《道路勘测设计》第4章横断面设计课后习题和答案解析

1、WORD格式.资料第四章横断面设计4-1某新建三级公路，设计速度 V 30km/h,路面宽度B= 7m,品&拱iG=2%,路 肩bj=0.75m,路肩横坡ij=3%。某平曲线转角a=34 50'08"，半径R = 150m, 缓和曲线Ls=40m,力口宽值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程之 差： K7+030; K7+080; K7+140; K7+16Q解：已知：JD= K7+086.42, a =345008" R = 150m, Ls = 40m平曲线要

2、素计算：402/ 、=0.44( m)24R24 150ls q W_ 2240R234040_一_22240 150-19.99 (m)专业.整理(R + p) tg % + q = (150 + 0 44) tg 34 50 08 +19.99 = 67.19( m)22nL 二180R： Lsji180150 34 50 0840 = 131.21(m)34 50 08E =(R p) sec - R =(150 0.44) sec -150 -7.67(m)22D=2T-L=2 67.19 -131.21 =3.17(m)(2)主点里程桩号计算JD K7+086.42-T-67.19Z

3、H K7+019.23+Ls+40HYK7+059.23ZHK7+019.23+L/2+131.21/2QZK7+84.84ZHK7+019.23+L +131.21HZK7+150.44-Ls-40YHK7+110.44超高过渡及加宽计算：新建三级公路，设计速度 V= 30km/h,无中间带，超高过渡采用采用内边线旋转，加宽线性过渡，路基边缘为设计高程，加宽值b = 0.7m,超高"=3%,J2%临界断面距过渡段起点X0 =叵 Lc = 2% M 40 = 26.67m。 ih3%先计算加宽值：K7+030处，处于前一段缓和曲线上，则有：K7+030- (K7+019.23) =1

4、0.77(m)< x0 =26.67(m),由教材 P106 表 4-13一一x 10 77 一 一 .可得：bx =，b =父 0.7 = 0.19(m)；Lc40路基路面宽度为：7+2>0.75+0.19=8.69(m);K7+140处，处于后一段缓和曲线上，则有：K7+150.44- (K7+140) =10.44 (m)< x0 =26.67(m),由教材 P106表 4-13一一x 10 44可得： bx = - *b =父 0.7 = 0.18( m)；Lc40路基路面宽度为：7+2>0.75+0.18=8.68(m);HY(K7+059.23)、K7+08

5、O QZ(K7+084.834)、YH(K7+110.43)均位于圆曲线上，加宽过渡已完成，加宽值均为：b = 0.7m；路基路面宽度为：7+2>0.75+0.7=9.2(m)。计算超高值：K7+030处，x =10.77 cx0 = 26.67m, bx= 0.19m;姆卜=bj iJ -iGbJiG - bJ B ih 1-xLc= 0.75 3% -2%0.75 2% - 0.75 7 3% l107 = 0.07440,. B. c 7 i c h中=bJ iJ1G =0.75 3%2% = 0.09322h内=bJiJ - bJ bx iG =0.75 3% - 7 - 0.1

6、9 2% =0.0037ZH(K7+019.23)同 HZ(K7+150.43), x = 0 < x0 = 26.67m , bx = 0m ;h外=bj ij -iG =0.753% -2% = 0.0075B“ 7”h中=bjijiG =0.75 3%2% = 0.09322h内=bjij-bjiG =0.75 3% =0.0075HY(K7+059.23)、K7+08O QZ(K7+084.834)、YH(K7+110.43)均位于圆曲线上，四点x >xg = 26.67m ,且位于圆曲线上，位置h# =bJ iJ -iGbJ B ih -0.75 3% 2% 广0.75

7、7 3% = 0.255,. B .-7 h中=bjijih=0.75 3% - 3% =0.127522h内=bjij - bj b ih = 0.75 3% - 7 0.7 3% = -0.021K7+140处，x =10.43 <x0 = 26.67m , bx= 0.18m;h外=bj ij -iG bjiG - bj B ih xLc= 0.75 3% -2%0.75 2% - 0.75 7 3% 1103 -0.07240B7h中=bjij BiG =0.75 3% 7 2% = 0.093 22h内=bjij - bj bx iG =0.75 3% - 7 0.18 3%

