道路勘测设计作业

第四章作业21112214王雯钰4-1某新建三级公路，设计速度v=30km/h，路面宽B=7m，路拱iG=2％。路肩宽bJ=0.75m，路肩横坡iJ=3％。某平面曲线转角α=34˚50’08”，半径R=150m，缓和曲线Ls=40m，加宽值b=0.7m，超高ih=3％，交点桩号为K7+086.42,。试求平曲线上5个主点及下列桩号的路基路面宽度、横断面上的高程与设计高之差：①K7+030；②K7+080；③K7+140；④K7+160。解：设计速度为30km/h的三级公路，查表得其ih=3％R250m，在平曲线内侧（右）加宽。平曲线要素：切线增长值：mRLLqss988.19240223内移值：mRLRLpss444.02384243442切线长：mqpRT19.672tan)(缓和曲线角：22'3876479.28RLs曲线长：mLRLs17.1312180)2(主点桩号：ZH=JD-T=K7+086.42-67.19=K7+019.23HY=ZH+Ls=K7+019.23+40=K7+059.23QZ=ZH+L/2=K7+019.23+131.17/2=K7+084.82HZ=ZH+L=K7+019.23+131.17=K7+150.40YH=HZ-Ls=K7+150.40-40=K7+110.40桩号路基宽度超高值左右左中右ZHK7+019.234.254.2500.09250K7+0304.254.4400.09250.07HYK7+059.234,254.95-0.0210.12750.26K7+0804.254.95-0.0210.12750.26QZK7+084.824.254.95-0.0210.12750.26YHK7+110.404.254.95-0.0210.12750.26K7+1404.254.4400.09250.07HZK7+150.404.254.2500.09250K7+160.004.254.2500.092504-2某双车道公路，设计速度v=60km/h，路基宽度8.5m，路面宽度7.0m。某平曲线R=125m，Ls=50m,48'3251。平曲线内侧中心附近的障碍物距路基边缘3m。试检查该平曲线能否保证停车视距和超车视距。若不能保证，清除的最大宽度是多少？解：查表可得，停车视距为75m，超车视距一般值为350m，加宽b为1.5m。视点轨迹线半径mbRRs1235.12缓和曲线角33'27116479.28RLs曲线长度mLRLs46.1622180)2((1)LSLmSLl73.43)7546.162(21)(2114.10]})(1[6arctan{2llllRlS51.5))(2sin()22cos1(llRhS≦1.5+1.5+0.75+3=6.75m符合停车视距要求（2)ScL22sin)2sin()22cos1(LSlRhs超=63m6.75m不符合要求。最大清除宽度为63-6.75=56.25m.