地下建筑结构课程设计

1.1基本资料

结构断面图如图1所示。围岩级别为V级，容重18kN/m3,围岩的弹性

抗力系K0.15106kN/m2，衬砌材料为C45混凝土，弹性模量为

E3.35107Kpa容重25KN/m3

图1.1 结构断面图

2.计算作用在衬砌结构上的主动荷载 2.1隧道深浅埋的确定

坍落拱高度按下式计算：hq0.452s11i(Bt5)

Ⅴ级围岩，s=5；B>5，i=0.1

hq0.452410.1(14.865)14.299m

浅埋隧道分界深度：HP因为hq2.5hq35.748m

14.299mH34mHp35.748m，所以是浅埋隧道

2.2竖直和水平荷载

00H34m,B14.86m,40,0.624垂直力：取 tggtantang(tan2g1)tangtangtan0.839(0.7041)0.8392.744

0.8390.445tan1tan(tangtan)tangtantantang0.283

H34qH(1tan)1834(10.2830.445)435.567kN/mBt14.86水平力：

e1H18340.283173.196kN/me2h1812.74340.283238.094kN/m

11ee1e2173.196238.094205.645kN/m

223.半拱轴线长度

3.1衬砌的几何尺寸

内轮廓线半径：r16.65m，r22.5m

1090,21400

内径r1，r2所画圆曲线端点截面与竖直轴线的夹角：1拱顶截面厚度：d00.5m， 拱底截面厚度：dn0.6m。

拱轴线半径：

r1`r10.5d06.650.50.56.9m,r2`r20.5dn2.50.50.52.75m00拱轴线各段圆弧中心角：1109,231

3.2半拱轴线长度S及分段轴长

1090S1r6.913.126m 001801801`1310S2r2.751.487m 001801802`2半拱线长度：SS1S213.1261.48714.613m 将半拱轴线等分为10段，每段轴长为：S4.分块图：

S14.6131.4613m 10105各分块接缝（截面）中心几何要素 5.1与竖直轴夹角

i:

S18001.4613180011`12.14040

r16.921112.1404012.1404024.28080

32124.2808012.1404036.42120

43136.4212012.1404048.56160

54148.5616012.1404060.70200

65160.7020012.1404072.84240

76172.8424012.1404084.98280

87184.9828012.1404097.12320

S19SS191.461313.1260.026m

S118000.0261800091`109109.54170

r22.75S18001.461318002`30.44590

r22.75S18001.461318000109`109.5417139.98720

r22.75另一方面1210931140 角度闭合差0

5.2各截面中心点坐标计算

000x1r1sin16.90.21031.4511

x2r1sin26.90.41122.8373 x3r1sin36.90.59374.0966

x4r1sin46.90.74975.1727 x5r1sin56.90.87216.0174

x6r1sin66.90.95556.5929x7r1sin76.90.99626.8736x8r1sin86.90.99236.8467

x9r2sin9a12.750.94243.396.5216x10r2sin10a12.750.64303.935.6981

y1r1(1cos1)6.9(10.9776)0.1543y2r11cos26.910.91150.6104y3r11cos36.910.80471.3477

y4r11cos46.910.66182.3335 y5r11cos56.910.48943.5235

y6r11cos66.910.29504.8645y7r11cos76.910.08756.2966y8r11cos86.910.12407.7566

y9r1a2r2cos96.91.332.750.33459.1499

y10r1a2r2cos106.91.332.750.765910.3362

6计算基本结构的单位位移ik

截α 面 0 0 1 12.1404 2 24.2808 3 36.4212 4 48.5616 5 60.702 6 72.8424 7 84.9828 8 97.1232 sinα 0 0.2103 0.4112 0.5937 0.7497 0.8721 0.9555 0.9962 0.9923 cosα 1 0.9776 0.9115 0.8047 0.6618 0.4894 0.295 0.0875 -0.124 x 0 1.4511 2.8373 4.0966 5.1727 6.0174 6.5929 6.8736 6.8467 y 0 0.1543 0.6104 1.3477 2.3335 3.5235 4.8645 6.2966 7.7556 d 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1/I 96.154 96.154 96.154 96.154 96.154 96.154 96.154 96.154 96.154 y/I 0 14.8366 58.6924 129.5867 224.3754 338.7986 467.7411 605.4433 745.732 y^2/I 0 2.2893 35.8258 174.6441 523.5799 1193.7569 2275.3267 3812.2341 5783.5988 (1+y)^2/I 系数

