道岔课程设计

道岔课程设计道岔课程设计#/11lwntiti502tan cos2t1tlwntiti502tan cos2t1t2

sinw1538-——2sin68620252__682tan62025cosi0711.49 2sin2506202526846sin20336150923614452612150则由图的几何关系可得：DmX则由图的几何关系可得：DmX3X4COSw2050-452770cos20336 143.42050 452769.5143.4lp685mm一1820410mm22,15029046故满足条件。三、选择导曲线半径:1、求算R0由knp=1538+820/2=1948mm,取k=1948mm2Sl0ksinksinR0 0coscos1435 62501948sin11912 1948sin62025cos11912cos62025143581980.02319480.110.0058517644mm因圆曲线半径须取为10m整数倍，故取Rq180000mm

S 1435&R180000 180717.5mm222、用1中求得的民反求kISR0coscosygK - gsin1435180717.5cos1o19‘12''cos6o20‘25" 144sin6o20'25''2115.85mm2116mmKKmin1948又知一般要求K2~4m所以K=2116mm满足要求：3、检算选用的导曲线半径能否满足动能损失要求:1800003520.64km2/h21000V3.62R10003523.6218000020.53m/s31000V3.63RI10003533.63180000180.28m/s30.65km2/h2（可）20.65m/s（可）0 0.5m/s3（可）所以R取180m时符合要求:四、计算道岔主要几何尺寸：1、 道岔理论全长Ltl0cos R0sinsinKcossin11912 2116cos620256250cos11912 180000sin62025sin11912 2116cos620256248.3415731.42103.0624144mm2、 确定道岔实际全长:LQ qLtm264624145.52050828848mm3、 确定道岔后长2tan—21435丄62025tan22050 4 15009mm4、确定道岔前长:aLQb13839mm五、配轨计算：IpIp12500mmhl2Lqlj3 28848-12500-24 16324mm180717.53550113 2116-1538-8-2618016395mm15 16Lt10cosn1 2224144-6250cos11912 153882616338mm7 l8 q l7 l8 q l0S0tanRS色2264662501437tan119122050--28-1250026466250-■33.1115705.7 211616218mm取：Km21l,180j180717.5143535 501131802050-16-125002116l1 5324mm,l2 11000mml3 6895mm,l4 9500mml5 6838mm,l6 9500mml7 5218mm,l8 11000mm六、导曲线支距计算：自坐标原点开始令导曲线上各支以导曲线起点为坐标原点，以直线基本轨作用边为横坐标,自坐标原点开始令导曲线上各支距测点的横向座标Xi依次为2000mm整数倍。可得相应的支距yi为：yiyRocoscosi式中,i的值可以由下式计算sinisin 'R0终点处，sin(atan(1/9))=144/6250+x n/180717.5Xn=15793.5计算结果见下表:Xisinicosicos cosiyi00.023040.999735014420000.034110.9994180.00031681320140000.045170.9989790.00075586628160000.056240.9984170.00131777738280000.067310.9977320.002002754506100000.078370.9969240.00281105652120000.089440.9959920.003742966820140000.100510.9949360.004798849101115793.20.110430.9938840.0058512791201复核：终点处支距为： SK*sin =1435-2116*sin(atan(1/9))=1201.3269 ，计算结果正确。七、配置岔枕：2500 14351、轨枕伸出钢轨的长度：M25001435532.5mm2、 确定岔枕的间距ai⑴q段：I26464220502480mm5004954即按间距为500和495配置4根钢轨。⑵l0段:I 6250214423502206155366mm5065409(3)l0与l1之间：I12500265062502202203160mm54035152510⑷l1段：I6056844056245408572⑸l2段：l110008440210128535165302508⑹l2之后：=4850-532.5*2-1435*2/COS(ATAN(1/9))=897/2=897/2/tan(a/2)=8100