5. Example 4 - Drilled-Shaft Footing

TxDOT Project 5-5253-01

Strut-and-Tie Model Design Examples

for Bridges

Example 4 – Drilled-Shaft Footing

C.S. Williams D.J. Deschenes O. Bayrak

2

Example 4 Overview

A 3-dimensional STM is required to properly

design the drilled-shaft footing

Two load cases are considered

Load Case 1 – All 4 drilled shafts in compression

Load Case 2 – 2 drilled shafts in compression, 2 in

tension

6.50’DDS = 4.00’OH = 0.75’OH = 0.75’

h=

5.0

0’

Wcol = 7.50’4.25’ 4.25’

x

z

L Drilled ShaftC

DDS = 4.00’

L1 = 16.00’

sDS = 10.50’

3

Drilled-Shaft Footing Geometry

x

y

Wcol = 7.50’

Dc

ol=

6.2

5’

L1 = 16.00’

L2

= 1

6.0

0’

OH = 0.75’ OH = 0.75’DDS = 4.00’ DDS = 4.00’

DD

S=

4.0

0’

DD

S=

4.0

0’

6.5

0’

OH = 0.75’

OH = 0.75’

6.50’

8.00’ 8.00’

8.0

0’

8.0

0’

L Column &C

L FootingC

L ColumnC

L FootingC

4

Drilled shafts are assumed to behave as

pinned supports

Moment and axial force is transferred

between the column and the footing

Drilled-Shaft Footing Geometry

5

Material Properties

Concrete: f ’c = 3.6 ksi

Reinforcement: fy = 60 ksi

6

Design Procedure for Bent Cap

Separate B- and D-

Regions

Define Load Case

Develop Strut-and-Tie

Model

Proportion Shrinkage

and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

7


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

8

The entire drilled-shaft footing is a D-region

Separate B- and D-Regions

Wco

l=

7.5

0’

h=

5.0

0’

Wcol = 7.50’

x

z

D-Region

B-Region

9

Define Load Case 1

y x

z

Pu = 2849 k

Muyy = 9507 k-ft

Factored Axial Load and Moment

10

Analyze Structural Component

Pu = 2849 k

Muyy = 9507 k-ft

R1

R2 R3

R4

y x

z

Determine Reactions at Drilled Shafts

11

Pu = 2849 k

Muyy = 9507 k-ft

R1

R2 R3

R4

y x

z

sDS = 10.50’

L Drilled ShaftC



All drilled shafts are in

compression

12


Determine How Loads are Applied to the STM

Since there is no shear force, the moment and axial

load are the same at the column-footing interface as

they are at the interface of the B- and D-regions

Moment and axial load must be converted into point

loads that can be applied to the STM

4 point loads will be applied to the STM since forces are

flowing to 4 drilled shafts

Start with determining the

linear stress distribution at

the interface between

the B- and D-regions

13



2.35’

7.50’

5.15’

1549 psi

x

z

C

T

2849 k

9507 k-ft

x

y

2.25” Clear

No. 5 Stirrups

Wcol = 7.50’

12 – No. 11 Bars

Dco

l=

6.2

5’

10 –

No

. 11

Ba

rs

11 Equal Spaces

11 E

qu

al S

paces

Dc

ol=

6.2

5’

1.5

6’

1.5

6’

3.1

3’

1.72’ 3.44’ 2.35’

5.15’ 2.35’

1.5

8’

Wcol = 7.50’

3.44’x

y

1549 psi

Neu

tral A

xis

0.30’

A B

CD

C

T


14


Locations of 4 point loads

on the column section

Assumed column

reinforcement layout

Dco

l=

6.2

5’

1.5

6’

1.5

6’

3.1

3’

1.72’ 3.44’ 2.35’

5.15’ 2.35’

Wcol = 7.50’

3.44’x

y

1549 psi

Ne

utra

l Ax

is

A B

CD

C

T


15


The loading will cause 2 of

the loads to act downward

(compressive) and 2 to act

upward (tensile)

The downward

(compressive) loads act at

points A and D

Along the centroid of the

linear stress diagram

Quarter points of the dimension Dcol

The upward (tensile) loads act at points B and C

Location is based on the centroid of the column’s tension

face reinforcement

Each load acts at the centroid of 6 of the #11 bars


16


1.5

8’

Wcol = 7.50’

x

y

Centroid of 6 – No. 11 Bars


Column Bars

Considered to Carry

Forces in Ties BI and

CJ of STM

Ne

utra

l Ax

is

0.30’

