隧道工程课程设计

武汉理工大学课程设计初始条件某高速大路隧道通过III类围岩(即IV级围岩)，埋深H=30m，隧道围岩自然容重γ=23KN/m3，计算摩擦角ф=35o,变形模量E=6GPa,承受矿山法施工；衬砌材料承受C25喷射混h凝土，材料容重 22KN /m3，变形模量E 25GPa。hh隧道洞身设计隧道建筑界限及内轮廓图确实定该隧道横断面是依据两车道高速大路IV级围岩来设计的，依据《大路隧道设计标准》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1.0×2mH5.0m2L1.0mLL 1.5mRE1.0mLE —建筑限界右顶角宽度；取1.0mRh0.25m隧道净宽为1.0+1.0+7.50+1.50+1.0=12m设计行车速度为120km/h,建筑限界左右顶角高度均取1m；隧道轮廓线如以下图：图1 隧道内轮廓限界图依据标准要求，隧道衬砌构造厚度为50cm〔15cm35cm〕通过作图武汉理工大学课程设计得到隧道的尺寸如下：图2 隧道内轮廓图得到如下尺寸:R19.47m，R25.6m，R311.2m3隧道衬砌构造设计支护方法及衬砌材料〔JTG-2023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。复合式衬砌应符合以下规定：初期支护宜承受锚喷支护，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，锚杆宜承受全长粘结锚杆。断面，仰拱厚度宜与拱墙厚度一样。8.4.2-1，对于两车道IV级围岩：12-15cm2.5-3m，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：HRB20×2.5m，承受梅花型局部布设，承受C25喷射混凝土。初次衬砌：承受C2514.8cm。防水板：承受塑料防水板及无纱布，且无纺布密度为300g/m2,防水板应承受铺满的EVA板武汉理工大学课程设计武汉理工大学课程设计2.0mm150mm。二次衬砌：依据《大路隧道设计标准C25号模筑防水混凝土，厚度为35cm。14.8+0.2+35=50cm。隧道深浅埋确实定及围岩压力的计算隧道的内轮廓尺寸为B=12.37m，H=13.69mIV级围岩需要预留变形量，查《大路隧道设计标准》8.4.1IV级围岩需预留变形量50-80mm50cm10cm，故隧道的近似开挖尺寸为：B12.3720.520.10.0813.65mtH13.6920.520.10.0814.97mt由于是IV级围岩H 2.5hp qH—深浅埋隧道的分界深度ph 0.452S-1qh—等效荷载高度q—跨度影响系数；1i(Bt

5)i—围岩压力增减率，当B515mi=0.1th 0.452S-10.4524-1[10.1(13.655)]6.714mqH 2.5h 2.56.71416.785mp q埋深H30H 16.785m故为深埋隧道。p又Ht14.971.11.7B 13.65t可用公式qhq计算均布垂直压力：qhq236.714154.42kN/m2由于该隧道围岩级别为IV围岩水平均不压力为：e〔1-1q25.74-m6 3衬砌构造内力计算根本资料公里等级 高速大路围岩级别 IV级围岩容重 r=23KN/m3武汉理工大学课程设计弹性抗力系数 K=0.18×106KN/m变形模量 E6GPa材料容重 22KN/m3h材料变形模量 E h衬砌厚度 d=0.5m图3 衬砌构造断面图荷载确实定围岩压力确实定隧道的内轮廓尺寸为B=12.37m，H=13.69mIV级围岩需要预留变形量，查《大路隧道设计标准》8.4.1IV级围岩需预留变形量50-80mm50cm10cm，故隧道的近似开挖尺寸为：B12.3720.520.10.0813.65mtH13.6920.520.10.0814.97mt由于是IV级围岩Hp2.5hqHp—深浅埋隧道的分界深度hq0.452S-1hq—等效荷载高度武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计—跨度影响系数；1i(Bt

5)i—围岩压力增减率，当B515mi=0.1th 0.452S-10.4524-1[10.1(13.655)]6.714mqH 2.5h 2.56.71416.785mp q埋深H30H 16.785m故为深埋隧道。p又Ht14.971.11.7B 13.65t可用公式qh计算均布垂直压力：qqh 236.714154.42kN/m2q考虑一衬后的围岩释放变形折减系数0.4q0.4154.4261.768KN/m3衬砌自重1 1g d2

dn

(0.50.5)2211KN/m22全部垂直荷载q=61.768+11=72.768KN/m2围岩水平均布压力e=0.4×72.768=29.1KN/m2衬砌几何要素衬砌的几何尺寸r1

