aashto design pavement

8/13/2019 aashto design pavement

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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

3

4

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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1

2

3

4

5


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6

7

8

9

10

11


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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

aashto design pavement

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