Explanation Given: Design life is 20 years. Annual rate of traffic growth is 3.5 % . Sub grade k value is 50 lb / in 3 . 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 aggregate interlock joints (no dowels) and a concrete shoulder. 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 3 . Assume a slab with aggregate interlock joints and concrete shoulders having thickness of 6.5 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 6.5 inches thick slab by interpolation effective k can be obtained as, Equivalent stress = 260 lb / in 2 − [ ( 260 lb / in 2 − 243 lb / in 2 ) × ( 230 lb / in 2 − 200 lb / in 2 ) ( 300 lb / in 2 − 200 lb / in 2 ) ] = 260 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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