Fatigue and Fracture of Concrete Slabs for Airfields

Report
Fatigue and Fracture Behavior
of Airfield Concrete Slabs
Principal Investigators:
Jeffery Roesler, Ph.D., P.E.
Surendra Shah, Ph.D.
Graduate Research Assistants:
Cristian Gaedicke, UIUC
David Ey, NWU
Urbana-Champaign, November 9th, 2005
Outline
Objectives
Experimental Design
Experimental Results
2-D Fatigue Model
Finite Element Analysis
Application of FEM Model
Fatigue Model Calibration
Fatigue Model Application
Summary
The Future  Cohesive Zone Model
Research Objectives
Predicting crack propagation and failure under
monotonic and fatigue loading
Can fracture behavior from small specimens predict crack
propagation on slabs.
Load
CMOD
Three point bending beam
(TPB)
Beam on elastic foundation
Slab on elastic foundation
Research Objectives
Integrate full-scale experimental slab data and a 2-D
analytical fracture model (Kolluru, Popovics and Shah)
Check if the monotonic slab failure envelope controls the
fatigue cracking life of slabs as in small scale test
configuration.
Load vs. CMOD for 63 mm depth beam
Load vs. CMOD in 63 mm beam
2.5
2.5
2
1
Load (kN)
Load (kN)
2
1.5
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.5
1
Cycle 1
CMOD (mm)
Cycle 1024
0.5
Load vs. CMOD curve
Load (kN)
4.5
0
4
0
3.5
3
2-D Model
2.5
2
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
CMOD (mm)
Monotonic load
0.002
0.004
0.006
0.008
0.01
CMOD (mm)
Fatigue load
Experimental Design
Beam Tests
Simple supported beams:
2 beams, 1100 x 80 x 250 mm.
2 beams,
700 x 80 x 150 mm.
2 beams,
350 x 80 x 63 mm.
• The beams have a notch in the
middle whose length is 1/3 of the
beam depth.
Beams on clay subgrade:
2 beams, 1200 x 80 x 250 mm.
2 beams,
800 x 80 x 150 mm.
2 beams,
400 x 80 x 63 mm.
• The beams have a notch in the middle whose
length is 1/3 of the beam depth.
Experimental Design
Slab Tests
Large-scale concrete slabs on clay
subgrade:
2 slabs, 2010 x 2010 x 64 mm.
4 slabs, 2130 x 2130 x 150 mm.
• The load was applied on the edge through an
200 x 200 mm. steel plate.
• The subgrade was a layer of low-plasticity
clay with a thickness of 200 mm.
Concrete Mix
Standard Paving Concrete:
¾” limestone coarse crushed
aggregate, 100 mm slump and
Modulus of Rupture 650 psi at 28
days
Experimental Results
Results on Beams
Load vs CMOD in Simple Supported Beams
Load (kN)
Monotonic Load
•Full Load-CMOD curve.
•Peak Load.
•Critical Stress Intensity Factor (KIC) .
•Critical CTOD (CTODc).
•Compliance for each load Cycle (Ci).
8
7
6
5
4
3
2
1
0
0
Load vs. CMOD Curve for Beams on Elastic Foundation
