Explanation Given: The reliability of the pavement is 90 % . The overall standard deviation is 0.35 . The value of initial PSI is 4.6 The value of TSI is 2.5 The drainage coefficient is 1 The load transfer coefficient is 3.0 . The modulus of elasticity is 4 .5 million lb / in 2 . The thickness of slab is 10 inches . The modulus of rupture is 700 lb / in 2 The deign life of the fixed pavement is same as the design life of rigid pavement. 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 total equivalent load factor. T L EL = W 20 S + W 42 T …… (II) Here, T L EL is total equivalent load factor. Write the expression for the 18-kip equivalent single axle load. W 18 = T L EL × D L × 365 …… (III) Here, W 18 is the 18-kip equivalent single axle load and D L is the design life. 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 ] } ) ] …… (IV) 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 loss in serviceability. ΔPSI = Initial PSI − TSI …… (V) Write the expression for the durability of the pavement. N = W 18 T L EL × 365 …… (VI). Here, N is the design life of the pavement. Write the expression for the 18 -kips equivalent single axle load for flexible pavement log 10 W 18 = [ Z R S 0 + 9.36 [ log 10 ( SN+1 ) ] − 0.20 + log 10 [ Δ PSI / 2.7 ] 0.40 + [ 1094 / ( SN + 1 ) 5.19 ] + 2.32 log 10 M R − 8.07 ] ……. (VII) Here, Z R is the reliability, 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, M R is the soil resilient modulus of the subgrade and W 18 is the 18 -kips equivalent axle load. Write the expression for the current single axle load for rigid pavement. W C 18 = 18-kip single axle W 18 × P D L …… (VIII) Here, W C 18 is the current single axle load P D L is the proportional direction W 18 assumed to be in the design lane. Write the expression for the remaining single axle load for the rigid pavement. W R 18 = W 18 − W C 18 …… (IX) Here, W R 18 is the remaining single axle load. Write the expression for the remaining trucks. Remaining truck = W R 18 T L EL …… (X) Write the expression for the remaining single axle load for the flexible pavement. W R 18 = Remaining Truck × T L EL …… (XI) Write the expression for the resilient modulus. M R = 1500 × C B R …… (XII) Here, C B R is the California Bearing Ratio of the soil. 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 pavement slab thickness 10 inch and axle load 12 kips is 0.175 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 10 inch and axle load 24 kips is 0.441 Substitute 0.175 for W 12 S and 0.441 for W 24 T in equation (II). T L EL = 0.175 + 0.441 = 0.616 Refer Table 4.4 “Cumulative Percent Probabilities of Reliability, R , of the standard Normal Distribution and Corresponding Z R of the book “Principles of Highway Engineering and traffic, 6 th Edition” The value of Z R corresponding to 90 % reliability is − 1.282 . Substitute 4.6 for Initial PSI and 2.5 for TSI in equation (V). Δ PSI = 4.6 − 2.5 = 2.1 Refer Table 4.9 “Relationship between California Bearing Ratio and modulus of subgrade reaction” of the book “Principles of Highway Engineering and traffic, 6 th Edition” k = 100 lb / in 3 Substitute − 1.282 for Z R , 0

The structural number such that PCC and the flexible pavement have same life expectancy.

Principles of Highway Engineering and Traffic Analysis

6th Edition

ISBN: 9781118879078

Author: Mannering

Publisher: WILEY

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