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一、设计资料拟建一条铁路隧道,埋深35m,长2km,穿过花岗岩地带。为了满足使用需求,隧道净跨为12m,净高为8m。花岗岩单轴抗压强度为180MPa,内摩擦角度为50°,容重为30KN/m3,弹性抗力系数为1.0×106KN/m3。拟采用贴壁式直墙拱形衬砌结构进行支护,衬砌材料拟用C25混凝土浇筑,混凝土容重为24KN/m3,弹性模量E=2.8×104MPa,取厚度方向b=1m进行设计。请给出具体的设计方案。备注,如果是深埋隧道,采用普氏地压理论计算;如果是浅埋,采用松散体理论计算地压。二、初步设计方案拱圈矢跨比41/00lf拱顶厚度0.4md0拱脚及边墙厚度mdddcn6.05.10三、判断深埋或浅埋221max1tan)245(tan)245tan(aHhaa比较与max2H与H大小,若max2HH,则为深埋,反之为浅埋本题为浅埋四、几何尺寸计算00002000000100200011002000182(0.25)2(0.5)(0.5)0.50.5sin0.5cos0.5(1cosnnnnnnfflllfRfddddRdmfddRdmfdRRdmRRdmlRdRfmRdfR)000sin1(sin)212nncnnclldddhHdf五、主动荷载根据松散体地压理论,垂直均布荷载和水平均布荷载分别为)245(tan)()245(tan2)(]tan)245(tan21[21221102111hheheddahhqnh水平均布荷载取222eee六、结构内力分析1.基本结构的单位变位(1)基本结构拱脚刚性固定时拱圈的单位变位nnnnnnAAAnIIImbdIbdI00330012,12系数为)sincos2cossin(sin81)cossin(21sin3)cos1(2cossinsin223sin2)cos1(sinsincos132'2'23222110nnnnnnnnnnnnnnnnnnnnnBbBbBbB拱圈单位变位为)()()()('2'20220322110221120011nBbEARmBbEIRmBbEIRmBEIRn(2)墙顶单位变位(拱脚弹性固定系数)取纵向计算宽度mb1,则ccchIEk44墙顶单位变位按弹性地基长梁计算,当75.2时,为0,,223321221eeceeuukhekeukuu(3)基本结构的单位变位1222222211122112111112ffuaufaaa2.主动荷载作用下基本结构荷载变位(1)基本结构拱脚刚性固定时拱圈的载变位,系数为3443'23'2312312311)cos1(sin61)cossinsin43(41sin51sin31sin81sin61)coscos32(sin61)cossin(41nnnnnnnnnnnnnnnnAaAaAAaaAa5sin3sin5sin3sin)sincoscossin(sin81sin31sinsin31sin)cossin(21)cos1(43)cos1(sin81)sin2cossin9sin24151213'2332'53'23'544535nnnnnnnnnnnnnnnnnnnnnnnnBAbaAAa(代入载变位计算公式得enAaEARemAaEIRemAaEIRqnAaEARqmAaEIRqmAaEIReeqq)()()()()()('5'5025504244031022204211031'2'2主动荷载载变位为eqzeqz222111(2)墙顶载变位作用于边墙上的荷载,有水平均布荷载e及拱圈传来的弯矩、竖向力和水平力,后者为0000220228zzzzzVMfeHqlVefqlM基本结构墙顶的载变位,为ezzzzzezzzzzVHVMuVuHuVMuu03020010302001)()((3)主动荷载作用下基本结构载变位zzzzzzzufaa2201103.主动荷载作用下多余未知力计算由多余未知力计算公式,得主动荷载下多余未知力为21222111120121022122211221012201aaaaaaaXaaaaaaaXzzzzzz4.弹性抗力1σn作用下基本结构的载变位(1)基本结构拱脚刚性固定时拱圈载变位,系数为}sin21cos322sin32cossin22sin221cos)2861(22411{sin31}sin32)4cos(2)]4cos(1[cossin)2831(223{31]cos32)4cos(sincos)2861(2[sin313cossin)4cos(32214332232311nnnnnnnnnnnnnnnnnnnnnnnnKkKk弹性抗力1σn作用下基本结构拱脚刚性固定时拱圈载变位为)()cos21()()cos21(222042112031mKkEIRmKkEIRnnnn(2)墙顶载变位,此时墙顶的载变位,仅由拱圈弹性抗力传来的弯矩、竖向力及水平力引起,系数为)]4cos(22[cos31cos32cos32sin32sin3233231nnn弹性抗力1σn作用下墙顶荷载为)()(cos21)(cos21)(cos21)(cos21)(003220320220120VMnRMnRNnRHnRVnnnnnnnn)45(n拱圈弹性抗力1σn作用下,基本结构墙顶的载变位为03020010302001)()(VHVMVuHuVMuu(3)弹性抗力1σn作用下,基本结构载变位ufaa2201105.