FLAC3D学习（SIMWE论坛君之枫整理）
gen zone reflect dip 90 dd 270 ori 15 0 -15.75 ; 摩尔库仑模型 model mohr
macro niyan 'bulk 3.2825e3 shear 1.5150e3 fric 20 coh 0.07' macro meiceng 'bulk 2.1735e3 shear 8.8917e2 fric 19 coh 0.05' prop niyan range z =-13.5 0 prop meiceng range z =-15.5 -13.5 prop niyan range z = -30 -15.5 ;为开挖定义整体部分
range name in z -15.75 0 cyl end1 15 0 -15.75 end2 15 6 -15.75 rad 2.25 range name hangdao x 12.75 17.25 y 0 6 z -16.50 -15.75 ;进行加载
fix x y z range z -30.001 -29.999 app sxx=-20 range x -0.001 0.001 app sxx=-20 range x 29.999 30.001 app syy=-20 range y -0.001 0.001 app syy=-20 range y 5.999 6.001 app szz=-20 range z -0.001 0.001 solve （1） model null range in model null range hangdao
range name in z -15.75 0 cyl end1 15 0 -15.75 end2 15 6 -15.75 rad 2.25 range name hangdaoliangce1 x 12.749 12.751 y 0 6 z -16.50 -15.75 range name hangdaoliangce2 x 17.249 17.251 y 0 6 z -16.50 -15.75 range name hangdao x 12.75 17.25 y 0 6 z -16.501 -16.499 apply nstress -0.05 plane dip 90 dd 0 range in ; 给巷道底部加压力
app nstress -0.05 plane dip 90 dd 0 range hangdao ; 给巷道直墙部分（左侧）加压力
apply nstress -0.05 plane dip 90 dd 0 range hangdaoliangce2 ; 给巷道直墙部分（右侧）加压力
apply nstress -0.05 plane dip 90 dd 0 range hangdaoliangce1;对点进行监测 hist unbal
hist gp xdisp 12.75 0 -15.75 :监测左帮 hist gp zdisp 15 0 -13.5 ;监测顶板 hist gp zdisp 15 0 -16.5 ;监测底板 solve
问：第一个solve平衡了,是第二个solve不能平衡，这是为什么？如果我把支护力加大，
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FLAC3D学习（SIMWE论坛君之枫整理） 也不能平衡？
回答：我遇到的不收敛的原因，有两个：
1）检查你的模型是不是没有建好，相互之间是不是没有连接好； 2）仔细检查你的参数 ，是不是已经发生了破坏。 回答：力施加有问题，隧道底部隆起破坏，所以无法计算
问：我这个模型就是该模型就是要出现巷道底臌现象，如果按照你的意思，力应该怎样施加才对？
回答：如果无法收敛，应该用step命令，关键问题是要运行多少步。 3.问： 命令流如下： g 10 10 m e def distort loop i (2,jzones) loop j (2,jzones)
x(i,j) =x(i,j)+mag*(urand-0.5) y(i,j)=y(i,j)+mag*(urand-0.5) end_loop end_loop end
prop d 1000 sh 1e8 bu 2e8 其中：
x(i,j) =x(i,j)+mag*(urand-0.5)中的urand是否是随机生成数的命令？ 回答：
urand ：random number drawn from uniform distribution between 0.0 and 1.0； grand ：random number drawn from normal distribution: mean = 0.0; standard deviation = 1.0
4. 问：怎么能把一个PLOT的图像数据导出来以便用其他软件绘图？
回答：用set log on 命令，把数据导出来，转到excel里处理一下，然后用surfer或者什么作图软件绘制就行了。
5. 问：用命令建立模型后，如何显示点的坐标？ 回答：使用plo blo gro gpnum on命令 6.interface问题
问：我加了一个层面，初始平衡就出现了这么大的位移？是怎么原因？ ;No1
gen zone brick p0 0 0 0 p1 10 0 0 p2 0 10 0 p3 0 0 8 & size 8 8 6 rat 1.02 1.02 1.02
gen zone brick p0 0 0 9 p1 10 0 9 p2 0 10 9 p3 0 0 11 &
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FLAC3D学习（SIMWE论坛君之枫整理） size 8 8 2 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,6) range z 0 11 group layer1 range z 9 11
;interface 1 face range plane norm (0,0,1) origin (0,0,8) dist 0.01 ini z add -1.0 range group layer1 ;No2
gen zone brick p0 0 0 11 p1 10 0 11 p2 0 10 11 p3 0 0 13 & size 8 8 2 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,1) range z 11 13 group layer2 range z 11 13
;interface 2 face range plane norm (0,0,1) origin (0,0,10) dist 0.01 ini z add -1.0 range group layer2 ;No3
gen zone brick p0 0 0 13 p1 10 0 13 p2 0 10 13 p3 0 0 14 & size 8 8 1 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,1) range z 13 14 group layer3 range z 13 14
;interface 3 face range plane norm (0,0,1) origin (0,0,12) dist 0.01 ini z add -1.0 range group layer3 ;No4
gen zone brick p0 0 0 14 p1 10 0 14 p2 0 10 14 p3 0 0 15 & size 8 8 1 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,1) range z 14 15 group layer4 range z 14 15
interface 4 face range plane norm (0,0,1) origin (0,0,13) dist 0.01 ini z add -1.0 range group layer4 ;No5
gen zone brick p0 0 0 15 p1 10 0 15 p2 0 10 15 p3 0 0 17 & size 8 8 2 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,15) range z 15 17 group layer5 range z 15 17
interface 5 face range plane norm (0,0,1) origin (0,0,14) dist 0.01 ini z add -1.0 range group layer5 ;No6
