Design of individual members in a steel truss bridge?

AI Thread Summary
The discussion focuses on the design of a steel truss footbridge, specifically addressing the calculation of member sizes using beam analogy. The user has calculated various parameters, including self-weight and design loads, leading to preliminary member selections. A key challenge is determining the maximum design moment (M*) and ensuring it is less than the moment capacity (Mb) for each member. It is noted that if loads are applied only at the nodes, M* is derived solely from the self-weight of the members, but considerations for lateral buckling of the compression flange may require a more comprehensive analysis. The conversation emphasizes the importance of understanding the overall design concept and member restraint in the truss.
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Hi, I have to design a steel structured truss footbridge.
Im using beam anology to determine the correct member sizes.
Having trouble determining the maximum design moment (M*)

Here's what I have calculated thus far:

Q=6kN/m
G=1kN/m
50m long, 3m depth, 3.125m spacing

Self Weight Preliminary Design
Chords :150UC30.0
Tension-Web :90x90x6EA
Comp-Web :150PFC
therefore SW=4.48kN/m

W*=1.2(1+4.84)+1.5(6)
W*=16/2
W*=8kN/m per truss
M*=2500/3m depth
N*=833kN

After trial and error, all satisfy N*<Nt,Nc
Chord :200UC46.2
Tension-Web :125x125x12EA
Comp-Web :380PFC

From this point onwards, how do I solve for M*<Mb?
And do I solve M*<Mb individually for all members?:
M*<Mb,200UC46.2
M*<Mb,125x125x12EA
M*<Mb,380PFC
??
 
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M* was for the beam analogy, and it produced N*. Provided the loads are at the nodes only, then the only M is from self-weight of a member, and you need not check any further.
 
On second thoughts, if the compression flange at the top is unrestrained against lateral buckling, then you do need Mb for the whole truss. That's a hard one. It depends on your whole design concept and where restraint is.
 

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