workshop slides everfe
TRANSCRIPT
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EverFE Workshop
Sacramento, CAMarch 11, 2004
Bill Davids, PhD, PEUniversity of Maine
JPCP Is a Complex Structure
A 7-12 layer of concrete on a base, sub-base, soil
Subjected to a variety of axle loads and fatigue effects
Experiences seasonal and daily temperature changes
Sawn transverse joints every 12 15 (+/-)
Transverse joints often doweled for better load transfer
Adjacent slabs may be tied at longitudinal joints
Can experience substantial early-age shrinkage
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JPCP Is a Complex Structure
I-90 in Washington State
Dowel Retrofit
Contraction Joint
Failure Modes in JPCP
Panel Cracking
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Failure Modes in JPCP
Corner Break
Failure Modes in JPCP
TransverseJoint Faulting
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Failure Modes in JPCP
Shrinkage Cracking
Mechanistic-Empirical Designof JPCP
1. Estimate design
parameters(thickness, joint
spacing, etc.)
2. Predict responseunder axle loads,
temperature
changes, etc.
3. Assess effect ofstresses on
fatigue life and
durability
NotOK
4. Plans andSpecs, Bid,
Construct
OKConstructionProblems?
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Neg. gradient
Pos. gradient
2. Linear thermal gradients through the slab thickness
Predicting Response of JPCPUsually a Westergaard-type analysis
~slab~
wheel
1. Three critical wheel load positions are assumed
Edge Interior Corner
3. Slabs are founded directly on a dense liquid
4. Assumes an infinitely large slab, no joint load transfer
Essential for understanding pavement behavior Critical for developing rational design methods Important in forensic analysis of pavement failures
2. Predictions of pavement structural response are:
1. Limitations of Westergaard-type analysis are severe
3. Clear need exists for better JPCP analysis tools
Predicting Response of JPCP
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What is EverFE? Software for the 3D Finite Element (FE) analysis of JPCP
Incorporates specialized strategies for modeling importantresponse characteristics
Utilizes problem-specific solvers for efficiency
Integrated modeling software and graphical user interface
Intuitive model construction and result visualization
Allows the generation of models with varying complexity
Anatomy of an EverFE ModelBasic model characteristics: Up to nine slab/shoulder units Up to three base/subgrade layers Dense liquid supports model Dowels, ties, aggregate interlock
Loading: Multiple axle types Thermal gradients
Extensive post-processing: Slab stresses and displacements Dowel results
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Workshop Objectives
1. Familiarize you with EverFEs capabilities
Overview basic finite-element concepts Cover details of EverFE unique capabilities
2. Give you hands-on experience with the software
Generate and run models Increasing level of model complexity
3. Explain what EverFE can and cant do
Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
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Finite-Element Concepts
Mathematical definition: functional method forsolving partial differential equations
Our definition: well-established numericaltechnique for determining stresses, strains and
displacements in engineering structures
Finite-Element Concepts
Why is FEA so popular?
Applies to wide classes of problems
Excellent for irregular geometries
Easily treats different boundary conditions
Easily generalized for computer implementation
Easily handles spatially varying material properties
Well-suited to nonlinear and dynamic problems
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Finite-Element Concepts
Optimization in Mechanical Design
Analysis of a Welded Connection Analysis and Design of a Floor Slab
Structural Analysis of a Frame
Finite-Element ConceptsFE Procedure in a Nutshell:
Divide a structure into discrete inter-connected finiteelements that meet at nodes
Make each finite element responsible for defining anapproximate solution over its domain
Take the original governing differential equation and
re-cast it using the properties of the finite elements(the mathematically difficult part)
Solve the resulting system of equations for unknowndisplacements, recover stresses, etc.
