隧道工程课程设计.

.1初始条件某一级公路隧道通过IV类围岩(即III级围岩)，埋深H=20m，隧道围岩天然容重γ=25KN/m3，计算摩擦角ф=50o,变形模量E=10GPa，采用矿山法施工。；衬砌材料采用C25喷射混凝土，材料容重，变形模量。2隧道洞身设计2.1隧道建筑界限及内轮廓图的确定该隧道横断面是根据两车道一级公路III级围岩来设计的，根据《公路隧道设计规范》确定隧道的建筑限界如下：W—行车道宽度；取3.75×2mC—余宽；因设置检修道，故余宽取为0mJ—检修道宽度；双侧设置，取为1.0×2mH—建筑限界高度；取为5.0m—左侧向宽度；取为1.0m—右侧向宽度；取为1.0m—建筑限界左顶角宽度；取1.0m—建筑限界右顶角宽度；取1.0mh—检修道高度；取为0.25m隧道净宽为1.0+1.0+7.50+1.0+1.0=11.5m设计行车速度为80km/h,建筑限界左右顶角高度均取1m；隧道轮廓线如下图：图1隧道内轮廓限界图（单位cm）word教育资料 .根据规范要求，隧道衬砌结构厚度为45cm（一次衬砌为10cm和二次衬砌35cm）通过作图得到隧道的尺寸如下：图2隧道内轮廓图得到如下尺寸:3隧道衬砌结构设计3.1支护方法及衬砌材料根据《公路隧道设计规范》（JTG-2004），本设计为一级公路，采用复合式衬砌，复合式衬砌是由初期支护和二次衬砌及中间防水层组合而成的衬砌形式。复合式衬砌应符合下列规定：1初期支护宜采用锚喷支护，即由喷射混凝土，锚杆，钢筋网和钢筋支架等支护形式单独或组合使用，锚杆宜采用全长粘结锚杆。2二次衬砌宜采用模筑混凝土或模筑钢筋混凝土结构，衬砌截面宜采用连结圆顺的等厚衬砌断面，仰拱厚度宜与拱墙厚度相同。由规范8.4.2-1，对于两车道Ⅲ级围岩：初期支护：拱部边墙的喷射混凝土厚度为8-12cm，拱墙的锚杆长度为2.5-3m，锚杆间距为1.0-1.2m；二次衬砌厚度：拱墙混凝土厚度为35cm因此确定衬砌尺寸及规格如下：深埋隧道外层初期支护，根据规范规定，采用锚喷支护，锚杆采用普通水泥砂浆锚杆，规格HRB20×2.5m，采用梅花型局部布设，采用C25喷射混凝土。初次衬砌：采用C25喷射混凝土，厚度为9.8cm。防水板：采用塑料防水板及无纱布，且无纺布密度为300g/m2,防水板应采用铺满的EVA板防水层，厚度为2.0mm，搭接长度为150mm。二次衬砌：根据《公路隧道设计规范》，采用C25号模筑防水混凝土，厚度为35cm。整个衬砌厚度为9.8+0.2+35=45cm。word教育资料 .3.2隧道深浅埋的确定及围岩压力的计算隧道的内轮廓尺寸为B=12.10m，H=10.50m因为IV级围岩需要预留变形量，查《公路隧道设计规范》8.4.1知Ⅲ级围岩需预留变形量为20-50mm，衬砌厚度为45cm，又每侧超挖量为10cm，故隧道的近似开挖尺寸为：由于是Ⅲ级围岩—深浅埋隧道的分界深度—等效荷载高度—跨度影响系数；i—围岩压力增减率，当取i=0.1埋深故为深埋隧道。又可用公式计算均布垂直压力：因为该隧道围岩级别为Ⅲ围岩水平均布压力为：4衬砌结构内力计算4.1基本资料公里等级一级公路围岩级别Ⅲ级围岩容重r=25KN/m3弹性抗力系数K=0.18×106KN/m变形模量E=10GPaword教育资料 .材料容重材料变形模量衬砌厚度d=0.45m图3衬砌结构断面图4.2荷载的确定4.2.1围岩压力的确定经前面计算可得,可用公式计算均布垂直压力：Ⅲ级围岩不考虑一衬后的围岩释放变形折减系数4.2.2衬砌自重（1）全部垂直荷载q=82.125+9.9=92.025KN/m2（2）围岩水平均布压力e<0.15×92.025=13.80KN/m24.3衬砌几何要素4.3.1衬砌的几何尺寸内轮廓线半径：内径所画圆曲线端点截面与竖直轴线的夹角：word教育资料 .拱顶截面厚度，拱底截面厚度4.3.2半拱轴线长度S及分段轴长将半拱轴长度等分为8段，则4.4计算位移4.4.1单位位移用辛普生法近似计算，按计算列表进行，单位位移的计算列表见表4-1表4-1单位位移计算表截面αsinαcosαxyd1/Iy/Iy2/I(1+y)2/I积分系数1/300.0000.0001.0000.0000.0000.450131.6870.0000.000131.6871114.7600.2550.9671.5840.1990.450131.68726.2065.215189.3144229.5200.4930.8703.0560.8060.450131.687106.14085.549429.5162344.2800.6980.7164.3311.7560.450131.687231.243406.0621000.2354459.0400.8580.5145.3203.0120.450131.687396.6421194.6862119.6572573.5720.9590.2835.9824.5310.450131.687596.6752703.5344028.5714686.2090.9980.0666.2636.1070.450131.687804.2144911.3356651.4502798.8460.988-0.1546.1947.7000.450131.6871013.9927807.7379967.40748111.4830.931-0.3665.7739.2320.450131.6871215.73711223.68113786.8411　　　　　　∑1053.4983767.39622432.33731020.628　注：1I—-截面的惯性矩，I=bd3/12,b取单位长度2不考虑轴力影响单位位移值计算如下：计算精度校核：word教育资料 .