Topic 4 - Traffic Characterization Handouts

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					                          Traffic Characterization

                                 Dr. Christos Drakos

                                 University of Florida




Topic 4 – Traffic Characterization
1. Introduction
Traffic is the most important factor in pavement design;
thickness is based on the number of load repetitions (traffic)
1.1 Traffic Characterization Procedures
a.       Fixed Traffic:
     •      Thickness governed by single-wheel load (use the highest anticipated
            load for design)
     •      Used for heavy load / low volume pavements (i.e. airfields)
b.       Fixed Vehicle Traffic:
     •      Thickness governed by # of repetitions of a standard vehicle or axle-
            load
     •      Convert all traffic to 18-kip single axle loads
c.       Variable Traffic and Vehicle:
     •      Loads are divided into groups (load spectra) and the corresponding
            stresses and strains are used for design
     •      More appropriate for mechanistic design methods
Topic 4 – Traffic Characterization
Fixed Vehicle Traffic
• Design is based on the total number of passes of the standard
   axle load (18-kip Equivalent Single Axle Load – ESAL) during
   the design period
• Covert all traffic to the standard axle load (ESAL)

Nd = 1.365 × 10 −9 ( ε c )
                               −4.477

                                                Nd, Nf   = Load cycles to failure
Nf = 0.0796 ( ε t )            (E1 )
                      −3.291           −0.854
                                                         = ESAL


Basic premise:
Must determine how many 18-kip single axle loads would cause
the same damage as one X-kip load

           How many ESAL does a 24-kip axle amount to?




Topic 4 – Traffic Characterization
1.2 Equivalent Axle Load Factor (EALF) a.k.a. (LEF)
• Defines the damage per pass to pavement by the axle in
   question relative to the damage per pass of a standard 18-k
   axle
Load Equivalence Factor (LEF) Depends on:
• Type of pavement
• Thickness / structural capacity
• Terminal conditions (definition of failure)
   – 20% of lane area with fatigue cracking
   – ½ inch rutting
• Theoretical analysis Nf(18)/Nf(X)
• Based on experience (AASHO Road Test)
   – Table 6.4 (flexible pavements)
   – Table 6.7 (rigid pavements)
Topic 4 – Traffic Characterization
1.3 Theoretical Analysis to get LEF (Mechanistic)
                24-kip                       18-kip
     AC                           AC
                                                         For the same structure
     BASE                         BASE                   apply 24- & 18-kip load
     SUBGRADE                     SUBGRADE
∞                             ∞


• KENLAYER:
      – (18-Kip) εt = 200µε → Nf(18) = 1,612,000
      – (24-Kip) εt = 267µε → Nf(24) = 623,000
• So, we can get an equivalent damage factor
• Nf(18)/Nf(24) = 2.59
• It would take 2.59 18-kip load single axles to cause the same
  damage as one 24-kip axle




Topic 4 – Traffic Characterization

    1.3 Theoretical Analysis to get LEF (Mechanistic)
            No. of 18k Single Axle Load to cause specific damage
    LEF =
            No. of Xk Single Axle Load to cause specific damage

    Issues with theoretical analysis:
    • Does the LEF change if we modify structural configuration
       (thickness, modulus, etc.)?
    • Which one is more critical – fatigue cracking or rutting
       analysis?
    Due to the many factors that influence the LEF, it is almost
    impossible to select an appropriate a single value that applies
    to all situations. For a truly mechanistic design method, each
    load group should be analyzed separately.
Topic 4 – Traffic Characterization
1.3 AASHTO Equivalent Factors (Empirical)

Wx ⎡ L18 + L 2s ⎤
                       4.79   ⎡10 G βx ⎤                                     W18
   =                          ⎢ G ⎥[L 2 x ]4.33                   EALF =
W18 ⎢ L x + L 2 x ⎥
     ⎣            ⎦           ⎢10 β18 ⎥                                      Wx
                              ⎣        ⎦
Where:
Wx   =       axle applications inverse of equivalency factors
W18 =        No of 18kip single axle loads
Lx   =       axle load being evaluated (kips)
L18  =       18 (standard axle load in kips)
L2s  =       code for standard axle = 1 (single axle)
L2x  =       code for axle load being evaluated
          L2x = 1 for single axle
          L2x = 2 for tandem axle
          L2x = 3 for triple axle (added in the 1986 AASHTO Guide)




Topic 4 – Traffic Characterization
1.3 AASHTO Equivalent Factors (Empirical)

Wx ⎡ L18 + L 2s ⎤
                       4.79   ⎡10G βx ⎤                                      W18
   =                          ⎢ G ⎥[L 2 x ]4.33                   EALF =
W18 ⎢ L x + L 2 x ⎥
     ⎣            ⎦           ⎢10 β18 ⎥                                      Wx
                              ⎣       ⎦
Where:
pt   = "terminal" serviceability index (point at which the
     pavement is considered to be at the end of its useful life)

