3D的动力(爆破)分析问题FLAC.docVIP

3D的动力（爆破）分析问题 我刚在做一个隧道爆破掘进的模拟，想这样做：q0HG tZ9u!tIl1.先模拟原岩应力状态V0pvT\mo mohrv)p4FvX;岩体参数V%ws~c r}T!]1r7F+wpro bulk 4.2e6 she 9.23e5 fric 12 coh 1e4 ten 9.6e3 range z 18.1 17.2,j)_ BD0VEini density 1800 range z 18.1 17.2y(c:f4whhY!s3rdpro bulk 7.02e6 she 1.83e6 fric 7.2 coh 4.7e3 ten 4.7e3 range z 17.2 13.6 {+u(gHf T^k\ini density 1650 range z 17.2 13.6 R4kg8ek_3\bpro bulk 9.9e6 she 2.82e6 fric 29 coh 0 ten 0 range z 13.6 7.1 E l Ny U7T XK K+Ptini density 1900 range z 13.6 7.1f/IL;I)s1\g.Lpro bulk 4.37e7 she 1.46e7 fric 30 coh 3e4 ten 3e4 range z 7.1 6.4w0Ob#S6L~Gini density 1990 range z 7.1 6.4 Jq^!ec \!cprop bulk 4.95e7 she 1.65e7 fric 38 coh 5e4 ten 5e4 range z 6.4 -0.4Q6z!@/PzDini density 2100 range z 6.4 -0.4HB$O-g}gA#tMi\prop bulk 6.52e7 she 2.67e7 fric 40 coh 4e4 ten 4e4 range z -0.4 -9.1V_0]T5D ~Ee.h5iini density 2480 range z -0.4 -9.1PS]r0Vu0v4N/{prop bulk 6.78e7 she 3.12e7 fric 43 coh 5e4 ten 5e4 range z -9.1 -17.68lZ-n+W` }%[ini density 2510 range z -9.1 -17.60^#Gk)b1}Dh/z*Cprop bulk 6.53e8 she 3.37e8 fric 44 coh 5e6 ten 5e6 range z -17.6 -26.9mGJ@9p8zIini density 2520 range z -17.6 -26.9(cE0`G x*@4@0_ ];设定边界条件(上边界z自由)Y$o?W:e!Lmfix z range z -26.8 -270?cN9c Hmfix x range x -39.9 -40.1dl-Ralq D3i kfix x range x 39.9 40.1,d%l2mJP QXYRfix y range y -0.1 0.1]YXF ?Gfix y range y 39.9 40.19s@u @e[N:D3zset grav 0 0 -9.8pj;w#^J_?3L;施加地面面状荷载#O+L t6v)~gapply szz -50e3 range z 18 18.23Q9W9kY}b~,Wsol rat 1e-5u$e$G \3~xsave geom.sav~(THBDM$|2.模拟爆破掘进T BI_ JNy QnewL c-C1ayu)W,EuMrest geom.sav ]]^+d!B2W4@config dyn$qOfK^pxMset dyn on%u _q9nF,wini xdisp 0 ydisp 0 zdisp 0On3w ^*l;[ryini xvel 0 yvel 0 zvel 0kTG$[0hNah-~ini state 05D I:^3Nvdef setup#Jt S~R~ {omega = 2.0 * pi * freq!F!d0Q(P[j8PendF*?2^*] CT e,Esetup|e4GW){3YC;y向速度时程曲线z \\-a8mkdef wavegu aK|#Qwave = 1/2.718282^(1.5*pi*dytime)*ampl*sin(omega * dytime) ;大概震动1.3s1MhD ?z I8eendCSV^ M r2YXm;Jset freq = 40.625 ampl = 0.04765P `C\8V%XwU1RksetupQ8A:\/g:i i }5m;p1为隧道