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二次衬砌内力计算书.doc

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1、1二次衬砌内力计算书二次衬砌内力计算书一 基本资料:围岩级别级,=20kN/m 3,弹性抗力系数 K=0.410 6kN/m3 ,二次衬砌类型 C20混凝土 45cm,=23KN/m 3,弹性模量 Eh=2.7107kPa,设计时速 100km/m,结构断面如图 1所示。2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/m 3;跨度影响系数,=1+i(l m-5)=1+0.1(13.044575-5)=1.8044575mq=0.4524-1201.8044575=129.92094kPa考

2、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 42%进行折减,取为 54.5668kPa2.水平围岩压力:e=0.35q=0.2554.5668=13.6417kPa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。内轮廓线与外轮廓线相应圆心的垂直距离为:m= (2+0

3、+0.5 )(2+0)(1 2)d 2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: r 1=m+r 1+0.5d0=6.0833934mr2=m+r 2+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= =90/1803.141592656.0833934=9.555772m 1180 1S2= =7.259732/1803.141592

4、658.5759934=1.08663176m 2180 2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:3S= =10.64240376/8=1.33030047m8 3.各分块接缝中心几何要素:(1)与竖直轴夹角 i 1= 1= =12.52929038 1 180 2= 1+ 1=12.52929038 +12.52929038 =25.05858076 3= 1+ 2=12.52929038 +25.05858076=37.58787114 4= 1+ 3=12.52929038 +37.587871

5、14 =50.11716152 5= 1+ 4=12.52929038 +50.11716152=62.6464519 6= 1+ 5=12.52929038 +62.6464519 =75.17574228 S 1=7S-S 1=7*1.33030047-9.555772=-0.24366871m 7= 1+ =88.3720616249 12 180 8= 7+ =97.259732 2 180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=r1sin 1=6.0833934sin12.52929038 =1.31972334mx2=r1

6、sin 2=6.0833934sin25.05858076=2.57658888mx3=r1sin 3=6.0833934sin37.58787114=3.71073268mx4=r1sin 4=6.0833934sin50.11716152=4.66813602mx5=r1sin 5=6.0833934sin62.6464519=5.4031982mx6=r1sin 6=6.0833934sin75.17574228=5.880908576ma2=(8.5759934-6.0833934)sin90=2.4926x7=r2sin 7a2=8.5759934sin88.3720616249-a

7、2 =6.07963197mx8=r2sin 8a2=8.5759934sin97.259732-a2 =6.01434395my1=r1(1-cos 1)= 6.0833934(1-cos12.52929038 = 0.154007m)y2=r1(1-cos 2)= 6.0833934(1-cos25.05858076)= 0.606834my3=r1 (1-cos 3)= 6.0833934(1-cos37.58787114)=1.331456my4=r1(1-cos 4)= 6.0833934(1-cos50.11716152)=2.284608my5=r1(1-cos 5)= 6.08

8、33934(1-cos62.6464519)=3.409395my6=r1(1-cos 6)= 6.0833934(1-cos75.17574228)=4.638674ma1=(r2-r 1)cos 1=(8.5759934-6.0833934)cos90=0y7=r1- r2cos 7=5.15985-10.473cos94.0804=5.90507my8=r1- r2cos 8=5.15985-10.473cos100.995=7.1573m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。4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

9、进行计算。单位位移的计算见表 1。单位位移值计算如下: 11= = 993.2604=48.280110-601 1 1.26382.6107 12= 21= = 2479.1036=120.50410-6012 1.26382.6107 22= = 10526.95478=511.69110-6022 2 1.26382.6107计算精度校核为: 11+2 12+ 22=(48.2801+2120.504+511.691)10 -6=800.97910-6 ss= = 16478.42249=800.97810-6 (1+)2 1.26382.6107闭合差 =0表 1 单位位移计算表5截

10、面 sin cos x y d I 1/I y/I y2/I (1+y)2/I积 分 系 数0 0 0 1 0 0 0.45 0.00759 131.6872 0 0 131.6872 11 14.0338 0.2425 0.9702 1.2512340.154007 0.45 0.00759 131.687220.280743.123372 175.3721 42 28.0676 0.4705 0.8824 2.4277760.606834 0.45 0.00759 131.687279.9123148.49351 340.0054 23 42.1015 0.6704 0.7420 3.45

