Solving trusses with the Direct Stiffness Method

AI Thread Summary
To determine if a planar truss is solvable, the condition 2j ≤ m + r should be used, where j is the number of joints, m is the number of members, and r is the number of reactions. Filtering trusses based on the stricter condition 2j = m + r may result in discarding some valid trusses. The discussion emphasizes the importance of maintaining solvable configurations while randomly generating trusses. Implementing this criterion will help in efficiently identifying unstable or indeterminate trusses without extensive calculations. Proper filtering is crucial to ensure that good trusses are not lost in the process.
Diego Saenz
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I'm creating a computer implementation to solve planar trusses. And I'm not sure how to check if the truss is solvable or not. Can you help me with that?

In my implementation, the trusses are created randomly (needs to be this way), so i get a lot of unstable or indeterminate trusses. I want to discard bad trusses without calculating the determinant of the stiffness matrix or trying to solve them.

Is this condition enough to discard such trusses? If I filter the trusses with this criteria, will i lose some good ones? ----> 2j = m+r

j - number of joints
m - number of members
r - number of reactions

Thanks.
 
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Diego Saenz said:
If I filter the trusses with this criterion, will I lose some good ones? ----> 2j = m+r
Diego Saenz: Yes, you would lose some good ones. Instead, discard only the trusses having 2*j > m + r. Keep the trusses having 2*j ≤ m + r.
 
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