Pavement Design

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Flexible Pavement Stress Analysis

University of Florida

Topic 2 – Flexible Pavement Stress Analysis

Need to predict & understand stress/strain distribution within the pavement structure as they (σ & ε) relate to failure (mainly 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!

Topic 2 – Flexible Pavement Stress Analysis

1. Available Models

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

• Multilayer Elastic Theory (MLET) • 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 & Before: lab testing (MR) After: field testing (FWD)



Units Guidelines

• Stress: – Reported in psi:

• Strain: – Reported in με:

• Deflections: – Reported in mils:

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

Topic 2 – Flexible Pavement Stress Analysis

�� = ��

�� = 10�

�

��

�� = �� = 10��

�� = ��

1000

��

Topic 2 – Flexible Pavement Stress Analysis

2. One-layer System

2.1 Based on Boussinesq (1885) Half-space: infinite area & depth

σz

zσz

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 material stiffness (E)

�� = 3

2� �

1 + � �

� � �

� �

��

What does a load from a tire look like?

Topic 2 – Flexible Pavement Stress Analysis

Contact area

What shape do we use for stress analysis in MLET?

Topic 2 – Flexible Pavement Stress Analysis

Why a circle?

What shape do we use for stress analysis in MLET?

Topic 2 – Flexible Pavement Stress Analysis

• Where:

– q = intensity or uniform pressure

– a = radius

Topic 2 – Flexible Pavement Stress Analysis

2.2 One-layer Solutions (Foster & Ahlvin)

Figures 2.2 – 2.6

Developed charts to determine σz, σt, σr, τrz & w (=0.5) • Axisymmetric loading:

– σz = Vertical stress

– σr = Radial stress

– σt = Tangential stress

– τrz = Shear stress

– w = Deflection

• Pre-solved @ radial distances

02a 1a3a

Offset

D e p th

σz

σr σt

τrz

Topic 2 – Flexible Pavement Stress Analysis

2.2 One-layer Solutions (Foster & Ahlvin)

Charts follow similar outline

Stress (z) is expressed in pressure percentage

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

Topic 2 – Flexible Pavement Stress Analysis

2.2 One-layer Solutions (Foster & Ahlvin)

Charts follow similar outline

Stress (rz) is expressed in pressure percentage

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

Topic 2 – Flexible Pavement Stress Analysis

2.2.1 Vertical Stress

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

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 =1 Figure 2.2 (vertical stress distribution)

� = �

� =

9000

� � �� =

9000

� � 80 ≅ 6 ��

Topic 2 – Flexible Pavement Stress Analysis

2.2.1 Vertical Stress (cont.)

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

��

� 100% ≅ 33 �� = 33 � 80

100 = 26.4 ��

Topic 2 – Flexible Pavement Stress Analysis

2.2.2 Shear Stress

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

Find: –Shear Stress, rz @ z=6” & r=6”

First, we need to calculate the radius:

z/a = 6/6 =1

r/a = 6/6 =1 Figure 2.5 (shear stress distribution)

r=6”

z=6”rz

� = �

� =

9000

� � �� =

9000

� � 80 ≅ 6 ��

Topic 2 – Flexible Pavement Stress Analysis

2.2.2 Shear Stress (cont.)

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

��

� 100% ≅ 18 �� = 18 � 80

100 = 14.4 ��

Topic 2 – Flexible Pavement Stress Analysis

Deflection Profile

2.2.3 Deflection

Flexible Plate Rigid Plate

q qRubber Steel

Ground Reaction

Which max. deflection is higher?

�� ≅ 79% � ��

�� = 2 1 − ν� ��

�

�� = 1.5��

�

�� = �

1 − ν� ��

�

�� = 1.18��

� (ν=0.5) (ν=0.5)

Topic 2 – Flexible Pavement Stress Analysis

2.2.3 Deflection (cont.)

a = 6”

q = 80 psi

∞

h= 24” Pavement Structure

We assume the compressibility of pavement structure to be negligible

Basically:

For one-layer theory:

Get F from Fig 2.6

Subgrade

�� = � � �

��

�� ≡ ��

Topic 2 – Flexible Pavement Stress Analysis

2.2.3 Deflection (cont.)

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

Find: F=0.37

Topic 2 – Flexible Pavement Stress Analysis

3. Multilayer Elastic Theory

E1, ν1

E2, ν2

E3, ν3

∞

a = radius

q = pressure

Point A Point B

Assumptions (p. 60): • Each Layer

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

Properties @ A = Properties @ B

Same properties in all directions

Hooke’s Law

�� = 1

� �� − ν �� + ��

Topic 2 – Flexible Pavement Stress Analysis

3. Multilayer Elastic Theory (cont.)

Assumptions (cont.): • Load

– Circular – Vertical – Uniformly distributed

• Full friction between layers – Same z, rz, w, ur @ interface

• Each layer continuously supported

Point A Point B

E1, ν1

E2, ν2

E3, ν3

∞

a = radius

q = pressure

Why do we want full friction between layers?

Topic 2 – Flexible Pavement Stress Analysis

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

excessive settlement or collapse

4.1 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

Why use the strain?