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Finite-Element ConceptsSimple problem from structures/strength of materials
x
f(x)
Elastic rod of length L, elastic modulus E, area A, fixed ends
Governing differential equation: )(2
2
xfdx
udEA =
Finite-Element ConceptsFinite-element discretization and solution
element nodes
stress
exact solution
FE solution
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Finite-Element ConceptsHow does each element represent the solution?
1D linear element
nodal displ.
interpolated displ.
1D quadratic element
constant stress linearly varying
stress
Finite-Element ConceptsBasic Element Types in Structures and Solid Mechanics
2D Elements
Beam/Truss Element
Plate/Shell Elements
t
3D Elements
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Finite-Element ConceptsHistory of FE Modeling of Concrete Pavements
Earliest models treated slabs as plates on elastic solids
ILLISLAB, JSLAB, etc. released in late 1970s, early 1980sModeling of multiple slabs with 2D plate elements
Methods for handling joint load transfer
Researchers began using existing general-purpose 3D codesDetailed models of doweled joints
Treatment of slab-base interaction
EverFE was first released in 1998Development started in 1995, has continued until present
Finite-Element Concepts
Important Issues to Bear in Mind:
FEA is an approximate method
Model must closely mimic physical reality
Accurate material properties
Appropriate boundary conditions
Reasonable representation of loads
The proper elements need to be used in discretization
Sufficient mesh refinement is essential
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Finite-Element ConceptsWhat is the peak tensile stress in a large slab with:
40 kN wheel load applied at the edge, r= 228 mmSlab properties: t= 254 mm, E= 27,600 MPa, v= 0.20Subgrade k= 0.027 MPa/mm
40 kN
~ Very large slab ~
Finite-Element Concepts
1948 Westergaard Solution: max = 1.43 MPa
Finite-Element Solution:
Build model with quadratic solid elements
Represent load with a 405mm x 405mmsquare contact area (equivalent area to circle)
Critical questions:
How large a slab to model? How many elements to use in the model?
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Finite-Element ConceptsFinite-Element Solution:
Start with a large slab (5000mm x 5000mm)
Study the effect of mesh refinement on solution
2 x 2 elements
5000mm
5000mm
24 x 24 elements
increase #of elements
Examine the effect of model size on solution
Finite-Element Concepts
0 10 20 300.6
0.8
1
1.2
1.4
1.6
Westergaard
Number of Elements Along Edge(Only even number of elements used)
Max
imum
Stress(MPa)
Effect of Mesh Refinement on Results
2 elements throughthickness
1 element through
thickness
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Discretization
12 u 12
24 u 24
Stress (MPa)
1.48
1.43
2 Elements through thickness
What if we change our discretization slightly?
Load is centered in element: Element captures linearvariation in stress
Element cant see peakstress!
13 u 13
25u
25
1.23
1.33
Finite-Element Concepts
Finite-Element ConceptsEffect of Model Size on Results
1000 3000 5000 7000 90000.5
0.7
0.9
1.1
1.3
1.5
Max
imum
Stress(MPa)
Slab Size (mm)
12x12x2 elements for all runs
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Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
Generating an EverFE Model
16-noded interface element
20-noded brick element
8-noded dense liquid element
xy
z
beam elementsfor dowels and
transverse ties
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Generating an EverFE ModelExample Analysis
Single slab, 5000mm long x 3600mm wide x 250mm thick Founded on 125mm thick bonded CTB with E= 7000 MPa Single 120 kN, dual wheel axle located at edge
Plan
Elevation
120 kNaxle
5000mm
3600mm
slab
bonded CTB
Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
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Slab-Base InteractionThe base layer is rarely bonded to the slab
Slip (relative horizontal movement) between slab and base Vertical separation of slab and base may occur
z=1.01mm, max = 2.53 MPa
Unbonded Base
Consider the model we just solved
Bonded Base (as solved)
z=0.91mm, max = 1.42 MPa
Slab-Base InteractionEverFEs treatment of slip and vertical separation
Slab-base interface may be fully bonded or tensionless
Slab and base layer are meshed separately
1mm or 0.1 inslab
base
corresponding pairs of nodespermanently tied if base is bonded (linear)
released under tension if base is unbonded (nonlinear)
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Slab-Base InteractionWhat are typical values for kand 0?
