Explanation Given: The reliability of the pavement is 95 % . The overall standard deviation is 0.3 . The value of initial PSI is 4.7 The value of TSI is 2.5 The drainage coefficient is 1 The load transfer coefficient is 3.0 . The modulus of elasticity is 5,000,000 lb / in 2 . The thickness of slab is 11 inches . The modulus of rupture is 700 lb / in 2 The CBR value of soil is 25 . The original design life is 20 years . 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 total traffic. N = W 18 Total Traffic × T L EL × 365 × P D L …… (VIII) Here, N is the design life and P D L is the proportional direction W 18 assumed to be in the design lane. 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 11 inch and axle load 18 kips is 1.0 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 28 kips is 0.85 Substitute 1.0 for W 18 S and 0.85 for W 28 T in equation (II). T L EL = 1 + 0.85 = 1.85 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 95 % reliability is − 1.645 . Substitute 4.7 for Initial PSI and 2.5 for TSI in equation (V). Δ PSI = 4.7 − 2.5 = 2.2 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 = 290 lb / in 3 Substitute − 1.645 for Z R , 0.3 for S 0 , 2.2 for Δ PSI , 2.5 for TSI , 1 for C d , 3 for J , 5 × 10 6 lb / in 2 for E c , 700 lb / in 2 for S c ′ and 290 lb / in 3 for k in equation (IV)

The duration for which the pavement will last with the new loading and the additional lane.

PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.

7th Edition

ISBN: 9781119610526

Author: Mannering

Publisher: WILEY

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Chapter 4, Problem 23P

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The duration for which the pavement will last with the new loading and the additional lane.

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The duration for which the pavement will last with the new loading and the additional lane.

Solution Summary: The author calculates the duration of the pavement with the new loading and the additional lane.

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The duration for which the pavement will last with the new loading and the additional lane.

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