二次衬砌内力计算书2007

1、隧道结构力学计算课程设计专业：岩土与隧道工程班级：隧道一班学号：201221030117姓名：李绍峰一 基本资料：围岩级别级，=20kN/m3，弹性抗力系数 K=0.4106kN/m3 ,二次衬砌类型C20混凝土45cm，=23KN/m3，弹性模量Eh=2.7107kPa,设计时速100km/m,结构断面如图1所示。图1 衬砌结构断面（尺寸单位：cm）二 荷载确定： 1.竖向围岩压力： q=0.452s-1式中：s围岩类别，此处s=4; 围岩容重，此处=20kN/m3; 跨度影响系数， =1+i(lm-5) =1+0.1(13.044575-5)=1.8044575mq=0.4524-1201

2、.8044575=129.92094kPa考虑到初期支护承担大部分荷载，二次衬砌作为安全储备，故对围岩压力进行折减，对本隧道按60%进行折减，取为77.952564kPa2.水平围岩压力：e=0.25q=0.2577.952564=19.488141 kPa三 衬砌几何要素1.衬砌几何尺寸 内轮廓半径r1= 5.7074m,r2= 8.2m ,内径r1 , r2所画圆曲线终点截面与竖直轴的夹角1=90，2=98.421132, 拱顶截面厚度d0=0.4m,墙底截面厚度dn=0.8m此处墙底截面为自内轮廓半径为r2的圆心向内轮廓墙底做连线并延长至与外轮廓相交，其交点到内轮廓墙底间的连线。内轮廓线

3、与外轮廓线相应圆心的垂直距离为：m= d(r2+d0+0.5d)r2+d01-cos2-dcos2代入数值计算得: m=0.35490916m外轮廓线半径: R1=m+r1+d0=6.46230916mR2=m+r2+d0=8.95490916m拱轴线与内轮廓线相应的垂直距离为m=0.1759934m 拱轴线半径： r1=m+r1+0.5d0=6.0833934mr2 =m+r2+0.5d0= 8.5759934m拱轴线各段圆弧中心角1=90，2=7.2597322.半拱轴线长度S及分段周长S分段轴线长度:S1=1180r1=90/1803.141592656.0833934=9.555772

4、mS2=2180r2=7.259732/1803.141592658.5759934=1.08663176m半拱轴线长度：S= S1+ S2 = 9.555772+1.08663176=10.64240376m将半拱轴线等分为8段，每段长为：S = S8 =10.64240376/8=1.33030047m3.各分块接缝中心几何要素：（1）与竖直轴夹角i1=1=Sr1180=12.529290382=1+1=12.52929038+12.52929038=25.058580763=1+2=12.52929038+25.05858076=37.587871144=1+3=12.52929038+

5、37.58787114 =50.117161525=1+4=12.52929038+50.11716152=62.64645196=1+5=12.52929038+62.6464519=75.17574228S1=7S-S1=7*=-0.24366871m7=1 + S1r2180=88.37206162498=7 + Sr2180=97.259732另一方面8=90+7.259732=97.259732角度闭合差0（2）接缝中心点坐标计算x1=r1sin1=6.0833934sin12.52929038=1.31972334mx2=r1sin2=6.0833934sin25.05858076

6、=2.57658888mx3=r1sin3=6.0833934sin37.58787114=3.71073268mx4=r1sin4=6.0833934sin50.11716152=4.66813602mx5=r1sin5=6.0833934sin62.6464519=5.4031982mx6=r1sin6=6.0833934sin75.17574228=5.880908576ma2=（8.5759934-6.0833934）sin90=2.4926x7=r2sin7a2=8.5759934sin88.3720616249-a2 =6.07963197mx8=r2sin8 a2=8.57599

7、34sin97.259732-a2 =6.01434395my1=r1(1-cos1)= 6.0833934（1-cos12.52929038)= 0.1448746my2=r1(1-cos2)= 6.0833934(1-cos25.05858076)= 0.5725754my3=r1 (1-cos3)= 6.0833934(1-cos37.58787114)=1.2627982my4=r1(1-cos4)= 6.0833934(1-cos50.11716152)=2.1826my5=r1(1-cos5)= 6.0833934(1-cos62.6464519)=3.288197my6=r1(1

8、-cos6)= 6.0833934(1-cos75.17574228)=4.526926ma1=(r2-r1)cos1=） cos90=0y7=r1- r2cos7= 6.0833934-8.5759934cos88.3720616249 =5.83976my8=r1- r2cos8= 6.0833934-8.5759934cos97.259732 =7.16712m当然也可以直接从图2中量出xi,yi，以后计算中只取四位有效数字。图2 衬砌结构计算图示四 位移计算1.单位位移用辛普生法近似计算，按计算列表进行计算。单位位移的计算见表1。单位位移值计算如下：11= 0sM1EhIds SEh1