9.47m，r2

5.6m内径r，r

所画圆曲线端点截面与竖直轴线的夹角：1 2164o，258o拱顶截面厚度d00.5m，拱底截面厚度dn0.5m半拱轴线长度S及分段轴长S643.14(9.470.5)583.14(5.60.5)16.77m180 2 180 28段，则SS2.1m8S 2.1 8.410-8E 25106h计算位移单位位移用辛普生法近似计算，按计算列表进展，单位位移的计算列表见表4-1截面αsin截面αsinαcosαxyd1/Iy/I(1+y)2/I积分00.0000.0000.2630.5080.7160.8750.9711.0001.0000.9650.0001.5752.9924.2025.2346.1686.4290.000115.25030.50045.75061.00076.25091.50010.22120.8620.84230.6981.78640.4853.21750.2385.6106-0.0268.1760.5000.5000.5000.5000.5000.5000.50096.00096.00096.00096.00096.00096.00096.000781∑768.003547.072系0.0000.00096.000数1/3121.2164.689143.121480.83268.061325.7252171.456306.220745.1324308.832993.5131707.1772538.5603021.3224194.4424784.8966417.3108083.102206.7 0.95 6.00 10.72 0.50 96.00 1029.31506.7 0.95 6.00 10.72 0.50 96.00 1029.315-0.28887200222.0 0.84 4.97 13.02 0.50 96.00 1249.920-0.53082000011036.28313190.907416273.95818869.798129568.59337430.737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： S18.410-8768.0006.451210-511 E Ih Sy

8.410-83547.07229.795410-512 E Ih 22 Eh

y28.410-829568.593248.376210-5I计算精度校核： 6.451229.79452248.37610-5314.418210-511 12 22 S12

8.410-837430.737314.418210-5ss E Ih闭合差△=0载位移——主动荷载在根本构造中引起的位移每一块上的作用力(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i34-2。各集中力均通过相应图形的形心武汉理工大学课程设计QGEaqaQGEaqagae0.0000.0000.0000.0000.0000.0000.0000.0000.000114.61023.1007.4010.7880.7880.1120.112-18.191-0.829103.62223.10020.2200.7120.7120.306-73.779-16.447-6.18788.04923.10031.3880.6050.6050.475-53.270-13.976-14.90974.22323.10032.3460.5100.5100.490-37.854-11.781-15.83367.89323.10093.9990.4670.4671.423-31.672-10.776-133.71319.42923.10084.7810.1340.1341.283-2.594-3.084-108.7744-2Mo计算表p集中力S-Qaq-Gag-Gagbh0.000.00001.570.22541.420.61421.210.95001.020.97090.932.84350.262.56760123456武汉理工大学课程设计武汉理工大学课程设计i1i1轴力：sini1i1轴力：sin(QG)cosENoipiiii式中ΔxΔyi iΔx=x-i iΔy=y-i ixi-1yi-1Mo 和No 的计算见表4-3及表4-4。ip ip84.1200.2090.2091.27375.9260.3930.3931.14970.4170.7870.4170.7852.5430.34462.29823.10023.100-6.327-4.816-107.085857.123-22.421-9.067-87.239∑〔Q+G〕i-∑Ei-1xyΔxΔy-Δx∑i-1〔Q+G〕-Δy∑i-1EMoip0.0001.5312.9644.2105.1885.8326.1526.1435.8060.0000.0000.1941.5310.7651.4331.6761.2462.8690.9784.2520.6445.7640.3207.3098.817-0.009-0.3370.0000.1940.5710.9111.1931.3831.5121.5451.5080.0000.0000.0000.0000.0000.000137.7107.401264.43127.621375.58159.009472.90491.356563.896185.354606.425270.135659.870354.2550.0000.0000.0000.0000.000-18.908-197.338-4.226-316.885-329.481-25.163-753.684-367.318-70.398-1256.869-304.55-126.34-1863.92055-180.44-280.25-2439.077695.458-417.359-2969.207222.376-534.216-3399.774外荷载在根本构造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a、aaq e g内力按下式计算之：弯矩：MoMoip i1,p