96.154 1 128.1164 4 249.3646 2 529.9715 4 1068.4846 2 1967.5082 4 3306.963 2 5119.2747 4 7371.2167 2 9 109.5417 1139.9870 2

0.9424 -0.3345 0.643 -0.7659 6.5216 9.1499 0.5 96.154 879.7995 8050.0773 9905.8303 4 5.6981 10.3362 993.867 10272.8078  961.54 3953.60226813.8255 7 校核 0.5 96.154 12356.6958 35682.5707 35682.5707 计算得各单位位移如下：

11S11.46135961.544.194310 EI3.35107Sy1.46135123953.602517.24610 I3.35107E22Sy21.4613526813.8257116.96410 I3.35107E计算精度校核为：

1121222(4.119417.2462116.96)105155.6506105

S1y1.461353335682.5707155.650610 EI3.351072闭合差Δ = 0

7.计算主动荷载在基本机构中产生的位移7.1每一分块上的作用力

1p

和

2p

竖直力：Qiqbi

衬砌外缘相邻两截面之间的水平投影长度bi，由计算图式得：

b11.503m,b21.433m,b31.308m

b41.116m,b50.88m,b60.592m,b70,291m

校核：

bi7.123B7.125m 2水平压力：Eiehi

衬砌外缘相邻两截面之间的竖直向投影长度hi，由计算图式得：

h10.160m,h20.482m,h30.753m,h41.023m,h51.242m h61.381m,h71.483m,h81.518m,h91.447m,h101.289m

校核：

h10.778mH10.78m

i自重力：GdS` ，钢筋混凝土构件重度γ ' = 25kN /m3。

G0.51.46132518.2663KN

由上文计算得q435.5670.45196.005KN/m

e205.6450.4592.54KN/m

7.2各分块外力及力臂表

各外力对各分分块截面的力臂可由分块图量出计算得各分块外力如下表： 集中力(KN) 力臂(m) 截面 0 1 2 3 4 5 6 7 8 9 10 Q 0 G 0 E 0 0 0.6984 0.6147 0.4991 0.3605 0.2028 0.0415 0 0 0 0 0.7223 0.6851 0.6205 0.5125 0.393 0.2478 0.1002 -0.0579 -0.201 -0.4564 0 0.3297 0.469 0.5894 0.6881 0.7425 0.7685 0.7647 0.7275 0.6359 0.4217 294.5955 18.2663 14.8064 280.8752 18.2663 44.6043 256.3745 18.2663 69.6826 218.7416 18.2663 94.6684 172.4844 18.2663 114.9347 116.035 18.2663 127.7977 0 0 0 18.2663 140.4757 18.2663 133.9054 18.2663 119.2841 57.0375 18.2663 137.2368 -0.1181 ∑(°(i-1)p)∑¦¤∑(¦Α)∑¦Α°ip°ip°ip¦¤¦Α

外荷载在基本结构中产生的内力计算原理如下 弯矩：MipMi1,pxi轴力：Nip0i0(QG)yEQasin(QG)cosE

0iiQGagEaE

x,y为相邻两截面中心点的坐标增量，按下式计算：

xixixi1,yyiyi1

7.3

0Mip计算表：

截面 0 0 0 0 0 -205.745-13.1937 -4.8817 0 1 5 -12.5142 -20.9194 2 -172.654 -127.956-11.3342 -41.0709 3 5 4 -78.8563 -9.3615 -65.1413 5 -34.9798 -7.1787 -85.339 -Eae 0 0 E i1 0 0 0 0 0 0 Mip-223.8209 -870.3507 -1865.2116 312.8618 14.8064 -433.689 -6.7532 612.0033 59.4107 -770.6958 -43.8035 886.6441 129.0933 -954.1177 -127.2602 -3099.9486 1123.652 223.7617 -949.1488 -266.2764 -4442.8713 -454.1919 -5761.0564 6 -4.8155 -4.5264 -98.2125 1314.4027 338.6964 -756.4388 7 6.7361 -1.8303 -104.945 1448.704 466.4941 -406.6512 -668.0662 -6935.813 8 0 9 0 10 0 7.4

