高速公路隧道工程课程设计计算书

1、高速公路隧道工程课程设计计算书目录一基本资料1二荷载确定12.1 围岩竖向均布压力12.2 围岩水平均布力 2三衬砌几何要素23.1 衬砌几何尺寸23.2 半拱轴线长度S及分段轴长 S23.3 割分块接缝重心几何要素3四计算位移34.1 单位位移44.2 载位移一一主动荷载在基本结构中引起的位移 44.3 载位移一一单位弹性抗力及相应的摩擦力引起的位移 84.4 墙低（弹性地基上的刚性梁）位移 12五解力法方程13六计算主动荷载和被动荷载（% =1）分别产生的衬砌内力 13七 最大抗力值的求解 15八计算衬砌总内力16九衬砌截面强度检算（检算几个控制截面） 169.1 拱顶（截面0） 169.

2、2 截面（7） 189.3 墙低（截面8）偏心检查 18十内力图18基本资料高速公路隧道，结构断面如图1所示，围岩级别为V级，容重Y = 18kN/m3,围岩 的弹性抗力系数K =0.15Ml06kN/m3,衬砌材料C20混凝土，弹性模量Eh = 2.95父107kPa ,容重 Y=23kN /m3。图1衬砌结构断面二荷载确定2.1 围岩竖向均布压力：q=0.45M2s，8¥式中：s围岩级别，此处s=5；围岩容重，此处Y=18kN/itf；。 跨度影响系数，8=1+i(Bm-5),毛洞跨度 Bm=12.30m, Bm=515时，i=0.1 ,此 处：8=1+0.1 X ( 12.3-

3、5) =1.73所以，有：q =0.45父25r 18父1.73 =224.208kPa考虑到初期之处承担大部分围岩压力，而二次衬砌一般作为安全储备，故对围岩压力进行折减，对于本隧道按照 50%折减，即q = 50%xq =0.5x224.208 =112.104kPa2.2 围岩水平均布力：e =0.4 X q=0.4 X 112.104=44.8416kPa三衬砌几何要素3.1 衬砌几何尺寸内轮廓半径 r 1=5.7m , r2=8.2m ;内径一、r2所画圆曲线的终点截面与竖直轴的夹角中1=90。，匕=101.6。；截面厚度d=0.45m。此处墙底截面为自内轮廓半径r2的圆心向内轮廓墙底

4、做连线并延长至与外轮廓相交，具交点到内轮廓墙底间的连线。外轮廓线半径：R=5.7+0.45=6.15mR2=8.2+0.45=8.65m拱轴线半径：/=5.7+0.5 X0.45=5.925r2'=8.2+0.5X 0.45=8.425拱轴线各段圆弧中心角：。=90°8=11.6°3.2 半拱轴线长度S及分段轴长 S分段轴线长度：S1 =二1-二 r1 二里 二 5.925 = 9.30696m 180180S2 二二 r2 = 116 二 8.425 = 1.70571m180180半拱轴线长度为：S=Si+S2=9.30696+1.70571=11.01268m

5、将半拱轴线等分为8段，每段轴长为：11.01278=1.376585 m3.3 割分块接缝重心几何要素(1)与竖直轴夹角四S 180.2 1 - 11 ti1 =r =13.3118r1二?2 =* .1 =26.6236 ,二 3 =二21 =39.9354:4 =<3号=53.2473 , : 5 ',4F =66.5591：6 =： 5 . G =79.871 J：-2sL 180- =92.23822二, S 180. S =7 3-6 =7 0.32915 -9.3070 = 0.329125m,2 = =9.3617 2：二:8 =: 72 =92.23829.361

6、7 =101.6另一方面，/ =d+% =90口+ 11.6。=101.6。角度闭合差=0。各接缝中心点坐标可由图1中直接量出。四计算位移1h1 h2h3h4h5h5h7h3R6R7R8xR4八4 ，Ga17 G I'8G3图2衬砌结构计算图4.1单位位移用辛普生法近似计算，按计算列表进行。单位位移的计算见表1单位位移计算如下=sjMds 3 1=4.7469 10-8 1185.1852 = 5.5305 10为0 印 Eh I'12=6j曳“定竺£ 丫= 4.7469m 103540.2667= 1.6520m 10”0 Eh IEhI« O N、E =

