102
Footings 1

Footings - libvolume8.xyzlibvolume8.xyz/textile/btech/semester7/advancedfabricstructureand...Footings 4 Typically, bottom of footing must be located below frost line Footings must

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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

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

Wall Footing

5

Wall footings – enlargement of the bottom of the wall

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

Pile Cap

9

Pile caps – distribute column loads to groups of piles

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

Assumed Soil Pressure

11

Soil Pressure - Sandy Soil

12

Soil Pressure - Clayey Soil

13

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

17

Design of Wall Footings

18

Design of Wall Footings

19

Design of Wall Footings

20

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

Wall Footing Design

Examples

23

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

25

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

Design of Isolated Square Footings

39

Constant thickness

Design of Isolated Square Footings

40

Stepped

Design of Isolated Square Footings

41

Tapered

One-way Shear

42

'2

u

c w

Vd

f bφλ=

Same as for wall footings

One-way Shear

43

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

45

Two-way Shear

46

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

Column Dowels

56

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

Isolated Rectangular Footings

65

Footing Design Examples

66

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

72

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

= ∴

=

= >

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

= = ∴

=

= <

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

94

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

96

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

Example 12.5

102