Data reported by Rasmussen and Rozycki (2001):
Base Type
Rough HMA
Smooth HMARough Asphalt Stabilized
Smooth Asphalt StabilizedCement Stabilized
Granular
kSB(MPa/mm)
0.270
0.0680.200
0.0654.100
0.027
0
(mm)
0.250
0.5100.510
0.6400.025
0.510
Slab-Base InteractionQuick parametric study
Re-run our single-slab model with an unbonded base Let 0 = 1mm, vary kSB use say 0, 0.5, 1, 2, 5, 10, 50 Study the effect of varying kSB on peak tensile stress
kSB0.0
0.5
1.02.0
5.010.0
50.0
2.532.25
2.142.01
1.831.71
1.51
Notes
Shear transfer has a large effect on stress Slab and base maintained full contact Model remained linear
Approaches bonded solution for large kSB
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Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
Analysis of Thermal Gradients
Corners of slab
curl upward
slab
thickness
- T
+ T
Nighttime curling: top of slab cools relative to thebottom after a warm day
Weight of slab
pulls downward
Tension on topof slab
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Analysis of Thermal Gradients
Center of slab
lifts upward
slabthickness
T
- T
Daytime curling: top of slab heats relative to thebottom during a warm day
Weight of slabpulls downward
Tension on bottomof slab
Analytical solutions for stresses exist for simple cases
However, thermal gradients are often nonlinear
Slab-base interaction plays a significant role in response
Loss of contact between slab and base layer Shear stresses develop at slab-base interface
Analysis of Thermal Gradients
slabthickness
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EverFE idealizes gradients as linear, bilinear or tri-linear Equal vertical spacing assumed between each T
2Elements
3Elements
1Element
Temperature VariationsUsed in FE Analysis
Analysis of Thermal Gradients
Bilinear Gradient:
Specified TemperatureVariation
slabthickness
Analysis of Thermal Gradients
Trilinear Gradient:
1Element
Temperature VariationsUsed in FE Analysis
Specified TemperatureVariation
3Elements
2Elements
slabthickness
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Analysis of Thermal GradientsQuick parametric study
Re-run our single-slab model Consider positive (+5oC/-5oC) and negative (-5oC/+5oC) gradients Consider both bonded and unbonded base with no shear transfer
Results of Analyses:
Maximum Principal Stress (MPa)
Bonded Unbonded
Positive 1.45 0.94
Negative 1.22 0.86
Analysis of Thermal Gradients
Effect of thermal gradient nonlinearity
Re-run our single-slab model with nonlinear gradients, unbonded base
Positive gradient Negative gradient
Results:
1.71 MPa for positive (82% increaseover linear gradient)
0.47 MPa for negative (45% decreaseover linear gradientplus peak stress is at mid-thickness of slab!)
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Analysis of Slab ShrinkageShrinkage can be simulated as an equivalent thermal gradient
Example:
Consider a uniform shrinkage of -0.0001 mm/mm
Coeff. of thermal expansion = 1.1x10-5/oC
Equivalent T = -0.0001/1.1x10-5/oC = -9.09oC
-9.09oC Re-run our single-slab model assuming:
No slab-base shear transfer A rough HMA base (E= 2000 MPa, kSB = 0.27 MPa/mm,
0 = 0.25mm)
Analysis of Slab ShrinkageResults of Simulation
No slab-base shear transfer
x= +/-0.25mm at x = 0mm, 5000mm
No stresses are developed in slab
BOS Stresses
max = 0.32 MPa
With slab-base shear transfer
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Early-Age Effects Concrete pavements sometimes crack during curing
Primary causes are thermal and/or shrinkage gradientsthat occur prior to concrete gaining full tensile strength
Shrinkage cracks innew pavement
Early-Age EffectsSimple example of how this can be studied with EverFE
Re-run our single-slab model founded on CTB
Consider a negative (-5oC/+5oC) thermal gradient
Unbonded base with no shear transfer
Examine effect of curing time on ratio of slab stress:slab MOR
Assumptions:
MOR = E= (usual ACI equations, psi)
Assume these relationships are valid for cure times of 128 days Type I cement, published relationship between time and
Examine effect of curing time on ratio of slab stress:slab MOR
cf
6c
f
000,57c
f
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Early-Age EffectsDetails of Analysis Parameters
Age-strength relationship
1.0 5.5 11100 1.172.5 10.3 15230 1.60
4.0 13.8 17580 1.859.0 20.7 21530 2.27
28.0 27.6 24870 2.62
Age E MOR
days MPa MPa MPa
cf
Early-Age EffectsResults of Analysis
TOS stresses
Displaced shape 0 5 10 15 20 25 300.30
0.35
0.40
0.45
0.50
Maxstress/MOR
Time (days)
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Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
Dowel Joint Load TransferThe challenge: How do we model this?
separation ofslab and subgrade
dowel
wheel load
high stresson subgrade
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Dowel Joint Load Transfer Early models used springs at transverse joints
Other models used beams on elastic foundations
Primarily 2D models with
plate elements
Both 2D models with plateelements and 3D models
Dowel Joint Load Transfer
Challenges for idealizing dowels in 3D FE models:
Dowel-slab interaction and dowel looseness are difficult to treat
Conventional discretizations require slab and dowel nodes to coincide
slab mesh lines
Plan View
dowels
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Dowel Joint Load TransferOur solution is the embedded dowel element
solidelement
Beam element is constrained to
displace compatibly with theembedding solid element
dowels
slab mesh linesImmediate Benefit:
Dowel Joint Load Transfer
Specification of dowels in EverFE
Dowels can be equally spaced Dowels can be located in wheelpaths Dowels can be manually located by specifying y-coordinate Each row of slabs can have different dowel placements
Example of dowel placement
Start a new model with say 2 rows x 3 columns of slabs Go to dowel panel Try different methods of dowel placement
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Dowel Joint Load TransferTreatment of Dowel-Slab Interaction with EverFE
Rigorous treatment Either bonded or unbonded Can be severe nonlinearity
Dowel Looseness
gap
length
gap
Less rigorous treatment Model remains linear Allows intermediate bond levels
Dowel-Slab Support Modulus
Kz = modulus ofdowel support diameter
Dowel LoosenessSignificance:
Has been studied experimentally and numerically Small gaps (< 0.50mm) can greatly reduce joint load transfer
Treatment by EverFE:
Embedded element formulation is very advantageous Treated as a nodal contact problem Multiple embedded beam elements are used for each dowel
multipleelements
singleelement
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914 mm
1220
mm 12 - 6.35 mm dowels
10 kN
rubber pad
k= 0.09 MPa/mm
51mm
