Topic 3 - Flexible Pavement Stress Analysis

Flexible Pavement Stress Analysis

Dr. Christos Drakos

University of Florida

Topic 3 – Flexible Pavement Stress Analysis

Need to predict & understand stress/strain distribution within the pavement structure as they (σ & ε) relate to failure (cracking & rutting)Numerical Models• Need model to compute deflections (δ) and strains (ε)• Numerous models available with different:

– Capabilities– Underlying assumptions– Complexity– Material information requirements

IDEAL MODEL

What would be an ideal model?

Predicts Input Parameters• Stresses• Strains

• Static & dynamic loads• Material properties• Traffic• Environment

No current model meets these requirements!

However, can obtain reasonable estimates!


1. Available Models

Most widely used• Reasonable Results• Properties Relatively Simple to Obtain

• Multilayer Elastic Theory• Finite Element Methods• Viscoelastic Theory (time and temp.-dependent behavior)• Dynamic Analysis (inertial effects)• Thermal Models (temperature change)

How do we get E? Before & after construction

E & ν Before: lab testing (MR)After: field testing (FWD)

Falling Weight Deflectometer

• Small trailer• Dropping Weight• Geophones• Deflection Basin

Uses elastic theory to predict the deflection basin for the given load. Then iterates with different moduli configurations until the calculated deflection basin matches the measured.



2. Multilayer Elastic Theory

E1, ν1

E2, ν2

E3, ν3

∞

z1

z2

z3

a = radius

q = pressure

Point APoint B

Assumptions (p. 60):• Each Layer

– Continuous– Homogeneous– Isotropic– Linearly Elastic– Material is weightless & infinite in areal extend– Finite thickness (except last layer)

Properties @ A = Properties @ B

Same properties in all directions

Hooke’s Law

( )( )trzz σσνσE1

ε +−=


2. Multilayer Elastic Theory (cont.)

Assumptions (cont.):• Surface stresses

– Circular– Vertical– Uniformly distributed

• Full friction between layers• Each layer continuously supported

Point APoint B

E1, ν1

E2, ν2

E3, ν3

∞

z1

z2

z3

a = radius

q = pressure

Why do we want full friction between layers?

2inlbs

psi =

610−×==inin

nmicrostraiµε

Units Guidelines

• Stress:– Reported in psi:

• Strain:– Reported in µε:

• Deflections:– Reported in mils:

1000in

mils =

For homework, exams, and projects, you are expected to convert all of your answers to these units.



3. One-layer System3.1 Based on Boussinesq (1885)

Half-space: infinite area & depth

Z

σz

σz

X

P

r

zσz

225

2z z

P

zr1

12π3

σ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

=

Point load on an elastic half-space• Examine σ distribution along Z & X

Where:– σz = Vertical stress– r = Radial distance from load– z = Depth– P = Point load

Notice that the stress distribution is independent of E


3.2 One-layer Solutions (Foster & Ahlvin)

Figures 2.2 – 2.6Developed charts to determine σz, σt, σr, τrz & w (ν=0.5)

• Axisymmetric loading:– σz = Vertical stress– σr = Radial stress– σt = Tangential stress– τrz = Shear stress– w = Deflection

• Pre-solved @ radial distances

2a

q

z

r

σz

σr σt

τrz

aq

02a 1a3a

0

2a

1a

Offset

Dep

th


3.2 One-layer Solutions (Foster & Ahlvin)Charts follow similar outline

Depth (z) and offset (r) are expressed in radial ratios


3.2.1 Vertical StressGiven:– Load, P = 9000 lbs– Pressure, q = 80 psi

2a9000

AP

q×π

== in680

9000a ≅

×π=

aq

r=6”

z=6”σzFind:– Vertical Stress, σz @ z=6” & r=6”

First, we need to calculate the radius:

z/a = 6/6 =1

r/a = 6/6 =1Figure 2.2 (vertical stress distribution)


3.2.1 Vertical Stress (cont)z/a = 6/6 =1r/a = 6/6 =1

33%100q

z ≅×σ psi4.26

1008033

z =×

=σ


Deflection Profile

3.2.2 Deflection Flexible Plate Rigid Plate

q qRubber Steel

Ground ReactionWhich deflection is higher?

