View
19
Download
1
Category
Preview:
Citation preview
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
3
Design for load transfer to soil uses unfactored loads
Support structural members and transfer loads to the soil
Structural members are usually columns or walls
Structural design of footing is done with factored loads
Footings
4
Typically, bottom of footing must be located below frost line
Footings must be designed to prevent bearing failure, sliding and overturning
Footings must be designed to prevent excessive settlement or tilting
Excavation may be required to reach a depth where satisfactory bearing material is located
Isolated Square Footing
6
Isolated or single column square footing – loads relatively light and columns not closely spaced
Combined Footing
7
Combined footings – support two or more columns – heavily loaded columns; closely spaced columns; columns near property line
Mat Footing
8
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
Soil Pressure
10
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
Allowable Soil Pressure
14
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
Allowable Soil Pressure (Table 12.1)
15
Maximum Allowable Soil Pressure
Material Allowable Pressure, ksf
Rock 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
Design of Wall Footings
16
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)
Design of Wall Footings
21
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
'
'
2
2
c c w
u
c w
V f b d
Vd
f b
φ φλ
φλ
=
=
Design of Wall Footings
22
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
Example 12.1
24
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.
Example 12.1
26
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
soil weight: ( )
( )
Footing weight
12 in150 150 psf
12 in/ft
Soil weight
36 in100 300 psf
12 in/ft
=
=
Example 12.1
27
Effective soil pressure and required width of footing:
4000 psf 150 psf 300 psf 3550 psf
Width of footing required
20 k/ft 15 k/ft9.86 ft
3.55 ksf
Use 10 ft
eq = − − =
+=
Example 12.1
28
Factored bearing pressure for design of concrete:
( ) ( )1.2 20 k/ft 1.6 15 k/ft4.80 ksf
10 ftuq
+= =
Example 12.1
29
Compute design shear (at distance d from face of wall):
( )
( )( )
10 ft 6 in 8.5 in4.80 ksf 18.2 k
2 12 in/ft 12 in/ft
18, 200 lb18.46 in
0.75(1.0) 2 3000 ksi 12 in
Much larger than orginal assumption
Try a thicker footing - say 20 in thick
16.5 in
uV
d
d
= − − =
= =
≈
Example 12.1
30
( )
( )
20 in4000 psf 150 psf
12 in/ft
28 in 100 psf 3517 psf
12 in/ft
Width of footing required
20 k/ft 15 k/ft9.95 ft
3.517 ksf
Use 10 ft
eq
= − −
=
+=
Example 12.1
31
( )
( )( )
10 ft 6 in 16.5 in4.80 ksf 15.0 k
2 12 in/ft 12 in/ft
15,000 lb15.21 in
0.75 2 3000 ksi 12 in/ft
15.21 in 3.5 in 18.71 in
Use a 20 in thick footing
uV
d
h
= − − =
= =
≈ + =
Example 12.1
32
( )( )
( ) ( )( ) ( )
22
22
10 ft 6 in4.5 ft
2 12 in/ft
Compute moment on a one-foot-long strip
4.80 k/ft 4.5 ft48.6 k-ft/ft
2 2
12 in/ft 48,600 lb-ft/ft198.3 psi
0.9 12 in 16.5 in
u
u
wLM
M
bdφ
− =
= = =
= =
Example 12.1
33
Appendix Table 4.12, ρ = 0.00345 < 0.0136, section is tension controlled; φ = 0.9
( )( )( )2in
0.00345 12 in 16.5 in 0.68 ft
sA = =
Use No 7 at 10 in (As = 0.72 in2 / ft from Table A.6)
Example 12.1
34
Development length: 1
5 in side cover
0.875c 3 3 3.4375 3.5
2 2
