Compute the load equivalency factor LEF for a 76000 lb 3381

Compute the load equivalency factor (LEF) for a 76,000 lb (338.1 kN) tandem axle on a flexible pavement with the following design values: SN = 4, pt = 2.5

Solution

Solution:-

Given

SN = 4 , pt = 2.5

Tandem axle = 76000 lb

W₇₆/W₁₈ = ((L₁₈ + L_2s)/(L₇₆ + L_2x))^4.79 * (10^(G/b76) /10^(G/b18) ) * (L_2x)^4.33

Where

L₇₆ = 76

L₁₈ = 18

L_2x = 2

L_2s = 1

G = Serviceability loss factor

    = log((4.2 -2.5)/(4.2 – 1.5)) = - 0.2009

b₇₆ = curve slope factor

        = 0.4 + ( 0.081* (76 +2)^3.23/((4+1)^5.19 * 2^3.23) =3.03

G/b₇₆ = -0.2009/3.03 = -0.0663

b₁₈ = 0.4 + (0.081 * (18+1)^3.23/((4+1)^5.19 *1^3.23) = 0.6577

G/b₁₈ = -0.2009/0.6577 = -0.30546

W₇₆/W₁₈ = ((18+1)/(76+2))^4.79 *(10^(-0.0663)/10^(-0.30546) ) *(2)^4.33 = 0.0402

Load equivalency factor (LEF) = 1/0.0402 = 24.87   Answer