隧道工程课程设计计算书.doc

......目录一设计依据…………………………………………………………………2二设计原始资料……………………………………………………………2三隧道平纵面设计及结构计算………………………………………23.1平面位置的确定…………………………………………………23.2纵断面设计…………………………………………………………23.3.横断面设计…………………………………………………………23.4.二次衬砌结构计算………………………………………………33.4.1基本参数………………………………………………………33.4.2荷载确定………………………………………………………33.4.3计算位移………………………………………………………33.4.4解力法方程……………………………………………………123.4.5计算主动荷载和被动荷载分别产生的衬砌力……………123.4.6最大抗力值的推求……………………………………………133.4.7计算衬砌总力………………………………………………143.4.8衬砌截面强度检算……………………………………………153.4.9力图…………………………………………………………16....c ......家坪隧道设计计算书一设计依据（1）中华人民国行业标准《公路隧道设计规》（JTGD70-2004），：人民交通，2004年．（2）《公路隧道施工技术规》（JTGF60-2009）：人民交通，2009年．（3）中华人民国行业标准《公路工程技术标准》（JTGB01—2003）（4）中华人民国行业标准《公路隧道勘测规程》（JTJ063—85）进行设计和计算。二设计原始资料（1）公路等级：高速公路；（2）设计车速：80km/h；（3）隧道平纵曲线半径和纵坡。平纵曲线设计满足规要求，洞口外各有不小于3s行车速度行程长度围的平纵、线形保持一致。（4）隧道结构设计标准①设计使用期：100年；②设计安全等级：一级；③结构防水等级：二级；④区域地震基本烈度为Ⅷ度区，按Ⅸ度抗震烈度进行设防。（6）1:10000的家坪隧道区域地形图三隧道平纵面设计及结构计算3.1平面位置的确定（见家坪地形平面图）。3.2.纵断面设计做出推荐线路的纵断面图，见纵断面设计图T-1、T-2。3.3.横断面设计....c ......（1）根据课程设计原始资料的要求，做出隧道的建筑界限及轮廓线设计图见T-3。（2）做出各级围岩衬砌结构图，见Ⅴ级围岩衬砌结构设计图T-4。3.4.二次衬砌结构计算选取Ⅴ级围岩复合式衬砌的二次衬砌作为典型衬砌，做结构计算。3.4.1基本参数（1）围岩级别：Ⅴ级（2）围岩容重：（3）围岩弹性抗力系数：（4）衬砌材料为C25混凝土，弹性模量，容重。3.4.2荷载确定（1）围岩垂直均布压力按矿山法施工的隧道围岩荷载为：q=0.45×2=0.45×2[1+i(B-5)]=0.45×2×18.5×[1+0.1×(12.64-5)]=234.96考虑到初期支护承担大部分围岩压力，而对二次衬砌一般作为安全储备，故对围岩压力进行折减，对本隧道按30%折减，取为165.(2)围岩水平均布压力e=0.4q=0.4×165=663.4.3计算位移1.单位位移：（所有尺寸见下图3-1）....c ......半拱轴线长度s=11.36m将半拱轴线长度等分为8段，则s=s/8=1.42ms/E=0.4982×10计算衬砌的几何要素，拱部各截面与直轴之间的夹角和截面中心垂直坐标详见下表3.1.单位位移计算表表3.1....c ......截面（。）sincos△x△yd(m)1/I()y/I()/I()/I()00.00000.00001.00000.00000.00000.500096.00000.00000.000096.0000112.92380.22370.97471.42470.16140.500096.000015.49442.5008129.4896225.84770.43600.90002.77720.63730.500096.000061.180838.9905257.3521338.77150.62620.77963.98901.40360.500096.0000134.7456189.1289554.6201451.69530.78470.61984.99872.42160.500096.0000232.4736562.95811123.9053564.61920.90350.42865.75523.63960.500096.0000349.40161271.68212066.4853677.54300.97650.21576.22004.99590.500096.0000479.60642396.06563451.2784790.33531.0000-0.00596.36986.42190.500096.0000616.50243959.11685288.1216890.00001.00000.00006.24547.85180.500096.0000753.77285918.47337522.0189∑864.00002643.177614338.916020489.2712注：1.I—截面惯性矩，I=/12,b取单位长度。2.不考虑轴力影响。单位位移值用新普生法近似计算，计算如下：=≈=0.4982×10×864.0000=4.3044×10==≈=0.4982×10×2643.1776=1.2115×10=≈=0.4982×10×14338.9160=7.1436×10计算精度校核为：+2+=(0.43044+2×1.2115+7.1436)×10=9.9971×10==0.4982×10×20489.2712=1.0207×10闭合差=02．载位移—主动荷载在基本结构中引起的位移（1）每一楔块上的作用力竖向力：Q=....c ......