# 2 Pavement Design2(Materials)_Arnold_v1.pdf

#### Description

presents

Pavement Design Materials Characterisation

Austroads Material Types • Unbound Granular

(UB Gr)

• Bound Materials - Cement bound - Bitumen bound

(CT) (AC)

(SG)

Characteristics? Modulus (E) Poisson’ Poisson’s ratio (ν (ν) Isotropy (EV=EH) Performance Relationship

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Unbound Granular Materials Factors affecting stiffness (Modulus) 1. Particle interlock & Friction 2. Density & Water content

Unbound Granular Materials Factors affecting stiffness (Modulus) 3. Confinement & Stress Level

σ σ2 σ1

UB Granular NonNon-linear

E2

E1

EGR is Stress Dependent The greater the confining stress, the greater is EGR

ε

2

Implication of Stress Dependency EGR varies with location in the pavement Chipseal E = 500 MPa Unbound Granular

Asphaltic Concrete E = 200 MPa

E = 100 MPa

E = 70 MPa

Modelling for stress dependancy Sublayering EBT = 467 500 MPa

Check EBT achievability: EBT(max) = ESGx2hgr/125 = 60 x 2370/125

370

= 467 MPa Calculate modular ratio: R = (EBT/ESG)1/5

ESG= 60 MPa

= (467/60)1/5 = 1.507

3

370

Sublayering (cont.) EBT = 1.507 x 309 = 467

74mm

E4 = 1.507 x 205 = 309

74mm

E3 = 1.507 x 136 = 205

74mm

E2 = 1.507 x 90 = 136

74mm

E2 = 1.507 x 60 = 90

74mm

ESG= 60 MPa

APPENDIX C: Circly output for Example 2.1 Note: Sign convention for strains output: Positive output ==> Compressive strain Negative output ==> Tensile strain L TYPE ELASTIC CONSTANTS 1 CROSS-ANISOTROPIC EH=0.2334E+03 EV=0.4669E+03 F=0.3459E+03 VVH= 0.3500E+00 VH = 0.3500E+00 2 CROSS-ANISOTROPIC EH=0.1549E+03 EV=0.3097E+03 F=0.2295E+03 VVH= 0.3500E+00 VH = 0.3500E+00 3 CROSS-ANISOTROPIC EH=0.1027E+03 EV=0.2055E+03 F=0.1522E+03 VVH= 0.3500E+00 VH = 0.3500E+00 4 CROSS-ANISOTROPIC EH=0.6816E+02 EV=0.1363E+03 F=0.1010E+03 VVH= 0.3500E+00 VH = 0.3500E+00 5 CROSS-ANISOTROPIC EH=0.4522E+02 EV=0.9044E+02 F=0.6700E+02 VVH= 0.3500E+00 VH = 0.3500E+00 6 CROSS-ANISOTROPIC EH=0.3000E+02 EV=0.6000E+02 F=0.4140E+02 VVH= 0.4500E+00 VH = 0.4500E+00 ESA750-Full POINT C O O R D I N A T E S --------------------------------X Y Z 1 0.0000E+00 0.0000E+00 0.3700E+03 2 0.3300E+02 0.0000E+00 0.3700E+03 3 0.6600E+02 0.0000E+00 0.3700E+03 4 0.9900E+02 0.0000E+00 0.3700E+03 5 0.1320E+03 0.0000E+00 0.3700E+03 6 0.1650E+03 0.0000E+00 0.3700E+03 7 0.1980E+03 0.0000E+00 0.3700E+03 8 0.2310E+03 0.0000E+00 0.3700E+03 9 0.2640E+03 0.0000E+00 0.3700E+03 10 0.2970E+03 0.0000E+00 0.3700E+03 11 0.3300E+03 0.0000E+00 0.3700E+03

D I S P L A C E M E N T S ---------------------------------L UX UY UZ 6 -0.3619E-03 0.4935E-08 -0.7868E+00 6 -0.1287E-01 0.4308E-08 -0.7876E+00 6 -0.2543E-01 0.3487E-08 -0.7855E+00 6 -0.3805E-01 0.2498E-08 -0.7806E+00 6 -0.5058E-01 0.1383E-08 -0.7730E+00 6 -0.6267E-01 0.2043E-09 -0.7627E+00 6 -0.7389E-01 0.2628E-09 -0.7499E+00 6 -0.8371E-01 0.3259E-09 -0.7347E+00 6 -0.9166E-01 0.3939E-09 -0.7176E+00 6 -0.9740E-01 0.4672E-09 -0.6989E+00 6 -0.1008E+00 0.5464E-09 -0.6792E+00