闭合差4.4.2载位移——主动荷载在基本结构中引起的位移1）每一块上的作用力(竖向力Q、水平力E、自重力G),分别由下面各式求得，Qi=q×biEi=e×hiGi=(di-1+di)/2×△S×rh其中：bi——衬砌外缘相邻两截面间的水平投影长度hi——衬砌外缘相邻两截面间的竖直投影长度di——接缝i的衬砌截面厚度均由图3直接量得，其值见表4-2。各集中力均通过相应图形的形心表4-2载位移Mop计算表word教育资料 .截面投影长度集中力S-Qaq-Gag-GagbhQGEaqagae00.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.00011.5800.200145.40015.8202.7610.7900.7900.100-114.866-12.498-0.27621.4760.600135.82915.8208.2820.7380.7380.300-100.242-11.675-2.48531.2740.957117.24015.82013.2100.6370.6370.479-74.682-10.077-6.32140.9891.25091.01315.82017.2550.4950.4950.625-45.006-7.823-10.78450.6601.52360.73715.82021.0230.3300.3300.762-20.043-5.221-16.00960.2801.57025.76715.82021.6720.1400.1400.785-3.607-2.215-17.01370.0701.5936.44215.82021.990-0.035-0.0350.7970.2250.554-17.51580.4171.53938.37415.82021.244-0.209-0.2090.7708.0013.299-16.348续表4-2∑i-1（Q+G）∑i-1ExyΔxΔy-Δx∑i-1（Q+G）-Δy∑i-1EMoip0.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0001.5840.1991.5840.1990.0000.000-127.640161.2202.7613.0560.8061.4720.607-237.315-1.676-481.033312.86911.0434.3311.7561.2750.950-398.908-10.491-981.512445.92924.2545.3203.0120.9891.256-441.024-30.463-1516.611552.76241.5095.9824.5310.6621.519-365.928-63.052-1986.864629.31862.5326.2636.1070.2811.576-176.838-98.551-2285.088670.90684.2046.1947.700-0.0691.59346.292-134.138-2389.669693.168106.1945.7739.232-0.4211.532291.824-162.689-2265.5832）外荷载在基本结构中产生的内力块上各集中力对下一接缝的力臂由图直接量得，分别记以aq、ae、ag。内力按下式计算之：弯矩：轴力：式中Δxi、Δyi——相邻两接缝中心点的坐标增值。Δxi=xi-xi-1Δyi=yi-yi-1Moip和Noip的计算见表4-3及表4-4。表4-3载位移Noip计算表word教育资料 .截面αsinαcosα∑i（Q+G）∑iEsinα*∑i（Q+G）cosα*∑iENop00.0000.0001.0000.0000.0000.0000.0000.000114.7600.2550.967161.2202.76141.0742.67038.404229.5200.4930.870312.86911.043154.1599.610144.549344.2800.6980.716445.92924.254311.33217.364293.968459.0400.8580.514552.76241.509474.00821.354452.654573.5720.9590.283629.31862.532603.62717.685585.942686.2090.9980.066670.90684.204669.4385.567663.870798.8460.988-0.154693.168106.194684.923-16.330701.2538111.4830.931-0.366747.362127.439695.440-46.671742.1113）主动荷载位移计算过程见表4-4表4-4主动荷载位移计算表截面Mp01/Iy/I1+yMp0/IyMp0/IMp0(1+y)/I积分系数1/300.000131.6870.0001.0000.0000.0000.00011-127.640131.68726.2061.199-16808.513-3344.894-20153.40742-481.033131.687106.1401.806-63345.857-51056.761-114402.61823-981.512131