11、94011.331456 0.45 0.00759 131.6872175.3358 233.452 715.8109 44 56.1353 0.8304 0.5572 4.2845112.284608 0.45 0.00759 131.6872300.8537687.3328 1420.7275 25 70.1691 0.9407 0.3392 4.85386 3.409395 0.45 0.00759 131.6872448.97381530.729 2560.3641 46 84.2029 0.9949 0.1010 5.1334624.638674 0.45 0.00759 131.6

12、872610.85422833.554 4186.9496 27 94.0804 0.9975 -0.0712 5.1333035.905069 0.48 0.00922 108.5069 640.741 3783.62 5173.6086 48 100.995 0.9816 -0.1907 4.9676067.157295 0.65 0.02289 43.69595312.74482238.407 2907.5924 1 993.26042479.10410526.9516478.42249注:I截面惯性矩,I= ,b取单位长度。312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

13、移(1)每一楔块上的作用力竖向力:Qi=qbi式中:b i衬砌外缘相邻两截面之间的水平投影长度,由图 2量得b1=1.2854m,b2=1.2086m, b3=1.0598m, b4=0.8476m, b5=0.5849m, b6=0.2827m , b7=0.0271m ,b i=5,3006m= =5.30055m(校核)2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=0.1582m, h2=0.4652m, h3=0.7444m, h4=0.9792m, h5=1.1555m, h6=1.2628m, h7=1.2574m, h8=1.3048m,

14、 h i=7.3274m=H=7.3274m(校核)自重力:Gi= S h1+2式中:d i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、a g、a e。表 2载位移 计算表0Q E G aq ae ag0 0 0 0 0 0 0 0 0 0 0 01 1.2854 0.1582 120.0564 5.3298 13.0806 0.6085 0.2157 0.6354 -73.0543-1.14963-8.31

15、1442 1.2086 0.4652 112.8832 15.6726 13.0806 0.5381 0.3568 0.6171 -60.7425-5.59198-8.072063 1.0598 0.7444 98.9853 25.0788 13.0806 0.4355 0.4766 0.562 -43.1081-11.9526-7.351324 0.8476 0.9792 79.1658 32.9892 13.0806 0.3069 0.568 0.4733 -24.296 -18.7379-6.191075 0.5849 1.1555 54.6297 38.9288 13.0806 0.0

16、016 0.6255 0.3563 -0.08741 -24.35 -4.660636 0.2827 1.2628 26.4042 42.5437 13.0806 0.0036 0.6456 0.2181 -0.09506-27.4662-2.852897 0.0271 1.2574 2.5311 42.3618 13.5167 -0.1513 0.6164 0.0775 0.382961-26.1118-1.047548 0 1.3048 0.0000 43.9587 16.3725 0 0.588 -0.0467 0 -25.84770.764598hi截 面 力 臂 Qaq Eae Ga

17、gbi 集 中 力 60 0 0 0 0 0 00 0 1.2512 0.1540 0 0 -82.515133.1370 5.330 1.1765 0.4528 -156.64133 -2.41346 -315.977259.1009 21.002 1.0316 0.7246 -267.29485 -15.2188 -660.902371.1668 46.081 0.8251 0.9532 -306.25352 -43.9223 -1060.303463.4133 79.070 0.5693 1.1248 -263.84399 -88.9374 -1442.183531.1236 117.9

18、99 0.2796 1.2293 -148.50287 -145.054 -1766.154570.6084 160.543 -0.0002 1.2664 0.09056601 -203.311 -1996.150586.6562 202.905 -0.1657 1.2522 97.207167 -254.083 -2178.109 (+) xy -x (+) -y 内力按下式计算:弯矩: = -x i -y i Qaq Gag - Eae001, 1(+) 1轴力: = sin i - cos i0 (+) 式中:x i,y i 相邻两接缝中心点的坐标增值,按下式计算x i=xi-xi-1y

