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Flexible Pavement ESAL Equation

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

WxW18=[L18+L2sLx+L2x]4.79[10G/βx10G/β18][L2x]4.33\frac{W_x}{W_{18}} = \left[ \frac{L_{18} + L_{2s}}{L_x + L_{2x}} \right]^{4.79} \left[ \frac{10^{G/\beta_x}}{10^{G/\beta_{18}}} \right] \left[ L_{2x} \right]^{4.33}

where:

  • WW = axle applications inverse of equivalency factors (where W₁₈ = number of 18,000 lb (80 kN) single axle loads)
  • LxL_x = axle load being evaluated (kips)
  • L18L_{18} = 18 (standard axle load in kips)
  • L2L_2 = 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)
  • GG = log⁡(4.2−pt4.2−1.5)\log\left( \dfrac{4.2 - p_t}{4.2 - 1.5} \right), a function of the ratio of loss in serviceability at time, t, to the potential loss taken at a point where pₜ = 1.5
  • ptp_t = "terminal" serviceability index (point at which the pavement is considered to be at the end of its useful life)
  • β\beta = 0.4+(0.081(Lx+L2x)3.23(SN+1)5.19L2x3.23)0.4 + \left( \dfrac{0.081 (L_x + L_{2x})^{3.23}}{(\mathit{SN} + 1)^{5.19} L_{2x}^{3.23}} \right), function which determines the relationship between serviceability and axle load applications
  • SN\mathit{SN} = structural number

Example Calculation for a Single Axle

  • Assumptions: Single axle, 30,000 lb (133 kN), SN = 3, ptp_{t} = 2.5
  • Answer: (Table D.4, p. D-6, 1993 AASHTO Guide) = 7.9
  • Calculations
W30W18=[18+1L30+L2x]4.79[10G/β3010G/β18][L2x]4.33\frac{W_{30}}{W_{18}} = \left[ \frac{18 + 1}{L_{30} + L_{2x}} \right]^{4.79} \left[ \frac{10^{G/\beta_{30}}}{10^{G/\beta_{18}}} \right] \left[ L_{2x} \right]^{4.33}
Table 1.
where :W₁₈equalspredicted number of 18,000 lb (80 kN) single axle load applications
W₃₀equalspredicted number of 30,000 lb (133 kN) single axle load applications
L_xequalsL30 = 30
L_(2x)equals1 (single axle)
Gequalsserviceability loss factor
equalslog⁡(4.2−2.54.2−1.5)=−0.2009\log\left( \dfrac{4.2 - 2.5}{4.2 - 1.5} \right) = -0.2009
b₃₀equalscurve slope factor
equals0.4+(0.081(30+1)3.23(3+1)5.19(1)3.23)=4.3880.4 + \left( \dfrac{0.081 (30 + 1)^{3.23}}{(3 + 1)^{5.19} (1)^{3.23}} \right) = 4.388
andG/b₃₀equals-0.2009/4.388 = -0.04578
b₁₈equals0.4+(0.081(18+1)3.23(3+1)5.19(1)3.23)=1.22040.4 + \left( \dfrac{0.081 (18 + 1)^{3.23}}{(3 + 1)^{5.19} (1)^{3.23}} \right) = 1.2204
G/b₁₈equals-0.2009/1.2204 = -0.1646
ThusW30W18=[18+130+1]4.79[10−0.0457810−0.1646][1]4.33=0.1260\dfrac{W_{30}}{W_{18}} = \left[ \dfrac{18 + 1}{30 + 1} \right]^{4.79} \left[ \dfrac{10^{-0.04578}}{10^{-0.1646}} \right] \left[ 1 \right]^{4.33} = 0.1260
andW30W18≅12.6%\dfrac{W_{30}}{W_{18}} \cong 12.6\%
FinallyLEFequals10.1260=7.9365≅7.9\dfrac{1}{0.1260} = 7.9365 \cong 7.9(same as contained in 1993 AASHTO Guide, Appendix D)

Example Calculation for a Tandem Axle

  • Assumptions: Tandem axle, 40,000 lb (133 kN), SN = 5, ptp_{t} = 2.5
  • Answer: (Table D.5, p. D-7, 1993 AASHTO Guide) = 2.08
  • Calculations
W40W18s=[L18+L2sL40+L2x]4.79[10G/β4010G/β18][L2x]4.33\frac{W_{40}}{W_{18s}} = \left[ \frac{L_{18} + L_{2s}}{L_{40} + L_{2x}} \right]^{4.79} \left[ \frac{10^{G/\beta_{40}}}{10^{G/\beta_{18}}} \right] \left[ L_{2x} \right]^{4.33}
Table 2.
where :L40equals40 (tandem axle)
L18equals18 (single axle)
L2xequals2 (tandem axle)
L2sequals1 (single axle)
Gequalsserviceability loss factor
equalslog⁡(4.2−2.54.2−1.5)=−0.2009\log\left( \dfrac{4.2 - 2.5}{4.2 - 1.5} \right) = -0.2009
b40equalscurve slope factor
equals0.4+(0.081(40+2)3.23(5+1)5.19(2)3.23)=0.538240.4 + \left( \dfrac{0.081 (40 + 2)^{3.23}}{(5 + 1)^{5.19} (2)^{3.23}} \right) = 0.53824
andG/b40equals-0.2009/0.53824 = -0.37325
b18equals0.4+(0.081(18+1)3.23(5+1)5.19(1)3.23)=0.500060.4 + \left( \dfrac{0.081 (18 + 1)^{3.23}}{(5 + 1)^{5.19} (1)^{3.23}} \right) = 0.50006
G/b18equals-0.2009/0.50006 = -0.40175
ThusW40W18=[18+140+2]4.79[10−0.3732510−0.40175][2]4.33=0.48064\dfrac{W_{40}}{W_{18}} = \left[ \dfrac{18 + 1}{40 + 2} \right]^{4.79} \left[ \dfrac{10^{-0.37325}}{10^{-0.40175}} \right] \left[ 2 \right]^{4.33} = 0.48064
FinallyLEFequals10.48064=2.08\dfrac{1}{0.48064} = 2.08 (same as contained in 1993 AASHTO Guide nbsp;Appendix D)

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APA

Pavement Interactive. (2018, September 7). Flexible pavement ESAL equation. HeadLight. Retrieved [the date you read it], from https://pavementinteractive.org/reference-desk/design/design-parameters/flexible-pavement-esal-equation/

ASCE

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

BibTeX

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