一个费解的警告信息！

suking_2009

警告信息内容：Both solid model and finite element model boundary conditions have been
applied to this model.  As solid loads are transferred to the nodes or  
elements, they can overwrite directly applied loads.

1、这到底是什么意思？指的是我哪些地方错了？
2、solid model 和 finite element model有什么区别啊？

感谢各位的热心相助了~

Rainyboy

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没什么错，通俗的说就是你既在几何模型上指定了载荷又在有限元模型上指定了载荷，ANSYS怕这两个操作相互干扰，比如你既在A面上指定了压力P，又在A面的某个节点上指定了集中力F，那么F很可能就覆盖了压力P分到该节点上的分量，但是如果你在A面指定了压力p，又在B面的一个节点上指定了集中载荷F，一般就不会产生影响，就可以忽略这个警告。
纯属个人理解哈。

suking_2009

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呵呵，谢谢，我暂时赞同你的观点！哈哈~还麻烦你帮我看下这段命令流，进行了荷载的分步，问题又是大大的来了，我想是荷载施加的问题，但是~！我~稀里糊涂了。希望你点拨哈子~


fini
/cle
/units,si

h=26.5             !定义相关变量
h1=30.45
h2=15.48
h3=11
gm=1000

/prep7             !定义单元材料及材料属性

et,1,plane42       !外围混凝土单元
mp,ex,1,2.55e10
mp,nuxy,1,0.167    !假定泊松比
mp,dense,1,2400
keyopt,1,3,0
keyopt,1,5,0
keyopt,1,6,0

et,2,plane42       !浆砌石单元
mp,ex,2,1.15e10
mp,nuxy,2,0.16     !假定泊松比比混凝土小
mp,dense,2,2300
keyopt,2,3,0
keyopt,2,5,0
keyopt,2,6,0

et,3,plane42        !基岩单元
mp,ex,3,6e10
mp,nuxy,3,0.167
mp,dense,3,2660
keyopt,3,3,0
keyopt,3,5,0
keyopt,3,6,0

!定义关键点
!坝剖面
K,1,0,0
K,2,5,0
K,3,6,1
K,4,27.6,1
K,5,27.6,9.3
!材料分线
K,10,6,1.5
K,11,27.1,1.5
K,12,27.1,6.2
K,13,17.58198,6.2
K,14,3.38198,25
K,15,0.6,25
K,16,0.6,8
K,17,6,8
K,18,8.51815,18.2
K,19,0.6,18.2
K,20,0,25
K,21,27.6,6.2
!溢流面
!圆弧段
K,6,27.1,9.3
k,7,16.5011,10.16
!中间直线段
K,8,7.87419,21.66615
!顶部曲线段
K,9,0,25.98761
K,30,1.16549,26.5
K,31,5.16549,24.64157
!灌浆廊道
!圆心
K,22,2.5,3
K,23,2.5,5.25
K,24,2,5.25
K,25,5.5,5.25
K,26,5,5.25
K,27,5,3
K,28,3.75,5.25
k,29,22.1811,7.32

circle,28,1.25,,,180
circle,28,1.75,,,180
larc,6,7,29
larc,30,31,14,6.259
larc,9,30,15,1.32
nummrg,all

!地基
K,32,-0.6799,0.97128
K,37,-40.4299,0.97128
K,34,-40.4299,-52.02872
K,35,80.6,-52.02872
K,38,80.6,1

!定义面
a,15,14,18,19
a,19,18,13,12,11,10,17,16
a,9,20,15,14,18,13,12,21,5,6,7,8,31,30
a,20,15,19,16,17,10,11,12,21,4,3,2,1
a,22,27,26,25,36,24,23
asba,4,5
a,22,27,26,25,36,24,23
a,23,33,26,27,22
asba,4,5
a,37,34,35,38,4,3,2,1,32

wprota,,-90
wpoff,6,,8
asbw,2
wpoff,11.51815,,-1.8
asbw,8

wpoff,9.58185,,-4.7
asbw,6
wpoff,-21.1,,-0.5
asbw,10
wpoff,-6,,7
asbw,11
wpcsys,-1



esize,1                  !@浆砌石划分网格
mshape,0,2d
mshkey,1
allsel
asel,s,,,5
amap,5,18,19,16,39
amesh,5
allsel
asel,s,,,2
amap,2,39,17,40,13
amesh,2
allsel
asel,s,,,9
amap,9,40,10,11,12
amesh,9

