abaqus2020的python二次开发的分割PartitionCell的边界选取findAt问题

cdefgh

请教一下各位，关于abaqus的python二次开发命令PartitionCellByPlanePointNormal，发现对某一部件执行该命令（分割部件）的for循环中，固定选取一边作为法线方向，有的时候没有报错有的时候却出现报错，每次循环都要对该边进行重新定义对象变量才保证绝对不会报错，但是循环次数太多则会有效率问题，如何解决比较好。
我发现对一个立方体的三个方向进行10等份分割，尝试运行了下面三种代码都中途中断，不知是什么原因。

1. m = mdb.Model(name='cube')
   
2. s = m.ConstrainedSketch(name='square',sheetSize=100.0)
   
3. s.Line(point1=(0.0,0.0),point2=(10.0,0.0))
   
4. s.Line(point1=(10.0,0.0),point2=(10.0,10.0))
   
5. s.Line(point1=(10.0,10.0),point2=(0.0,10.0))
   
6. s.Line(point1=(0.0,10.0),point2=(0.0,0.0))
   
7. p=m.Part(name='cube',dimensionality=THREE_D,type=DEFORMABLE_BODY)
   
8. p.BaseSolidExtrude(sketch=s,depth=10.0)
   
9. e1 = p.edges.findAt(((0.5,0.0,0.0),))
   
10. e2 = p.edges.findAt(((0.0,0.5,0.0),))
    
11. e3 = p.edges.findAt(((0.0,0.0,0.5),))
    
12. for i in range(1,10):
    
13.     p.PartitionCellByPlanePointNormal(point=(i,0,0),normal=e1[0],cells=p.cells)
    
14.     p.PartitionCellByPlanePointNormal(point=(0,i,0),normal=e2[0],cells=p.cells)
    
15.     p.PartitionCellByPlanePointNormal(point=(0,0,i),normal=e3[0],cells=p.cells)

复制代码

1. m = mdb.Model(name='cube')
   
2. s = m.ConstrainedSketch(name='square',sheetSize=100.0)
   
3. s.Line(point1=(0.0,0.0),point2=(10.0,0.0))
   
4. s.Line(point1=(10.0,0.0),point2=(10.0,10.0))
   
5. s.Line(point1=(10.0,10.0),point2=(0.0,10.0))
   
6. s.Line(point1=(0.0,10.0),point2=(0.0,0.0))
   
7. p=m.Part(name='cube',dimensionality=THREE_D,type=DEFORMABLE_BODY)
   
8. p.BaseSolidExtrude(sketch=s,depth=10.0)
   
9. for i in range(1,10):
   
10.     e1 = p.edges.findAt(((0.5,0.0,0.0),))
    
11.     e2 = p.edges.findAt(((0.0,0.5,0.0),))
    
12.     e3 = p.edges.findAt(((0.0,0.0,0.5),))
    
13.     p.PartitionCellByPlanePointNormal(point=(i,0,0),normal=e1[0],cells=p.cells)
    
14.     p.PartitionCellByPlanePointNormal(point=(0,i,0),normal=e2[0],cells=p.cells)
    
15.     p.PartitionCellByPlanePointNormal(point=(0,0,i),normal=e3[0],cells=p.cells)

复制代码

1. m = mdb.Model(name='cube1')
   
2. s = m.ConstrainedSketch(name='square',sheetSize=100.0)
   
3. s.Line(point1=(0.0,0.0),point2=(10.0,0.0))
   
4. s.Line(point1=(10.0,0.0),point2=(10.0,10.0))
   
5. s.Line(point1=(10.0,10.0),point2=(0.0,10.0))
   
6. s.Line(point1=(0.0,10.0),point2=(0.0,0.0))
   
7. p=m.Part(name='cube',dimensionality=THREE_D,type=DEFORMABLE_BODY)
   
8. p.BaseSolidExtrude(sketch=s,depth=10.0)
   
9. for i in range(1,10):
   
10.     e1 = p.edges.findAt(((i,0.0,0.0),))
    
11.     e2 = p.edges.findAt(((0.0,i,0.0),))
    
12.     e3 = p.edges.findAt(((0.0,0.0,i),))
    
13.     c1 = p.cells.findAt(((i,0.0,0.0),))
    
14.     c2 = p.cells.findAt(((0.0,i,0.0),))
    
15.     c3 = p.cells.findAt(((0.0,0.0,i),))   
    
16.     p.PartitionCellByPlanePointNormal(point=(i,0,0),normal=e1[0],cells=c1)
    
17.     p.PartitionCellByPlanePointNormal(point=(0,i,0),normal=e2[0],cells=c2)
    
18.     p.PartitionCellByPlanePointNormal(point=(0,0,i),normal=e3[0],cells=c3)

复制代码

hwy87

代码有问题吧   一般 要需要可以联系QQ664887646 讨论哦

NY-Cxj

以最后一段代码为例：做以下尝试，如有问题可以加Q：2812468512
    e1 = p.edges.findAt(((i,0.0,0.0),))
    c1 = p.cells.findAt(((i,0.0,0.0),))
    p.PartitionCellByPlanePointNormal(point=(i,0,0),normal=e1[0],cells=c1)
    e2 = p.edges.findAt(((0.0,i,0.0),))
    c2 = p.cells.findAt(((0.0,i,0.0),))
    p.PartitionCellByPlanePointNormal(point=(0,i,0),normal=e2[0],cells=c2)
    e3 = p.edges.findAt(((0.0,0.0,i),))
    c3 = p.cells.findAt(((0.0,0.0,i),))   
    p.PartitionCellByPlanePointNormal(point=(0,0,i),normal=e3[0],cells=c3)

duncanusn

e1 = p.edges.findAt(((0.5,0.0,0.0),))
e2 = p.edges.findAt(((0.0,0.5,0.0),))
e3 = p.edges.findAt(((0.0,0.0,0.5),))

這個用找出來的edgeArray記得用sets存起來，下次訪問的時候透過p.sets['set-1'].edges[0]去抓取，就不會跑掉，試試看唄

cdefgh

我后来用建立基准轴的方法解决了问题，感谢各位解答