0Np1.0576 -102.1961 1524.0078 603.7309 40.9958 -880.8434 -7876.7991 3.6715 -85.1504 1542.2741 744.2066 501.3933 -1037.6473 -8494.532 8.3367 -50.3021 1560.5404 878.112 1285.105 -1041.7043 -8293.0967 计算表：

sini cosi Σ（Q+G) 0 截面 0 0 E 0 1 siniQG0 cosiE Nip 0 0 0 14.4753 54.1553 103.878 51.322 197.5059 422.5374 1 0.2103 0.9776 2 0.4112 0.9115 3 0.5937 0.8047 4 0.7496 0.6618 5 0.8720 0.4894 6 0.9555 0.2950 312.8618 14.8064 65.7973 612.0033 59.4107 251.6612 886.6441 129.0933 526.4154 1123.652 223.7617 842.3656 148.0887 694.2769 1314.4027 338.6964 1146.2726 165.7418 980.5308 1448.704 466.4941 1384.2322 137.6163 1246.6159 7 0.9962 0.0875 1524.0078 603.7309 1518.1685 52.7992 1465.3693 1622.6545 8 0.9923 -0.1240 1542.2741 744.2066 1530.3705 -92.284 9 0.9424 -0.3345 1560.5404 878.112 1470.6506 -293.7221 1764.3727 10 0.6430 -0.7659 1578.8067 997.3961 1015.1076 -763.9065 1779.0141

基本结构中，主动荷载产生的弯距校核为：

0M10,qqBB7.125x10196.0057.1255.69812982.4415KNm242e20M10H46.27116.20845376.9627KNm e20M10qGi(x10xiagi)135.8848KNm

000M10MMq10e10g2928.44155376.9627135.88488495.2889KNm 0从Mp计算表中得出：M10p8293.0967KNM

0闭合差8495.28898293.09678495.2889100%2.38%5%(可以）

7.5主动荷载计算表 截M面 0 0 oip1/I 96.154 0 y/I 1+y 1 0 MipI0 Mp×y/I 0 Mp*(1+y)/I 0 系数 1 4 2 4 2 4 2 4 2 4 1 1 -223.8209 96.154 14.837 1.154 -21521.24 -3320.727 -24841.968 2 -807.3507 96.154 58.692 1.610 -77629.875 -47385.276 -125015.151 3 -1865.211 96.154 129.587 2.348 -179347.269 -241706.315 -421053.584 4 -3099.948 96.154 224.375 3.334 -298071.981 -695550.967 -993622.948 -1932435.416 5 -4442.871 96.154 338.798 4.524 -427199.163 -1505236.252 -3248626.467 6 -5761.056 96.154 467.740 5.865 -553947.731 -2694678.736 -4866139.725 7 -6935.813 96.154 605.442 7.297 -666905.096 -4199234.628 -6631355.981 8 -7876.799 96.154 745.731 8.756 -757384.529 -5873971.452 -8290254.841 9 -8494.532 96.154 879.798 10.150 -816781.923 -7473472.918 -9039634.886 10 -8293.096 96.154 993.865 11.336 -797413.144 -8242221.741 Σ -4206166.71 -26852426 校核 -31058592.71 -31058592.71 主动位移计算如下：