.687231.2432.756-129252.583-226967.536-356220.12044-1516.611131.687396.6424.012-199718.354-601551.682-801270.03625-1986.864131.687596.6755.531-261644.675-1185512.023-1447156.69846-2285.088131.687804.2147.107-300916.979-1837699.993-2138616.97227-2389.669131.6871013.9928.700-314688.937-2423104.814-2737793.75148-2265.583131.6871215.73710.232-298348.374-2754352.193-3052700.5681　　　　∑-1438629.862-7696895.378-9135525.240　△1p=△S/Eh×∑Mp0/I=6.39×10-8×（-1438629.862）=-9192.84×10-5△2p=△S/Eh×∑Mp0y/I=6.39×10-8×(-7696895.378)=-49183.16×10-5计算精度校核△Sp=△1p+△2p△Sp=△S/Eh×∑Mp0(1+y)/I因此，△Sp=6.39×10-8×（-9135525.240）=-58376.006×10-5△1p+△2p=-（9192.84+49183.16）×10-6=-58376.000×10-5闭合差△≈0.000。word教育资料 .4.4.3载位移——单位弹性抗力及相应的摩擦力引起的位移1）各接缝处的抗力强度按假定拱部弹性抗力的上零点位于与垂直轴接近450的第3截面，α3=44.28°=αb;最大抗力位于第5截面,α5=76.25°=αh;拱部各截面抗力强度，按镰刀形分布，最大抗力值以上各截面抗力强度按下式计算：σi=σh（cos2αb-cos2αi）/（cos2αb-cos2αh）计算得，σ3=0，σ4=0.5436σh，σ5=σh。边墙截面弹性抗力计算公式为：σ=σh[1-(yiˊ/yhˊ)2]式中yiˊ——所求抗力截面与外轮廓线交点到最大截面抗力截面的垂直距离；yhˊ——墙底外边缘cˊ到最大抗力截面的垂直距离。(yiˊ和yhˊ在图3中可量得)y6ˊ=1.634m;y7ˊ=3.198m;y8ˊ=4.776m;则有：σ6=σh[1-（1.634/4.776）2]=0.8777σhσ7=σh[1-（3.198/4.776）2]=0.5516σhσ8=0;按比例将所求得的抗力绘在图4上。word教育资料 .图4结构抗力图2）各楔块上抗力集中力按下式近似计算：式中，——楔块i外缘长度，由图3量得。的方向垂直于衬砌外缘，并通过楔块上抗力图形的形心。3）抗力集中力与摩擦力之合力按近似计算:式中μ——围岩与衬砌间的摩擦系数。取μ=0.2,则=1.0198其作用方向与抗力集中力的夹角为β=arctanμ=11.301°。由于摩擦阻力的方向与衬砌位移方向相反，其方向朝上。的作用点即为与衬砌外缘的交点。word教育资料 .将的方向线延长，使之交于竖直轴。量取夹角ψk(自竖直轴反时针方向量度)。将分解为水平与竖向两个分力：RH=RisinψkRV=Ricosψk以上计算例入表4-5中，并参见图3。表4-5弹性抗力及摩擦力计算表截面σ(σh)(σi-1+σi)/2△S外(σh)R(σh)ψksinψkcosψk30.00000.0000.00000.00000.0000.0001.00040.54360.2721.64750.456762.3700.8860.46451.00000.7721.64751.296776.9160.9740.22660.87770.9391.64751.577489.4141.0000.01070.55160.7151.64751.2007103.7990.971-0.23980.00000.2761.64750.4634116.1260.898-0.440续表4-5RH(σh)RV(σh)vhRi(σh)0.0000.000　　0.0000.4050.2120.2120.4050.4481.2630.2940.5051.6681.2721.5770.0160.5213.2451.5471.166-0.2860.2354.4111.1770.416-0.2040.0314.8270.4544）计算单位抗力图及其相应的摩擦力在基本结构中产生的内力弯矩轴力式中rKi----力Ri至接缝中心点K的力臂，由图3量得，计算见表4-6和表4-7word教育资料 .表4-6Mσ0计算表截面号R4=0.4567σhR5=1.29674σhR6=1.5774σhR7=1.2007σhR8=0.4634σhMoσ(σh)r4i-R4r4ir5i-R5r5ir6i-R6r6ir7i-R7r7ir8i-R8r8i40.9000-0.411　　　　　　　　-0.41152.5557-1.1670.9894-1.283　　　　　　-2.45064.0766-1.8622.5820-3.3481.0605-1.673　　　　-6.88375.4556-2.4924.1180-5.3402.6538-4.1860.8364-1.004　　