POINT C O O R D I N A T E S ---------------------------------X Y Z 1 0.0000E+00 0.0000E+00 0.3700E+03 2 0.3300E+02 0.0000E+00 0.3700E+03 3 0.6600E+02 0.0000E+00 0.3700E+03 4 0.9900E+02 0.0000E+00 0.3700E+03 5 0.1320E+03 0.0000E+00 0.3700E+03 6 0.1650E+03 0.0000E+00 0.3700E+03 7 0.1980E+03 0.0000E+00 0.3700E+03

N O R M A L S T R A I N S ---------------------------------L XX YY ZZ 6 -0.3789E-03 -0.6886E-03 0.1129E-02 6 -0.3795E-03 -0.6866E-03 0.1126E-02 6 -0.3820E-03 -0.6801E-03 0.1118E-02 6 -0.3822E-03 -0.6688E-03 0.1101E-02 6 -0.3751E-03 -0.6522E-03 0.1074E-02 6 -0.3557E-03 -0.6302E-03 0.1032E-02 6 -0.3213E-03 -0.6025E-03 0.9761E-03

THICKNESS INTERFACE 0.7400E+02 ROUGH 0.7400E+02

ROUGH

0.7400E+02

ROUGH

0.7400E+02

ROUGH

0.7400E+02

ROUGH

INFINITE

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Sublayering (cont.) What Noifsublayering Eunderlying>EBT ? EGR = 500 EGR = 500 Cemented Layer ECT = 2000 MPa

ESG= 60 MPa

Self Evaluation Exercise 2

5

SE Exercise 2.1 450

264

290 600

189 300

187

136

120

97

78

70

SG = 50

SG = 50

600

Ev(lim it ) = 50 x 2  450  R=   50 

0.2

= 1.55

50 x 1.55 = 78

125

300

= 1392 Ev(lim it ) = 50 x 2  264  R=   50 

125

= 264

0.2

= 1.39

50 x 1.39 = 70

Determination of EGR 1. Presumptive Values (Normal Standard) Cover (mm) 40 &< 75 100 125 150 175 200 >250

Modulus of cover material (MPa) 1000 2000 3000 4000 5000 350 350 350 330 280 350 350 340 260 210 350 350 290 220 160 350 340 240 160 150 350 290 190 150 150 350 240 150 150 150 350 190 150 150 150 310 150 150 150 150

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Determination of EGR 1. Presumptive Values (High Standard) Cover 40 &< 75 100 125 150 175 200 >250

Modulus of cover material (MPa) 1000 2000 3000 4000 5000 500 500 500 480 400 500 500 480 380 300 500 500 410 310 230 500 480 340 240 210 500 410 270 210 210 500 340 210 210 210 500 270 210 210 210 450 210 210 210 210

2. Repeated Load Triaxial (RLT) test

σ3

εR

Deviator Stress

εT

Vert. Stress

Deviator Stress

σ3 εR εT

Resilient Modulus =

Long. Strain

Deviator Stress Resilient Strain

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Repeated Load Triaxial (RLT) test apparatus - obtain resilient moduli for design but later discuss a more important use to ensure materials will resist rutting

RLT Outputs Current draft TNZ T/15 test - 6 stage permanent deformation test Permanent strain [%]

1 0.8 0.6

F

0.4

E

0.2

B

A

C

D

0 0

50,000

100,000 150,000 200,000

250,000 300,000

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RLT Outputs – Resilient Modulus Show outputs in spreadsheet Exercise 1 – Determine Relationship between resilient modulus and bulk stress

Concepts (cont.) σ3

εR

Deviator Stress

εT

Vert. Stress

Deviator Stress

σ3 εR εT

Resilient Modulus =

Long. Strain

Deviator Stress Resilient Strain

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Pass 2

Pass 3

Pass 4

Stress

Pass 1

Pass 5

Pass 6

Element

Axial deformation

Time (t)

axial

Compressive Resilient deformation Permanent deformation Time (t)

Time (t) Permanent deformation Resilient deformation Expansive

Stresses on an Element in the Pavement σd σ3

where: σ1

= major principal stress [kPa]

σ3

= minor principal (confining) stress [kPa]

σd = cyclic deviator stress [kPa]

σ3

σ3

σ2

σ1

σ1

= σd+ σ3

Bulk _ Stress = σ 1 + σ 3 + σ 2 σ 3 = σ 2 = cell _ pressure

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Deviator Stress (q)

