Timber rafter ceiling joist connection 8

[Calc1]

I am using method of joints to calculate the tension force in ceiling joist but the result is 2 times the computer run or any roof framing book table, anyone has any suggestions, for example if uniform load is 54 lb/ft for LD=1 on the rafter(Horz) and the span is 17 ft with 1 ft overhang each side, the reaction at the wall = 459 lbs, the ceiling joist tension force=T=1377 lbs now if you use any framing book, they recommend 6-16d nail which capacity is 1/2 around 650 lbs which is close to computer run result, am I missing something here?

 

[SacreBleu]

Another way to understand this is to calculate the compression in the top chord or rafter as the vector equivalent of the reaction at support, then calculate the tension in bottom chord/ceiling joist as the vector equivalent of the top chord/rafter.

If you use the method of joints approach for a truss with few panel points or for this rafter/ridge board and tensioned ceiling joist example, you just aren't getting the true picture. Or: envision a truss that has its first panel point very close to the reaction. In that case, you will get closer to the correct numbers, as opposed to the truss with few panel points. In both trusses, the bottom chord tension should be equal, in reality.

 

[apsix]

TTK/SacreBlue
"apply the UDL directly to the rafter you will get the higher (double) tension force in the ceiling joist"
If this is the result you got there is a problem somewhere.
I analysed the pinned truss with both the UDL and 459 lb load at the apex and obtained identical tie tensions (the lower value). This result appears to be backed-up by the framingdesign.com results.
I agree the rafter results are different, the UDL case obviously has shear and moment, however the average of the varying axial compression is equal to the apex point load case (by my software results).

What you are telling me is that the tie tension force is identical for the same total load, whether applied as a UDL or single midspan point load. I'm convinced that can't be the case.

Cheers

 

[Calc1]

using method of section and take a moment about the apex, then ceiling joist tension force will be the lower one, which is the same as computer run, I think you should draw the FBD.

 

[pba]

I'm coming in late to this one, as I've enjoyed the discussion so far!

For a symetrical rafter arrangement, supported at the ends only (and neglecting cantiler eaves overhangs), half of the total vertical load (due to rafter load only) is the magnitude of the end reactions.

Now the member orientation at the support is the ceiling joist is horizontal and the rafter is inclined. Assuming axial load only in a pinned truss, the ceiling joist can not contribute. The vertical component of load in the rafter is thus equal to the support reaction. Given the geometry as 4V:12H the horizontal component of force is 12/4 times the vertical force. The only member which is present to resist this force is the ceiling joist so it will take THREE TIMES the vertical reaction force.

By my way of thinking it does not matter if the rafter is loaded with a udl or point loads as the ceiling joist tension force is a function of geometry and support reaction only. The only way I can think of which would change this is if the ceiling joist is otherwise loaded or if the entire weight is applied as a point load at the ridge.

Now, who agrees with me?

 

[1254BW]

Calc1,
Finally you got it right and you will agree with apsix.
I guess apsix try to explain the FBD at the support.
Imagine a king post truss - two truss elements at each roof slope (three node points).
Then, ceiling tension = 3/4 x Reaction x 12/4.

 

[miecz]

Seeing this long discussion, I decided to model this up and, to my surprise, the tension in the ceiling joist is indeed 688.5 lbs. Seems that the rafter carries some of the 459 lb reaction in shear. If you put the total force (918#)at the ridge line, the shear in the rafter is zero and the tension in the ceiling joist is 1377 lbs.

 

[SacreBleu]

jmiec,
That may be true due to the way your software handles, this, but the unescapable truth is that the tension in the ceiling joist is 1377 lbs no matter what. The problem with the way statics is taught in college is: You aren't informed of the "real-life" considerations when doing those textbook examples, nor is the professor savvy of the pros/cons of different methods.

 

[Structr]

I agree with PBA - your explanation is very clear. I drew this thing out - including the 4:12 slope triangle. It is really obvious that in order to solve this problem, first turn off the computer, get a pencil, calculator, and common sense.
Reaction = 54x17/2=459 lbs
Tension in ceiling joist = 459x3=1377 lbs (by proportional triangles).
I am amazed that more than one computer program computes anything so basic incorrectly!

 

[bylar]

I have read all the above comments and at first I thought that it was just a misunderstanding of the facts. It seemed to be a simple static’s problem and I think that Reaction = 54x17/2=459 lbs
Tension in ceiling joist = 459x3=1377 lbs (by proportional triangles).

So I too ran it thru STAAD and got half the value.
Now I am utterly confused. I believe it is 1377 but I don’t understand how STAAD gets half that value.

I guess that is why I distrust computer programs.

 

[Structr]

It would be interesting to contemplate that if there ever was a failure, the plantiff's engineer would really seize upon this fact.

