lecture3-AbaqusStandard中的非线性分析.ppt

,Abaqus/Standard中的非线性分析,第三讲,Dassault Systmes,2008,Introduction to Abaqus/Standard and Abaqus/Explicit,概述,非线性结构力学求解平衡方程非线性输入文件非线性悬臂梁分析的输出习题3：非线性静力学,非线性结构力学,Dassault Systmes,2008,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性结构力学,一些材料非线性的例子,非线性的来源 材料 非线性：非线性弹性塑性材料损伤失效机制等等。注意：如果预定义了温度或场变量，则材料与温度或场变量的相关性不引入非线性。,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性结构力学,自接触的例子：例子手册，压缩减震器,边界条件 非线性：接触问题在分析过程中边界条件变化。严重不连续形式的非线性。,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性结构力学,几何 非线性：大绕度和大变形大旋转结构不稳定（屈曲）预载荷效应,几何非线性例子：聚合物键盘罩,求解平衡方程,Dassault Systmes,2008,Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,典型的非线性问题具有所有三种形式的非线性。在方程中必须包括非线性项。一般的，每个自由度的非线性方程是耦合的。静态平衡的基本表达式为：由单元应力引起的加在节点上的内力I，与外力 P，必须平衡，即：,(Eq.3.1),Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,为了求解非线性平衡问题，在Abaqus/Standard中使用基于牛顿-拉普森技术的增量迭代法。假定前一步载荷增量的解，u0，为已知。假定在第i 次迭代之后，得到近似解ui。设ci+1为离散平衡方程的精确解和当前解之差（Equation 3.1），所以有：将方程3.2的左手边在近似解ui 附近，以泰勒级数方式展开，可以得到,(Eq.3.2),(Eq.3.3),Introduction to Abaqus/Standard and Abaqus/Explicit,略去高阶项，方程可以写为 其中 为切线刚度。解的下一次近似为注意：如果载荷与位移相关（比如，旋转的表面压力），刚度矩阵中包含载荷刚度的贡献。,求解平衡方程,Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,网格中的静力平衡施加规定的“载荷增量”。迭代，直到每个节点的所有节点力的和非常小。满足平衡后，更新模型状态。回到第1步，施加下一个载荷增量。,Introduction to Abaqus/Standard and Abaqus/Explicit,单自由度例子非线性弹簧求解u(P)或 P(u)典型的，在分析步中，载荷从0递增到PFINAL。时间经常从0变到1。,求解平衡方程,Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,牛顿-拉普森求解技术第一次迭代(i=1)假定前面收敛增量步的解u0，P0，为已知的。在当前增量步中，将一个小的增量P，载荷施加到结构上。Abaqus基于u0处的切线刚度K0 确定位移修正c1；前一增量步结束时，总载荷PTOTAL和内力间的关系为：,Introduction to Abaqus/Standard and Abaqus/Explicit,Abaqus更新模型的状态为u1，形成K1 并计算 I1。总载荷PTOTAL与内力 I1的差称为残差，R1:R1=PTOTAL-I1.如果 R1 在模型的每个自由度上都非常小(在容差范围之内)，结构就是平衡的。默认的容差R1必须小于结构对时间平均力的0.5%。Abaqus自动计算时间平均力。如果迭代不能得到收敛的解，Abaqus执行另外的迭代，以找到收敛的解。,求解平衡方程,Introduction to Abaqus/Standard and Abaqus/Explicit,第二次迭代(i=2)基于更新的刚度K1，计算新的位移纠正 c2，并且把新的残差R2与容差进行比较，察看在u2 处是否得到收敛解，。,求解平衡方程,Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,该过程将一直重复，直到力的残差在允许的容差之内。每次迭代i需要：形成切线刚度Ki。求解系统方程组，得到位移修正ci+1。修正位移的估计值：ui+1=ui+ci+1。基于ui+1 计算内力向量Ii+1。进行平衡收敛判断：是否Ri+1 在容差之内?是否,Introduction to Abaqus/Standard and Abaqus/Explicit,求解平衡方程,一般的，每个分析步(*STEP)需要几个增量步。,非线性输入文件,Dassault Systmes,2008,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性输入文件,*HEADING CANTILEVER BEAM EXAMPLE-LARGE DISPLACEMENT*NODE 1,0.,0.11,200.,0.*NGEN1,11,1*ELEMENT,TYPE=B211,1,3*ELGEN,ELSET=BEAMS1,5,2,1*BEAM SECTION,SECTION=RECT,ELSET=BEAMS,MATERIAL=MAT150.,5.