弹性抗力nσ=1作用下多余未知力计算由多余未知力计算公式,得弹性抗力1σn作用下多余未知力为21222111120121022122211221012201aaaaaaaXaaaaaaaX6.计算弹性抗力nσ拱脚弹性抗力计算公式为nzpnznzpnnnnuuuXXXXXXuXfuuXuuku22211121211)(sin联立上述公式,得拱脚弹性抗力nσ为ttkuXfuuXutkuXfuuXutznnnzzzz1sin])([sin])([21211212117.计算总的多余未知力nznzXXXXXX2221118.计算拱圈截面内力将拱圈分成10等分,计算各截面内力)()()()(cos0021002MMyXXMNNXNzz其中2/2/)(cossin)(2200eyqxMeyqxNzz)()(00MN、为拱圈弹性抗力使基本结构拱圈内任一截面产生的轴力及弯矩,仅45才存在。9.计算边墙截面内力将边墙分为10等分,计算各截面内力及变形)(55)(88)(88)(77)(772)(663)(66)(552)(2)(1))(2)(4)(2)(2xhdcxhdcxxhdcxhdcxcxhdcxhdcxcxhdcxhdcxccccccccQQMMQQQMMMxkheQQkMMkxkhekeQQkMMky其中2002100XHQVfXXMMVVpcppcpc000000000HHHMMMVVVzpzpzpxexxexexxexxxxsin)sin(coscos)sin(cos墙底无扩基时取0de,这时的dM及dQ很小,可以略去其影响10.计算结果拱圈内力计算结果截面)/(mkNMi拱顶0000487.18630314669.64917210.0900451195.1592052990.7155118740.03223596449.11408634693.14063420.18009023910.31841061.4252262080.128682644338.81431114762.5176930.27013535815.47761592.1233924370.288558577167.88915294874.52658940.36018047720.63682122.8043535690.510568343-44.814285315023.87057350.45022559625.796026493.4625920140.792913074-273.92149665203.39476960.54027071630.955231794.0927743011.13330503-488.50116225404.33644470.63031583536.114437094.6897942841.528986137-653.25620455616.63238880.72036095441.273642395.2488145221.976750329-729.89523115829.27328390.81040607346.432847695.7653054722.472969532-678.84845976030.666932拱脚0.90045119351.592052996.2350821893.013623058-852.24582836258.503327边墙内力计算结果截面kNQ/墙顶0-1204.6090975427.8365483227.6702013.56567417.0041710.51863769413.866448625437.1720271564.8983782.49012822.3902521.037275388520.69064745446.507505484.90944481.39032719.1601831.555913083605.24249195455.842984-86.226954220.54979813.1142242.074550777492.23956935465.178462-304.24131320.0369167.3716352.593188471324.22140525474.513941-320.9312975-0.2419973.1641763.111826165174.36496665483.849419-249.2262774-0.3337360.63123573.63046385968.859032475493.184898-158.0891329-0.330388-0.57943384.1491015547.7949482285502.520376-81.16221635-0.288291-0.93955194.667739248-19.686375575511.855855-29.04851127-0.240537-0.855208墙底5.1
本文标题:地下结构课程设计公式
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