SME
Transportation Engineering · Problem type
ESALs from an axle load
Use this for “Trucks with a 100 kN single axle cross a road 500 times a day. By the fourth-power rule, how many 80 kN ESALs does it carry in 20 years?”
Chart
A log scale: each grid line up is 10 times the one below.
Axle load P = 100 kN
Standard axle load Pₛ = 80 kN
Trucks per day N = 500
Design life in years Y = 20
This is an example. Type your own values to start a new problem.
LEF
=
(
P (kN)
÷
Pₛ (kN)
)
4
Ed
=
LEF
×
N
ESAL
=
Ed
×
365
×
Y
Pₛ is the standard axle, N the trucks that cross each day and Y the design life in years.
Formulas
LEF
=
(
P
Pₛ
)
4
2.4414
=
(
100
80
)
4
Ed
=
LEF
×
N
1,220.7 = 2.4414 × 500
ESAL
=
Ed
×
365
×
Y
8,911,133 = 1,220.7 × 365 × 20
Assumptions
•
Each truck has one single axle at the load P; the standard axle is 80 kN (18 kips).
•
No traffic growth: the same number of trucks every day of the design life.
•
The fourth-power rule approximates the AASHO Road Test results; the full AASHTO equivalency factors also depend on SN and the end condition.
Step-by-step
WE KNOW
P = 100 kN, Pₛ = 80 kN, N = 500, Y = 20
FIND
load equivalency factor (LEF), ESALs per day (Ed), total ESALs (ESAL)
Step 1 · Find load equivalency factor (LEF)
Use LEF = (P ÷ Pₛ)⁴
Damage
grows
with
the
fourth
power
of
the
load:
one
pass
of
P
does
as
much
as
this
many
standard
axles.
LEF
=
(
100
80
)
4
LEF
=
1.25
4
LEF = 2.4414
Step 2 · Find ESALs per day (Ed)
Use Ed = LEF × N
Each truck counts as LEF standard axles.
Ed = 2.4414 × 500
Ed = 1,220.7
Step 3 · Find total ESALs (ESAL)
Use ESAL = Ed × 365 × Y
The same ESALs every day, 365 days a year, for the design life.
ESAL = 1,220.7 × 365 × 20
ESAL = 445,555.5 × 20
ESAL = 8,911,133
Check
Put the numbers back in. Both sides should match.
2.4414
=
(
100
80
)
4
✓
1,220.7 = 2.4414 × 500 ✓
8,911,133 = 1,220.7 × 365 × 20 ✓
Topic check
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