gen zone brick p0 0 0 17 p1 10 0 17 p2 0 10 17 p3 0 0 18 & size 8 8 1 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,1) range z 17 18 group layer6 range z 17 18
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FLAC3D学习（SIMWE论坛君之枫整理）
interface 6 face range plane norm (0,0,1) origin (0,0,16) dist 0.01 ini z add -1.0 range group layer6 ;No7
gen zone brick p0 0 0 18 p1 10 0 18 p2 0 10 18 p3 0 0 20 & size 8 8 2 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,1) range z 18 20 group layer7 range z 18 20
;interface 7 face range plane norm (0,0,1) origin (0,0,17) dist 0.01 ini z add -1.0 range group layer7 ;No8
gen zone brick p0 0 0 20 p1 10 0 20 p2 0 10 20 p3 0 0 24 & size 8 8 3 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,17) range z 20 24 group layer8 range z 20 24
;interface 8 face range plane norm (0,0,1) origin (0,0,19) dist 0.01 ini z add -1.0 range group layer8 ;No9
gen zone brick p0 0 0 24 p1 10 0 24 p2 0 10 24 p3 0 0 29 & size 8 8 4 rat 1.02 1.02 1.02
gen zone reflect dip 90 dd 90 origin(0,0,0) range z 24 29 group layer9 range z 24 29
;interface 9 face range plane norm (0,0,1) origin (0,0,23) dist 0.01 ini z add -1.0 range group layer9
;----------边界条件------------------------ fix x range x -20.1 -19.9 fix x range x 19.9 20.1 fix y range y -0.1 0.1 fix y range y 9.9 10.1 fix z range z -0.1 0.1
;-----------初始条件------------------------- apply szz -26e6 range z 27.9 28.1
initial szz -1.08e6 grad 0 0 2.6e4 range z 0 28 set gravity 0 0 -10 initial density 2.6e3
initial sxx -45.8e6 grad 0 0 2.6e4 range z 0 28 initial syy -45.8e6 grad 0 0 2.6e4 range z 0 28 ;--------------------------------------------- model mohr
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FLAC3D学习（SIMWE论坛君之枫整理）
;-----------材料属性-------------------------- ;-----------No1-----------------------
pro bulk 8e9 shear 7e9 fric 25 coh 7.5e6 ten 2.6e6 range z 0 8 ;-----------No2-------------------------
pro bulk 7.8e9 shear 3.5e9 fric 30 coh 6.5e6 ten 2.4e6 range z 8 10 ;------------No3---------------------
pro bulk 7e9 shear 3.0e9 fric 22 coh 5.5e6 ten 2.2e6 range z 10 12 ;------------No4-----------------------------
pro bulk 6e9 shear 2.3e9 fric 26 coh 5.5e6 ten 2.0e6 range z 12 13 ;-----------No5-----------------------------------
pro bulk 6.2e9 shear 2.0e9 fric 28 coh 5.2e6 ten 1.2e6 range z 13 14 ;-------------------No6------------------------
pro bulk 5.8e9 shear 2.5e9 fric 35 coh 6.0e6 ten 1.3e6 range z 14 16 ;-------------------No7------------------------
pro bulk 9e9 shear 3.2e9 fric 30 coh 10.3e6 ten 1.2e6 range z 16 17 ;-------------------No8------------------------
pro bulk 8.1e9 shear 3.1e9 fric 26 coh 8.0e6 ten 1.5e6 range z 17 19 ;-------------------No9------------------------
pro bulk 7.6e9 shear 2.6e9 fric 23 coh 8.5e6 ten 1.8e6 range z 19 23 ;-------------------No10------------------------
pro bulk 7.5e9 shear 2.1e9 fric 26 coh 8.0e6 ten 1.9e6 range z 23 28 ;int 1 prop ks 8e10 kn 8e10 fric 30 dil 6 tension 2e7 bslip=on ;int 2 prop ks 8e10 kn 8e10 fric 30 dil 6 tension 2e7 bslip=on ;int 3 prop ks 8e10 kn 8e10 fric 35 dil 8 tension 2e7 bslip=on int 4 prop ks 9e11 kn 9e11 fric 32 dil 5 tension 2e7 bslip=on int 5 prop ks 9e11 kn 9e11 fric 32 dil 5 tension 2e7 bslip=on int 6 prop ks 8e11 kn 8e11 fric 33 dil 5 tension 2e7 bslip=on ;int 7 prop ks 8e10 kn 8e10 fric 33 dil 5 tension 2e7 bslip=on ;int 8 prop ks 8e10 kn 8e10 fric 33 dil 5 tension 2e7 bslip=on ;int 9 prop ks 8e10 kn 8e10 fric 33 dil 5 tension 2e7 bslip=on hist unbal step 5000
apply szz -26e6 range z 27.9 28.1
initial szz -1.08e6 grad 0 0 2.6e4 range z 0 28 set gravity 0 0 -10 initial density 2.6e3
initial sxx -45.8e6 grad 0 0 2.6e4 range z 0 28 initial syy -45.8e6 grad 0 0 2.6e4 range z 0 28
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