隧道内力计算课程设计曲墙式衬砌计算

1、3 拱形曲墙式衬砌结构计算3.1 基本资料：公路等级 围岩级别 围岩容重 弹性抗力系数 变形模量 衬砌材料 材料容重 变形模量 二衬厚度山岭重丘高速公路V级T s=20KN/m3K=0.18 X 106 KN/mE=1.5GPaC25 喷射混凝土丫 h=22 KN/m5Eh=25GPad=0.45m图2衬砌结构断面(单位：cm)3.2 荷载确定：3.2.1 围岩竖向压力根据公路隧道设计规范的有关计算公式及已知的围岩参数，代入公式S-1q=0.45 X 2 X 丫 X 其中：S围岩的级别，取S=5;容一一围岩容重，取丫 =20 KN/mf;宽度影响系数，由式=1+i (B-5)计算，其中，B为隧

2、道宽度，B=11.93+2X 0.45+2 X0.10=13.03m,式中0.10为一侧平均超挖量;B5 时，取 i =0.1 ,=1+0.1*(13.03-5)=1.803所以围岩竖向荷载(考虑一衬后围岩释放变形取折减系数0.4)q=0.45X16X 20X1.803*0.4 =259.632*0.4 kN/m3=103.85 kN / m33.2.2 计算衬砌自重g=1/2*(d 0+dn ) * 丫 h=1/2 X (0.45+0.45) X 22=9.9 kN/m3根据我国复合式衬砌围岩压力现场量测数据和模型实验，并参考国内外有关资料，建议V级围岩衬砌承受80%-60%J围岩压力，为安

3、全储备这里取：72.70 kN / m31)全部垂直荷载q= 72.70+g=82.60 kN /m32)围岩水平均布压力e=0.4 X q=0.4 乂 82.60=33.04 kN / m33.3 衬砌几何要素3.3.1 衬砌几何尺寸内轮廓线半径： r 1 =7.000 m , r 2 = 5.900 m内径n, r2所画圆曲线的终点截面与竖直轴的夹角：a 1=70.3432 , a 2 =108.7493 拱顶截面厚度d0 =0.45 m , 拱底截面厚度dn=0.45m。3.3.2 半拱轴线长度S及分段轴长 SS =12.363 m将半拱轴长度等分为 8 段，则 S=S/8=12.363

4、/8=1.545 m S/Eh =1.545/0.25 X 108 =6.18 X 10-8 m3.3.3 各分块截面中心几何要素各分块截面与竖直轴的夹角及截面中心点的坐标可以由图 3 直接量得，具体数值见表 2-1 。闭合差 =0.0001 X 10-603.4.1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3-1表表3-1单位位移计算表截cos axyd1/Iy/Iy2/I(1+y) 2/i积分系数1/3面asin a00.0000.0001.0000.0000.0000.45131.6870.0000.000131.6871114.4720.2500.9681.531

5、0.1940.45131.68725.5474.956187.7384228.9430.4840.8752.9640.7650.45131.687100.74177.067410.2352343.4150.6870.724.2101.6760.45131.687220.708369.906943.0094457.8870.8470.5325.1882.8690.45131.687377.8111083.9391971.2482571.9160.9510.3105.8324.2520.45131.687559.9342380.8403632.3964684.1940.9950.1016.1525

6、.7640.45131.687759.0454375.1376024.9152796.4720.994-0.1136.1437.3090.45131.687962.5027034.9289091.61948108.7490.947-0.3215.8068.8170.45E131.6871053.4981161.086 10237.299 12691.1593570.348 20157.368 28351.5631注：1.I 截面惯性矩，I=bd3/12,b取单位长度2 .不考虑轴力的影响。单位位移值计算如下：5 11=AS/EhXE 1/I=6.18 X10-8X 1053.498=65.10

