Explanation Given: The load transfer coefficient is 2.8 . The drainage coefficient is coefficient is 1 . The modulus of rupture is 695 lb / in 2 . The modulus of elasticity is 6.0 × 10 6 lb / in 2 . The design life is 40 years. The CBR value is 20 . The overall standard deviation is 0.6 . The initial PSI is 4.2 . Number of 30000 lb triple axles are 400 . Number of 20000 lb tandem axles are 900 . Number of 10000 lb single axles are 1200 . The thickness of the rigid pavement is 11 inch . Formula Used: Write the expression for the total axle load factor. W n = A F n × Number of axles …… (I) Here, W n is the total axle load factor, A F n is the axle load factor for a particular load. Write the expression for the 18-kip equivalent single axle load. log 10 W 18 = [ Z R S 0 + 7.35 [ log 10 ( D+1 ) ] − 0.06 + log 10 [ Δ PSI / 3 ] 1 + [ 1.624 × 10 7 ( D + 1 ) 8.46 ] + ( 4.22 − 0.32 TSI ) log 10 ( S c ′ C d [ D 0.75 − 1.132 ] 215.62 J { D 0.75 − [ 18.42 ( E c k ) 0.25 ] } ) ] …… (II) Here, Z R is the reliability of the standard normal distribution, S 0 is the overall standard deviation, Δ PSI is the loss in serviceability from the time the pavement is new until it reaches its TSI, TSI is the terminal serviceability index, C d is the drainage coefficient, J is load transfer coefficient, E c is the modulus of elasticity, k is the modulus of subgrade reaction, S c ′ is the modulus of rupture and W 18 is the 18 -kips equivalent axle load. Write the expression for the soil resilient modulus. M R = 1500 × CBR …… (III) Here, M R is the soil resilient modulus and CBR is the California Bearing Ratio. Write the expression for the loss in serviceability. ΔPSI = Initial PSI − TSI …… (IV) Here, ΔPSI is the loss in serviceability. Write the expression for the total equivalent load. T L EL = W 10 s + W 20 Tandem + W 30 T r i p l e …… (V) Write the expression for the design lane 18-kip equivalent single axle load. Des W 18 = T L EL × D L × 365 × P D L …… (VII) Here, Des W 18 is the design lane 18-kip equivalent single axle load, P D L is the proportional direction W 18 assumed to be in the design lane and D L is the design life. Calculation: Refer Table 4.6 “Axle load equivalency factors for Rigid pavement, Single Axles, and TSI = 2.5 ” of the book “Principles of Highway Engineering and traffic, 6 th Edition.” The axle load equivalency factor corresponding to the pavement slab thickness 11 inch and axle load 10 kips is 0.080 . Substitute 0.080 for A F 10 and 1200 for Number of axles in equation (I). W 10 S = 0.080 × 1200 = 96 Refer Table 4.7 “Axle load equivalency factors for Rigid pavement, Tandem Axles, and TSI = 2.5 ” of the book “Principles of Highway Engineering and traffic, 6 th Edition.” The axle load equivalency factor corresponding to pavement slab thickness 11 inch and axle load 20 kips is 0.203 . Substitute 0.203 for A F 20 and 900 for Number of axles in equation (I). W 20 T = 0.203 × 900 = 182.7 Refer Table 4.8 “Axle load equivalency factors for Rigid pavement, Triple Axles, and TSI = 2.5 ” of the book “Principles of Highway Engineering and traffic, 6 th Edition.” The axle load equivalency factor corresponding to pavement slab thickness 11 inch and axle load 30 kips is 0