0.4
0.6
0.8
1
1.2
CMOD (mm)
Load vs. CMOD for Beam on Elastic Foundation
10
9
8
B63
7
B150
6
B250
Load (kN)
7
Load (kN)
0.2
6
5
4
3
Compliance vs. load cycle
5
4
3
2
2
1
1
0
0
0
0.5
1
1.5
2
CMOD (mm)
2.5
3
3.5
0
0.2
0.4
0.6
CM OD (mm)
0.8
1
Beam FEM Setup
Small Beam
UIUC Testing
Monotonic Results and Crack Length
FSB 63
From FEM
0.03
y = 0.0006x - 0.008
R2 = 0.9972
Compliance
0.025
0.02
0.015
0.01
2.5
0.005
0
0
10
20
30
40
50
60
2
load (KN)
Crack length (mm)
From Testing
1.5
FSB 63b
FSB 63c
1
Peak Load
2.5
0.5
2
Load (KN)
0
0
1.5
FSB 63B
FSB 63C
1
0.5
0
0
0.01
0.02
0.03
CMOD (mm)
0.04
0.05
0.06
10
20
30
crack length (m m )
40
50
60
Experimental Results
Results on Beams
Load vs. CMOD Fatigue Cycles in 63 mm beam
2.5
Cycle 1
Cycle 1024
Cycle 10240
Cycle 45568
2
Load (kN)
Fatigue Load in FSB
•Load vs. CMOD curves
•Compliance vs. number of
cycles
•Peak Load.
•Stress Intensity Factor (KI) .
•Compliance for each load
Cycle (Ci).
1.5
1
0.5
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
CMOD (m m )
Experimental Results
Results on Slabs
Monotonic Load
•Full Load-CMOD curve.
•Peak Load.
•Compliance for each load Cycle
(Ci).
≈ 5 mm
Load vs CMOD for 150 mm slabs
Load vs CMOD for 63 mm depth slabs
140
Edge notch Specimen A
120
60
1/3 depth notch
Edge notch Specimen B
Edge notch
50
Unnnotched (Estimated CMOD)
Load (kN)...
Load (kN)
...
100
80
60
40
30
20
40
10
20
0
0
0
0
0.2
0.4
0.6
0.8
CMOD (mm)
1
1.2
1.4
0.2
0.4
0.6
CMOD (mm)
0.8
1
1.2
Experimental Results
Results on Slabs
Compliance vs. number of Cycles
Compliance (mm/kN)
Fatigue Load
•Load vs. CMOD curves
•Compliance vs. number of cycles
•Peak Load.
•Stress Intensity Factor (KI) .
•Compliance for each load Cycle (Ci).
0.1
0.09
0.08
0.07
0.06
0.05
0.04
U1
L2
U2
L3
0.03
0.02
0.01
0
0
Crack Length vs Number of Cycles (T2 slab)
10,000
20,000
30,000
40,000
50,000
60,000
70,000
Number of Cycles
Crack Length (mm)…...
60
45
30
15
0
0
10000
20000
30000
40000
Number of Cycles
50000
60000
2-D Fatigue Model
2-D Fatigue Model
(Kolluru, Popovics and Shah, 2000)
Monotonic Test of TPB
w
P
Fatigue Test of TPB
P
a
Pc
*C
t
s
Pmax
*C(loop)
i
Ci
Cu(1)
½ Pc
Cu(2) Cu(3)
Pmin
CMOD
6sa g 2 (a0 / w)
E1 
2
Ci w t
E2 
6sac g 2 (aeff / w)
Cu
( loop#)
2
wt
CMOD
*Secant compliance
E1  E2
Relation between load and effective
crack length aeff is obtained !!
PP
P
P c Pc
Pmax
PmaxA
Pc
B
C
O
O
D
O’
a0
aeff
P
O’min
a0 a0
a\failurea\failure
aeffaeff
2-D Fatigue Model
2-D Fatigue Model
O’-O: no crack growth, linear
part of the load-CMOD curve.
O-B: Crack Deceleration Stage,
