二广高速连怀段公路路基路面工程综合设计（K53+200～K54+400）毕业设计计算书

毕业设计计算书题目： 二广高速公路连怀段路基路面工程综合设计 （K53+200-K54+400） 学 生 姓 名： 学 号： 班 级： 专业（全称）： 土木工程（道路方向）指 导 教 师： 二广高速公路连怀段路基路面工程综合设计（K53+200K54+400）摘 要本设计为二广高速连怀段公路路基路面工程综合设计（K53+200K54+400） ，全长1.2公里，双向六车道，设计车速为 80km/h，路基宽度为 32m。设计的依据为国家颁布的最新技术标准，设计规范，施工技术规范，同时也参考了其他相关资料。 此次毕业设计主要包括：路线纵断面设计、路基横断面设计、边坡稳定性分析、边坡防护设计、挡土墙设计、路基路面排水设计以及路面结构设计等。设计中采用了骨架内植草防护；挡土墙选用的是仰斜式路堤挡土墙设计；排水设计对排水沟尺寸设计进行了水文计算；水泥路面和沥青路面拟定干湿和干燥两种状态各三种方案，并通过仔细比较选出最优方案；在本路段设置了三个盖板涵用于排水，一个盖板通道用于机耕通道；设置了一座桥来减少房屋和农田的占用。在计算机编程方面，我利用C语言编制了平曲线要素测设计算程序。通过本次毕业设计，在专业知识方面得到了综合训练和提高，增强了独立分析和解决问题的能力，取得了很好的效果。 关键词：高速公路；设计；路基；水泥混凝土路面；沥青路面；排水工程；挡土墙3ABSTRACTIn this design, our task is to develop the general design of subgrade and pavement of an expressway, which is from Lian hao te to Ji huai high speed . The expressway is 1.2 kilometres from K53+200 to K54+400, the designed speed limit is 80 km/h, and the wide of the subgrade is 34.5m with six lanes of two ways. The design basis are the lasted technical standards design and construction code, At the same time , some related information are also refereMy prime contents of the design project include: the design of vertical section and cross-section, stability analysis of road foundation, side slope protection, retaining wall design, drainage design for subgrade and pavement, pavement structure design, ect. In this design, the subgrade slopes are pretected with grass planted in the framework; retaining walls used the batterly retaining wall; the size of drainage ditch is determined through by hydrological calculation in drainage design; each there structure of cement and alsphat pavement are developed distingsishly under the wet and dry conditions, the optimal pavement structure is obtained in terms of careful comparison; In this set of three cover sections of culvert for drainage, a cover for tractor access channel;and set up one bridges to reduce the occupation of houses and farmland.During this graduation designing, the expertise is a comprehensive training, and the ability of independent analysis and solving problems were improved and enhanced, so good effects have been abtaind. Key words: Expressway; Design; Subgrade; Cement concrete pavement; Asphalt pavement; Drainage engineering; Retaining wall 目录第一章 线形设计11.1 平面线形设计11.2 纵断面设计21.3 平纵线形组合设计3第二章 边坡稳定性分析42.1 汽车荷载当量换算42.2 简化Bishop 法求稳定系数K5第三章 4.5 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）36第五章 水泥混凝土路面375.1 水泥混凝土路面设计总则375.2 结构组合设计原则375.3 交通量计算395.4 交通参数分析405.5路面结构组合设计405.6路面结构计算415.7 水泥混凝土路面结构设计及方案比选原则与总结62第六章 沥青路面设计656.1据交通量确定累计标准轴次656.2 沥青路面结构组合设计666.3沥青路面结构计算686.4沥青路面方案比选79参考文献1致 谢3城南学院 二广高速连怀段路基路面工程综合设计 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社超高最小半径；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: K53+669.543，圆曲线半径R=3000m，转角=9202.3(Y)，缓和曲线 Ls1=0，Ls2=340。计算：切线角：0=28.6479Ls/R =28.6479340/3000 =3.247切线增值：q=Ls/2=340/2=170m内移值：p=Ls/24R= 340/243000 =1.605m切线长：T=(R+p)tg/2+q=（3000+1.605）0.1644+170=405.12m曲线总长：L=(-20)R/180+Ls2=(9202.3-23.247)3000/180+340=658.726m切曲差：D=2T-L=254.803+405.12-658.726=10.8m各主点桩号：圆缓点桩号为：YH=HY+Lh-2Ls=K53+413.921+658.726-340 =K53+733.181缓直点桩号为：HZ=YH+Ls= K53+733.181+340=K54+073.181曲中点桩号为：QZ=HZ-Lh/2= K54+073.181-658.726/2= K53+574.818交点桩号：JD=QZ+= K53+669.44（校核）。平曲线计算无误。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80Km/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，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K53+200-K54+400共1200m设置1个变坡点。变坡点的桩号为K53+690。 