Eqa5.1

W0 =

FlexibleRigid W%79W ⋅≅

Eqa18.1

W0 =

( )E

qa12W

2

0ν−

=( )

E2qa1

W2

0ν−π

=


3.2.2 Deflection (cont.)

a = 6”

q = 80 psi

∞

h1= 4”

h2= 8”

h3= 12”

Pavement Structure

How can we use one-layer theory to estimate the deflection of the system?

We can assume the pavement structure to be incompressible

Basically:

A

Asurface δδ ≡

For this case (assuming one-layer):

FE

aqA ×

×=δ Get F from Fig 2.6



Given:z/a=24/6=4r/a=0

Find:F=0.37



071.037.02500

680=

×=w 0071.037.0

25000680

=×

=w

w=71 mils (High) w=7.1 mils (Low)

a = 6”

q = 80 psi

∞

h1= 4”

h2= 8”

h3= 12”

A

• Examine two cases:

Clay Dense SandE=2,500 E=25,000

Subgrade quality is very important in pavement design


• Purpose of the pavement structure:– Protect the subgrade; reduce stresses to a tolerable level to prevent

excessive settlement or collapse4.1 Vertical Stress• Vertical stress on top of subgrade; important in pvt design as

it accounts for permanent deformation (rutting)• Allowable σz depends on E of the subgrade material

4. Stresses & Strains for Design

– To combine the effect of stress (σ) and stiffness (E)

– Effect of horizontal stress is relatively small; vertical strain caused primarily by vertical stress

Vertical compressive strain (εc) used as a design criterion

( )( )trzz σσνσE1

ε +−=Eσz≅ εc

aq

∞

h1

h2

E1

E2

E3


4.2 Tensile Strain

Horizontal ‘principal’ strain (εt) used as a design criterion.

aq

∞

h1

h2

E1

E2

E3

ε

• Tensile strain at the bottom of AC layer; used in pvt design as the fatigue cracking criterion

• Two types of strain:– Overall minor principal strain, ε3– Horizontal ‘principal’ strain, εt (not an actual principal strain)


4.2.1 Overall Principal Strains• Based on all 6 components of normal and shear stresses – σx, σy, σz, τxy, τxz, τyz− Solve cubic equation to get σ1, σ2, & σ3

− Then calculate principal strains ( )( )2133 σσνσE1

ε +−=

Minor principal strain (ε3) considered to be tensile strain because tension is negative

aq

ACε3

What is the orientation of ε3?

Minor principal strain (ε3) does not always act on the horizontal planeε

3


4.2.1 Horizontal ‘Principal’ Strain• Based on the horizontal normal and shear stresses only – σx, σy, τxy

• Horizontal ‘principal’ strain (εt) is slightly lower than the minor principal strain (ε3)–

• Maximum horizontal strain on the X-Y plane• Always acts on the horizontal plane• Used by the program KENLAYER to predict fatigue failure

aq

ACεt

2

x y x y 2t xy2 2

ε + ε ε − ε⎛ ⎞ε = − + γ⎜ ⎟

⎝ ⎠

tε≥ε3


Developed solutions for:• Vertical deflections (flexible & rigid)• Vertical stresses (limited # of cases)

− σ & δ highly dependent on stiffness ratio E1/E2

Notice the importance of stiffness ratio in reducing stresses.