Stresses on an Element in the Pavement

q = σ1 −σ 3

p=

1 (σ 1 + 2σ 3 ) 3 s re St

(pmin, qmin)

(pmax, qmax)

ath p s

Mean Normal Stress (p)

Bulk _ Stress = σ 1 + σ 3 + σ 2

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RLT Outputs – Resilient Modulus Show outputs in spreadsheet Exercise 1 – Determine Relationship between resilient modulus and bulk stress

Determination of EGR (cont.) 3. Back-calculation from FWD test Measure actual deflection bowl under applied load

AC

1500 2135 1840

Model pavement in CIRCLY with ‘seed’ moduli

GR

215 300 273

CIRCLY calculates theoretical deflection bowl

SG

60 50 68

Compares theoreticalwith actual bowl Ajust E and Repeat until satisfactory agreement is achieved

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Eiso vs Eaniso When using back-calculated moduli from FWD results FWD ⇒ ELMOD ⇒ Isotropic moduli (Eiso) CIRCLY ⇒ requires Anisotropic moduli (Eaniso) Single relationship not possible. Depends on materials and stress environment Tonkin & Taylor suggest: For SG materials (ν=0.45): Eiso = 0.67 Eaniso(V) For GR materials (ν=0.35): Eiso = 0.75 Eaniso(V)

FWD Outputs Refer to spreadsheet Determine an appropriate model for CIRCLY from FWD data

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Performance Relationship No Performance Relationship for UB currently in Austroads

K N=  ε

exp

Current approach is to control deformation by means of: - good quality materials and - sound construction specs

Summary: Austroads Characterisation of UB Gr. E-range Poisson’ Poisson’s Ratio Anisotropic Performance Rel: Rel: Stress Dependent:

350 – 500 MPa ν = 0.35 EV = 2EH None Sublayered

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Austroads Material Types • Unbound Granular

(UB Gr)

• Bound Materials - Cement bound - Bitumen bound

(CT) (AC)

(SG)

Cemented Materials Cemented

3-5%

Modified

1½-3%

Subgrade Cementation (3-5%) Hard, rigid, inflexible Failure mode: Cracking

Subgrade Modification (1½-3%) Stiffer than UB, flexible Behaves like UB

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Determination of ECT 1. Presumptive Values Base Quality crushed rock 4-5% cement Typical ECT (MPa) MPa)

Crushed Rock 2-4% cement

5000

3500

Subbase quality natural gravel 4-5% cement

2000

Performance Relationship NL (CT )

K = RF CT  με

  

exp 12

N = Number of load repetitions to failure ε = critical horizontal tensile strain in the bottom of the layer

K CT =

113000 + 191 E0.804

exp =12

RF = Reliability Factor

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Reliability Factor Allows for different project reliabilities Project reliability is the probability that the constructed pavement will perform as predicted Possible reasons for nonnon-compliance: compliance: Material characterisation Design model vs. constructed reality Limitation of performance prediction models

APPENDIX B : PROJECT RELIABILITY Table 6.7 Suggested Reliability Factors (RF) for Cemented Materials Fatigue

Desired Project Reliability 80% 4.7

85% 3.3

90% 2.0

95%

97.5%

1.0

0.5

Table 6.13 Suggested Reliability Factors (RF) for Asphalt Fatigue

Desired Project Reliability

Table 2.1 Typical Project Reliability Levels

Project Reliability (%)

Motorway

95 – 97.5

Urban arterial > 7,000 vpd Rural strategic > 2,500 vpd

90 – 97.5

80%

85%

90%

95%

97.5%

Urban other

85 – 95

2.5

2.0

1.5

1.0

0.67

Rural other

80 - 90

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Life Phases of CT materials Pre-cracking phase

Post-cracking phase

Cover: 175mm AC…

… or 230mm UB Gr.

Cemented

Cracked

Modelled as Cemented E=2000-5000MPa, ν= 0.2 Unsublayered

Modelled as “Unbound” E=500MPa, ν= 0.35 Unsublayered

“Life 1” governed by CT strain criterion

“Life 2” governed by SG strain criterion

Summary: Austroads Characterisation of CT E-range Poisson’ Poisson’s Ratio Isotropic Performance relationship

NL( CT )

2000 -5000 ν = 0.2 EV = EH

K  = RF CT   με 

12

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Austroads Material Types • Unbound Granular

(UB Gr)

• Bound Materials - Cement bound - Bitumen bound

(CT) (AC)

(SG)

Factors affecting AC modulus Increase in: - Temperature - Binder viscosity - Loading rate - Air voids - Aggregate angularity - Age of AC - Cracking - Binder content - Stress level

Effect on EAC decrease increase increase decrease increase increase decrease In. then dec. No change

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Determination of EAC Presumptive Values Lab. Samples; 25° 25°C; 5% Air voids Binder (Austr.) Austr.)