A B

CD

Dco

l=

6.2

5’

1.5

6’

1.5

6’

3.1

3’

1.72’ 3.44’ 2.35’

1.5

8’

Wcol = 7.50’

3.44’x

y



Column Bars

Considered to Carry


CJ of STM

Ne

utra

l Ax

is

0.30’

A B

CD


17


The 4 point loads must be equivalent to the

factored axial load and moment acting on

the footing

18



Sum of loads

acting at A and D Sum of loads

acting at B and C

1.72’

Wcol = 7.50’

Ne

utra

l Ax

is

0.30’

A B

CD

19


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

Final STM the following slides will explain the

details of its development

20

R2 = 259.5 k

FA = 1763.6 k

R3 = 259.5 k

FD = 1763.6 k

FB = 339.1 kFC = 339.1 k

R1 = 1165.0 k

R4 =

1165.0 k

339.1

k

33

9.1

k

A

BC

D

E

F

G

H

IJ

y

x

z

Develop Strut-and-Tie Model

Then, determine the distance

from this strut to the top face

of the footing

First, determine the distance from

these horizontal ties to the bottom

face of the footing

21


Determine Distance from Horizontal Ties to Bottom of Footing

The ties coincide with the centroid of the horizontal

reinforcement running in the x- and y-directions

Bars oriented in the x-direction will be placed directly on top

of the bars oriented in the y-direction

The distance from the tension ties to the bottom of

the footing is 5.4”

x

z

4.0” Clear Cover

5.4”

No. 11 Bars

No. 11 Bar

22


Determine Distance from Horizontal Strut to Top of Footing

Several options were considered

1. Position strut at the top surface of the footing (Adebar,

2004; Adebar and Zhou, 1996)

Does not allow Nodes A and D to be triaxially confined

within the footing (effective triaxial confinement should be

guaranteed in order to use the strength check that will be

introduced later)

Results in a large height of the STM smaller forces in the

horizontal ties at the bottom of the footing

2. Assume the total depth of the strut is h/4 centroid of

the strut positioned h/8 from the top surface of the

footing (Park et al., 2008; Windisch et al., 2010; Paulay

and Priestley, 1992)

Based on the depth of the flexural compression zone of an

elastic column at a beam-column joint of a moment frame

Rationale is questionable

23



3. Position strut based on the depth of the compression

stress block determined from a flexural analysis of the

footing

Not accurate to treat the D-region as a beam

4. Assume the position of the strut coincides with the

location of the top mat of reinforcement

Justified if horizontal ties exist within the STM near the top

surface of the footing (methodology will be used for Load

Case 2)

Not justified otherwise

24



25



(Not Drawn to Scale)

Numbers in parenthesis

correspond to the numbering

of the options presented on

the previous slides y

z

h=

60

.0”

4.9”

Top Mat of

Steel (4) DA

Top of

Footing (1)

7.5”

h/8 (2)

26



Solution:

Chose a reasonably conservative value

(considering height of STM)

Location should be deep enough into the footing

that triaxial confinement is guaranteed

Strut is located 0.1h = 6.0 in. below the top surface of the footing

h=

60

.0”

y

z

DA

Top of

Footing (1)

4.9”

Top Mat of

Steel (4)

7.5”

h/8 (2)

6.0”

0.1h

27



Chosen location is not significantly different from the position of the top mat of steel offers consistency with the STM for Load Case 2

Total Height of STM = 60.0 in. – 5.4 in. – 6.0 in. = 48.6 in.

(Not Drawn to Scale)

Place struts and ties to model flow of forces

28


Add horizontal ties since drilled shafts

will tend to “push away” from one another possible diagonal orientation

is not feasible for construction y

x

z

Vertical ties are needed to carry the

upward (tensile) loads


29


Loads will tend to flow from the downward

(compressive) forces to the nearest drilled shafts add diagonal struts

y

x

z


30


Diagonal struts are needed to equilibrate

the forces in the vertical ties y

x

z


31


Part of compressive the force will also flow

to the other drilled shafts

y

x

z

Remainder of the struts are placed to ensure

equilibrium at each node

32


Add a horizontal strut to equilibrate forces

in the y-direction y

x

z

Place struts in the plane of the horizontal

ties to achieve equilibrium

Performing a linear-elastic analysis of the chosen

STM results in the member forces shown

33

R2 = 259.5 k

FA = 1763.6 k

R3 = 259.5 k

FD = 1763.6 k

FB = 339.1 kFC = 339.1 k

R1 = 1165.0 k

R4 =

1165.0 k

339.1

k

33

9.1

k

A

BC

D

E

F

G

H

IJ

y

x

z


Alternative valid STM

A direct load transfer from Node A to Node F is more likely represented by Strut AF of the

chosen STM (previous slide)

