Flexible Pavement Stress Analysis - FIT Staffwebstaff.fit.ac.cy/eng.ma/acet406notes/Topic 3 -...

Flexible Pavement Stress Analysis

Dr. Antonis Michael

Frederick University

Notes Courtesy of 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–––

IDEAL MODEL

Predicts Input Parameters

•• Strains

••••


1. Available Models

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

• Multilayer Elastic Theory• Finite Element Methods• Viscoelastic Theory ( • Dynamic Analysis (inertial effects)• Thermal Models (temperature change)

Falling Weight Deflectometer

• Small trailer• Dropping Weight• Geophones• Deflection Basin

Uses elastic theory to predict the deflection basin for the given load. Then



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––– Linearly – Material is – Finite thickness (except last layer)


2. Multilayer Elastic Theory (cont.)

Assumptions (cont.):• Surface stresses

– Circular– Vertical– Uniformly distributed

• Full • 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?

2in

lbspsi =

610−×==in

innmicrostraiµε

Units Guidelines

• Stress:– Reported in psi:

• Strain:– Reported in µε:

• Deflections:– Reported in mils:

1000

inmils =

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



3. One-layer System

3.1 Based on Boussinesq (1885)Half-space:

Z

σz

σz

X

P

r

zσz

2

2

52

zz

P

z

r1

1

2π

3σ

+

=

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

Where:– σz = – r = – z = – P =


3.2 One-layer Solutions (Foster & Ahlvin)

Figures 2.2 – 2.6

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

• Axisymmetric loading:

– σz =

– σr =

– σt =

– τrz =

– w =

•

2a

q

z

r

σz

σr σt

τrz

a

q

02a 1a3a

0

2a

1a

Offset

Depth


3.2 One-layer Solutions (Foster & Ahlvin)

Charts follow similar outline

Depth (z) and offset (r) are


3.2.1 Vertical Stress

Given:– Load, P = 9000 lbs– Pressure, q = 80 psi

a

q

r=6”

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

First, we need to


3.2.1 Vertical Stress (cont)

z/a = 6/6 =1r/a = 6/6 =1


Deflection Profile

3.2.2 Deflection

Flexible Plate Rigid Plate

q qRubber Steel

Ground Reaction


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

Basically:

A

For this case (assuming



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



a = 6”

q = 80 psi

∞

h1= 4”

h2= 8”

h3= 12”

A

• Examine two cases:

Clay Dense Sand

E=2,500 E=25,000

Subgrade quality


• Purpose of the pavement structure:– Protect the subgrade; reduce

4.1 Vertical Stress

• Vertical stress on top of subgrade; important in pvt design as it accounts

• Allowable σz depends on

4. Stresses & Strains for Design

– To combine

– Effect of

Vertical compressive

εc

a

q

∞

h1

h2

E1

E2

E3


4.2 Tensile Strain

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

a

q

∞

h1

h2

E1

E2

E3

ε

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

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


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 σσνσE

1ε +−=

Minor principal strain (ε3) considered to be

a

q

AC Minor principal strain (ε3)


4.2.1 Horizontal ‘Principal’ Strain

• Based on the horizontal normal and shear stresses only – σx, σy, τxy

• Horizontal ‘principal’ strain (εt) is

–• Maximum • Always acts on the horizontal plane• Used by the program KENLAYER to predict fatigue failure

a

q

AC


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

− σ & δ highly dependent on

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 – Vertical

a

q

∞

h1 E1

E2

• Flexible

• Rigid

5.1.1 Surface Deflections


5.1.2 Surface Deflections Example

a=6”

q=80 psi

∞

6”E1=50,000 psi

E2=10,000 psi

Given:h1/a=6/6=1E1/E2=5


5.1.3 Interface Deflections ExampleF

h1/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

• 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


5.2 Two-Layer Vertical Stress

a=6”

q=80 psi

∞

h1E1=500,000 psi

E2=5,000 psi

Maximum allowable σc for clay = 8 psi

Given:σc/q=0.1E1/E2=100


5.2 Critical Tensile Strain

a=6”

q=80 psi

∞

6”E1=200,000 psi

E2=10,000 psi

e = εt= critical tensile strainGiven:E1/E2=20h1/a=1

εt

Str

ain

Fact

or,

Fe


6. Failure Criteria

6.1 Fatigue Cracking Model

• Based on Miner’s cumulative damage concept– Amount of damage expressed as a damage ratio predicted/allowable

load repetitions

( ) ( )2 3f f

f 1 t 1N f E− −= ε

( ) 5f

d 4 cN f−= ε

f1 =

f2 & f3 =

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=

( ) 4.4779

d 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 parameterson pavement responses

• Variables to be considered:– Layer – Layer


7.1 Effect of HMA Thickness

Tensile Strain (εt)••

Compressive Strain (εc)•


7.2 Effect of Base Thickness

Tensile Strain (εt)•



7.3 Effect of Base Modulus


•



7.4 Effect of Subgrade Modulus



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