Approx. NZ equivalent

Max particle size

10mm

14mm

20mm

Class 170

80/100

3500

3700

4000

Class 320

50/60

4500

5000

5500

Class 600

30/40

6000

6500

7000

Performance Relationship

NL ( AC)

K  = RF AC   με 

exp 5

N = Number of load repetitions to failure ε = critical horizontal tensile strain in the bottom of the layer

K AC =

6918(0.856 VB + 1.08) 0.36 Smix

Smix = EAC exp = 5

VB = Volume of binder, expressed as %

20

Summary: Austroads Characterisation of AC E-range Poisson’ Poisson’s Ratio Isotropic Performance relationship

NL (CT )

1500 -> ν = 0.4 EV = EH

 K AC  = RF   µε 

5

Austroads Material Types • Unbound Granular

(UB Gr)

• Bound Materials - Cement bound - Bitumen bound

(CT) (AC)

(SG)

21

Subgrade (SG) • Modelled as an unbound material • Anisotropic

EV = 2EH

• Poisson’s ratio: - Cohesive soils - Granular soils

ν = 0.45 ν = 0.35

Determination of Modulus (ESG) 1. RLT test 2. Presumptive value based on CBR • E = 10CBR • This is at best a crude approximation • Evidence indicates for many NZ subgrade soils the factor 10 ranges from about 4 to 13, with most on the lower end of the range

• Upper limit for ESG is 150 MPa

22

Determination of Modulus (ESG) ESG can be improved by stabilisation Restrict binder application rates to “modification” Modification can achieve an improvement factor of up to 3

Design Modulus Use 10-%tile value EDESIGN = EMEAN - 1.28(std. dev.)

23

Self Evaluation Exercise 3

SE Exercise 3 (solution) Mean CBR = 6.26 Design CBR = 6.26 – 1.28(1.72) = 4.06 say 4 Edesign = 10(CBR) = 40 MPa

24

Performance Relationship NL ( SG )

exp  9300 K SG  7  =  με  

N = Number of load repetitions to failure ε = critical vertical compressive strain in the top of the subgrade

KSG = 9300

exp= 7

Comparison with “old” old” TNZ strain criterion NL( SG)

 9300   =   με 

7

NL ( SG)

 21000   =   με 

4.348

25

• Isotropic (E V=EH) • Poisson’s ratio ν=0.2 • No sublayering

Asphaltic Concrete

Stiffness affected by: • Temperature • Volume of bitumen & air voids • Aggregate content & properties • Bitumen properties • Age of mix • Rate of loading • Important input parameter • Design Moisture Content • Determine representative SG modulus statistically: ESG = Emean – 1.3(std. dev)

• Isotropic (E V=EH) • Poisson’s ratio ν=0.4 • No sublayering

Elastic Characterisation • Anisotropic (E V=2E H) • Poisson’s ratio ν=0.35 • Sublayered E Shear modulus: f = v 1+ ν (h/125) E BT ≤ E SG x 2

Performance Criterion

Modulus: Values & Determination • Currently no performance Presumptive Values: • Basecourse: 400-500MPa criterion. • Subbase: 150-350MPa • Rely on sound material Determination: and construction • Lab.: RLT test specifications. • Back-analysis • Loadman • E=10CBR 12 Presumptive values: K  N = RF CT  • Base quality (4-5%): 5000MPa  µε  • Subbase “ (4-5%): 2000MPa 113000 Determination: K CT = 0.804 + 191 • Lab.: Flexural test E • Back-analysis • UCS⇒E correlation 5 Presumptive values: K  N = RF AC  • 1500 MPa ⇒  µε  Determination: 6918(0.856VB + 1.08) • Lab: Flexural test K AC = 0 .36 • Back-analysis Smix • Shell nomographs V expressed as a % B

• Anisotropic (E V=2E H) • Poisson’s ratio: Cohesive (clay):ν=0.45 Non-Cohesive: ν=0.35

7

 9300  N =    µε 

• E=10CBR (≤150MPa) Determination: • Lab.: RLT test • Back-analysis (FW D, B/B) • Field (Loadman, Scala, DCP)

Self Evaluation Exercise 4 Answers: 4.1 (a) KCT = 400 4.1 (b) FAC = 4568 4.2

µεAC = 288 µεCT = 126.5

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