昨天练习遇到一问题，给beam加载加不上，说没有定义单元，
可是我有定义单元类型啊 beam tapered 44 (3d)::l
不知道咋回事::l:-d作者: sfszj 时间: 2003-9-11 12:42
没用过这个单元，看看这个例子如何？来自校验手册
Test Case
A tapered cantilever plate of rectangular cross-section is subjected to a load F at its tip. Find the maximum deflection δ and the maximum principal stress σ1 in the plate.
Figure 34.1. Beam Problem Sketch
Material Properties
E = 30 x 106 psi
υ = 0.0
Geometric Properties
L = 20 in
d = 3 in
t = 0.5 in
Loading
F = 10 lbs
Analysis Assumptions and Modeling Notes
The problem is solved first using quadrilateral shell elements (SHELL63) and then using tapered beam elements (BEAM44). For the quadrilateral shell elements (used in triangular form), nodal coupling is used to ensure symmetry. For the beam elements, the area and Y dimension of the beam are not used and are input as 1.0. Node 12 is arbitrarily located at Z = 1.0 in order to define the orientation of the beam.
Results Comparison
Target ANSYS Ratio
SHELL63 Deflection, in -0.042667 作者: sfszj 时间: 2003-9-11 12:42
示意图：作者: sfszj 时间: 2003-9-11 12:42
命令流：
/COM,ANSYS MEDIA REL. 7.1 (03-13-2003) REF. VERIF. MANUAL: REL. 7.1
/VERIFY,VM34
/PREP7
/TITLE, VM34, BENDING OF A TAPERED PLATE (BEAM)
! INTROD. TO STRESS ANALYSIS, HARRIS, 1ST PRINTING, PAGE 114, PROB. 61
! PLATE ELEMENTS (SHELL63)
ANTYPE,STATIC ! STATIC ANALYSIS
ET,1,SHELL63,2
R,1,.5 ! THICKNESS = 0.5
MP,EX,1,30E6
MP,NUXY,1,0 ! POISSON'S RATIO IS ZERO
N,1
N,8,20,-1.5
FILL
N,11
N,18,20,1.5
FILL
E,1,2,12
E,2,3,12
E,13,12,3
E,3,4,14
E,14,13,3
EGEN,3,2,2,5
CP,1,UZ,2,12 ! COUPLE APPROPRIATE DEGREES OF FREEDOM
CP,2,ROTY,2,12
CPSGEN,6,1,1,2 ! GENERATE 6 SETS OF EQUATIONS
OUTPR,ALL,ALL
D,8,ALL,,,18,10
D, ALL,ROTX,0 ! REMOVE "TORSIONAL" DEGREES OF FREEDOM
F,1,FZ,-10
FINISH
/SOLU
SOLVE
FINISH
/POST1
ETABLE,STRS,S,1 ! STORE S1(TOP) FOR SHELL63
ESORT,STRS ! SORT ELEMENTS BASED ON S1(TOP)
*GET,SMAX,SORT,,MAX ! GET MAXIMUM S1 AS SMAX
PRNSOL,DOF ! PRINT NODAL DISPLACEMENTS
LFT_NODE = NODE (0,0,0)
*GET,DEFL,NODE,LFT_NODE,U,Z
*DIM,LABEL,CHAR,2,2
*DIM,VALUE_C1,,2,3
LABEL(1,1) = 'DEFLECTI','MX_PRIN_'
LABEL(1,2) = 'ON (in) ','STRS psi'
*VFILL,VALUE_C1(1,1),DATA,-.042667,1600
*VFILL,VALUE_C1(1,2),DATA,DEFL,SMAX
*VFILL,VALUE_C1(1,3),DATA,ABS(DEFL/.042667 ) ,ABS( SMAX/1600 )
SAVE,TABLE_1
FINISH
/CLEAR, NOSTART
/PREP7
/TITLE, VM34, BENDING OF A TAPERED PLATE (BEAM)
! TAPERED BEAM ELEMENTS (BEAM44)
ANTYPE,STATIC ! STATIC ANALYSIS
ET,1,BEAM44
*DO,I,1,10 ! CREATE DO LOOP FOR REAL CONSTANTS
R,I, 1,1,.003125*(I-1),.25,1
RMORE,1,1,.003125* I ,.25,1
*ENDDO
RMOD,1,3,.5E-3 ! GIVE FREE END A POSITIVE MOMENT OF INERTIA
RLIST
MP,EX,1,30E6
MP,GXY,1,30E6/2.6
N,1
N,11,20
FILL
N,12,,,1 ! NODE 12 FOR ALIGNING BEAM AXES
NGEN,10,1,12 ! NODES 12 TO 21 ARE COINCIDENT
E,1,2,12
EGEN,10,1,1,,,,,1 ! GENERATE ELEMENTS WITH REAL CONSTANT INCREASED BY 1
D,11,ALL
D,1,UY,,,10,,ROTX,ROTZ
F,1,FZ,-10
OUTPR,ALL,ALL
FINISH
/SOLU
SOLVE
FINISH
/POST1
ETABLE,STRS,NMISC,1 ! STORE SMAX (MAXIMUM STRESS) FOR BEAM44
ESORT,STRS ! SORT ELEMENTS BASED ON SMAX (MAXIMUM STRESS)
*GET,SMAX,SORT,,MAX ! GET MAXIMUM STRESS AS SMAX
PRNSOL,DOF ! PRINT NODAL DISPLACEMENTS
LFT_NODE = NODE (0,0,0)
*GET,DEFL,NODE,LFT_NODE,U,Z
*DIM,LABEL,CHAR,2,2
*DIM,VALUE_C2,,2,3
LABEL(1,1) = 'DEFLECTI','MX_PRIN_'
LABEL(1,2) = 'ON (in) ','STRS psi'
*VFILL,VALUE_C2(1,1),DATA,-.042667,1600
*VFILL,VALUE_C2(1,2),DATA,DEFL,SMAX
*VFILL,VALUE_C2(1,3),DATA,ABS(DEFL/.042667 ) ,ABS( SMAX/1600 )
SAVE,TABLE_2
RESUME,TABLE_1
/COM
/OUT,vm34,vrt
/COM,------------------- VM34 RESULTS COMPARISON ---------------
/COM,
/COM, | TARGET | ANSYS | RATIO
/COM,
/COM,RESULTS USING SHELL63:
/COM,
*VWRITE,LABEL(1,1),LABEL(1,2),VALUE_C1(1,1),VALUE_C1(1,2),VALUE_C1(1,3)
(1X,A8,A8,' ',F12.6,' ',F12.6,' ',1F5.3)
/NOPR
RESUME,TABLE_2
/GOPR
/COM,
*VWRITE,LABEL(1,1),LABEL(1,2),VALUE_C2(1,1),VALUE_C2(1,2),VALUE_C2(1,3)
(1X,A8,A8,' ',F12.6,' ',F12.6,' ',1F5.3)
/COM,-----------------------------------------------------------
/OUT
FINISH
*LIST,vm34,vrt作者: gussds 时间: 2003-9-13 22:12
正在考虑变截面梁的问题，正好也在分析这个算例。用shell单元没什么好说的。对于用beam44单元，它只是准确定义了平面内的抗弯刚度，对于梁在片面内的宽度并没有按实际情况取为了1。这在仅考虑弯曲变形的情况下是可以的。如果想准确定义实际的宽度，好象还是有点麻烦，因为j点到上表面的厚度是根据i点来定义的，见附图。好象还有点麻烦，应该有办法的，正在考虑中。作者: 有谁共鸣 时间: 2003-11-27 11:33
强烈支持，多讨论大家分析之中遇到的问题
