Workshop Slides EVERFE

7/27/2019 Workshop Slides EVERFE

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EverFE Workshop

Sacramento, CAMarch 11, 2004

Bill Davids, PhD, PEUniversity of Maine

[email protected]

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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Corner Break


TransverseJoint Faulting


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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


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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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


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 ~


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


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 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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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











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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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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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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


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


Bilinear Gradient:

Specified TemperatureVariation

slabthickness


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


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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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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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



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


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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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:


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)











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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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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

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