Explanation Given: The reliability of the pavement is 95 % . The overall standard deviation is 0.45 . The value of initial PSI is 4.5 The value of TSI is 2.5 The drainage coefficient is 1.0 The load transfer coefficient is 3.0 . The modulus of subgrade reaction is 190 lb / in 3 . The design life is 20 year . The modulus of elasticity is 6 × 10 6 lb / in 2 . The thickness of slab is 12 inches . The modulus of rupture is 600 lb / in 2 Numbers of trucks traveling per day is 1227.76 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 16 S + W 20 S + W 35 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 change in the design life. Δ N = N 0 − N …… (VII) Here, Δ N is the change in design life and N 0 is the original design life of the pavement. 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 12 inch and axle load 16 kips is 0.599 Substitute 0.599 for AF 16 S and 1227.76 for number of axles in equation (I). W 16 S = 0.599 × 1227.76 = 735.428 The axle load equivalency factor corresponding to pavement slab thickness 12 inch and axle load 20 kips is 1.590 Substitute 1.590 for AF 20 S and 1227.76 for number of axles in equation (I). W 20 S = 1.590 × 1227.76 = 1952.1384 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 change in design life of the rigid pavement when the modulus of rupture is changed.

PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.

7th Edition

ISBN: 9781119610526

Author: Mannering

Publisher: WILEY

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Chapter 4, Problem 19P

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