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1、【精品文档】如有侵权,请联系网站删除,仅供学习与交流高速公路路基路面综合设计方案.精品文档.CHENGNAN COLLEGE OF CUST 毕业设计(论文)题目: 湖南省永州至蓝山高速公路 K78+600K80+700段路基路面综合设计摘要 本次毕业设计主要任务是对永州至蓝山高速公路K78+600-K80+700段进行路基路面综合设计。该路段全长2100m,双向四车道,设计车速100Km/h,路基宽26m,属于比较典型的山区高速公路。我们所做的工作主要包括:路线设计、路基工程设计、路基工程设计和桥梁涵洞的布设四部分。整个设计过程严格按照最新规范,手册进行,设计过程力求原理、方法和公式计算准确
2、,对应的图纸、表格满足相关编制办法的要求。此外,英文翻译及计算机程序设计也是本次设计的重要组成。本次设计不但让我们熟悉了公路设计的基本原理、方法和设计步骤,而且也让我们对大学四年所学知识进行了一次全面的总结。关键词:毕业设计;永州至蓝山高速公路;路基路面;翻译;编程Abstract The main task of this graduation design is the comprehensive design of subgrade and pavement of Yongzhou to Lanshan Expressway K68+100-K70+200. The section of
3、 full-length 2100m, two-way four driveway, design speed of 100Km/h, the roadbed width 26m, belongs to the typical mountainous expressway. Our work mainly includes: route design, subgrade engineering design, design of subgrade and Bridge Culvert layout four. The whole design process is strictly in ac
4、cordance with the latest specification manual, design process, to principle, method and accurate calculating formula, the corresponding drawings, form to meet the relevant measures for requirements. In addition, English translation and computer program design is the design of the important. The desi
5、gn not only makes us familiar with the basic principles, methods and steps of design of highway design, but also let us to the university four years learning conducted a comprehensive summary.Keywords: graduation design; Yongzhou to Lanshan Expressway Subgrade and pavement; translation; programming;
6、目录第一章 线形设计61.1 平面线形设计61.2 纵断面设计71.3 平纵线形组合设计8第二章 边坡稳定性分析92.1 汽车荷载当量换算92.2 简化Bishop 法求稳定系数K.10第三章 挡土墙设计与验算.213. 1 设计资料.213. 2 墙背土压力计算.223.3 墙身截面性质计算。.26第四章 排水设计.29.4. 1 路基排水设计规定.294. 2 排水设施的设置.304. 3 路基排水综合设计.304. 4 路堑边沟验算.304.5 拦水带设计验算沟渠横断面设计按照最佳横断面法(即固定了纵坡和允许最大流速)31第五章 水泥混凝土路面325.1 水泥混凝土路面设计总则325.2
7、 结构组合设计原则325.3 交通量计算325.4 交通参数分析335.5 路面结构组合设计335.6 路面结构计算335.7 水泥混凝土路面结构设计及方案比选原则与总结54第六章 沥青路面设计576.1 据交通量确定累计标准轴次576.2 沥青路面结构组合设计576.3 沥青路面结构计算586.4 沥青路面方案比选77参考文献78致 谢81 第一章线形设计1.1 平面线形设计 道路平面线形设计,是根据汽车行驶的力学性质和行驶的轨迹要求,合理地确定各线形的几何参数,保持线形的连续性和均衡性,避免采用长直线,并注意使线形与地形、地物、环境和景观等协调。在设计中注意直线的长度符合规范要求,对于同反
8、向曲线间的直线要满足直线最小长度要求。规范规定当设计速度60Km时,同向直线最小长度以不小于设计速度的六倍为宜。对于反向曲线间的直线不应小于设计速度的两倍为宜。对于圆曲线半径的选择应遵循如下原则:在地形条件许可的情况,应力求是半径尽可能接近不设超高最小半径;选取半径时,最大半径值一般不应超过10000m.1.1.1 平曲线要素的计算已知:JD1桩号为: K79+002.560,圆曲线半径R=1600m,转角= 393651.7(Y),缓和曲线Ls=260。计算:切线角:0=28.6479Ls/R =28.6479260/1600 =4.655切线增值:q=Ls/2=260/2=130m内移值:
9、p=Ls/(24R)= 260/(241600) =1.76m切线总长:T=(R+p)tg/2+q=(1600+1.76)0.827+130=1454.655m曲线总长:Lh=(20)R/180+2Ls=(393651.724.655)1600/180+520=1366.257m超距:D=2TL=21454.6551366.257=1383.053m各主点桩号:直缓点桩号:ZH=JDT= K79+002.5601454.655= K77+547.905缓圆点桩号:HY=ZH+Ls= K77+547.905+260= K77+807.905圆缓点桩号为:YH=HY+Lh2Ls= K77+807.
