Footings
1
Acknowledgement
This Powerpoint presentation was
prepared by Dr. Terry Weigel,
University of Louisville. This work
and other contributions to the text
by Dr. Weigel are gratefully
acknowledged.
2
Footings
Support structural members and transfer
loads to the soil
Structural members are usually columns or
walls
Design for load transfer to soil uses
unfactored loads
Structural design of footing is done with
factored loads
3
Footings
Footings must be designed to prevent bearing
failure, sliding and overturning
Footings must be designed to prevent
excessive settlement or tilting
Typically, bottom of footing must be located
below frost line
Excavation may be required to reach a depth
where satisfactory bearing material is
located
4
Wall Footing
Wall footings – enlargement of the bottom of
the wall
5
Isolated Square Footing
Isolated or single column square footing –
loads relatively light and columns not
closely spaced
6
Combined Footing
Combined footings – support two or more
columns – heavily loaded columns; closely
spaced columns; columns near property line
7
Mat Footing
Mat or raft foundation – continuous concrete
slab supporting many columns; soil strength
relatively low; large column loads; isolated
spread footings would cover more than 50
percent of area; reduce differential
settlement
8
Pile Cap
Pile caps – distribute column loads to groups
of piles
9
Soil Pressure
Soil pressure is assumed to be uniformly
distributed beneath footing if column load
is applied at the center of gravity of the
footing
Footings supported by sandy soils
Footings supported by clayey soils
Footings supported eccentric loads
10
Assumed Soil Pressure
11
Soil Pressure - Sandy Soil
12
Soil Pressure - Clayey Soil
13
Allowable Soil Pressure
Actual soil pressure is based on unfactored
loads
Allowable soil pressure may be determined by
a geotechnical engineer
When soil exploration is not feasible, values
provided by building codes may be used
Factor of safety is typically 3
14
Allowable Soil Pressure (Table 12.1)
Maximum Allowable Soil Pressure
Material
Rock
Allowable Pressure, ksf
20% of ultimate
strength
Compact coarse or fine sand, hard
clay or sand clay
8
Medium stiff clay or sandy clay
6
Compact inorganic sand and silt
mixtures
4
Loose sand
3
Soft sand clay or clay
2
Loose inorganic sand-silt mixtures
1
Loose organic sand-silt mixtures,
muck or bay mud
0
15
Design of Wall Footings
Generally, beam design theory is used
Shear strength almost always controls
footing depth
Compute moment at the face of the wall
(concrete wall) or halfway between wall
face and its centerline (masonry walls)
16
Design of Wall Footings
17
Design of Wall Footings
18
Design of Wall Footings
19
Design of Wall Footings
20
Design of Wall Footings
Shear may be calculated at distance d from
face of the wall
Use of stirrups is not economical – set d so
that concrete carries all the shear
Vc  2 f c' bw d
d
Vu
2 f c' bw
21
Design of Wall Footings
Design a 12-in wide strip
Section 15.7 of ACI Code:
Depth of footing above bottom
reinforcement not less than 6 in
for footings on soil and not less
than 12 in for footings on piles
Minimum practical depth of footing is 10 in
and 16 in for pile caps
22
Wall Footing Design
Examples
23
Example 12.1
Design a wall footing to support a 12-in. wide
reinforced concrete wall with a dead load
of 20 k/ft and a live load of 15 k/ft. The
bottom of the footing is to be 4 foot
below final grade, the soil weighs 100
lb/ft3 the allowable soil pressure is 4 ksf.
The concrete strength is 3,000 psi and the
steel is Grade 60.
24
Example 12.1
25
Example 12.1
Assume a footing thickness of 12 in. With a
minimum cover of 3 in., this gives a d value
of about 8.5 in. Compute the footing
weight and
Footing weight
soil weight:
 12 in 

 150   150 psf
 12 in/ft 
Soil weight
 36 in 

 100   300 psf
 12 in/ft 
26
Example 12.1
Effective soil pressure and required width of
footing:
qe  4000 psf  150 psf  300 psf  3550 psf
Width of footing required
20 k/ft  15 k/ft
 9.86 ft
3.55 ksf
Use 10 ft
27
Example 12.1
Factored bearing pressure for design of
concrete:
qu 
1.2  20 k/ft   1.6 15 k/ft 
10 ft
 4.80 ksf
28
Example 12.1
Compute design shear (at distance d from
face of wall):
6 in
8.5 in 
 10 ft
Vu  