10 in5 in one-half c-c spacing of bars
2
3.5 in 04.0 Use 2.5
0.875 in
t e s
b
bb b
b
b tr
b
c
duse c in
c
c K
d
ψ ψ ψ λ= = = =
= ←
= + ≅ + = ← =
= = ←
+ += = ∴
Example 12.1
35
( )
( )( )
'
2,
2
,
3
40
3 60,000 psi 1 32.86 diameters
40 2.53000 psi
0.68 in /ft32.86 31.03 diameters
0.72 in /ft
31.03 0.875 in 27.15 in
yd t e s
b trb c
b
s requiredd
b s provided
d
f
c Kd fd
A
d A
ψ ψ ψ
λ= =
+
=
= =
= =
l
l
l
Example 12.1
36
( ) ( )10 ft 12 in/ft6 in 3 in 51 in 27.15 in
2− − = >
Available length for development
Example 12.1
37
( )( )( ) 20.0018 12 in 20 in 0.432 in / ft
sA = =
Temperature and shrinkage steel
Use No 5 at 8 in (As = 0.465 in2 / ft)
Design of Isolated Square Footings
38
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 – one-way shear and two-way shear
Two-way Shear
44
ACI Code Section 11.11.1.2 states that critical section is at a distance d/2 from face of support
Two-way Shear
47
'
'
'
4
42
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
2
c c o
c c o
c
c
sc c o
o
V f b d
V f b d
dV f b d
b
λ
λβ
β
αλ
=
= +
⇒
= +
<- ACI Code Equation 11-33
<- ACI Code Equation 11-35
<- ACI Code Equation 11-34
Two-way Shear
48
αs = 40 for interior columns
αs = 30 for exterior columns
αs = 20 for corner columns
Flexural Design – Isolated Square Footings
49
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
Flexural Design – Isolated Square Footings
50
Reinforcing steel areas required to resist moment are often less than minimum required steel:
Code Section 10.5.4 states that minimum area and maximum spacing need only be equal to values required for temperature and shrinkage steel
,min
'
,min
200
3
s w
y
c
s w
y
A b df
fA b d
f
=
=
Flexural Design – Isolated Square Footings
51
Maximum steel spacing may not exceed three times the footing thickness or 18 in.
Load Transfer from Column to Footing
52
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
Load Transfer from Column to Footing
53
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
Load Transfer from Column to Footing
54
Bearing strength permitted at the base of the column ->
Bearing strength permitted on the footing is the same value multiplied by ->
See ACI Code Section 10.14.1
'
10.85 cf Aφ
2
1
2A
A≤
Definition of A1 and A2
55
A2 is the area of footing geometrically similar to and concentric with the column
A1 is the area of the column
Excess Bearing Load
57
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
Development Length for Dowels
58
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
Splice Length for Dowels
59
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
Splice Length for Dowels
60
These dowels must extend into the footing for a distance not less than the development length for dowels
Insufficient Development or Splice Length
61
Use a larger number of smaller dowels
Use a deeper footing
Add a cap or pedestal to the footing
Column Uplift
62
Development length must be those for tension
Splice requirements are those found in ACI Code Section 12.17
Isolated Rectangular Footings
63
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
Isolated Rectangular Footings
64
ACI Code Section 15.4.4.2 Reinforcement in band width 2
Reinforcement in short direction 1γ
β= =
+
β 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
Example 12.2
67
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.