grease and drinking straw
Laboratory Tests of Hammons (1997)
unbonded CTB
Dowel Looseness
VerticalDisplacement(mm)
0.2
0.4
0.6experimentalno CTB
Distance from Joint (mm)
0-200 100-100-400
model, no looseness
model, gap = 0.08 mm
Dowel Looseness
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Distance from Joint (mm)
0-200 100-100-400
Ve
rticalDisplacement(mm)
0.2
0.4
0.6
model, k= 0.09 MPa/mm, gap = 0.08mm
experimentalwith CTB
Dowel Looseness
model, k= 0.07 MPa/mm,gap = 0.08 mm
Dowel LoosenessExample for 2-slab system:
Slabs are 4600mm long x 3600mm wide x 250mm thick
Founded directly on dense liquid, k = 0.03 MPa/mm
E= 28000 MPa, = 0.20, density = 0
Center an 80-kN axle with 2 wheels transversely, left of joint
Set linear aggregate interlock stiffness to 0
Use 11 evenly spaced 32mm diameter dowels at the joint
Choose dowel looseness, de-select bonded, Emb = 225 mm
Set GapB to 125mm (1/2 embedded length)
We will vary GapA: (0 to 0.4mm in 0.05mm increments)
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Dowel LoosenessResults of Analysis
Gap l u LTE(mm) (mm) (mm) (%) (MPa)
0.00 0.467 0.467 100 0.8650.05 0.528 0.528 81 1.0190.10 0.578 0.384 66 1.1210.15 0.622 0.344 55 1.267
0.20 0.646 0.323 50 1.3090.30 0.660 0.310 47 1.3230.40 0.664 0.306 46 1.326
Dowel-Slab Support ModulusBackground:
More traditional method of idealizing dowel-slab interaction
Dowel-slab interface idealized with distributed springsResults in a linearly elastic model
Can specify varying degrees of bond and dowel locking
Example:
Consider the same example we just analyzed
Specify dowel-slab support modulus in lieu of dowel loosenessVary modulus from 1 to a very large value, say 1x10-6
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Dowel-Slab Support ModulusResults of Analysis
Kz l u LTE(MPa) (mm) (mm) (%) (MPa)
1e6 0.474 0.471 99 0.9121e4 0.505 0.457 90 1.182
5000 0.517 0.447 86 1.223500 0.612 0.357 58 1.310
100 0.771 0.200 26 1.3621 0.969 0.004 0 1.417
Dowel Misalignment/Mislocation
Inaccurately cut transverse joints mislocated dowelsImproperly placed dowels dowel misalignment
Elevation View
z
Actual
position q
s
Intendedposition
x
Plan View
Intendedposition
ActualPosition qr
y
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Dowel Misalignment/MislocationTreatment by EverFE
Embedded dowel element permits implementation Straightforward when dowel-slab support modulus is specified A different solver must be used when modeling looseness
Example with EverFE
Consider the same example we just analyzed Vary x from 0 100mm with Kz = 2000 (LTE = 79% at x = 0) Study effect of x on response
Dowel Misalignment/MislocationResults of Analysis:
0 78.6 1.26 3789 135.720 78.6 1.26 3787 136.1
40 78.6 1.26 3783 137.260 78.2 1.27 3770 139.580 77.9 1.27 3745 143.7
100 77.3 1.27 3703 150.2
x LTE Dowel Bearing(mm) (%) (MPa) Shear Stress
(N) (MPa)
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Transverse Ties
Can be independently specified for each longitudinal joint
Modeled with same embedded elements used for dowels
Can model tie-slab support and restraint moduli
Assumed evenly spaced along each joint
First tie is placed at tie spacing from left-hand joint
Transverse Ties
4600mm (typ)
3600mm
1800mm
Model Properties 250mm slab on dense liquid 12-32mm dowels give 80% LTE
at transverse joint Tied shoulder
13mm diameter, 750mm long ties Corner axle load and thermal
gradient considered in analyses
Example to Illustrate Tie Effectiveness
80 kN axle
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Transverse Ties
650 670 690 710 730 750
0.7
0.9
1.1
1.3
1.5
Tie Spacing (mm)
Maxim
um
PrincipalStress(MPa)
Slab stress with NO ties:
Axle load: 1.33 MPa Thermal: 0.746 MPa Axle+thermal: 1.39 MPa
Axle load
Axle + thermal
Thermal
Transverse TiesObservations and Conclusions:
Ties can dramatically reduce slab stresses due to corner loads
Tie effectiveness strongly depends on its proximity to joint
700 mm spacing Max. stress = 0.719 MPa
710 mm spacing Max. stress = 1.38 MPa
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Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
Aggregate InterlockThe challenge: How do we model this?