  4.80 ksf   18.2 k
12 in/ft 12 in/ft 
 2
18, 200 lb
d
 18.46 in
0.75(1.0) 2 3000 ksi 12 in 


Much larger than orginal assumption
Try a thicker footing - say 20 in thick
d  16.5 in
29
Example 12.1
 20 in 
qe  4000 psf  
 150 psf  
 12 in/ft 
 28 in 

 100 psf   3517 psf
 12 in/ft 
Width of footing required
20 k/ft  15 k/ft
 9.95 ft
3.517 ksf
Use 10 ft
30
Example 12.1
6 in
16.5 in 
 10 ft
Vu  


  4.80 ksf   15.0 k
12 in/ft 12 in/ft 
 2
15,000 lb
d
 15.21 in
0.75 2 3000 ksi 12 in/ft 


h  15.21 in  3.5 in  18.71 in
Use a 20 in thick footing
31
Example 12.1
10 ft
6 in

 4.5 ft
2
12 in/ft
Compute moment on a one-foot-long strip
wL  4.80 k/ft  4.5 ft 
Mu 

 48.6 k-ft/ft
2
2
12 in/ft  48, 600 lb-ft/ft 

Mu

 198.3 psi
2
2
 bd
0.9 12 in 16.5 in 
2
2
32
Example 12.1
Appendix Table 4.12, r = 0.00345 < 0.0136,
section is tension controlled;  = 0.9
2
in
As   0.0034512 in 16.5 in   0.68
ft
Use No 7 at 10 in (As = 0.72 in2 / ft from
Table A.6)
33
Example 12.1
Development length:
 t  e  s    1
cb  5 in  side cover
db
0.875
 3.4375  use cb  3.5 in
cb  3   3 
2
2
10 in
 5 in  one-half c-c spacing of bars
cb 
2
cb  K tr 3.5 in  0
 4.0  Use 2.5

0.875 in
db
34
Example 12.1
3 f y  t e s


'
d b 40  f c cb  K tr
db
d
3 60,000 psi 1
 32.86 diameters
40 3000 psi 2.5
2


A

0.68
in
/ft 
s ,required
d

   32.86  
  31.03 diameters
2
d b  As , provided 
 0.72 in /ft 
d   31.03 0.875 in   27.15 in
35
Example 12.1
Available length for development
10 ft 12 in/ft   6 in  3 in  51 in  27.15 in
2
36
Example 12.1
Temperature and shrinkage steel
As   0.0018 12 in  20 in   0.432 in 2 / ft
Use No 5 at 8 in (As = 0.465 in2 / ft)
37
Design of Isolated Square Footings
Most isolated square footings have a constant
thickness
For very thick footings, it may be economical
to step or taper footing
Two types of shear must be considered – oneway shear and two-way shear
38
Design of Isolated Square Footings
Constant thickness
39
Design of Isolated Square Footings
Stepped
40
Design of Isolated Square Footings
Tapered
41
One-way Shear
Same as for wall footings
d
Vu
2 f c' bw
42
One-way Shear
43
Two-way Shear
ACI Code Section 11.11.1.2 states that critical
section is at a distance d/2 from face of
support
44
Two-way Shear
45
Two-way Shear
46
Two-way Shear
Vc  4 f c' bo d
<- ACI Code Equation 11-33

4 
Vc   2    f c' bo d <- ACI Code Equation 11-35
c 

 c  ratio of the length of the long side of the
column to the length of the short side of
the column bearing on the footing

sd 
'
Vc   2 

f

c bo d <- ACI Code Equation 11-34
bo 

47
Two-way Shear
s = 40 for interior columns
s = 30 for exterior columns
s = 20 for corner columns
48
Flexural Design – Isolated Square
Footings
Flexural reinforcement is required in two
directions
The values of d for the layers of steel in
the two directions will be different
For square footings, design using the value of
d for the upper layer is typical
For square footings supporting non-square
columns, moments are larger in the
shorter direction of the column
49
Flexural Design – Isolated Square
Footings
Reinforcing steel areas required to resist
moment are often less than minimum
required steel:
200
As ,min 
As ,min 
fy
bw d
3 f c'
fy
bw d
Code Section 10.5.4 states that minimum
area and maximum spacing need only be
equal to values required for temperature
and shrinkage steel
50
Flexural Design – Isolated Square
Footings
Maximum steel spacing may not exceed three
times the footing thickness or 18 in.
51
Load Transfer from Column to
Footing
All forces at the base of the column must be
transferred to the footing
Compressive forces must be transferred by
bearing
Tensile forces may be transferred by
reinforcement or mechanical connectors
52
Load Transfer from Column to
Footing
Columns transfer loads directly over the area
of the column
Load transfer into the footing may by
assumed to occur over an effective area
which may be larger than the column area
For the same strength of concrete, the
footing can support more bearing load
than can the column
53
Load Transfer from Column to
Footing
Bearing strength permitted at the base of
the column ->  0.85 f c' A1
Bearing strength permitted on the footing is
the same value multiplied by -> A2
A1
2
See ACI Code Section 10.14.1
54
Definition of A1 and A2
A1 is the area of the column
A2 is the area of footing geometrically similar
to and concentric with the column
55
Column Dowels
56
Excess Bearing Load
Excess bearing load can be carried by dowels
or column bars extended into footing
ACI Code Section 15.8.2 requires that the
dowel area not be less than 0.005 times
the gross cross-sectional area of the
column
57
Development Length for Dowels
Development length of dowels must be
sufficient to transfer column force to
footing
Development length of dowels may not be less
than the length required if bearing stress
was not exceeded
58
Splice Length for Dowels
ACI Code does not permit splicing of No 14 or
No 18 bars
ACI Code Section 15.8.2.3 does permit No 14
or No 18 bars to be spliced to No 11 (or
larger) dowels in footings
These dowels must extend into the column
not less than the development length for
the No 14 or No 18 bar, or the
compression lap splice length for the
dowels, whichever is larger
59
Splice Length for Dowels
These dowels must extend into the footing
for a distance not less than the
development length for dowels
60
Insufficient Development or
Splice Length
Use a larger number of smaller dowels
Use a deeper footing
Add a cap or pedestal to the footing
61
Column Uplift
Development length must be those for
tension
Splice requirements are those found in ACI
Code Section 12.17
62
Isolated Rectangular Footings
Square footings are more econonical than
rectangular footings
Long direction steel is uniformly distributed
along short direction
Short direction steel is non uniformly
distributed along long direction
63
Isolated Rectangular Footings
ACI Code Section 15.4.4.2
Reinforcement in band width
2


Reinforcement in short direction   1
 is the ratio of the length of the footing in
the long direction to the length in the
short direction
Remaining steel is distributed uniformly
throughout the two portions of the
footing outside the band
64
Isolated Rectangular Footings
65
Footing Design Examples
66
Example 12.2
Design a square column footing for a 16-in.
square tied interior column that supports
loads of D = 200 k and L = 160 k. The
column is reinforced with eight No 8 bars,
the bottom of the footing is 5 foot below
final grade, the soil weighs 100 lb/ft3 the
allowable soil pressure is 5 ksf. The
concrete strength is 3,000 psi and the
steel is Grade 60.
67
Example 12.2
Assume a footing thickness of 24 in. with a
minimum cover of 3 in., this gives a d value
of about 19.5 in. Compute the footing
weight and
Footing weight
soil weight:
 24 in 

 150   300 psf
 12 in/ft 
Soil weight
 36 in 

 100   300 psf
 12 in/ft 
68
Example 12.2
Effective soil pressure and required area of
footing:
qe  5000 psf  300 psf  300 psf  4400 psf
200 k  160 k
A
 81.82 ft 2
4.40 ksf
Use 9 ft x 9 ft
69
Example 12.2
Factored bearing pressure for design of
concrete:
qu 
1.2  200 k   1.6 160 k 
81 ft
2
 6.12 ksf
70
Example 12.2
Depth required to resist punching shear:
bo  4(16  19.5)  142 in