34


A

BC

D

E

F

GIJ

y

x

z

H

35


Performing a linear-elastic analysis of the STM

should result in the reactions at the drilled

shafts that were previously calculated

Recommendation:

Develop the STM within a structural analysis

software program

The STM can easily be modified and checked

Ensure the 25° rule is satisfied

36


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

Ties EF and GH have the same force

The reinforcement required for Tie FG will be used for

Tie EH to maintain symmetry

R2

FA

R3

FD

FB

FC

R1

R4

A

BC

D

E

F

G

H

IJ

y

x

z

Proportion Ties

37

Reinforcement for Horizontal Ties

Use #11 bars

38


Proportion Ties

Use #11 bars

39


Proportion Ties

Proportion Ties

TxDOT practice allows the reinforcement of the

horizontal ties to be placed within a 45° distribution

angle from the drilled shafts (TxDOT Bridge Design

Manual – LRFD, 2009)

For simplicity, the tie reinforcement will be placed

within the 4-ft diameter of the shafts

40


x

z

4.00’

45°45°

Length over which bars

could be spaced = 4.78’

339.1

k

33

9.1

k

A

BC

D

E

F

G

H

IJ

y

x

z

R2

FA

R3

FD

FB

FC

R1

R4

Proportion Ties

Ties BI and CJ have the same force

Column reinforcement extended into the footing

carries the forces in these ties

41

Reinforcement for Vertical Ties

Use #11 column bars

42


Proportion Ties

Recall that the upward loads on the STM were each

located at the centroid of 6 bars of the column’s

tension face reinforcement

43


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

44

Perform Strength Checks

Nodes within the 3-dimensional STM have very

complex and largely unknown geometries

The value of attempting to define the nodal geometries is

limited

A conservative alternative procedure is needed

A simplified nodal strength check procedure was

developed after conducting a literature review

See TxDOT Project 5-5253-01 implementation report for

details

45


Proposed Procedure

Limit compressive bearing stress to νf’c, where ν is the concrete efficiency factor of

the STM provisions

Neglect triaxial confinement factor, m

Ensure all nodes are triaxially confined within the

footing

Prevents need to define nodal geometries

R2 = 259.5 k

FA = 1763.6 k

R3 = 259.5 k

FD = 1763.6 k

FB = 339.1 kFC = 339.1 k

R1 = 1165.0 k

R4 =

1165.0 k

A

BC

D

E

F

G

H

IJ

y

x

z


46

Check Critical Bearings

Perform bearing checks at Nodes A and D

and Nodes E and H

CTT CTT

CCC CCC

47

Perform Nodal Strength Checks Check Bearing at Nodes E and H

Dc

ol=

6.2

5’

3.1

3’

3.1

3’

3.44’x

y

A B

CD

48

Perform Nodal Strength Checks Check Bearing at Nodes A and D

The bearing areas of Nodes A and D are the

shaded regions shown

49

Perform Nodal Strength Checks Check Bearing at Nodes A and D

Strength check procedure is satisfied and all

nodal strengths are adequate

50


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

51

Crack control reinforcement is not required

for footings

Satisfy shrinkage and temperature

reinforcement requirement per AASHTO LRFD

Article 5.10.8

Use #11 bars on the bottom face

Use #7 bars on all other faces

Proportion Shrinkage and

Temperature Reinforcement

52



where As = area of reinforcement in each direction and each face

(in.2/ft) b = least width of component section (in.)

h = least thickness of component section (in.)

fy = specified yield strength of reinforcing bars < 75 ksi

The spacing limit of 12 in. controls

53


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

Horizontal ties must be developed at the nodes

directly above the drilled shafts (Nodes E, F, G, and H)

Determine equivalent square area for the cross-

section of the drilled shafts

54

Provide Necessary Anchorage for Ties Horizontal Ties

A

BC

D

E

F

G

H

I J

55

Since nodal geometries

were not determined, the

critical development

section is conservatively

taken at the inner edge of

this equivalent square

area

Providing a clear cover of

3 in., the use of 90-degree

hooks is adequate for

proper anchorage


x

z

4.00’

3.54’ (42.5”)

A A

3” Clear

Cover

Available Length

= 51.3” > 19.8”

Section A-A

Critical

Section

A

BC

D

E

F

G

H

I J

Ties BI and CJ should be anchored at Nodes I and J

56

Provide Necessary Anchorage for Ties

Vertical Ties

Use 90-degree hooks to anchor the ties

Nodes I and J are smeared nodes

A

BC

D

E

F

G

H

I J

57


Vertical Ties

Available length for development cannot be defined

x

z

3” Min.

