隧道工程电子版

1、二次衬砌结构计算二次衬砌结构计算一、基本资料：所设计的公路等级为高速公路，设计车速为100Km/h，围岩类别为W级，容重 7= 21.5KN/m3，围岩的弹性抗力系数为K = 1.5 x 1Q5kN/m，衬砌材料为C25混凝土，弹性模量 E. = 2.95 x107 KPa，容重y. = 23KN / m3。二、荷载确定：1、围岩竖向均布压力：q = 0. 4 5 齐s -围岩类别，此处s = 4 ；y围岩容重，此处7 = 21.5KN / m3 ； 跨度影响系数， =1 + iJm -5）， 毛洞跨度 lm=3.75 x 2 + 0.75 x 2 + 0.5 +1.0 + 0.1 x 2 =

2、 10.7，lm 在 5-15 之间，取 i=0.1，故有 1 + 0.1x （11.7 - 5） = 1.67则 q = 0.45 x 24-1 x 25 x1.67 = 150.3KPa考虑到初期支护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，对本隧道按照35%折减，即：q = 35%q = 35% x 150.3 = 52.605kP， 围岩水平均布力：e=0.2q=0.2x 52.605=10.521Kpa2、计算位移：（1）单位位移：（所以尺寸见图）半拱轴线长度S = 10.8348（m）一e S 10.8348将半拱轴线长度分为8段，则庭=-=一-一 =

3、1.35435m88S_ 1.35435耳2.95 x107=0.4591x 10-7 m / KPa计算衬砌几何要素，拱部各截面与垂直轴线之间的夹角和截面中心垂直作坐标见表-1单位位移计算表截面sincosX(m)Y(m)D(m)中中中中中中中0001000.35113.6250.2355670.9718581.34280.1530.35227.250.4578750.8890172.61670.60410.35340.8750.6544120.7561383.75651.33020.35454.50.8141170.5807014.70372.29410.35568.1250.9280.3

4、72585.40983.446 40.35681.750.9896520.1434895.83864.7280.35795.3750.995603-0.093685.9696.07320.3581090.945517-0.325576.08577.41310.35表一1续表1/1 (m 4 )21 (m 3)件 (m 2)(1+22 山(m 2)279.88340.00000.00000.0000279.883464.9896.552371.967279.8834171.233102.1299720.1713279.8834376.555495.631519.718279.8834647.79

5、01474.2013037.029279.8834970.3563327.0875533.46279.88341327.4316262.6249183.02279.88341689.68410331.59714014.00279.88342063.020515393.27719809.852518.95067310.35937393.097954189.2153注：1.截面惯性矩，1=, b取单位长度。JL2不考虑轴力影响。.单位位移移植用辛普森法近似计算，计算如下：5 =， d 笠- = 0.4591 x 10-7 x2518.9504 = 1.1565x 10-40 E s E I5性 m

6、 dq 2= 0.4591 x 10-7 x 7310.359 = 3.356x 10-40 E 2 s E IM 2AS 寸 y 25 =，2d 牝 一 匕=0.4591x10-7 X37393.0979 = 17.167 x10-4220 E s E I511 + 2512 +522 = 1.1565 x10-4 + 2 x 3.356 x10-4 +17.167 x10-4 = 25.0355x10- 45 =竺 +y = 0.4591 x 10-7 x 54189.2153 = 25.0355 x 10-4 h闭合差A = 0（2）载移一主动荷载在基本结构中引起的位移1）每一楔块上的作

7、用力竖向力：Q = qbi i横向力：Ei =eh自重力：G =幺x As x yi 2h式中：q衬砌外缘相邻两截面之间的水平投影长度;h衬砌外缘相邻两截面之间的竖直投影长度;id,接缝i截面厚度。作用在各楔块上的力均列入表一2中2）外荷载在基本结构中产生的内力按下式计算 弯矩：轴力:M 0 = M 0 -Ax （Q + G）-Ay e-Qa Ga Ea ipi1, pi*i1N。= sin 甲.（Q + G ）cos 甲.Eiiii-1截集中力力臂面QGEa qa gaeQ. a qG. a g0000000001201.4610.90251.60970.64010.65370.312112