侧向力：E=eh自重力：G=算式中:b——衬砌外缘相邻两截面之间的水平投影长度，图中量得；h——衬砌外缘相邻两截面之间的垂直投影长度，图中量得；d——为接缝的衬砌截面厚度。作用在各楔块上的力均列入下表3.2：载位移计算表表3.2截面集中力力臂QGQGE00.00000.00000.00000.00000.00000.00000.00000.00001251.702016.523211.40360.68440.70780.3275172.264811.69512238.952016.523233.63280.58380.66280.4723139.500210.95163214.098016.523254.15520.47310.58430.5931101.28989.65454178.381016.523271.93720.34030.47610.683960.70317.86675133.722016.523286.07440.16720.34380.740122.35835.6807682.127016.523295.8528-0.00260.19420.7587-0.21353.2088726.469016.5232100.7284-0.17220.03740.7392-4.55800.618080.000016.523297.13120.0000-0.09120.71420.0000-1.5069（2）外荷载在基本结构中产生的力力按下算式计算弯矩：=-轴力：=sin，的计算结果见下表3.3.表3.4：E(Q+G)E△x△y△x(Q+G)△yE0.00000.00000.00000.00000.00000.00000.00000.00003.73470.00000.00001.42470.16140.00000.0000-187.694615.8848268.225211.40361.35250.4759362.77465.4270-722.2327....c ......32.1194523.700445.03641.21180.7663634.620134.5114-1534.428049.1979754.321699.19161.00971.0180761.6385100.9770-2514.811263.7037949.2258171.12880.75651.2180718.0893208.4349-3533.078072.72351099.4710257.20320.46481.3563511.0341348.8447-4468.675674.45841198.1212353.05600.14981.4260179.4786503.4579-5222.130569.37111241.1134453.7844-0.12441.4299-154.3945648.8663-5784.4665载位移计算表表3.3载位移计算表表3.4截面sincos(Q+G)Esin(Q+G)cosE00.00001.00000.00000.00000.00000.00000.000010.22370.9747268.225211.403659.989911.114748.875220.43600.9000523.700445.0364228.323240.5308187.792430.62620.7796754.321699.1916472.368377.3347395.033640.78470.6198949.2258171.1288744.8817106.0731638.808750.90350.42861099.4710257.2032993.3489110.2456883.103360.97650.21571198.1212353.05601169.915276.15661093.758671.0000-0.00591241.1134453.78441241.0921-2.65561243.747781.00000.00001257.6366550.91561257.63660.00001257.6366（3）主动荷载位移计算结果见表3.5：主动荷载位移计算表表3.5截面1/Iy/I/I·y/I·(1+y)/I00.000096.00000.00000.00000.00000.00001-187.694696.000015.4944-18018.6863-2908.2160-20926.90222-722.232796.000061.1808-69334.3422-44186.7763-113521.11843-1534.428096.0000134.7456-147305.0868-206757.4199-354062.50674-2514.811296.0000232.4736-241421.8711-584627.2030-826049.07405-3533.078096.0000349.4016-339175.4891-1234463.1100-1573638.59906-4468.675696.0000479.6064-428992.8602-2143205.4300-2572198.29027-5222.130596.0000616.5024-501324.5258-3219455.9725-3720780.49838-5784.466596.0000753.7728-555308.7812-4360173.4886-4915482.2698....c ......∑-2300881.6426-11795777.616-14096659.259则：=≈=-0.4982××2300881.6426=-0.1146=≈=-0.4982××11795777.616=-0.5877计算精度校核：+=-0.1146-0.5877=-0.7023==-0.4982××14096659.259=-0.7023闭合差：=03.