Vehicle and Traffic
Consideration
•Highways
•Airports
4/7/2015
CEE 320
1
Steve Muench
Load Quantification For Highways


Equivalent Single Axle Load (ESAL)
 Converts wheel loads of various magnitudes and repetitions
("mixed traffic") to an equivalent number of "standard" or
"equivalent" loads
 Based on the amount of damage they do to the pavement
 Commonly used standard load is the 18,000 lb. equivalent
single axle load
Load Equivalency
 Generalized fourth power approximation will give ESAL
Factors
4
load



  relative
 18 , 000 lb. 
damage factor
Wheel Configuration (Trucks)
4/7/2015
3
Typical Load Equivalent Factor LEFs
6
ESALs per Vehicle
5.11
5
4
3
1.85
2
1.35
1
0.0007
0.10
Car
Delivery Truck
0
Loaded 18-Wheeler
Loaded 40' Bus
Loaded 60'
Articulated Bus
Notice that cars are insignificant and thus usually
ignored in pavement design.
4/7/2015
4
Load Quantification For Airports

Equivalent Single Wheel Load (ESWL)
 Converts wheel loads of various magnitudes
Configurations, and repetitions (“Different
Gears") to an equivalent number of "standard"
or "equivalent" loads
 Based on the amount of damage they do to
the pavement
Mixed Highway Traffic Analysis
Equivalent Axle Load Computation
Example



LEF (ESAL) Example
Assume a logging truck has three axles:
Truck tractor



Trailer


Steering axle (single axle) = 14,000 lb (62.2 kN)
Drive axle (tandem axle) = 34,000 lb (151.1 kN)
Pole trailer axle (tandem axle) = 30,000 lb (133.3 kN)
The total equivalent damage by this truck (pt = 3.0, SN
= 3): is
 Steering axle @ 14,000 lb=0.47 ESAL, Drive axle @
34,000 lb=1.15 ESAL, Pole axle @ 30,000 lb=0.79
ESALTotal=2.41 ESAL
Highway Axle Configuration
Sample Conversion
Traffic Growth
Traffic Growth
Traffic Growth
Equivalent single wheel load for
Dual Tire (Metric)
Log
ESWL 
10
Log
 Z 

0 . 301 Log 
10
d 2
P
10
 2S 
Log 10  d 2 


Where:
P is the wheel load Kg,
S is the center to center distance between the two
wheels in cm,
d is the clear distance between two wheels in cm,
and
z is the desired depth in cm.
Equivalent single wheel load for
Dual Tire
Example:
Find ESWL at depths of 5cm, 20cm and 40cm for a
dual wheel carrying 2044 kg each. The center to
center tire spacing is 20cm and distance between the
walls of the two tires is 10cm.
P
P
d
d/2
s
2S
Equivalent single wheel load for
Dual Tire

Solution:
 For desired depth z=40cm, which is twice the tire spacing,
ESWL = 2P=2*2044 = 4088 kg.
 For z=5cm, which is half the distance between the walls of
the tire, ESWL = P = 2044kg.
 For z=20cm,
0 . 301
Log
ESWL 
10
Log
Log
2044 
10
Log
 20 


10  10 2 


 2 * 20

10  10 2

Therefore, ESWL = antilog(3.511)= 3244.49 kg



 3 . 511
Equivalent single wheel load for
Dual Tire (Pound)
P
s

P
1 ( S d / 100 )
Pd/2
Pd/2
d
d
sd
d/2
h =30 in
Where:
Pd is the dual wheel load lb,
A
0
Sd is the center to center distance between the two
wheels in inches,
d is the clear distance between two wheels in
inches, and
z is the desired depth in inches.
Equivalent single wheel load for
Dual Tire (Pound)
P
P
P
s
s( A)
s(0)



P
1 ( S
2

P
d
2
/ 2h )
d
d
1 S 2 d 2 h 2  S 2 d

P
[1 ( S d / 100 )][ 1 S t / 100 ]
dt
Equivalent single wheel load
(CBR) Procedure
Nose
Axis of Fuselage
Tail
6’ 3’’
9’ 9’’
46’’
18’ 2’’
64’’
Boeing 747
6’ 3’’
CBR vs Thickness
40
30
Pavement
Thickness 20
Inches
Thickness
10
0
20
40
60
CBR %
80
100
ESAL Tables

See Handouts