1pMpS1.461354206166.7118347.675810 EI3.3510702ps00M2MPSdsEIEyMI0p

1.4613526852426117132.686910

3.35107计算精度校核：

1p2p183476.7581061171326.869106135480.3627105

3pMp1yS1.4613531058592.71135480.362710 IE3.351070闭合差为0

8.解主动荷载的力法方程

8.1墙底位移计算

单位弯距作用下的转角：1主动荷载作用下的转角：

121256410 363Kd0.15100.50450M106.4108293.0967530758.9110 ,p18.2解力法方程

衬砌矢高fy1010.3362m 力法方程系数如下：

a111114.19436410568.19105

a1212f1(17.24610.336264)105678.8105

a2222f21116.96410.33622641056954.5105

a101pap18347.68530758.19105549105.9105

a202pfap117132.710.3362530758.191055603155105

将各系数代入力法方程，得出立法方程式如下：

68.19105X1p678.8105X2p549105.91050678.810X1p6954.510X2P560315510解方程得：X1p5550

a12a20a10a221149.2708KNm 2a11a22a12 X2pa10a21a20a11693.5149KNm

a22a11a12a129.主动荷载下的各截面的内力

0MipX1pyX2PMip 计算公式为：

0NipX2pcosNip

计算过程列入下表：

表9.1 主动荷载产生的衬砌内力计算表 截面 0 1 2 3 4 5 6 7 8 9 10

表9.2 主动荷载产生的衬砌内力计算表

0 Nip0 MipX1p 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 1149.2708 X2pyi 0 107.0093491 423.321495 934.6500307 1618.317019 2443.59975 3373.603231 4366.785919 5378.624158 6345.591984 7168.308709 Mip 1149.2708 1032.459249 765.241595 218.7092307 -332.3607809 -850.0007498 -1238.182369 -1419.756281 -1348.904142 -999.6692165 24.48280938 0 -223.8209 -807.3507 -1865.2116 -3099.9486 -4442.8713 -5761.0564 -6935.813 -7876.7991 -8494.532 -8293.0967 截面 0 1 2 3 4 5 6 7 8 9 X2p 693.5149 693.5149 693.5149 693.5149 693.5149 693.5149 693.5149 693.5149 693.5149 693.5149 x2p*cosα 693.5149 677.9801662 632.1388314 558.07144 458.9681608 339.4061921 204.5868955 60.68255375 -85.9958476 -231.9807341 Nip 693.5149 729.3021662 829.6447314 980.60884 1153.245061 1319.936992 1451.202796 1526.051854 1536.658652 1532.391966 0 51.322 197.5059 422.5374 694.2769 980.5308 1246.6159 1465.3693 1622.6545 1764.3727 10 1779.0141 693.5149 -531.1630619 1247.851038 10.单位弹性反力及相应摩擦力作用下基本结构产生的位移