5. Two-layer Theory (Burmister)


5.1 Two-Layer Deflections• In one-layer theory we assumed that all layers could be

represented as one– δsurface = δtop of the subgrade

• For two-layer theory we have:– Vertical Surface Deflection– Vertical Interface Deflection

a

q

∞

h1 E1

E2

• Flexible

• Rigid

22

max 5.1 FEqa

W =

22

max 18.1 FEqa

W =

5.1.1 Surface Deflections

Why use E2 for surface deflection?• E2 accounts for most of the deflection (see following example)• F2 takes into account the stiffness ratio


5.1.2 Surface Deflections Examplea=6”

q=80 psi

∞

6”E1=50,000 psi

E2=10,000 psi

Given:h1/a=6/6=1E1/E2=5Find:F2=0.6

milsW

FEqa

W

⋅≅=

==

43"0432.0

6.010000

)80(65.15.1

max

22

max


5.1.3 Interface Deflections ExampleF

h 1/a

Offset

a=6”

q=80 psi

∞

6”E1=50,000 psi

E2=10,000 psi

Given:h1/a=6/6=1 ;r/a=0E1/E2=5

Find:F=0.83

milsW

FEqa

W

⋅≅=

==

40"0398.0

83.010000

)80(6

2

• For the same example as above


5.1.4 Surface Vs Interface Deflections

Compare the results from the example:• Surface deflection = 43 mils• Interface deflection = 40 mils Top layer compression = 3 mils

%7100433

≅×=– Top Layer

%931004340

≅×=– Subgrade Layer

Compression percentages:


5.2 Two-Layer Vertical Stressa=6”

q=80 psi

∞

h1E1=500,000 psi

E2=5,000 psi

What thickness do we have to use to protect the subgrade?

Maximum allowable σc for clay = 8 psi

Given:σc/q=0.1E1/E2=100 Fig 2.15

Find:a/h1=1.15

"2.515.16

h1 ==


5.2 Critical Tensile Straina=6”

q=80 psi

∞

6”E1=200,000 psi

E2=10,000 psie = εt= critical tensile strainGiven:E1/E2=20h1/a=1 Fig 2.21

Find:Fe=1.2

t e

t

q 80F 1.2

E 200000in

0.00048 480in

ε = =

ε = = µε

εt

Stra

in F

acto

r, F

e


6. Failure Criteria6.1 Fatigue Cracking Model• Based on Miner’s cumulative damage concept

– Amount of damage expressed as a damage ratio predicted/allowableload repetitions

( ) ( )2 3f f

f 1 t 1N f E− −

= ε

( ) 5f

d 4 cN f−

= ε

f1 =Laboratory to field shift factor

f2 & f3 =Determined from fatigue tests on lab specimens

6.2 Rutting Model• Allowable number of load repetitions related to εc on top of

the subgrade– Does not account for failure in other layers

f4 & f5= Predicted performance to field observation shift factors

( ) 4.4779d cN 1.365 10

−−= × ε

( ) ( )3.291 0.854

f t 1N 0.0796 E− −

= ε


7. Sensitivity Analysis• Sensitivity analyses illustrate the effect of various parameters

on pavement responses• Variables to be considered:

– Layer thicknesses h1 & h2– Layer moduli E1, E2, & E3


7.1 Effect of HMA Thickness

Tensile Strain (εt)• Critical thickness where εt is max• Above hcr, increasing h1 effectively

reduces εt

hcr

Compressive Strain (εc)• Increasing h1 effectively reduces εc

when base is thin


7.2 Effect of Base Thickness

Tensile Strain (εt)• Increase in h2 does not

significantly decrease εt especially when h1 is large

Compressive Strain (εc)• Significant decrease of εc when h1

is low


7.3 Effect of Base Modulus

Tensile Strain (εt)• Increase in E2 significantly

decreases εt when E1 is low• Limits bending

Compressive Strain (εc)• Small decrease of εc when E1 is low


7.4 Effect of Subgrade Modulus

Tensile Strain (εt)• Minimal effect on εt

Compressive Strain (εc)• As expected, E3 has great effect on εc independent of what E1 might be