7、4.7469 10, 18662.4491 = 8.7086 104EhI计算精度校核为：5 2、12 、22 =1.2566 10-S (y 1)、ss=x - =4.7469 1026928.1676Eh I3 =1.2566 10闭合差=0。4.2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：Qi =qbi式中：bi衬砌外缘相邻两截面之间的水平投影长度，由图 2量得:b1 =1.3171m,b2 =1.3367m,b3 =1.1960m,b4 = 0.9749m,b5 =0.7136m,b6 =0.4121m,b7 =0.0804m,b8 = 0.0101m.Bb

8、= 6.1409m =6.1506m(校核)h1 =0.1608m,h2 =0.4824m,h3 =0.783Sm,h4 = 1.0352m,h5 =1.2362m,h6 =1.3668m,h7 =1.4171m,h8 = 1.3970m.7单位位移计算表表1截回(°)sincosx (m)y(m)d(m)I1I_yI2 y T(1+y)2 I0001000.450.00759131.68720.00000.0000131.6872113.311810.23030.97311.36680.16080.450.00759131.687221.17533.4050177.4428226.

9、623630.44810.8942.65320.63320.450.00759131.687283.384452.7990351.2549339.935440.64190.76683.79891.38690.450.00759131.6872182.6370253.2993750.2606453.247260.80120.59844.74362.38190.450.00759131.6872313.6658747.12071506.1396566.559070.91750.39785.43713.56780.450.00759131.6872469.83371676.27282747.6276

10、679.870890.98440.17595.8294.88430.450.00759131.6872643.20003141.58184559.6690792.238280.9992-0.03915.91956.25110.450.00759131.6872823.19015145.84386923.91138101.60.9796-0.20115.74867.61790.450.00759131.68721003.18027642.12689780.1745E0.068341185.18523540.266718662.449126928.1676bd 3注：1. I 截面惯性矩，I =

11、晅,即单位长度。122.不考虑轴力的影响b图3 内力Mi0、NP计算图示自重力：Gi =diS h =14.3901式中：di接缝i的衬砌截面厚度，本设计为等厚度衬砌；作用在各楔体上的力均列入表2,各集中力均通过相应的图形的行心。(2)外荷载在基本结构中产生的内力。楔体上各集中力对下一接缝的力臂由图 2中量得，分别记为aq,ae,ag内力按照下式计算（见图3）。弯矩：MiP =Mi0p 跳工（Q+G） EQaq Qag QaeXi ±i 1轴力：NiP =sinaL (Q G)-cosai q E式中：Axi, Ayi -相邻两接缝中心点的坐标增值为二X- %y T'支Mi；

12、,、；的计算见表2及表3.载位移N；计算表表3截回sincosE (Q+G)SEsina(Q+G)cosa 2 EN0 p00.00001.00000.00000.00000.00000.00000.000010.23030.9731173.25277.210539.90017.016632.883520.44810.8940337.492328.8421151.230325.7849125.445430.64190.7668485.958863.9934311.936949.0702262.866840.80120.5984609.6391110.4135488.442866.0714422

13、.371450.91750.3978704.0266165.8467645.944465.9738579.970660.98440.1759764.6148227.1362752.686839.9532712.733670.9992-0.0391788.0181290.6812787.3877-11.3656798.753380.9796-0.2011803.5405353.3249787.1483-71.0536858.2019#载位移M 0的计算表p截面集中力力臂-Qaq-Qag-Eae£(q+g)i J_E E x y-xZ (Q+G)i _J_-yZ Ei_LMpQGEaqa

14、gae0000000000000000-1158.862614.24777.21050.65330.67340.3015-103.7849-9.5944-2.17400.00000.00000.68340.04020.00000.0000-115.55332149.849414.247721.63160.57290.63320.4422-85.8487-9.0216-9.5655173.11027.21051.28640.4724-222.6890-3.4063-446.08443134.076414.247735.15130.45230.55280.5628-60.6427-7.8761-1

15、9.7832337.207328.84211.14570.7538-386.3384-21.7412-942.46604109.290214.247746.42000.31160.44220.6533-34.0548-6.3003-30.3262485.531363.99340.94470.9950-458.6815-63.6735-1535.5022579.997414.247755.43320.15080.31160.7035-12.0636-4.4396-38.9972609.0692110.41350.69351.1859-422.3895-130.9393-2144.3315646.

