Optimizing strength to weight in balsa wood deck arch truss bridge

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The discussion focuses on optimizing the strength-to-weight ratio in a balsa wood deck arch truss bridge. A suggestion is made to use short diagonal braces instead of long ones to reduce weight and achieve more equilateral triangles. An analysis of both options for a single cell is recommended to evaluate their effectiveness. The conversation emphasizes the importance of design choices in structural efficiency. Overall, the optimization of truss design is crucial for enhancing performance.
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Hey, so I'm designing a deck arch truss bridge from 1/4" x 1/4" balsa wood sticks and would like a few tips on how to optimize my design further. There are some constraints: no more than 1 layer of lamination (so 2 total), length minimum (whole thing is around 14 inches), and so on. The bridge is designed to support the highest weight at the center by a 2x2x1 inch wood block without bending too much.

I like the design I have currently, but want to go further and use some actual math to back it. Research online says a good rise to span ratio of the arch is around 1:4 to 1:6.5, but is there any math I could run through to optimize this using formulas of the strength of an arch to find a specific(ish) number that I can use?

The bridge is judged based on its strength but also its weight, so this must be accounted for as well. Also, are there any improvements I can do to the trusses and their configurations? If you have any questions or need clarification, I'd be happy to answer.

If anyone has free software recommendations for testing and modelling this type of bridge, please let me know. I'm currently using AutoCAD and SOLIDWORKS but I feel I could get better simulation results with other software.

Thanks!
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I notice you have the diagonal braces in each cell on the long diagonal. I would have gone for the short diagonal, both to save weight and to make the triangles closer to equilateral.
Perhaps you could analyse those two options, just for one 'cell'.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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