高程：188，竖曲线半径R=15000m，i1=2.041%,i2=-3.239%， w=i1-i2=-5.28%凸形曲线 曲线长L1=Rw=150005.28%=792m 切线长T=L/2=7922=396m 外距E=T/(2R1)= 396（215000）=5.227m1.2.2 竖曲线要点桩号及高程的计算竖曲线一：起点桩号= K53+690-T1= K53 +294终点桩号= K53+690+T1= K54+086变坡点一对应桩号设计高程182.77m竖曲线一起点设计高程179.918m竖曲线一终点设计高程175.174m1.3 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，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位于K53+690，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。2 第二章 边坡稳定性分析取K54+140段横断面做路堤边坡稳定性验算。路基左侧填土高度15.44m，顶宽32m，路基填土为粘性土，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 K54+140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4 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=41.8+(4-1)1.3+0.6=11.7m故按双向布八辆车，布满行车道时，h=(4550)/(1811.712.8)=0.82m2.2 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=15.44m，由坡角向下做垂线，量取路堤高H=15.44+0.82=16.26m得D点；（3）由C 点向右引水平线，在水平线上截取4.5H=73.17m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。图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*cos14.197.72138.96-0.330.9455-45.85664.75579.23224.0921.12380.16-0.190.982-72.230177.1580.32733.9432.08577.44-0.050.9986-28.872269.0878.689641.1210.7192.60.040.99947.70489.7522.386551.514.73265.140.070.997618.559123.5529.92864.0243.04774.720.160.9877123.955361.0179.411074.2450.16902.880.290.9563261.835420.7481.094284.4154.48980.640.440.8988431.481456.9779.274194.9450.52909.360.570.8192518.335423.7680.9369105.6637.64677.520.7070.7071479.006315.7280.0437118.5818.57334.260.8480.5299283.452155.7690.93081977.37mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=1.8K3=1.921K=1K2=1.8K3=1.9210.79170.860.865448181.86167.41166.374329.51250.4660.89340.930.935909288.18276.02275.1144250.4660.97530.980.986471356.58352.83352.5464250.4661.01801.01.009103110.15111.05111.1265250.4661.03021.011.014581148.98151.10151.2775250.4661.06221.021.026513414.61427.96429.055250.4661.09141.031.02664459.792486.56488.8101250.4661.10381.01271.005536485.80529.52533.3250.4661.08480.960.957472465.23522.04527.1161250.4661.03650.890.878605381.80444.61450.4502250.4660.92500.740.73561266.67329.17335.3627250.4663559.713798.33820.5331.80021.921.932128mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1.932K2=1.933K3=1.933K=1.932K2=1.933K3=1.9330.865900.8659450.865945166.286166.27166.278829.51250.4660.936170.9361960.936196275.037275.03275.0303250.4660.986540.9865460.986546352.521352.51352.5194250.4661.009041.0090431.009043111.132111.13111.1332250.4661.014481.0144751.014475151.291151.29151.2932250.4661.026291.0262721.026272429.147429.15429.1557250.4661.026241.0262121.026212489.000489.01489.0181250.4661.004921.0048731.004873533.622533.65533.6517250.4660.956680.9566130.956613527.549527.58527.589250.4660.877620.8775410.877541450.951450.99450.9968250.4660.734430.7343320.734332335.897335.94335.9459250.4663822.443822.63822.6121.933091.93311.933179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=1.933。(2) 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Wisincoswi*sinWi*tanci*li*cos14.076.322138.96-0.190.981784-26.402464.7553679.9172224.0117.4313.2-0.050.998749-15.66145.951280.0996934.0126.44475.920.070.99754733.3144221.778780.0032741.078.24148.250.160.98711723.7269.084521.124351.5311.76211.680.190.98178440.219298.6428830.0425964.1833.8608.40.290.957027176.436283.514480.0074374.438.64695.520.420.907524292.1184324.112379.8621284.7740.44727.920.540.841665393.0768339.210780.2948495.4333.72606.960.670.742361406.6632282.843480.62042108.0216.81302.550.810.58643245.0655140.988394.063351568.551mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=1.634K3=1.744K=1K2=1.634K3=1.74430.860.8932440.9275980.931016161.963155.96155.3922250.4660.9754490.984490.985389231.740229.61229.4027250.4661.0301671.017511.016251292.944296.58296.9561250.4661.0616771.0327471.02986984.968287.348387.59248250.4661.0703241.035971.032552120.230124.217124.6285250.4661.0921671.0397321.034515332.844349.630351.3934250.4661.1032441.0273041.019749366.169393.237396.1509250.4661.0933050.9956670.985954383.704421.331425.4819250.4661.0545810.9334380.921386344.652389.381394.4749250.4660.963890.8174340.802863243.857287.548292.7667250.4662563.072734.862754.241.634041.743551.755913mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1.756K2=1.757K3=1.757K=1.756K2=1.757K3=1.75730.860.9313630.9313910.931391155.3343155.3295155.3295250.4660.985480.9854880.985488229.3814229.3797229.3797250.4661.0161231.0161131.016113296.9935296.9966296.9966250.4661.0295771.0295531.02955387.6173387.6193987.61939250.4661.0322061.0321771.032177124.6704124.6739124.6739250.4661.0339861.0339421.033942351.5734351.5883351.5883250.4661.0189821.0189191.018919396.4491396.4737396.4737250.4660.9849680.9848860.984886425.9078425.9431425.9431250.4660.9201630.9200620.920062394.9994395.0428395.0428250.4660.8013840.8012620.801262293.307293.3518293.3518250.4662756.2342756.3992756.3991.7571841.757291.75729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。(3) 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*cos14.428.48172.14-0.420.90752-72.298880.21724100.281424.1423.6479.08-0.280.96-134.142223.251399.3634.0436.12733.24-0.140.99015-102.654341.6898100.005341.0711.48233.07-0.050.99874-11.6535108.610626.7165451.516.65338-0.020.9998-6.76157.50837.492561.4116.52335.280.040.999213.4112156.240530.9951774.0652.561066.970.140.99015149.3758497.20888.4403384.1559.691211.670.280.96339.2676564.638287.64894.3764.441308.130.410.91208536.3333609.588687.6879104.7857.761172.530.5450.83843639.0289546.39988.16993115.545.52924.060.680.73321628.3608430.61288.71867127.327548.10.820.57236449.442255.414691.92158133.732.4750.040.920.3919146.036823.3186432.160822473.748mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=1.966K3=2.083K=1K2=1.966K3=2.0830.7118040.807970.813563253.579223.3972221.861829.04250.4660.829520.893630.89736388.913361.0114359.5117250.4660.9249120.956960.958831477.553461.5572460.6599250.4660.9754490.986890.987563138.733137.1238137.0314250.4660.990480.995050.995326196.874195.9687195.9163250.4661.017841.008681.008148183.954185.6243185.7223250.4661.0553921.023331.021472554.911572.2935573.3378250.4661.090481.026361.02264598.164635.5284637.8451250.4661.1031461.009261.003809632.080690.8737694.6306250.4661.0924060.967610.960361580.8911655.8058660.7607250.4661.0500920.894390.885339494.557580.652586.5897250.4660.9544840.766720.75581363.899453.0111459.5546250.4660.8206380.609980.59773767.605290.952192.81586250.4664864.115152.8475173.4221.966292.0830122.091329mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=2.091K2=2.092K3=2.092K=2.091K2=2.092K3=2.0920.8139230.813960.813968221.7638221.7516221.751629.04250.4660.8975990.897620.897629359.4157359.4038359.4038250.4660.9589510.95890.958966460.6024460.5952460.5952250.4660.9876060.98760.987612137.0254137.0247137.0247250.4660.9953430.995340.995345195.9129195.9125195.9125250.4661.0081141.008111.00811185.7286185.7294185.7294250.4661.0213521.021331.021337573.4051573.4134573.4134250.4661.0224011.022371.022371637.9947638.0133638.0133250.4661.0034581.003411.003414694.8736694.9038694.9038250.4660.9598950.959830.959837661.0818661.1218661.1218250.4660.8847570.884680.884684586.9756587.0236587.0236250.4660.7551090.755020.755021459.9817460.035460.035250.4660.5969490.596850.59685192.9382992.9535592.95355250.4665174.7615174.9285174.9282.0918712.0919382.091938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Fs3=2.092（4）确定第四个过坡脚滑动面，在辅助线上做圆4，半径R=28.97m，将结果列入表中，验算其稳定性：图2-6 