道路工程作业

1、 12土木2班 周莉莉 2012454014 1.已知平原区某一级公路有一弯道，偏角左=163642，半径R=1000m，JD=K7+153.63。要求：（1）计算曲线主点里程桩号；解：经查相关资料，取缓和曲线长度Ls=85m.P=Ls2/24R=852/(24*1000)=0.301q=Ls/2-Ls3/(240*R2)=85/2-853/(240*10002)=42.50T=(R+p)tg(/2)+q=(1000+0.301)tg(16.61/2)+42.5=188.517L=/180R+Ls=374.899E=(R+p)sec(/2)-R=(1000+0.301)sec(16.61/2)

2、-1000=10.902J=2T-L=2*188.517-374.899=2.135主点里程桩号计算：JD=K7+153.63ZH = JD T = K7+153.63 188.517 = K6+965.11HY = ZH + Ls = K7+153.63 +85 = K7+238.63QZ = ZH + L/2= K7+153.63+374.899/2 =K7+341.080HZ = ZH + L = K7+153.63+374.899 =K7+528.529YH = HZ Ls = K7+153.63 85=K7+68.632.平原区某公路有两个交点间距为371.82m，JD1=K15+3

3、85.63，偏角1=201952（右偏），半径R1=700m，JD2为右偏，2=170532，R2=850m，试按S型曲线计算LS1、LS2长度，并计算两曲线主点里程桩号。解：（1）计算确定缓和曲线长度Ls1、Ls2：令两曲线的切线长相当，则取T1=371.82/2=185.91m 按各线形要素长度1：1：1计算Ls1： Ls1=R/2=20.1952/180700/2=123.37 取Ls1=124m 则经计算得，q=Ls1/2Ls1/240R2=61.984p= Ls12/24R Ls14/2384R3 =0.915T1=(R+p)tan(a/2)+q=187.57m 185.91m187

4、.57-185.91=1.66m，即T1计算值偏长切线长度与缓和曲线长度的增减有近似1/2的关系， LS1=124-21.66=120.68m，取Ls1=120m。（q=Ls1/2Ls1/240R2=60.00p= Ls12/24R Ls14/2384R3 =0.857T1=(R+p)tan(a/2)+q=185.58m )则计算得，T1= 185.58m T2=371.82-T1=371.82-185.58=186.24m按1：1：1计算Ls2： Ls2=R/2=17.0532PI/180850/2=126.50(q=Ls2/2Ls2/240R2=63.24p= Ls22/24R Ls24/

5、2384R3 =0.784T2=(R+p)tan(a/2)+q=191.02m )计算切线长T2得，T2=191.02m191.02-185.91=5.11取Ls2=126.5-25.11=116.28(q=Ls2/2Ls2/240R2=58.13p= Ls22/24R Ls24/2384R3 =0.663)T2=(R+p)tan(a/2)+q=185.89m )计算得， T2=185.89m185.91-185.89=0.02取Ls2=116.28+20.02=116.32JD1曲线要素及主点里程桩号计算L1=/180R+Ls=366.73E1=(R+p)sec(/2)-R=(700+0.8

6、57)sec(20.1952/2)-700=11.88J1=2T-L=2*185.58-366.73=4.43得：R1=700 Ls1=120 1=20.1952T1=185.58 L1=366.73 E1=11.88 J1=4.43主点里程桩号计算：JD1=K15+385.63(ZH = JD T = K15+385.63 185.58 = K15+200.05HY = ZH + Ls = K15+385.63+120 = K15+505.63QZ = ZH + L/2= K15+385.63+366.73/2 = K15+567.00HZ = ZH + L = K15+385.63+366

7、.73= K15+752.36YH = HZ Ls = K15+385.63 120= K15+265.63)JD1=K15+385.63ZH1= K15+200.05HY1= K15+505.63QZ1=K15+567YH1= K15+265.63HZ1= K15+752.36JD2里程桩号计算：JD2 = JD1+ 交点间距 - J1 = HZ1 + 曲线间直线长度 + T2JD2 = JD1+ 371.82 - J1= K15+385.63+ 371.82 -4.43 = K150+753.02JD2里程桩号计算：JD2 = K15+753.022.平原区某公路有两个交点间距为371.8

8、2m，JD1=K15+385.63，偏角1=201952（右偏），半径R1=700m，JD2为右偏，2=170532，R2=850m，试按S型曲线计算LS1、LS2长度，并计算两曲线主点里程桩号。JD2曲线要素及主点里程桩号计算L2=/180R+Ls=369.31E2=(R+p)sec(/2)-R=（850+0.663)sec(17.0532/2)-850=10.17J2=2T-L=2*185.89-369.31=2.47 R2=850 Ls2=116.32 2=17.0532 T2=185.89 L2=369.31 E2=10.17 J2=2.47(ZH = JD T = K15+753.0

9、2 185.89 = K15+567.13HY = ZH + Ls = K15+753.02+116.32= K15+869.34QZ = ZH + L/2= K15+753.02+369.31/2 = K15+937.68HZ = ZH + L = K15+753.02+369.31= K15+1122.33YH = HZ Ls = K15+753.02116.32= K15+636.7)JD2 = K15+753.02ZH2= K15+567.13 HY2= K15+869.34QZ2= K15+937.68YH2= K15+636.7HZ2= K15+1122.33某二级公路一路段有三个

10、变坡点，详细资料如下：变坡点桩号 设计高程 竖曲线半径 K12+450 172.513 5000 K12+950 190.013 4000 K13+550 173.513 3000试计算K12+700K13+300段50m间隔的整桩号的设计高程值。 解：第一个与第二个变桩点：i1=(190.013-172.513)/500=3.5%第二个与第三个变桩点：i2=(173.513-190.013)/600=-2.75%=i2- i1= - 0.0275-0.035= - 0.06250，为凸形。第二个变桩点曲线长 L = R=40000.0625=250m切线长：T=L/2=125m外距：E=T2

11、/2R=1.95竖曲线起点QD（K12+950）- 125= K12+825竖曲线终点ZD（K12+950）+ 125= K13+75计算设计高程：上半程：二级公路最大纵坡规定为：6%设若第一个桩左-6%，则L = R=50000.095=475mT=L/2=237.5m第一个桩竖曲线最大桩号：K12+687.5 （K12+450+237.5） 设若第三个桩右+6%，则L = R=3000*0.0875=262.5mT=L/2=131.25m第三个桩竖曲线最小桩号：K13+ 418.75（K13+ 550-131.25）故K12+700K13+300段内K12+825K13+75段只考虑只考虑

12、第二个变桩点设置的竖曲线，其余高程由切线求得K12+700高程： 172.513 +（700-450）*3.5%=181.263K12+750高程：172.513 +（750-450）*3.5%=183.013K12+800高程：172.513 +（800-450）*3.5%=184.763K12+850：横距x1= Lcz QD = 12850 1282525m竖距y1=x12/2R=25*25/(2*4000)=0.078切线高程 HT = H1 + i1（ Lcz - BPD）= 190.013 + 0.035(12850 - 12950)= 186.5m 设计高程 HS = HT -

13、y1 = 186.5 - 0.078=186.42m K12+850：横距x1= Lcz QD = 12850 1282525m竖距y1=x12/2R=25*25/(2*4000)=0.078切线高程 HT = H1 + i1（ Lcz - BPD）= 190.013 + 0.035(12850 - 12950)= 186.500m 设计高程 HS = HT - y1 = 186.5 - 0.078=186.420m K12+900：横距x1= Lcz QD = 12900 1282575m竖距y1=x12/2R=75*75/(2*4000)=0.7切线高程 HT = H1 + i1（ Lcz

14、 - BPD）= 190.013 + 0.035(12900 - 12950)= 188.263m 设计高程 HS = HT - y1 = 188.263 - 0.7=187.563m K13横距x2= ZD-Lcz = 13075 1300075m竖距y2=x22/2R=75*75/(2*4000)=0.7切线高程 HT = H1 + i1（ Lcz - BPD）= 190.013 - 0.0275(13000 - 12950)= 188.638m 设计高程 HS = HT - y1 = 188.638 - 0.7=187.938m K13+50横距x2= ZD-Lcz = 13075 1305025m竖距y2=x22/2R=25*25/(2*4000)=0.078切线高程 HT = H1 + i1（ Lcz - BPD）= 190.013 - 0.0275(13050 - 12950)= 187.263m 设计高程 HS = HT - y1 = 187.263 - 0.078=187.185m K13+100 高程: 1