abaqus出现警告信息，请教高手指点

Posted 2023-05-04

提交任务后，任务运行完成了，但是没有结果，还是初始状态，也没有错误信息，只有警告信息。求各位大侠帮忙指点一下，我该从那些地方下手修改模型？我是个新手，正在学习abaqus中。提示如下警告信息：（1）The request for mises output will be replaced by a request for s outputIntegration and section point output variables will not be output for deformable elements that are declared as rigid using the *rigid body option（2）8 elements are distorted. Either the isoparametric angles are out of the suggested limits or the triangular or tetrahedral quality measure is bad. The elements have been identified in element set WarnElemDistorted.（3）Boundary conditions are specified on inactive dof of 1041 nodes. The nodes have been identified in node set WarnNodeBCInactiveDof.（4）Each of the nodes in node set WarnNodeBcAtRbSlaveNode have prescribed boundary conditions and belong to rigid bodies but are not rigid body reference nodes. The boundary conditions at these nodes will not constrain the motion of the rigid bodies. See the status file for further details.（5）Boundary conditions are defined at the nodes contained in node set WarnNodeBcIntersectKinCon. In addition the nodes are also part of a surface involved in kinematic contact. The kinematic contact constraint will be overridden by the boundary conditions in case of a conflict. Penalty contact may be used instead.（6）The slave surface nodes in node set WarnNodeRbMassRatio2fix-Step1 have much larger masses than the nodes on the rigid master surface. Significant contact noise may result with the kinematic contact algorithm when the slave nodes have significantly greater mass than the master nodes. Suggested workarounds include using mass-scaling to adjust the ratio of nodal masses or using the penalty contact algorithm. See the status file for further details.万分感激~~

计算完成还是初始状态检查一下有没有输出场变量和历史变量。 参考技术A 这个挺简单的，检查下你的模型，重点是网格划分后颜色不一样的地方。本回答被提问者采纳

请教STM32F4 的DSP库 FFT运算的问题.求高手指点

参考技术A 代码如下；

int32_t main(void)

for(index=0;index<2048;index++)

testInput_f32_10khz[index]= 100*sin(3.14*index/512);


arm_cfft_f32(&arm_cfft_sR_f32_len1024, testInput_f32_10khz, ifftFlag, doBitReverse);

arm_cmplx_mag_f32(testInput_f32_10khz, testOutput, fftSize);

arm_max_f32(testOutput, fftSize, &maxValue, &testIndex);

for(index=0;index<1024;index++)

printf("index%d =%d\r\n",index,(u32)testOutput[index]);



while(1); /* main function does not return */


1. 如果2048个点，我直接取正弦波的值，那么打印出来是

index2 =72203
index1022 =72647
貌似是对称的。。。
for(index=0;index<2048;index++)

testInput_f32_10khz[index]= 100*sin(3.14*index/512);


2.但是如果2048个点，我按照下面方式
for(index=0;index<2048;index++)


if(index%2==0)
testInput_f32_10khz[index]= 100*sin(3.14*index/512);
else
testInput_f32_10khz[index]= 100*cos(3.14*index/512);

那么测试出来是
index1022 =102399
不对称。。。

3.如果我虚部为0，实部位正弦值：
for(index=0;index<2048;index++)


if(index%2==0)
testInput_f32_10khz[index]= 100*sin(3.14*index/512);
else
testInput_f32_10khz[index]= 0;


出来是
index2 =51212
index1022 =51212
是对称的。。本回答被提问者和网友采纳