# Flexible Pavement ESAL Equation

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

- Canonical URL: https://pavementinteractive.org/reference-desk/design/design-parameters/flexible-pavement-esal-equation/
- Last updated: 2018-09-07
- Review: reviewed by licensed professional engineers on the previous Pavement Interactive site, where reviewers were not recorded by name

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

$$\frac{W_x}{W_{18}} = \left[ \frac{L_{18} + L_{2s}}{L_x + L_{2x}} \right]^{4.79} \left[ \frac{10^{G/\beta_x}}{10^{G/\beta_{18}}} \right] \left[ L_{2x} \right]^{4.33}$$

where:

- $$W$$ = 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)
- $$L_{18}$$ = 18 (standard axle load in kips)
- $$L_2$$ = 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)
- $$G$$ = $$\log\left( \dfrac{4.2 - p_t}{4.2 - 1.5} \right)$$, a function of the ratio of loss in serviceability at time, t, to the potential loss taken at a point where pₜ = 1.5
- $$p_t$$ = "terminal" serviceability index (point at which the pavement is considered to be at the end of its useful life)
- $$\beta$$ = $$0.4 + \left( \dfrac{0.081 (L_x + L_{2x})^{3.23}}{(\mathit{SN} + 1)^{5.19} L_{2x}^{3.23}} \right)$$, function which determines the relationship between serviceability and axle load applications
- $$\mathit{SN}$$ = structural number

## Example Calculation for a Single Axle

- Assumptions: Single axle, 30,000 lb (133 kN), SN = 3, $$p_{t}$$ = 2.5
- Answer: (Table D.4, p. D-6, 1993 AASHTO *Guide*) = 7.9
- Calculations

$$\frac{W_{30}}{W_{18}} = \left[ \frac{18 + 1}{L_{30} + L_{2x}} \right]^{4.79} \left[ \frac{10^{G/\beta_{30}}}{10^{G/\beta_{18}}} \right] \left[ L_{2x} \right]^{4.33}$$

**Table 1.**

| where : | W₁₈ | equals | predicted number of 18,000 lb (80 kN) single axle load applications |
| --- | --- | --- | --- |
|  | W₃₀ | equals | predicted number of 30,000 lb (133 kN) single axle load applications |
|  | L_x | equals | L30 = 30 |
|  | L_(2x) | equals | 1 (single axle) |
|  | G | equals | serviceability loss factor |
|  |  | equals | $$\\log\\left( \\dfrac{4.2 - 2.5}{4.2 - 1.5} \\right) = -0.2009$$ |
|  | b₃₀ | equals | curve slope factor |
|  |  | equals | $$0.4 + \\left( \\dfrac{0.081 (30 + 1)^{3.23}}{(3 + 1)^{5.19} (1)^{3.23}} \\right) = 4.388$$ |
| and | G/b₃₀ | equals | -0.2009/4.388 = -0.04578 |
|  | b₁₈ | equals | $$0.4 + \\left( \\dfrac{0.081 (18 + 1)^{3.23}}{(3 + 1)^{5.19} (1)^{3.23}} \\right) = 1.2204$$ |
|  | G/b₁₈ | equals | -0.2009/1.2204 = -0.1646 |
| Thus |  |  | $$\\dfrac{W_{30}}{W_{18}} = \\left[ \\dfrac{18 + 1}{30 + 1} \\right]^{4.79} \\left[ \\dfrac{10^{-0.04578}}{10^{-0.1646}} \\right] \\left[ 1 \\right]^{4.33} = 0.1260$$ |
| and |  |  | $$\\dfrac{W_{30}}{W_{18}} \\cong 12.6\\%$$ |
| Finally | LEF | equals | $$\\dfrac{1}{0.1260} = 7.9365 \\cong 7.9$$(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, $$p_{t}$$ = 2.5
- Answer: (Table D.5, p. D-7, 1993 AASHTO *Guide*) = 2.08
- Calculations

$$\frac{W_{40}}{W_{18s}} = \left[ \frac{L_{18} + L_{2s}}{L_{40} + L_{2x}} \right]^{4.79} \left[ \frac{10^{G/\beta_{40}}}{10^{G/\beta_{18}}} \right] \left[ L_{2x} \right]^{4.33}$$

**Table 2.**

| where : | L40 | equals | 40 (tandem axle) |
| --- | --- | --- | --- |
|  | L18 | equals | 18 (single axle) |
|  | L2x | equals | 2 (tandem axle) |
|  | L2s | equals | 1 (single axle) |
|  | G | equals | serviceability loss factor |
|  |  | equals | $$\\log\\left( \\dfrac{4.2 - 2.5}{4.2 - 1.5} \\right) = -0.2009$$ |
|  | b40 | equals | curve slope factor |
|  |  | equals | $$0.4 + \\left( \\dfrac{0.081 (40 + 2)^{3.23}}{(5 + 1)^{5.19} (2)^{3.23}} \\right) = 0.53824$$ |
| and | G/b40 | equals | -0.2009/0.53824 = -0.37325 |
|  | b18 | equals | $$0.4 + \\left( \\dfrac{0.081 (18 + 1)^{3.23}}{(5 + 1)^{5.19} (1)^{3.23}} \\right) = 0.50006$$ |
|  | G/b18 | equals | -0.2009/0.50006 = -0.40175 |
| Thus |  |  | $$\\dfrac{W_{40}}{W_{18}} = \\left[ \\dfrac{18 + 1}{40 + 2} \\right]^{4.79} \\left[ \\dfrac{10^{-0.37325}}{10^{-0.40175}} \\right] \\left[ 2 \\right]^{4.33} = 0.48064$$ |
| Finally | LEF | equals | $$\\dfrac{1}{0.48064} = 2.08$$ (same as contained in 1993 AASHTO Guide nbsp;Appendix D) |