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Tied Rafter, Basic Determinacy Question

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?

RE: Tied Rafter, Basic Determinacy Question

This is definitely a stable structure. I don't remember how to use those formulas for checking determinacy, but are the supports NOT considered joints? So there are only 3 joints?

DaveAtkins

RE: Tied Rafter, Basic Determinacy Question

As long as your continuous chords can transfer moment across the joints where they connect to the tension ties, this is absolutely stable.

RE: Tied Rafter, Basic Determinacy Question

r = 9 (4 pins and 1 roller)
n = 3 (3 individual members)
r = 3n therefore statically determinate.

RE: Tied Rafter, Basic Determinacy Question

It's stable. In your determincy check you are equating your rafters to bars (two force members). This is incorrect as rafters are beams and can carry bending moment

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

I think you're over-complicating it.

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

The three rods are pinned together and they are stable as a single piece. The single piece has one pin and one roller support. Therefore is determined structure.

RE: Tied Rafter, Basic Determinacy Question

(OP)
Thanks everyone.

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.

Quote (Shotzie)

r = 9 (4 pins and 1 roller)
n = 3 (3 individual members)
r = 3n therefore statically determinate.

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

Interesting question. I suspect the diagram does not refer to actual rafters, but to some form of load carrier. The rafters in my former cabin had that tie on every third one, but the load was on the rafters from snow and the peak had no load. I suspect my ties were in compression, but somewhat unclear as to stress distribution pattern since eaves tied the lower ends from moving outward. Did the distorted rafters have an "S" shape? Unbalanced loadings, right to left, would have made an even more complicated displacement pattern. Then make one rafter weaker than the other, even more questions.

RE: Tied Rafter, Basic Determinacy Question

(OP)

Quote (JGard1985)

This is incorrect as rafters are beams and can carry bending moment

You will have bending moment in the rafters at the connection of the tension tie.

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

to work as a classic truss you need a tie member joining the two reaction points.

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

Quote (StrEng007)

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.

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

(OP)

Quote (rb1957)

to work as a classic truss you need a tie member joining the two reaction points.

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.

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

(OP)

Quote (BeFEA)

As others pointed out, your structure is statically determinate. It is geometrically stable and it has the minimum number of necessary connections.

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.

Quote (BeFEA)

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

Quote (oldestguy)

The rafters in my former cabin had that tie on every third one, but the load was on the rafters from snow and the peak had no load. I suspect my ties were in compression, but somewhat unclear as to stress distribution pattern since eaves tied the lower ends from moving outward.

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.

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