*ELEMENT STIFFNESS MATRIX - how is the matrix stored

[riccardoPietro]

Hi everybody,

I carried out the stiffness matrix of each element of my Abaqus 2D model.
The element that I am using are CPE6 (triangular elements with 6 nodes). My output is the following:

** ELEMENT NUMBER 416 STEP NUMBER 1 INCREMENT NUMBER 1
** ELEMENT TYPE CPE6
*USER ELEMENT, NODES= 6, LINEAR
** ELEMENT NODES
** 10, 102, 11, 1240, 1415, 1416
1, 2
*MATRIX,TYPE=STIFFNESS
1413.8091460764 ,
496.07100182906 , 928.14849090294
870.46664095379 , 150.43516802503 , 5335.5950818700
194.34617505079 , 303.21631276313 , -93.452227151622 , 1918.9767742134
-420.37330773502 , 9.3775720898828 , 944.46693565659 , -216.47757685401
1761.4892799466 ,
-34.533434935878 , -3.9209018649905 , -172.56656982825 , 349.05885390710
-590.96958576891 , 1105.2513244617
-3638.2627776032 , -632.03626727816 , -3526.8420084713 , -778.49922107987
89.217973346962 , 3.1362181024602 , 12476.475942846
-807.68029538120 , -1303.4262216275 , -602.85519297682 , -1204.7404879390
3.1362181024602 , 38.748903096538 , -140.64031041303 , 5720.5579976796
9.9638184977461 , 9.9781567635673 , -3811.5686474461 , 867.38684836901
-3911.1176273159 , 664.22706885316 , 3193.3267763427 , 48.603908773072
11567.075752573 ,
9.9781567635672 , 19.658019977084 , 691.74282026597 , -1432.6065313643
839.87109695621 , -1430.2885999271 , 48.603908773072 , -132.54562804269
-140.68437350723 , 5471.7356733092
1764.3964798104 , -33.825631429378 , 187.88199743703 , 26.696001665700
1536.3167461008 , 130.70630357742 , -8593.9159064615 , 1499.4356718955
-7047.6800726519 , -1449.5116092516 , 12153.000755765
141.81839667367 , 56.324299849310 , 26.696001665700 , 66.095078419676
-44.937724525628 , -58.849579673199 , 1499.4356718955 , -3118.5945631669
-1449.5116092516 , -2495.9529339521 , -173.50073645767 , 5550.9776985232





So, my question is: which order of degrees of freedom were used by abaqus to store the matrix?

I'll try to better explain: the matrix is depending on nodal displacements and during its generation it is depending on the order of nodal displacement storage:

e.g. looking at the element connectivity (element nodes), I would say that the order is: ux_10, uy_10, ux_102, uy_102 ... where ux_10 is the x-displacement of node 10 and so on...

does anyone know if it is actually built in such way?

thank you

Riccardo

 

[Mustaine3]

1. You say output, but the keywords say input.
2. Since when uses a stiffness matrix displacements?
3. Abaqus Analysis User's Guide 32.15.1 User-defined elements

 

[riccardoPietro]

Hi Mustaine, thank you for your answer,

but:
- this is my output, since it is the result of my job solution (launch of the .inp file)... the keyword to make the stiffness matrices appear were added in the .inp file.. this one is not an input file, but a .mtx file. Or, one could make this matrix to be written in the .dat file (user choice)
- the matrix is not depending on the entity of displacements, but on the order in which they are stored (the matrix elements would be always the same, but the order of storage of the degrees of freedom affects the order of the rows of my matrix)

in the documentation I can only find that, for each node, the order of storage is x and y, but nothing is reported about the order over the nodes.

I would assume that the order is the same as the connectivity is shown, but I don't have any evidence. Am I right?

thanks

Riccardo

 

[testing]

@riccardoPietro: The section of the documentation that Mustaine pointed you to specific discusses order of nodes and DoF for the matrices. Default assumes the element DoF is ordered first by node, then by DoF at node. However, you can explicitly tell Abaqus what order they are in when you define the element. I recommend reading through the user element documentation again.

 

[Mustaine3]

@riccardoPietro
I was seeing the keywords and the data and this means to me that you've tried to use that in the input file.