field validation and parametric study of a thermal crack spacing model

Field Validation and Parametric Study of a Thermal Crack Spacing

Model

David H. Timm - Auburn UniversityVaughan R. Voller - University of Minnesota

Presented at the Annual Meeting of the Association of Asphalt Paving Technologists

Lexington, KentuckyMarch 10 – 12, 2003

Cracking Characteristics• Thermal cracking common in cold

climates• Features

– Transverse cracks– Regular spacing

0

0.25

0.5

0.75

1

0 50 100 150 200 250 300 350 400 450 500

Location, ft

Dis

tanc

e A

cros

s La

nes

Crack Spacing

Focus of thisStudy is the question

What features control the spaces betweenCracks?

Model Stress Profile in Thermally Cooled Asphalt Layer on Granular

Base

E, , , H,

E, , , c,

Modeled in Two ways

Finite Difference Code--FLAC

x

50x250 mm

Grid Element Sizes

63x315 mm

313x1563 mm

Asphalt Concrete (Elastic Model)

z

Granular Base (Mohr Coulomb Model)

1-D Semi-Analytical ModelElastic Layer with Elastic-Plastic

Restraint

catan q=kux

Timm, Guzina and VollerInt J Solids and Structures, 2002

xt

2/xxwHku

xxtanHc

dxd

tx

tx

Form of Stress Profile

Curling Stress

Rate of StreesIncrease T

1E

Distance fromfree end

Comparison of Models

Crack Spacing from Stress Curve

St

xc

1

xCracking

may occurCrackingwill notoccur

SlidingOn Rigid Base

xtanHc

Hx

1

St

x

Crack Spacing from Stress Curve

xcxc

Average Spacing = 1.5·Xc

• Validate thermal crack spacing model with field data

• Perform sensitivity analysis on length scale– Help guide future laboratory work– Develop more complete understanding– Identify how material selection will

affect spacing

Objectives

• Field Validation– 4 similar sections at Mn/ROAD

• Parametric Study– 10 input variables

• Layer 1– Stiffness, Poisson, Density, Thickness, Thermal

Coef.• Layer 2

– Stiffness, Poisson, Density, Cohesion, Friction Angle

Scope

E, , , H,

E, , , c,

1. Select MnROAD sections2. Analyze thermal crack spacing by

section3. Analyze in situ thermal conditions4. Gather material property data for

model5. Simulate pavement, determine

spacing6. Compare predictions to measured7. Assess validity

Field Validation Methodology

• Similar thickness designs• Identical binders• Common subgrade• Different base layers

MnROAD Sections

150 155 160231

838

102 102

711838

0

200

400

600

800

1000

1200

Cell 1 Cell 2 Cell 3 Cell 4

Dep

th B

elow

Pav

emen

t Sur

face

, mm

HMAC

Class 4 G.B.

Class 6 G.B.

Class 5 G.B.

Class 3 G.B.

LEGEND

Average Crack Spacing

0

0.25

0.5

0.75

1

0 100 200 300 400 500

Location, ft

0

0.25

0.5

0.75

1

0 100 200 300 400 500

Location, ft

0

0.25

0.5

0.75

1

0 100 200 300 400 500

Location, ft

Avg Spacing

Cell 1: 12 m

Cell 2: 8 m

Cell 3: 13 m

Cell 4: 9 m0

0.25

0.5

0.75

1

0 100 200 300 400 500

Location, ft

-35

-30

-25

-20

-15

-10

-5

0

5

10

0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72

Time, hr

Tem

pera

ture

, C

D = 0.03048D = 0.054864D = 0.158496D = 0.192024D = 0.292608D = 0.445008D = 0.597408D = 0.902208D = 1.207008D = 1.511808D = 2.426208

Depth, m

Top of pavement

Bottom of pavement

Feb 1 Feb 2 Feb 3

Temperature Cycling

• Backcalculation• Laboratory testing as part of

Mn/ROAD project• Derived values

– Thermal coefficient = fn (Volumetrics)• Model ‘tuned’ with friction and

cohesion

Material Property Data

E, , , H,

E, , , c,

Resulting Friction and Cohesion

Cell Friction Angle, o Cohesion, kPa1 30 102 50 153 35 104 25 10

Mohr-Coulomb Properties of Material DirectlyBeneath HMA

Model Comparison

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14 16

Measured Average Spacing, m

Pred

icte

d Sp

acin

g, m

Line of Equality

Cell 1

Cell 2

Cell 3

Cell 4

• Crack spacings pass reasonableness check

• Recently, model has been used to predict other crack spacing phenomenon

Model Assessment

0

1

2

3

4

5

6

7

0 20 40 60 80 100 120 140Location, m

Dis

tanc

e Acr

oss

Pave

men

t, m

Average Spacing = 12 mStandard Deviation = 4.88 m

TiN Coating

Curling Stress

Rate of StressIncrease

Max stress

Factors that Influence Stress Profile

• Uniform temperature change• 2-layer structure• 10 input parameters varied from low,

