Determine the required slab thickness, according to the AASHTO Alternate Design Method, for a rigid pavement being designed near Jacksonville, Florida, for the following input data:
Estimated future traffic load = 10,000,000 ESALs
Design reliability = 95%
Overall standard deviation = 0.39
Initial serviceability index = 4.5
Terminal serviceability index = 2.5
Effective subgrade modulus = 250 lb/in2
Mean concrete modulus of rupture = 700 lb/in2
Mean concrete elastic modulus = 4,200,000 lb/in2
Joint spacing = 20 ft
Base modulus = 25,000 lb/in2
Base thickness = 6 inches
Slab width = 12 ft supported by asphalt concrete shoulders
What will be an ideal response?
Solution:
Rigid pavement type: jointed plain concrete pavement (JPCP)
according to Table 20.14, Jacksonville, FL has:
mean annual wind speed = WIND = 8.1 mph,
mean annual temperature = TEMP = 68°F,
mean annual precipitation = PRECIP = 52.8 in
assume slab thickness D = 11.5 in
effective positive TD = 0.962 – 52.181/D + 0.341 WIND +0.184 TEMP – 0.00836 PRECIP = 11.25
According to Table 16 (Page 28), the required thickness of the concrete slab is
11.9 inches.
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