The only zero-force members are CD and FE. Member BD bears 2lbs in compression (hence the buckling right before failure) and CF holds 2lbs in tension. At 8:32 you can see that point b is not in equilibrium if Fbd=0. When analyzing joint C you represented Fac as a compression force, but you had previously determined that it was a tension force. When you reverse its direction you find that Fcf=0 does not satisfy equilibrium. I hope this helps.
I agree. Point B Is in vertical equilibrium but absolutely not in horizontal equilibrium. Just looking at the moment from the support (1Lb X L/2)/H gives the reaction force in BD.
If I'm looking at the efficiency of the structure (strength divided by weight) the lower weight of the structure would account for the loss of the vertical support members. While the structure would be lighter it would also likely be weaker (as the video showed), thus giving a lower strength to weight ratio.
I'm not an engineer, just a curious mind. I'm making plans for building a modest 2 story home. I have done construction for quite a while and I don't like it when trades, start putting holes in floors members to run wire, plumbing, or duct. I want to design my own floor trusses and I have a good idea how to test them before I turn in my plans to an engineer for approval and or revision.
In my book testing should be done to verify calculations whenever possible. Its a theme that is the basis of my videos... The engineer should verify any and all truss designs... Many a structure has failed because of small errors or miscalculations.
This will depend on the materials you use (as well as how you connect your members together). A given material will have a tension member capacity and a compression member capacity -- these known values tell you how much force, in tension or compression respectively, a member can sustain before failure.
The only zero-force members are CD and FE. Member BD bears 2lbs in compression (hence the buckling right before failure) and CF holds 2lbs in tension.
At 8:32 you can see that point b is not in equilibrium if Fbd=0.
When analyzing joint C you represented Fac as a compression force, but you had previously determined that it was a tension force. When you reverse its direction you find that Fcf=0 does not satisfy equilibrium.
I hope this helps.
I agree. Point B Is in vertical equilibrium but absolutely not in horizontal equilibrium. Just looking at the moment from the support (1Lb X L/2)/H gives the reaction force in BD.
F(ab) goes down right? it is already proven in this equation 6:49
so why does you calculate it goes up in the other equation? 7:35
because it is in tension. one end goes to the support (Point A), while an equal and opposite force goes to joint B. Newton's third law
I thought bending should be considered as well. Also glued wood joint is not a pinned truss structure.
How did you account for the weight difference when you removed the vertical support structures?
If I'm looking at the efficiency of the structure (strength divided by weight) the lower weight of the structure would account for the loss of the vertical support members. While the structure would be lighter it would also likely be weaker (as the video showed), thus giving a lower strength to weight ratio.
I'm not an engineer, just a curious mind. I'm making plans for building a modest 2 story home. I have done construction for quite a while and I don't like it when trades, start putting holes in floors members to run wire, plumbing, or duct. I want to design my own floor trusses and I have a good idea how to test them before I turn in my plans to an engineer for approval and or revision.
In my book testing should be done to verify calculations whenever possible. Its a theme that is the basis of my videos... The engineer should verify any and all truss designs... Many a structure has failed because of small errors or miscalculations.
But how are you supposed to calculate when it breaks ??? Ive been searching all day. I need to calculate when it will break
This will depend on the materials you use (as well as how you connect your members together). A given material will have a tension member capacity and a compression member capacity -- these known values tell you how much force, in tension or compression respectively, a member can sustain before failure.
5:19
Nice