Stable crack growth, nonlinear
part of the load-CMOD curve
until peak load.
B-D: Crack Acceleration Stage
Post peak load-CMOD.
a
P
Pc
Pmax A
C
O
D
O’
a0
a\crit
a\failure
-3
-5
C
O
D
B
C
a0
a\crit
Da
C1( a  a0 ) n1
DN
B
a0
0.4Nf Nf
a
Log(Da/DN)
-7
a\failure
a\crit
B
# cycles
a\failure
Da
C 2(DKI ) n 2
DN
Where:
C1, n1, C2, n2 are constants
Da = incremental crack growth between DN
DN = incremental number of cycles
DK =stress intensity factor amplitude of a load cycle
a
Finite Element Analysis
FEM Mesh
Computation of the Stress
Intensity Factor & Ci(a)
GI  Lim
Da 0
1
1
Fc  (vc  vd )  Lim
Fe  (vc  vd )
2Da
Da 0 2Da
An indirect method was used to calculate
KI, called “Modified Crack Closure Integral
Method. (Rybicki and Kanninen, 1977)
Element-1
Fc
a
Y, v
c
O’
Element-2
X, u
O e
f
d
b
Da
Element-3
Da
Element-4
Da
Finite Element Analysis
Relation between Crack length, Compliance and
CMOD
0.25
CMOD vs. Crack Length
CMOD (mm)
Normal equations for TPB Beams are not
applicable. FEM Modelation is required.
y=-1E-7x 2+2.9E-4x
0.2
0.15
0.1
0.05
FEM
Poly. (FEM)
The CMOD increases its value with the
increase of the Crack Length.
0
0
1000
1500
2000
Crack length (mm)
Compliance vs. Crack Length
1.8
Normalized compliance
The normalized compliance at the midslab
edge predicted using the FEM model
shows a quadratic behavior.
500
y = 1E-07x 2 - 1E-06x + 1.0228
1.5
1.2
0.9
0.6
FEM
0.3
Poly. (FEM)
0
0
400
800
1200
Crack length (mm)
1600
2000
Finite Element Analysis
KI/P (mm-3/2)
Relation between Stress
Intensity Factor and Crack
Length
0.002
0.0015
y = 5E-10x 2 - 9E-07x + 0.0014
0.001
FEM-1
0.0005
FEM-2
y = 2E-06x + 0.0007
Lineal (FEM-1)
Polynomial (FEM-2)
0
0
400
800
1200
1600
2000
Crack length (mm)
0.25
y=-1E-7x 2+2.9E-4x
Relation between CMOD
and Crack Length
CMOD (mm)
0.2
0.15
0.1
0.05
FEM
Poly. (FEM)
0
0
500
1000
1500
Crack length (mm)
2000
Application of FEM Model
Step 1: Experimental Relation between CMOD
and the Displacement
Experimental relation
between CMOD
measurements and
displacement.
CMOD vs Displacement
0.2
y = 0.0005x3 - 0.0037x2 + 0.0123x
R2 = 0.9851
CMOD (mm)
0.16
0.12
0.08
0.04
0
0
2
4
6
Displacement (mm)
8
10
Application of FEM Model
Step 2: Determination of the Load vs. CMOD curves
Load vs CMOD Curve
Edge notch Specimen A
120
Edge notch Specimen B
Unnnotched (Estimated CMOD)
100
Load (kN)
The relation
between CMOD
and displacements allows to
estimate the
CMOD for the
unnotched
specimen
140
80
60
40
20
0
0
0.2
0.4
0.6
0.8
CMOD (mm)
1
1.2
1.4
Application of FEM Model
Step 3: Estimation of the Crack Length