 

[1254BW]

It is amazing! that all forgot the FBD the first thing we learned in analsis class 101.
Case 1: one truss element at each slope UL(54plf) applied at top chord; convert it to nodal CL(229.5#), Reaction =459#.
tension at bottom chord = (459 - 229.5) x 12/4 = 688.5#
Case 2: two truss element at each slope UL(54plf) applied at top chord; convert it to nodal CL(114.75#), Reaction =459#.
tension at bottom chord = (459 - 114.75) x 12/4 = 1032.75#
Case 3: one truss element at each slope TL(54plfx17'=918#) applied at the ridge; no nodal CL, Reaction =459#.
tension at bottom chord = 459 x 12/4 = 1377#
Case 4: two truss element at each slope TL(54plfx17'=918#) applied at the ridge; no nodal CL, Reaction =459#.
tension at bottom chord = 459 x 12/4 = 1377#
Calc1, run each case with your favorite program and verify; how the puter program do more than manual calc and be amazed by it's capacity!!
yep, time to time we need to brush our brain . . .

 

[1254BW]

oops forgot spell check! posting again
It is amazing! that all forgot the FBD the first thing we learned in analysis class 101.
Case 1: one truss element at each slope UL(54plf) applied at top chord; convert it to nodal CL(229.5#), Reaction =459#.
tension at bottom chord = (459 - 229.5) x 12/4 = 688.5#
Case 2: two truss element at each slope UL(54plf) applied at top chord; convert it to nodal CL(114.75#), Reaction =459#.
tension at bottom chord = (459 - 114.75) x 12/4 = 1032.75#
Case 3: one truss element at each slope TL(54plfx17'=918#) applied at the ridge; no nodal CL, Reaction =459#.
tension at bottom chord = 459 x 12/4 = 1377#
Case 4: two truss element at each slope TL(54plfx17'=918#) applied at the ridge; no nodal CL, Reaction =459#.
tension at bottom chord = 459 x 12/4 = 1377#
Calc1, run each case with your favorite program and verify; how the puter program do more than manual calc and be amazed by it's capacity!!
yep, time to time we need to brush our brain . . .

 

[Structr]

I have to keep repeating this over and over (it's pathetic, I need a life)...When you are doing your cases , you are getting an incorrect tension in bottom chord, due to the conversion of UDL to nodal loads. Merely a schoolboy exercise.

 

[SlideRuleEra]

Don't throw out your software yet! Tension in the bottom chord really is 688.5 lb. Rather than try to explain in words here, have put a short, zipped PowerPoint presentation on my website at the link below. The file, named "Truss Analysis" is near the bottom of my home page. After unzipping, the presentation is in manual mode (keep hitting the enter key) so that you can go thru it step-by-step. Several of my continuing education PE review students did not believe it either, but it is true. Will discuss as requested.

http://www.SlideRuleEra.net

 

[miecz]

Leaving the error prone software out of this, take a free body to one side of the ridge line, then sum moments about the ridge line. Then 459lb*8.5ft-54plf*8.5ft*8.5ft/2-8.5 ft*(4/12)*T=0, and T=688.5 lb.

 

[SlideRuleEra]

I neglected to say that PowerPoint offered is merely a graphic REPRESENTATION of a hand solution using ONLY the method of joints. - it is NOT a software solution.

http://www.SlideRuleEra.net

 

[pba]

SlideRuleEra,

End reaction of 459 lb is agreed. Why divide by two in the analysis? That is where I would diverge from your logic/

 

[bylar]

I am ashamed for myself and my colleagues.

As was pointed out in prior posts if we take a free body diagram and take the moments around the ridge we have the load acting down and the reaction acting up and the joist (bottom chord) counteracting the results of these two moments and that is how we get the 688#

I am ashamed because I have been working for over 40 years and have gotten away from the fundamentals of engineering and statics.

What also upsets me is that so many others apparently have fallen into the same trap.

 

[TTK]

Well I'm also not too big to admit when I am wrong...

In my initial haste to answer the post quickly, I solved the problem assuming the rafter was carrying axial load only (truss approach) but then later condemned those who tried to use a pure truss analysis method.....my apologies.

As the simple FBD noted above shows, the rafter takes a significant portion of the gravity loads in shear since it is a frame member and not a truss member. The rafter axial load is greatly reduced, which also reduces the tension force in the ceiling joist.

My lesson learned....slow down and analyze the problem instead of trying to post a "quick" reply!

 

[SlideRuleEra]

pba - The 459 Lb is not divided by two, the vertical forces of +459 Lb and - 229.5 Lb are summed. This is done at each joint. The example that we are working on has the disadvantage of being symmetrically loaded, making it difficult to really follow the steps used in the method of joints.

Take a look at this slight difference in loading (54 lb/ft UDL & 300 lb point load)

I have placed another zipped graphic PowerPoint method of joints solution on my website. It is titled "Truss Analysis With Point Load". The results are somewhat surpising.

http://www.SlideRuleEra.net