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Wood Roof - Thrust 9

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Buleeek

Structural
Sep 5, 2017
98
Hi everyone,

I'm analyzing the TFEC Bulletin No. 2018-12 "Behavior of statically determinate and indeterminate rafters". If I use a collar tie (or strut) system (with no horizontal reaction at the bottom of the rafter)can I assume there will actually be NO horizontal force at the wall top plate? I know that collar ties should not be used to carry the thrust, but struts yes (located higher than top plate level in this case).

I am not sure what situation occurs in a real life. I can't assume that the rafter has no thrust at the bottom, since it is resting on the wall plate and is always notched/nailed/connected.

What do you usually do in such situation? That assumption determines a wall design and how much thrust it will carry.

Thanks,
 
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If you still need something to do, running all with the same stiffness would be interesting, maybe 2x10's ?, now that I finally got past the sign convention. That sign "unconvention", least from what I was used to, was a big surprise to me. But like I say, in the end I'm glad that was a sign convention problem and that the final result didn't contradict my intuition... at least not too much.
Thanks again for doing all that analysis work. Really nice of you.

“What I told you was true ... from a certain point of view.” - Obi-Wan Kenobi, "Return of the Jedi"
 
The structure is statically determinate if supports are pin/roller. Reactions and member forces are not dependent on member stiffness. You don't need a computer to determine the tie force or the reactions. Not sure how many times I have to say this.

My sign convention for member force has always been that tension is '+' and compression is '-', but as stated by others, sign conventions are variable. The program really should state the convention being used to avoid confusion.

BA
 
BA,

Isn't that nice when your hand calculated predictions are confirmed by the computer analysis? On the other hand, when there is deviation, albeit small, don't you interested in knowing the cause/reason (other than rounding obviously)? I don't think learning is one way street, practicing is another matter though.
 
One question about about the static from RISA. Shouldn't the tie be modeled as "pinned" (only tension/compression)?
Thanks,
 
Yes, you can use "axial force only" (truss) member, as usual is the case, for which will give you the exact solutions as hand calculation.
 
I hope this sign convention makes sense.

f_ujhd6t.png
 
ax1e said:
If you still need something to do, running all with the same stiffness would be interesting, maybe 2x10's ?

:) Wish granted. But a general shape rather than 2x10s. Note the roof slope has changed to 1V:2H (6':12') to be more realistic. There is still self weigh in the models that I don't know how to get ride of it at this moment. I'll try later.

M1_akabvd.png


m2_hbw9hr.png
 
retired13,

When specify BLC (Basic Load Combinations) remove any "gravity Y" load. That will eliminate the self weight load.
 
Buleeek,

Appreciate! It works :)
 
Does this spreadsheet make sense?

Enter span, pitch and load on sheet 1

sheet 1 rafter tie at variable height roller supports
sheet 2 rafter tie at variable height fixed supports
sheet 3 frame without rafter tie and two fixed supports
sheet 4 truss on roller supports

“What I told you was true ... from a certain point of view.” - Obi-Wan Kenobi, "Return of the Jedi"
 
 https://files.engineering.com/getfile.aspx?folder=a11f7254-e016-4383-baeb-c2b126b75230&file=RAFTER_TIE_FRAMES_&_TRUSS.xlsx
For those who do not have a DropBox account, see below:

RafterTieFrame_oty5ob.png


BA
 
Something is wrong with that sketch. With span = 20 and W = 100, R can't be 1000, so which is correct?

Edit: Okay, I see: should be w = 100, so W = 2000, then R = 1000 would be correct.
Then M@B = 1000*4.5 - 100(4.5)^2 = 3487.5 (ok)
Tie Force = WL/8h = 2000*20/8(3.333-1.5) = 2728 (ok)

BA
 
That's just the first worksheet. There are 4 worksheets in the downloads I provided, each with various support conditons.

But there seems to be a problem with balance of horizontal forces at B.
AB = 0
BCh = 870 compression * 12/12.65 = 875 left
BDh = 2727 tension to right
2727-875 = unbalance of 1902
What's going on with that?

“What I told you was true ... from a certain point of view.” - Obi-Wan Kenobi, "Return of the Jedi"
 
On the sketch, Rafter Tie is 2727 tension, which I believe is correct, but then shows rafter tie of 825. What is the reason for this?

BA
 
ax1e said:
That's just the first worksheet. There are 4 worksheets in the downloads I provided, each with various support conditons.

Okay, I see that now, but it is only the first sheet that pertains to this thread because the top of stud wall can't act like a pin. So on the first sheet, how do we interpret the two values shown for rafter tie force, namely 2727 and 825?

BA
 
Note, if both ends pinned, you will have horizontal reaction at supports.
 
I was looking for the tie force by sum horizontal forces at B
If BC is determined from sum vertical loads at C, gives BC axial load of 275/4*12.65 = 870 compression
Horizontal component is 870/12.65*12 = 825
[highlight #FCE94F]Then sum of horiz forces at B yields tie force of 825 tension
So you can't have the 2727 you get from sum of moments at C[/highlight]

“What I told you was true ... from a certain point of view.” - Obi-Wan Kenobi, "Return of the Jedi"
 
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