xi

(QG)yi

EQaq

Gag

Eae武汉理工大学课程设计表4-3载位移No 计算表截α截αsinαcosα∑i〔Q+G〕∑Eisinα*∑icosα*∑iENop691.5001.0007106.750122.0000.95880.848-0.026-0.288-0.530面〔Q+G〕00.0000.0001.0000.0000.0000.0000.0000.000115.2500.2630.965137.7107.40136.2227.14029.082230.5000.5080.862264.43127.621134.20923.799110.410345.7500.7160.698375.58159.009269.02941.176227.853461.0000.8750.485472.90491.356413.61144.290369.321576.2500.9710.238563.896185.354547.73644.056503.680606.425270.135606.218-7.071613.289659.870354.255631.872-102.095733.967740.093430.181627.634-227.961855.595主动荷载位移4-4截Mp01/I96.00截Mp01/I96.00096.00096.00032.676-72353.692 -121264.788-193618.480443.869-120659.391255.2526466.764278.309489.8171∑面00.0001面00.0001-18.9082-316.885y/I1+yMp0/IyMp0/IMp0(1+y)/I积分系数1/30.0001.0000.0000.0000.000118.6241.194-1815.183-352.146-2167.329473.4401.765-30420.982-23272.051-53693.0342-753.6896.00160.89406-1256.896.00275.426904-1863.996.00408.192502-2439.096.00553.347904-2969.296.00701.660704-3399.796.00846.437402-346171.792-466831.183-178936.78-760839.215-939776.002-234151.56 -1349649.59-1583801.150 44-285043.89 -2083385.83-2368429.725 28-326378.29 -2877677.39-3204055.681 46-1083146.7-6059744.06-7142890.869540武汉理工大学课程设计武汉理工大学课程设计△=△S/E×∑M0/I=8.4×10-5×〔-1083146.795〕=-9098.426×10-51p h p△=△S/E×∑M0y/I=8.4×10-8×(-6059744.064)=-50901.850×10-52p h p计算精度校核△=△△Sp 1p+ 2p△=△S/E×∑M0(1+y)/ISp h p因此，△=8.4×10-5×〔-7142890.860〕=-60000.276×10-5Sp△△=-〔9098.426+50901.850〕×10-6=-60000.276×10-51p+ 2p闭合差△=0.000。载位移——单位弹性抗力及相应的摩擦力引起的位移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4503α=45.75°=α3 b5α=76.25°=α;5 h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：σ=σ〔cos2α-cos2α〕/〔cos2α-cos2α〕i h b i b h计算得，σ0， σ=0.5847σ， σσ。3 4 h 5 h边墙截面弹性抗力计算公式为：σ=σ[1-(yˊ/yˊ)2]h i h式中yˊ——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；iyˊ——墙底外边缘cˊ到最大抗力截面的垂直距离。(yˊyˊ3h i hyˊ=2.47m; yˊ=4.93m; yˊ=6.87m;6 7 8则有：σ6=σh[1-〔2.47/6.87〕2]=0.8707σhσ7=σh[1-〔4.93/6.87〕2]=0.4850σhσ8=0;4武汉理工大学课程设计4各楔块上抗力集中力R，i按下式近似计算：i

i1

S /21 i外式中，S ——楔块i外缘长度，由图3量得。i外R的方向垂直于衬砌外缘，并通过楔块上抗力图形的形心。i抗力集中力与摩擦力之合力Ri按近似计算:

RR” 12i i式中μ——围岩与衬砌间的摩擦系数。取μ=0.2,RR” 12则 i i

=1.0198R”i其作用方向与抗力集中力的夹角为β=arctanμ=11.301位移方向相反，其方向朝上。Ri

的作用点即为R，与衬砌外缘的交点。i武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计R的方向线延长，使之交于竖直轴。量取夹角ψ(自竖直轴反时针方向量度)。将Ri k i分解为水平与竖向两个分力：R=R

sinψ

R=R

cosψH i k V i k4-53。截面截面(σi-1+σ△S外(σ)hR(σ)hψsinψkkcosψk)/2i0.00002.16702.16702.16702.16702.1670330.00000.00040.58470.29251.00000.79260.87070.93570.48500.67880.00000.2430.00000.0000.0001.0000.646164.6760.9040.4281.751079.9260.9850.1752.067095.1760.996-0.0901.4980110.4260.937-0.3490.5359125.6760.812-0.5834-5RR(σ)H hR(σ)V hvhR(σ)i h0.0000.5841.7242.0591.4040.4350.0000.2760.306-0.186-0.523-0.3130.2760.5830.396-0.127-0.4390.5842.3084.3675.7706.2060.0000.6341.7172.0271.4690.525计算单位抗力图及其相应的摩擦力在根本构造中产生的内力Mo弯矩 i

iKiNo sinR cosR轴力 i i V i H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4-7武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计截面号R=0.6461σ4R=1.7510截面号R=0.6461σ4R=1.7510σR=2.0670σR=1.4980σR=0.5359σMo σh5h6h7h8hσ( h)r4i-Rr44ir5i-Rr55ir6i-R6r6ir7i-R7r7ir8i-R8r8i40.82362.56234.05365.36547.2631-0.532-0.5325-1.6560.7563-1.324-2.9806-2.6197-3.4678-4.6932.463-4.310.828-1.71-8.64443013.684-6.452.372-4.900.785-1.17-15.9921953695.364-9.393.872-8.002.456-3.680.936-0.50-26.278475305234-7N0计算表σ面αsinαcosαΣR(σV)ΣRH(σh)sinαΣR(Vσ)cosαΣR(σh)Noσ(σh)hh号461.00000.87440.48530.27630.58400.24160.2834-0.0418576.00000.97010.24260.58262.30800.56520.55990.0054-0.0256 91.5000 0.9997 0.3962 4.36660.3960-0.11080.5068-0.291 -0.1267 107.0000 0.9566 5.7704-0.1211-1.68191.5607-0.529 -0.4398 122.0000 0.8486 6.2057-0.3727-3.28292.9102截H45截H45602截面号M0截面号M0σ〔σ〕hy/I(1+y)M01/IM0y/IM01/I(1+y)/Iσσσ〔σ〕 〔σ〕h h〔σ〕h积分系数1/34-0.53296.0005-2.980275.424408.1996.00023.869-51.084-146.561-197.64525.252-286.059-1216.324-1502.3834武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计6.764-829.808-4783.016-5612.82428.309-1535.889-11225.813-12761.70349.817-2522.188-22238.134-24760.3221Σ-3857.256-27288.612-31145.8686-8.64496.0007-15.999-26.27396.000896.000553.3446-8.64496.0007-15.999-26.27396.000896.000553.344701.664846.4321σ h σ△=△S/E×∑M0y/I=8.4×10-5×(-27288.612)=-229.2243×10-52σ h σ校核为：△+△=-(32.4010+229.2243)×10-11=-261.6253×10-61σ 2σ△S=△S/E×∑M0(1+y)/I=8.4×10-11×(-31145.868)=-261.6253×10-6σ h σ闭合差△=0。墙底〔弹性地基上的刚性梁〕位移单位弯矩作用下的转角：β=1/(KI)=-96.000/0.18×106=-533.33×10-61 8主动荷载作用下的转角：β=βM0=-3399.774×533.33×10-6p 18p

=-1813201.467×10-6单位抗力及相应摩擦力作用下的转角：β=βM0=533.33×10-6×(-26.273)=-14012.179×10-6σ 18σ解力法方程衬砌矢高 f=y=11.776m8计算力法方程的系数：a=δ+β=〔64.512+533.33〕×10-6=597.842×10-611 11 1a=δ+fβ=〔297.954+11.776×533.33〕×10-6=6578.448×10-612 12 1a=δ+f2β=〔2483.762+8.817×11.7762×533.33〕×10-6=76442.860×10-622 22 1a=△+β+(△+β)×σ10 1p p 1σ σ h=-(90984.26+1813201.467+324.010σ+14012.179σ)×10-6h h=-(1904185.727+14336.189σ)×10-6ha=△+fβ+(△+fβ)×σ20 2p p 2σ σ h=-(509018.50+11.776×1813201.467+2292.243σ+11.776×14012.179σ)h h=-(21861278.98+167299.163σ)×10-6h以上将单位抗力图及相应摩擦力产生的位移乘以σ倍，即被动荷载的载位移。h求解方程：武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计X=(aa-aa)/(aa-a2)1 1220 2210 1122 12=〔594.5350-1.778σ〕h其中：X