7、62 X10-66 12=AS/ E hXEy/I=6.18 乂 10-8 X 3570.348=220.6475 乂 10-67 22=AS/ E hXEy2/I=6.18 乂 10-8 X 20157.368=1245.7253 乂 10-6计算精度校核：8 11+2 6 12+ 6 22= (65.1062+2*220.6475+1245.7253 ) X 10-6_ _ _-6=1752.1265X 105s=AS/ EhXE(1+y) 2/1=6.18 X 10-8 X 28351.563=1752.1266 X 10-63.4.2载位移一一主动荷载在基本结构中引起的位移1）每一块上

8、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,Q =q*b iE =e*h iG =（ d i-i +di）/2* AS* r h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3-2。各集中力均通过相应图形的形心。31,086 ,U34的泊汶1 mmmooRsnLnI prug/2口修匚:4 口1匚4一厂AC日Q.-T7UGTQ5TQ3 T Q5 T 31图3衬砌结构计算图示（单位：cmj）表3-2载位移Mp计算表1投影长度集中力S面bhQGEaqagae-Qaq-Gag00.0000.0000

9、.0000.0000.0000.0000.0000.0000.0000.00011.5310.194126.461 15.2966.4100.7650.7590.097-96.742-11.60921.4340.571118.448 15.296 18.8660.7170.6990.285-84.928-10.69231.2450.911102.837 15.296 30.0990.6230.5940.455-64.067-9.08640.9781.19380.78315.29639.4170.4890.4510.597-39.503-6.89850.6441.38353.19415.2964

10、5.6940.3220.2790.692-17.129-4.26760.3111.51225.68915.29649.956-0.1550.1190.7563.982-1.82070.0091.5450.74315.29651.047-0.004-0.0460.7730.0030.70480.0001.5080.00015.29649.8240.000-0.2090.7540.0003.197续表3-2-Gae22 i-1yA xAy-A xE i-1(Q+G-yi-1 EMip(Q+GI2i-1 Ex0.0000.0000.0000.0000.0000.0000.0000.0000.0000

11、.000-0.6220.0000.0001.5310.1941.5310.1940.0000.000-108.973-5.377141.7566.4102.9640.7651.4330.571-203.136-3.660-416.766-13.695275.50025.2764.2101.6761.2460.911-343.273-23.026-869.913-23.532393.63355.3755.1882.8690.9781.193-384.973-66.062-1390.881-31.620489.71194.7925.8324.2520.6441.383-315.374-131.09

12、7-1890.368-37.767558.201140.486 6.152 5.7640.3201.512-178.624-212.415-2317.013-39.459599.185190.443 6.143 7.309-0.0091.5455.393-294.234-2644.606-37.568615.224241.489 5.806 8.817-0.3371.508207.330-364.166-2835.8132）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：Mp MOi,p 为（Q G）yiE Q?aq Gag

13、 Eaei 1i 1轴力：Nip sin i （Q G） cos i E式中Axi、Ayi 相邻两接缝中心点的坐标增值。A Xi =xi- x i-iA yi =yi - y i-iMip和Nip的计算见表3-2及表3-3。表3-3载位移Noip计算表EiEsin a * Ei (Q+GCOS a *EiE面asin aCOS aEi (Q+G00.0000.0001.0000.0000.0000.0000.0000.000114.4720.2500.968141.7566.41035.4256.20629.219228.9430.4840.875275.50025.276133.32722.

14、119111.208343.4150.6870.726393.63355.375270.53540.224230.311457.8870.8470.532489.71194.792414.78450.391364.393571.9160.9510.310558.201140.486530.62743.609487.018684.1940.9950.101599.185190.443596.11119.266576.844796.4720.994-0.113615.224241.489611.304-27.218638.5228108.7490.947-0.321630.519291.31459

15、7.060-93.636690.6963）主动荷载位移计算过程见表3-4表3-4主动荷载位移计算表数1/3面Mp01/Iy/i1+yMp0/IyMp0/IMp0(1+y)/I00.000131.6870.0001.0000.0000.0000.00011-108.973131.68725.5471.194-14350.405-2783.979-17134.38442-416.766131.687100.7411.765-54882.723-41985.283-96868.00623-869.913131.687220.7082.676-114556.441-191996.594-306553.

