Recent content by comhedoisl

After saying that, I wonder if that would show on the “global stiffness matrix”… Maybe for the type of truss of the exercise (equal rods equally spaced, sharing a common node), the global stiffness matrix is always symmetrical? Or always proportional to identity matrix? I’ll try investigate that

Thanks for running some cases, Retrograde. Symmetry is key here... But it’s not that trivial as it seems to be. I think the reasoning brought by JStephen and Tomfh is one path to solve the exercise, if you apply it to the symmetry of “stiffness distribution”.

There’s a reason for Node “O” to move up, and that would be compatibility. Each member deforms (Fi)*L/EA, Fi being the axial force in each member. Using the values from the example: F1 = 4 kN (compression) F2 = 6.8 kN (tension) F3 = 2.4 kN (compression) So, deformation of each member will be...

This would be second order analysis... Although it is not stated in the question, we may assume a linear elastic analysis, not considering second order (P following node in deformed configuration*). *Assuming linear elastic analysis and displacements small enough that there's no significant...

Yes, I think that should be assumed, as it’s not stated in the question.

No… as I posted previously, considering an example of 3 equal rods, but not equally spaced, the displacement of node O (common node) does not align with the direction of the force.

Hi Nick. That may help answer the displacement aligned with force bit. I’m not sure how to do it with more than 3 rods, even with 3 may be difficult, I havent tried yet. But the “displacement value independent of angle” bit would still be pending if using only displacement compatibility. Thanks.

OK, I think that makes sense. It's clearer now for me. Thank you, Tomfh.

Tomfh, here's one example showing a situation where the perpendicular components to force P sum to zero, but the displacement of node O is not aligned with P. The components with same direction as P sum to P. Equilibrium holds. The truss in this case is not symmetrical: all the members are...

Thank you all for the good discussion. Tomfh, I like the reasoning and seems pretty convincing, but I still want to understand something... Wouldn't this 'sum of perpendicular components sum to zero' always occur, no matter if the truss is symmetric or not? The resultant of the forces in node O...

That’s what the exercise want us to prove or show why. The full displacement of a node in a truss do not always follow the direction of the force applied to that node. Why would that be the case in the truss from the example? Is it because of symmetry? It seems to be related to symmetry in...

Thank you all for your inputs. The question does not mention about small deflections, but I think it just take it for granted. This assumption should be taken if needed. It also feels right that the deflection should be in the direction of the force P. I just couldn’t properly explain (in...

Hi all. This is an interesting theoretical question from a book of Feodosiev - Advanced Stress and Stability Analysis. It reads like this: "A plane truss consisting of n>2 equal and equally spaced rods, connected in a common node, O. The force P acts in the plane of the truss. Show that the...