Download - 02 - Stresses in Flexible Pavements
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Stresses in Pavements
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Flexible Pavements
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Stresses Due to Point Load
z
r
R
P
3
z 2 3
2
r 2 3
t 2
P 3z2 R R
P 3r z R1 22 R R R z
P R z1 22 R R z R
z
r
t
Note: Picture and equations in textbook are incorrect
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Deflections Due to Point Load
z
r
R
P
2
z 3
2
0
P 1 2 1 zu2 E R R
P 1u @ z 0
ER
uz
Note: Picture and equations in textbook are incorrect
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Stresses Due to Circular Load
o
2a
z
3
z o 2 2
3
or 2 2 2 2
3
ot 2 2 2 2
z1a z
2 1 z z1 22 a z a z
2 1 z z1 22 a z a z
Note: equations are only valid along load centerline
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Stresses Due to Circular Load
o
2a
z
3
z o 2
3
or 2 2
3
ot 2 2
z a11 z a
2 1 z a z a1 22 1 z a 1 z a
2 1 z a z a1 22 1 z a 1 z a
Note: equations are only valid along load centerline
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Hooke’s Law
z z r t
r r z t
t t r z
1E
1E
1E
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Strains Due to Circular Load
3
oz 2 2
3
or 2 2
3
ot 2 2
1 2 z a z a1 2E 1 z a 1 z a
1 2 1 z a z a1 22E 1 z a 1 z a
1 2 1 z a z a1 22E 1 z a 1 z a
Note: equations are only valid along load centerlineNote: equations in textbook are incorrect
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Deflections Due to Circular Load
2oz z 2
z
a 1 1d dz 1 2 1 z a z aE 1 z a
2o
o
2 a 1d @ z 0
E
Note: equations are only valid along load centerline
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Boussinesq Example
6"
18"E = 10 ksi
A
B
9000-lb dual wheelwith 90-psi tires
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Boussinesq Example
6"
18"E = 10 ksi
A
B
9000-lb dual wheelwith 90-psi tires
o = 90 psi11.3"
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Rigid Loading
o
z 22 1 r a
P
o 2
Pwhere is the average pressure on the platea
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Rigid LoadingP
2 2o
o r
a 1 P 1d u
2E 2Ea
do
textbook equation
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Rigid vs. Flexible Loading
2o
o
2 a 1d
E
Flexible Plate
2o
o
a 1d
2E
Rigid Plate
2o
flexibleo
rigido
a 12d E
d 2
oa 12 E
rigidoflexible
o
d4 1.27 0.79d 4
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Foster & Ahlvin
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Ahlvin & Ulery
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Ahlvin and Ulery (1962)
z
r
t
z
p
p 2 1 2 = function
p 2 1 2 E = modulus
pa 1 z 1E a
A B
A C F E
A D E
A H (deflection)
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Multiple Wheel Loads
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Multiple Wheel Loads
L RA A L A Rr ,z r ,z
A
z
rL rR
L R
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Donald Burmister
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Burmister’s Solution
E1, 1
E2, 2
h1 E1, 1
E2, 2
h1
2aq
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Burmister’s Solution
Surface deflection
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Burmister’s SolutionVertical Stress
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Burmister Example
E1 = 500 ksih1 = 6"
E2 = 10 ksi
A
9000-lb dual wheelwith 90-psi tires
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Burmister Example
E1 = 500 ksih1 = 6"
E2 = 10 ksi
A
o = 90 psi11.3"
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Odemark’s Method
E1, 1
E2, 2
h1E2, 2
E2, 2
he
A A
Note: only valid at or below the layer interface
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Odemark’s Method
E2, 2
h1
E2, 2
he
31 1
21
h EStiffness1
3e 2
22
h EStiffness1
A A
Note: only valid at or below the layer interface
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Odemark’s Method
3 3e 2 1 1
2 22 1
h E h E1 1
2213e 1 2
2 1
1Eh hE 1
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Odemark’s Method
13e 12
Eh hE
1 2If we assume
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Odemark’s Method
13e 12
Eh f hE
f 0.8 for 2-layer systemf 0.9 for n-layer system(f 1.0 for first interface)
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Odemark Example
E1 = 500 ksih1 = 6"
h2 = 12" E2 = 50 ksi
E3 = 10 ksi
A
B
9000-lb dual wheelwith 90-psi tires
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Odemark Example
A
B
o = 90 psi
E1 = 500 ksih1 = 6"
h2 = 12" E2 = 50 ksi
E3 = 10 ksi
11.3"
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Odemark Example
A
o = 90 psi
E1 = 500 ksih1 = 6"
E2 = 50 ksi
11.3"
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Odemark Example
o = 90 psi
E2 = 50 ksi
31eh 1.0 6" 10 12.9"
A
11.3"
E2 = 50 ksi
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Odemark Example
A
B
o = 90 psi
E1 = 500 ksih1 = 6"
h2 = 12" E2 = 50 ksi
E3 = 10 ksi
11.3"
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Odemark Example
A
B
o = 90 psi
h2 = 12" E2 = 50 ksi
E3 = 10 ksi
31eh 1.0 6" 10 12.9"
11.3"
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Odemark Example
B
o = 90 psi
h2 = 24.9" E2 = 50 ksi
E3 = 10 ksi
11.3"
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Odemark Example
B
o = 90 psi
E3 = 10 ksi
3e,2h 0.8 24.9" 5 34.1"
11.3"