Clear Cover

ldh = 19.8”

No. 11 Column Bar

Node I Available

Length = ?

60.0”

58


Vertical Ties

Geometry Cannot

be Defined

Standard TxDOT design practice specifies hooked

anchorage for the column bars extending into the

footing

Years of successful practice

90-degree hooks are specified in the current design

based on the success of this standard practice

59


Vertical Ties

Load Case 2

The same design procedure is now

followed for Load Case 2

61


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

Previously Performed

62


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

63

Define Load Case 2

y x

z

Pu = 1110 k

Muyy = 7942 k-ft

Factored Axial Load and Moment

64

Pu = 1110 k

Muyy = 7942 k-ft

R1

R2 R3

R4

y x

z

sDS = 10.50’

L Drilled ShaftC



2 drilled shafts are in

compression and 2

are in tension


65

Determine How Loads are Applied to STM

Use exactly same

procedure as with

Load Case 1

4.41’ 3.09’

1.5

8’

1.47’ 2.94’ 3.09’

Dco

l=

6.2

5’

1.5

6’

1.5

6’

3.1

3’

Wcol = 7.50’

2.94’x

y

1106 psi

0.30’

Ne

utra

l Ax

is



Column Bars

Considered to Carry


CJ of STMA B

CD

C

T

1.47’

Wcol = 7.50’

0.30’

Ne

utra

l Ax

is

A B

CD

66


Determine How Loads are Applied to STM

Sum of loads

acting at A and D Sum of loads

acting at B and C

Final STM the following slides will explain the

details of its development

67


R2 = 100.7 k

FA = 1026.8 k

R3 = 100.7 k

FB = 471.8 kFC = 471.8 k

R1 = 655.7 k

47

1.8

k

47

1.8

k

A

BC

D

E

F

GIJ

H

K

L

M

N

10

0.7

k 10

0.7

k

FD = 1026.8 k

R4 = 655.7 k

y

x

z

Horizontal ties are needed near

the top surface of the footing

Determine location of horizontal

ties at the bottom of the STM

68


Determine Distance from Lower Ties to Bottom of Footing

Again taken as 5.4” to coincide with the centroid of

the horizontal reinforcement

Determine Distance from Upper Ties to Top of Footing

x

z 4.9”

No. 7 Bars

No. 7 Bar

4.0” Clear Cover

The ties coincide with the centroid of the #7 bars of the top mat of steel 4.9” from the top surface of

the footing


69


Place horizontal ties as was done for

Load Case 1 y

x

z

Vertical ties are needed to carry the

tensile loads and drilled-shaft reactions


70


Loads will tend to flow from the downward

(compressive) forces to the nearest drilled shafts add diagonal struts

y

x

z

Envision each set of vertical ties as a non-contact lap splice connect with a diagonal strut


71


Add a tension ring near the top of the

footing to equilibrate the diagonal struts

y

x

z


72


Diagonal struts are needed to equilibrate

the forces in the vertical ties y

x

z

Remainder of the struts are placed to ensure

equilibrium at each node

73


Place struts in the plane of the upper

horizontal ties to achieve equilibrium y

x

z

Place struts in the plane of the lower

horizontal ties to achieve equilibrium

Performing a linear-elastic analysis of the chosen

STM results in the member forces shown

74


R2 = 100.7 k

FA = 1026.8 k

R3 = 100.7 k

FB = 471.8 kFC = 471.8 k

R1 = 655.7 k

47

1.8

k

47

1.8

k

A

BC

D

E

F

GIJ

H

K

L

M

N

10

0.7

k 10

0.7

k

FD = 1026.8 k

R4 = 655.7 k

y

x

z

75


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

A

BC

D

E

F

GIJ

H

K

L

M

N

y

x

z

R2

FA

R3

FD

FB

FC

R1

R4

76

Proportion Ties

Comparing the STMs for the 2 load

cases, the forces in the bottom

horizontal ties of Load Case 1 govern


Determine reinforcement needed for

horizontal ties along the top of the footing Tie LM has the largest force

Use #7 bars

77


Proportion Ties

Can the bars provided to satisfy the shrinkage and

temperature reinforcement requirement carry the tie

force?

x

y

Bars Considered to

Carry Force in Tie

s ≈ 11”

Shrinkage and temperature

reinforcement is spaced at about

11” (satisfies 12” limit)

Reinforcement considered to

carry the tie force are positioned

directly above the drilled shafts

3 bars are needed 4 bars are

provided

Since the force in Tie LM was the

largest, enough bars are provided

for all the ties along the top of the

STM

78

Proportion Ties Reinforcement for Horizontal Ties

47

1.8

k

47

1.8

k

A

BC

D

E

F

GIJ

H

K

L

M

N

y

x

z

R2

FA

R3

FD

FB

FC

R1

R4

79

Proportion Ties

Forces in Ties BI and CJ govern for

the current load case


Use #11 column bars

80


Proportion Ties

Longitudinal column reinforcement extended into

the footing can carry the tie forces

A

BC

D

E

F

GIJ

H

K

L

M

N

y

x

z

R2

FA

R3

FD

FB

FC

R1

R4

10

0.7

k 10

0.7

k

81

Proportion Ties

Reinforcement must be provided to

carry forces in Ties FL and GM


Use the #9 bars of the drilled shafts

82


Proportion Ties

20 – No. 9 Bars

All longitudinal bars within the drilled shafts will be

extended into the footing

Only those properly anchored at Nodes L and M can be

considered to carry the forces in Ties FL and GM

83


Proportion Ties

A

BC

D

E

F

GIJ

H

K

L

M

N

y

x

z

R2

FA

R3

FD

FB

FC

R1

R4

100.7

k 10

0.7

k

To maintain symmetry, the 4 bars indicated by

circles will be considered to carry the forces in the ties must be properly anchored

84


Proportion Ties

Figure: Assumed Drilled-Shaft Reinforcement Layout

4.0

0’

20 – No. 9 Bars

No. 3 Spiral

R2 = 100.7 k

FA = 1026.8 k

R3 = 100.7 k

FB = 471.8 kFC = 471.8 k

R1 = 655.7 k

A

BC

D

E

F

GIJ

H

K

L

M

NFD = 1026.8 k

R4 = 655.7 k

y

x

z

85

Perform Strength Checks The proposed strength check procedure required the

compressive bearing forces to be checked

The bearing forces of Load Case 1 control

86

Satisfy shrinkage and temperature

reinforcement requirement per AASHTO LRFD

Article 5.10.8

Necessary shrinkage and temperature

reinforcement was already determined when

considering Load Case 1



87


Separate B- and D-

Regions

Define Load Case


Model


and Temperature

Reinforcement

Analyze Structural

Component

Provide Necessary

Anchorage for Ties

Perform Strength

Checks

Proportion Ties

A

BC

D

E

F

GIJ

H

K

L

M

N

Proper anchorage was determine for all the ties

when Load Case 1 was considered with the

exception of the horizontal ties along the top of the

STM and Ties FL and GM

88


A

BC

D

E

F

GIJ

H

K

L

M

N

Ties KL, LM, MN, and KN must be properly anchored

at Nodes K, L, M, and N

Nodes are smeared

Diagonal struts will create large extended nodal zones

89


90

The critical development

section is conservatively

taken at the inner edge of

the equivalent square area

of the drilled shafts

Providing a clear cover of 3

in., the use of straight bars is

adequate to properly

anchor the #7 bars


x

z

4.00’

3.54’ (42.5”)

A A

3” Clear

CoverAvailable Length =

51.3” > 26.6”

Section A-A

Critical

Section

A

BC

D

E

F

GIJ

H

K

L

M

N

Ties FL and GM should be anchored at Nodes L and M

Nodes L and M are smeared nodes geometry cannot be

defined

91


Vertical Ties

Available length for development cannot be

determined

Similar reasoning as used for Ties BI and CJ is applied

180-degree hooks will be used to anchor the 4 bars

extending into the footing from the drilled shafts

4.0

0’

92


Vertical Ties

180-degree hooks

ldh = 15.8 in.