0 0

εc

∞

E3�� = 1

� �� − ν �� + �� ≅

��

Topic 2 – Flexible Pavement Stress Analysis

4.2 Tensile Strain

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

∞

• 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)

Topic 2 – Flexible Pavement Stress Analysis

4.2.1 Overall Principal Strains

• Based on all 6 components of normal and shear stresses – σx, σy, σz, xy, xz, yz − Solve cubic equation to get σ1, σ2, & σ3

− Then calculate principal strains

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

ACε3

What is the orientation of ε3?

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

�� = 1

� �� − ν �� + ��

Topic 2 – Flexible Pavement Stress Analysis

4.2.2 Horizontal ‘Principal’ Strain

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

• 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 most analytical tools to predict fatigue failure

ACεt �� = �� + ��

2 −

�� − ��

�

+ ��

�� ≥ ��

Topic 2 – Flexible Pavement Stress Analysis

Deflection Profile

5 Deflection revisited – Multilayer theory

Flexible Plate Rigid Plate

q qRubber Steel

Ground Reaction

�� ≅ 79% � ��

�� = 2 1 − ν� ��

�

�� = 1.5��

�

�� = �

1 − ν� ��

�

�� = 1.18��

� (ν=0.5) (ν=0.5)

Topic 2 – Flexible Pavement Stress Analysis

5.1 Multilayer deflection

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 compressibility of pavement structure to be negligible

Basically:

For this case (assuming one-layer):

Get F from Fig 2.6�� = � � �

��

�� ≡ ��

Topic 2 – Flexible Pavement Stress Analysis

5.1 Multilayer deflection (using single layer theory)

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

Find: F=0.37

Topic 2 – Flexible Pavement Stress Analysis

5.1 Multilayer deflection using single layer theory (cont.)

w=71.0 mils (High) w=7.1 mils (Low)

a = 6”

q = 80 psi

∞

h1= 4”

h2= 8”

h3= 12”

• Examine two cases:

Clay Dense Sand

E=2,500 psi E=25,000 psi

Subgrade quality is very important in pavement design

� = 80 � 6

2,500 0.37 = 0.071 � =

80 � 6

25,000 0.37 = 0.0071

Topic 2 – Flexible Pavement Stress Analysis

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.

6. Two-layer Theory (Burmister, 1943)

�� = 1.18 ��

��

�� = 1.5 ��

��

Topic 2 – Flexible Pavement Stress Analysis

6.1 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

∞

h1 E1

• Flexible plate:

• Rigid plate:

6.1.1 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

Topic 2 – Flexible Pavement Stress Analysis

6.2 Surface Deflections Example

a=6”

q=80 psi

∞

6”E1=50,000 psi

E2=10,000 psi

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

Find: F2=0.6

�� = 1.5 6 � 80

10,000 0.6

�� = 0.0432 ��≅ 43 ��

Topic 2 – Flexible Pavement Stress Analysis

6.3 Interface Deflections Example F

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=0 E1/E2=5

Find: F=0.83

• For the same example as above

�� = 6 � 80

10,000 0.83

�� = 0.0398 ��≅ 40 ��

Topic 2 – Flexible Pavement Stress Analysis

6.4 Surface Vs Interface Deflections

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

– Top Layer

– Subgrade Layer

Compression percentages:

= 3

43 � 100 ≅ 7%

= 40

43 � 100 ≅ 93%

Topic 2 – Flexible Pavement Stress Analysis

6.5 Two-Layer Vertical Stress

a=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=0.1 E1/E2=100

Fig 2.15

Find: a/h1=1.15

ℎ� = 6

1.15 = 5.2 ��

Topic 2 – Flexible Pavement Stress Analysis

6.6 Critical Tensile Strain

a=6”

q=80 psi

∞

6”E1=200,000 psi

E2=10,000 psi

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

Fig 2.21

Find: Fe=1.2

εt

S tr

a in

F a c to

r, F

�� = �

�� =

200,000 1.2

�� = 0.00048 = 480 ��

Topic 2 – Flexible Pavement Stress Analysis

7. Failure Criteria

7.1 Fatigue Cracking Model(s)

• Allowable number of load repetitions related to εt at the bottom of asphalt layer

C = Laboratory to field shift factor

f1 & f2 =Determined from fatigue tests on lab specimens

7.2 Rutting Model(s)

• Allowable number of load repetitions related to εc on top of the subgrade – Does not account for failure in other layers

f3 & f4= Predicted performance to field observation shift factors

�� = � 1

��

�� 1

��

�� = 0.0796 1

��

�.�� 1

��

�.��

�� = 1.365 � 10 ��

��

�.�� = ��

��

Topic 2 – Flexible Pavement Stress Analysis

8. 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

Topic 2 – Flexible Pavement Stress Analysis

8.1 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

Topic 2 – Flexible Pavement Stress Analysis

8.2 Effect of Base Thickness

Tensile Strain (εt) • Increase in h2 does not

significantly decrease εt especially when AC layer is thick (8”)

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

layer is thin (2”)

Topic 2 – Flexible Pavement Stress Analysis

8.3 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

Topic 2 – Flexible Pavement Stress Analysis

8.4 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