aggregateinterlock
wheel load
Interaction of two rough crack surfaces Seasonal joint opening significantly affects load transfer
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Aggregate InterlockUsual FE Treatment of Aggregate Interlock:
Springs at Transverse Joints
Simple, traditional approach Model remains linear No effect of joint opening
Coulomb Friction
Shear depends on normal stress Any joint opening => no shear
EverFEs Two-Phase Model
aggregate particles
crack
Relies on Walravens Model
concrete is two-phase medium
aggregate particles are rigid spheres paste is rigid-plastic
cracks follow aggregate boundaries
particles bear on paste, at point of slip
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( )Fy pu x yA A=
( )Fx pu y xA A= +
Particle Equilibrium:
pu
Fx
Fy
pu pu
=
aggregate
particle
deformed
paste
embedment
crack
opening
EverFEs Two-Phase Model
EverFEs Two-Phase Model
Two-phase model parameters
1) pu = ultimate strength of cement paste
2) = paste-aggregate coefficient of friction (0.4 0.5)
ccpuf0.8= Walraven suggests
fcc = 1.25fc (units are MPa)
3) aggregate volume fraction (usually 0.7 0.8)
4) Maximum aggregate size (typically 18 or 20 mm)
5) Initial joint opening (seasonally variable)
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EverFEs Two-Phase ModelInitial joint opening is a critical parameter
Greatly affects nonlinear aggregate interlock model Affects contact between joint faces
initial jointopening
04
8
12
16
20
0 0.5 1.0
ShearStress
Relative Vertical Displacement
increasing
jointo
pening
directeffect
EverFEs Two-Phase Model
Tests by Colley and Humphrey (1967) Finite Element Idealization
Zero
StiffnessTwo-Phase
Model
2743 mm
1219mm
Loading PL
Joint FillerPre-cracked
178m
m
229m
m
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EverFEs Two-Phase Model
178 mm Slab
1 2 3
Joint Opening (mm)
0
20
40
60
80
100
Two-phasemodel
Experimental
data
020
40
60
80
100
1 2 3
Joint Opening (mm)
229 mm Slab
LTE(%)
EverFEs Two-Phase ModelExample for 2-slab system:
Slabs are 4600mm long x 3600mm wide x 250mm thick
Found directly on dense liquid, k = 0.03 MPa/mm
E= 28000 MPa, = 0.20, density = 0
Center an 80-kN axle with 2 wheels transversely, left of joint
No dowels at the joint
Specify nonlinear aggregate interlock model
pu = 50 MPa = 0.4 volume fraction = 0.75 Dmax = 20mm
Examine effect of joint opening on response
Same asdoweledmodel
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0.1 0.489 0.486 99 0.8630.5 0.519 0.456 88 1.2271.0 0.568 0.406 71 1.271
1.5 0.632 0.343 54 1.3012.0 0.707 0.267 38 1.3303.0 0.860 0.114 13 1.3864.0 0.963 0.012 1 1.414
JointOpening l u LTE
(mm) (mm) (mm) (%) (MPa)
EverFEs Two-Phase ModelResults of analysis:
EverFEs Two-Phase Model
Practical use of two-phase model:
Recent research has validated this type of model
However, the model is not perfect:
It assumes no fracture of coarse aggregate(Walraven suggests scaling down pu to account for this)
EverFE does not account for smooth surface at sawcut
(Will tend to overestimate joint shear transfer)
Estimating parameters is difficult
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Linear Aggregate Interlock ModelSprings at transverse joints
Simple approach Model remains linear Joint opening has no effect
Example for 2-slab system
0.0 0.974 0.000 0 1.4180.1 0.647 0.328 51 1.3600.5 0.538 0.436 81 1.2901.0 0.518 0.456 88 1.238
10.0 0.493 0.481 97 0.997100.0 0.488 0.486 99 0.852
JointStiffness l u LTE
(MPa/mm) (mm) (mm) (%) (MPa)
Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
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46
Effect of Slab-Base ShearTransfer and Dowel Locking