10、905+13666.257520 = K78+654.162缓直点桩号为:HZ=YH+Ls= K78+654.162+260= K78+914.162曲中点桩号为:QZ=HZLh/2= K78+914.1621366.257/2=K78+231.0335交点桩号:JD=QZ+= K79+002.560(校核)。验算与交点桩号一样,设计合格。1.2 纵断面设计1.2.1 纵坡设计的原则纵断面设计首先要注意坡度的选择符合各级道路规定的最大坡度。本次设计公路为100Km/h,根据规定允许最大坡度为6%,合成坡度不能大于10.5%。其次为了保证排水,防止水渗入路基影响稳定性,应设置不小于0.3%的纵坡
11、,最小合成坡度不宜小于0.5%。相邻竖曲线衔接时应注意:(1)同向竖曲线:特别是两同向凹曲线间如果直线坡段不长,应合并为单曲线或复曲线形式的竖曲线,避免出现断背曲线。(2)反向竖曲线:反向竖曲线间应设置一段直线坡段,直线坡段的长度一般不小于设计速度的3秒行程。以使汽车从失重渡到增重有一个缓和段。本组设计从K78+600- K80+700共2100m设置5个变坡点。变坡点的桩号为K78+640 ,设计高程为285.22,竖曲线半径R1=571.1442m,i1= 3.98%,i2=-0.940%, w1= i1i2 = 4.92%为凸形曲线。竖曲线一: 曲线长 L1=R1w1=574.14423
12、%=40.848m 切线长 T1=L1/2=40.8482=20.424m 外距 E1=T/(2R1)=20.424 (2571.1442)=0.365m变坡点2的桩号为K78+980,设计高程为282.02,竖曲线半径R2=12000m,i1=-0.940%,i2=2.830%,w1= i1i2=-3.77%为凹曲线。竖曲线二: 曲线长 L2=R2w1=120003.77%=452.4m 切线长 T1=L1/2=452.42=226.2m 外距 E1=T/(2R1)=226.2 (212000)=2.13m变坡点3的桩号为K79+380,设计高程为293.3408,竖曲线半径R3=20000
13、m,i1=2.830%,i2=1.156%,w1= i1i2=1.674%为凸曲线。竖曲线三: 曲线长 L2=R2w1=200001.674%=334.8m 切线长 T1=L1/2=334.82=167.4m 外距 E1=T/(2R1)=167.4 (220000)=0.7m变坡点4的桩号为K80+020,设计高程为300.7388,竖曲线半径R4=10000m,i1=1.156%,i2=-1.032%,w1= i1i2=2.188%为凸曲线。竖曲线四: 曲线长 L1=R4w1=100002.188%=218.8m 切线长 T1=L1/2=218.82=109.4m 外距 E1=T/(2R1)
14、=109.4 (210000)=0.598m变坡点5的桩号为K80+580,设计高程为294.9586,竖曲线半径R5=3000m,i1=-1.032%,i2=4.525%,w1= i1i2=-5.557%为凹曲线。竖曲线五: 曲线长 L1=R5w1=30005.557%=166.71m 切线长 T1=L1/2=218.82=83.355m 外距 E1=T/(2R1)=83.355 (23000)=1.158m1.2.2 竖曲线要点桩号及高程的计算竖曲线一:起点桩号 = K78+640T1= K78+619.576终点桩号 = K78+640T1= K78+660.424变坡点一对应桩号设计高
15、程 = 284.85竖曲线一起点设计高程 = 283.78m竖曲线一终点设计高程 = 285.23m竖曲线二:起点桩号 = K78+980T= K78+753.8终点桩号 = K78+980T= K79+206.2变坡点一对应桩号设计高程 = 282.0218竖曲线一起点设计高程 = 284.12m竖曲线一终点设计高程 = 288.41m竖曲线三:起点桩号 = K79+380T= K79+212.6终点桩号 = K79+380T= K79+547.4变坡点一对应桩号设计高程 = 293.3408竖曲线一起点设计高程 = 288.64m竖曲线一终点设计高程 = 295.25m竖曲线四:起点桩号
16、= K80+020T= K79+910.6终点桩号 = K80+020T= K80+129.4变坡点一对应桩号设计高程 = 300.7388竖曲线一起点设计高程 = 299.41m竖曲线一终点设计高程 = 299.58m竖曲线五:起点桩号 = K80+580T= K80+496.645终点桩号 = K80+580T= K80+663.355变坡点一对应桩号设计高程 = 294.9586竖曲线一起点设计高程 = 295.86m竖曲线一终点设计高程 = 298.82m1.3 平纵线形组合设计尽管平纵线形设计均是按照标准进行设计,但若平纵线形组合不好,不仅有限于其优点的发挥,而且会加剧两方面的缺点,