Vu 2  81.0 ft   2.96 ft 
d
2

2
  6.12  442.09 k
442, 090 lb

0.75 4 3000 psi 142 in 
 18.95 in  19.5 in Ok
442, 090 lb
d
 40 19.5 in

0.75 
 2  3000 psi 142 in 
 142 in

 10.12 in  19.5 in Ok
71
Example 12.2
72
Example 12.2
Depth required to resist one-way shear:
Vu1   9 ft  2.208 ft  6.12 ksf   121.62 k
d

121, 620 lb

0.75 2 3000 psi 108 in 
 13.71 in  19.5 in Ok
73
Example 12.2
Flexural design
wL  6.12 ksf  9 ft  3.83 ft 
Mu 

 404 k-ft
2
2
12 in/ft  404, 000 lb-ft 

Mu

 131.2 psi
2
2
 bd
0.9 108 in 19.5 in 
2
2
74
Example 12.2
Appendix Table 4.12, r = 0.00225 < rmin
200
r
 0.0033 
60, 000 psi
3 3000 psi
r
 0.00274
60, 000 psi
As   0.0033108 in 19.5 in   6.95 in 2
Use nine No 8 (As = 7.07 in2)
75
Example 12.2
Development length:
 t  e  s    1
cb  bottom cover  3.5 in 
cb  one-half center-to-center bar spacing  6 in
cb  Ktr 3.5 in  0

 3.5  Use 2.5
db
1.0 in
76
Example 12.2
3 f y  t e s


'
db 40  f c cb  K tr
db
d
3 60, 000 1
 32.86 diameters
40 3000 2.5
2


A


6.95
in
s , required
d
 32.30 diameters

   32.86  
2 
db  As , provided 
 7.07 in 
d  32.30 1.0 in   32.30 in
77
Example 12.2
Available length for development
 9 ft 12 in/ft   16 in  3 in  43 in  32.30 in
2
2
78
Example 12.3
Design for load transfer for the column and
footing in Example 12.2. The strength of
the sand-lightweight concrete (different
from Example 12.2) in the column is 4 ksi.
79
Example 12.3
Bearing force at the column base:
1.2  200 k   1.6 160 k   496 k
Design bearing force at the column base:
  0.85 f  A1  0.65  0.85 4 ksi 16 in 
'
c
2
 566 k  496 k Ok
80
Example 12.3
Design bearing
force in the
footing
concrete:
108 in 
2
16 in 
2
 6.75  Use 2
  0.85 f  A1
'
c
A2

A1
0.65  0.85 3 ksi 16 in   2 
2
 848.6 k  496 k Ok
Minimum dowel area:
0.005 16 in   1.28 in 2
2
81
Example 12.3
Dowel development length into the column
d

0.02db f y
 f c'

0.02  0.75 in  60, 000 psi 
 0.85
4000 psi
 16.74 in
Dowel development length into the footing
d

0.02db f y
 f c'

0.02  0.75 in  60, 000 psi 
1.0 
3000 psi
 16.43 in
82
Example 12.3
Development length must not be less than:
d
 0.0003db f y  0.0003  0.75 in  60, 000 ksi 
 13.50 in
d
 8.0 in
83
Example 12.4
Design for load transfer for a 14-in. square
column to a 13 ft square footing if Pu =
800 k. Normal weight concrete is used in
both the column and the footing. The
concrete in the column is 5 ksi and in the
footing is 3 ksi. The column is reinforced
with eight No 8 bars.
84
Example 12.4
Bearing force at the column base = 800 k
Design bearing force at the column base:
  0.85 f  A1  0.65  0.85 5 ksi 14 in 
'
c
2
 541.5 k  800 k No good
85
Example 12.4
Design bearing force in the footing concrete:
156 in 
2
14
in


