《道路勘測(cè)設(shè)計(jì)》第4章橫斷面設(shè)計(jì)課后習(xí)題和答案解析

1、WORD格式.資料第四章橫斷面設(shè)計(jì)4-1某新建三級(jí)公路，設(shè)計(jì)速度 V 30km/h,路面寬度B= 7m,品&拱iG=2%,路 肩bj=0.75m,路肩橫坡ij=3%。某平曲線轉(zhuǎn)角a=34 50'08"，半徑R = 150m, 緩和曲線Ls=40m,力口寬值b= 0.7m,超高ih=3%,交點(diǎn)樁號(hào)為K7+086.42。試求平曲線上5個(gè)主點(diǎn)及下列樁號(hào)的路基路面寬度、橫斷面上的高程與設(shè)計(jì)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平曲線要

2、素計(jì)算：402/ 、=0.44( m)24R24 150ls q W_ 2240R234040_一_22240 150-19.99 (m)專業(yè).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)主點(diǎn)里程樁號(hào)計(jì)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超高過(guò)渡及加寬計(jì)算：新建三級(jí)公路，設(shè)計(jì)速度 V= 30km/h,無(wú)中間帶，超高過(guò)渡采用采用內(nèi)邊線旋轉(zhuǎn)，加寬線性過(guò)渡，路基邊緣為設(shè)計(jì)高程，加寬值b = 0.7m,超高"=3%,J2%臨界斷面距過(guò)渡段起點(diǎn)X0 =叵 Lc = 2% M 40 = 26.67m。 ih3%先計(jì)算加寬值：K7+030處，處于前一段緩和曲線上，則有：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路基路面寬度為：7+2>0.75+0.19=8.69(m);K7+140處，處于后一段緩和曲線上，則有：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路基路面寬度為：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)均位于圓曲線上，加寬過(guò)渡已完成，加寬值均為：b = 0.7m；路基路面寬度為：7+2>0.75+0.7=9.2(m)。計(jì)算超高值：K7+030處，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內(nèi)=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內(nèi)=bjij-bjiG =0.75 3% =0.0075HY(K7+059.23)、K7+08O QZ(K7+084.834)、YH(K7+110.43)均位于圓曲線上，四點(diǎn)x >xg = 26.67m ,且位于圓曲線上，位置h# =bJ iJ -iGbJ B ih -0.75 3% 2% 廣0.75

7、7 3% = 0.255,. B .-7 h中=bjijih=0.75 3% - 3% =0.127522h內(nèi)=bjij - bj b ih = 0.75 3% - 7 0.7 3% = -0.021K7+140處，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內(nèi)=bjij - bj bx iG =0.75 3% - 7 0.18 3%

8、= 0.0039K7+160處，位于直線段上，已無(wú)超高。經(jīng)計(jì)算，各樁號(hào)超高加寬如表1所示：表1各樁號(hào)超高加寬表樁號(hào)加寬值路基路向?qū)挾韧饩壋咧芯€超高內(nèi)緣超高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某雙車道公路，設(shè)計(jì)速度 V= 60km/h,路基寬度8.5m,路面寬度7.0m。某平曲線R= 125m Ls=50m, a =51° 3248"。曲線內(nèi)側(cè)中心附近的障礙物距路基邊緣3nl試檢查該平曲線能否保證停車視距和超車視距？若不能保證，清除的最大寬度是多少？ 解：平曲線要素計(jì)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)視距檢查對(duì)平原微丘區(qū)，采用三類加寬：b=1.5 m,查教材P112表4-17知停車視距$=75 m,查教材P114表4-20知超車視距 夕=350 m“=51o32 / 48" =0.89966 rad軌跡半徑:Rs= R-B/2+1.

11、5 =125-7/2+1.5=123 m緩和曲線角：B 0= Ls/2R = 50/250 = 0.2 rad偏移量：e=125-123 = 2 m軌跡線圓曲線長(zhǎng)：L' = RsX (a-20) = 123X 0.89966 0.2 >2) =61.458 m軌跡線緩和曲線長(zhǎng)：l = Ls e B V0 2 >0.2 = 49.6 m軌跡線曲線長(zhǎng)：L= L +21 = 61.458+2 X49.6 = 160.658 m按停車視距控制時(shí)：L>ST>L',用教材P126式5-39計(jì)算,1160.658-75l'=2(L S) =2 = 42.829

12、ma = arctan'一 一 i 6Rs -l U J J=arctan | 獨(dú)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 ,滿足要求。按超車視距控制時(shí)：L<SC用教材P126式5-40計(jì)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，不滿足視距要求，需清除一定距離清除距離 h=hmax h=63.504-6.75=56.754 mR = 1000m ,4-3已知平原區(qū)某一級(jí)公路有一彎道，偏角 "左=16：36'42"，半 交點(diǎn)里程樁號(hào)JD=K7+153.63,求：緩和曲線長(zhǎng)度；平曲線幾何要素；平曲線主點(diǎn)里程樁號(hào)解：取平原區(qū)一級(jí)公路，設(shè)計(jì)速度為 V = 100km/h確定緩和曲線長(zhǎng)度：由旅客感覺(jué)舒適得：Ls =0.036V-= 0.036父"0- = 36m ；R1000由行駛時(shí)間不過(guò)短：Ls = =83.3mLs 5=83.3m1.2 1.2查教材P42表2-3可知，取該平曲線中緩和曲線長(zhǎng)為：Ls =85m平曲線要素計(jì)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