附录A 衬砌内力计算

发布时间 : 星期三 文章附录A 衬砌内力计算更新完毕开始阅读

?11??12??SE?SE??1IyI??21.273.35?101.273.35?1077?555.556?21.061?10?6

?6?1670.372?63.324?10

?22??SE?yI?1.273.35?107?8637.556?327.454?10?6

计算精度校核为:角度闭合差为零。

?33??SE?(1?y)I2?1.273.35?107?12533.856?475.164?10?6

?6?11?2?12??22?(21.061?2?63.324?327.454)?10?6?475.163?10

闭合差为零 ②载位移

根据《公路隧道设计规范》Ⅴ级围岩中二次衬砌承担70%的外荷载,则q=223.074kpa,e=89.232kpa.

每一分块上的作用力: 竖向力:

Ei?ehiaG,由计算图示得bi如下:bi?xi?xi?1

水平力:Ei?ehi,由计算图示得hi如下:hi?yi?yi?1

自重力:G=d3△S3r=1.2730.6324.5=18.669KN.

各分块上集中力对下一截面的力臂由计算图示量得,分别计为

aQ、aE、aG,由上述公式得,各外力及其力臂如下表: 表 A2 外力及相应力臂

截面 Q 集中力 G E aQ 力臂 aG aE 0 0.000 0.000 1 282.100 18.669 0.000 0.000 0.000 0.000 0.000 9.090 0.632 0.632 0.051 119.205 2 274.825 3 260.461 4 239.380 5 212.125 6 175.715 7 128.330 8 64.746 9 10 18.669 18.669 18.669 18.669 18.669 18.669 18.669 18.669 18.669 27.037 44.286 60.392 74.942 85.843 100.731 110.050 113.052 109.564 0.616 0.616 0.584 0.584 0.537 0.537 0.475 0.475 0.394 0.394 0.288 0.288 0.145 0.145 (0.006) (0.156) 0.151 215.105 0.248 288.607 0.338 341.192 0.420 374.885 0.481 392.348 0.564 397.089 0.617 393.750 0.633 387.809 0.614 384.892

组合计算图示,计算各分块如下:

表A3

截面 ?Ea M0p计算表

?E E ?QaQ?Ga G 0 (178) (169) (152) (128) (101) (69) (37) (9) 0 0 0 (12) (11) (11) (10) (9) (7) (5) (3) 0 3 ? G)?(Qi?1 ??x?(Q?G)??y?Ei?1 i?1i?1 0 M p0 1 2 3 4 5 6 7 8 9 10 0 (0) (4) (11) (20) (31) (41) (57) (68) (72) (67) 0 0 301 594 873 1131 1362 1557 1704 1787 1806 0 0 9 36 80 141 216 302 402 512 625 0 0 (371) (694) (937) (1076) (1073) (895) (494) 20 564 0 0 (3) (18) (54) (118) (208) (340) (496) (649) (768) 0 (191) (749) (1635) (2785) (4120) (5519) (6854) (7925) (8625) (8893)

表A4

截sinα cosα 面 0 0.000 1 0.160 2 0.316 3 0.464 1.000 0.987 0.949 0.886 Np0计算表

E cos?i? 0Np?(Q? G) ?E g)sin?i? (Q? 0.000 300.769 594.263 873.393 0.000 9.090 36.127 80.413 0.000 48.146 187.801 405.098 0.000 8.973 34.276 71.240 0.000 39.173 153.526 333.858 4 0.600 0.800 5 0.720 0.694 6 0.830 0.558 7 0.939 0.344 8 0.994 0.111 9 0.992 (0.129) 10 0.933 (0.361) 1131.442 1362.236 1556.620 1703.619 1787.034 1805.703 1824.372 140.805 215.747 301.589 402.320 512.370 625.422 734.986 678.477 112.680 980.846 149.717 1292.249 168.142 1599.619 138.416 1776.028 56.779 1790.669 (80.536) 1701.370 (265.306) 565.796 831.129 1124.107 1461.204 1719.248 1871.206 1966.676

M10q??q0B2(x10?B4)??223.074?16.7572(7.017?16.7574)??5285.604M10e??00e2H2??89.2322?7.839??3048.8992

M10g???Gi(x10?xi?agi)??384.892M10p?M10q?M10e?M10g??5285.604?3048.899?384.892??8719.3950000从A3得到M80p=-8686139 闭合差??8893?87198719?100%?1.99%

③主动荷载位移

表A5主动荷载位移计算表

截面 0Mp I0yMp I 0(1?y) M p M 0 p 1/I y/I 1+y I0 1 2 3 4 5 6 7 8 9 10 0 (191) (749) (1635) (2785) (4120) (5519) (6854) (7925) (8625) (8893) 56 56 56 56 56 56 56 56 56 56 56 0 6 22 50 88 134 188 250 319 389 458 1.000 1.102 1.405 1.901 2.578 3.418 4.380 5.509 6.742 8.009 9.237 0 (10591) (41601) (90809) (154728) (228916) (306606) (380778) (440256) (479155) (494044) 0 (1079) (16843) (81834) (244156) (553476) (1036278) (1716810) (2527951) (3358371) (4069336) 0 (11670) (58444) (172643) (398885) (782392) (1342884) (2097588) (2968207) (3837526) (4563379) 积分系数1/3 1 4 2 4 2 4 2 4 2 4 1 ∑ (2380474) (11522024) (13902498)

?1p??2p??s0M1MEIM2MEI0pds??sE?sE??MI0p??0p1.273.35?101.277?2380474??90244.835?10?6?s00p

7ds?yMI??3.35?10?11522024??436805.089?10?6计算精度校核:

?1p??2p??(90244.835?436805.089)?10?SE(1?y)MI0p?6??527049.924?10?6

?6?sp????1.273.35?107?13902498??527049.924?10

闭合差△=0

五 载位移——单位弹性抗力及摩擦力引起的位移 (1)各接缝处的抗力强度度 抗力上零点假设在接缝5,?最大抗力值假定在接缝7,?4?46.08=?b ?69.912??h。。7

最大抗力值以上各级面抗力强度按下式计算: 则?5?i?cos?b?cos?icos?b?cos?h2222?h

?0,?6?0.47?h

最大抗力值一下各截面抗力强度按下式计算:

?i?(1?(yiyh))?h

2式中:yi——所考察截面外缘点到h点的垂直距离;

yh——墙脚外缘点到

h点的垂直距离.

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