19、 i=yi-yi-1、 的计算见表 2及表 300表 3载位移 计算表0截 面 sin cos (Q+G) E sin (Q+G)cos E0 0.0000 1.0000 0 0 0 0 01 0.2425 0.9702 133.1370 5.3298 32.2850 5.1707 27.11432 0.4705 0.8824 259.1009 21.0023 121.9103 18.5323 103.37803 0.6704 0.7420 371.1668 46.0812 248.8473 34.1903 214.65704 0.8304 0.5572 463.4133 79.0704 38

20、4.7979 44.0607 340.73725 0.9407 0.3392 531.1236 117.9992 499.6269 40.0307 459.59626 0.9949 0.1010 570.6084 160.5430 567.6902 16.2158 551.47457 0.9975 -0.0712 586.6562 202.9048 585.1692 -14.4379 599.60718 0.9816 -0.1907 603.0288 246.8635 591.9595 -47.0826 639.0421基本结构中,主动荷载产生弯矩的校核为:=-q = -93.4 (4.967

21、6- )=-1147.2413082(84) 10.60112 10.60114= - =- =-888.593408 22 33.692 7.2632= -G i(x8-xi+agi)08= - 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=-13.0806 (4.9676-1.2512+0.6354)-13.0806 (4.9676-2.4278+0.6171)-13.0806 (4.9676-3.4594+0.562)-13.0

22、806 (4.9676-4.2845+0.4733)-13.0806 (4.9676-4.8539+0.3563)-13.0806 (4.9676-5.1335+0.2181)-13.5161 7(4.9676-5.133+0.0775)-16.3725(-0.0467)=-145.2998= =-1147.2413-888.5934-145.2998=-2181.13450808+08+08另一方面,从表 2中查 =-2178.108708闭合差 = =0.14%|2181.13452178.1087|2181.1345(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截 面 (1

23、+y) 积 分 系 数 1/30 0 131.6872 0.0000 1.0000 0 0 0.0000 11 -82.515363 131.6872 20.2807 1.1540 -10866.2206 -1673.4724 -12539.6930 42 -315.97666 131.6872 79.9123 1.6068 -41610.0953 -25250.4234 -66860.5187 23 -660.90228 131.6872 175.3358 2.3315 -87032.3995 -115879.8522 -202912.2518 44 -1060.3031 131.6872

24、300.8537 3.2846 -139628.3926 -318996.1392 -458624.5318 25 -1442.1825 131.6872 448.9738 4.4094 -189917.0371 -647502.2094 -837419.2466 46 -1766.1535 131.6872 610.8542 5.6387 -232579.8913-1078862.3793 -1311442.2706 27 -1996.1501 108.5069 640.7410 6.9051 -216596.1487-1279015.1852 -1495611.3339 48 -2178.

25、1087 43.6959 312.7448 8.1573 -95174.5281 -681192.1873 -776366.7153 1 -980152.8369-3901230.9827 -4881383.8196M 1I yI I I (1+)I 1P= = - 980152.8369= -47642.967510-6010 0 1.26382.6107 2P= = - 3901230.9827= -189629.835210-6020 0 1.26382.6107计算精度校核 1P+ 2P=-(47642.96751+18962.98352)=-237272.8027 2P= = - -

26、4881383.8196=-237272.802710-6 ( 1+) 0 1.26382.6107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= 42.1015 = b最大抗力值假定在接缝 5, 5=70.1691= h i= h2 2 2 2 查表 1算得: 3=0, 4= 0.5512 h, 5= h最大抗力值以下各截面抗力强度按下式计算: i=(1- ) h22式中: 所考察截面外缘点到 h点的垂直距离 28墙脚外缘点到 h点的垂直距离2由图 2中量得:=1.2628m, = 2.5202m, =3.8250m, 6

27、7 8则: 6=(1- ) h=0.8910 h1.262823.82502 7=(1- ) h=0.5659 h2.520223.82502 8= 0按比例将所求抗力绘于图 2中。(2)各楔体上抗力集中力 按下式近似计算:=( ) 12 外式中 ; 楔体 i外缘长度,可通过量取夹角,用弧长公式求得, 的方向垂直 外 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= 1+2式中:围岩与衬砌间的摩擦系数,此处取 =0.2则:Ri= = 1.0198 (R i的作用点为 与衬砌外缘的交点) 1+0.92 将 Ri的方向线延长,使之交于竖直轴,量取夹