16、03544-1390.881131.687377.8113.869-183161.260-525489.656-708650.91625-1890.368131.687559.9345.252-248937.362-1058481.662-1307419.02446-2317.013131.687759.0456.764-305121599-2063838.62627-2644.606131.687962.5028.309-348260.938-2545439.197-2893700.13548-2835.813131.687 1161.086 9.817-37344

17、0.375-3292623.783-3666064.1581E-1454730.326-7713271.529 -9168001.855产S/EhX!2M/I=6.18 X10-8X (-1454730.326 ) = - 89902.334 X1Q-6 2p = AS/ E hX 12 Moy/I=6.18 X 10-8X (-7713271.529)= -476680.181 X 10-6 计算精度校核 sp=Z 1p+2 2p sp=AS/ EhXEMP0(1+y)/I因止匕， sp=6.18 X10-8X (-9168001.855 ) = -566582.515.515 X10-6

18、ip+A2p= - (89902.334+476680.181 ) X 10-6= -566582.515 X10-6 闭合差 =0.000。3.4.3 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，a 3=43.4150 =ab;最大抗力位于第 5 截面 ,a 5=71.9159 =a 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：(T i = (T h(COS2 a b-COS2a i ) /(COS2 a b-COS2h)计算得，(T3=0,(T 4=0.5682 (T h

19、,(T5=(Th o边墙截面弹性抗力计算公式为：(7=(7h1-(y i / y h )2式中yi 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；/ / 一 /一/ 4 ._. .一 ,一yh 墙底外边缘C 到最大抗力截面的垂直距离。 (y i 和 yh 在图 3 中可量得 )/ / _ .一 / y6 =1.559m; y 7 =3.152m; y 8 =4.707m;则有：(T6= h 1- (1.559/4.707 ) 2= 0.8903 口(T7=(Th 1-(3.152/4.707 ) 2= 0.5516 口(T 8=0;按比例将所求得的抗力绘在图 4 上。2)各楔块上抗力

20、集中力Ri按下式近似计算：R i 11Si#/2式中，外楔块i外缘长度，由图3量得Ri， 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Ri按近似计算:式中 围岩与衬砌间的摩擦系数。取仙=0.2,则Ri Ri。12 =1.0198 R其作用方向与抗力集中力的夹角为B=arctg n =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。R的作用点即为R；与衬砌外缘的交点。将R的方向线延长，使之交于竖直轴。量取夹角巾k（自竖直轴反时针方向量度）将R分解为水平与竖向两个分力：H Ri Sin 巾 kRv= Ri coS 巾 k以上计算例入表3-5中，并

21、参见图3。表3-5弹性抗力及摩擦力计算表截回(T i ( (Th)(T i-1 + (Ti)/2 $外(Th)R( b h)小ksin。k30.00000.0000.00000.00000.0000.00040.56820.2841.59870.463264.4390.90251.00000.7841.59871.278477.1090.97560.89030.9451.59871.540989.2121.00070.55160.7211.59871.1754101.0800.98180.00000.2761.59870.4497111.8880.928续表3-5截回cos 4 kRH( bh

22、)R/( (Th)vhR( (Th)31.0000.0000.0000.00040.4310.4180.2000.2000.4180.45450.2231.2460.2850.4851.6641.25460.0141.5410.0210.5063.2051.5117-0.1921.153-0.2260.2804.3581.1538-0.3730.417-0.1680.1134.7760.441rr九 /、.4）巾舁串仪阮力囹及贝和丹的摩撩力仕是不结悯中厂生的内力弯矩MoRKi轴力N o siniRv COS iRh式中r Ki -力R至接缝中心点K的力臂，由图3量得，计算见表3-6和表3-7。

