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1、Course Project Road Pavement EngineeringHighway School of Changan UniversityXXXAdvisor: XXX2016-xx-xx1 Part I Data Collection The project is segment A of Shijiazhuang to Zhengzhou grade 1 highway. 1.1 Climate data Natural Zone: II4 Annual Rainfall: 570mm; HT: 37C; LT: -7C Frost-thaw: Max frozen dept
2、h of 60cm; Frozen index of 1023C on average and 1540C in Max 1.2 Geological and subgrade conditions Soil Type: alluvial soil(as clay soils of low liquid limit) Embankment: 0.6m of height ； Water table: 2.7m from the roadbed surface in mid-wet state； Subgrade width: 24.5m=4*3.75+2*2.5(Hard shoulder)+
3、2*0.75(soil shoulder)+3.0(central median) 2 1.3 Traffic data Table 1. Traffic combination N O. Classificati on Vehicle models Daily traffic flow(veh/d) Vehicle conversion coefficient 1 Medium size Mitsubishi T653B 320 1.5 2 Medium size JieFang CA390 220 1.5 3 Large size JAC HF150 180 2 4 Large size
4、CNHTC JN162 230 2 5 Trailer XJ HQP40 140 3 6 Trailer DFM EQ155 90 3 Note: NO. 5 & 6 is vision traffic, only considered in the calculation of the pavement structure. Part II Asphalt Pavement Design 1.1 Traffic Analysis ESALs calculation Specific calculating process is in excel, and applied equations
5、are: Deflection and asphalt layer bottom tensile stress as design criteria 3 Tensile stress at the bottom of semi-rigid base as design criteria Accumulative equivalent ESALs Increasing ratio for every five years are 1=8%,2=7%, 3=5% so the total increasing ratio G can be calculated as: G1= (1+8%)5-1/
6、 8%=5.87 G2= (1+8%)5*(1+7%)5-1/ 7%=8.45 G3= (1+8%)5*(1+7%)5*(1+5%)5-1/ 5%=11.39 G=G1+G2+G2=25.71 According to specification for design of highway asphalt pavement clause 3.1.7, Ne= Ne1 =365*8534*25.71*0.5=40.04msa (Deflection and asphalt layer bottom tensile stress as design criteria) Ne2=365*82980*
7、25.71*0.5=389.3msa4 (Tensile stress at the bottom of semi-rigid base as design criteria) 1.2Subgrade Modulus From the table 5.1.4-1 in JTG D50-2006, consistency is between 0.95- 1.10, the corresponding MR is less than 40 MPa and for extremely heavy traffic subgrade MR min =40MPa, so MR=40MPa. 2. Com
8、bination Design 2.1 Asphalt surface course & Base and subbase courses Asphalt surface and Base and subbase material and thickness selection limit is from the table below in JTG D50-2006:56 2.2 Summary Three combination schemes are Scheme1: (Asphalt + asphalt + granular ) 4cmAC13+5cmAC16+7cmAC25+20cm
9、AM20+30cm graded macadam Scheme2 : (Asphalt + HBM + Granular) 5cmAC10+7cmAC16+10cmSMA20+30cm cement stabilized base+15cm graded macadam Scheme3:(Asphalt + Asphalt + HBM) 4cmAC10+7cmAC16+18cmSMA20+30cm limestone soil And choose scheme2 to check whether it meets requirement.7 3. Material proportion pr
10、operties 3.1 Asphalt surface Materials The corresponding material properties can be found in the Table E.1& 2:8 Namely: type E C (20C)MPa E C (15C)PMa Splitting strength(MPa) AC10 1500 2100 1.5 AC16 1300 1800 0.99 3.2 Base and subbase materials Type E C (deflection)MPa E C (tensile stress)PMa Splitt
11、ing strength(PMa) Cement macadam 1600 4000 0.55 SMA 1500 1900 0.8 Graded macadam 450 - - 4. Thickness Design 4.1 Determine design criteria Determine Design Deflection L d:Equation given A C =1(1 for expressway & class I, 1.1 for class II, and 1.2 for others) A S =1(1 for AC and 1.1 for others) A b =
12、1(1 for semi-rigid base, and 1.6 for flexible base) So L d=600*4004*10 4 *1*1*1=18.1(0.01mm) Determine Tensile Stress at bottom of each layer Equation given AC10 layer: K s =0.09N e 0.22 /A c =0.09* .22 /1=4.23 , R = sp /k s =1.5/4.23=0.36MPa10 AC16layer: K s =0.09N e 0.22 /A c =0.09* .22 /1=4.23 ,
13、R = sp /k s =0.9/4.23=0.21MPa SMA20layer: K s =0.09N e 0.22 /A c =0.09* .22 /1=0.55 , R = sp /k s =0.9/4.23=0.21MPa Cement stabilized base layer: K s =0.35N e 0.22 /A c =0.35* .22 /1=16.47 R = sp /k s =0.55/16.47=0.03MPa 4.2 Determine calculated criteria BISAR Solution for surface resilient deflecti
14、on ： BISAR Solution for layer bottom tensile stresses 11 Layer1 Layer2 Layer3 Layer4 4.3 Finalize the thickness design Compare the design criteria and the calculated criteria: Correction of BISAR solution for deflection: Assuming ls=ld=18.1(0.01mm), then F=1.63*(18.1/(2000*10.65)0.38* (40/0.7)0.36=0.476 Therefore, corrected BISAR solution is Ls=33.03*0.476= 15.72(0.01mm) 40cm, there is no need to set another layer. Finial thickness design of scheme2:
