《交通工程学》课后习题参考答案

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《交通工程学》习题解习题2-1解:⑴小时交通量:hQ/2493195190210195201205220219232217208201辆⑵5min高峰流率:hQ/27845602325辆⑶15min高峰流率:hQ/26841560)220219232(15辆⑷15min高峰小时系数:929.04671249315PHF习题2-2解:已知:%26.131326.0082.03086.17082.086.1730,/h1500C,/d50000AADT3.13.11xKx辆辆设计小时交通量:hKAADTDHV/66301326.050000100辆车道数:42.4150066301CDHVn该道路需修6车道。注:此题5.0DK。如果6.0DK,3.5n。习题2-3解:1000606100Q辆/h车头时距:6.31000/3600/3600Qhts/辆车头间距:206.36.3206.3tshVhm/辆车流密度:5020/1000/1000shK辆/km第一辆车通过时间:2.12024VSth习题2-4解:stntii5)3.56.47.44.53.59.42.51.58.47.40.52.50.59.41.58.4(1611161hkmsmtnsVniiS/72/2080100161hkmVnViit/16.726.1154161)9.673.786.767.669.675.732.696.700.756.760.722.690.725.736.700.75(1611161习题3-1解:已知:t东=2.0min,t西=2.0min,X东=29.0辆,Y东=1.5辆X西=28.6辆,Y西=1.0辆1、先计算向东行情况:hkmtlvqYtthttYXq/67.66608.12min8.1525.75.10.2/5.451min/525.7225.16.28东东东东东东东西东西东辆辆2、再计算向西行情况:hkmtlvqYtthttYXq/27.6460867.12min867.15.70.10.2/450min/5.7220.10.29西西西西西西西东西东西辆辆习题3-3解:根据浮动车调查法计算公式:辆)被测试车超越的车(辆的速度超越的车以辆的速度超越的车其中以辆被测试车超越的车超越观测车(空间平均车速)辆133.0/60133.0/80174.0/100173.07.0-/3.78064.05064.0224017705/224070570517303xhkmxhkmxhkmxxhkmtlvhqYtthttYXqcccccccacac习题3-4解:总停驶车辆数=28+25+38+33=124辆总延误=124×15=1860辆•s每辆停车的平均延误=总延误/停车辆数=1860/113=16.46s交叉口引道上每辆车的平均延误=总延误/引道上总交通量=1860/(113+119)=8.02s停车的百分数=停车辆数/引道上交通量=113/232=48.7%取置信度90%,则K2=2.70,于是停车百分比的容许误差=%07.11232487.070.2)487.01(取置信度95%,则K2=3.84,于是停车百分比的容许误差=%2.13232487.084.3)487.01(习题4-2解:已知:畅行速度hkmVf/82;阻塞密度kmKj/105辆;速度与密度为线性关系模型。⑴最大流量:因5.5221052jmKK辆/km412822fmVVkm/h∴5.2152415.52mmmVKQ辆/h。⑵此时所对应的车速:41mVVkm/h。习题4-4解:已知:N=56,09.3561731Nfkmgjjj车辆到达数jk实测频数jfjjfk)(jkpNkpFjj)(01234011141190112833360.04550.14060.21720.22370.17282.5487.87312.16312.5279.6775678≥9∑5321056251814801730.10680.05500.02430.00940.00325.9813.0801.3600.5250.180对于泊松分布,把jF小于5的进行合并,并成6组,可算出932.056145.56981.55677.99527.1211163.1214421.10112222226122NFfjjj由DF=6-2=4,取05.0,查表得:2205.0488.9可见此分布符合泊松分布。习题4-5解:已知:交通流属泊松分布,则车头时距为负指数分布。交通量hQ/1200辆,sQ/31360012003600辆。⑴车头时距st5的概率:19.035)5(etehP⑵车头时距st5时出现的次数:19.035)5()5(ehPhP∴次数为:8.22619.01200(次/h)。