Post on 03-Jun-2018
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AASHTO Design Equation
(Rigid Pavements)
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Pavement Performance
p
tt
Wp 1.54.5 4.54.5
pt
Initial
Construction
terminal serviceability
Wt
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Rigid Performance
7.35
18 0.875 1 D
7
18 8.46
1.62 101.0
1
D
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Design Equation
1966
t10
10 18 10 7
8.46
plog
1.5l
4.5
4.5og W 7.35log 1 0.06
1.62 101.01
D
D
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Design Equation
t10
10 18 10 7
8.46
4.5 plog4.5 1.5
log W 7.35log D 1 0.061.62 10
1.0D 1
0.75
tt 10
0.75
4c
D 1.132f4.22 0.32p log
690 18.42DE k
1972
Concrete Tensile Strength (psi)
Concrete elastic modulus (psi)
Effective modulus of subgrade reaction (psi/in)
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Design Equation
t10
10 18 10 7
8
R 0
.46
4.5 plog4.5 1.5
log W 7.35log D 1 0.061.62 10
D
Z S
1.01
0.75
t 10
0.75
4c
c dS C
215.6
D 1.1324.22 0.3
3 J
2p log
18.42DE k
1993
Concrete MOR (psi) Drainage Coefficient
Joint Load Transfer Coefficient
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Drainage Coefficient, Cd
Quality< 1% 1-5% 5-25% > 25%
Excellent 1.25-1.20 1.20-1.15 1.15-1.10 1.10Good 1.20-1.15 1.15-1.10 1.10-1.00 1.00
Fair 1.15-1.10 1.10-1.00 1.00-0.90 0.90
Poor 1.10-1.00 1.00-0.90 0.90-0.80 0.80Very Poor 1.00-0.90 0.90-0.80 0.80-0.70 0.70
Percentage of Time Material Approaches Saturation
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Load Transfer Coefficient, J
Asphalt Shoulders Tied PCC Shoulders
Dowels No
Dowels
Dowels No
DowelsJPCP
3.2 3.8 - 4.4 2.5 - 3.1 3.6 - 4.2JRCP
CRCP 2.9 - 3.2 2.3 - 2.9
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Standard Deviation, So
Source Flexible Rigid
AASHO Road Test Sn 0.35 0.25
AASHO Road Test So 0.45 0.35
Typical Range for So 0.40 0.50 0.35 0.40
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Design Equation
t10
10 18 R 0 10 7
8.46
4.5 plog4.5 1.5
log W Z S 7.35log D 1 0.061.62 10
1.0D 1
0.75
c dt 10
0.75
4c
D 1.132S C4.22 0.32p log
215.63 J 18.42DE k
1993
Effective modulus of subgrade reaction (psi/in)
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Calculating k1. Assume a trial slab thickness, D
2. Determine seasonal values for MR and ESB
3. Convert MR and ESB to composite k values
4. Adjust k values for shallow bedrock (if needed)
5. Convert k values to damage factors (uf)
6. Calculate an average damage factor (uavg)
7. Convert uavg into a year-round keffvalue8. Adjust keff for loss of support over time
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28
Burmisters Solution
Flexible Plate
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29
Burmisters Solution
Rigid Plate
1.18
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Plate Load Test
2a = 30"
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Burmisters Solution
o 2
2
1.18qaw FE
2
o 2
q Ek
w 1.18aF
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1
2
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5
6
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1
2
3
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Design Equation
t10
10 18 R 0 10 7
8.46
4.5 plog4.5 1.5
log W Z S 7.35log D 1 0.061.62 10
1.0D 1
0.75
c d
0.75
t 10
4c
D 1.132S C4.22 0.32p log
215.63 18.42DE k
J
1993
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Design Equation
t18 4.22 0.32p0.75
4c
1W
18.42D
E k
t4.22 0.32p
0.75
418 c
1 18.42D
W E k
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Convert k to Damage Factor
b
0.75i
4c i
18.42u D
E k
tb 4.22 0.32 p
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Convert Damage Factor to k
ceff 4
1 b0.75
Ek
18.42
D u
tb 4.22 0.32 p
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Simplified Version
3.42
0.250.75i iu D 0.39 k
6
c t
Let E 5 10 psi and p 2.5
4
0.2920.75
eff
D u
k 0.39
Equation in the textbook is wrong
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1
2
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2
1
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Loss of Support, LS
Foundation LS
Cement Treated Granular Base 0 1
Cement Aggregate Mixture 0 1
Asphalt Treated Base 0 1Bituminous Stabilized Mixtures 0 1
Lime Stabilized Soil 1 3
Unbound Granular Materials 1 3
Natural Subgrade Materials 2 3
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1
2
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Design Equation
t10
10 18 R 0 10 7
8.46
4.5 plog
4.5 1.5log W Z S 7.35log D 1 0.06
1.62 101.0
D 1
0.75
c dt 10
0.75
4c
D 1.132S C4.22 0.32p log
215.63 J 18.42D
E k
1993
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Procedure Using Equation
Assume a trial slab thickness, D
Calculate k = f(D,MR,Esb,Dsb,Dsg,LS)
Calculate W18 = f(D,k,ZR,So,pt,Sc,Ec,J)
W18 = W18
STOP
Yes
No
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Procedure Using NomographAssume a trial slab thickness, D
Calculate k = f(D,MR,Esb,Dsb,Dsg,LS)
Find D = f(W18,k,ZR,So,pt,Sc,Ec,J)
D = Dtrial
STOP
Yes
No