After my September post, I continued working on my truss program pursuing a couple new goals.
While working, I drew a truss as a personal example. It’s an odd layout, one most likely never seen in practice or in the field. I realized that the truss might make an interesting puzzle for fellow structural engineers.
We know that statically determinate trusses can be solved using the method of joints and/or the method of sections but solving this example might benefit from a couple of extra insights.
The answer is illustrated in the following, a static diagram and a live plot:
Only the top and bottom chords are loaded!
Yes, it turns out that Members 3, 4 and 8 are all zero force members.
By symmetry, Members 3 and 4 must carry the same force. The sum of the vertical components of Members 3 and 4 must equal the vertical external load at Node 1, which is zero (nonexistent). Therefore, Members 3 and 4 are zero force members.
There is no external load at Node 3 or Node 4. Therefore, Member 8 is a zero force member.
This is a good place to discuss the odd layout:
Trusses must only be loaded at nodes (at ‘panel points’); this is because all truss members are axial members, no bending allowed.
Therefore, this truss can only be loaded at the ridge, the midspan top chords, amd the midspan bottom chord.
A roof system would need smaller spaced support points at say 5′ or less, and a ceiling would certainly require closer spaced support than the 10′ spacing in our truss as given.
Finally, Member 8 would work better as a vertical member connecting Nodes 1 and 5.