8. Computer Program KENLAYERProgram should be on a disk at the back of your textbook

8.1 System• Elastic multi-layer analysis system• Elastic theory assumptions apply

– Load Circular uniformly distributed


8.2 LoadsCircular, uniform pressurePARAMETER ACTUAL LOAD

LOAD=0 Single wheel

LOAD=1 Dual wheelX

Y X – Longitudinal (direction of traffic)Y – Transverse

LOAD=2 Dual tandemX

Y

Yw

Yw

Xw


8.3 Material Properties• Material types

– 1 = Linear elastic– 2 = Nonlinear elastic– 3 = Linear viscoelastic– 4 = Combination of 2 & 3

t

ε

ε

σ1

23

8.4 Input/Output• Program LAYERINP creates the input file• Run KENLAYER to perform the analysis• Default name for the output file is LAYER.TXT


8.5 KENLAYER Example 1a = 4.5”

q = 100 psi

E3=10,000 psi ; ν3=0.5

E1=500,000 psi ; ν1=0.4

E2=50,000 psi ; ν2=0.5

∞

h1= 6”

h2= 12”

Given:• Three-layer system• Uniform circular load• Elastic material

Calculate:• Maximum deflection• Critical tensile strain• Critical compressive strain

Where would the critical/maximum values occur?– Maximum deflection δmax @ z=0– Critical tensile strain εt @ bottom of AC layer– Critical compressive strain εc @ top of subgrade


8.5 KENLAYER Example 1 (cont.)Procedure:• Create input file

– LAYERINP.EΧΕ• Run the analysis

– KENLAYER.EXE• Retrieve the output

– LAYER.TXT

Output format:• Single wheel load is analyzed in axisymmetric space• Sign convention:

– Positive (+) = Compression– Negative (-) = Tension

Is there a way to find out?


Given:• Three-layer system• Dual wheel load• Elastic material

∞

8”

4”

14” 4”q=100 psia=4 in

E1=500,000 psiν1 =0.4

E2=15,000 psiν2 =0.5

E3=5,000 psiν3 =0.5

Plane of Symmetry

x

xx

x

x

xx

x

x x x

xCheck output

8.6 KENLAYER Example 2

Calculate:1. δmax2. εt3. εc

Where would the critical/maximum values occur?


8.6 KENLAYER Example 2 (cont.)Procedure:• Create input file

– LAYERINP.EΧΕ• Run the analysis

– KENLAYER.EXE• Retrieve the output

– LAYER.TXT

Output format:• Dual wheel load is analyzed in spatial coordinates• Sign convention remains the same:

– Positive (+) = Compression– Negative (-) = Tension


8.6 KENLAYER Example 2 (cont.)

Which strain considered critical for cracking & rutting?

Output format:• Results for each point (X,Y) at each requested depth (z)• Principal stresses and strains

δ σz σ1 σ2 σ3 εz ε1 ε3 εh

τ

σσ1σ3 σ2

Principal Stresses act onplanes where τ = 0

τmax2

31max

σστ −=


Output @ Location (0,7,12.05)

σ1 = 6.72 psiσ2 = 2.04 psiσ3 = 1.47 psi

σ1

σ3

σ2

( )( )

( )( ) µε1.472.040.46.725000

1ε

σσ0.4σE1

ε

z

yxzz

⋅=+−=

+−=

993

Can we use the principal stresses to calculate vertical strain?

= 992.2 µε (output)

8.6 KENLAYER Example 2 (cont.)


8.7 KENOUTKenout is a data manipulation program geared to help you post process your dataProcedure:• Rename the KENLAYER output file (LAYER.TXT) to something

relevant to your problem (i.e. Example2)• Run the Kenout.exe program• The program then asks for the filename to be read (Example2

– no .txt extension needed)• Then it prompts you to give a new filename to store the

reduced data (i.e. Ex2 – again, no extension needed)

Output format:• Original file – Example2• Kenout – KENOUT.EXE• Modified file – Ex2

Topic 3 - Flexible Pavement Stress Analysis

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