17、造成行车上的危险,也就不能获得最优的立体线形、平纵线形的合理设计。因此平纵组合设计要注意以下几点要求:(1)当竖曲线与平曲线组合时,竖曲线宜包含在平曲线之内,切平曲线应稍长与竖曲线。这种布置的优点是:当车辆驶入凹行曲线的顶点之前,即能清楚的看到平曲线的始端,辩明转弯的走向,不致因判断错误而发生交通事故。(2)要保持平曲线和竖曲线大小的均衡,这样有利于驾驶员视觉上的均衡。(3)要选择适当的合成坡度。本段设计中竖曲线变坡点的起点和终点基本落在平面曲线的缓和曲线内,平纵组合符合设计要求。 第二章 边坡稳定性分析取K75+920段横断面做路堤边坡稳定性验算。路基左侧填土高度12.49m,顶宽26m,路
18、基填土为粘性土,土粘聚力为20kPa,内摩擦角为25,边坡坡度采用1:1.5。由于填方边坡要受到路基顶部车辆荷载的影响,在进行稳定性验算时要先化为换算土柱高度。容重为=18KN/m3,荷载为公路I级。其横截面初步拟定如图1所示:图2-1 K21+940横断面图2.1 汽车荷载当量换算 将车辆荷载换算成土柱高(当量高度)。车辆按最不利情况排列,即假设一辆车停在硬路肩上,另两辆以最小间距d=0.6m与它并排。按以下公式换算土柱高度为:式中: N横向分布并列的车辆数,因为按最不利布载,中线每边各布3 辆,取N=4; Q每一辆重车的重力(标准车辆荷载为550KN); L前后轮最大轴距,按公路工程技术标
19、准(JTG B012003)规定对于标准车辆荷载为12.8m;r路基填料的容重;B荷载横向分布宽度,表示如下:B=Nb+(N-1)m+d式中:b后轮轮距,取1.8m;m相邻两辆车后轮的中心间距,取1.3m;d轮胎着地宽度,取0.6m则:B=Nb+(N-1)m+d=31.8+(3-1)1.3+0.6=8.6m故按双向布6辆车,布满行车道时,h=(3550)/(188.612.8)=0.83m2.2 简化Bishop 法求稳定系数K2.2.1 最危险圆弧圆心位置的确定以坡脚为坐标原点,按4.5H 法初定滑动圆心辅助线: (1)由表查得:=26, =35,以坡角为圆心将AB线逆时针旋转26,再以B点
20、为圆心,BC为基线,旋转35,两直线交于F点;(2)量得坡角到路面的距离h=12.49m,由坡角向下做垂线,量取路堤高H=12.49+0.83=13.32m得D点;(3)由C 点向右引水平线,在水平线上截取4.5H=53.28m得E点;(4)连接点E、F得直线EF,即为滑动圆心辅助线;(5)绘出五条不同的位置的滑动曲线;(6)将圆弧范围土体分成若干段;(7)算出滑动曲线每一分段中点与圆心竖曲线之间的偏角; sin=式中:X分段中心距圆心竖直线的水平距离,圆心竖曲线左侧为负,右侧为正; R滑动曲线半径m 最危险滑动面圆心的确定如图2所示:图2-2 最危险圆弧圆心位置的确定图示 2.2.2 用简化
21、毕绍普法求稳定系数FS通过计算选中的五个圆心点对应的安全系数Ki,得到最小值K 和对应的圆心点,再进行验算。计算Ki 时,采用简化bishop 法,并假设滑动面通过坡脚。简化bishop 法需要迭代,先假设一个K 值进行反复带入计算(具体迭代过程见表)。简化毕肖普法的计算公式如下: -第i土条底滑面的倾角;-第i土条垂直方向外力;-系数,按计算;-第i土条滑弧面所在地基土层的内磨檫角和粘结力;-滑动圆弧全长; -第i土条宽度; -第i土条路堤部分的重力; -第i条土地基部分的重力; U-地基平均固结度,根据地基情况,此处取U=1。 (1)在圆心辅助线上取圆心点O1,作半径为R1=29.51m圆
22、弧滑动面,对滑动面范围内的土体按断面形式将滑动面分为11个土条,如图2-3 验算其稳定性,计算结果见表2-1。2-3 圆心为O1的滑动面示意图表2-1 土坡稳定性计算表土条号li面积Wisincoswi*sinWi*tanci*li*cos14.197.72138.96-0.330.9455-45.85664.75579.23224.0921.12380.16-0.190.982-72.230177.1580.32733.9432.08577.44-0.050.9986-28.872269.0878.689641.1210.7192.60.040.99947.70489.7522.386551
23、.514.73265.140.070.997618.559123.5529.92864.0243.04774.720.160.9877123.955361.0179.411074.2450.16902.880.290.9563261.835420.7481.094284.4154.48980.640.440.8988431.481456.9779.274194.9450.52909.360.570.8192518.335423.7680.9369105.6637.64677.520.7070.7071479.006315.7280.0437118.5818.57334.260.8480.529
