路基路面课程设计毕业论文含英文版.docx

course projectroad pavement engineering highway school of changan university xxx advisor: xxx 2016-xx-xxpart i data collectionthe project is segment a of shijiazhuang to zhengzhou grade 1 highway. 1.1 climate datanatural zone: ii4annual rainfall: 570mm; ht: 37c; lt: -7c frost-thaw: max frozen depth of 60cm; frozen index of 1023c on average and 1540c in max 1.2 geological and subgrade conditionssoil 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)+2*0.75(soil shoulder)+3.0(central median) 1.3 traffic datatable 1. traffic combinationno.classificationvehicle modelsdaily traffic flow(veh/d)vehicle conversion coefficient1medium sizemitsubishi t653b 3201.52medium sizejiefang ca390 2201.53large sizejac hf15018024large sizecnhtc jn16223025trailerxj hqp4014036trailerdfm eq155903note: no. 5 & 6 is vision traffic, only considered in the calculation of the pavement structure.part ii asphalt pavement design1.1 traffic analysisesals calculation specific calculating process is in excel, and applied equations are:deflection and asphalt layer bottom tensile stress as design criteria tensile stress at the bottom of semi-rigid base as design criteria accumulative equivalent esalsincreasing ratio for every five years are1=8%,2=7%, 3=5% so the total increasing ratio g can be calculated as:g1= (1+8%)5-1/ 8%=5.87 g2= (1+8%)5*(1+7%)5-1/ 7%=8.45g3= (1+8%)5*(1+7%)5*(1+5%)5-1/ 5%=11.39g=g1+g2+g2=25.71according 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*25.71*0.5=389.3msa(tensile stress at the bottom of semi-rigid base as design criteria)1.2subgrade modulusfrom 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 mrmin=40mpa, so mr=40mpa.2. combination design2.1 asphalt surface course & base and subbase coursesasphalt surface and base and subbase material and thickness selection limit is from the table below in jtg d50-2006:2.2 summarythree combination schemes arescheme1: (asphalt + asphalt + granular )4cmac13+5cmac16+7cmac25+20cmam20+30cm graded macadamscheme2 : (asphalt + hbm + granular) 5cmac10+7cmac16+10cmsma20+30cm cement stabilized base+15cm graded macadam scheme3:(asphalt + asphalt + hbm)4cmac10+7cmac16+18cmsma20+30cm limestone soiland choose scheme2 to check whether it meets requirement.3. material proportion properties3.1 asphalt surface materials the corresponding material properties can be found in the table e.1& 2:namely:typeec(20c)mpaec(15c)pmasplitting strength(mpa)ac10150021001.5ac16130018000.93.2 base and subbase materialstypeec(deflection)mpaec (tensile stress)pmasplitting strength(pma)cement macadam160040000.55sma150019000.8graded macadam450-4. thickness design4.1 determine design criteriadetermine design deflection ld: equation givenac=1(1 for expressway & class i, 1.1 for class ii, and 1.2 for others) as=1(1 for ac and 1.1 for others)ab=1(1 for semi-rigid base, and 1.6 for flexible base)so ld =600*4004*104*1*1*1=18.1(0.01mm)determine tensile stress at bottom of each layerequation given ac10 layer: ks=0.09ne0.22/ac=0.09*400400000.22/1=4.23 , r=sp/ks=1.5/4.23=0.36mpaac16layer:ks=0.09ne0.22/ac=0.09*400400000.22/1=4.23 , r=sp/ks=0.9/4.23=0.21mpasma20layer:ks=0.09ne0.22/ac=0.09*400400000.22/1=0.55 , r=sp/ks=0.9/4.23=0.21mpacement stabilized base layer:ks=0.35ne0.22/ac=0.35*400400000.22/1=16.47 r=sp/ks=0.55/16.47=0.03mpa4.2 determine calculated criteriabisar solution for surface resilient deflection：bisar solution for layer bottom tensile stresses layer1layer2layer3layer44.3 finalize the thickness designcompare 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:part iii cement concrete pavement design1. project statementnatural zone ii4pavt width: 7msubgrade soil: clay with a low liquid limitwater table: 1.2m beneath the roadbed surfaceaggregate: ganitepavt type: jpcpps=80kn, pm=150knadtt=4500saannual increasing rate: 5%2. design data2.1 traffic analysiswheel-track coefficient:design period: 20ys, and safety level: class-ii, pavt width=7m, according to jtg d40-2011 figure=0.62, so accumulative equivalent esals can be calculated as: ne=ns（1+gr)t-1365/gr=4500(80/100)16(1+5%)20-13650.62/5%=947803sane is between 0.03msa1msa belonging to moderate traffic level.2.2 subgrade modulusaccording to jtg d40-2011 appendix e, reference mr=90mpaand humid coefficient= 0.75so mr=90*0.75=67.5mpa 3. combination design3.1 concrete surface courseaccording to jtg d40-2011, concrete slab recommended thickness select 240mm3.2 base and subbase coursesaccording to jtg d40-2011, available materials for base arelight or moderategraded crushed stonescement-stabilized, lime-ash-stabilized crushed stonesand available materials for subbase arelight or moderateunscreened crushed stones, graded crushed stones, or no subbaseconstruction thicknessgraded crushed stones/ unscreened crushed stones or gravels100-200therefore, 200mm graded crushed stones as base layer and no subbase.3.3 summarylayermaterialthickness(mm)surfaceppc230subasegraded crushed stones2003.4 other design parameterreliability : for class ii highway, safety level is ii and target reliability is 85 percent, reliability index is 1.04.and from table below the variance level is moderate, so reliability coefficient is 1.13.temperature gradient is 88 c/m based on table below.4. joint design4.1 transverse jointstransverse joint spacing: 4.5m and no dowel bars are used at the transverse joints 4.2 longitudinal jointslongitudinal joint spacing: 3.5m and tie bars are used at the longitudinal joints. besides, the size of tie bar is 14*700*600 based on table below from jtg d40-2011.5. material proportion properties5.1 concrete materialsrupture strength is 4.5mpa, mr=29gpa, poissons ratio is 0.15, thermal coefficient is 1e-5traffic classplain cement concete (pcc)moderate=4.55.2 base and subbase materialsmr=300mpa based on jtg d40-2011 appendix e6. calculation and analysis of load-induced stress6.1 load-induced fatigue stress because concrete slab on granular base can be considered as single-layer plate system on elastic foundation, so the equation for load-induced fatigue stress is kr=0.87 (0.870.92, tire bar) kc=1.05(checked in table as follow) kf= ne =(94.78104)0.057=2.191and in ,equivalent modulus of unbound materials: et=130mpa ex= (hi2ei)/ hi2=h12e1/h12=300mpa hx= hi=h1=0.20m =0.86+0.26ln( hx )=0.86+0.26ln(0.20）=0.442 et=(ex/e0)e0=(300/67.5)0.44267.5=130.51mpabending stiffness: 34.17mnm dc=echc3/(12(1-vc2)）=290000.243/(12(1-0.152）=34.17mnmr=1.21 (dc/ et )1/3=0.775mso ps=1.4710-3r0.70hc-2ps0.94=1.4710-30.7750.700.24-2800.94=1.313mpathus pr = kckrkfps =1.052.1910.871.313=2.627mpa6.2 load-induced ultimate stresspm=1.4710-3r0.70hc-2pm0.94=1.4710-30.7750.700.24-21500.94=2.541mpap,max=krkcpm=0