Only considering the two load cases presented

x

y

16.00’16.0

0’

90-Degree

Hooks

180-Degree

Hooks

93

Reinforcement Layout

Anchorage of Vertical Ties

x

z

No. 11 Bars

No. 9 Bars

(Only Hooked Bars

are Shown)

No. 11

Bar

4.0” Clear

A

A

5.0

0’

0.33’ 0.33’0.33’ 0.33’1.67’ 1.67’ 1.67’ 1.67’

0.75’ 0.75’

16.00’

13 Eq. Spa. = 4.00’(No. 11 Bars)

13 Eq. Spa. = 4.00’(No. 11 Bars)

7 Eq. Spa. = 6.50’(No. 11 Bars)

7.50’

94


Elevation View (Main Reinforcement)

x

z

0.50’ 0.50’15 Eq. Spa. = 15.00’

(No. 7 Bars)

4.0” Clear3.0” Clear

5E

q. S

pa. =

4.0

5’

(No

. 7

Bars

)

Location of No. 11

Bar of Bottom Mat

No. 7 Bars

No. 7

Bars

A

A

No. 7 Bar

95


Elevation View (Shrinkage and Temperature Reinforcement)

96


Section A-A (Main Reinforcement)

Section A-A (Shrinkage and Temperature Reinforcement)

4.0” Clear

0.75’ 0.75’10 Eq. Spa. = 4.00’

(No. 11 Bars)

10 Eq. Spa. = 4.00’

(No. 11 Bars)

7 Eq. Spa. = 6.50’

(No. 11 Bars)

5.0

0’

No. 11 Bar

3.0” Clear

No. 7 Bars

No. 7 Bar

No. 7 Bars 4.0” Clear

Location of No. 11

Bar of Bottom Mat

0.50’ 0.50’15 Eq. Spa. = 15.00’

(No. 7 Bars)

No. 7 Bars

y

z

y

z

x

y

3.0” End

Cover

16.00’

0.50’0.50’15 Eq. Spa. = 15.00’

(No. 7 Bars – Side Face Reinforcement)

0.75’ 0.75’7 Eq. Spa. = 6.50’

(No. 11 Bars)

13 ES = 4.00’

(No. 11 Bars)

13 ES = 4.00’

(No. 11 Bars)

16

.00

’

0.50’

0.50’

15

Eq

. S

pa. =

15

.00

’

(No

. 7

Bars

–S

ide

Fac

e R

ein

forc

em

en

t)0.75’

0.75’

10

ES

= 4

.00

’

(No

. 11

Ba

rs)

10

ES

= 4

.00

’

(No

. 11

Ba

rs)

7E

q.

Sp

a. =

6.5

0’

(No

. 11

Bars

)

97


Plan View (Bottom-Mat Reinforcement)

x

y16.00’

16

.00

’

17 Eq. Spa. = 15.26’ (No. 7 Bars)

17

Eq

. S

pa

. =

15

.26

’ (N

o. 7

Ba

rs)

4.0” Side

Cover

15

Eq

. S

pa

. =

15

.00

’

(No

. 7

Ba

rs –

Sid

e F

ac

e R

ein

forc

em

en

t)

15 Eq. Spa. = 15.00’

(No. 7 Bars – Side Face Reinforcement)

3.0” End

Cover

0.50’0.50’

0.50’

0.50’

98


Plan View (Top-Mat Reinforcement)

99

References

AASHTO LRFD Bridge Design Specifications, 5th ed., 2010. American Association of State Highway and Transportation

Officials, Washington, D.C., 2010.

Adebar, Perry. “Discussion of ‘An evaluation of pile cap design methods in accordance with the Canadian design

standard’.” Canadian Journal of Civil Engineering 31.6 (2004): 1123-126.

Adebar, Perry, and Luke (Zongyu) Zhou. “Design of Deep Pile Caps by Strut-and-Tie Models.” ACI Structural Journal

93.4 (1996): 437-48.

Park, JungWoong, Daniel Kuchma, and Rafael Souza. “Strength predictions of pile caps by a strut-and-tie model

approach.” Canadian Journal of Civil Engineering 35.12 (2008): 1399-413.

Paulay, T., and Priestley, M. J. N. Seismic Design of Reinforced Concrete and Masonry Buildings. New York: John

Wiley and Sons, 1992, 768 pp.

Texas Department of Transportation Bridge Design Manual - LRFD. Revised May 2009. Texas Department of

Transportation, 2009. <http://onlinemanuals.txdot.gov/txdotmanuals/lrf/lrf.pdf>.

Windisch, Andor, Rafael Souza, Daniel Kuchma, JungWoong Park, and Túlio Bittencourt. Discussion of “Adaptable

Strut-and-Tie Model for Design and Verification of Four-Pile Caps.” ACI Structural Journal 107.1 (2010): 119-20.

Documents

5. Example 4 - Drilled-Shaft Footing