16、198114.247761.28950.02010.16080.7236-0.9286-2.2910-44.3491703.3142165.84670.39201.3166-275.6992-218.3537-2685.953179.013214.247763.5450-0.18090.01010.70351.6305-0.1439-44.7039763.7600227.13620.09041.3668-69.0439-310.4497-3108.664081.132314.247762.6437-0.2211-0.06030.65330.25030.8591-40.9251787.02082

17、90.6812-0.16081.3668126.5529-397.3030-3419.22987基本结构中，主动荷载产生弯矩的校核为:. 0M 8qB B言一/12.301212.3012=-112.104 (5.7486 ) = -1843.25872408e44.841620- 7.8893 =-1395.4942 ,M 8g =一，Gi(x8 - xi - agi) =190.1785M0d =M0a+M0e+M 8a = -1843.2587 -1395.4942 -1395.4942= -3428.9314pq e g另一方面,从表2中得到M0p =-3421.1228闭合差3428

18、.9314 -3421.93143428.9314100% =0.23%(3)主动荷载位移(计算过程见表 4) 主动荷载位移计算表截 面一 0Mp1I_yI(1+y)0MpI0MpyI0Mp(1+y)I00.000131.6870.0001.0000.0000.0000.0001-115.553131.68721.1751.161-15216.891-2446.876-17663.7672-446.084131.68783.3841.633-58743.619-37196.459-95940.0783-942.466131.687182.6372.387-124110.745-172129.1

19、92-296239.9364-1535.502131.687313.6663.382-202206.056-481634.605-683840.6615-2144.331131.687469.8344.568-282381.100-1007479.288-1289860.3876-2685.953131.687643.2005.884-353705.752-17276057577-3108.664131.687823.1907.251-409371.392-2559021.506-2968392.8978-3419.230131.6871003.1808.618-45

20、0268.942-3430103.772-3880372.7132-1896004.496-9417616.702-11313621.1980二1pS M1 M0)EhIds2ps M2 M0dsEhp =4.66639 10” (-1897334.6519)=-0.0885yMEhp =4.66639 10(-9423842.3315)=0.439811计算精度校核："p 2P = -0.0885 0.4398 = -0.5283spEh(1 y)MpI= 4.66639 10”(-11321176.9834)=-0.5283闭合差： o 04.3载位移一一单位弹性抗力及相应的摩擦

21、力引起的位移(1)各接缝处的抗力强度抗力上零点假定在接缝 3, 口3 =39.9354o = 0(b最大抗力值假定在接缝6,川=79.8709最大抗力值以上各截面抗力强度按下式计算： 22cos ： b -cos : i -i -22hcos ： b - cos 二 h查表 1,算得 二一3 二0 ,二4 = 0.4 1 * 7 二5 ,=0.7.7 165=用了 ;最大抗力值以下各截面抗力强度按下式计算：12二 i 二(1-与)二 hYh式中：y；一所考察截面外缘点到h点的垂直距离；yh 一墙角外缘点到h点的垂直距离。由图 2 中量得：y7 =1.4171m, y8 = 2.8241m;12

22、则：；二7 =(1-4)二 h =0.7482二 h,二 8 二0； yh按比例将所求得抗力绘于图2上。(2)各楔体上抗力集中力R；按下式计算：R =('+、§外2式中：§外一表示楔体i外缘长度，可以通过量取夹角，用弧长公式求得，R的方向垂直于衬砌外缘，并通过楔体上抗力图形的形心。(3)抗力集中力玉摩擦力的合力 R按下式计算：R =R;历谑式中：R 一围岩于衬砌间的摩擦系数，此处取 0 =0.2 0则：R =1.0198R;其作用 方向与抗力集中力的夹角a =arctanN =11.0399° ；由于摩擦力的方向与衬砌位移方向相 反，其方向向上。将Ri得方

