Topic 5 - Material Characterization

Topic 5 – Material Characterization

Dr. Christos Drakos

University of Florida


1. Introduction

What material properties have we used up to now?

• Most paving materials are not elastic, but experience some permanent deformation after each load application

• If the load is small compared to the strength of the material and is repeated for a large number of times, the deformation is almost recoverable and can be considered elastic

Pavement design inputs covered in previous topics:• Predicted stresses and strains – load magnitude • Traffic – load cycles


Resilient Modulus (MR)

Time

Load

Time

Def

orm

atio

n

εr

εpr

dRM

εσ

=

Type and duration of loading is supposed to simulate that occurring in the field


1. Surface Course• Layer that comes to contact with traffic; normally contains the highest

quality material2. Base

• Provides additional load distribution and contributes to drainage and frost resistance

3. Subbase (optional)• Functions primarily as structural support but it can also help with

drainage and frost action4. Subgrade

• Native soil – the only non-engineered material in the pvt structure

Typical Flexible Pavement Structure

SUBBASE COURSE (Optional)

BASE COURSE

SUBGRADE (SOIL)

AC Decreasing Stiffness


Subgrade performance generally depends on:1. Load bearing capacity; subgrade must be able to support

loads passed on from the pavement structure2. Moisture content; it tends to affect a number of subgrade

properties including load bearing capacity3. Shrinkage and/or swelling; some soils shrink or swell

depending upon their moisture contentRemedies for poor subgrade conditions:1. Remove and replace; subgrade soil can be removed and

replaced with higher quality fill2. Stabilization; adding an appropriate binder – such as lime,

portland cement or emulsified asphalt – can increase subgrade stiffness and/or reduce swelling tendencies

3. Additional base layers; include a subbase, or increase thickness of base


Subgrade Seasonal Variations

Normal MR

50,000 psiFrozen MR

Thaw MR

FreezeTime

ThawTime

RecoveryTime

NormalTime

Total Time = 12 Months

MR

Time


2. Design Resilient Modulus (Base & Subgrade)

2.1 Correlations

Maybe there is information already available

Also:• MR=1500(CBR)• MR=1155+555(R)


• Basically a penetration test• Piston penetrates soil at constant rate – 0.05 in/min• Pressure is recorded• Take the ratio to the bearing capacity of a standard rock• Range: 0 (worst) – 100 (best)

2.1.1 California Bearing Ratio (CBR)

CBR =Pressure to cause 0.1” penetration to the sample

Pressure to cause 0.1” penetration for standard rock



PistonDeflection dial (loading)

Deflection dial (swelling)

Proctor

6” Ø

Sample 4.5”

Surcharge (confinement)


5 or lessOH

15 or lessCH LL > 50%

10 or lessMH

5 or lessOL

15 or lessCL LL < 50%

15 or lessML

Fine-grained soils

5 - 20SC

10 - 40SM

10 - 40SP

20 - 40SW

20 - 40GC

20 - 60GM

30 - 60GP

40 - 80GW

Coarse-grained soils

CBR RangeUSC Soil TypeGeneral Soil Type



• Resistance value of a soil determined by stabilometer• Closed-system Triaxial test• Apply vertical pressure• Measure horizontal pressure induced in the fluid

2.1.2 Stabilometer (R-value)

115.2100100

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−×⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

h

v

pp

D

R

pv & ph = Vertical and horizontal pressure respectivelyD2 = displacement of stabilometer fluid to increase ph from 5 to

100 psi, measured in revolutions of a calibrated pump handle


2.1.2 Stabilometer (R-value)

Pressure Gauge TestingHead

BottomPlunger

SampleFluid under pressure

160 psi

160 psi

Apply the vertical pressure and then measure the resulting horizontal pressure

115.2100100

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−×⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

h

v

pp

D

R


2.2 In-situ Testing (Plate Loading Test)

• Circular plate 30 in Ø; series of plates used to minimize bending

• Apply load at constant rate to reach 10 psi • Pressure held constant until the deflection increases no more

than 0.001 in/min for three consecutive minutes• Use average dial reading to determine the deflection