1和2

抗力上零点假定在接缝4，448.56160b 最大抗力值假定在接缝6，672.84240h；

cos2bcos2iih22cosbcosh

BH弧最大抗力值以上各截面抗力强度按下式计算：

所以计算可得：40,50.5655h,6h

y'i2i12hy'hAH弧最大抗力值以下各截面抗力强度按下式计算：

式中：

y'iy'h

--所考察截面外缘点到h点的垂直距离； --墙角外缘点到h点的垂直距离。

````y71.6179,y83.0867,y94.3763,yh5.2107

由图测量可得：

1.61792715.21072h0.9036h3.086720.6491h12h则：85.21074.37632915.21072h0.2946h,100按比例将所求得的抗力绘于图1上。

10.1各楔块上的力 各楔块上抗力集中力

R'i

按下式近似计算：

R'i(i1i2)Si外

式中：Si外--楔快i外缘长度，可通过量取夹角，用弧长公式求得，直于衬砌外缘，并通过楔块上抗力图形的形心。

抗力集中力与摩擦力的合力

R'i的方向垂

R'i：

RiR'i12

式中：--围岩与衬砌间的摩擦系数，此处取=0.2。

2RR'10.21.0198R'i ii则：

其作用方向与抗力集中力

RiRiR'i的夹角arctan11.3099。由于摩擦阻力的方向

R'i与衬砌位移的方向相反，其方向向上。画图时，也可取切向：径向=1：5的比例求出合力将

Ri的方向。的作用即为与衬砌外缘的交点。

的方向线延长，使之交于竖直轴，量取夹角k,

RHRisink将

Ri分解为水平与竖直两分力：

RVRicosk

10.2分解得到的水平力跟竖直力

截i面 4 0 i1i2S 0 Ri k 0 sink cosk 0 1 Rh 0 Rv 0 1.515 0 5 0.5655 0.2828 1.515 0.4368 65.965 0.9133 0.4073 6 1 0.7828 1.515 1.2093 78.716 0.9807 0.1957 -0.0026 0.398971799 0.177925808 1.185968931 0.236635616 1.470523105 -0.00384979 1.17317334 -0.249729491 0.667534527 -0.292999353 0.162049878 -0.159786137 7 0.9036 0.9518 1.515 1.4705 90.15 1 0.9781 -0.2082 8 0.6491 0.7764 1.515 1.1995 102.017 0.9157 -0.4019 9 0.2946 0.4719 1.515 0.7290 113.698 0 10 0.1473 1.515 0.2276 135 0.7121 -0.7021 10.3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

Rr

轴力：NsinRvcosR弯矩：Mioi0iiih

k式中：ri--力Ri至接缝中心点i的力臂，由分块图量得。

表10.1结构内力图1

R5=0.4368 R6=1.2093 R7=1.4705 R8=1.1995 R9=0.7290 R10=0.2276 截r5i -R5r5r6i -R6r6r7i r8i r9i -R9r9i r10i -R10r1-R7r7i -R8r8i 面 i i 0i 0.52-0.220 0 0 0 0 0 0 0 0 0 -0.2277 5 11 76 1.97-0.860.70-0.840 0 0 0 0 0 0 0 -1.7137 6 93 46 22 92 3.39-1.482.15-2.600.7867 13 13 5 60 4 -2.053.58-4.342.2434.71 8 73 95 08 6 5.84-2.554.90-5.923.6379 31 23 16 75 4 6.55-2.865.88-7.124.78410 92 51 94 21 2

-1.1560 0 4 -3.299-0.9860.822 2 0 -5.348-2.6972.2488 8 4 -7.035-4.2483.5416 2 1 0 0 0 0 0 0 0 0 -5.2438 -10.6833 0.845-0.6166 0 0 -17.1427 9 2.222-0.223-1.6199 0.9805 -23.1135 1 2 表10.2 结构内力图2

截面 α sini cosi 5 60.702 0.8721 0.4894 0.177925808 0.398971799 0.1552 6 72.8424 0.9555 0.295 0.414561425 1.584940729 0.3961 7 84.9828 0.9962 0.0875 0.410711635 3.055463835 0.4092 8 97.1232 0.9923 -0.124 0.160982144 4.228637174 0.1597 Rv RhsiniRvcosiRh 0.1953 0.4676 0.2674 Ni0 -0.0401 -0.0715 0.1418 -0.5244 0.6841 0.9424 -0.3345 -0.132017209 4.896171701 -0.1244 -1.6378 1.5134 9 109.5417 0.643 -0.7659 -0.291803346 5.058221579 -0.1876 -3.8741 3.6865 10 139.9872 10.4单位抗力及相应摩擦力产生的载位移

截0Mi0Mi0Mi 1/I y/I (1+y) Mi0y/I (1y)/I 系数 /I 面 5 -0.2276 96.1538 338.7979 4.5235 -21.88460488 -77.11040529 -98.99501017 4 6 -1.7137 96.1538 467.7402 5.8645 -164.7787671 -801.5663124 -966.3450794 2 7 -5.2438 96.1538 605.4420 7.2966 -504.2112964 -3174.816849 -3679.028146 4 8 -10.683 96.1538 745.7304 8.7556 -1027.211045 -7966.637984 -8993.849029 2 9 -17.143 96.1538 879.7977 10.1499 -1648.364593 -15082.37119 -16730.73579 4 10 -23.113 96.1538 993.8649 11.3362 -2222.402779 -22971.19961 -25193.60239 1 Σ -4434.741461 -37948.26733 -42383.00879 校核 -42383.00879 1s0s00M1MMS1.46135ds4434.741519.344710 7EIEI3.351000M2MyMS1.461352ds22971.1996165.533710 0EIEI3.35107校核为：

12184.8785105s (1y)Mi0S1.4613542383.0088184.8785107EI3.3510闭合差0

10.5墙底位移

单位抗力及相应摩擦力作用下的转角：M1011479.26391011.解单位弹性反力及摩擦力作用下的力法方程

`05

力法方程：

a11X1a12X2a10a21X1a22X2a20

衬砌矢高fy1010.3362m

a111114.19436410568.19105

a1212f1(17.24610.336264)105678.8105

a2222f21116.96410.33622641056954.5105

`a111498.6068105a22f`15455.5012105X1求解方程为：

X2

a1a21a11a22.71612a11a22a12a12a2a1a225.05842a11a22a1211.2单位弹性反力及摩擦力作用下各截面的内力