medium, and high• Maximum tensile stress curves

plotted and evaluated– Maximum Stress– Rate of Stress Increase– Curling Stress

Parametric Investigation Methodology

Input ParametersLayer Input Units Low Medium

(Baseline)High

1 E1 Pa 5*109 1.4*1010 3*1010

unitless

0.15 0.20 0.25

kg/m3 2,200 2,300 2,400H1 cm 7.6 15 30 /C 1.33*10-5 2.15*10-5 2.97*10-5

2 E2 Pa 5.5*107 5.5*108 5.5*109

unitless

0.35 0.4 0.45

kg/m3 1,800 2,000 2,200c2 kPa 0, 0.1, 1, 10, 70, 140 2 20 40 60

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

E1 = 3*1010 Pa

E1 = 1.4*1010 Pa

E1 = 5.0*109 Pa

Distance From Free End, m

Maximum Tensile Stress, PaFLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

E1 = 3*1010 Pa

E1 = 1.4*1010 Pa

E1 = 5.0*109 Pa

Distance From Free End, m

Maximum Tensile Stress, Pa

HMAC Stiffness (E1)

1

TEx

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

Maximum Tensile Stress, Pa

1=0.15

1 =0.20

1 =0.25

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

Maximum Tensile Stress, Pa

1=0.15

1 =0.20

1 =0.25

HMAC Poisson Ratio (1)

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

Maximum Tensile Stress, Pa

H1 = 0.076 mH1 = 0.152 m

H1 = 0.305 m

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

Maximum Tensile Stress, Pa

H1 = 0.076 mH1 = 0.152 m

H1 = 0.305 m

HMAC Thickness (H1)

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

Maximum Tensile Stress, Pa

1=2.97*10-5

1=2.15*10-5

1=1.33*10-5

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

Distance From Free End, m

Maximum Tensile Stress, Pa

1=2.97*10-5

1=2.15*10-5

1=1.33*10-5

HMAC Thermal Coeff. (1)

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

E2=5.5*107 Pa

E2=5.5*109 Pa

E2=5.5*108 Pa

Distance From Free End, m

Maximum Tensile Stress, PaFLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

E2=5.5*107 Pa

E2=5.5*109 Pa

E2=5.5*108 Pa

Distance From Free End, m

Maximum Tensile Stress, Pa

Base Stiffness (E2)

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

c2=140 kPac2=70 kPa

c2=10 kPa

c2=1 kPac2=.1 kPac2=0 kPa

Distance From Free End, m

Maximum Tensile Stress, PaFLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

c2=140 kPac2=70 kPa

c2=10 kPa

c2=1 kPac2=.1 kPac2=0 kPa

Distance From Free End, m

Maximum Tensile Stress, Pa

Base Cohesion (c2)

As c getsLargeOnly elasticresistance

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

2 = 60o

2 = 40o

2 = 20o

Distance From Free End, m

Maximum Tensile Stress, PaFLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

2 = 60o

2 = 40o

2 = 20o

FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06FLAC (Version 3.30)

10 20 30 40 50 60 70

.500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

5.000

(10 )+06

2 = 60o

2 = 40o

2 = 20o

Distance From Free End, m

Maximum Tensile Stress, Pa

Base Friction Angle (2)

Note: c = 10 kPa

Curling Stress

Rate of StressIncrease

Max stress

Factors that Influence Stress Profile

Relative Influence on Each CriteriaInput

ParameterMaximum

StressRate of Stress

IncreaseCurlingStress

E1 3 1 --

2 -- --

1 -- -- --

H1 -- -- 3

1 3 1 --

E2 -- 3 --

-- -- --

2 -- -- --

c2 -- 3 3

2 -- 2 --

• Model compared favorably to field data• Model is sensitive to base material

properties• Model is simple, yet provides length

scale to thermal cracking problem• Key input parameters are…

– Stiffnesses of HMAC and Base– Thermal coefficient– Frictional properties of Base material

Conclusions

• Further validation with field sections– Model has compared favorable to other

types of cracking• Incorporate a fracture mechanics

model to simulate crack propagation• Examine viscoelastic constitutive

models

Recommendations

• Plan mitigation strategies– Saw and seal– Material selection

• Assess probability and expectation of cracking

Potential Uses of Model

• Dr. Bojan Guzina

• Minnesota Department of Transportation– Minnesota Road

Research Project

Acknowledgements

Thank You!

Questions?