8、8.95457.1262191.123210.90254.7460.55850.62740.4512106.74236.84023170.00410.90257.63930.44430.53660.543275.53275.85024142.10810.902510.1410.31530.44710.641544.8064.874505105.936110.902512.12330.18010.33130.675619.0793.61199664.332910.902513.48370.0070.18010.68110.45031.9635719.56410.902514.1528-0.141

9、40.0260.6638-2.76630.283M 0i, pN0的计算见表一3、一4 ip载位移计算表一 2817.508510.902514.0823-0.3088-0.11320.6191-5.4066-1.234载位移计算表M 0表一3i, pEaeZ(Q + G )i-1Z Ei-1AxA yAxE(Q + G )i-1A Zei-1M 0p0.00000.00000.00000.00000.00000.00000.00000.00000.50240.00000.00001.34280.1530.00000.0000-72.49382.1414212.36291.60971.2739

10、0.4511144.03210.2024-283.28454.1497414.38866.35571.13980.7261253.01045.3150-594.62416.5055595.095113.9950.94720.9639304.314517.7615-955.87088.1905748.105624.1360.70611.1523287.676938.28157-1307.933429.1837864.944236.25930.42881.2816205.366065.3320-1595.68609.3805940.179649.7530.13041.345272.456695.2

11、8731778.13568.71844970.64663.89580.11671.3399-87.9258123.01381827.50964载位移计算表No表一4截面sin中cosZ(Q + G )Z E .sin( Z (Q +(:os OZE G )N 0p0010000010.23550.9718212.36251.6097-0.9538-0.038850.063320.45780.8890414.38826.3557-0.29740.9973188.851330.65440.7561595.294713.995-0.99930.1416389.460840.81410.5807748

12、.305224.1360.56970.5430608.892650.9280.372865.143836.2593-0.93410.1306802.804760.98960.1439940.379249.743-0.86390.8665930.523870.9956-0.0936970.845763.8958-0.09340.4854966.622380.9455-0.3255999.256777.97810.2282-0.8463944.53863）主动荷载位移计算结果见表一5主动荷载位移计算表表一5M 0P1了yIM o pIM oyIPM o (1 + y )I截面00279.88340

13、0001-72.4938279.883464.989-20289.81122-4711.299568-25001.110792-283.285279.8834171.233-79286.62903-48507.65479-127794.28383-594.624279.8834376.555-166425.4148-223908.678-390334.09284-955.871279.8834647.79-267532.3695-619203.5455-886735.9155-1307.93279.8834970.356-366068.8526-12691618946

14、-1595.69279.88341327.431-446606063-2564769.08671778.136279.88341689.684497670.63743004487.2733502157.91181827.51279.88342063.021511489.61163770189.8514281679.463Z-337048.85122491021.8422153972.991则:A = j 曳M0ds =翌Z也=-0.4591x10-7 x 337048.8512 = 0.01547 ipE,p E IA = js M2 M0 ds As Z y M =

15、 -0.4591 x 10-7 x 2491021.842 = 0.011442 p0 E I p E IhA1 +A 2 =(0.01547+0.01144)=0.02691As (1+ y)M0A =p = -0.4591 x 10-7 x 2153972.991 = -0.02690sp E I闭合差A=0载位移一单位弹性抗力图及相应的摩擦力引起的位移1).各接缝处的抗力强度抗力上零点假定在接缝3处，平3 = 40.875 = % ；最大抗力值假定在接缝6处，甲=81.75 =甲；最大抗力值以上各截面抗力强度按下式计算:os (- C QS(/ (20(-ib0. 55 J1由计算可得:

16、b 3 = 0，b4 = 0.4255b, b5 = 0.7855b,b =b最大抗力值下各截面抗力强度按下式计算: b i-(yi - yh) _式中y；所求抗力截面与外轮廓的交点至最大抗力截面的垂直距离，量得yh墙 底外边缘C至最大抗力截面的垂直距离量得y = 1.3426m，y = 2.0125m8气=1一（1.34262-2.01252）气=0.555气，。8 = 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气按下式近似计算：R=G , 1 +b）/2xAS.外式中：AS,外一楔块i的外缘长度。3）抗力集中力与摩擦力之合力Ri按下式计算：R R 顼1 + r 2其中，u为