⑶车头时距st5时车头间隔的平均值h:sdtetdtehtt81555习题4-6解:λ=Q/3600=720/3600=0.5(辆/s)P(h≥2)=e-0.4=0.67每小时出现的次数为:720*0.67=482.4次/h习题4-8解:(1)直行车流的车头时距h服从参数s/3136001200辆的负指数分布,车头时距超过6s的概率为13.036)6(6etehPt1小时内,次要车道能通过的车辆数为:令)36)1(36(khkPPk=)())((keke36136361136003600eekPQkk次==257辆/h或者直接根据P103式(4-44)013600hheeQ次=257辆/h(2)直行车流的车头时距h服从参数s/3136001200辆的移位负指数分布,车头时距超过6s的概率为:189.0351-)6(6etehPt)(1小时内,次要车道能通过的车辆数为:令)36)1(36(khkPPk=)())((1361136keke36001kPQkk次=269辆/h或者直接根据P104式(4-50))1)(311(1200)1)(1(3600135)(0eeeeQhh次=269辆/h习题4-9解:已知:Q=1500辆/h,每个收费站服务量为600辆/h。1.按3个平行的M/M/1系统计算s/36536003/1500辆,s/613600600辆,1656/136/5,系统稳定。辆5)1(n,辆17.4nq,辆/36snd,辆/301sdw而对于三个收费站系统辆1535n,辆5.12317.4d,辆/36sd,辆/30sw2.按M/M/3系统计算s/12536001500辆,s/613600600辆256/112/5,16532/5N,系统稳定。04494.0625.15625.61)6/51(!3)25(!)25(1)0(2023kkkP辆5.3)6/51(04494.03!3)2/5(24q辆65.25.3qn辆/4.812/55.3sqw习题4-10解:已知:V1=50km/h,Q1=4200辆/h,V2=13km/h,Q2=3880辆/h,V3=59km/h,Q3=1950辆/h,t=1.69h1.计算排队长度k1=Q1/V1=4200/50=84辆/km,k2=Q2/V2=3880/13=298.5辆/kmVw=(Q2–Q1)/(k2–k1)=(3880–4200)/(298.5–84)=–1.49km/hL=(0×1.69+1.49×1.69)/2=1.26km2.计算阻塞时间⑴排队消散时间t′排队车辆为:(Q1–Q2)×1.69=(4200–3880)×1.69=541辆疏散车辆率为:Q2–Q1=1950–3880=–1930辆/h则排队消散时间:hQQQQt28.0193054169.1)(2321'⑵阻塞时间:t=t′+1.69=0.28+1.69=1.97h习题5-1解:已知:dvehAADT/45000,大型车占总交通量的30%,6.0DK,12.0K,平原地形。查表5-3,7.1HVE8264.0)17.1(3.011)1(11HVHVHVEPf取设计速度为100km/h,二级服务水平,71.0)/(2CVhpcuCB/2000,0.1Wf,0.1Pf一条车道的设计通行能力:hvehfffNCVCCPHVWB/5.117318264.01171.02000)/(21车道数:5.526.012.05.11734500021001001DKKCAADTn故该高速公路修成6车道。习题5-2解:已知:L1=300m、R=0.286、VR=0.560、V=2500pcu/hL2=450m、R=0.200、VR=0.517、V=2900pcu/h第一段:计算非约束情况下的交织车速SW及非交织车速SnWdcbRdnWWLNVVaSS/)/()1()3048.0(147.801.24或非约束情况下型式B的常数值如下:abcdSW0.11.20.770.5SnW0.022.01.420.95hkmSW/08.74750/)3/2500()56.01()3048.0(1.0147.801.245.077.02.15.0hkmSnW/15.81750/)3/2500()56.01()3048.0(02.0147.801.2495.042.1295.0利用式(5-8)计算5.349.1)]08.7415.81(011.0)750/57.71(56.0703.0085.0[3)](011.0)/57.71(703.0085.0[maxWWnWRWNSSLVNN核查交织区段诸限制值:30001400WV,19003.8333/2500/NV,8.056.0RV5.0286.0R,760750L确定服务水平:查表5-10hkmSW/8008.74,属于二级,hkmSnW/8617.81,属于二级。第二段:计算非约束情况下的交织车速SW及非交织车速SnWhkmSW/51.67450/)3/2900()517.01()3048.0(1.0147.801.245.077.02.15.0hkmSnW/34.69450/)3/2900()517.01()3048.0(02.0147.801.2495.042.1295.0利用式(5-8)计算5.376.1)]51.6734.69(011.0)450/57.71(517.0703.0085.0[3)](011.0)/57.71(703.0085.0[maxWWnWRWNSSLVNN核查交织区段诸限制值:30001500WV,190067.9663/2900/NV,8.0517.0RV5.0200.0R,760450L确定服务水平:查表5-10hkmSW/7251.67,属于三级,hkmSnW/7734.69,属于三级。习题5-3北解:已知T=60s,三相式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