Saturday, 18. March 2006, 14:07:20
Large Lateral Deflection of Unequal Stiffness Springs
两弹簧系统,受力F (F=5*sqrt(2)) 作用,求系统的应变能以及节点位移。
此问题的求解是通过增加质量,和
slow dynamics技术(包括估计临界阻尼)。组合单元(COMBIN40)用于提供x, y方向的阻尼。阻尼系数分别为 cx = 2 * sqrt(Kx * m) ,cy = 2 * sqrt(Ky * m) 这里 m 是任意值,这里假定为1, Kx 和 Ky 在求解之前不能得到,所以用 `Ky = k2 = 1N/cm` 和 `Kx = Ky/2 = 0.5 N/cm` 来近似。
注意到cx和cy的定义都是临界阻尼的定义式,根据临界阻尼的性质,当时间无限长时,单自由度系统的自由振动响应趋于0。kx和ky是如何选择的呢?如果更换别的值,得出的结果就完全不对了。 对此疑问很大
由于对载荷的抵抗是变形位置的函数,所以进行大变形分析。使用 POST1 获得结果
/COM,ANSYS MEDIA REL. 10.0 (05/31/2005) REF. VERIF. MANUAL: REL. 10.0
/VERIFY,VM9
/PREP7
/TITLE, VM9, LARGE LATERAL DEFLECTION OF UNEQUAL STIFFNESS SPRINGS
/COM, REF: G.N. VANDERPLAATS, "NUMERICAL OPTIMIZATION TECHNIQUES FOR
/COM, ENGINEERING DESIGN", PP 72-73, MCGRAW-HILL, 1984
ET,1,COMBIN14,,,2 ! UX AND UY DOF ELEMENT
ET,3,COMBIN40,,,,,,2 ! ALL MASS IS AT NODE J, UX DOF ELEMENT
ET,4,COMBIN40,,,2,,,2 ! ALL MASS IS AT NODE J, UY DOF ELEMENT
R,1,1 ! SPRING STIFFNESS = 1
R,2,8 ! SPRING STIFFNESS = 8
/COM USE COMBIN40 MASS, K, AND DAMPING C, TO APPROX. CRITICAL DAMPING
R,3,,1.41,1 ! C = 1.41, M = 1
R,4,,2,1 ! C = 2, M = 1
N,1
N,2,,10
N,3,,20
N,4,-1,10
N,5,,9
LOCAL,11,0,0,0,0,45 ! 定义一个局部坐标系,将 x 轴移动45度
NROTAT,2 ! 将节点 2 的节点坐标系旋转到当前的局部坐标系,这样即将施加的45度载荷就与局部坐标系x轴一致了。
E,1,2 ! ELEMENT 1 IS SPRING ELEMENT WITH STIFFNESS 1
REAL,2
E,2,3 ! ELEMENT 2 IS SPRING ELEMENT WITH STIFFNESS 8
TYPE,3
REAL,3
E,4,2 ! ELEMENT 3 IS COMBINATION ELEMENT WITH C = 1.41
TYPE,4
REAL,4
E,5,2 ! ELEMENT 4 IS COMBINATION ELEMENT WITH C = 2
NSEL,U,NODE,,2
D,ALL,ALL
NSEL,ALL
FINISH
/SOLU
ANTYPE,TRANS ! 瞬态动力学分析 FULL TRANSIENT DYNAMIC ANALYSIS
NLGEOM,ON ! 大变形 LARGE DEFLECTION
KBC,1 ! STEP BOUNDARY CONDITION
F,2,FX,7.071068 ! 载荷在旋转后的节点坐标系内 FORCE IS IN ROTATED NODAL COORDINATE SYSTEM
AUTOTS,ON
NSUBST,20
OUTPR,,20
OUTPR,VENG,20
TIME,15 ! ARBITRARY TIME FOR SLOW DYNAMICS
SOLVE
FINISH
/POST1
SET,,,,,15 ! USE ITERATION WHEN TIME = 15
ETABLE,SENE,SENE ! 应变能 STORE STRAIN ENERGY
SSUM ! 相加 SUM ALL ACTIVE ENTRIES IN ELEMENT STRESS TABLE
*GET,ST_EN,SSUM,,ITEM,SENE
PRNSOL,U,COMP ! PRINT DISPLACEMENTS IN GLOBAL COORDINATE SYSTEM
*GET,DEF_X,NODE,2,U,X
*GET,DEF_Y,NODE,2,U,Y
*DIM,LABEL,CHAR,3,2
*DIM,VALUE,,3,3
LABEL(1,1) = 'STRAIN E','DEF_X (C','DEF_Y (C'
LABEL(1,2) = ', N-cm ','m) ','m) '
*VFILL,VALUE(1,1),DATA,24.01,8.631,4.533
*VFILL,VALUE(1,2),DATA,ST_EN ,DEF_X,DEF_Y
*VFILL,VALUE(1,3),DATA,ABS(ST_EN/24.01), ABS(8.631/DEF_X), ABS(DEF_Y/4.533 )
/COM
/OUT,vm9,vrt
/COM,------------------- VM9 RESULTS COMPARISON ---------------------
/COM,
/COM, | TARGET | ANSYS | RATIO
/COM,
*VWRITE,LABEL(1,1),LABEL(1,2),VALUE(1,1),VALUE(1,2),VALUE(1,3)
(1X,A8,A8,' ',F10.3,' ',F10.3,' ',1F5.3)
/COM,----------------------------------------------------------------
/OUT
FINISH
*LIST,vm9,vrt