Example 12.2
68
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
soil weight: ( )
( )
Footing weight
24 in150 300 psf
12 in/ft
Soil weight
36 in100 300 psf
12 in/ft
=
=
Example 12.2
69
Effective soil pressure and required area of footing:
2
5000 psf 300 psf 300 psf 4400 psf
200 k 160 k81.82 ft
4.40 ksf
Use 9 ft x 9 ft
eq
A
= − − =
+= =
Example 12.2
70
Factored bearing pressure for design of concrete:
( ) ( )2
1.2 200 k 1.6 160 k6.12 ksf
81 ftuq
+= =
Example 12.2
71
Depth required to resist punching shear:
( )( )( )
( )( )
( )
22
2
4(16 19.5) 142 in
81.0 ft 2.96 ft 6.12 442.09 k
442,090 lb18.95 in 19.5 in Ok
0.75 4 3000 psi 142 in
442,090 lb
40 19.5 in0.75 2 3000 psi 142 in
142 in
10.12 in 19.5 in Ok
o
u
b
V
d
d
= + =
= − =
= = <
=× +
= <
Example 12.2
73
Depth required to resist one-way shear:
( )( )( )
( )( )
1 9 ft 2.208 ft 6.12 ksf 121.62 k
121,620 lb13.71 in 19.5 in Ok
0.75 2 3000 psi 108 in
uV
d
= =
= = <
Example 12.2
74
( ) ( ) ( )
( )( )( ) ( )
22
22
6.12 ksf 9 ft 3.83 ft404 k-ft
2 2
12 in/ft 404,000 lb-ft131.2 psi
0.9 108 in 19.5 in
u
u
wLM
M
bdφ
= = =
= =
Flexural design
Example 12.2
75
Appendix Table 4.12, ρ = 0.00225 < ρmin
( ) ( ) ( ) 2
2000.0033
60,000 psi
3 3000 psi0.00274
60,000 psi
0.0033 108 in 19.5 in 6.95 insA
ρ
ρ
= = ←
= =
= =
Use nine No 8 (As = 7.07 in2)
Example 12.2
76
Development length:
1
bottom cover 3.5 in
one-half center-to-center bar spacing 6 in
3.5 in 03.5 Use 2.5
1.0 in
t e s
b
b
b tr
b
c
c
c K
d
ψ ψ ψ λ= = = =
= = ←
= =
+ += = ∴
Example 12.2
77
( )
( )
'
2,
2
,
3
40
3 60, 000 1 32.86 diameters
40 2.53000
6.95 in32.86 32.30 diameters
7.07 in
32.30 1.0 in 32.30 in
yd t e s
b trb c
b
s requiredd
b s provided
d
f
c Kd fd
A
d A
ψ ψ ψ
λ= =
+
=
= =
= =
l
l
l
Example 12.2
78
( ) ( )9 ft 12 in/ft 16 in3 in 43 in 32.30 in
2 2− − = >
Available length for development
Example 12.3
79
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.
Example 12.3
80
Bearing force at the column base:
( ) ( )1.2 200 k 1.6 160 k 496 k+ =
Design bearing force at the column base:
( ) ( ) ( ) ( )2'
10.85 0.65 0.85 4 ksi 16 in
566 k 496 k Ok
cf Aφ =
= >
Example 12.3
81
Design bearing force in the footing concrete:
( )( )
( )
( )( )( ) ( )
2
2
' 21
1
2
108 in6.75 Use 2
16 in
0.85
0.65 0.85 3 ksi 16 in 2
848.6 k 496 k Ok
c
Af A
Aφ
= ∴
=
= >
Minimum dowel area: ( )2 20.005 16 in 1.28 in=
Example 12.3
82
( ) ( )( )'
0.02 0.02 0.75 in 60,000 psi16.74 in
0.85 4000 psi
b y
d
c
d f
fλ= = =l
Dowel development length into the column
( ) ( )( )'
0.02 0.02 0.75 in 60,000 psi16.43 in
1.0 3000 psi
b y
d
c
d f
fλ= = =l
Dowel development length into the footing
Example 12.3
83
( ) ( )0.0003 0.0003 0.75 in 60, 000 ksi
13.50 in
8.0 in
d b y
d
d f= =
=
=
l
l
Development length must not be less than:
Example 12.4
84
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.
