EALF

(redirected from Equivalent Axle Load Factor)
AcronymDefinition
EALFEquivalent Axle Load Factor (traffic engineering term)
EALFElectronic Annual Learning Forum (Canada)
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References in periodicals archive ?
Generally, this is described by the load equivalency factor (LEF, also equivalent axle load factor), where an axle load is said to be equivalent (producing equal pavement wear) to a number of applications of a reference (standard) axle load.
The equivalent axle load factor [F.sub.j] defines the damage to a pavement by pass of the actual axle in question relative to the damage by pass of a standard axle load (Huang 2003).
When combined, the formulas (3) and (4) yield the following formula for calculating the equivalent axle load factor:
For this reason the parameters K in Eq (5) will be reduced and the formula for equivalent axle load factor will take a simpler fOrm:
It can be rearranged to obtain the following formula for calculating the equivalent axle load factor:
1 that the form of fatigue criteria have significant effect on equivalent axle load factor. The exponent n is constant in the French method (n = 16) and nearly constant in the AASHTO 1993 method (n [approximately equal to] 4).
The equivalent axle load factors are the basis for determining the design traffic and, as a consequence, for designing of road pavements.
The AASHTO Guide for Design of Pavement Structures (1993) uses standard axle load of 80.1 kN (18-kip) and presents a set of tables which define the equivalent axle load factors for rigid pavements in relation to actual axle load, axle configuration, concrete slab thickness and terminal value of [p.sub.t]--Present Serviceability Index.
Kuo and Lin (2001) used finite element model and developed regression formulas for equivalent axle load factors for rigid pavements.
Concept for calculation of equivalent axle load factors based on the fatigue criteria
Calculated values of equivalent axle load factors [F.sub.j]
Figs 1 and 2 present calculated values of equivalent axle load factors [F.sub.j] and exponents n in relation [F.sub.j] = [([Q.sub.j]/[Q.sub.s]).sup.n].
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