8、= 0.0039K7+160处，位于直线段上，已无超高。经计算，各桩号超高加宽如表1所示：表1各桩号超高加宽表桩号加宽值路基路向宽度外缘超高中线超高内缘超高ZH K7+019.2308.50.00750.0930.0075K7+0300.198.690.0740.0930.0037HY K7+059.240.79.20.2550.1275-0.021K7+0800.79.20.2550.1275-0.021QZ K7+084.830.79.20.2550.1275-0.021YH K7+110.430.79.20.2550.1275-0.021K7+1400.188.680.0720.0930

9、.0039HZ K7+150.4308.50.00750.0930.0075K7+16008.50004-2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, a =51° 3248"。曲线内侧中心附近的障碍物距路基边缘3nl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：平曲线要素计算:L；502/ 、p = 0.8 3( m)24R24 125lsLs3_ _ 2240R35050_ _ _7 22240 125= 24.97 (m)51 32 48T = ( R p)

10、 tg q =(125 0.83) tg - 24.97 = 85.73( m)nJi。*L=R： Ls =- 125 51 32 4850 = 126.42(m)18018051 32 48E =(R p) sec- -R =(125 0.83) sec-125 =77.34(m)D =2T -L = 285.73-162.46 = 9( m)(2)视距检查对平原微丘区，采用三类加宽：b=1.5 m,查教材P112表4-17知停车视距$=75 m,查教材P114表4-20知超车视距 夕=350 m“=51o32 / 48" =0.89966 rad轨迹半径:Rs= R-B/2+1.

11、5 =125-7/2+1.5=123 m缓和曲线角：B 0= Ls/2R = 50/250 = 0.2 rad偏移量：e=125-123 = 2 m轨迹线圆曲线长：L' = RsX (a-20) = 123X 0.89966 0.2 >2) =61.458 m轨迹线缓和曲线长：l = Ls e B V0 2 >0.2 = 49.6 m轨迹线曲线长：L= L +21 = 61.458+2 X49.6 = 160.658 m按停车视距控制时：L>ST>L',用教材P126式5-39计算,1160.658-75l'=2(L S) =2 = 42.829

12、ma = arctan'一 一 i 6Rs -l U J J=arctan | 独6 二6 1232 -42.829 /42.829 )+ - f49.6 I 49.6 ) jj= arctan0.175354= 0.17359rad, -2,0:.hmax = Rs 1 -cos一2一Sin I 2 -、l -l= 3.818 1.847 =5.665mh =1.5 1.5 0.75 3 = 6.75mhmax <h ,满足要求。按超车视距控制时：L<SC用教材P126式5-40计算, l ,49.6.、=arctan = arctan=0.067108rad6Rs6x1

13、232%hmax =Rs 1 -CoS-2&) J 二sin l sin -222二 S -Ls=3.818 18.522 41.464=63.504mhmax小，不满足视距要求，需清除一定距离清除距离 h=hmax h=63.504-6.75=56.754 mR = 1000m ,4-3已知平原区某一级公路有一弯道，偏角 "左=16：36'42"，半 交点里程桩号JD=K7+153.63,求：缓和曲线长度；平曲线几何要素；平曲线主点里程桩号解：取平原区一级公路，设计速度为 V = 100km/h确定缓和曲线长度：由旅客感觉舒适得：Ls =0.036V-= 0.036父"0- = 36m ；R1000由行驶时间不过短：Ls = =83.3mLs 5=83.3m1.2 1.2查教材P42表2-3可知，取该平曲线中缓和曲线长为：Ls =85m平曲线要素计算：Ls852p =24R24 1000=0.30( m)3lsLs一22 240R852853852= 42.45( m)240 1000(R p) 1g - q =(1000 0.30) tg 16364242.45 = 188.48(m)22jiL =180R，" Lsji1801000 16 36 42 85 = 374.93(m)16 36 42E =(R