23、向线延长，使之交于竖直轴，量取夹角，将 R分解为水平和竖直两个分力:Rh =Rsin¥k,RV =Rcos中k；以上计算结果列入表5中。弹性抗力及摩擦力计算表表5截面a(%)1，、2（5）S外R)中ksinW kcosW kRh9h)Rv（）300.00000.00000.000000.00001.00000.00000.000040.41270.20641.42890.300760.22620.86800.49660.26100.149350.77150.59211.42890.862871.95290.95080.30980.82030.267361.00000.88571.42

24、891.290784.85200.99600.08971.28540.115870.74820.87411.35521.208097.43170.9916-0.12931.1979-0.156280.00000.37411.41330.5392108.63130.9476-0.31950.5110-0.1723（4）计算单位抗力及其相应的摩擦力在基本结构中产生的内力弯矩：Mi0r =-2轴力：Ni0y =sin“i£ Rv-cos%E R式中：b 一力Rj至接缝中心点ki的力臂，由图2量得。计算见表6及表7.N0计算表表7截面asin久cos«工R。）sin«Z

25、RV )R RHG)cosaZ RH。)N0 INcr(ah)453.24730.80120.59840.14930.11960.26100.1562-0.0365566.55910.91750.39780.41660.38221.08130.4302-0.0479679.87090.98440.17590.53240.52412.36680.41630.1078792.23830.9992-0.03910.37620.37593.5646-0.13940.51528101.60000.9796-0.20110.20390.19974.0756-0.81961.0193M(0计算表截面R4=0

26、.3007R5=0.8628R6=1.2907R7=1.2080R8=0.5392M*G)4i-R4r4i(%)r5i-R5r5i5)r6i-RUi )r7i- R7r7i )r8i- R8i (J)40.4824-0.1451-0.145151.8593-0.55910.6533-0.5636-1.122763.1959-0.96102.0201-1.74290.6935-0.8951-3.598974.4321-1.33273.3567-2.89612.0603-2.65910.7638-0.9227-7.810685.5376-1.66514.6029-3.97123.4070-4.39

27、722.1407-2.58600.9749-0.5257-13.1453#17(5)单位抗力及相应摩擦力产生的载位移。计算见表 8.截面Mh)1I_yI(1 + y)一 0M仃I一 0MgyIMW(1+y)I4-0.1451131.6872313.66583.3819-19.1020-45.4990-64.60105-1.1227131.6872469.83374.5678-147.8492-527.4964-675.34566-3.5989131.6872643.20005.8843-473.9342-2314.8367-2788.77087-7.8106131.6872823.19017.

28、2511-1028.5520-6429.5815-7458.13358-13.1453131.68721003.18028.6179-1731.0632-13187.0665-14918.12972-3400.5006-22504.4801-25904.9807单位抗力及相应摩擦力产生的载位移计算表表80sM1M0EhIM0-上二Eh''：ic =ds=4.66639 10(-3400.5006)-1.5868 10”sM2M0-2 d0 EhI,.金EhyM0 一上= 4.66639 10上(-22504.4801)= -0.0011校核为：-：2? =-0.0012.S (

29、1 y)M°-o：s； = '、 =4.66639 10(-25904.9807) =-0.0012-EhI闭合差： 0 0；4.4墙低(弹性地基上的刚性梁)位移单位弯矩作用下的转角：二='"8726 =8.7791 10”KI8 0.15 10主动荷载作用下的转角：W =M；二=-3421.1228 8.7791 10“ = -3.0035单位抗力及相应摩擦力作用下的转角：一：_ =M：_可=J3.1453 8.7791 10%=-1.1540 10- ac 80 a五解力法方程衬砌矢高f 二 y =7.6179m计算力法方程的系数为：11 =、11T7