*MATERIAL,NAME=MAT1*ELASTIC2.E5,.3*BOUNDARY1,1,6*AMPLITUDE,NAME=RAMP0.0,0.0,0.5,0.3,1.0,1.0,Introduction to Abaqus/Standard and Abaqus/Explicit,*RESTART,WRITE,FREQ=3*STEP,NLGEOM,INC=25APPLY POINT LOAD*STATIC0.1,1.0,0.001,1.0*CLOAD,AMPLITUDE=RAMP11,2,-1200.*NODE PRINT,FREQ=1U,RF*EL PRINT,FREQ=10S,E*NODE FILE,FREQ=5U*END STEP,和线性输入的主要不同,最小时间增量,最大时间增量,分析步的时间周期,建议的初始时间增量,先前定义的载荷幅值函数,和线性输入主要的不同,非线性输入文件,Introduction to Abaqus/Standard and Abaqus/Explicit,分析步和过程输入*STEP,NLGEOM,INC=25NLGEOM:包括所有由以下原因引起的几何非线性效应：大挠度、大旋转、大变形。预载荷（初始应力）。载荷刚度。如果上面列出的项不重要，应用NLGEOM选项得到的结果同没有应用NLGEOM选项得到的结果类似，但是求解的费用更高。INC=25:在本例中允许的最大增量为25：如果在施加全部载荷之前达到了最大增量数量，程序将会中止。保证程序的运行时间不会太长用户可以重新启动分析。默认值为100。,非线性输入文件,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性输入文件,对于所有的瞬态过程，都有类似的时间增量，它们是*STATIC*DYNAMIC*HEAT TRANSFER*VISCO*COUPLED TEMPERATURE-DISPLACEMENT*SOILS*MODAL DYNAMIC(只允许固定时间增量)*COUPLED THERMAL-ELECTRIC物理（真实）时间或名义时间的度量取决于过程和时间相关过程或率相关行为。,非线性悬臂梁分析的输出,Dassault Systmes,2008,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,状态(.sta)文件总结分析的过程显示计算过程中使用的自动时间增量。在作业运行的同时，可以检查状态文件。在每个成功的增量之后，记录一行。,SUMMARY OF JOB INFORMATION:STEP INC ATT SEVERE EQUIL TOTAL TOTAL STEP INC OF DOF IF DISCON ITERS ITERS TIME/TIME/LPF TIME/LPF MONITOR RIKS ITERS FREQ 1 1 1 0 3 3 0.100 0.100 0.1000 1 2 1 0 2 2 0.200 0.200 0.1000 1 3 1 0 2 2 0.350 0.350 0.1500 1 4 1 0 2 2 0.575 0.575 0.2250 1 5 1 0 3 3 0.913 0.913 0.3375 1 6 1 0 2 2 1.00 1.00 0.08750,Introduction to Abaqus/Standard and Abaqus/Explicit,自动的时间增量试探算法（基于多年的经验）控制时间积分的精度。在静力学分析中，基于迭代次数达到收敛。容易得到收敛解（比最大允许迭代数量少很多）：增加迭代步长不容易收敛或发散：减小增量步长否则：保持同样的增量步长自动的时间增量步长控制工作很好。如果没有特殊的原因，用户不要改变它。提示：对于高度非线性问题，推荐初始的时间增量为总时间增量的一小部分（比如10%）。,非线性悬臂梁分析的输出,Introduction to Abaqus/Standard and Abaqus/Explicit,信息(.msg)文件包括：所有的收敛控制：利用*CONTROLS选项覆盖默认值不是经常需要关于具体模型特征的细节：非默认模型特征使用NLGEOM参数重启动文件的写出频率所有的迭代细节,非线性悬臂梁分析的输出,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,求解器信息：数值奇异：这表明在线性方程组的求解过程中造成了大量数位的丢失，以至于结果是不可靠的。最常见的原因是静力应力分析过程中没有约 束的刚体模式。零主元：在线性方程组求解过程中，当没有相应的刚 度抵抗所施加的力时，会出现这样的信息。最常见的原因是没有约束的刚体模式和过度约束的自由度。负特征值：负特征值表明刚度矩阵是非正定的；比如，超过结构的屈曲载荷。,Numerical singularities,Zero pivots,Negative eigenvalues,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,有用的修理故障信息：最大残差的位置过度变形的位置接触条件变化的位置,Introduction to Abaqus/Standard and Abaqus/Explicit,S T E P 1 S T A T I C A N A L Y S I S SMALL DISPLACEMENT ANALYSIS AUTOMATIC TIME CONTROL WITH-A SUGGESTED INITIAL TIME INCREMENT OF 0.100 AND A TOTAL TIME PERIOD OF 1.00 THE MINIMUM TIME INCREMENT ALLOWED IS 1.000E-03 THE MAXIMUM TIME INCREMENT ALLOWED IS 1.00 CONVERGENCE TOLERANCE PARAMETERS FOR FORCE CRITERION FOR RESIDUAL FORCE FOR