24、9283.452155.7690.93081977.37mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=1.8K3=1.921K=1K2=1.8K3=1.9210.79170.860.865448181.86167.41166.374329.51250.4660.89340.930.935909288.18276.02275.1144250.4660.97530.980.986471356.58352.83352.5464250.4661.01801.01.009103110.15111.05111.1265250.4661.03021.011
25、.014581148.98151.10151.2775250.4661.06221.021.026513414.61427.96429.055250.4661.09141.031.02664459.792486.56488.8101250.4661.10381.01271.005536485.80529.52533.3250.4661.08480.960.957472465.23522.04527.1161250.4661.03650.890.878605381.80444.61450.4502250.4660.92500.740.73561266.67329.17335.3627250.46
26、63559.713798.33820.5331.80021.921.932128mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1.932K2=1.933K3=1.933K=1.932K2=1.933K3=1.9330.865900.8659450.865945166.286166.27166.278829.51250.4660.936170.9361960.936196275.037275.03275.0303250.4660.986540.9865460.986546352.521352.51352.5194250.4661.009041.0090
27、431.009043111.132111.13111.1332250.4661.014481.0144751.014475151.291151.29151.2932250.4661.026291.0262721.026272429.147429.15429.1557250.4661.026241.0262121.026212489.000489.01489.0181250.4661.004921.0048731.004873533.622533.65533.6517250.4660.956680.9566130.956613527.549527.58527.589250.4660.877620
28、.8775410.877541450.951450.99450.9968250.4660.734430.7343320.734332335.897335.94335.9459250.4663822.443822.63822.6121.933091.93311.933179用迭代法试算假定计算结果Fs 与假定接近,故得土坡的稳定安全系数Fs1=1.843。一、 在滑动圆弧圆心辅助线上取同一圆心点O2,作半径为R2=30.86m的圆弧滑动面,按断面形式将滑动面分为10条,如图2-4验算其稳定性,计算结果见表2-2。图2-4 圆心为O2的滑动面示意图表2-2 土坡稳定性计算表土条号li面积Wisin
29、coswi*sinWi*tanci*li*cos14.076.322138.96-0.190.981784-26.402464.7553679.9172224.0117.4313.2-0.050.998749-15.66145.951280.0996934.0126.44475.920.070.99754733.3144221.778780.0032741.078.24148.250.160.98711723.7269.084521.124351.5311.76211.680.190.98178440.219298.6428830.0425964.1833.8608.40.290.957027
30、176.436283.514480.0074374.438.64695.520.420.907524292.1184324.112379.8621284.7740.44727.920.540.841665393.0768339.210780.2948495.4333.72606.960.670.742361406.6632282.843480.62042108.0216.81302.550.810.58643245.0655140.988394.063351568.551mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=1.634K3=1.744
31、K=1K2=1.634K3=1.74430.860.8932440.9275980.931016161.963155.96155.3922250.4660.9754490.984490.985389231.740229.61229.4027250.4661.0301671.017511.016251292.944296.58296.9561250.4661.0616771.0327471.02986984.968287.348387.59248250.4661.0703241.035971.032552120.230124.217124.6285250.4661.0921671.0397321