-13.02286.6259-3.0265.5229-7.1624.1930-6.6142.4307-2.9180.8712-0.404-20.124表4-7Nσ0计算表截面号αsinαcosαΣRV(σh)ΣRH(σh)sinαΣRV(σh)cosαΣRH(σh)Noσ(σh)459.04000.85730.51490.21180.40460.18150.2083-0.0268573.57200.95900.28340.50531.66760.48460.47270.0119686.20900.99780.06690.52153.24490.52030.21700.3033798.84600.9882-0.15290.23514.41100.2323-0.67450.90688111.48300.9309-0.36530.03104.82700.0289-1.76331.79225）单位抗力及相应摩擦力产生的载位移计算过程见表4-8。截面号Mσ0（σh）1/Iy/I(1+y)Mσ01/I（σh）Mσ0y/I（σh）Mσ0(1+y)/I（σh）积分系数1/34-0.411131.6872396.6424.012-54.127-163.032-217.15925-2.450131.6872596.6755.531-322.653-1461.941-1784.59446-6.883131.6872804.2147.107-906.387-5535.307-6441.69227-13.022131.68721013.9928.700-1714.805-13204.005-14918.80648-20.124131.68721215.73710.232-2650.074-24465.491-27115.5571　　　　Σ-4240.312-31508.650-35748.952　word教育资料 .表4-8单位抗力及相应摩擦力产生的载位移计算表△1σ=△S/Eh×∑Mσ01/I=6.39×10-8×(-4240.312)=-270.960×10-6△2σ=△S/Eh×∑Mσ0y/I=6.39×10-8×(-31508.650)=-2013.403×10-6校核为：△1σ+△2σ=-(270.960+2013.403)×10-6=-2284.363×10-6△Sσ=△S/Eh×∑Mσ0(1+y)/I=6.39×10-8×(-35748.952)=-2284.358×10-6闭合差△≈0。4.4.4墙底（弹性地基上的刚性梁）位移1）单位弯矩作用下的转角：β1=1/(KI8)=131.6872/0.18×106=731.596×10-62）主动荷载作用下的转角：βp=β1M8p0=731.596×(-2265.583)×10-6=-1657491.46×10-63）单位抗力及相应摩擦力作用下的转角：βσ=β1M8σ0=731.596×10-6×(-20.124)=-14722.64×10-64.5解力法方程衬砌矢高f=y8=9.232m计算力法方程的系数：a11=δ11+β1=（67.32+731.596）×10-6=798.913×10-6a12=δ12+fβ1=（240.74+9.232×731.596）×10-6=6994.83×10-6a22=δ22+f2β1=（1433.43+9.2322×731.596）×10-6=63787.228×10-6a10=△1p+βp+(△1σ+βσ)×σh=-(91928.4+1657491.46+270.960σh+14722.64σh)×10-6=-(1749419.86+14993.6σh)×10-6a20=△2p+fβp+(△2σ+fβσ)×σh=-(491831.6+9.232×1657491.46+2013.403σh+9.232×14722.64σh)=-(15793792.76+137932.82σh)×10-6以上将单位抗力图及相应摩擦力产生的位移乘以σh倍，即被动荷载的载位移。求解方程：X1=(a12a20-a22a10)/(a11a22-a122)=（548.873-4.140σh）其中：X1p=548.873,X1σ=-4.140X2=(a12a10-a11a20)/(a11a22-a122)word教育资料 .=（187.412+2.616σh）其中：X2p=187.412,X2σ=2.6164.6计算主动荷载和被动荷载（σh=1）分别产生的衬砌内力计算公式为：和计算过程列入表4-9和表4-10中。表4-9主、被动荷载作用下衬砌弯矩计算表截面MopX1pyX2p*y[Mp]Moσ(σh)X1σ(σh)X2σ*y(σh)[Moσ](σh)00.000548.8730.0000.000548.8730.000-4.1400.000-4.1401-127.640548.8730.19937.295458.5280.000-4.1400.521-3.6202-481.033548.8730.806151.054218.8940.000-4.1402.108-2.0323-981.512548.8731.756329.095-103.5430.000-4.1404.5940.4544-1516.611548.8733.012564.485-403.253-0.411-4.1407.8793.3285-1986.864548.8734.531849.164-588.827-2.450-4.14011.8535.2636-2285.088548.8736.1071144.525-591.690-6.883-4.14015.9764.9537-2389.669