∆=

pk

k = Modulus of subgrade reactionp = pressure on the plate (10 psi)∆ = deflection of the plate


2.2 In-situ Testing (Plate Loading Test)

Reaction (Steel Beam)

∆

Pressure GaugeDeflection Dial @ 1/3 Points

Hydraulic Jack

∆=

pk

Figure 7.36 – Correlation of k with MR


2.3 Laboratory Testing (Triaxial Test)


2.3 Laboratory Testing (Triaxial Test)

Sample

= Confining Pressure (σ2,σ3)= Deviator Stress (σd)

σ1

σ2

σ3

In Triaxial cell (cylinder): σ2=σ3

Deviator Stress:Axial stress in excess of the confining pressure in Triaxial cell

31 σσσ −=d

What can affect the results?

Level of confinement

State of confinement defined by the first invariant

321 σσσθ ++=

σ3

σdσ1


• For Granular subgrade:– MR = function (confinement)– k1 & k2 experimentally determined values

2.3.1 Triaxial Test – Granular Material

Run Triaxial test at certain levels of confining pressure and vary the deviator stress. (Example 7.2 Huang)

log θ, psi

log

MR,

psi

k2

xk1

21

kR kM θ×=


2.3.1 Triaxial Test – Granular MaterialSample Conditioning1. Set the confining pressure to 5 psi, and apply a deviator

stress of 5 psi, and then 10 psi, each for 200 repetitions2. Set the confining pressure to 10 psi, and apply a deviator

stress of 10 psi, and then 15 psi, each for 200 repetitions3. Set the confining pressure to 15 psi, and apply a deviator

stress of 15 psi, and then 20 psi, each for 200 repetitions

Resilient Modulus TestAfter sample conditioning the resilient modulus test follows a constant confining pressure-increasing deviator sequence, and the results are recorded at the 200th repetition of each deviator stress


2.3.1 Triaxial Test – Granular MaterialTest Procedure1. Set the confining pressure to 20 psi, and apply a deviator

stress of 1, 2, 5, 10, 15 and 20 psi2. Reduce the confining pressure to 15 psi, and apply a

deviator stress of 1, 2, 5, 10, 15 and 20 psi3. Reduce the confining pressure to 10 psi, and apply a

deviator stress of 1, 2, 5, 10, and 15 psi4. Reduce the confining pressure to 5 psi, and apply a deviator

stress of 1, 2, 5, 10, and 15 psi5. Reduce the confining pressure to 1 psi, and apply a deviator

stress of 1, 2, 5, 7.5, and 10 psi; stop the test after 200 repetitions of the last deviator stress level or when the specimen fails



(psi) (psi) (mils) (µε) (psi) (psi)1 0.264 66 15151.5 612 0.496 124 16129.0 625 1.184 296 16891.9 65

10 2.284 571 17513.1 7015 3.428 857 17502.9 7520 4.420 1105 18099.5 801 0.260 65 15384.6 462 0.512 128 15625.0 475 1.300 325 15384.6 50

10 2.500 625 16000.0 5515 3.636 909 16501.7 6020 4.572 1143 17497.8 651 0.324 81 12345.7 312 0.672 168 11904.8 325 1.740 435 11494.3 35

10 3.636 909 11001.1 4015 3.872 968 15495.9 451 0.508 127 7874.0 162 0.988 247 8097.2 175 2.224 556 8992.8 20

10 3.884 971 10298.7 2515 5.768 1442 10402.2 301 0.636 159 6289.3 42 0.880 220 9090.9 55 2.704 676 7396.4 8