9、I =1.330300472.7107751.8458=28.765910-612=21= 0sM1M2EhIds SEhyI =1.330300472.71071061.7291=40.6221510-622= 0sM22EhIds SEhy2I =1.330300472.71073671.4288=140.4702310-6计算精度校核为：11+212+22=（28.7659+240.62215+140.47023）10-6=250.4804310-6SS= SEh(1+y)2I = 1.330300472.71076546.73279=250.4804310-6闭合差= 0表1 单位位移

10、计算表注：I截面惯性矩，I=bd312,b取单位长度。不考虑轴力的影响。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=1.36399m,b2=1.30423m, b3=1.18658m, b4=1.01493m, b5=0.79525m, b6=0.53573m , b7=0.24678m , b8=0.01477mbi=6.46226mB2=6.46229m(校核)水平压力：Ei=ehi式中：衬砌外缘相邻两截面之间的垂直投影长度，由图2量得h1=0.14556m, h2=0.43097m,

11、h3=0.69904m, h4=0.93852m, h5=1.13863m, h6=1.28952m, h7=1.38272m, h8=1.40043m, hi=7.42539mH=7.42543m(校核)自重力：Gi=di-1+di2Sh式中：di接缝i的衬砌截面厚度。注：计算G8时，应使第8个楔体的面积乘以h作用在各楔体上的力均列入表2中，各集中力均通过相应图形的形心。（2）外荷载在基本结构中产生的内力楔块上各集中力对下一接缝的力臂由图2中量得，分别记为aq、ag、ae。表2 载位移Mp0计算表 i-1 (Q+G) i-1 Exy-x i-1 (Q+G)y i-1 EMP000000000

12、01.31970.144900-76.7074118.69532.83671.25690.4277-149.18815-1.21325-295.3500233.244711.23551.13410.6902-264.5228-7.75474-624.2267339.621924.85850.95740.9198-325.15404-22.8648-1015.2813434.056743.14850.73511.1056-319.07506-47.705-1412.2484513.176665.33830.47771.2387-245.14446-80.9345-1758.9176574.160

13、690.46860.19871.3128-114.08572-118.767-2005.6718614.8982117.4153-0.06531.327440.1528537-155.857-2137.4790内力按下式计算：弯矩：Mip0 = Mi-1,p0 - xi i-1 (Q+G) -yi i-1 E Qaq Gag - Eae轴力：Nip0 = sini i (Q+G) - cosi i E式中：xi，yi 相邻两接缝中心点的坐标增值，按下式计算xi=xi-xi-1yi=yi-yi-1Mip0、Nip0 的计算见表2及表3表3 载位移Np0计算表截面sincos(Q+G)Esin(Q

14、+G)cosENp00010000010.21694 0.97619 118.69532.836725.74962.769122.980520.42354 0.90588 233.244711.235598.789510.178088.611630.60998 0.79242 339.621924.8585207.161719.6983187.463440.76736 0.64122 434.056743.1485333.076527.6677305.408850.88819 0.45948 513.176665.3383455.797430.0216425.775860.96672 0.25

15、586 574.160690.4686555.049823.1469531.902970.99960 0.02841 614.8982117.4153614.65003.3357611.314480.99198 -0.12637 639.6637144.7071634.5358-18.2863652.8221基本结构中，主动荷载产生弯矩的校核为：M8q0 = -qB2(x8-B4) = -77.95256412.924582（6.0143-12.924584）=-1402.0201M8e0 = -e2H2=-19.48814127.425432=-537.2589M8g0 = -Gi(x8-x

16、i+agi)= - G1(x8-x1+ag1) G2(x8-x2+ag2) G3(x8-x3+ag3) G4(x8-x4+ag4) G5(x8-x5+ag5) G6(x8-x6+ag6) G7(x8-x7+ag7) G8ag8=-12.3688（6.0143-1.3197+0.6573）-12.8813（6.0143-2.5766+0.6058）-13.8803 （6.0143-3.7107+0.5575）-15.3183 （6.0143-4.6681+0.4684）-17.1282（6.0143-5.4032+0.3579）-19.2225 （6.0143-5.8809+0.2305）-21

17、.5004 （6.0143-6.0796+0.0917）-23.6141(-0.0378)=-205.11575M8P0=M8q0+M8e0 +M8g0= -1402.0201-537.2589-205.11575=-2144.39475另一方面，从表2中查M8P0 = -2137.4790闭合差 =|2144.39475-2137.479|2144.39475=0.32%（3）主动荷载位移计算过程见表4表4 主动荷载位移计算表1P= 0sM1MP0EhIds SEhMP0I = - 1.330300472.7107448651.6278= -22105.239710-62P= 0sM2MP0

