Topic 3 - Flexible Pavement Stress Analysis
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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!
Topic 3 – Flexible Pavement Stress Analysis
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.
Topic 3 – Flexible Pavement Stress Analysis
Topic 3 – Flexible Pavement Stress Analysis
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 +−=
Topic 3 – Flexible Pavement Stress Analysis
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.
Topic 3 – Flexible Pavement Stress Analysis
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
3.2 One-layer Solutions (Foster & Ahlvin)Charts follow similar outline
Depth (z) and offset (r) are expressed in radial ratios
Topic 3 – Flexible Pavement Stress Analysis
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)
Topic 3 – Flexible Pavement Stress Analysis
3.2.1 Vertical Stress (cont)z/a = 6/6 =1r/a = 6/6 =1
33%100q
z ≅×σ psi4.26
1008033
z =×
=σ
Topic 3 – Flexible Pavement Stress Analysis
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
=
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
3.2.2 Deflection (cont.)
Given:z/a=24/6=4r/a=0
Find:F=0.37
Topic 3 – Flexible Pavement Stress Analysis
3.2.2 Deflection (cont.)
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
Topic 3 – Flexible Pavement Stress Analysis
• 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
Topic 3 – Flexible Pavement Stress Analysis
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)
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – 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.
5. Two-layer Theory (Burmister)
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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:
Topic 3 – Flexible Pavement Stress Analysis
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 ==
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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?
Topic 3 – Flexible Pavement Stress Analysis
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?
Topic 3 – Flexible Pavement Stress Analysis
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
Topic 3 – Flexible Pavement Stress Analysis
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 −=
Topic 3 – Flexible Pavement Stress Analysis
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.)
Topic 3 – Flexible Pavement Stress Analysis
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
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