548.8737.7001443.072-397.724-13.022-4.14020.1432.9818-2265.583548.8739.2321730.18813.478-20.124-4.14024.151-0.113表4-10主、被动荷载作用下衬砌轴力计算表截面NopcosaX2pcosφ［Np］Noσ(σh)X2σcosφ(σh)［Nσ］(σh)00.0001.000187.412187.4120.0002.6162.616138.4040.967181.228219.6320.0002.5302.5302144.5490.870163.083307.6320.0002.2762.2763293.9680.716134.175428.1430.0001.8731.8734452.6540.51496.412549.066-0.0271.3461.3195585.9420.28353.002638.9440.0120.7400.752word教育资料 .6663.8700.06612.391676.2610.3030.1730.4767701.253-0.154-28.820672.4330.907-0.4020.5058742.111-0.366-68.635673.4761.792-0.9580.8344.7计算最大抗力值首先求出最大抗力方向内的位移。由式：并考虑接缝5的径向位移与水平方向有一定的偏离，因此将其修正如下计算过程列入表3-11。表4-11最大抗力位移修正计算表截面号Mp/IMσ/Iyiy5-yiMp/I(Y5-Yi)Mσ/I(Y5-Yi)积分系数1/3072279.549-545.1980.0004.531327498.634-2470.2934160382.314-476.6440.1994.332261576.184-2064.8212228825.600-267.5360.8063.725107375.359-996.57343-13635.33259.7331.7562.775-37838.045165.75824-53103.299438.2903.0121.519-80663.911665.76245-77541.042693.0504.5310.0000.0000.0001　　　　Σ621438.869-5000.847　位移值为：δhp=6.39×10-8×621438.869×0.9590=3808.184×10-5δhσ=6.39×10-8×(-5000.847)×0.9590=-30.645×10-5则可得最大抗力σh=δhp/(1/K-δhσ)=3808.184×10-5/[1/(0.18×106)+30.645×10-5]=122.06word教育资料 .3.8计算衬砌总内力按下式进行计算：M=Mp+σhMσN=Np+σhNσ计算过程列入表4-12表4-12衬砌总内力计算表截面号MpMσ[M]M/IMy/INpNσ[N]e积分系数1/30548.873-505.34143.5325732.6590.000187.412319.309506.7210.085911458.528-441.79816.7302203.156438.428219.632308.772528.4040.031742218.894-247.978-29.083-3829.874-3086.879307.632277.857585.490-0.049723-103.54355.366-48.177-6344.348-11140.678428.143228.605656.748-0.073444-403.253406.2482.994394.3181187.686549.066160.998710.0650.004225-588.827642.38353.5567052.62331955.447638.94491.761730.7060.073346-591.690604.55512.8651694.14610346.153676.26158.130734.3910.017527-397.724363.897-33.827-4454.588-34300.335672.43361.583734.016-0.04614813.4776-13.816-0.3389-44.6231-411.961673.476101.820775.296-0.00041　　　Σ3881.0635900.60　　　　　根据拱顶切开点之相对转角和相对水平位移应为零的条件来检查。式中：=6.39×10-8×3881.06=24.80×10-5βa=M8β1=-0.3389×731.596×10-6=-24.794×10-5闭合差：word教育资料 .△=(24.80-24.794)/24.80=0.024%式中：=6.39×10-8×35900.60=229.405×10-5fβa=9.232×（-24.794）×10-5=-228.898×10-5闭合差：△=（229.405-228.898）/229.405=0.221%3.9检验截面强度检算几个控制截面：3.9.1截面0e=0.0859<0.2d=0.2×0.45=0.09m,而则满足3.9.2截面1e=0.0317<0.2d=0.2×0.45=0.09m,而则满足3.9.3截面7e=0.00461<0.2d=0.2×0.45=0.90m,word教育资料 .而则满足3.9.4墙底(截面8)偏心检查e=0.0004m