《道路勘測設(shè)計(jì)》第4章橫斷面設(shè)計(jì)課后習(xí)題及答案

1、第四章橫斷面設(shè)計(jì)4-1某新建三級公路，設(shè)計(jì)速度V=30km/h,路面寬度B=7m,路拱iG=2%,路肩bj=0.75m,路肩橫坡ij=3%。某平曲線轉(zhuǎn)角久=34二5008,半徑R=150m,緩和曲線Ls=40m,力口寬值b=0.7m,超高ih=3%,交點(diǎn)樁號為K7+086.42。試求平曲線上5個主點(diǎn)及下列樁號的路基路面寬度、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=40ms平曲線要素計(jì)算：I2402p=0=0.44(m)24R24150ls233= 19.99( m)Ls4040

2、ZT二一2240R2240150T=(R+p)tg%+q=(150+044)tg345008+19.99=67.19(m)22L=一R：Ls=131.21(m)180180E=(Rp)sec-R=(1500.44)sec345008-150=7.67(m)22D=2T-L=267.19-131.21-3.17(m)(2)主點(diǎn)里程樁號計(jì)算JDK7+086.42-T-67.19ZHK7+019.23+Ls+40HYK7+059.23ZHK7+019.23+L/2+131.21/2QZK7+84.84ZHK7+019.23+L+131.21HZK7+150.44-Ls-40

3、YHK7+110.44超高過渡及加寬計(jì)算：新建三級公路，設(shè)計(jì)速度V=30km/h,無中間帶，超高過渡采用采用內(nèi)邊線旋轉(zhuǎn)，加寬線性過渡，路基邊緣為設(shè)計(jì)高程，加寬值b=0.7m,超高ih=3%,臨界斷面距過渡段起點(diǎn)=iGI=2%父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)=10.77(m)x0=26.67(m),由教材P106表4-13一一x1077可得：bx=*b=m0.7=0.19(m)；Lc40路基路面寬度為：7+2X0.75+0.19=8.69(m);K7+140處，處于后一段緩和曲線上，則有：K7+150.4

4、4-(K7+140)=10.44(m)x0=26.67(m),由教材P106表4-13一一x1044可得：bx=b=父0.7=0.18(m)；Lc40路基路面寬度為：7+2X0.75+0.18=8.68(m);HY(K7+059.23)、K7+08OQZ(K7+084.834)、YH(K7+110.43)均位于圓曲線上，加寬過渡已完成，加寬值均為：b=0.7m;路基路面寬度為：7+2X0.75+0.7=9.2(m)。計(jì)算超高值：K7+030處，x=10.77xO=26.67m,bx=0.19m;h外=bjij-iGbjiG-bjBihl-xLc=0.753%2%0.752%一0.7573%11

5、07=0.07440,.B.-7八一h中=bjijiG=0.753%-2%=0.09322h內(nèi)=bjij-bjbxiG=0.753%-7-0.192%=0.0037ZH(K7+019.23)同HZ(K7+150.43),x=0x0=26.67m,且位于圓曲線上，位置h外=bjij-iGbjBih-0.753%-2%0.7573%-0.255B7h中=bjijih=0.753%3%=0.127522h內(nèi)=bJiJ-bJbih=0.753%-70.73%=-0.021K7+140處，x=10.43STL,用教材P126式5-39計(jì)算1 八l (L -S)=2160.658-75 =42.829ml

6、a=arctanJ一叫二 arctan 比6 1232-42.82942.82949.649.6：|=arctan0.175354=0.17359radhmax=Rs1-cos22-sin-l-l=3.8181.847=5.665mh=1.51.50.753=6.75mhmaxh,滿足要求。按超車視距控制時：LSC用教材P126式5-40計(jì)算, l ,49.6 .、=arctan =arctan =0.067108rad6Rs f hmax =Rs 1 -CoS6x123:工、 ：工 S-Lssin 一 l sin-222fot=3.81818.52241.464=63.504mhmaxh,不

7、滿足視距要求，需清除一定距離。清除距離h=hmaxh=63.504-6.75=56.754m4-3已知平原區(qū)某一級公路有一彎道,偏角 a左=16：3642,半徑 R = 1000m,交點(diǎn)里程樁號JD=K7+153.63求:緩和曲線長度；平曲線幾何要素;平曲線主點(diǎn)里程樁號。解：取平原區(qū)一級公路，設(shè)計(jì)速度為V =100km/ h確定緩和曲線長度:由旅客感覺舒適得:Ls、，3) “3-0.036- =0.036 -00-1000由行駛時間不過短:LsV 1001.2 1.2-83.3mLs (min) = 83.3mLs =85m查教材P42表2-3可知，取該平曲線中緩和曲線長為:平曲線要素計(jì)算:22Ls85-p=0.30(m).3ls Ls852 二2 240R224R241000853852=42.45(m)2401000T=(Rp)tgq=(10000.30)tg16364242.45=188.48(m)22JTL 二R： Ls 二1801801000 16 3642 85 = 374.93(m)163642E=(Rp)sec-R=(10000.30)sec-1000=10.9(m)22D=2T-L=2188.48-374.93=2.03(m)主點(diǎn)里程樁號計(jì)算JDK7+153.6