18、EhIds SEhyMP0I = - 1.330300472.71071413982.5874= -69667.470410-6计算精度校核1P + 2P = -（22105.2397+69667.4704）10-6=-91772.710110-6SP = SEh（1+y）MP0I = - 1.330300472.71071862634.2152=-91772.710110-6闭合差 =03.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抗力上零点假定在接缝3，3= 37.58787114=b最大抗力值假定在接缝5，5= 62.6464519=hi = cos2b-cos2

19、icos2b-cos2hh查表1算得：3=0，4= 0.5201h，5=h最大抗力值以下各截面抗力强度按下式计算：i = （1-yi2yh2）h式中： yi2所考察截面外缘点到h点的垂直距离yh2墙脚外缘点到h点的垂直距离由图2中量得： y6 = 1.15408m， y7 = 2.5368m， y8 = 3.93724m，则： 6 = （1 - 1.1540823.937242）h =0.91408h7 = （1 - 2.536823.937242）h =0.58487h8 = 0按比例将所求抗力绘于图2中。（2）各楔体上抗力集中力Ri按下式近似计算：Ri= （i-1-i2）Si外式中 ；Si

20、外楔体i外缘长度,可通过量取夹角，用弧长公式求得，Ri的方向垂直于衬砌外缘，并通过楔体上抗力图形的形心。（3）抗力集中力和摩擦力的合力Ri按下式计算：Ri = Ri 1+2式中：围岩与衬砌间的摩擦系数，此处取=0.2则：Ri = Ri 1+0.92 = 1.0198Ri （Ri的作用点为Ri 与衬砌外缘的交点）将Ri的方向线延长，使之交于竖直轴，量取夹角k,将Ri分解为水平和竖直两个力：RH = Ri sink， RV = Ri cosk表5 弹性抗力及摩擦力计算表（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：M10 = -Rjrji轴力：N10 = siniRv - cosi

21、RH式中：rji力Rj至接缝中心点ki的力臂，有图中量得计算见表表6及表7表6 M0计算表表7 N0计算表 （5）单位抗力及相应的摩擦力产生的载位移计算见表8表8 单位抗力及相应的摩擦力产生的载位移计算表1= 0sM1M0EhIds SEhM0I = - 1.330300472.7107702.7718= -34.625810-62= 0sM2M0EhIds SEhyM0I = - 1.330300472.71073748.07=-184.668910-6计算精度校核1+2= -（34.6258+184.6689）10-6= -219.294710-6s= SEh（1+y）M0I = - 1.

22、330300472.71074450.84410-6 =-219.294710-6闭合差 =04.墙底（弹性地基上的刚性梁）位移单位弯矩作用下的转角：a= 1KI8 = 10.410623.4375=58.593810-6主动荷载作用下的转角： ap0 = M8P0a =-2137.47958.5938 10-6 = -125242.9102 10-6单位抗力及相应摩擦力作用下的转角：a0 = M80a = -13.966758.5938 10-6 = -818.3613 10-6五 解力法方程衬砌矢高f = y8 = 7.16712m计算力法方程的系数为：a11 = 11 + a = (28

23、.7659+58.5938)10-6=87.359710-6a12 = 12 + f a = (40.62215+7.1671258.5938)10-6=460.570910-6a22 = 22 + f2 a = (140.47023+7.16712258.5938)10-6 =3150.293610-6a10 = 1p+ ap0 + (1+ a0)h = -(22105.2397+125242.9102+34.6258h+818.3613h)10-6= -(147348.1499+852.9871h)10-6a20 = 2p+ fap0 + (2+ fa0)h =-(69667.4704+7

24、.16712125242.9102+184.6689h+7.16712818.3613h)10-6= -(967298.43695+6049.9625h)10-6以上将单位抗力及相应的摩擦力产生的位移乘以，即为被动荷载的载位移。求解方程为：X1 = a22a10-a12a20a122-a11a22 =3150.2936-147348.1499-852.9871h-460.5709(-967298.43695-6049.9625h)460.57092- 87.35973150.2936 = 296.1223-1.57375h式中：X1P =296.1223，X1 = -1.57375X2 = a

25、11a20-a12a10a122-a11a22= 87.3597-967298.43695-6049.9625h-460.5709(-147348.1499-852.9871h )460.57092- 87.35973150.2936=263.7571 +2.1505h式中：X2P = 263.7571，X2 =2.1505六 计算主动荷载和被动荷载（h = 1）分别产生的衬砌内力计算公式为：MP=X1P+yX2P+MP0NP=X2Pcos+NP0M= X1+yX2+M0N= X2cos+ N0计算过程列入表9，表10中表9 主、被动荷载作用下衬砌弯矩计算表表10 主、被动荷载作用下衬砌轴力计算表七 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