3. pavement stresses
DESCRIPTION
Highway structural designTRANSCRIPT
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Pavement Pavement Response Response
Under LoadUnder Load
20122012
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Stresses Stresses Distribution Distribution
Wheel Load
Load Distribution in Pavements
150 psi
3 psi
Wearing C.
Base
Sub-base
Sub-grade
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Pavement Structure
Not Drawn to Scale
Wheel Load
Vertical Stress
σS
σB
σSG
σSB
Subgrade
Wheel Load
Pavement Structure
Not Drawn to Scale
Horizontal Stress
Subgrade
Stresses Stresses Distribution Distribution 3
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PressPressureure p = Tire Pressure = Contact
Pressure Tire Pressure
P
a = P/π p
• Type of Load ( Static, and Dynamic)• Tangential Forces ( Acceleration, and Braking)
p
Contact Area Contact Area is circle with radius = a
Loads and Loads and Contact Area Contact Area Tandem
Tridem Single
Dual
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Axles TypesAxles Types
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The stresses, strains, and deflection are the pavement response to the applied load. Stress is a force load per unit area, and strain is the change in dimension. Pavement stresses, strains, and deflection are caused by traffic loading, daily or seasonal temperature and moisture variations and by any change in the conditions of pavement support. The theory used to calculate stress, strains, and deflections in pavement system is the multi-layer theory.
H1, E1, μ1
H2, E2, μ2
H3, E3, μ3
Hn, En, μn
P
Interface 1
Interface 2
Interface 3σz
σr
σt
dz
dr
Multi-Layer Multi-Layer Elastic TheoryElastic Theory
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Load Load RepresentatRepresentation ion
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Single Load Single Load Location is defined by z & r
Z1Z2
Pa
2
1
p
V. Stress σ1 > σ2
V. deflection Δ1 > Δ2
Z
Pa
2 1
p
V. Stress σ1 > σ2
V. deflection Δ1 > Δ2
r
Dual LoadDual Load
Z
Ps
1
p
r
P
r
p1 2
σ1 =σ due to load1+σ due to load2
Δ1 =Δ due to load1+Δ due to load2
Load Load RepresentatRepresentation ion
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Tandem Tandem defined by z/a and r/a
σ1 = ∑σ due to loadi
Δ1 = ∑Δ due to loadi
P PPP
S1S2
St
1
How to Calculate Stress & Deflection How to Calculate Stress & Deflection
One Layer TheoryOne Layer Theory
Stress
Deflection
For any system of loads
Stress
Deflection
Two Layers TheoryTwo Layers Theory Under single load
Load Load RepresentatRepresentation ion
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E1 =Ep= α
E2 ( Subgrade)
P
ap
No deflection in the pavement structure …..all deflection is in the subgrade
So, the layered system is composed of only ONE LAYER (Subgrade)
This layer follows the same assumption of the Elastic Multilayer Theory
Vertical Stress (Vertical Stress (σσ) & Vertical Deflection () & Vertical Deflection (ΔΔ))
E1 =Ep= α
E2 ( Subgrade)
P
ap
1) Using (z/a, r/a) Figure (1) m= σ/p σ
2) Using (z/a, r/a) Figure (2) F Δ= pa*F/ E2
Single LoadSingle Load
One Layer Theory(Boussinsq Theory)
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Dual LoadDual Load
Z
Ps
p
P
p12
σ total = σ 1+ σ 2
σ1 (z/a, r/a)…….m1 = σ 1 / p
σ 2 (z/a, r/a)…….m2 = σ 2 / p
Δ total = Δ 1+ Δ 2
Δ 1 (z/a, r/a)……F1…. Δ1 = paF1/Esubgrade
Δ 2 (z/a, r/a)……F2…. Δ2 = paF2/Esubgrade
Same procedure for Tandem Load …………. 4 times
One Layer Theory(Boussinsq Theory)
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ExampleExample
Z=14 inch
E2=5000 psi
P
a=7 p=50psi
r=7 1 2
z1/a = 2 , r1/a = 0 ……….m1 = 30%
m1= σ1 / p ……. σ1 = 0.3*50 = 15 psi
z2/a = 2 , r2/a = 1.0 ………m2 = 20%
m2= σ2 / p ……. σ2 = 0.2*50 = 10 psi
z1/a = 2 , r1/a = 0 ……….f1 = 0.7
Δ1 = paF1/Esub= 50*7*0.7/5000 = 0.049
z2/a = 2 , r2/a = 1 ……….f2 = 0.57
Δ2 = paF2/Esub = 50*7*0.57/5000 =0.04
One Layer Theory(Boussinsq Theory)
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Vertical Stress Vertical Stress ((σσ) )
m(%)
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Vertical Vertical Deflection (Deflection (ΔΔ))
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E pav. = E1
E Subgrade = E2
P
ap
1- Same assumption as in the
Multilayer theory
2- In this theory, pavement
deflection is considered 3- Deflection at depth Z in a two layer system given by: Δ = 1.18 paf/Esub (Rigid Pav.)
