542 k
Example 2
Pu =683 k < N1, bearing stress is adequate. The
minimum area of dowels is required.
0.005 A1  0.005 * 18 in   1.62 in 2
2
Use minimum number of bars is 4, so use 4 # 8 bars
placed at the four corners of the column.
Example 2
The development length of the dowels in compression
from ACI Code 12.3.2 for compression.
ld 
0.02d b f y
fc

0.021 in 60000 psi 
 18.97 in  Use 19 in
4000 psi
The minimum ld , which has to be greater than 8 in., is
ld  0.0003d b f y  0.00031 in 60000 psi   18 in  8 in
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
ld
db
fy

f ydb
 ld 
20 f c

60000 psi 1.0 in 

 47.4 in
20 f c
20 4000 psi
There is adequate development length provided.
ld 
L
2
 cover 
c
2

144 in
2
 3 in 
18 in
2
 60 in
Example 2
The development length, ld for the #6 bars
ld
db
fy

f ydb
 ld 
25 f c

60000 psi 0.75 in 

 28.5 in
25 f c
25 4000 psi
There is adequate development length provided.
ld 
L
2
 cover 
c
2

102 in
2
 3 in 
18 in
2
 39 in
Example 2 - Final design
23 #6
12 #8