A NONLINEAR PROBLEM 5.000E-03 CRITERION FOR DISP.CORRECTION IN A NONLINEAR PROBLEM 1.000E-02 INITIAL VALUE OF TIME AVERAGE FORCE 1.000E-02 AVERAGE FORCE IS TIME AVERAGE FORCE ALTERNATE CRIT.FOR RESIDUAL FORCE FOR A NONLINEAR PROBLEM 2.000E-02 CRITERION FOR ZERO FORCE RELATIVE TO TIME AVRG.FORCE 1.000E-05 CRITERION FOR RESIDUAL FORCE WHEN THERE IS ZERO FLUX 1.000E-05 CRITERION FOR DISP.CORRECTION WHEN THERE IS ZERO FLUX 1.000E-03 CRITERION FOR RESIDUAL FORCE FOR A LINEAR INCREMENT 1.000E-08 FIELD CONVERSION RATIO 1.00,非线性悬臂梁分析的输出,Introduction to Abaqus/Standard and Abaqus/Explicit,CONVERGENCE TOLERANCE PARAMETERS FOR MOMENT CRITERION FOR RESIDUAL MOMENT FOR A NONLINEAR PROBLEM 5.000E-03 CRITERION FOR ROTATION CORRECTION IN A NONLINEAR PROBLEM 1.000E-02 INITIAL VALUE OF TIME AVERAGE MOMENT 1.000E-02 AVERAGE MOMENT IS TIME AVERAGE MOMENT ALTERNATE CRIT.FOR RESIDUAL MOMENT FOR A NONLINEAR PROBLEM 2.000E-02 CRITERION FOR ZERO MOMENT RELATIVE TO TIME AVRG.MOMENT 1.000E-05 CRITERION FOR RESIDUAL MOMENT WHEN THERE IS ZERO FLUX 1.000E-05 CRITERION FOR ROTATION CORRECTION WHEN THERE IS ZERO FLUX 1.000E-03 CRITERION FOR RESIDUAL MOMENT FOR A LINEAR INCREMENT 1.000E-08 FIELD CONVERSION RATIO 1.00 VOLUMETRIC STRAIN COMPATIBILITY TOLERANCE FOR HYBRID SOLIDS 1.000E-05 AXIAL STRAIN COMPATIBILITY TOLERANCE FOR HYBRID BEAMS 1.000E-05 TRANS.SHEAR STRAIN COMPATIBILITY TOLERANCE FOR HYBRID BEAMS 1.000E-05 SOFT CONTACT CONSTRAINT COMPATIBILITY TOLERANCE FOR PP0 5.000E-03 SOFT CONTACT CONSTRAINT COMPATIBILITY TOLERANCE FOR P=0.0 0.100 DISPLACEMENT COMPATIBILITY TOLERANCE FOR DCOUP ELEMENTS 1.000E-05 ROTATION COMPATIBILITY TOLERANCE FOR DCOUP ELEMENTS 1.000E-05,非线性悬臂梁分析的输出,Introduction to Abaqus/Standard and Abaqus/Explicit,TIME INCREMENTATION CONTROL PARAMETERS:FIRST EQUILIBRIUM ITERATION FOR CONSECUTIVE DIVERGENCE CHECK 4 EQUILIBRIUM ITERATION AT WHICH LOG.CONVERGENCE RATE CHECK BEGINS 8 EQUILIBRIUM ITERATION AFTER WHICH ALTERNATE RESIDUAL IS USED 9 MAXIMUM EQUILIBRIUM ITERATIONS ALLOWED 16 EQUILIBRIUM ITERATION COUNT FOR CUT-BACK IN NEXT INCREMENT 10 MAXIMUM EQUILIB.ITERS IN TWO INCREMENTS FOR TIME INCREMENT INCREASE 4 MAXIMUM ITERATIONS FOR SEVERE DISCONTINUITIES 12 MAXIMUM CUT-BACKS ALLOWED IN AN INCREMENT 5 MAXIMUM DISCON.ITERS IN TWO INCREMENTS FOR TIME INCREMENT INCREASE 6 CUT-BACK FACTOR AFTER DIVERGENCE 0.2500 CUT-BACK FACTOR FOR TOO SLOW CONVERGENCE 0.5000 CUT-BACK FACTOR AFTER TOO MANY EQUILIBRIUM ITERATIONS 0.7500 CUT-BACK FACTOR AFTER TOO MANY SEVERE DISCONTINUITY ITERATIONS 0.2500 CUT-BACK