Example 12.4
85
Bearing force at the column base = 800 k
Design bearing force at the column base:
( ) ( )( )( )2'
10.85 0.65 0.85 5 ksi 14 in
541.5 k 800 k No good
cf Aφ =
= <
Example 12.4
86
Design bearing force in the footing concrete:
( )( )
( )
( )( )( ) ( )
2
2
2
1
' 21
1
2
156 in11.14 Use 2
14 in
0.85
0.65 0.85 3 ksi 14 in 2
649.7 k 800 k No good
c
A
A
Af A
Aφ
= = ∴
=
= <
Example 12.4
87
Design dowels to resist excess bearing force:
( )( )
2
2 2
800 k 541.5 k 258.5 k
258.5 k4.79 in
0.9 60 k
0.005 14 in 0.98 in
sA
− =
= =
=
Use eight No 7 bars (As = 4.80 in2)
Example 12.4
88
( ) ( )( )
( ) ( )
'
0.02 0.02 0.875 in 60, 000 psi14.85 in
1 5000 psi
0.0003 0.0003 0.875 in 60,000 ksi
15.75 in
8.0 in
b y
d
c
d b y
d
d f
f
d f
λ= = =
= =
= ←
=
l
l
l
Dowel development length into the column
Example 12.4
89
( ) ( )( )
( ) ( )
'
0.02 0.02 0.875 in 60, 000 psi19.42 in
1.0 3000 psi
0.0003 0.0003 0.875 in 60,000 ksi
15.75 in
8.0 in
b y
d
c
d b y
d
d f
f
d f
λ= = = ←
= =
=
=
l
l
l
Dowel development length into the footing
Example 12.5
90
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.
Example 12.5
91
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
soil weight: ( )
( )
Footing weight
24 in150 300 psf
12 in/ft
Soil weight
60-24 in100 300 psf
12 in/ft
=
=
Example 12.5
92
Effective soil pressure and required area of footing:
( )( ) ( )
2
2
2
4000 psf 300 psf 300 psf 3400 psf
185 k 150 k98.5 ft
3.40 ksf
Use a footing 7'-0" x 14'-0" 98.0 ft
1.2 185 k 1.6 150 k4.71 ksf
98.0 ft
e
u
q
A
A
q
= − − =
+= =
=
+= =
Example 12.5
93
Depth required to resist one-way shear. Take b = 7 ft.
( )( )( )
( )( )( )
1 7 ft 4.625 ft 4.71 ksf 152.49 k
152,490 lb19.14 in
0.75 1 2 4000 psi 84 in
19.14 4.5 in 23.64 in
uV
d
h
= =
= =
= + =
Example 12.5
95
Depth required to resist punching shear: ( )
( )( )( )
( )( )( )
( )
22
2
4 18 19.5 in 150 in
98.0 ft 3.125 ft 4.71 ksf 415.58 k
415,580 lb14.60 in 19.5 in Ok
0.75 1 4 4000 psi 150 in
415,580 lb
40 19.5 in0.75 2 4000 psi 150 in
150 in
8.11 in 19.5 in Ok
o
u
b
V
d
d
= + =
= − =
= = <
=× +
= <
Example 12.5
97
( ) ( ) ( )
( ) ( )( ) ( )22
14 ft 9 in6.25 ft
2 12 in/ft
6.25 ft6.25 ft 7 ft 4.71 ksf 643.9 k-ft
2
12 in/ft 643,900 lb-ft268.8 psi
0.9 84 in 19.5 in
u
u
M
M
bdφ
− =
= =
= =
Flexural design (steel in long direction)
Example 12.5
98
Appendix Table 4.13, ρ = 0.00467
( ) ( ) ( ) 20.00467 84 in 19.5 in 7.65 insA = =
Use ten No 8 (As = 7.85 in2)
Example 12.5
99
( ) ( ) ( )
( ) ( )( ) ( )22
7 ft 9 in2.75 ft
2 12 in/ft
2.75 ft2.75 ft 14 ft 4.71 ksf 249.3 k-ft
2
12 in/ft 249,300 lb-ft52.0 psi
0.9 168 in 19.5 in
u
u
M
M
bdφ
− =
= =
= =
Flexural design (steel in short direction)
Too low for Table A.13
Example 12.5
100
( ) ( ) ( ) 2
2000.0033
60,000 psi
3 4000 psi0.00316
60,000 psi
0.0033 168 in 19.5 in 10.81 insA
ρ
ρ
= = ←
= =
= =
Use 18 No 7 (As = 10.82 in2)
Example 12.5
101
Reinforcement in band width 2 2 2
Reinforcement in short direction 1 2 1 3β= = =
+ +
Use 2/3 x 18 = 12 bars in band width
Recommended