        ⎡ 4.2 − p t ⎤           function of the ratio of loss in serviceability at time, t, to
G = Log ⎢
        ⎣ 4.2 − 1.5 ⎥
                                the potential loss taken at a point where pt = 1.5
                    ⎦

           ⎡ 0.081(L x + L 2 x )3.23 ⎤    function which determines the relationship
β = 0 .4 + ⎢                   3.23 ⎥     between serviceability and axle load applications
           ⎣ (SN + 1) L 2 x
                      5.19
                                     ⎦
Topic 4 – Traffic Characterization
1.4 AASHTO Equivalent Factors – Example 1
Calculate the LEF for a 30,000 lb single-axle load. The structural number (SN) is equal to three
(3) and the terminal serviceability is 2.5.
                                                                                L18 := 18    L2s := 1
W 18     =          predicted number of 18-kip single axle load applications
                                                                                                              Lx := 30       p t := 2.5
Wx       =          predicted number of 30-kip single axle load applications
Lx       =          L    30   = 30                                                                            L2x := 1       SN := 3
L2x      =          1 (single axle)

                                               ⎡ 0.081( L18 + L2s) 3.23 ⎤                                 ⎡ 0.081( Lx + L2x) 3.23 ⎤
         ⎛ 4.2 − p t ⎞         β 18 := 0.4 +   ⎢                         ⎥                 β x := 0.4 +   ⎢                         ⎥
G := log ⎜                                     ⎢ ( SN + 1) 5.19 ⋅ L 3.23 ⎥                                ⎢ ( SN + 1) 5.19 ⋅ L 3.23 ⎥
         ⎝ 4.2 − 1.5 ⎠                         ⎣                   2s    ⎦                                ⎣                   2x ⎦



                                                                       ⎛ G      ⎞
                                                                4.79   ⎜ βx
             W 18                Wx      ⎛ L18 + L2s ⎞                 ⎜ 10     ⎟ ( L ) 4.33
 EALF                                    ⎜ L +L                        ⎜ G      ⎟ 2x
             Wx                  W 18
                                         ⎝ x 2x ⎠
                                                                       ⎜ β 18
                                                                       ⎝ 10     ⎠
                                                                −1
         ⎡              ⎛ G                    ⎞           ⎤
         ⎢ L + L 4.79 ⎜ β x                                ⎥
           ⎛ 18 2s ⎞ ⎜ 10
 EALF := ⎢ ⎜                                   ⎟ ( L ) 4.33⎥                    EALF = 7.935
         ⎢ ⎝ Lx + L2x ⎠ ⎜ G                    ⎟ 2x ⎥
         ⎢              ⎜ β 18                             ⎥
         ⎣              ⎝ 10                   ⎠           ⎦




Topic 4 – Traffic Characterization
1.5 AASHTO Equivalent Factors – Example 2
Calculate the LEF for a 40,000 lb tandem-axle load. The structural number (SN) is equal to f ive
(5) and the terminal serviceability is 2.5.
                                                                                L18 := 18      L2s := 1
W 18     =          predicted number of 18-kip single axle load applications
                                                                                                             Lx := 40        p t := 2.5
Wx       =          predicted number of 40-kip tandem axle load applications
Lx       =          L    40   = 40                                                                           L2x := 2        SN := 5
L2x      =          2 (tandem axle)

                                               ⎡ 0.081( L18 + L2s) 3.23 ⎤                                 ⎡ 0.081( Lx + L2x) 3.23 ⎤
        ⎛ 4.2 − p t ⎞          β 18 := 0.4 +   ⎢                         ⎥                 β x := 0.4 +   ⎢                         ⎥
G := log⎜                                      ⎢ ( SN + 1) 5.19 ⋅ L 3.23 ⎥                                ⎢ ( SN + 1) 5.19 ⋅ L 3.23 ⎥
        ⎝ 4.2 − 1.5⎠                           ⎣                   2s    ⎦                                ⎣                   2x ⎦



                                                                       ⎛ G      ⎞
                                                            4.79       ⎜ βx
             W 18                Wx   ⎛ L18 + L2s ⎞                    ⎜ 10     ⎟ ( L ) 4.33
 EALF                                 ⎜
             Wx                  W 18
                                      ⎝ Lx + L2x ⎠                     ⎜ G      ⎟ 2x
                                                                       ⎜ β 18
                                                                       ⎝ 10     ⎠
                                                                −1
         ⎡              ⎛ G                    ⎞            ⎤
         ⎢ L + L 4.79 ⎜ β x                                 ⎥
           ⎛ 18 2s ⎞ ⎜ 10
 EALF := ⎢ ⎜                                   ⎟ ( L ) 4.33 ⎥                   EALF = 2.081
         ⎢ ⎝ Lx + L2x ⎠ ⎜ G                    ⎟ 2x ⎥
         ⎢              ⎜ β 18                              ⎥
         ⎣              ⎝ 10                   ⎠            ⎦
Topic 4 – Traffic Characterization
2. Computation of Design ESALs
ESAL = Equivalent Single Axle Load
ESALs = Cumulative ESALs for all vehicles over the entire design
period (we can also calculate ESAL for specific vehicle type)

