Tied Rafter, Basic Determinacy Question
Tied Rafter, Basic Determinacy Question
(OP)
I'm having trouble illustrating the determinacy of this structure.
This is what I'm getting from determinacy checks if I treat it like a truss:
Treating the rafter as a continuous member,
b=3 (3 bars)
r=3 (3 support reactions)
b+r=6
2(j)=2(5)=10
6<10, unstable. This isn't correct, I believe this rafter is in fact stable.
Treating the rafter as two individual segments on each side,
b=5 (5 bars)
r=3 (3 support reactions)
b+r=8
2(j)=2(5)=10
8<10, unstable. Again, I don't believe this is correct.
Is there an exception that I'm missing?
This is what I'm getting from determinacy checks if I treat it like a truss:
Treating the rafter as a continuous member,
b=3 (3 bars)
r=3 (3 support reactions)
b+r=6
2(j)=2(5)=10
6<10, unstable. This isn't correct, I believe this rafter is in fact stable.
Treating the rafter as two individual segments on each side,
b=5 (5 bars)
r=3 (3 support reactions)
b+r=8
2(j)=2(5)=10
8<10, unstable. Again, I don't believe this is correct.
Is there an exception that I'm missing?






RE: Tied Rafter, Basic Determinacy Question
DaveAtkins
RE: Tied Rafter, Basic Determinacy Question
RE: Tied Rafter, Basic Determinacy Question
n = 3 (3 individual members)
r = 3n therefore statically determinate.
RE: Tied Rafter, Basic Determinacy Question
You will have bending moment in the rafters at the connection of the tension tie.
Jeff
Pipe Stress Analysis Engineer
www.xceed-eng.com
RE: Tied Rafter, Basic Determinacy Question
you have three reactions, three equations of equilibrium, therefore statically determinate.
because of your reactions the inclined members are doing a bunch of bending, and are not axially loaded struts.
another day in paradise, or is paradise one day closer ?
RE: Tied Rafter, Basic Determinacy Question
RE: Tied Rafter, Basic Determinacy Question
I threw this into a computer model, and this is what I found.
If this is considered a truss, it will not be stable. Classic truss members are two force members only and we need to assume the relative moment at each joint is low in comparison to the axial force. This approach for determinacy was correct:
b=5 (5 bars)
r=3 (3 support reactions)
b+r=8
2(j)=2(5)=10
To make this a stable truss, I would need to provide a pin reaction in lieu of the roller and also add a truss member connecting the two supports (horizontal member). Only then would I be able to consider this a truss with 2-force (uniaxial) members.
Therefore, this cannot be treated as a truss. It will have to be treated as a generic structure in which an internal moment can develop at the location of the collar tie as many of you have pointed out. The following determinacy check would be applicable.
In summary, this will only be considered stable if the generic structure approach is utilized. A bending moment in the rafter will have to develop. What's interesting is that most trusses are constructed with continuous chords, they aren't segmented as the way any structural analysis technique will show. As long as loads are only applied at the joints, a truss theory approach can be used to determine the stability.
RE: Tied Rafter, Basic Determinacy Question
RE: Tied Rafter, Basic Determinacy Question
Thanks, you are correct.
However, my problem was about how to treat this as a truss with a load placed only at the pin. In this case, I was referring to "Truss" as 2-force members only. I see that it would not be stable in that "theoretical" situation.
RE: Tied Rafter, Basic Determinacy Question
as a classic truss, the inclined members react axial load. the upper tension tie will have zero load in it. you need a tie at the reaction points to carry the horizontal component of the RH inclined member over the the LH reaction point.
another day in paradise, or is paradise one day closer ?
RE: Tied Rafter, Basic Determinacy Question
Theoretically, a truss should not have continuous chords. Practically, the moment in the chords is zero when the structure is shaped like a stable "truss". The equilibrium of joints shows that there are no bending moments in such elements. Yours would not be a stable "truss" if you had full pinned connections in every joint.
As others pointed out, your structure is statically determinate. It is geometrically stable and it has the minimum number of necessary connections.
RE: Tied Rafter, Basic Determinacy Question
That would provide:
b=3
r=3 (if you have 1 pin and 1 roller, if you have 2 pins this turn statically indeterminate)
2(j)=6
Statically determinate.
If you ALSO had a collar tie:
b=6
r=4 (you NEED two pins in this situation)
2(j)=10
Statically determinate.
RE: Tied Rafter, Basic Determinacy Question
Thats exactly what I'm saying in my response above. This cannot be treated as a truss in the classic (2-force) member only approach.
Theoretically, a truss should not have continuous chords.
I agree with this statement. Physically, most trusses are constructed a continuous members, but the classic approach was to ignore any internal moment so long as loads were placed at joints. This was a method used before computers were able to do a more precise analysis. Moments, if any, are considered to be negligible in comparison to axial loads. With that being said, a truss doesn't have to assume to develop internal moment just because the chord is continuous. The rational is up to the engineer who is designing it.
RE: Tied Rafter, Basic Determinacy Question
No, those ties are in tension. The intermediate frames differ from the idealized framing in the OP sketch by removing the tie and replacing the bottom right roller with a pin. This second pin is laterally restrained every 4 ft (US Code) with those ties. Without those ties, the top of the wall will barrel out.