表11.1 单位弹性反力及摩擦力作用产生的衬砌内力计算表

截面 0 0 -5.058377356 1 0 -5.058377356 2 0 -5.058377356 3 0 -5.058377356 4 0 -5.058377356 5 -0.22761648 -5.058377356 6 -1.7137287 -5.058377356 7 -5.24376254 -5.058377356 8 -10.68331315 -5.058377356 9 -17.14266436 -5.058377356 10 -23.11349798 -5.058377356 Mi0 x1σ x2σyi 0 0.41908822 1.657883666 3.660435478 6.337928463 9.570041113 13.21227898 17.10195001 21.06468309 24.85168701 28.07375024 Mi -5.058377356 -4.639289136 -3.40049369 -1.397941878 1.279551107 4.284047277 6.440172927 6.799810118 5.322992581 2.650645295 -0.098125101 表11.2 单位弹性反力及摩擦力作用产生的衬砌内力计算表

截面 0 0 1 0 2 0 3 0 4 0 5 -0.0401 6 -0.0715 7 0.1418 8 0.6841 9 1.5134 10 3.6865 12最大抗力值的求解

Ni0x2σ 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 2.716061051 x2σcosαi 2.716061051 2.655221283 2.475689648 2.185614328 1.797489204 1.329240278 0.80123801 0.237655342 -0.33679157 -0.908522422 -2.080231159 Ni 2.716061051 2.655221283 2.475689648 2.185614328 1.797489204 1.289140278 0.72973801 0.379455342 0.34730843 0.604877578 1.606268841 首先求出最大抗力方向内的位移。

hp由式：

hp(y6yi)MipSds0yhsin60EIEI

sMM(yy)MS6iiiihdsyahsin60EIEIsMipMih12.1最大抗力位移修正计算表

截M/I ip面 Mi/I yi M/I(y6yi)y6yi Mip/I(y6yi) i积分系数 0 110506.8077 -486.3824381 0 4.8645 537560.366 -2366.00737 1 4 2 4 2 4 2 1 99274.9278 -446.0854938 0.1543 4.7102 467604.7649 -2101.151893 2 73580.92259 -326.9705472 0.6104 4.2541 313020.6028 -1390.965405 3 21029.73372 -134.4174882 1.3477 3.5168 73957.36756 -472.7194226 4 -31957.76739 123.0337602 2.3335 2.531 -80885.10926 311.3984472 5 -81730.84133 411.9276228 3.5235 1.341 -109601.0582 552.3949422 6 -119055.997 619.2473968 4.8645 0

位移值为：

0 0 Σ 909891.8833 -4203.68226 hp4.3621108909891.88330.95553792.408105h4.3621104203.682260.955517.52081085

则可得最大抗力:hhp1hK208.5171

13.计算X1和X2

X1X1pX1n1149.2708(5.0584208.5171)94.5126X2X2pX2n693.51492.7161208.51711259.860114.计算衬砌截面总的内力

MiMipMih按下式进行计算：

NiNipNih

计算过程列入表：

表14.1 衬砌总内力计算表

Ni Mi Mi Ni Mip Nip 截面 0 1149.2708 -5.058377356 94.51262504 693.5149 2.716061051 1259.860073 1 1032.459249 -4.639289136 65.08813416 729.3021662 2.655221283 1282.961207 2 765.241595 -3.40049369 56.18051355 829.6447314 2.475689648 1345.868356 3 218.7092307 -1.397941878 -72.78555511 980.60884 2.185614328 1436.3468 4 -332.3607809 1.279551107 -65.55249528 1153.245061 1.797489204 1528.052296 5 -850.0007498 4.284047277 43.29636295 1319.936992 1.289140278 1588.744784 6 -1238.182369 6.440172927 104.7038107 1451.202796 0.72973801 1603.365649 7 -1419.756281 6.799810118 -1.879597364 1526.051854 0.379455342 1605.174781 8 -1348.904142 5.322992581 -238.9691678 1536.658652 0.34730843 1609.078399 9 -999.6692165 2.650645295 -446.9643475 1532.391966 0.604877578 1658.519284 10 24.48280938 -0.098125101 4.022047922 1247.851038 1.606268841 1582.785558 截面 0 1 2 3 4 5 6 7 8 9 Mi/I Mi*yi/I 9087.752408 0 6258.474438 965.6826058 5401.972457 3297.363988 -6998.611068 -9432.028137 -6303.124546 -14708.34113 4163.111822 14668.7245 10067.6741 48974.20067 -180.7305158 -1137.987766 -22977.8046 -178206.6613 -42977.34111 -393238.3734 积分系数 0.333333333 1.333333333 0.666666667 1.333333333 0.666666667 1.333333333 0.666666667 1.333333333 0.666666667 1.333333333 10 Σ 386.7353771 3997.374204 0.333333333 -59029.4877 -609995.1434 15.衬砌截面强度简算 15.1拱顶转角为0

s0MiMiSdsaa0 EIEIMiS257.492105I式中：E

5aM101257.411110

闭合差：257.4919257.4111257.4919100%0.0313%

15.2拱顶相对水平位移为0

s0MiyiMiyiSdsfafa0 EIEIMiyiS2660.8534105I式中：E

fa2660.6523105闭合差：2660.85432660.65232660.8543100%0.00756%

15.3 h点的位移

s0MiyindsyiahEIk

式中：

(y6yi)MiSsin6139.0114105EIhK

139.01141050

16.内力图：