Combining sub-matrices into global matrix

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The discussion focuses on combining five element matrices into a global stiffness matrix for truss analysis. Participants confirm that overlapping values in the global matrix should be summed to create a single value, such as combining AA1 and AA2. A Fortran coding example illustrates how to add contributions from triangular elements into the global matrix. The final stage of the problem involves using determinants to solve for specific values, with a participant expressing appreciation for the clarity provided in the responses. Overall, the conversation emphasizes the importance of correctly summing overlapping matrix values and understanding the determinant method for final calculations.
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Homework Statement



Combine five element matrices into single matrix forming a global stiffness matrix for simple truss analysis (five elements, four nodes).

Homework Equations



-See attached image-

The Attempt at a Solution



Where the four sub-matrices overlap in the global matrix two or three values/numbers occupy the same position in the global matrix (two separate values for AA, three separate values for BB, and so on). Should I simply add these values to form a single value for the global matrix? For example: if AA1 = -10, and AA2 = +5 should I simply combine them as AA = -10+5 = -5?

Is it really as simple as that?

Thanks.
 

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Yes, that is the way you build a global matrix from individual elemental matricies.
 
Thank you very much. I'm assuming your post is based on my AA = AA1 + AA2 and not just the matrix image I posted! I very much appreciate your confirmation.
 
Suppose you had a triangular element with nodes 2,6,8 at the corners. You would add it into the global matrix as follows where A could be a 100X100 matrix. a is the 3X3 contribution from the element. Below would be the coding in Fortran.
The first indice is the row, the second is the column.

A(2,2)=A(2,2) + a(1,1)
A(2,6)=A(2,6) + a(1,2)
A(2,8)=A(2,8) + a(1,3)

A(6,2)=A(6,2) + a(2,1)
A(6,6)=A(6,6) + a(2,2)
A(6,8)=A(6,8) + a(2,3)

A(8,2)=A(8,2) + a(3,1)
A(8,6)=A(8,6) + a(3,2)
A(8,8)=A(8,8) + a(8,3)
 
In my last equation there is a typo.

a(8,3) should be a(3,3).
 
Thank you. It was kind of you to post that, and I appreciate it. Apologies for not replying more promptly but I had a couple of deadlines to meet and my mind was away from the matrix problem for a day or two.

Now my mind is back on this issue I am much more confident but stuck at the final stage. I’ve attached an example someone else produced which shows three stages. I can get from stage one to stage two but not two to three. I don’t know where the 160083 came from, or the three bracketed numbers. Any help with this would be very welcome!
 

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The problem is solved by using determinants where for a1 you divide the determinant of the stiffness matrix with the load vector substituted for the first column by the determinant of the original stiffness matrix. All numbers are evenly divisible by 3 so it is factored out.
 
Thank you once again. Your reply to me is just the sort of thing I want to read in the textbooks, yet it is nearly always missing and so I often learn little from them. Perhaps I've just been reading the wrong books!
 
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