Explanation Given: Design life is 20 years. Annual rate of traffic growth is 3.5 % . Sub grade k value is 50 lb / in 2 . Stabilized sub-base is of 6 in . Modulus of rupture of the concrete is 600 lb / in 2 . P i = 4.5 P t = 2.5 S c ′ = 650 lb / in 2 E c = 5 × 10 6 lb / in 2 J = 3.2 C d = 1.0 S a = 0.3 R = 9.5 % Traffic volume data on the highway indicate that the AADT during the first year of operation is 24 , 000 with the following vehicle mix and axle loads. Passenger cars = 50 percent 2-axle single-unit trucks ( 12 , 000 lb / axle ) = 40 percent 3-axle single-unit trucks ( 16 , 000 lb / axle ) = 10 percent The pavement has doweled joints and no concrete shoulders. Formula used: The design equivalent single axle load is given by, ESAL = G jt × AADT i × 365 × N i × F Ei ....... (I) Here, G jt is the growth factor for a given growth rate r and design period n , AADT is the first annual average daily traffic for axle category i , N i is the number of axles on each vehicle in category i and F Ei is the load equivalency factor for axle category i . The total ESAL is given by, ESAL = ESAL 1 + ESAL 2 ....... (II) Here, ESAL 1 is the equivalent single axle load for 2-axle single-unit trucks and ESAL 2 is the 3-axle single-unit trucks . The stress ratio is given by, Stress ratio = Equivalent stress S c ′ ....... (III) Here, S c ′ is the working stress of the concrete. The damage percent is given by, Damage percent = E S A L 2 N ...... (IV) Here, N is the number of repetitions of load. Calculation: The design equivalent single axle load for 2-axle single-unit trucks is calculated as, Substitute 28.28 for G jt , 24 , 000 for AADT, 2 for N i and 0.40 for F Ei in equation (I). ESAL 1 = 28.28 × 24 , 000 × 365 × 2 × 0.40 = 79.27 × 10 6 The design equivalent single axle load for 3-axle single-unit trucks is calculated as, Substitute 28.28 for G jt , 24 , 000 for AADT, 3 for N i and 0.40 for F Ei in equation (I). ESAL 2 = 28.28 × 24 , 000 × 365 × 3 × 0.40 = 29.73 × 10 6 The design headwater depth is calculated as, Substitute 79.27 × 10 6 for ESAL 1 and 29.73 × 10 6 for ESAL 2 in equation (II). ESAL = 79.27 × 10 6 + 29.73 × 10 6 = 109 × 10 6 From table 20.22, "Design k values for untreated and Cement-Treated sub-bases" of book "Traffic and highway engineering" for 6 in untreated sub-base k = 230 lb / in 2 . Assume a slab with aggregate interlock joints and concrete shoulders having thickness of 8 inches . From table 20.24, "Equivalent stress values for single axles and tandem axles" of book "Traffic and highway engineering" equivalent stress values can be obtained and for 8 inches thick slab by interpolation effective k can be obtained as, Equivalent stress = 242 lb / in 2 − [ ( 242 lb / in 2 − 225 lb / in 2 ) × ( 230 lb / in 2 − 200 lb / in 2 ) ( 300 lb / in 2 − 200 lb / in 2 ) ] = 242 lb / in 2 − 5

The minimum depth of concrete pavement required for the design period of 20 years.

Traffic and Highway Engineering

5th Edition

ISBN: 9781305156241

Author: Garber, Nicholas J.

Publisher: Cengage Learning

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