CMOD vs. Crack Length
1600
Crack Length (mm)…..
The crack length is
estimated using
this modified
equation from the
FEM model and the
CMOD
y = 231084x3 - 32098x2 + 3357.4x
R2 = 0.9622
1400
1200
1000
800
600
400
200
0
0
0.05
0.1
0.15
CMOD (mm)
0.2
0.25
Application of FEM Model
Step 4: Estimation of the Normalized
Compliance (FEM)
Normalized Compliance vs Crack Length
1.7
2
y = 1E-07x - 1E-06x + 1.0228
1.6
Normalized Compliance
The normalized
compliance
obtained from the
FEM Model for
different crack
length is multiplied
by the experimental
initial compliance
2
R = 0.9018
1.5
1.4
1.3
1.2
1.1
1
0
500
1000
Crack Length, a (mm)
1500
2000
Application of FEM Model
Step 5: Experimental Compliance vs. Crack
Length curves
Compliance vs Crack Length
0.0015
Edge notch Specimen A
Edge notch Specimen B
Compliance (mm/kN)
0.00125
Unnnotched (Estimated CMOD)
0.001
0.00075
0.0005
0.00025
0
0
250
500
750
1000
1250
Crack Length, a (mm)
1500
1750
2000
Fatigue Model Calibration
Step 1: The Compliance is measured for each
fatigue load cycle of slab T2
Compliance vs Number of Cycles
Compliance (mm/kN)..
0.0005
0.0004
0.0003
0.0002
0.0001
0
0
10000
20000
30000
40000
Number of Cycles
50000
60000
Fatigue Model Calibration
Step 2: The Crack length is obtained for each
cycle using the FEM Model for slab T2.
Crack Length (mm)-log scale …….
Crack Length vs Number of Cycles
2
1.6
1.2
0.8
0.4
0
0
10000
20000
30000
40000
Number of Cycles
50000
60000
Fatigue Model Calibration
Step 3: The Critical Crack acrit is Critical Number
of Cycles Ncrit is obtained for slab T2.
Crack Length vs Number of Cycles
60
Crack Length (mm)
This point of
critical crack
length is a
point of
inflexion in the
curve
45
acrit = 41 mm
30
15
Ncrit = 25000
0
0
10000
20000
30000
40000
Number of Cycles
50000
60000
Fatigue Model Calibration
Step 4: The two sections of the model are
calibrated
Crack Growth Rate vs Crack Length
0
Da
C1(a  a0 ) n1 C1(a  a0 )  12.25
DN
-0.5
-1
Log (D a/D N)
Different fatigue
equations apply
for crack length
bigger or smaller
than acrit
•Log C1 = 17.6
-1.5
-2
-2.5
Da
C 2(DKI ) n 2  C 2 * (DKI ) 6.657
DN
-3
-3.5
0
200
400
acrit = 41 mm
600
800
1000
1200
1400
Crack Length, a (mm)
•Log C2 = -15.0
1600
1800
2000
Fatigue Model Application
Estimation of N1
N1 is he required number of cycles to achieve acrit
This fatigue equation allows to predict crack propagation for
any number of cycles N<N1
•Log C1 = 17.6
If we have an unnotched slab, a0 = 0
Da
C1( a  a0 ) n1  C1( a  a0 )  12.25
DN
N
a
n1  C1(a  a )  12.25dN  a
C
1( a  a0 )
0
0