=594.5350,

=-1.7781p 1σX=(aa-aa)/(aa-a2)2 1210 1120 1122 12=〔243.8255+2.341σ〕h其中：X

=243.8255,

=2.3412p 2σ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：

M Xp 1p

yX2p

MopN Xp 2p和

cosNopM X 1

yX2

Mo4-94-10

N X

cosNo6659-1816.51594.534.251036.74-185.2315260-2345.27594.535.761405.41-345.3235408-2812.86594.537.301782.12-436.2025917-3169.32594.538.812149.80-424.9715796截面MopX截面MopX1pyX*y2p[M]pMo(σ)σ hX(σ)X*y(σ2σ h1σ h)[Mo](σσ)h00.0001-18.8092-315.5453-747.5654-1240.4656780.000-1.7780.000-1.7780.000-1.7780.454-1.3240.000-1.7781.7910.0130.000-1.7783.9242.146-0.532-1.7786.7164.406-2.980-1.7789.9545.196-8.644-1.77813.4943.072-15.999-1.77817.110-0.666-26.273-1.77820.641-7.410594.530.0050594.530.1954594.530.7655594.531.6756594.532.860.000594.53547.302623.028186.527465.517408.652255.621699.53553.604武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计4-10主、被动荷载作用下衬砌轴力计算表1.000243.826243.8260.0002.3412.3410.946230.644269.1440.0002.2142.2140.827201.721330.1350.0001.9371.9370.650158.591410.0650.0001.5231.5230.428104.293497.336-0.0421.0010.9600.17542.650569.2970.0050.4090.415-0.090-21.997603.4200.507-0.2110.296-0.349-85.094642.1761.561-0.8170.744截面0截面01NopcosaXcosφ2p［N］pNo(σ)Xcosφ(σ2σ)h［N］(σ)σ hσh23456780.00038.500128.414251.474393.043526.647625.417727.270822.162-0.583-142.199679.9632.910-1.3651.545首先求出最大抗力方向内的位移。由式：s (y y)M h php E I s(yh

y)Mh E I并考虑接缝5的径向位移与水平方向有肯定的偏离，因此将其修正如下

s(y5

y)Mi

sinhp 5p E I 5

s(y5

y)Mi

sinh 5 E I 53-11。4-11最大抗力位移修正计算表武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计546.45591546.45591146.-158-12.-1194.00-5480.50320.75843890329.-30029.2456.23620286.-27412.0yiy-yiM/Ip(Y5-Yi)M/Iσ(Y-Y)5 i积分系数1/30.0005.616-439691.51-2314.9294078292.6491078292.649182044.791261302.619333662.07947058.9665-24392.38490.2185.398442877.783-1941.05320.8424.774292658.7048.08841.7863.830128925.7641082.11723.2042.41217026.2251399.54045.6160.0000.0000.0001Σ207860.245-1782.359截面号截面号pM/Iσ5-234.140-174.3341.694282.537580.240684.266δ=8.4×10-8×207860.245×0.97=1693.645×10-5hpδ=8.4×10-8×(-1782.359)×0.97=-14.972×10-5hσ则可得最大抗力σ=δ/(1/K-δ)=1693.645×10-5/[1/(0.18×106)+14.972×10-5]h hp hσ=109.074计算衬砌总内力按下式进展计算：M=MσMp h σN=N+σNp h σ4-124-12衬砌总内力计算表截面Mp截面MpMσ[M]M/IMy/I号02805.1200.00011160.697 1351.0512pσ1/3243.826202.340446.1650.06551269.144191.401460.5440.02634330.135167.399497.534-0.02502积分系NN[N]e数武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计武汉理工大学课程设计000322000322-243247.4.32.2766026-386399.12.6.63227442-383393.9.69.6353316-240234.-5.9.365407587.46-8.0-0.69897272Σ410.065131.608541.6730.00344497