载位移——单位弹性抗力图及相应的摩擦力引起的位移（1）各接缝处的弹性抗力强度抗力上零点假设在接缝3处，=38.7715=；最大抗力值假定在接缝6处，=77.5430=；最大抗力值以上各截面抗力强度按下式计算：===算出:=0,=0.3985,=0.7556,=;....c ......最大抗力值以下各截面抗力强度按下式计算：=式中：—所考察截面外缘点到h点的垂直距离；—墙脚外缘点到h点的垂直距离。由图中量得：=1.4813m，=2.9097m==0.7408=0按比例将所求得的抗力绘在分块图上。（2）各楔块上抗力集中力按下式近似计算=式中：—楔块外缘长度。（3）抗力集中力与摩擦力之合力R按下式计算：=其中，为围岩与衬砌间的摩擦系数，取0.2，则=1.0198，其作用放向与抗力集中力的夹角==11.3099.由于摩擦阻力的方向与衬砌位移方向相反，其方向朝上。将的方向线延长，使之交于垂直轴，量取夹角（自竖直轴反时针方向量），将分解为水平和竖向两个分力：=sin....c ......=将以上结果列入表3.6中。弹性抗力及摩擦力计算表表3.6截面()(+)/2()()sincos()()40.39850.19931.49320.303458.79040.85530.51820.25950.157250.75560.57711.49320.878770.22350.94100.33840.82690.297361.00000.87781.49321.336782.69960.99190.12711.32580.169970.74080.87041.49321.325494.94650.9963-0.08621.3205-0.114380.00000.37041.43500.5420104.5680.9678-0.25150.5246-0.1363（4）计算单位抗力图及其相应的摩擦力在基本结构中产生的力弯矩：=-轴力：=-式中：—力至接缝中心点K的力臂。计算结果见表3.7表3.8：计算表表3.7截面=0.3034=0.8787=1.3367=1.3254=0.5420()()()()()()40.51150.1552　　　　　　　　-0.155251.94520.59020.67620.5942　　　　　　-1.184463.34611.01522.10981.85390.72580.9702　　　　-3.839374.64331.40883.50243.07762.15922.88620.80071.0612　　-8.433885.80181.76034.80584.22293.56164.76082.2362.96361.0230.5545-14.2620....c ......截面（）sincos()sin()()cos()()451.69530.78470.61980.15720.12340.25950.1608-0.0375564.61920.90350.42860.45450.41071.08640.4657-0.0550677.54300.97650.21570.62440.60972.41220.52030.0894790.33531.0000-0.00590.51010.51013.7327-0.02180.5319890.00001.00000.00000.37380.37384.25730.00000.3738计算表表3.8（5）单位抗力及相应摩擦力产生的载位移，计算见表3.9：单位抗力及相应摩擦力产生的载位移计算表表3.9截面()1/Iy/I/Iy/I(1+y)/I4-0.155296.0000232.4736-14.8982-36.0774-50.97555-1.184496.0000349.4016-113.6977-413.8140-527.51176-3.839396.0000479.6064-368.5694-1841.3360-2209.90547-8.433896.0000616.5024-809.6435-5199.4496-6009.09318-14.262096.0000753.7728-1369.1495-10750.2878-12119.4373　∑-2675.9582-18240.9648-20916.9231=≈=-0.4982××2675.9582=-1.3332×=≈....c ......=-0.4982××18240.9648=-9.0876×校核为：+=-（1.3332+9.0876）×=-1.0421×=-0.4982××20916.9231=-1.0421×闭合差：=04.墙底（弹性地基上的刚性梁）单位弯矩作用下的转角：==×96=64×主动荷载作用下的转角：==-5784.4665×64×=-3.7021单位抗力及相应摩擦力作用下的转角：==-14.2620×64×=-0.00913.4.4解力法方程衬砌计算矢高f==7.8518(m)计算力法方程的系数：==4.2085×10+64×=6.82085×==1.2875×10+7.8518×64×=5.1539×10以上将单位抗力图及相应摩擦力产生的位移乘以倍，即被动荷载的载位移。....c ......