28、角 k,将 Ri分解为水平和竖直两个力:RH=Ri sin k,R V=Ri cos k表 5弹性抗力及摩擦力计算表截 面 ( n) R( n)3 0 0 0 0 0 0 0 0 04 0.5512 0.2756 1.2983 0.364896 60.4286 0.8697 0.4935 0.31737 0.180085 1.0000 0.7756 1.2983 1.026899 74.4625 0.9635 0.2679 0.98937 0.275076 0.8910 0.9455 1.2983 1.251848 88.4964 0.9997 0.0262 1.25142 0.032857

29、0.5659 0.7285 1.2589 0.935203101.3099 0.9806 -0.1961 0.91704 -0.18348 0.0000 0.2830 1.3048 0.376503101.3099 0.9806 -0.1961 0.36919 -0.073812+ () 外 () ()(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:=-R jrji01轴力:9= sin iR v - cos iR H01式中:r ji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 计算表04 0.6533 -0.2384 -0.23845 1.9125

30、 -0.6979 0.6533 -0.6709 -1.36876 3.1197 -1.1384 1.9125 -1.9639 0.6533 -0.8178 -3.92017 4.1825 -1.5262 3.0892 -3.1723 1.8736 -2.3454 0.6181 -0.5780 -7.62198 5.1575 -1.8820 4.2113 -4.3246 3.0758 -3.8503 1.8308 -1.7122 0.5917 -0.2228-11.9918截 面 R=0.3649 ()R=1.0269 R=1.2518 R=0.9352 R=0.3765 () () () ()

31、 ()表 7 计算表0截 面 sin cos4 56.1353 0.8304 0.5572 0.1801 0.1495 0.3174 0.1768 -0.02735 70.1691 0.9407 0.3392 0.4552 0.4282 0.5924 0.2010 0.22726 84.2029 0.9949 0.1010 0.4880 0.4855 0.6253 0.0632 0.42237 94.0804 0.9975 -0.0712 0.3046 0.3038 0.4419 -0.0314 0.33538 100.9950 0.9816 -0.1907 0.2308 0.2265 0.3

32、680 -0.0702 0.2967()sin()()cos()()(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截 面 ( 1+y) 积 分 系 数1/34 -0.238389 131.68724 300.8537253.284608 -31.3928 -71.7203 -103.11308 25 -1.368745 131.68724 448.9738364.409395 -180.246 -614.531 -794.77696 46 -3.920126 131.68724 610.8542375.638674 -516.231 -239

33、4.63 -2910.856 27 -7.621913 108.50694 640.7409856.905069 -827.031 -4883.67 -5710.7028 48 -11.99178 43.695949 312.7448048.157295 -523.992 -3750.37 -4274.3596 1 -1882.78 -10225.3 -12108.0721I I I I ( 1+) I() 1 = = - 1882.782041=-91.517710-6010 0 1.26382.6107 2 = = - 10225.29016=-497.027810-6020 0 1.26

34、382.6107计算精度校核 1 + 2 =-(91.5177+497.0278)=-588.5454510 -610 2 = = - 12108.072210-6 =-588.5454510-6 ( 1+) 0 1.26382.6107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:= 43.6959 = 87.3919 10-6= 18 10.2106主动荷载作用下的转角:= =-2178.1087 87.3919 10-6 = -190349.0561 10-6008单位抗力及相应摩擦力作用下的转角:= = -11.9918 87.3919 10-6= -1047.9

35、845 10-60 08 五 解力法方程衬砌矢高f = y8 = 7.1573m计算力法方程的系数为:a11 = 11+ a = (48.2801 + 87.3919)10 -6=135.67210-6a12 = 12+ fa = (120.504+7.157387.3919)10 -6=745.994010-6a22 = 22+ f2a = (511.691+7.1573 287.3919)10-6 =4988.510910-6a10 = 1p + + ( 1 + ) h0 0= -(47642.9675+190349.0561+91.5177 h+1047.9845 h)10-6= -(

