1负特征值问题 THE SYSTEM MATRIX HAS 8 NEGATIVE EIGENVALUES. 负特征值是非线性分析的必然产物。所以不必大惊小怪,甚至久而久之,对于你熟悉的问题,你都会视而不见了。若出了问题,可先检查下有没有伴随的 numerical sigularity(数值奇异)和 Zero pivot(零主元)产生。如果没有,可以参考这几个方面: 1).刚体...
在使用通用有限元软件(如Abaqus,lsdyna,ansys)进行隐式分析计算(或静力分析,或动力学初始状态求解)时,对于复杂装配体模型,大家或多或少会遇到以下警告信息: “***WARING:THE SYSTEM MATRIX HAS *NEGTIVE EIGENVALUES.”即警告:系统矩阵出现了负特征值。往往产生这样的警告后,计算便很难收敛了。但也有例外,在接触...
system matrix 系数矩阵; 实用场景例句 全部 Q : WARNING : THE SYSTEM MATRIX HAS 1 NEGATIVE EIGENVALUES. 一般在什么情况下会发生系统矩阵出现负特征值? 互联网 Presents a torsional vibration model during the flameout of a cylinder with the system matrix method. ...
1 负特征值问题:THE SYSTEM MATRIX HAS 8 NEGATIVE EIGENVALUES.负特征值是非线性分析的必然产物。所以不必大惊小怪,甚至久而久之,对于你熟悉的问题,你都会视而不见了。若出了问题,可先检查下有没有伴随的numerical sigularity(数值奇异)和 Zero pivot(零主元)产生。如果没有,可以参考这几个方面: 1).刚体位...
***WARNING: THE SYSTEM MATRIX HAS 1 NEGATIVE EIGENVALUES. NEGATIVE EIGENVALUES MEAN THAT THE SYSTEM MATRIX IS NOT POSITIVE DEFINITE: FOR EXAMPLE, A BIFURCATION (BUCKLING) LOAD MAY HAVE BEEN EXCEEDED. ***WARNING: SOLVER PROBLEM. NUMERICAL SINGULARITY WHEN PROCESSING NODE SPLINE-2-1.98 D....
the system matrix has 31 negative eigenvalues.怎么解决 我来答 1个回答 #热议# 公司那些设施可以提高员工幸福感?ji...2@163.com 2017-04-18 知道答主 回答量:4 采纳率:0% 帮助的人:2236 我也去答题访问个人页 展开全部 这就是英语 已赞过 已踩过< 你对这个回答的评价是? 评论 收起 ...
***WARNING: THE SYSTEM MATRIX HAS 1NEGATIVE EIGENVALUES. 下面的应力应变关系曲线是一个负斜率的例子: *Material, name=Steel *Plastic 418.,0. 780.,0.095 500,0.15 另外,如果材料是不可压缩性的(例如金属材料),在弹塑性分析中使用二次完全积分单元(C3D20)容易产生体积自锁。如果使用二次减缩积分单元(C3D...
***WARNING: THE SYSTEM MATRIX HAS 1 NEGATIVE EIGENVALUES. In an eigenvalue extraction step the number of negative eigenvalues is equal to the number of eigenvalues below the current shift point. This may be used to check that eigenvalues have not been missed. Note: the lanczos eigensolver ...
At\(E_0 (0,0)\)The Jacobian matrix has eigenvalues strictly negative (which are\(-l\)and\(-d\)), thus\(E_0\)is stable in nature. At\(E_1 (1,0)\)the Jacobian matrix has eigenvalues\(l-1\)and\(\frac{\beta }{1+a}-d\), thus\(E_1\)is stable node if\(\beta < d(...
This result has already been found in [78, 81]. Then, the eigenvalues of the M=C^TC matrix are equal to (1, 1, 1) and the operator {\mathfrak {m}}_{12}[C]=2, showing a maximal violation of the Bell inequalities. Similarly, the resulting concurrence is still maximal,...