Low: kSB 0 (bond-breaker)
Intermediate: kSB = 0.035 MPa/mm, = 0.60 mm (ATB)
High: kSB = 0.416 MPa/mm, = 0.25 mm (HMAC)0
0
Prior studies have identified critical parameters:
Interface shear stiffness (kSB) and elastic limit ( ) Dowel-slab bond (dowel locking)
Interface properties for given base types (Zhang and Li 2001):
0
250 mm thick slab 150 mm thick base 32 mm dowels,no looseness
Material properties:
E= 28,000 MPa= 0.20= 2,400 kg/m3
= 1.1x10-5/oC
Parametric Study FE Model
4600 mm
3600mm
doweled joints
60,300 DOF3,024 brick elements slab
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Parametric Study Loads
UniformShrinkage
T
10 oC
+ Gradient+ Shrinkage
+ T T
6 oC
14 oC
Gradient+ Shrinkage
T T
14 oC
6 oC
Slab Displacements and Stresses
Displacementsdue to
T T
Max. principalstresses due to T T
500 XMagnification
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Selected Maximum Stresses (kPa)
DowelType
T
+ T T
T T
LoadCase
T+ T T
T T
Low Int. High
Degree of Slab/BaseShear Transfer
Locked
Unlocked
0870688
159973818
5941180991
0872689
118906669
5911510547
StressLocation
BottomBottom
Top
BottomBottom
Top
Discussion of Results for TT
Negative prestrain gradientproduces curling, tension on top
Dowel restraint uniformlyincreases tension
Shear stresses at bottom of slabdecreasetension on top of slab =
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Effect of Dowel Restraint for + TT
0 400 800 1200 1600 20000.9
1.1
1.3
1.5
Dowel Axial Restraint Modulus (MPa)
Max.
PrincipalStress(MPa)
High Slab/BaseShear Transfer
IntermediateSlab/BaseShear Transfer
Discussion of Results for + TT
Positive prestrain gradientproduces tension on bottom
Shear stresses at bottom of slabincreasetension on bottom of slab
Dowel restraint restricts relative
slip between slab and base
With high base/slab sheartransfer, restricted slip decreasestension due to base-slab shear
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Workshop Topics Introduction
Overview of Finite-Element Concepts
Generation and Solution of a Simple Model
Slab-Base Interaction
Analysis of Thermal Gradients and Slab Shrinkage
Modeling Dowel Joint Load Transfer
Modeling Aggregate Interlock Joint Load Transfer
Example of a More Complex Simulation
Obtaining EverFE and Program Architecture
Obtaining EverFE
1. Get a cashiers check for $5000 made out to Bill Davids
Go to http://www.civil.umaine.edu/EverFE Download EverFE2.23.exe Run EverFE2.23.exe on your computer You can now run EverFE using the new
desktop icon, or from the Programs menu Questions to [email protected]
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Program ArchitectureBasic architecture of software
User Interface
(Tcl/Tk/vtk)
FE meshing code
(compiled C++)
FE solver
(compiled C++)
What you see
Nonlinear agg.
interlock
(compiled C++)
What does the hard work
Program Architecture
Directory structure
Top-level directory
Aggregate interlock data
Project definitions/results
Help file and manual
Finite-element solver
Tcl/Tk code
Tcl/Tk libraries
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Program ArchitectureHow Project Data is Stored
Each project has a file with a .prj extension, and a subdirectory The .prj file is a placeholder to allow the project to be recognized The subdirectory contains project definition, FE input/output
Why is this important?
These files are simple ASCII text files, but can get large If you want to archive a project to save disk space, you simplymove the .prj file and entire subdirectory to another storage device
At any time in the future, you can copy the .prj file andsubdirectory back to EverFE2.23/data, and it will be recognized
Thank You