32、.034515332.844349.630351.3934250.4661.1032441.0273041.019749366.169393.237396.1509250.4661.0933050.9956670.985954383.704421.331425.4819250.4661.0545810.9334380.921386344.652389.381394.4749250.4660.963890.8174340.802863243.857287.548292.7667250.4662563.072734.862754.241.634041.743551.755913mi(=cos+si
33、n*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1.756K2=1.757K3=1.757K=1.756K2=1.757K3=1.75730.860.9313630.9313910.931391155.3343155.3295155.3295250.4660.985480.9854880.985488229.3814229.3797229.3797250.4661.0161231.0161131.016113296.9935296.9966296.9966250.4661.0295771.0295531.02955387.6173387.6193987.61939250.
34、4661.0322061.0321771.032177124.6704124.6739124.6739250.4661.0339861.0339421.033942351.5734351.5883351.5883250.4661.0189821.0189191.018919396.4491396.4737396.4737250.4660.9849680.9848860.984886425.9078425.9431425.9431250.4660.9201630.9200620.920062394.9994395.0428395.0428250.4660.8013840.8012620.8012
35、62293.307293.3518293.3518250.4662756.2342756.3992756.3991.7571841.757291.75729用迭代法试算假定计算结果Fs 与假定接近,故得土坡的稳定安全系数Fs2=1.757。二、 在滑动圆弧圆心辅助线上取同一圆心点O3,作半径为R3=29.04m 的圆弧滑动面,按断面形式将滑动面分为13个土条,如图2-5计算其稳定性,计算结果如下表:图2-5 圆心为O3的滑动面示意图 表2-3 土坡稳定性计算表土条号li面积Wisincoswi*sinWi*tanci*li*cos14.428.48172.14-0.420.90752-72.2
36、98880.21724100.281424.1423.6479.08-0.280.96-134.142223.251399.3634.0436.12733.24-0.140.99015-102.654341.6898100.005341.0711.48233.07-0.050.99874-11.6535108.610626.7165451.516.65338-0.020.9998-6.76157.50837.492561.4116.52335.280.040.999213.4112156.240530.9951774.0652.561066.970.140.99015149.3758497.2
37、0888.4403384.1559.691211.670.280.96339.2676564.638287.64894.3764.441308.130.410.91208536.3333609.588687.6879104.7857.761172.530.5450.83843639.0289546.39988.16993115.545.52924.060.680.73321628.3608430.61288.71867127.327548.10.820.57236449.442255.414691.92158133.732.4750.040.920.3919146.036823.3186432.160822473.748mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=1.966K3=2.083K=1K2=1.966K3=2.0830.7118040.807970.813563253.579223.3972221.861829.04250.4660.829520.893630.89736388.913361.0114359.5117250.4660.9249120.9569