17、围岩与衬砌间的摩擦系数，本计算中取0.2。则，R = 1.0198R,其作用方 向与抗力集中力的夹角P= arctan r = 11.3099。由于摩擦角的方向与衬砌位移方向相 反，其方向向上。将R,的方向延长，使之交于竖直轴。量取夹角w k。将R,分解为水平与竖直两个分力:R = R sinwR = R cosw以上计算列入表一6中弹性抗力及摩擦角计算表表一6截面号(bb +oi-1)/2iAS外R Gi /)w ksinwkcosw kR(b )Rv （气）40.42550.22281.35440.3017603256.74360.1937000540.9810607980.0584509

18、90.2960452250.78550.60551.35440.820089266.4254-0.436724423-0.899595341-0.358153-0.7377484610.892751.35441.209140680.1534-0.9990847620.042774279-1.20803390.0517201270.74690.873451.35441.1830006891.825-0.658491849-0.752587859-0.7789963-0.8903119800.373451.35440.5058006897.954-0.535101371-0.844787856-0

19、.2706546-0.4272943.计算单位抗力图在基本结构中产生的内力弯矩 M 0 =Z Rr ibi ki轴力 No = siipZ R - ccqsp R c切s式中 L 力R至接缝中心点K的力臂，由图上量取。N 0计算表b计算见表一7和表一9M 0计算表表一7截面号R = 0.3321b 4hR = 0.8346b 5hR = 1.1719b 6hR = 1.1701b 7hR8 = 0.5011b hM 0 / 、 b (b )hr4 iR r 4 4ir5i-R5 rr6i-ir7 iR rr8i-R8r8i40.4510.14970.1497851.70330.56560.57

20、060.47621.0418886962.86430.95121.66021.3850.59230.69413.0309533274.10281.36253.32172.77221.80112.11070.09650.11296.3584544485.03241.67124.43.67223.263.8202.1612.52850.29010.145311.83784925截面号Wsin中cos 中MVsin wM R! RVHCos w RN 0 (bb hH454.50.81410.58070.29600.29170.05840.998270.2070568.1250.9280.3725-

21、0.7377-0.6726-0.3580.9365-0.5511681.750.98960.14340.05170.0516-1.20800.35480.2245795.3750.9956-0.0936-0.8903-0.7772-0.77890.7116-0.959381090.9455-0.3255-0.4272-0.4144-0.27060.9635-0.4921表一 85).单位抗力产生的载位移，计算见表9单位抗力产生的在位移计算表表一9截面号M 0(b)1I2IM 0aIM 0yIM 0(1 + Y ) 5I40.14978279.8834647.7941.9209356597.02

22、59862138.946921951.0418887279.8834970.356291.6073491011.0029421302.61029163.0309533279.88341327.431848.31352044023.3813974871.69491776.3584544279.88341689.6841779.62584710743.7787312523.40458811.837849279.88342063.0213313.21749724421.731627734.949096274.68514940296.9206546571.6058A = Mi M0 ds a SMa

23、=-0.4591 X10-7 X 6274.6852 = 0.288 x 10-3 PT a E T 后 0 hA2aMJ 0 E Ih-AS V yM 0M0 ds 百乙 y = -0.4591 x 10-7 x 40296.92 = -1.850 x 10-3校核为:A +A =-(0.288 + 1.85)x 10-3 =-2.138 x 10-31a 2aAsa(1 + y)M 0a =-0.4591 x 10-7 x 46571.6 = -2.1381x 10-3I闭合差 A = 0(4)墙底(弹性地基上的刚性梁)位移 单位弯矩作用下的转角：例=白=15 110 x 279.8834

24、 = 186.6 x 10-5 s主动荷载作用下的转角：P0 =M0 f- =-1827.51x186.6x10-5 =-3.399 apsp a单位抗力及相应摩擦力作用下的转角：P 0 =M0 f- = 11.8378 x 186 x 10-5 = 0.022 aaa4.解立法方程衬砌计算失高f = Y = 7.4131m8a11 =811 + f- =(11.565 + 186.6)x 10-5 = 1.982 x 10-3a12 药修 + f f- =(33.56 + 7.4131x186.6)x10-5 = 1.417x10-2a22 =522 + f 2 f- =(171.67 +