23、表3-6Mr 0计算表截面号R4=0.4632 hR5=1.2784 6 hr 4i-R4r 4ir 5i-R5r 5iR=1.5409 h R=1.1754 6 hr 6i-R6r6ir 7i-R7r7iR8=0.4497(Tr 8i-R8r8ih M(h)40.5455-0.253-0.25352.0714-0.950.72090-0.920-1.88063.5733-1.652.27053-2.9020.8280-1.276-5.83374.9634-2.293.76994-4.8192.3729-3.650.879-1.03-11.80886.1756-2.865.15815-6.59

24、53.8727-5.962.423-2.841.052-0.473-18.745表3-7nJ计算表截面号asin aCOS a2 R(b h)2 RH()sin 呆 2R/( (T h)COs a 2RH( b h)No(r ( b h)457.88670.84670.53200.19990.41780.16920.2223-0.0531571.91590.95040.31100.48501.66400.46100.5175-0.0565684.19370.99480.10190.50623.20480.50360.32660.1770796.47150.9937-0.11190.28044.

25、35830.2786-0.48750.76618108.7493 0.9472-0.32050.11274.77550.1068-1.53061.63745）单位抗力及相应摩擦力产生的载位移计算过程见表3-8。表3-8单位抗力及相应摩擦力产生的载位移计算表截面号M，0(CT h)1/Iy/I(1+y)M 01/I(CT h)Mt 0y/I(CT h)M(1+y)/I(CT h)积分系数1/34-0.253131.6872377.8113.869-33.271-95.455-128.72625-1.880131.6872559.9345.252-247.564-1052.641-1300.205

26、46-5.833131.6872759.0456.764-768.160-4427.678-5195.83827-11.808131.6872962.5028.309-1554.912-11364.855-12919.76348-18.745131.68721161.086 9.817-2468.447-21764.304-24232.74412-3760.404-26826.852-30587.2483二 S/Eh XI2M01/1=6.18 X 108X(-3760.404)= -232.3930 X 10-6 2。= S/ EhXEMt0y/I=6.18 X10-8X (-26826.8

27、52)= - 1657.8995乂 10-6校核为： +2。= -(232.3930+ 1657.8995)乂 10-6=-1890.2925 乂 10-6 S. = S/ EhXE ML0(1+y)/I=6.18 X10-8X (-30587.248)=-1890.2919 乂 10-6 闭合差 =0.0006 X 10-60o综上计算得结构抗力图如图4i 01图4结构抗力图3.4.4墙底（弹性地基上的刚性梁）位移1）单位弯矩作用下的转角：0i=1/（KI 8）= 131.6872 /0.18 X 106=731.5956 X 10-62）主动荷载作用下的转角：B p=B iMp0=-283

28、5.813 X 731.5956 X 10-6 = -2074668.313 乂 10-63）单位抗力及相应摩擦力作用下的转角：B 尸 B0=731.5956 X 10-6x （-18.745） = - 13713.7595 X 10-63.5 解力法方程衬砌矢高 f=y8=8.817m计算力法方程的系数：aii=i+B 产(65.1062+731.5956) X 10-6=796.7018 X 10-6ai2= 5 12+f B 1= (220.6475+8.817*731.5956 ) 乂 10-6=6671.1259 X 10-6a22= 6 22+f2 B 尸(1245.7253+8.

29、817*8.817*731.5956 ) X 10-6=58119.5934 X 10-6 a10=A 1p+ 0 p+( 1 0 + 0 J x h h-6=-(89902.334+2074668.313+232.3930 6 h+13713.7595 6 h) X10-6=-(2164570.647+13946.1525 6 h) X 10a20=A2p+f B p+( A2 ,r+f B 7)X (T h=-(476680.181+8.817*2074668.313+1657.8995o h+8.817*13713.7595 o h)=-(18769030.7+122572.117 6