30、=5.5305 10” 8.7791 104 =9.3322 10 工:1212 fM =1.6520 104 7.6179 8.7791 10 =6.8531 10，_24_2_ _4_2：22 = 22 f 7=8.7086 107.61798.7791 10 -5.1818 102222a：1。= 1p *（七 噢）二h 二 一3.0920 -1.1699 102入：2。= 2p 1啜（2三f 2>入=-23.3198-8.8964： h以上将单位抗力及相应摩擦力产生的位移乘以仃卜，即为被动荷载的载位移。求解 方程为：X1 = - 22-210 -122。=294.21 -2.47

31、43； 12 - -11 - 22式中:X1p = 294.21, X1- = -2.4743;>11 > 2n -<-12 1nX2 :10=411.12 2.0441；-，2-111,22式中：X2p =411.12, X,- =2.0441 p2 i.，以上解的X1,X2值应带入原方程，校核计算。六 计算主动荷载和被动荷载（仃h=1）分别产生的衬砌内力计算公式为:Mp "p yX2P M0Np =X2pC0S：N；rrrMX可 yX2. MN二=X2二 cos 二，N 0计算过程列入表9、10。.主、被动荷载作用下衬砌弯矩计算表表9截面一 0M pX1pX2p

32、LyMpMo （西）X16 (j)X2Jy g)MJ 防）00.0000294.20730.0000294.20730-2.47430.0000-2.47431-115.6492294.207366.1080244.66610-2.47430.3287-2.14562-446.4538294.2073260.3209108.07440-2.47431.2943-1.18003-943.2406294.2073570.1817-78.85160-2.47432.83490.36064-1536.7437294.2073979.2457-263.2907-0.1451-2.47434.86882.

33、24945-2146.0126294.20731466.7924-385.0128-1.1227-2.47437.29283.69586-2687.9363294.20732008.0314-385.6977-3.5989-2.47439.98393.91067-3110.7260294.20732569.9496-246.5691-7.8106-2.474312.77772.49288-3421.1228294.20733131.86794.9524-13.1453-2.474315.5715-0.0480主、被动荷载作用下衬砌轴力计算表表10截回N0 pX2p cosa p pNpKl0

34、Nor (ah)XzJJcosaBh)Na(a)00.0000411.1196411.119602.04412.0441132.8835400.0605432.944001.98911.98912125.4454367.5409492.986301.82741.82743262.8668315.2465578.113301.56741.56744422.3714246.0140668.3854-0.03651.22321.18665579.9706163.5434743.5140-0.04790.81310.76526712.733672.3159785.04950.10780.35960.4

35、6737798.7533-16.0748782.67860.5152-0.07990.43538858.2019-82.6761775.52581.0193-0.41110.6083七最大抗力值的求解首先求出最大抗力方向内的位移。考虑到接缝5的径向位移与水平方向有一定的偏离，因此修正后有:二 , ASM p ,、.p =p =-Z(y5-yi)sm«5EhI,AS_ Mk, 、.5 = =1 (y5 -yi)sm«5EhI计算过程列入表11,位移值为：hp =4.66639 10$ (-212096.5914) = 9.8973 10工h二=4.66639 10$ (-20

36、38.576036尸-0.95128 10”最大抗力值为:9.8973 10'.hp-hp46 0.95128 100.15 10= 97.2277最大抗力位移修正计算表表11截回MpIMbI。)(Y5 -Yi)Mp7p(Y5-Yi)牛(Y5 - Yi) (h)038743.3464-325.83284.8843189234.1269-1591.4649132219.4048-282.54894.7235152188.3587-1334.6199214232.0210-155.38924.251160501.7445-660.57513-10383.750547.49033.4974-

37、36316.1291166.09254-34672.0249296.22052.5024-86763.2752741.26215-50701.2794486.69151.3165-66748.2343640.72936-50791.4607514.97940002212096.5914-2038.5760八计算衬砌总内力按下式计算衬砌总内力:=M p 二hM 三=Np 二hN士计算过程列入表12.21计算精度的校核为以下内容:根据拱顶切开点的相对转角和相对水平位移应为零的条件来检查:jSx. MEh=0式中：_2S % M = 4.66639 10(5320.483) = 一2.4827 1。" EhI'-a =M8二=0.283 8.7791 104 =2.4827 10”闭合差：=0f-oEhI式中:MyI= 4.