FACTOR AFTER PROBLEMS IN ELEMENT ASSEMBLY 0.2500 INCREASE FACTOR AFTER TWO INCREMENTS THAT CONVERGE QUICKLY 1.500 MAX.TIME INCREMENT INCREASE FACTOR ALLOWED 1.500 MAX.TIME INCREMENT INCREASE FACTOR ALLOWED(DYNAMICS)1.250 MAX.TIME INCREMENT INCREASE FACTOR ALLOWED(DIFFUSION)2.000 MINIMUM TIME INCREMENT RATIO FOR EXTRAPOLATION TO OCCUR 0.1000 MAX.RATIO OF TIME INCREMENT TO STABILITY LIMIT 1.000 FRACTION OF STABILITY LIMIT FOR NEW TIME INCREMENT 0.9500,非线性悬臂梁分析的输出,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,PRINT OF INCREMENT NUMBER,TIME,ETC.,EVERY 1 INCREMENTS RESTART FILE WILL BE WRITTEN EVERY 3 INCREMENTS THE MAXIMUM NUMBER OF INCREMENTS IN THIS STEP IS 25 LARGE DISPLACEMENT THEORY WILL BE USED LINEAR EXTRAPOLATION WILL BE USED CHARACTERISTIC ELEMENT LENGTH 40.0 DETAILED OUTPUT OF DIAGNOSTICS TO DATABASE REQUESTED PRINT OF INCREMENT NUMBER,TIME,ETC.,TO THE MESSAGE FILE EVERY 1 INCREMENTS EQUATIONS ARE BEING REORDERED TO MINIMIZE WAVEFRONT COLLECTING MODEL CONSTRAINT INFORMATION FOR OVERCONSTRAINT CHECKS COLLECTING STEP CONSTRAINT INFORMATION FOR OVERCONSTRAINT CHECKS,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,INCREMENT 1 STARTS.ATTEMPT NUMBER 1,TIME INCREMENT 0.100 CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 1 AVERAGE FORCE 1.251E+03 TIME AVG.FORCE 1.251E+03 LARGEST RESIDUAL FORCE-4.637E+03 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-1.84 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.-1.84 AT NODE 11 DOF 2 FORCE EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.AVERAGE MOMENT 7.200E+03 TIME AVG.MOMENT 7.200E+03 LARGEST RESIDUAL MOMENT 28.8 AT NODE 9 DOF 6 LARGEST INCREMENT OF ROTATION-1.382E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION-1.382E-02 AT NODE 11 DOF 6 ROTATION CORRECTION TOO LARGE COMPARED TO ROTATION INCREMENT.CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 2 AVERAGE FORCE 37.8 TIME AVG.FORCE 37.8 LARGEST RESIDUAL FORCE 0.215 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-1.84 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.-1.007E-02 AT NODE 11 DOF 1 FORCE EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.AVERAGE MOMENT 7.200E+03 TIME AVG.MOMENT 7.200E+03 LARGEST RESIDUAL MOMENT-0.346 AT NODE 5 DOF 6 LARGEST INCREMENT OF ROTATION-1.382E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION 5.898E-07 AT NODE 11 DOF 6 THE MOMENT EQUILIBRIUM EQUATIONS HAVE CONVERGED,0.005 6.25,.005 0.2,0.005 36,0.005 36,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,4次或更少次数的迭代（如果这样的情况再次出现，t可以增加）,CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 