25、7.41312 x 186.6)x 10-5 = 0.1043a =A +f0 +( +f0 )a =-0.01547-3.399-(0.288x 10-3 + 0.022)xa101 p ap- hh1aaa= -3.3835 - 0.02228aha20 =A 2 + f Po += 25.186 - 0.163bh+ f P0 )b =-0.0144-7.4131x3.399-(1.85x10-3+ 7.4131 x0.022)b2bab把上式代入公式求X1, X 2的值得:a a -a a0.1043x(-3.3835-0.02228b )-0.0147x(-25.186-0.163b

26、 )X = 22 1012 20 =hh1 a2 - a a0.01472 - 0.001982x10-4 x 0.1043=365.2 *.1屋2h式中： X = 365.2X =-8.113 TOC o 1-5 h z 1b1b HYPERLINK l bookmark56 o Current Document 0.001982 x(-25.186 - 0.163bJ-0.0147 x(-3.385 - 0.02228bJ x 20.01472 - 0.001982 x 0.1043=20.94 + 0.136bh式中：广20.94X = 0.1362b5.计算主动荷载和被动荷载分别产生的

27、衬砌内力计算公式为：M=X + X Y + M 0,N=X cosa+ N 0P1P2 PPp2 ppM=X + X Y + M 0,N=X cosa+ N 0bb1b2bb2bb计算过程详见表一10和表一11主动荷载和弹性抗力作用下结构的弯矩MM计算表表一10截面M 0PX1PX 2 PYm PM0 X b1bX Y2 bm b00365.20365.20-8.1130-8.1131-72.4938365.23.20382295.910020-8.1130.020808-8.0921922-283.285365.212.64985494.5648540-8.1130.0821576-8.03

28、08423-594.624365.227.854388-201.5696120-8.1130.1809072-7.9320934-955.871365.248.038454-542.6325460.14978-8.1130.3119976-7.6512225-1307.93365.272.168-870.5621.0418887-8.1130.4687-6.6024116-1595.69365.299.00432-1131.485683.0309533-8.1130.643008-4.43903971778.136365.2127.172812270.508816.3584544-8.1130

29、.8259552-0.9285981827.51365.2155.230312347.9403111.837849-8.1131.00818164.73303 06主、被动荷载作用下衬砌轴力N N计算表表11截面N 0PX cos 92 PLn J pN 0bX cos 9 2bn jb0020.9420.9400.136-0.864150.063320.34949270.41279200.1321648-0.83582188.851318.61566207.4669600.120904-0.7533389.460815.832734405.2935300.1028296-0.62014608

30、.892612.159858621.052460.2070.0789752-0.23775802.80477.78968810.59438-0.55110.050592-0.78716930.52383.013266933.537070.22450.01957040.21667966.6223-1.959984964.66232-0.9593-0.01273-0.72978944.5386-6.81597937.72263-0.4921-0.044268-0.03066.最大抗力值的推力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:5 竺 官(y

31、)siwhpE J6-yi65二皂兰(y)sin甲h_E J S 6计算过程见表一12表一12最大抗力位移修正计算表M /1M /1Y - Y(Y - Y Yb (Y - Y截面6-r-p6ii II0102213.4177-2270.6940244.728483265.0388-10735.84135182820.30249-2264.870214.575378902.8839-10361.78121226467.13286-2247.6993644.1239109147.8092-9269.2874063-56415.98834-2220.0611583.3978-191690.2452-

32、7543.3238034-151873.8419-2141.4500282.4339-369645.7439-5212.0752225-243655.8525-1847.9052391.2816-312269.3405-2368.2753546-316684.0592-1242.413328000Z97710.4023-45490.58434位移值为8 =d =AS/ExSM /I（Y -Y）hp6 pP 6 i=0.4591 x 10-7 x 97710.4032=0.0044868 =8 =AS/ExZM /I（Y -Y）ha6aa 6 i=-0.4591 x 10-7 x -45490.

33、58=-2.088 x 10 -3最大抗力值为：ah =聪 /（1/K-8）= = 0.1615/（1/1 x10 5 + 0.004486） = 35.92计算过程详见表一13衬砌总内力的计算表表一13截面MPa MTMTNPa N h PLn Je0365.2-291.4189673.7810420.94-65.5554739-2354.752622-0.081332821295.91002-290.67153665.238483470.412792-65.38733902-2348.713218-0.00223036294.564854-288.4678446-193.903207.46696-64.89161261-2330.9067250.0831877953-201.569612-284.9207806-486.4904405.29353-64.09369107-2302.2453830.2113112694-5