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Stresses in Flexible Pavement Systems

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Stresses in Flexible Pavement Systems

Multi-layered Elastic Theory

• Pavement behavior under wheel loads is characterized by considering it to be a homogeneous half-space subjected to a circular load of radius “a” and uniform pressure “p”

Multi-layered Elastic System

Assumptions

• Homogeneous Material Properties• Finite Layer Thickness• Layers Infinite in Lateral Directions• Isotropic Layers• Full Friction Between Layers• No Surface Shearing Forces• Solutions Characterized by E, µ

Assumptions

• Most Not Valid• Reasonable for Small Strains

Stresses

• Normal Stresses (Perpendicular)–σz, σt, σr

• Shear Stresses (Parallel)– τrt = τtr

– τzt = τtz

– τrz = τzr

Strains

One Layer System

• Boussinesq Equations– Stress, Strains, Deflections

• Assumes– Homogeneous– Isotropic– Elastic Media– Point Load at Surface

Vertical Stress Distribution

σc

r

z

Boussinesq’s Formula

Note: σz is not a functionof E (materials stiffness)

One Layer Theory

• Need Theory for Circular Load– Foster & Ahlvin

• Charts, pp. 49-51• µ = 0.5

– Waterway Experiment Station• Equations & Tables

One Layer Elastic Equations

Example

• Given– a = 5 inches– z = 10 inches– r = 20 inches

• Find– Radii for Functions A and B

Example

• z/a = 10/5 = 2• r/a = 20/5 = 4

Example

• A = 0.01160• B = -0.00401

Example

• Given– Load = 9000 lbs.– Tire Inflation = 80 psi– E = 10,000 psi–µ = 0.4

• Find–σz1, σz2,– z1, z2

1

2r1 = 0

r2 = 6 in.z2 = 6 in.

z1 = 0

Burmister Influence Curves

Available Solutions

• Vertical Deflections– Rutting

• Interface Deflections– Rutting in Layers

• Tensile Strain at Bottom of HMA– Fatigue Analysis

Influence Values

Example 1

1• Given (Flexible)– p = 80 psi– E1 = 500,000 psi– E2 = 10,000 psi– a = 6 inches

• Find–∆max

6”

Example 2

1• Given (Flexible)– p = 80 psi– E1 = 50,000 psi– E2 = 10,000 psi– a = 6 inches

• Find–∆max

6”

Interface Deflections

• Deflections at Layer Intersections

∆s =pa

E2F

Interface Deflections

Example 3

• Given– p = 80 psi– E1 = 500,000 psi– E2 = 10,000 psi– a = 6 inches–∆max = 0.022

• Find HMA Contribution to ∆max

∆max 6”

∆s

Tensile Strains

• εt at Bottom of HMA Controls Fatigue Cracking

• For a Single Tire

εt =p

E1Fe

Strain Factors

Example 4

• Given– P = 9000 lbs.– p = 67.7 psi– E1 = 150,000 psi– E2 = 15,000 psi

• Find– εt

εt

8”