Flexible Pavement Stress Analysis

Transcription

1 Flexible Pavement Stress Analysis Dr. Antonis Michael Frederick University Notes Courtesy of Dr. Christos Drakos, University of Florida 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

2 1. Available Models Multilayer Elastic Theory Finite Element Methods Viscoelastic Theory ( Dynamic Analysis (inertial effects) Thermal Models (temperature change) Most widely used Reasonable Results Properties Relatively Simple to Obtain Falling Weight Deflectometer Small trailer Dropping Weight Geophones Deflection Basin Uses elastic theory to predict the deflection basin for the given load. Then

3 2. Multilayer Elastic Theory a = radius = pressure E 1, ν 1 z 1 E 2, ν 2 z 2 Point A Point B E 3, ν 3 z 3 Assumptions (p. 60): Each Layer Continuous Linearly Material is Finite thickness (except last layer) 2. Multilayer Elastic Theory (cont.) a = radius = pressure E 1, ν 1 z 1 E 2, ν 2 z 2 Point A Point B E 3, ν 3 z 3 Assumptions (cont.): Surface stresses Circular Vertical Uniformly distributed Full Each layer continuously supported Why do we want full friction between layers?

4 Units Guidelines Stress: Reported in psi: Strain: Reported in µε: lbs psi = 2 in in µε = microstrain = 10 6 in Deflections: Reported in mils: in mils = 1000 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 Boussines (1885) Point load on an elastic half-space Examine σ distribution along Z & X Z σ z σ z σ z r P z X σ z = 3 2π Where: σ z = r = z = P = Half-space: 1 2 r 1 + z 5 2 P 2 z

5 3.2 One-layer Solutions (Foster & Ahlvin) Developed charts to determine σ z, σ t, σ r, τ rz & w (ν=0.5) Figures Axisymmetric loading: 2a σ z = σ r = σ t = τ rz = w = σ r τ rz σ z σ t z Depth 0 1a 2a a r 3a 2a 1a 0 Offset 3.2 One-layer Solutions (Foster & Ahlvin) Charts follow similar outline Depth (z) and offset (r) are

6 3.2.1 Vertical Stress Given: Load, P = 9000 lbs Pressure, = 80 psi Find: Vertical Stress, σ z=6 & r=6 σ z r=6 a z=6 First, we need to Vertical Stress (cont) z/a = 6/6 =1 r/a = 6/6 =1

7 3.2.2 Deflection Flexible Plate Rigid Plate Rubber Steel Deflection Profile Ground Reaction Deflection (cont.) A a = 6 = 80 psi h 1 = 4 h 2 = 8 h 3 = 12 Pavement Structure How can we use one-layer theory to estimate the deflection of the system? We can assume For this case (assuming Basically:

8 3.2.2 Deflection (cont.) Given: z/a=24/6=4 r/a= Deflection (cont.) a = 6 = 80 psi h 1 = 4 h 2 = 8 Examine two cases: Clay Dense Sand E=2,500 E=25,000 h 3 = 12 A Subgrade uality

9 4. Stresses & Strains for Design 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 σ z depends on Vertical compressive To combine a Effect of E 1 h 1 E 2 h 2 ε c E Tensile Strain Tensile strain at the bottom of AC layer; used in pvt design as the Two types of strain: Overall minor Horizontal principal strain, ε t (not Horizontal principal strain (ε t ) used as a design criterion. a ε E 1 h 1 E 2 h 2 E 3

10 4.2.1 Overall Principal Strains Based on all 6 components of normal and shear stresses σ x, σ y, σ z, τ xy, τ xz, τ yz Solve cubic euation to get σ 1, σ 2, & σ 3 1 Then calculate principal strains ε 3 = ( σ3 ν( σ1 + σ2 )) E Minor principal strain (ε 3 ) considered to be a AC Minor principal strain (ε 3 ) Horizontal Principal Strain Based on the horizontal normal and shear stresses only σ x, σ y, τ xy Horizontal principal strain (ε t ) is Maximum Always acts on the horizontal plane Used by the program KENLAYER to predict fatigue failure a AC

11 5. Two-layer Theory (Burmister) Developed solutions for: Vertical deflections (flexible & rigid) Vertical stresses (limited # of cases) σ & δ highly dependent on 5.1 Two-Layer Deflections In one-layer theory we assumed that all layers could be represented as one a δ surface = δ top of the subgrade For two-layer theory we have: Vertical Vertical Surface Deflections Flexible Rigid h 1 E 1 E 2

12 5.1.2 Surface Deflections Example a=6 =80 psi E 1 =50,000 psi 6 E 2 =10,000 psi Given: h 1 /a=6/6=1 E 1 /E 2 = Interface Deflections Example For the same example as above a=6 =80 psi F E 1 =50,000 psi 6 E 2 =10,000 psi Given: h 1 /a=6/6=1 ;r/a=0 E 1 /E 2 =5 h 1 /a Offset

13 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 =80 psi E 1 =500,000 psi h 1 E 2 =5,000 psi Maximum allowable σ c for clay = 8 psi Given: σ c /=0.1 E 1 /E 2 =100

14 5.2 Critical Tensile Strain a=6 =80 psi ε t E 1 =200,000 psi E 2 =10,000 psi e = ε t = critical tensile strain Given: E 1 /E 2 =20 h 1 /a=1 6 Strain Factor, F e 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 f2 f3 ( ) ( ) ( ) ( ) N = f ε E N f 1 t Rutting Model = f ε d 4 c ( ) f5 f 1 = f 2 & f 3 = Allowable number of load repetitions related to ε c on top of the subgrade Does not account for failure in other layers N d = εc N = ε E f t 1 ( ) f 4 & f 5 =

15 7. Sensitivity Analysis Sensitivity analyses illustrate the effect of various parameters on pavement responses Variables to be considered: Layer Layer 7.1 Effect of HMA Thickness Tensile Strain (ε t ) Compressive Strain (ε c )

16 7.2 Effect of Base Thickness Tensile Strain (ε t ) Compressive Strain (ε c ) 7.3 Effect of Base Modulus Tensile Strain (ε t ) Compressive Strain (ε c )

17 7.4 Effect of Subgrade Modulus Tensile Strain (ε t ) Compressive Strain (ε c )