esize,1                  !外围混凝土划分网格
mshape,0,2d
mshkey,1
asel,s,,,8
amesh,8
asel,s,,,12
amap,12,42,43,41,4
amesh,12
asel,s,,,6
amap,6,43,1,2,3
amesh,6
asel,s,,,10
amap,10,44,16,15,20
amesh,10

esize,1
mshape,0,2d
mshkey,0
asel,s,,,13
amesh,13

esize,2                !基岩划分网格
mshape,0,2d
mshkey,1
asel,s,,,4
amap,4,37,34,35,38
amesh,4

esize,0.1               !现浇拱划分网格
mshape,0,2d
mshkey,1
asel,s,,,7
amap,7,24,23,26,25
amesh,7

esize,0.5                 !内部C10划分网格
mshape,0,2d
mshkey,1
asel,s,,,1
amesh,1

esize,1                 !溢流面划分网格
mshape,0,2d
mshkey,0
asel,s,,,3
amesh,3


!加载及求解
allsel
FINISH  
/SOL
SAVE

!施加约束
nsel,all
nsel,s,loc,x,-40.4299
nsel,r,loc,y,-52.02872,0.97128
d,all,ux
nsel,all
nsel,s,loc,x,80.6
nsel,r,loc,y,-52.02872,0.97128
d,all,ux
nsel,all
nsel,s,loc,x,-40.4299,80.6
nsel,r,loc,y,-52.02872
d,all,all

!施加重力
TIME,1
ALLSEL
acel,,9.8

SOLVE


TIME,2
allsel
sfl,41,pres,gm*9.8*h1
sftran
allsel
sfl,39,pres,gm*9.8*h2
sftran

nsel,all
nsel,s,loc,x,0
sfgrad,pres,,y,0,-gm*9.8
sf,all,pres,gm*9.8*(h1+1)

nsel,all
nsel,s,loc,x,27.6
nsel,r,loc,y,1,9.3
sfgrad,pres,,y,1,-gm*9.8
sf,all,pres,gm*9.8*h2

nsel,all
nsel,s,loc,y,9.3
sfgrad,pres,,x,27.6,1
sf,all,pres,gm*9.8*(h2-9.3)

nsel,all
nsel,s,loc,x,11.7611,16.5011
nsel,r,loc,y,10.16,16.48
sfgrad,pres,,y,16.48,-gm*9.8
sf,all,pres,

nsel,all
nsel,s,loc,x,16.5011,22.1811
nsel,r,loc,y,7.32,10.16
sfgrad,pres,,y,10.16,-gm*9.8
sf,all,pres,gm*9.8*(h2-10.16)

nsel,all
nsel,s,loc,x,22.1811,27.1
nsel,r,loc,y,7.32,9.3
sfgrad,pres,,y,9.3,-gm*9.8
sf,all,pres,gm*9.8*(h2-9.3)

nsel,all
nsel,s,loc,x,0
nsel,r,loc,y,0,11
sfgrad,pres,,y,11,-770*9.8
sf,all,pres,

esel,all
esel,s,,,411,416,1
sfgrad,pres,,x,0,-24777
sfe,all,3,pres,0,308201
esel,all
sfe,411,4,pres,0,0.24*gm*9.8*(h1-h2)+9.8*gm*h2-1524.22*2
esel,all
sfe,405,4,pres,0,0.24*gm*9.8*(h1-h2)+9.8*gm*h2-1524.22*3
esel,all
esel,s,,,381,404,1
sfgrad,pres,,x,0,-1524.22
sfe,all,2,pres,0,0.24*gm*9.8*(h1-h2)+9.8*gm*h2-1524.22*4

SOLVE

Rainyboy

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瞬态动力学？第一秒施加重力，第二秒施加水压？为什么呢？

suking_2009

Rainyboy 发表于 2010-11-1 09:30

                               
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回复 suking_2009 的帖子

瞬态动力学？第一秒施加重力，第二秒施加水压？为什么呢？

就是创建荷载步，是为了消除由重力产生的位移场，因为这个位移是之前就已经发生了的，是初始位移。我的命令流没有写完~正在思考中~lcoper命令还未添加就已经出现很多错误了