Rigid Pavement ESAL Equation

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At first glance, this equation looks quite complex - it is.

\frac{W_{\mathrm{x}}}{W_{18}} = \left[\frac{L_{18}+L_{2\mathrm{s}}}{L_{\mathrm{x}}+L_{2\mathrm{x}}}\right]^{4.62}\left[\frac{10^{G/\beta_x}}{10^{G/\beta_{18}}}\right]\left[L_{2\mathrm{x}}\right]^{3.28}

Where:

axle applications inverse of equivalency factors (where W₁₈ = number of 18,000 lb (80 kN) single axle loads)

L_{x}

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)

$\log\left(\dfrac{4.5-p_t}{4.5-1.5}\right)$ a function of the ratio of loss in serviceability at time, t, to the potential loss taken at a point where $p_{t}$ = 1.5

p_{t}

"terminal" serviceability index (point at which the pavement is considered to be at the end of its useful life)

$1.00+\left(\dfrac{3.63(L_x+L_{2x})^{5.20}}{(D+1)^{8.46}L_{2x}^{3.52}}\right)$ function which determines the relationship between serviceability and axle load applications D = Slab Depth in inches

Example Calculation

Assumptions: Single axle, 30,000 lb (133 kN), D = 7 in., $p_{t}$ = 2.5
Answer: (Table D.13, p. D-15, 1993 AASHTO Guide) = 7.7
Calculations

\frac{W_{\mathrm{x}}}{W_{18}} = \left[\frac{L_{18}+L_{2\mathrm{s}}}{L_{\mathrm{x}}+L_{2\mathrm{x}}}\right]^{4.62}\left[\frac{10^{G/\beta_x}}{10^{G/\beta_{18}}}\right]\left[L_{2\mathrm{x}}\right]^{3.28}

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

L_{x}

L₃₀ = 30

L_{2x}

1 (single axle)

serviceability loss factor

\log\left(\dfrac{4.5-2.5}{4.5-1.5}\right) = -0.1761

b₃₀

curve slope factor

1.00+\left(\dfrac{3.63(30+1)^{5.20}}{(7+1)^{8.46}(1)^{3.52}}\right) = 5.7298

and

G/b₃₀

-0.1761/5.7298 = -0.03073

b₁₈

1.00+\left(\dfrac{3.63(18+1)^{5.20}}{(7+1)^{8.46}(1)^{3.52}}\right) = 1.3709

G/b₁₈

-0.1761/1.3709 = -0.12845

Thus,

\frac{W_{30}}{W_{18}} = \left[\frac{18+1}{30+1}\right]^{4.62}\left[\frac{10^{-0.03073}}{10^{-0.12845}}\right][1]^{3.28} = 0.1305

and

$\dfrac{W_{30}}{W_{18}}\cong 13.05\%$ of W₁₈ loads allowable with a 30,000 lb. single axle

Finally,

LEF

$\dfrac{1}{0.1305} = 7.67\cong 7.7$ (same as contained in 1993 AASHTO Guide âcol Appendix D)

Teaching this? Figures and worked examples are free to use with credit. Course reading lists

Cite this page

APA

Pavement Interactive. (2018, September 19). Rigid pavement ESAL equation. HeadLight. Retrieved [the date you read it], from https://pavementinteractive.org/reference-desk/design/structural-design/rigid-pavement-esal-equation/

ASCE

Pavement Interactive. (2018). “Rigid Pavement ESAL Equation.” HeadLight, <https://pavementinteractive.org/reference-desk/design/structural-design/rigid-pavement-esal-equation/> ([the date you read it]).

BibTeX

@misc{pavint-rigid-pavement-esal-equation,
  author = {{Pavement Interactive}},
  title = {Rigid Pavement ESAL Equation},
  year = {2018},
  publisher = {HeadLight},
  howpublished = {\url{https://pavementinteractive.org/reference-desk/design/structural-design/rigid-pavement-esal-equation/}},
  note = {Accessed [date]}
}

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