二次襯砌內(nèi)力計算書2007

1、隧道結(jié)構(gòu)力學(xué)計算課程設(shè)計專業(yè)：巖土與隧道工程班級：隧道一班學(xué)號：201221030117姓名：李紹峰一 基本資料：圍巖級別級，=20kN/m3，彈性抗力系數(shù) K=0.4106kN/m3 ,二次襯砌類型C20混凝土45cm，=23KN/m3，彈性模量Eh=2.7107kPa,設(shè)計時速100km/m,結(jié)構(gòu)斷面如圖1所示。圖1 襯砌結(jié)構(gòu)斷面（尺寸單位：cm）二 荷載確定： 1.豎向圍巖壓力： q=0.452s-1式中：s圍巖類別，此處s=4; 圍巖容重，此處=20kN/m3; 跨度影響系數(shù)， =1+i(lm-5) =1+0.1(13.044575-5)=1.8044575mq=0.4524-1201

2、.8044575=129.92094kPa考慮到初期支護(hù)承擔(dān)大部分荷載，二次襯砌作為安全儲備，故對圍巖壓力進(jìn)行折減，對本隧道按60%進(jìn)行折減，取為77.952564kPa2.水平圍巖壓力：e=0.25q=0.2577.952564=19.488141 kPa三 襯砌幾何要素1.襯砌幾何尺寸 內(nèi)輪廓半徑r1= 5.7074m,r2= 8.2m ,內(nèi)徑r1 , r2所畫圓曲線終點截面與豎直軸的夾角1=90，2=98.421132, 拱頂截面厚度d0=0.4m,墻底截面厚度dn=0.8m此處墻底截面為自內(nèi)輪廓半徑為r2的圓心向內(nèi)輪廓墻底做連線并延長至與外輪廓相交，其交點到內(nèi)輪廓墻底間的連線。內(nèi)輪廓線

3、與外輪廓線相應(yīng)圓心的垂直距離為：m= d(r2+d0+0.5d)r2+d01-cos2-dcos2代入數(shù)值計算得: m=0.35490916m外輪廓線半徑: R1=m+r1+d0=6.46230916mR2=m+r2+d0=8.95490916m拱軸線與內(nèi)輪廓線相應(yīng)的垂直距離為m=0.1759934m 拱軸線半徑： r1=m+r1+0.5d0=6.0833934mr2 =m+r2+0.5d0= 8.5759934m拱軸線各段圓弧中心角1=90，2=7.2597322.半拱軸線長度S及分段周長S分段軸線長度:S1=1180r1=90/1803.141592656.0833934=9.555772

4、mS2=2180r2=7.259732/1803.141592658.5759934=1.08663176m半拱軸線長度：S= S1+ S2 = 9.555772+1.08663176=10.64240376m將半拱軸線等分為8段，每段長為：S = S8 =10.64240376/8=1.33030047m3.各分塊接縫中心幾何要素：（1）與豎直軸夾角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角度閉合差0（2）接縫中心點坐標(biāo)計算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當(dāng)然也可以直接從圖2中量出xi,yi，以后計算中只取四位有效數(shù)字。圖2 襯砌結(jié)構(gòu)計算圖示四 位移計算1.單位位移用辛普生法近似計算，按計算列表進(jìn)行計算。單位位移的計算見表1。單位位移值計算如下：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計算精度校核為：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閉合差= 0表1 單位位移

10、計算表注：I截面慣性矩，I=bd312,b取單位長度。不考慮軸力的影響。2.載位移主動荷載在基本結(jié)構(gòu)中引起的位移（1）每一楔塊上的作用力豎向力：Qi=qbi式中：bi襯砌外緣相鄰兩截面之間的水平投影長度，由圖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)水平壓力：Ei=ehi式中：襯砌外緣相鄰兩截面之間的垂直投影長度，由圖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接縫i的襯砌截面厚度。注：計算G8時，應(yīng)使第8個楔體的面積乘以h作用在各楔體上的力均列入表2中，各集中力均通過相應(yīng)圖形的形心。（2）外荷載在基本結(jié)構(gòu)中產(chǎn)生的內(nèi)力楔塊上各集中力對下一接縫的力臂由圖2中量得，分別記為aq、ag、ae。表2 載位移Mp0計算表 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內(nèi)力按下式計算：彎矩：Mip0 = Mi-1,p0 - xi i-1 (Q+G) -yi i-1 E Qaq Gag - Eae軸力：Nip0 = sini i (Q+G) - cosi i E式中：xi，yi 相鄰兩接縫中心點的坐標(biāo)增值，按下式計算xi=xi-xi-1yi=yi-yi-1Mip0、Nip0 的計算見表2及表3表3 載位移Np0計算表截面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基本結(jié)構(gòu)中，主動荷載產(chǎn)生彎矩的校核為：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，從表2中查M8P0 = -2137.4790閉合差 =|2144.39475-2137.479|2144.39475=0.32%（3）主動荷載位移計算過程見表4表4 主動荷載位移計算表1P= 0sM1MP0EhIds SEhMP0I = - 1.330300472.7107448651.6278= -22105.239710-62P= 0sM2MP0