30、h) X10-6以上将单位抗力图及相应摩擦力产生的位移乘以6h倍，即被动荷载的载位移。求解方程：2X1 = (a 12a20 - a 22a10)/( a 11a22 - a 12 )=(329.6545-3.972其中： X 1p=329.65446,X 1X2 = (a 12a10 - a 11a20)/( a=(285.0995+2.2059其中： X 2p=285.0995,Xh)(T= -3.972211a22 - a 12 )h)2 17= 2.56483.6计算主动荷载和被动荷载(Th =1 )分别产生的衬砌内力计算公式为：MpX1p yX2pM opNpX2p ?cosN op

31、X 1 yX2X2 ?cosNo计算过程列入表3-9 和表 3-10 中。表3-9主、被动荷载作用下衬砌弯矩计算表0 截面X1py3yMpML(川X,( b h)X2y( b h)M(门)00.000329.6540.0000.000329.6540.000-3.9720.000-3.9721-108.973329.6540.19455.309275.9900.000-3.9720.498-3.4742-416.766329.6540.765218.101130.9900.000-3.9721.962-2.0103-869.913329.6541.676477.827-62.4320.000-

32、3.9724.2990.3274-1390.881329.6542.869817.950-243.276-0.253-3.9727.3593.1345-1890.368329.6544.2521212.243-348.471-1.880-3.97210.9065.0546-2317.013329.6545.7641643.314-344.045-5.833-3.97214.7844.9797-2644.606329.6547.3092083.792-231.160-11.808-3.97218.7462.9678-2835.813329.6548.8172513.7227.564-18.745

33、-3.97222.614-0.102表3-10主、被动荷载作用下衬砌轴力计算表截面MpcosaX2pcos 4NM，( (Th)X2cos()(b h)：Nd ( b h)00.0001.000285.099285.0990.0002.5652.565129.2190.968276.054305.2720.0002.4832.4832111.2080.875249.490360.6980.0002.2442.2443230.3110.726207.095437.4050.0001.8631.8634364.3930.532151.558515.951-0.0531.3631.3105487.0

34、180.31088.498575.516-0.0570.7960.7406576.8440.10128.842605.6870.1770.2590.4367638.522-0.113-32.133606.3880.766-0.2890.4778690.696-0.321-91.639599.0581.637-0.8240.8133.7计算最大抗力值首先求出最大抗力方向内的位移 由式：hps (yhy)MpE I三(yh y)ME I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s (y5yi)Mp .hpsin 5E I(y5yi)M sin计算过程列入表3-11。表3-11最

35、大抗力位移修正计算表截面号Mp/IM /Iyiy5-y iM/I(Y5-Yi)M,/I(Ys-Yi)1/3043411.273-523.0220.0004.252184584.732-2223.8904136344.399-457.4970.1944.058147485.573-1856.5232217249.693-264.6370.7653.48760149.681-922.79043-8221.46243.0601.6762.576-21178.487110.92324-32036.321412.7342.8691.383-44306.233570.81245-45889.112665.

36、5614.2520.0000.0000.00012351442.298-4598.224位移值为：6 hp=6.18 X 10-8 X 351442.298 乂 0.9506=2064.6209 乂 10-56 h 产6.18 X 10-8 X (-4598.224) 乂 0.9506= -27.0132 乂 10-5则可得最大抗力h = 6hp/(1/K- 。)=2064.6209 X 10-5/1/(0.18 X 106)+27.0132 X 10-5=74.889883.8计算衬砌总内力按下式进行计算：M=ip+(rh M.N=N+(rhN。计算过程列入表3-12表3-12衬砌总内力计算

37、表截分系面MMMM/IMy/INpN,Ne数1/号3329.65-297.432.214242.21285.0221.6506.70.06300.0001440449983836275.99-260.115.812082.50305.2214.6519.90.0301404.0064076437250224130.99-150.4-19.5-2568.9-1965.2360.6193.9554.6-0.032209808585498959352-62.43-37.9-4996.6-8374.4437.4161.0598.4-0.06324.4884244885205303534-243.2234.72-8.55-1126.6-3232.4515.9113.2629.2-0.01427606989851570836-348.4378.5030