Question

A flexible pavement is to be designed to last 10 years. The initial PSI is 4.2 and the TSI (the final PSI) is determined to be 2.5. The subgrade has a soil resilient modulus of 15,000 lb/in2 . Reliability is 95% with an overall standard deviation of 0.35. For design, the daily car, pickup truck and light van traffic is 40,000 and the daily truck traffic consists of 1500 passes of single-unit trucks with two single axles and 325 passes of tractor semi-trailer trucks with single, tandem, and triple axles. The axle weights are:

• Cars, pickups, light vans = two 2 kip single axles

• Single-unit truck = 8 kip single axle and 24 kip single axle

• Tractor semi-trailer truck = 10 kip single axle, 18 kip tandem axle, and 42 kip triple axle.

Answer 1

we got D2=3.2 inches. . we will sound it off to hio Ebcz.3.2 would be conservatoire estimate and allow for variations in constructions, rounding off to Dext number and safe practise),

substituting weget 109.000) =-1.60520-35/19-swi] -020 +10910 [1.772.75 +2.32109 6:5000 -8.07 0.404 1094 (51+195.19 solne for SN. Then compare this value with the assumed value' (2358 lose to u), This nalue is very close to the value we have assumed. Otherwise we hame to assume the other sirahe and repeat One more iteration, Then for the calculated SN (napprox 3.5) we have to find The value of the De=? einen a,-0-00.-(surface course-hot mix) ag=0.20 - C base course-soilcement a3=0.11 (93=0.11-subbase -Crushed - ston ally = Mzal eginen), Diapo hio inches Dz =? D3= 10 inches SN=aID, +Q2D, M. & Q3 D3M3 9.5 20.04x4 +0:20Y D2(1:0) +0:11 x10x1 D2 = 3.2 inchy.

-0.20 Then we have to use the egn. 6 log.Wię = ZRS0 +9:36 (10960 (sN+!) -6.20 +log.[OPSI 29) 0.4D +[1094 (EN +y%) +2.32log me for Wie = 18 lipeg single axle load=16000 lb FR = reliability = -1.645 so = Brerall Standard deviation =0.35 (given) SN: Structural Number (?) DOSI = difference of initiay PSI and find PS2] MR = soil resilent modulus of fubgrade . in eblin2 = 15000eblina (ginen). .=1.71

and final Ps] =27 with an initial PSI of 4.2 [ginen in Question]. APSI = 707 TSN = Qi DitQLD, M2 +93 DsM3. 1-0 where 21,92,93 : Stsuctural layer coefficients of wearing surface, base, and - Subbase layers respectively Di, D2,D3 : Thickness of the wearing surface, base &subbase layers in inches sespectively M2/M3: dsianage coefficients for the base BH and Subbase respectively (ginen) Di= 4 inches M2 c . z = 10 inchos I m 2 Gal (given) Dz = ? ginen From standard tables. We can find a values for ginens type of panement components, I coefficient of a = wearing surface > for HM & Cginen) = 0.HH a = coeff.foot Bare for soil cement (given) = 0.20 az - Subbage : Crushed stone = Oill

for semi tractor tsucks. Note the compositerely small effect of cars and other light vehicles in terms of the 18 kip ESAL This small effect underscores the nonlinear relationship between the axle loads and the pavement damage for example From table: Atle load equivalency for tandem-arles with SN=41 a 36kip single axle load has 14-4 times the impact on parement acan le kip single axle Icad (twice the weight has lu lu times the impact. This is the meaning of this). Cinens the computed 18 kip ESAL, the daily traffic on this highway produce an is kip ESAL total of 19 684.85 [0.00 OuX 40000+ 2.931 *1500 +0-836*325). = 4,684.85 Nept Traffic (total axle accumulatrons) ores The ginen 10-year design period will be to 4,684.85 X365x10 =17099702-518k ipecae

[from table] Ciren this, 18 kip equivalent Single arte load for cats, pickups, and light vans is & kip Single axle equivalent -0.0009- - This gives an 18-kip ESAL total of 0.0004 for each vehide - For Single unit tsucks). skip Single axle equivalent = 0.041 (from 1 table] 24-kip single axle equivalent = 2.890 (from table) This gives an 18-kip ESAL total of 2.931 [0.oul+ 2.890] for single unit trucks. For tractor sersi trailer tsucles. table of 10 kip sing-axle equivalent = 0.102 18 kip tandem-arle equivalent = 0.092[From 42 kip tsiple-axle equivalent = tandem anderen 0.644 [from table: Axle-load Equival- -ency factors for Flexible parements, Triple-axles . This gives an 18 kip ESAL of total of 0.838 [0.102 +0.092 +0.604]

Solution * Because the axle - load equivalency factors from standard tables. axle load-equirelency factors are a function of the stsuctural number (sw), we have to find assume that an Stsuctual Number (SN), so that to start the problem. Swe will calculate and will solve to assine at a structural numbes at the end of the problem, then we have to check to male sure that it is consistents with ous assumed value). Let we assume a typical assumption ie, SN1 = 4. Table: for axle load equivalency Factor for flexible pavements, single axles, TST=2.5 Axle load SN. Parement structual kips 4 Cassumed) Number 2 --> 0.0002 .22 - 2.09 h = 0.003 24 - 2.89 6 +0.013 $ 0. 10 0.102 12 +0.213 14 +0. 388 16 nyo.645 36 – 14.4 18 - 10 ..20 1.47 26 - 3.91 & - 5.91 6 -6.8 32 - sig 34 - 11.3 38 - 18.1 uo- 22.5