Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

abaqus monitor problem

Status
Not open for further replies.

han_cyuan

Structural
Sep 20, 2023
31
You have entered a zero or negative density value. Abaqus allows non-physical data; however, most applications expect a positive density value and greater than 1.00000e-36 .

4 nodes associated with rigid bodies have boundary conditions prescribed at nodes other than the reference node. These boundary conditions will be transferred to the associated rigid body reference node.The reference nodes and the dependent nodes have been identified in node set WarnNodeOverconBoundRB.

Nodes belonging to 2 RIGID BODIES have boundary conditions prescribed at nodes other than the reference node. These boundary conditions indicate the rigid bodies cannot rotate about certain axes. Zero rotational boundary conditions have been added to these reference nodes.The reference nodes have been identified in node set WarnNodeOverconBoundRBRot.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 4.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 5.

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 4

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 5

Solver problem. Zero pivot when processing D.O.F. 4 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_4_1_1_1_1.

Solver problem. Zero pivot when processing D.O.F. 5 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_5_1_1_1_1.

The system matrix has 1 negative eigenvalues.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 4.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 5.

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 4

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 5

Solver problem. Zero pivot when processing D.O.F. 4 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_4_1_1_1_1.

Solver problem. Zero pivot when processing D.O.F. 5 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_5_1_1_1_1.

The system matrix has 1 negative eigenvalues.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 4.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 5.

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 4

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 5

Solver problem. Zero pivot when processing D.O.F. 4 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_4_1_1_2_1.

Solver problem. Zero pivot when processing D.O.F. 5 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_5_1_1_2_1.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 4.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 5.

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 4

Solver problem. Zero pivot when processing node 1 (assembly) D.O.F. 5

Solver problem. Zero pivot when processing D.O.F. 4 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_4_1_1_3_1.

Solver problem. Zero pivot when processing D.O.F. 5 of 1 nodes. The nodes have been identified in node set WarnNodeSolvProbZeroPiv_5_1_1_3_1.

Solver problem. Zero pivot when processing node ASSEMBLY.1 D.O.F. 4.
 
Replies continue below

Recommended for you

1. Make sure that density is correctly defined.
2. Apply boundary conditions only to reference nodes of rigid body constraints, not to the nodes belonging to the rigid body node/element set.
3. The rest requires information about the model.
 
mass_f67st6.png
load_bqx06u.png
run_ewxbfk.png
gra_tkxws7.png
bcs_lsxbmx.png
%E6%9C%AA%E5%91%BD%E5%90%8D_tujllm.png




Thank you for your reply. The model I want to build is a single-mass car. The mass block is connected to a negligible wheel by a spring, so the density of the wheel is set to 0.
When I assume that the car body and wheels are rigid bodies, when I set the movement conditions of the vehicle, are they imposed on the reference point of the rigid body?
 
Better use very low but not zero density. And apply boundary conditions directly to the reference point. Usually, Abaqus can transfer them automatically to the ref point but it's better to avoid this since it may not work properly in some cases.
 
Will setting the boundary conditions directly at the reference point create excessive constraints?
According to your suggestion, I should change the loading gravity, inertial force, and displacement conditions to the reference point. Will it be beneficial to the analysis?
 
Check the documentation chapter Abaqus Introduction & Spatial Modeling --> Spatial Modeling --> Rigid Body Definition, paragraphs Kinematics of a Rigid Body --> Boundary Conditions and Defining Loads on Rigid Bodies. There it's all explained.
 
I would like to ask whether it is recommended to use an interaction-rigid body or a part-discrete rigid body for rigid bodies. Which of these two should I choose?
 
A detailed answer to this exact question is provided in the documentation chapter "What is the difference between a rigid part and a rigid body constraint?". In your case, I would probably use rigid body constraint.
 
I would like to ask how to express the establishment of a wheel. It can be ignored or become a point.
 
fft_yuehsu.jpg

I now use fft to convert the acceleration of the car body into frequency. There is only the frequency of the car, not the frequency of the bridge. How can I correct my model?
My vehicle currently moves from the left to the right of the bridge using hard contact and frictionless force. Do I need to correct the interaction?
 
11_ppyefo.png
22_po542p.png

I would like to ask, the red line is the acceleration of the center of mass of the car body. How can I modify it to be the same as the blue line, so that there is an acceleration response from the beginning?
I would also like to ask what is the difference between field output and history output? What if they both take the acceleration of the same point?
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor