拱桥计算例题

dhd12hf例题1：ml150mf30mhd7.031/20mkN32/19mkN3/24mkN，，，，，试确定拱轴系数m。md7.0，，j0fmjcos2.2402.5142.8143.1423.5003.8934.3240.730570.721910.713190.704400.695540.686630.67765解：假定514.2假mmkNgd/8.307.0247.020mddffj09733.372191.027.027.03cos220mddfhj96251.272191.027.027.009733.3cos22mkNdhhgjdj/55929.932472191.07.01996251.2207.0cos21038.38.3055929.93ggdj计m大于半级令142.3假mmkNgd/8.307.0247.020mddffj10346.370440.027.027.03cos220mddfhj95658.270440.027.027.010346.3cos22mkNdhhgjdj/02511.942470440.07.01995658.2207.0cos21053.38.3002511.94ggdj计m故取164.022.814-3.142089.0-计假mm142.3m例题2：某空腹无铰拱桥，ml400mf80514.2mmkNMj10000，拱圈厚度，，md9.0拱轴系数，半跨恒载对拱脚截面产生的弯矩值，试分别计算恒载作用下拱顶截面、拱跨四分之一截面、拱脚截面的N、Q、M。计算结果小数点后保留三位有效数字。查表已知各数据如下：336314.0121)(0890.11f，2)(16703.9f72191.0cosj69198.0sinj，，，94042.0cos4l34001.0sin4l，，解：mdffj125.8)cos1(20kNfMHjg769.1230260.012d011.0)(0890.1121f009.0)(16703.92fkNHSg418.1311mfys733.2125.8336314.01mymfyll747.12)1(2144（在y1坐标系中坐标）在y坐标系中坐标为：986.0733.2747.14myl1、不计弹压内力计算kNNd769.1230'kNHNlgl744.130894042.0769.1230cos4'4kNHNjgj879.170472191.0769.1230cos'2、弹压内力计算SyxkNNd418.13''sykNQd0''mkNMd671.36733.2418.13''SyxsykNNl619.1294042.0418.13''4kNQl562.434001.0418.13''4mkNMl230.13986.0418.13''4kNNj687.972191.0418.13''kNQj285.969198.0418.13''mkNMj350.72)733.2125.8(418.13''3、恒载作用下的总内力计kNNNNddd351.1217418.13769.1230'''kNQQQddd000'''mkNMMMddd671.36671.360'''kNNNNlll125.1296619.12744.1308''4'44kNQQQlll562.4562.40''4'44mkNMMMlll230.13230.130''4'44kNNNNjjj192.1695687.9879.1704'''kNQQQjjj285.9285.90'''mkNMMMjjj350.72350.720'''拱脚截面（需计算剪力）最不利内力计算步骤：1、查表P480、P325得弯矩影响线图形及、、，绘制M、H、V影响线图；2、查表P325得弯矩影响线峰值及截面号，查表P240得相应截面H影响线竖标值，查表P213得相应截面V影响线竖标值；3、计算最不利M及相应H、V；4、查表P592得、，计算N及Q。注：1、多车道内力=单车道内力X车道数X多车道折减系数；2、其它截面无需计算剪力，轴向力N可查表P480得，查表P695得相应截面N影响线竖标值，计算N。3、每个截面均应计算最大正弯矩及相应的N、Q和最小负弯矩及相应的N、Q。MHVjcosjsinN桥梁墩台计算以最大水平力控制设计，计算内容为最大水平力及相应的M和V，例题3-3-3：查表P480得：maxHflflH2212742.0)03675.009067.0(（为任意截面最大正弯矩对应的+最小负弯矩对应的）HH查表P240得H影响线最大竖标值（24#截面）为fl23265.0查表P325得M影响线最大竖标值（24#截面）为22005.0)01405.001905.0(llM相应的：llV5.0)33378.016622.0(l03515.0查表P213得V影响线最大竖标值（24#截面）为5.0kNH73.565105023265.0270105012742.09.72max相应的：mkNM28.5735003515.027050005.09.72kNV5.3325.0270505.09.7例题3：（课程设计）某悬链线无铰拱桥，ml350mf70514.2m，，拱轴系数集度，试计算拱跨四分之一截面最大正弯矩和最小负弯矩以及相应的轴向力。，2/0.3mkN拱圈厚度，md9.0计算荷载为公路-Ⅱ级车道荷载，人群荷载解：mdffj125.7)cos1(20mfys396.2125.7336314.01mdllj623.35sin0公路-Ⅱ级车道荷载标准值：mkNqK/875.7kNPK225260.012d015.0)(0890.1121f012.0)(16703.92fmymfyll532.12)1(2144（在y1坐标系中坐标）在y坐标系中坐标为：864.0396.2532.14myl+-1、最大正弯矩及相应的轴向力计算查表得弯矩影响线图形（P489、P332）++256.1100887.02lMmkNqK/875.7kNPK225195.7/04040.02flH275.1234458.0lV影响线峰值：111.205926.0l弯矩：相应：1H位于12#截面687.0/13750.0fl（P241）相应：V83678.0（P215）(1)不计弹压内力计算：mkNM616.563111.2225256.11875.7max相应：1H相应：VkNH236.211687.0225195.7875.71kNV941.28483678.0225275.12875.7相应：NkNHNl619.224cos41说明：本例中无须求V相应的(2)弹压内力计算：HyxsykNHH137.3111mkNM710.2864.0137.3kNN950.294042.0137.3kNQ067.134001.0137.3(3)最大正弯矩时总内力：mkNM326.566710.2616.563kNN669.221950.2619.224kNQ067.1067.10+-2、最小负弯矩及相应的轴向力计算查表得弯矩影响线图形（P489、P332）++083.1301031.02lM565.15/08739.02flH537.515542.0lV影响线峰值：046.102936.0l弯矩：相应：1H位于18’#截面035.1/20709.0fl（P241）相应：V32163.0（P215）(1)不计弹压内力计算：mkNM379.338)046.1(225)083.13(875.7max相应：1H相应：VkNH449.355035.1225565.15875.71kNV971.11532163.0225537.5875.7相应：NkNHNl968.377cos41(2)弹压内力计算：HyxsykNHH269.5111mkNM552.4864.0269.5kNN955.494042.0269.5kNQ792.134001.0269.5(3)最小负弯矩时总内力：mkNM827.333552.4379.338kNN923.342955.4968.337kNQ792.1792.10例题4：某悬链线无铰拱桥，ml40mf8514.2makPE7100.3，，拱轴系数右拱脚右移0.02m，求由于相对水平位移而引起的左拱脚及拱顶的N、Q、M。EIlf222099621.0467.0mI72191.0cosj69198.0sinj，，，左拱脚左移0.01m，fys336314.0，，4722221027.167.0100.3840099621.0099621.0EIlfmh03.002.001.0kNhX22.2361027.103.042222Xyxmfys69.28336314.0336314.0kNNj53.17072191.022.236kNQj46.16369198.022.236mkNMj33.1254)69.28(22.236kNNd22.2360dQmkNMd43.63569.222.236例题5：某悬链线无铰拱桥，，，拱轴系数，ml40mf8514.2makPE7100.3EIl333100032.0467.0mI72191.0cosj，，，EIl322099621.069198.0sinj，，左拱脚向左发生水平位移0.02m，沉降0.01m，右拱脚向右发生水平位移0.02m，沉降0.04m，求由于拱脚相对位移而引起的左拱脚N、Q、M。fys336314.0，1、拱脚相对水平位移内力2Xyxmh04.002.002.0kNhX31410274.104.0422247222210274.167.0100.3840099621.0099621.0EIlfmkNMj4.1667)8336214.08(314kNNj7.22672191.0314kNQj3.21769198.03143XyxmV03.001.004.0kNVX2.9410184.303.0433347333310184.367.0100.340100032.0100032.0EIlmkNMj4.188420221.94kNNj2.6569198.0221.94kNQj6872191.0221.942、拱脚相对垂直位移内力拱脚位移总内力：mkNM8.35514.18844.1667kNN5.1612.657.226kNQ3.285683.217