strut and time method
TRANSCRIPT
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Development of the Strut-and-Tie Method for A endix A of
James K. Wight
the Building Code (ACI 318-08)
University of Michigan
Strut and Tie Modeling
be designed by idealizing the concrete
and reinforcement as an assembly of
axially loaded members, inter-
connected at nodes, to form a truss
capable of carrying loads across aregion or member.
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Components of Strut and Tie Models
Steps to Build Strut & Tie Model
-
Compute forces or distribution of
stresses on boundary
Represent stress distributions as forces
forces across the member or D-region
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Concept of D-Regions
(force discontinuities)
h
h 2h
Concept of D-Regions(geometric discontinuities)
h2h1
1 2
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Basic Requirements
Model approximates stress flow
Define component dimensions and
strengths
Define and factors
Select reinforcement details
Modeling Stress Flow
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Modeling Stress Flow in D-Regions(Dapped Beam)
How to Select the CorrectStrut-and-Tie Model
Some researchers suggest using a finite
element model to determine stress
trajectories, then selecting a STM to
model the stress flow.
,required amount or reinforcement is
close to an ideal model.
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Required Definitions for Code
Geometric rules to follow when creating
a strut-and-tie model.
Component strengths for determining
members sizes and final geometry of
.
Other Codes with Rules for Useof Strut-and-Tie Models
AASHTO LRFD Specification
Canadian Code for Design of Concrete
Structures (CSA Standard, 2004)
ecommen a ons orPractical Design of Structural Concrete
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Strength of Compression Struts
(what to consider)
Longitudinal cracking due to transverse
tension strain
Transverse tension forces
Sustained loads
Reinforcement grid crossing strut
Confinement by concrete or steel
Lateral Expansion of Strut
(bottle-shaped strut)
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Effective Compressive Strength of Struts
ns uF F
0.75 =
ns cu cF f A=
min. cross-sectional area of strutcA =
s .cu c=
AASHTO and CSA Evaluation ofEffective Concrete Strength
Trans. tension strain,1= (s + 0.002) cot2s
1
0.850.8 170
cs c c
ff f
=
+
s = angle between strut and tie (steel)
s = strain in steel tie
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FIP Recommendations fors in
Cracked Struts
0.80 - struts with longitudinal cracking
(splitting), but crossed by minimum
reinforcement grid
0.75 - struts crossed by normal width
0.60 - struts crossed by wide cracks
ACI Recommended s Values for Struts
1.0 prismatic shape (constant width) over its,
in a B-region
0.75 inclined (bottle-shaped) strut crossedby minimum reinforcement grid
0.60 inclined bottle-shaped strut notcrossed by minimum reinforcement grid;
where accounts for lightweight concrete 0.40 struts in flexural tension zones
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Struts in a Flexural Tension Zone
A
A
Sect. A-A
Minimum Reinforcement Grid (fc
6000 psi)
Strut centerline2
12
1
sii
i
A in 0.003b s
s (A 4)
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Reinforcement Grid
(only horizontal bars)
Min. Reinf. Grid in other Codes
AASHTO 0.003 EW
CSA 0.002 EW
FIP 0.001 EF, EW
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Modification for Higher
Strength Concrete
(41 MPa), must calculate required
amount of transverse reinforcement
crossing strut.
This procedure can result in a
significant difference when compared to
requirements of ACI Eq. (A-4).
Modification for HigherStrength Concrete
=
2 longitudinal/1 transverse
equ re ransversetie capacity = 50%
of strut strength
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Tie Dimensions
Full width (out or plane) of member
Width (in plane) of tie is function ofeffective compression strength ofconcrete in nodes where tie is anchored
dimensions
Tie - Dimensions and Strength
Asfy
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Strength of Ties
Strength = As fy, where = 0.75*
Anchorage of ties at nodes is a major
concern
*A constant value of = 0.75 is to be used for
sizing the strut-and-tie model, but the use of
= 0.75 to also select the reinforcement mayneed further examination within the appropriate
Code subcommittee.
Anchorage Check
Definition of Extended Nodal Zone
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Nodal Zone Shape and Dimensions
Width of compression face is same as
width of strut connecting to nodal zone
(smaller allowable strength governs)
Height (width of face perpendicular to
tie force) of nodal zone is equal to tie
force divided by effective compressivestrength of the concrete in the node
Node Shape and Size
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Effective Strength of Nodal Zones
Function of type of members connected
to the node
Possible combinations are CCC, CCT
and CTT
Strength can be enhanced by addition
of confinement reinforcement
Examples of CCC and CCT Nodes
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Examples of CCT and CTT Nodes
Examples of CCT and CTT Nodes
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Strength of Nodes
nn uF F
0 75 0.75 =
nn cu nF f A=
n uA area of node face to force F=
n0.85
cu cf f =
Recommended n Values (Nodes)
1.0 - CCC node
. -
0.6 - CTT node
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Use of STM in the ACI Code
- .However,
Is listed as alternate procedure inseveral sections of the code (e.g.corbels, short shear walls).