2
A2

A1
  0.85 f  A1
'
c
 11.14  Use 2
A2

A1
0.65  0.85  3 ksi 14 in   2 
2
 649.7 k  800 k No good
86
Example 12.4
Design dowels to resist excess bearing force:
800 k  541.5 k  258.5 k
258.5 k
As 
 4.79 in 2
0.9  60 k 
0.005 14 in   0.98 in 2
2
Use eight No 7 bars (As = 4.80 in2)
87
Example 12.4
Dowel development length into the column

0.02db f y

0.02  0.875 in  60, 000 psi 
 14.85 in
1 5000 psi
d  0.0003d b f y  0.0003  0.875 in  60, 000 ksi 
d
 f
'
c
 15.75 in 
d
 8.0 in
88
Example 12.4
Dowel development length into the footing

0.02db f y

0.02  0.875 in  60, 000 psi 
 19.42 in 
1.0  3000 psi
d  0.0003d b f y  0.0003  0.875 in  60, 000 ksi 
d
 f c'
 15.75 in
d
 8.0 in
89
Example 12.5
Design a rectangular footing for an 18-in.
interior square column for D = 185 k and
L = 150 k. The long side of the footing
should be twice the length of the short
side. The normal weight concrete
strength for both the column and the
footing is 4 ksi. The allowable soil
pressure is 4000 psf and the bottom of
the footing is 5 ft below grade.
90
Example 12.5
Assume a footing thickness of 24 in. with a
minimum cover of 3 in., this gives a d value
of about 19.5 in. Compute the footing
weight and
Footing weight
soil weight:
 24 in 

 150   300 psf
 12 in/ft 
Soil weight
 60-24 in 

 100   300 psf
 12 in/ft 
91
Example 12.5
Effective soil pressure and required area of
footing:
qe  4000 psf  300 psf  300 psf  3400 psf
185 k  150 k
A
 98.5 ft 2
3.40 ksf
Use a footing 7'-0" x 14'-0"  A  98.0 ft 2 
1.2 185 k   1.6 150 k 
qu 
 4.71 ksf
2
98.0 ft
92
Example 12.5
Depth required to resist one-way shear. Take
b = 7 ft.
Vu1   7 ft  4.625 ft  4.71 ksf   152.49 k
d

152, 490 lb

0.75 1 2 4000 psi 84 in 
 19.14 in
h  19.14  4.5 in  23.64 in
93
Example 12.5
94
Example 12.5
Depth required to resist punching shear:
bo  4 18  19.5 in   150 in

Vu 2  98.0 ft 2   3.125 ft 
d
d

415,580 lb

2
  4.71 ksf   415.58 k
0.75 1 4 4000 psi 150 in 
 14.60 in  19.5 in Ok
415,580 lb
 40  19.5 in

0.75 
 2  4000 psi 150 in 
 150 in

 8.11 in  19.5 in Ok
95
Example 12.5
96
Example 12.5
Flexural design (steel in long direction)
14 ft
9 in

 6.25 ft
2
12 in/ft
 6.25 ft 
M u   6.25 ft  7 ft  4.71 ksf  
  643.9 k-ft
 2 
12 in/ft  643,900 lb-ft 

Mu

 268.8 psi
2
2
 bd
0.9  84 in 19.5 in 
97
Example 12.5
Appendix Table 4.13, r = 0.00467
As   0.00467 84 in 19.5 in   7.65 in 2
Use ten No 8 (As = 7.85 in2)
98
Example 12.5
Flexural design (steel in short direction)
7 ft
9 in

 2.75 ft
2 12 in/ft
 2.75 ft 
M u   2.75 ft 14 ft  4.71 ksf  
  249.3 k-ft
 2 
12 in/ft  249,300 lb-ft 

Mu

 52.0 psi
2
2
 bd
0.9 168 in 19.5 in 
Too low for Table A.13
99
Example 12.5
200
r
 0.0033 
60, 000 psi
3 4000 psi
r
 0.00316
60, 000 psi
As   0.0033168 in 19.5 in   10.81 in 2
Use 18 No 7 (As = 10.82 in2)
100
Example 12.5
Reinforcement in band width
2
2
2



Reinforcement in short direction   1 2  1 3
Use 2/3 x 18 = 12 bars in band width
101
Example 12.5
102