ESALs = (ADT0)(T)(TF)(G)(Y)(D)(L)*365
• Where:
   –   ADT0         =   Initial Average Daily Traffic
   –   T            =   Percent Trucks (decimal)
   –   TF           =   Truck Factor (decimal)
   –   G            =   Growth Factor
   –   Y            =   Design Period
   –   L            =   Lane Distribution Factor (decimal)
   –   D            =   Directional Distribution Factor (decimal)




Topic 4 – Traffic Characterization
2. Computation of Design ESALs
Design ESALs:
              m
ESALs = ∑ ESAL (i)                 m = vehicle types           Must calculate ESALs for
                                                               each vehicle type
              i=1

ESAL(i) = (ADT0)(T)(Ti)(TFi)(G)(Y)(D)(L)*365
   New terms:
   Ti = Distribution of specific type of truck within all trucks (decimal)
   TFi = Truck factor for the specific truck type (decimal)

2.1 Average Daily Traffic (ADT)
Unless otherwise stated, ADT is in all lanes & both directions
Also, ADT includes:
   • Cars
   • Single-unit trucks & buses
   • Multiple-unit trucks
Topic 4 – Traffic Characterization
2.2 Average Daily Truck Traffic (ADTT) or (T)
• Minimum traffic information required for pavement design;
  everything else can be found in tables
• Very important in pavement design
   – Effort to collect actual data
   – Table 6.9; guide to distribution of truck types among total amount of
     trucks

2.2.1 Example:
• 4000 ADT ; 20% Trucks; Rural System/Principal
• Find the # of 2-axle, 4-tire trucks


# 2-axle, 4-tire trucks = (4000) (0.2) (365)

                          ADT     T




Topic 4 – Traffic Characterization
Topic 4 – Traffic Characterization

2.2 Average Daily Truck Traffic (ADTT) or (T)
• Minimum traffic information required for pavement
  design; everything else can be found in Tables
• Very important in pavement design
     – Effort to collect actual data
     – Table 6.9; guide to distribution of truck types among total
       amount of trucks

2.2.1 Example:
• 4000 ADT ; 20% Trucks; Rural System/Principal
• Find the # of 2-axle, 4-tire trucks


# 2-axle, 4-tire trucks = (4000) (0.2) (365) (0.6) =175,000/year

                            ADT     T        Table 6.9




Topic 4 – Traffic Characterization
2.3 Truck Factor (TF)
Sum of ESALs divided by the number of trucks weighed (count
of trucks, not axles)
             ESALs
    TF =                                       What is the importance of TF?
           # of Trucks
For the same ESALs; if TF increases      Less # of trucks
If less # of trucks produce the same ESALs      More severe loads

•    Single TF can be applied to all trucks (weighed average); or
     separate for each truck type if the growth rates are different
•    Table 6.10 (Truck Factors)
     – If we use all trucks, we do not have to calculate ESALs
        for each truck type
Topic 4 – Traffic Characterization




Topic 4 – Traffic Characterization

2.4 Directional Distribution (D)
• Usually assume D = 0.5
        Where could that be different?
2.5 Lane Distribution (L)
• Function of ADT & # of lanes (Table 6.14)
        Outer traffic
        Center traffic        •      We design for Outer Lane, but
                                     everything is built the same
        Inner traffic
                              •      Inner Lane usually under-
                                     loaded
Topic 4 – Traffic Characterization

2.6 Growth Rate Factor (G)
• Assuming a yearly rate of growth (r)

          G=
                 1
                 2
                         [
                   × 1 + (1 + r )
                                 Y
                                     ]
• Asphalt Institute (Table 6.13); (G)(Y) combined




Topic 4 – Traffic Characterization

2.7 Example 1
4-lane Rural/Principal
4000 ADT                     Determine the ESALs for
20% Trucks                   2-axle, 6-tire trucks
20-year design; r=4%

ESAL(i) = (ADT0)(Ti)(TFi)(G)(Y)(D)(L)*365
• Distribution of trucks (Table 6.9)
   – Ti = 10% (for 2-axle, 6-tire) * 20% (trucks) = 0.02
• Truck factor (Table 6.10)
   – TFi = 0.25
• Growth (Table 6.13)
   – GY = 29.78
• Lane Distribution (Table 6.14)
   – L = 0.94
                     ESAL(i)=4000*0.02*0.25*29.87*0.5*0.94*365
                            = 102,175 ESALs/20 years
Topic 4 – Traffic Characterization

2.8 Example 2
4-lane Rural/Principal
4000 ADT                  Determine the total ESALs
20% Trucks
20-year design; r=4%

ESALs = (ADT0)(T)(TF)(G)(Y)(D)(L)*365

• Trucks
   – T = 20% (trucks) = 0. 2
• Truck factor (Table 6.10)
   – TF = 0.38

ESALs = 4000*0.2*0.38*29.87*0.5*0.94*365
      = 1.55*106 ESALs/20 years