Currently required for shear strength
design of deep beams.
640 k (includes member weight)
All member widths = 20 in.
Example: ACI Concrete International
Magazine, May 2003
dv60 in.
.
53 in. 107 in. 212 k428 k
16 in. 16 in.
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640 k
All member widths = 20 in.
Beam Dimensions and Initial Truss Model
dv60 in.
.
53 in. 107 in. 212 k428 k
16 in. 16 in.
Check Max. Allowable Shear Force
=.
Max. All. Shear Force = 10
4000(max) 0.75 10 20 54 512 k (o.k.)
1000
c w
n
f b d
V
= =
Assumed thatd0.9h
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212 k428 k
Splitting of Node 2
dv
=
?
1
2
3
4 51 = 42.2
o2 = ? 2
49.7 in. 50.2 in. 50.2 in.
10 in.428 k 212 k
Establish Truss Geometry: Start withleft portion of the beam
Assumes heights of Nodes 1 & 2 = 15 in.
Thus, dv= 60 in. 2(7.5 in.)
= 45 in. (1140 mm)
From eometr :
tan 1 = (45in./49.7in.),
And thus 1 = 42.2 deg.
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212 k428 k
Initial Truss Geometry
dv
=
45
in.
1
2
3
4 51 = 42.2
o2 = ? 2
49.7 in. 50.2 in. 50.2 in.
10 in.428 k 212 k
Establish Equilibrium at Node 1
12
F14Fy = 428 - F12sin 1 = 0
F12 = 637 k
428 kFx = F14 F12cos 1 = 0
F14 = 472 k
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Geometry and dimensions of Node 1
and Strut 1-2
14
b1
1
Establish/Check Dimensions at Node 1
fcu(1) = (0.85) n fc'= 0.750.850.804
1
428( ) 1.34 ksi (o.k.)
20 16w b
Rf base
b= = =
l
l
= 2.04 ksi
1414
472 11.6 12 in.(1) 20 0.75 2.72w cu
Fwb f
= = =
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Establish/Check Width and Strength of
Strut 1-2
( ) ( )12 14 1 1 1
12
cos sin
8.89 10.8 19.7 in.
bw w
w
= +
= + =
l
(1-2) 0.85 0.85 0.75 4 2.55 ksicu s cf f = = =
12(1-2) (1-2)(1-2) 0.75 2.55 19.6 20 757 k 637 k (o.k.)
ns cu w
ns
F f w bF
= = =
Modified Truss Geometryand Member Forces
Thus, height of Node 1 = 12 in.
Because Node 2 is a CCC node, assume
it has a total height of 10 in.
, v= = .
Reestablish truss geometry and member
forces!
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212 k428 k
Final truss geometry and member forces
dv
=
49
in.
434 k 217 k
217 k
1
2
3
4 51 = 44.6
o 2 = 44.3o
2
212
k
49.7 in. 50.2 in. 50.2 in.
10 in.428 k 212 k
Select Reinforcement for Tie 1-4
214434
( .) 9.64 in.sF
A req d = = =.
y
Select 13 No. 8 bars,
As = 10.3 in.2
3 in.
20 in.
3 in.
3 in.
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Check Anchorage at Node 1
Tie 1-4
a = 6 in./tan1a = 6.09 in.
b1 = 16in. a
Critical section
1 6in.
Length 22.1 in.
Check Anchorage at Node 1
0.02 0.02 1 60, 0001.0 in.
e yf
d
= = l
b
1 4000
19.0 in. (> 8d and > 6 in.)
c
dh
f
=l
Although dh is less than 22 in., this would be a tight fit if
only 90 hooks were used. In-plane 180 hooks could be
used for some bars to partially relieve this rebar detailingproblem. The use of mechanical anchorage devices could
also be considered.
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Minimum Reinforcement Crossing Strut 1-2
Strut centerline2
12
1
sii
i
A in 0.003b s
s
Minimum Reinforcement CrossingStrut 1-2 for s = 0.75
and width of cracks crossing the strut.
Because this is a deep beam, I recommend
that you also satisfy ACI Code Sections
11.7.4 and 11.7.5.
, ,the minimum skin reinforcement requirements
of ACI Code Section 10.6.7 must be satisfied.
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2 #4
per layer
3 in.Final design
in left span
Four #4 legs6 at 8 in.
3 at 3 in.
20 in.
13 #8
212 k428 k
Comments on right half of truss
dv
=
49
in.
434 k 217 k
217 k
1
2
3
4 51 = 44.6
o 2 = 44.3o
2
212
k
49.7 in. 50.2 in. 50.2 in.
10 in.428 k212 k
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22in.60in.
Analysis of fan-shaped struts 2-4
and 3-5, and tie 3-4
2
49in.
25o
25o
3
5
104 in.
22in. six stirrups at s = 10 in.
4
Section A
Final design of longitudinal andtransverse steel
5 at 10 in. 15 at 6 in.
2 in. 2 in.
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Gracias
Preguntas?