Post on 01-Jan-2016
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
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
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
Analyze Structural Component
Determine Reactions at Drilled Shafts
All drilled shafts are in
compression
12
Analyze Structural Component
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
Analyze Structural Component
Determine How Loads are Applied to the STM
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
Analyze Structural Component
14
Determine How Loads are Applied to the STM
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
Analyze Structural Component
15
Determine How Loads are Applied to the STM
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
Analyze Structural Component
16
Determine How Loads are Applied to the STM
1.5
8’
Wcol = 7.50’
x
y
Centroid of 6 – No. 11 Bars
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
Centroid of 6 – No. 11 Bars
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
Analyze Structural Component
17
Determine How Loads are Applied to the STM
The 4 point loads must be equivalent to the
factored axial load and moment acting on
the footing
18
Analyze Structural Component
Determine How Loads are Applied to the STM
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
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
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
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
Determine Distance from Horizontal Strut to Top of Footing
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
Develop Strut-and-Tie Model
Determine Distance from Horizontal Strut to Top of Footing
25
Develop Strut-and-Tie Model
Determine Distance from Horizontal Strut to Top of Footing
(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
Develop Strut-and-Tie Model
Determine Distance from Horizontal Strut to Top of Footing
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
Develop Strut-and-Tie Model
Determine Distance from Horizontal Strut to Top of Footing
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
Develop Strut-and-Tie Model
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
Place struts and ties to model flow of forces
29
Develop Strut-and-Tie Model
Loads will tend to flow from the downward
(compressive) forces to the nearest drilled shafts add diagonal struts
y
x
z
Place struts and ties to model flow of forces
30
Develop Strut-and-Tie Model
Diagonal struts are needed to equilibrate
the forces in the vertical ties y
x
z
Place struts and ties to model flow of forces
31
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
A
BC
D
E
F
GIJ
y
x
z
H
35
Develop Strut-and-Tie Model
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
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
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
Reinforcement for Horizontal Ties
Proportion Ties
Use #11 bars
39
Reinforcement for Horizontal Ties
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
Reinforcement for Horizontal Ties
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
Reinforcement for Vertical Ties
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
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
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
Perform Strength Checks
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
Perform Strength Checks
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
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
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
Proportion Shrinkage and
Temperature Reinforcement
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
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
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
Provide Necessary Anchorage for Ties Horizontal Ties
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
Provide Necessary Anchorage for Ties
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
Provide Necessary Anchorage for Ties
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
Provide Necessary Anchorage for Ties
Vertical Ties
Load Case 2
The same design procedure is now
followed for Load Case 2
61
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
Previously Performed
62
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
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
Analyze Structural Component
Determine Reactions at Drilled Shafts
2 drilled shafts are in
compression and 2
are in tension
Analyze Structural Component
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
Centroid of 6 – No. 11 Bars
Centroid of 6 – No. 11 Bars
Column Bars
Considered to Carry
Forces in Ties BI and
CJ of STMA B
CD
C
T
1.47’
Wcol = 7.50’
0.30’
Ne
utra
l Ax
is
A B
CD
66
Analyze Structural Component
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
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
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
Place struts and ties to model flow of forces
69
Develop Strut-and-Tie Model
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
Place struts and ties to model flow of forces
70
Develop Strut-and-Tie Model
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
Place struts and ties to model flow of forces
71
Develop Strut-and-Tie Model
Add a tension ring near the top of the
footing to equilibrate the diagonal struts
y
x
z
Place struts and ties to model flow of forces
72
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
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
Develop Strut-and-Tie Model
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
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
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
Reinforcement for Horizontal Ties
Determine reinforcement needed for
horizontal ties along the top of the footing Tie LM has the largest force
Use #7 bars
77
Reinforcement for Horizontal Ties
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
Reinforcement for Vertical Ties
Use #11 column bars
80
Reinforcement for Vertical Ties
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
Reinforcement for Vertical Ties
Use the #9 bars of the drilled shafts
82
Reinforcement for Vertical Ties
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
Reinforcement for Vertical Ties
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
Reinforcement for Vertical Ties
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
Proportion Shrinkage and
Temperature Reinforcement
87
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
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
Provide Necessary Anchorage for Ties
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
Provide Necessary Anchorage for Ties
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
Provide Necessary Anchorage for Ties Horizontal Ties
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
Provide Necessary Anchorage for Ties
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
Provide Necessary Anchorage for Ties
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
Reinforcement Layout
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
Reinforcement Layout
Elevation View (Shrinkage and Temperature Reinforcement)
96
Reinforcement Layout
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
Reinforcement Layout
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
Reinforcement Layout
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.