36、)10-6237992.0236 +1139.5022a20 = 2p + + ( 2 + ) h0 0=-(189629.8352+7.1573190349.0561+497.0278 h+7.15731047.9845 h)10-6= -( )10-61552015.1344+7997.7672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=221012202121122=4988.5109(237992.0236-1139.5022)745.9940(1552015.13447997.7672)745.99402 135.6724988.5109= 2

37、44.6500 2.3433式中:X 1P =244.6500,X 1 =-2.3433X2=112012102121122=11135.672(1552015.13447997.7672)745.9940(237992.0236-1139.5022 )745.99402 135.6724988.5109=274.5329+1.9537式中:X 2P =274.5329,X 2 =1.9537六 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:=1+2+0=2+0 = 1 +2 +0 = 2 + 0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截 面0

38、 0 244.650 0 244.65 0 -2.3433 0 -2.34331 -82.5154 244.650 42.27994716 204.41 0 -2.3433 0.3009 -2.04242 -315.9767 244.650 166.5959162 95.27 0 -2.3433 1.1856 -1.15773 -660.9023 244.650 365.5286085 -50.72 0 -2.3433 2.6013 0.25804 -1060.3031 244.650 627.2000526 -188.45 -0.2384 -2.3433 4.4634 1.88175 -14

39、42.1825 244.650 935.9911149 -261.54 -1.3687 -2.3433 6.6609 2.94896 -1766.1535 244.650 1273.468725 -248.03 -3.9201 -2.3433 9.0626 2.79927 -1996.1501 244.650 1621.135694 -130.36 -7.6219 -2.3433 11.5367 1.57158 -2178.1087 244.650 1964.912991 31.45 -11.9918 -2.3433 13.9832 -0.3519()M X Xy ()y()()表 10主、被

40、动荷载作用下衬砌轴力计算表截 面0 0 274.5329 274.533 0 1.9537 1.95371 27.1143 266.3389 293.453 0 1.8954 1.89542 103.3780 242.2459 345.624 0 1.7239 1.72393 214.6570 203.6920 418.349 0 1.4496 1.44964 340.7372 152.9790 493.716 -0.0273 1.0887 1.06145 459.5962 93.1340 552.730 0.2272 0.6628 0.89006 551.4745 27.7295 579.2

41、04 0.4223 0.1973 0.61977 599.6071 -19.5347 580.072 0.3353 -0.1390 0.19628 639.0421 -52.3598 586.682 0.2967 -0.3726 -0.0759 Xcos () cos() ()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,因此修正后有:12 hP = 5P= (5)5 h = 5 = (5)5计算过程列入表 11中,位移值为: hP= 139779.76450.9407 =63914.655410-61.26382.6107 h = - 152

42、3.03290.9407=-696.410710-61.26382.6107最大抗力值为:= = 91.1230= 163914.655410610.2106+696.4107106表 11最大抗力修正位移计算表截 面 (y5-yi) 积 分 系 数1/30 32217.2840 -308.5827 3.4094 109841.4490 -1052.0804 11 26918.7930 -268.9602 3.2554 87631.1215 -875.5700 42 12545.7455 -152.4580 2.8026 35160.2170 -427.2730 23 -6679.6610 3

43、3.9709 2.0779 -13879.9253 70.5894 44 -24816.8630 247.8024 1.1248 -27913.6872 278.7249 25 -34441.6640 388.3312 0 0 0 4 139779.7645 -1523.0329MM()(-)()()八 计算衬砌总内力按下式计算衬砌总内力: M=+ N=+ 计算过程列入表 12中表 12衬砌总内力计算表截 面 M N e 积 分 系 数1/30 244.6500 -213.5285 31.1215 274.5329 178.0270 452.5599 0.0688 4098.3011 0.00

44、00 11 204.4146 -186.1111 18.3034 293.4532 172.7134 466.1666 0.0393 2410.3291 371.2072 42 95.2693 -105.4957 -10.2264 345.6239 157.0898 502.7137 -0.0203 -1346.6890 -817.2168 23 -50.7237 23.506684 -27.2170 418.3490 132.0886 550.4376 -0.0494 -3584.1306 -4772.1139 44 -188.4531 171.47065 -16.9824 493.7162 96.7134 590.4296 -0

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