3 AVERAGE FORCE 37.7 TIME AVG.FORCE 37.7 LARGEST RESIDUAL FORCE-2.281E-06 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-1.84 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.3.349E-05 AT NODE 11 DOF 2 THE FORCE EQUILIBRIUM EQUATIONS HAVE CONVERGED AVERAGE MOMENT 7.200E+03 TIME AVG.MOMENT 7.200E+03 LARGEST RESIDUAL MOMENT 1.523E-05 AT NODE 7 DOF 6 LARGEST INCREMENT OF ROTATION-1.382E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION 3.637E-07 AT NODE 11 DOF 6 THE MOMENT EQUILIBRIUM EQUATIONS HAVE CONVERGED ITERATION SUMMARY FOR THE INCREMENT:3 TOTAL ITERATIONS,OF WHICH 0 ARE SEVERE DISCONTINUITY ITERATIONS AND 3 ARE EQUILIBRIUM ITERATIONS.TIME INCREMENT COMPLETED 0.100,FRACTION OF STEP COMPLETED 0.100 STEP TIME COMPLETED 0.100,TOTAL TIME COMPLETED 0.100,0.005 0.2,0.005 36,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,没有增加,INCREMENT 2 STARTS.ATTEMPT NUMBER 1,TIME INCREMENT 0.100 CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 1 AVERAGE FORCE 75.6 TIME AVG.FORCE 56.7 LARGEST RESIDUAL FORCE 0.861 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-1.84 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.-2.013E-02 AT NODE 11 DOF 1 FORCE EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.AVERAGE MOMENT 1.440E+04 TIME AVG.MOMENT 1.080E+04 LARGEST RESIDUAL MOMENT-1.38 AT NODE 5 DOF 6 LARGEST INCREMENT OF ROTATION-1.382E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION 3.914E-06 AT NODE 11 DOF 6 THE MOMENT EQUILIBRIUM EQUATIONS HAVE CONVERGED,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,4个连续增量步的迭代次数等于或小于4：t=1.5told,CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 2 AVERAGE FORCE 144.TIME AVG.FORCE 90.9 LARGEST RESIDUAL FORCE-6.928E-05 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-1.84 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.1.701E-04 AT NODE 11 DOF 2 THE FORCE EQUILIBRIUM EQUATIONS HAVE CONVERGED AVERAGE MOMENT 1.600E+04 TIME AVG.MOMENT 1.160E+04 LARGEST RESIDUAL MOMENT 1.218E-04 AT NODE 7 DOF 6 LARGEST INCREMENT OF ROTATION-1.382E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION 1.804E-06 AT NODE 11 DOF 6 THE MOMENT EQUILIBRIUM EQUATIONS HAVE CONVERGED TIME INCREMENT MAY NOW INCREASE TO 0.150 ITERATION SUMMARY FOR THE INCREMENT:2 TOTAL ITERATIONS,OF WHICH 0 ARE SEVERE DISCONTINUITY ITERATIONS AND 2 ARE EQUILIBRIUM ITERATIONS.TIME INCREMENT COMPLETED 0.100,FRACTION OF STEP COMPLETED 0.200 STEP TIME COMPLETED 0.200,TOTAL TIME COMPLETED 0.200,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,INCREMENT 3 STARTS.ATTEMPT NUMBER 1,TIME INCREMENT 0.150 CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 1 AVERAGE FORCE 133.TIME AVG.FORCE 105.LARGEST RESIDUAL FORCE 3.02 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-2.75 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.