7.5 3.260 815 9202.5 10.510 4.440 1110 9009.0 13

1

20

15

10

5

Resilient Modulus, MR

Stress Invariant, θ

Confining Pressure, σ3

Deviator Stress, σd

Recoverable Deformation

Recoverable Strain, εr

(4) Distance GagenDeformatioeRecoverabl=εr

rd

εσ

=RM

3d 3σ+σ=ϑ



y = 3686x0.3511

1000

10000

100000

1 10 100

Stress Invariant, θ (psi)

Res

ilien

t Mod

ulus

, MR

(psi)

351.0R 3686M θ×=


2.3.2 Triaxial Test – Cohesive Material

Run Triaxial tests at certain values of deviator stress and varythe confining pressure. (Example 7.3 Huang)

σd, psi

MR,

psi

k2

k1

k3

k4

)( 231 dR kkkM σ−+=

)( 241 kkkM dR −+= σ2kd <∀σ

2kd >∀σ

• For Fine-grained (cohesive) subgrade:– MR = function (deviator stress, σd)– k1, k2, k3 & K4 experimentally determined values


• To determine subgrade MR, Asphalt Institute suggests:– Confining stress σ3 = σ2 = 2 psi– Deviator stress σd = σ1 - σ3 = 6 psi

• For granular material:– For example use data from Fig 7.8

• For fine-grained material:– For example use data from Fig 7.9

2.3.3 Asphalt Institute Subgrade Characterization


5.9449123960 35.0

12

=⋅=

⋅= kR kM θ

2.3.4 Granular material example

k1 = 3960

k2 = 0.35


2.3.5 Fine-grained material example

2kd >σ

5910)2.56(3885600)( 241

=−+=−+= kkkM dR σ


3. Hot-Mix Asphalt3.1 Structural Layer Coefficient• Used in AASHTO design procedure• Describes the quality of the material• Function of MR & position in the pavement

3.2 Marshall Test• Used in the Marshall Mix Design procedure• Performed on cylindrical specimens• Measure Stability (fail load) and Flow (deformation)

3.3 Cohesiometer Test• Used to measure the cohesion of cemented materials • Apply load at controlled rate until failure

Why?


3. Hot-Mix Asphalt (cont.)


4. Bases4.1 Untreated Granular Base


4.2 Stabilized Granular Base

Bituminous Treated Cement Treated


5. Sub-Bases


6. DrainageImportant to keep water away from the pavement structure

6.1 Detrimental effects of water:• Reduces strength of unbound material & subgrade• Causes pumping, faulting, cracking & shoulder deterioration• Pore-pressure increase Pumping of fines Loss of

support– Load goes over saturated base & soil– Pore-pressure increases– Water is incompressible; moves up

• Causes heaving – swelling of soils• Pore-pressure within the AC layer causes stripping, durability

cracking• Frost action (in Northern climates)


6.2 Sources of water:i. Seepageii. Raise of water tableiii. Infiltrationiv. From proximity of water table

– Capillary moisture held in the pores from surface tensionv. Vapor movement

– Movement of water associated with fluctuating temperature and pressure


6.2 Sources of water:


6.3 Protecting the Pavement Structure:Need to minimize availability of water

• Impervious surface/shoulders. How? (sealants)• Drainage to remove water quickly (Ditch)• Drainage layer for subsurface water

Design phase / Construction

Three drainage installations for subsurface water:1. Drainage layer or blanket2. Longitudinal drain3. Transverse drain

Can be performed after construction


6.4 Drainage Deficiencies for Pavements with Ditch6.4.1 Typical Pavement with a Ditch


6.4.2 Water infiltrating from rutted shoulder

6.4.3 Water infiltrating due to debris caused ponding


6.4.4 Water infiltrating due to differential settlement


6.5 Subsurface Drainage

Base Course as drainage layer

6.5.1 Drainage layer / Blanket


6.5.2 Longitudinal drainage


6.5.3 Transverse drainageSame concept as longitudinal drainage, but in this case the drainage runs across the lanes

Where would we use the transverse drain?

Topic 5 - Material Characterization

Documents

plate loading

triaxial test

triaxial cell

3 laboratory

situ testing

deviator stress

confining

deflection