18、EhIds SEhyMP0I = - 1.330300472.71071413982.5874= -69667.470410-6計算精度校核1P + 2P = -（22105.2397+69667.4704）10-6=-91772.710110-6SP = SEh（1+y）MP0I = - 1.330300472.71071862634.2152=-91772.710110-6閉合差 =03.載位移單位彈性抗力及相應(yīng)的摩擦力引起的位移（1）各接縫處的抗力強(qiáng)度抗力上零點假定在接縫3，3= 37.58787114=b最大抗力值假定在接縫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強(qiáng)度按下式計算：i = （1-yi2yh2）h式中： yi2所考察截面外緣點到h點的垂直距離yh2墻腳外緣點到h點的垂直距離由圖2中量得： y6 = 1.15408m， y7 = 2.5368m， y8 = 3.93724m，則： 6 = （1 - 1.1540823.937242）h =0.91408h7 = （1 - 2.536823.937242）h =0.58487h8 = 0按比例將所求抗力繪于圖2中。（2）各楔體上抗力集中力Ri按下式近似計算：Ri= （i-1-i2）Si外式中 ；Si

20、外楔體i外緣長度,可通過量取夾角，用弧長公式求得，Ri的方向垂直于襯砌外緣，并通過楔體上抗力圖形的形心。（3）抗力集中力和摩擦力的合力Ri按下式計算：Ri = Ri 1+2式中：圍巖與襯砌間的摩擦系數(shù)，此處取=0.2則：Ri = Ri 1+0.92 = 1.0198Ri （Ri的作用點為Ri 與襯砌外緣的交點）將Ri的方向線延長，使之交于豎直軸，量取夾角k,將Ri分解為水平和豎直兩個力：RH = Ri sink， RV = Ri cosk表5 彈性抗力及摩擦力計算表（4）計算單位抗力及其相應(yīng)的摩擦力在基本結(jié)構(gòu)中產(chǎn)生的內(nèi)力彎矩：M10 = -Rjrji軸力：N10 = siniRv - cosi

21、RH式中：rji力Rj至接縫中心點ki的力臂，有圖中量得計算見表表6及表7表6 M0計算表表7 N0計算表 （5）單位抗力及相應(yīng)的摩擦力產(chǎn)生的載位移計算見表8表8 單位抗力及相應(yīng)的摩擦力產(chǎn)生的載位移計算表1= 0sM1M0EhIds SEhM0I = - 1.330300472.7107702.7718= -34.625810-62= 0sM2M0EhIds SEhyM0I = - 1.330300472.71073748.07=-184.668910-6計算精度校核1+2= -（34.6258+184.6689）10-6= -219.294710-6s= SEh（1+y）M0I = - 1.

22、330300472.71074450.84410-6 =-219.294710-6閉合差 =04.墻底（彈性地基上的剛性梁）位移單位彎矩作用下的轉(zhuǎn)角：a= 1KI8 = 10.410623.4375=58.593810-6主動荷載作用下的轉(zhuǎn)角： ap0 = M8P0a =-2137.47958.5938 10-6 = -125242.9102 10-6單位抗力及相應(yīng)摩擦力作用下的轉(zhuǎn)角：a0 = M80a = -13.966758.5938 10-6 = -818.3613 10-6五 解力法方程襯砌矢高f = y8 = 7.16712m計算力法方程的系數(shù)為：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以上將單位抗力及相應(yīng)的摩擦力產(chǎn)生的位移乘以，即為被動荷載的載位移。求解方程為：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 計算主動荷載和被動荷載（h = 1）分別產(chǎn)生的襯砌內(nèi)力計算公式為：MP=X1P+yX2P+MP0NP=X2Pcos+NP0M= X1+yX2+M0N= X2cos+ N0計算過程列入表9，表10中表9 主、被動荷載作用下襯砌彎矩計算表表10 主、被動荷載作用下襯砌軸力計算表七 最大抗力值的求解首先求出最大抗力方向內(nèi)的位移考慮到接縫5的徑向位移與水平方向有一定的偏離，因此修