-3.764E-02 AT NODE 11 DOF 1 FORCE EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.AVERAGE MOMENT 2.518E+04 TIME AVG.MOMENT 1.613E+04 LARGEST RESIDUAL MOMENT-4.47 AT NODE 5 DOF 6 LARGEST INCREMENT OF ROTATION-2.071E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION 1.722E-05 AT NODE 11 DOF 6 THE MOMENT EQUILIBRIUM EQUATIONS HAVE CONVERGED,t=1.5told,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 2 AVERAGE FORCE 252.TIME AVG.FORCE 145.LARGEST RESIDUAL FORCE-7.965E-04 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-2.75 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.5.629E-04 AT NODE 11 DOF 2 THE FORCE EQUILIBRIUM EQUATIONS HAVE CONVERGED AVERAGE MOMENT 2.798E+04 TIME AVG.MOMENT 1.706E+04 LARGEST RESIDUAL MOMENT 7.461E-04 AT NODE 7 DOF 6 LARGEST INCREMENT OF ROTATION-2.070E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION 5.967E-06 AT NODE 11 DOF 6 THE MOMENT EQUILIBRIUM EQUATIONS HAVE CONVERGED TIME INCREMENT MAY NOW INCREASE TO 0.225 ITERATION SUMMARY FOR THE INCREMENT:2 TOTAL ITERATIONS,OF WHICH 0 ARE SEVERE DISCONTINUITY ITERATIONS AND 2 ARE EQUILIBRIUM ITERATIONS.TIME INCREMENT COMPLETED 0.150,FRACTION OF STEP COMPLETED 0.350 STEP TIME COMPLETED 0.350,TOTAL TIME COMPLETED 0.350 RESTART INFORMATION WRITTEN IN STEP 1 AFTER INCREMENT 3,4 or fewer,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,INCREMENT 4 STARTS.ATTEMPT NUMBER 1,TIME INCREMENT 0.225 CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 1 AVERAGE FORCE 1.528E+03 TIME AVG.FORCE 490.LARGEST RESIDUAL FORCE-4.550E+03 AT NODE 11 DOF 1 LARGEST INCREMENT OF DISP.-5.95 AT NODE 11 DOF 2 LARGEST CORRECTION TO DISP.-1.82 AT NODE 11 DOF 2 FORCE EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.AVERAGE MOMENT 4.853E+04 TIME AVG.MOMENT 2.493E+04 LARGEST RESIDUAL MOMENT-344.AT NODE 9 DOF 6 LARGEST INCREMENT OF ROTATION-4.477E-02 AT NODE 11 DOF 6 LARGEST CORRECTION TO ROTATION-1.371E-02 AT NODE 11 DOF 6 MOMENT EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.,t=1.5told,Introduction to Abaqus/Standard and Abaqus/Explicit,非线性悬臂梁分析的输出,CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION 2 AVERAGE FORCE 281.TIME AVG