圆心为O4的滑动面示意图表2-4 土坡稳定性计算表土条号li面积Wisincoswi*sinWi*tanci*li*cos14.428.88180.26-0.440.897998-79.314484.0011699.2287524.1424.6499.38-0.310.950737-154.808232.711198.4012334.0437.64764.09-0.170.985444-129.895356.065999.5298541.0711.84240.36-0.090.995942-21.6324112.007826.6414451.517.34352-0.040.9992-14.08164.03237.4699961.4151.241040.170.050.99874952.0085484.719230.981274.0660.121220.440.190.981784231.8836568.72587.6929584.1566.121342.240.330.943981442.9392625.483886.1854694.3764.841316.250.470.882666618.6375613.372584.85955104.7854.841113.250.620.784602690.215518.774582.50873115.540.32818.50.740.672607605.69381.42181.38543127.316.97344.540.880.474974303.1952160.555676.280772544.839mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=1K2=2.048K3=2.176K=1K2=2.048K3=2.1760.692950.7978810.80377264.417229.6458227.963228.97250.4660.806270.8801990.884349410.668376.1787374.4138250.4660.906220.9467620.949038502.740481.2145480.0607250.4660.954000.9754630.976668145.334142.1368141.9615250.4660.980560.9900980.990634205.496203.5172203.4072250.4661.022041.0101261.009457504.57510.5307510.8692250.4661.070321.0250171.022473613.289640.3975641.9903250.4661.097761.0190691.014652648.291698.3525701.3926250.4661.101680.989610.983319633.784705.563710.0768250.4661.073520.9256760.917378560.103649.5612655.437250.4661.017440.8409860.831081454.870550.3142556.8727250.4660.885050.6752080.66343267.595350.7606356.988250.4665211.165538.1735561.4332.047732.1762372.185377mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtanK1=2.185K2=2.186K3=2.186K=2.185K2=2.186K3=2.1860.8041580.8042010.804201227.8531227.841227.84128.97250.4660.8846220.8846520.884652374.298374.2852374.2852250.4660.9491880.9492040.949204479.9849479.9765479.9765250.4660.9767470.9767560.976756141.9499141.9487141.9487250.4660.9906690.9906730.990673203.4203.3992203.3992250.4661.0094131.0094081.009408510.8915510.894510.894250.4661.0223061.0222871.022287642.0955642.1072642.1072250.4661.0143611.0143291.014329701.5939701.6161701.6161250.4660.9829040.9828590.982859710.3764710.4095710.4095250.4660.9168310.916770.91677655.828655.8713655.8713250.4660.8304280.8303560.830356557.3105557.3589557.3589250.4660.6626530.6625670.662567357.4062357.4525357.4525250.4665562.9885563.165563.162.1859882.1860562.186056 用迭代法试算假定计算结果Fs 与假定接近，得土坡的稳定安全系数，Fs4=2.186。 （5）确定第五个过坡脚滑动面，在辅助线上做圆5，半径R=28.93m，将结果列入表中，验算其稳定性：图2-7 圆心为O5的滑动面示意图表2-5 土坡稳定性计算表土条号li面积Wisincoswi*sinWi*tanci*li*cos14.429.08184.32-0.450.893029-82.94485.8931298.6796624.1425.08509.12-0.330.943981-168.01237.249997.7020334.0438.28777.08-0.190.981784-147.645362.119399.1601941.0712.56254.99-0.10.994987-25.499118.825326.6159151.517.7359.31-0.050.998749-17.9655167.438537.453161.4152.481065.340.040.999242.6136496.448430.9951774.0661.441247.230.170.985444212.0291581.209288.0198684.1567.881377.960.330.943981454.7268642.129486.1854694.3766.641352.790.450.893029608.7555630.400185.85577104.7856.961156.290.60.8693.774538.831184.128115.542.84869.650.730.683447634.8445405.256982.6971127.321.34433.250.870.493052376.9275201.894579.184112581.608mi(=cos+sin*tan/K)(Wi*tan+c*li*cos)/miRtan