Present Pavement

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PAVEMENT DESIGN THICKNESS USASSTHO METHOD 1993





Design ProcedureFlexible Pavement

1. Determine Traffic Plan ( W18 )2. Determine Reliability Value ( % )

3. Overall Standard Deviation ( S0 )

4. Resilient Modulus ( Mr )

5. Design Serviceability loss ( ∆PSI = IPo – IPt )6. Determine Structural Number ( SN )

7. Calculate the Thickness for each layer based SN



EQUATION



1. TRAFFIC





2. RELIABILITY

Reliability is the probability that any particular type of scombination of distress manifestations) will remain below or w permissible level during the life design.



Relation Between Reliability and C





3. Determine Standart Normal Dev



But the range So Values providII in AASHTO Guide are base

values identified bellow :0,30 – 0,40 for Rigid Paveme

0,40 – 0,50 for Flexible Pave



4. Determine Resilient Modulus ( M





5. Design Serviceability Loss( ∆PS



Present Serviceability Index ( P0 )

The Serviceability of a pavement is defined as its abilitythe type of traffic ( automobiles and trucks ) which use facility

primary measure of serviceability is the Present Serviceability(Po).

It ranges from 0 ( impossible road ) to 5 ( Perfect Road )

Since some considerations must also be given to the selection should be recognized that Po values observed at the AASHO Rwere 4,2 for Flexible Pavement and 4,5 for Rigid Pavement.



Terminal Serviceability Index ( Pt )

Terminal Serviceability Index ( Pt ) is based on the lowest inwill be tolerated before rehabilitation , resurfacing, or reconstr

becomes necessary .

2,5 or higher is suggested for major highway design.

2,0 for highways with lesser traffic volumes.

For relatively minor highways where economic dictate that capital outlay be kept at a minimum, it is suggested that this iaccomplished by reducing the design period of the total traffirather than by designing for a terminal serviceability less th



Design Serviceability Loss ∆PSI = Po - P



6. Structural Number

To Determine Structural Number can use equatiofrom graph based on ASSTHO’s Method.

Equation to Determine SN





Designing Layer Thickness





Layer Coefficients a1




a2 = 0,249(log10EBS)-0,977

With EBS is a function of net only moisture but also for the str(ϴ).

Values for stress state within the base course vary with the submodulus and thickness of the surface layer .






a3 = 0,227(log10ESB)-0,839











EXAMPLE



NEW ROAD DESIGN EXAMPLE

Planning Principal Arterial road and road life are 13 years. Distribu

is 50 % . That road planned support present traffic 2,5 10

and gfactor is 4% . One direction of road has 2 line.

The other parameters can follow :

Resilient Modulus ( Mr ) = 5.700 psi

Asphalt Concrete ( EAC ) = 400.000 psi

Granular Base ( EBS ) = 30.000 psi

Granular Subbase ( ESB ) = 11.000 psi

Drainage Coefficient for Base m2 = 1,2

Drainage Coefficient for Subbase m3 = 1,2

Determined the thickness each Pavement Layer ?



SOLUTION





ESAL Road Life ( 13 years )

W18 = W18 x Growth Factor

Growth Factor = 16,63 ( see Table bellow )

W18 = 1.125.000 x 16,63

= 18.337.500





Layer Coefficients a

Asphalt Concrete a1 = 0,42 ( see figure on bellow )

Granular Base a2 = 0,14 ( see figure on bellow )

Granular Subbase a3 = 0,083 ( see figure on bellow )









Design Serviceability Loss

Po = 4 ( able to design) and Pt = 2,5 ( Major Highway )

∆PSI = 4 – 2,5

= 1,5

Reliability



y

R is taken 90 %

Determining Thickness D1




R= 90%

So= 0.40 ( flexible pavement )

W18 = 18.337.500EBS = 30 Ksi

ΔPSI = 1,5

Solution SN1= 4,4

D1= SN1/a1= 4,4/ 0,42 = 10,476 in

=26,609 cm ~ 27 cm = 10,62 in

SN1= D1.a1= 10,62 . 0,42 = 4,46

EBS = 30 ksi

a1= 0,42; D1=? SN







R= 90%


Wt = 18.337.500

ESB = 30 Ksi

ΔPSI = 1,5

Solution SN2= 6

SN1*= 4,4 ; D1= 10,62 in

SN2= SN1+D2 . a2 . m2

D2= (SN2-SN1)/(a2 . m2)= (6 - 4,4)/(0,14 . 1,20)= 9,524 in=24,19 cm ~ Taken 26 cm = 10,23 in

SN2= SN1+ D2 .a2 . M2 = 4,4 + 10,23 . 0,14 . 1,20= 6,11

a3= 0,083; E3= 11 ksi

a2= 0,14; D2=?SN

a1= 0,42; D1=10,62inch







R= 90%


Wt = 18.337.500

Mr = 5,7 Ksi

ΔPSI = 1,5

Solution SN3= 7,1

SN1= 4,4 ; SN2= 6

SN3= SN1+SN2+D3.a3.m3

D3= (SN3-SN2-SN1)/(a3.m3)= (7,1-6-4,4)/(0,083.1,2)

= minus (Not Use Subbase)

Mr Subgrade = 5,7 ksi

a2=0.14; D2=10,23 inchSN

a1= 0,42; D1=10,62 inch

a3=0,083 ; D3=?





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Present Pavement