6.8 kN is intended to be the tension in member 2. It's a little off due to the slight change in slope.
Reactions at the bottom of member 1 have a resultant aligned with the member. They are a little off due to change in slope, but there is no shear in member 1.
There are no moments anywhere...
The members are not modeled as beams. They are hinged at each end, so they must act as axial rods. If as you say, horizontal member #2 carried most of the applied load, it would stretch by delta = FL/AE which would be inconsistent with the strain in Members 1 and 3.
Congratulations comhedoisl, you were correct.
Duh!! Thinking about it last night, I realized that Node "O" had to deflect upwards since member 3 is in compression and is pinned at the top. Now we have the proof thanks to Retrograde and goutam-freela. Thanks guys!
I suggest you check it using an appropriate software package. I don't have one myself, but I believe goutam-freela does. If not, someone else will certainly have one. Failing that, it can be checked by hand calculation.
Just had another look at your diagram. Member 3 has 2.4 kN compression downward. Member 1 has a vertical component of 2.4 kN upward. This suggests that your diagram is out of scale. There is no reason for Node "O" to move up or down. If it does not move up or down, then it simply moves to...
That structure was indeterminate. You did not analyze it as an indeterminate structure. Your sketch shows that Members 1, 2 and 3 rotated after loading; and that changes the reactions. After loading, P should be at Node "O" in its final position, not its original unloaded position...
Of course 1. and 2. would be true. But not as a result of elements being symmetrical. The given postulates hold true for any stable arrangement of spokes, whether symmetrical or unsymmetrical, whether odd or even in number.
Every spoke is pin connected at both ends, but the central pin...
That simply means that the k term is zero for both P and R. What I said before is still true.
They are 2D vectors of the form P = ai+bj and the vector sum P+R = 0.
We have a structure resisting a single applied load; that load can be represented by a vector in the form P = ai+bj+ck. That expression provides both magnitude and direction of the load P.
It seems to me that the resultant of all of the reactions must be equal and opposite to the applied...
Cutting the span to 25' is an excellent solution, but not one which ET members were at liberty to suggest. But good for you; let's hope the client or architect agrees.
Kootk's comments are interesting; perhaps a separate thread is warranted as it is an important subject.
You could make the truss symmetrical.
You could drop the bottom chord and reduce headroom.
You could reduce roof slope.
You could raise the ridge.
You could change to a steel truss.
Can't think of anything else you could do to move things forward.
Both of the selected designs appear much too shallow, but they should be checked by calculation, taking into account the actual dead and live load.
Truss programs used by fabricators should be capable of providing a reliable answer.