Δ = 1.5 paf/Esub (Flexible Pav.)
Δ = vertical deflection under CL of the applied load
p = tire pressure (psi)
a = √P/πp
Esub = Subgrade modulus of elasticity
f = two layer deflection factor
Two Layer Theory ( Burmister Theory)
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4- This theory is applied only to one load and r/a = 0
5- When Z/a = 0 F = 1.0 for any E2 / E1 ratio
6- If it is required to calculate deflection in two layers system under dual or dual tandem load system, the thickness conversion method described earlier can be used to convert the two layers to one layer system.
Two Layer Theory ( Burmister Theory)
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Layer EquivalencyLayer Equivalency
It is the conversion of a thickness of a layer of material with known modulus to an equivalent thickness of another material with known modulus using the formula
Zsub. = Z pav. 3√E1/E2
E pav. = E1
E Subgrade = E2
P
ap
E Subgrade = E2
P
ap
E Subgrade = E2
Two Layer Theory ( Burmister Theory)
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a = √15000 / π*100 = 7.0
Two layerTwo layer
z/a = 2 , E2 / E1 = 1/100 ….f’ = 0.13
Δ = K paf /Esub= 1.5 *100*7*0.13/1000= 0.137
One layerOne layer
z/a = 2 , r/a = 0 ……….f1 = 0.69
Δ = paF1/Esub = 100*7*0.69/1000 =0. 483
ConversionConversion
Z = Z pav. 3√E1/E2 = 143√100000/1000 =65
Δ = paF1/Esub = 100*7*0.17/1000 =0. 12
Z=14
E2=1000 psi
P = 15000lb
p=100psi
E1=100000 psi
ExampleExample
Two Layer Theory ( Burmister Theory)
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Vertical Stress Vertical Stress ((σσ) )
m(%)
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Vertical Vertical Deflection (Deflection (ΔΔ))
Z/a
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22Use of Multilayer theory in pavement evaluation
- Pavement evalaution ………..to find E1 & E2
- Since the two layer theory ( or multilayer thoery, in general) links between the pavement deflection and its characteristics, i.e. E1 & E2, then if we can measure the deflection, it will be easy to back calculate E1 & E2.
- Plate Bearing Test is used to measure the deflection of subgrade under given load.
P
a = 15Rigid plate
How to get E1 & E2 Using the Plate Bearing Test
1- Put the Plate on the top of subgrade and measure the deflection ( Δ) at the top of subgrade ( i.e. z = 0.0)
p = P / π (15)2
using the two layer theory
Δ = K p a F’ / E2
Δ is measured, K = 1.18, F’ = 1.0 (for z/a = 0.0), and p & a is known ………then get E2
P
a = 15Rigid plate
Subgrade
Pavement Evaluation
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How to get E1 & E2 Using the Plate Bearing Test
2- Put the Plate on the top of the base ( or pavement materials) and measure the deflection ( Δ) at the top of subgrade using the two layer theory
Δ = K p a F’ / E2 …… get F’
Known F’ and z/a ...get E2 / E1 …get E1
Z=choused
E2
P
E1
ExampleExample
A certain flexible pavement consists of 2-in bituminous surface, 7-in crushed stone base course, and 9-in sandy subbase. A 30-in rigid plate was used to determine the load- deflection characteristics. The following results were obtained
1- subgrade deflection = 0.10 in at 20 psi
2- Pavement deflection = 0.10 in at 98 psi
Solution:
step (1)
Δ = 0.1 = K p a F’ / E2
0.1 = 1.18 × 20 ×15 ×1 / E2
E2 = 3540 psi
2 in7 in9 inSubgrade
18 in Pavement
)E1 (
E2
30 in
Pavement Evaluation
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Given Req.
SWLMultiwheel load
P , ZP, ZP, σ max. P, Δ max. Z , σ max. Z, Δ max
σ
Δ
ZZPP
ChartChart, EquationChartChart, EquationChartDrawn Relation (Δ , P)
ChartChart, EquationDrawn Relation (σ , Z) Drawn Relation (Δ , Z) Drawn Relation (σ , P) Drawn Relation (Δ , P)
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step (2)
Δ = 0.1 = K p a F’ / E2
0.1 = 1.18 × 98 ×15 ×F’ / 3540
F’ = 0.204 ( using Chart with z / a = 18/15 = 1.2 and f’ = 0.204 )
E2 / E1 = 1 / 80 …… E1 = 80 × 3540 = 283200 psi
Z=18 E2=3540
psi
E1
Given : Z, p, Δ max., E2
Req. : P
1- Assume P a = √ P / π p
2- Calculate Δ
3- If Δ = Δ max. ……..P req. = P
4- if not repeat 3 to 4 times to draw the shown relation and find Preq at Given Δ
Pavement Evaluation
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