1
DN
1

Da C1(a  a ) n1  C1(a  a )  12.25
0
0
acrit
1
N1 
da
a
C1(a  a ) n1  C1(a  a )  12.25
0
0
0
Fatigue Model Application
Estimation of N2
N2 is he required number of cycles to achieve afailure
This fatigue equation allows to predict crack propagation for
any number of cycles N1<N<N2
Da
C 2(DKI ) n 2  C 2 * (DKI ) 6.657
DN
•Log C2 = -15.0
N
a  C 2(DKI ) n2  C 2 * (DKI ) 6.657 dN  acrit
N1
DN
1

Da C 2(DKI ) n 2  C 2 * (DKI ) 6.657
a fa ilu re
1
N2 
da
n
2

6
.
657
C 2(DKI )  C 2 * (DKI )
acrit
NTOTAL  N1  N 2  61,000
Summary
Currently, empirical fatigue curves don't
consider crack propagation. Fracture
mechanics approach has clear advantages to
predict crack propagation.
Monotonic tests are failure envelope for
fatigue.
Mechanics of model work but model
coefficients need to be improved.
A Cohesive Zone Model has greater potential
to give a more conceptual and accurate
solution to cracking in concrete pavements
Tasks Remaining
Fatigue crack growth prediction of beams on
elastic foundation (NWU)
Complete model calibration on remaining slabs
Several load (stress) ratios
Tridem vs. single pulse crack growth rates
Write final report
Current Model Limitations
Crack propagation assumed to be full-depth
crack across slab
pre-defined crack shape
Need geometric correction factors for all
expected slab sizes, configurations, support
conditions
Need further validation/calibration with other
materials and load levels
Fracture Mechanics Size Effect
• Size Effect Method (SEM) log 
N
Strength Theory
► Energy concept
► Equivalent elastic crack model
►Two size-independent fracture
parameters: Gf and cf
Quasi-brittle
B ft 
N 
1  D D0
LEFM
1
2
log D
Bazant ZP, Kazemi MT. 1990, Determination of fracture energy, process zone length and brittleness
number from size effect, with application to rock and concrete, International Journal of Fracture, 44,
111-131.
• Two-Parameter Fracture Model (TPFM)
– Equivalent elastic crack model
– Two size-independent fracture parameters : KI and CTODc
Jenq, Y. and Shah, S.P. 1985, Two parameter fracture model for concrete, Journal of Engineering
Mechanics, 111, 1227-1241.
What is the Cohesive
Zone Model?
Modeling approach that defines cohesive
stresses around the tip of a crack

Traction-free macrocrack
Bridging zone
Cohesive stresses are related to
the crack opening width (w)
Crack will propagate, when  = ft
Microcrack zone
w
wf
ft
How can it be applied to
rigid pavements?
The cohesive stresses are defined by a cohesive
law that can be calculated for a given concrete
Concrete properties

Cohesive law
ft
 ft
Gf
GF  G f
w1
v4
Cohesive
Elements
Cohesive Elements are
located in Slab FEM model
u4
w
wf
v3
wn
u3
wt
v1
v2
u1
Cohesive
Finite Element
u2
Why is CZM better for
fracture?
The potential to predict slab behavior under
monotonic and fatigue load
Cohesive laws
The cohesive relation is a MATERIAL
PROPERTY
Predict fatigue using a cohesive relation
that is sensitive to applied cycles,
overloads, stress ratio, load history.
Allows to simulate real loads

ft
 ft
Gf
GF  G f
w1
Load vs CMOD for 63 mm depth slabs
60
1/3 depth notch
Edge notch
Compliance vs. number of Cycles
40
Compliance (mm/kN)
Load (kN)...
50
30
20
10
0
0
0.2
0.4
0.6
CMOD (mm)
0.8
1
0.1
0.09
0.08
0.07
0.06
0.05
0.04
U1
L2
U2
L3
0.03
0.02
1.2
0.01
0
0
10,000
20,000
30,000
40,000
50,000
60,000
Cohesive
Elements
70,000
Number of Cycles
Monotonic and Fatigue
Slab behavior
Cohesive
Finite Element
w
wf
Proposed Ideas
Laboratory Testing and Modeling of Separated
(Unbonded) Concrete Overlays
Advanced Concrete Fracture Characterization
and Modeling for Rigid Pavement Systems
Laboratory Testing and Modeling of
Separated Concrete Overlays
Concrete Overlay
Existing Concrete
Pavement
hol
he
- Asphalt Concrete Bond Breaker ~ 2”
Bond
Breaker
Laboratory Testing and Modeling of
Separated Concrete Overlays
New PCC
New PCC
AC Interlayer
AC Interlayer
Old PCC
Old PCC
Rigid Support
Support
Cohesive elements
Advanced Concrete Fracture
Characterization and Modeling for Rigid
Pavement Systems
Concrete properties

Cohesive law
ft
 ft
Gf
GF  G f
w1
v4
Cohesive
Elements
Cohesive Elements are
located in Slab FEM model
u4
w
wf
v3
wn
u3
wt
v1
v2
u1
Cohesive
Finite Element
u2

similar documents