At first glance, this equation looks quite complex – it is.
Where: | W | = | axle applications inverse of equivalency factors (where W18 = number of 18,000 lb (80 kN) single axle loads) |
Lx | = | axle load being evaluated (kips) | |
L18 | = | 18 (standard axle load in kips) | |
L2 | = | code for axle configuration 1 = single axle 2 = tandem axle 3 = triple axle (added in the 1986 AASHTO Guide) x = axle load equivalency factor being evaluated s = code for standard axle = 1 (single axle) |
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G | = | ![]() |
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pt | = | “terminal” serviceability index (point at which the pavement is considered to be at the end of its useful life) | |
b | = | ![]() D = Slab Depth in inches |
Example Calculation
- Assumptions: Single axle, 30,000 lb (133 kN), D = 7 in., pt = 2.5
- Answer: (Table D.13, p. D-15, 1993 AASHTO Guide) = 7.7
- Calculations
where : | W18 | = | predicted number of 18,000 lb (80 kN) single axle load applications |
W30 | = | predicted number of 30,000 lb (133 kN) single axle load applications | |
Lx | = | L30 = 30 | |
L2x | = | 1 (single axle) | |
G | = | serviceability loss factor | |
= | ![]() |
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b30 | = | curve slope factor | |
= | ![]() |
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and | G/b30 | = | -0.1761/5.7298 = -0.03073 |
b18 | = | ![]() |
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G/b18 | = | -0.1761/1.3709 = -0.12845 | |
Thus, | ![]() |
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and | ![]() |
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Finally, | LEF | = | ![]() |