求解方程：式中：，式中：，3.4.5计算主动荷载和被动荷载分别产生的衬砌力计算公式为：计算过程见下表3.10、3.11：截面[][]00.0000462.67240.0000462.67240.0000-2.65900.0000-2.65901-187.6946462.6724109.5610384.53870.0000-2.65900.3467-2.31222-722.2327462.6724432.6097173.04930.0000-2.65901.3691-1.28993-1534.4280462.6724952.7866-118.96900.0000-2.65903.01520.35634-2514.8112462.67241643.8217-408.3171-0.1552-2.65905.20212.38795-3533.0780462.67242470.6200-599.7857-1.1844-2.65907.81863.97536-4468.6756462.67243391.2986-614.7046-3.8393-2.659010.73224.23407-5222.1305462.67244359.2907-400.1674-8.4338-2.659013.79562.70288-5784.4665462.67245329.93038.1361-14.2620-2.659016.8673-0.0536主、被动荷载作用下衬砌弯矩计算表表3.10主、被动荷载作用下衬砌轴力计算表表3.11....c ......截面·cos[]·cos[]00.0000678.8164678.81640.00002.14822.1482148.8752661.6208710.49600.00002.09382.09382187.7924610.9049798.69730.00001.93331.93333395.0336529.2389924.27250.00001.67481.67484638.8087420.75981059.5685-0.03751.33161.29415883.1033290.96271174.0660-0.05500.92080.865861093.7586146.42531240.18400.08940.46340.552771243.7477-3.97251239.77520.5319-0.01260.519481257.63660.00001257.63660.37380.00000.37383.4.6最大抗力值的推求首先求出最大抗力方向的位移。考虑到接缝6处的径向位移与水平方向有一定的偏移，因此将其修正如下：计算过程见下表3.12：最大抗力位移修正计算表表3.12截面[]/I[]/I[]()/I[]()/I044416.546-255.26024.9015217707.6998-1251.1580136915.7118-221.97514.7401174984.1653-1052.1839216612.7314-123.83124.264270840.0091-528.04103-11421.023934.20123.4979-39949.5996119.63254-39198.4435229.24312.4799-97208.2200568.49995-57579.423381.62911.2619-72659.4736481.57786-59011.646406.4643-0.09445570.6994-38.3702∑259285.2804-1700.0430位移值为：....c ......·=0.4982××259285.2804×0.9765=0.01261·=-0.4982××1700.0430×0.9765=-8.271×最大抗力值为：3.4.7计算衬砌总力按下式计算衬砌总力：计算过程见下表3.13：衬砌总力计算表表3.13截面[][]e0462.6724-374.647788.0247678.8164302.6825981.49890.08978450.36640.00001384.5387-325.794758.7439710.4960295.01511005.51110.05845639.4160910.20172173.0493-181.7481-8.6989798.6973272.40101071.0983-0.0081-835.0907-532.20333-118.969050.1975-68.7715924.2725235.98631160.2588-0.0593-6602.0684-9266.66334-408.3171336.4621-71.85501059.5685182.33531241.9038-0.0579-6898.0833-16704.39855-599.7857560.1205-39.66521174.0660121.99011296.0561-0.0306-3807.8587-13859.08246-614.7046596.5712-18.13351240.184077.88041318.0644-0.0138-1740.8126-8696.92577-400.1674380.8281-19.33931239.775273.17871312.9540-0.0147-1856.5712-11922.714688.1361-7.55810.57811257.636652.66281310.29940.000455.4944435.7308Σ-7595.2081-59636.0551计算精度的校核：根据拱顶切开点之间相对转角和相对水平位移应为零的条件来检查。由表3.13得，-0.4982××7595.2081=-3.6996×....c ......0.5781×64×=3.6996×闭合差：0-0.4982××59636.0551=-2.9049×7.8518×3.6996×=2.9049×闭合差：03.4.8衬砌截面强度检算检算几个控制截面：（1）拱顶（截面0）0.0897m<0.45d=0.45×0.5=0.225m故偏心距符合规要求。又有：=0.0897m,亦符合规要求。0.0897/0.45=0.1993，可得：0.7016.785>2.4式中：—混凝土极限抗压强度，取。符合规要求。（2）墙底（截面8）偏心检查符合规要求。其他截面偏心距均小于。3.4.9力图将力计算结果按比例尺绘制弯矩图及轴力图，如图2所示:....c ..........c ..........c