3 Bent Pavilion Build

[scoots_36]

Hello All,

I am in the planning/design phase of a 20x20 3 bent pavilion for my backyard. 

The original plan was for the bents to have a structural ridge beam on top of a kingpost that would sit on the tie beams. When I ran the Midspan point loaded beam calculator in the tool box, I realized that my tie beams would not be able to to support the load on the kingposts.

My next thought was to use kingpost trusses by adding top chords and struts to distribute the load more evenly. (See my not-to-scale drawing attached.) In my limited knowledge through research on the forum, my understanding is that by using trusses, it would take the kingpost out of compression and into tension while transferring the load through the top chords down to the posts. Is that correct?

If it is, do I still need to calculate the load on what will now be the bottom chord? And if so, using what method? Also, how would I calculate what size the top chords would need to be to handle the load?

Thanks in advance for any help you can give!!




 

[Don P]

That is correct.

There is an excellent kingpost truss article written by the TFG for the Nat'l Park Service. Throw all those terms in a search and it should pop up.

If there is not some other load sitting on it the bottom chord could be a cable. It is in tension and if large that is worth checking. I suspect you'll be fine in tensile strength by the time you have a 2x6 worth of material across it.

Shear of the heeljoint and its design is really the place to start. That plane in red wants to shear off, that's usually the toughest joint in a truss, restrain that thrust.


 
Look at the crushing faces of that joint, check for compression in the top and bottom chords. the angle of force to grain , it is neither perp or parallel to grain. Use the Hankinson formula to interpolate between those two published allowable strengths for your species and grade. The result will give you the allowable compressive strength in crushing at that angle to grain. Dang right I'm lazy, bisect that heeljoint angle and do that math once!

A shear failure there would really ruin anyone's day under there, they are snap-whump failures. Bury a healthy piece of allthread, nuts and washers or similar in there. Check for vertical shear in the remaining section between notch and post if you are like mine here. I stuck the post tenon up through the top chord as well. Nothing wrong with belt, suspenders, and baling twine.



 

The kingpost needs to be pinched between the top chords and then extend down to the bottom chord. I would make it around 1/4-1/2" short in the shoulders at the bottom chord, ratchet strap a slight upward crown on the bottom chord so that when things shrink it ends up flat and just a tension element, not some quasi kind of rope, kind of beam.


 

This set was lifted up under the double 2x12 structural ridge beam, it sits on the flat top. Then you can see where i lifted inside of framed pockets in the wall and persuaded jack studs underneath to make the trusses take the ridge load.





I had to prove the bottom chord was a tension element with this one, that's a 2 piece bottom chord with a spline through the kingpost, the span was longer than I could saw;


 


Top chord, is kind of a column, kind of a beam  :D... it is the quasi element here so here goes. In a pure theoretical truss the only loads would be applied at the nodes, or panel points, depending on who you are talking to. That is, to be "correct" there could be a ridge beam sitting on top of the kingpost and another on each side over the web strut intersections. "IF" that is the case, "THEN" the design of the top chord is a simple column, pure compression, buckling is the failure mode and check.

In the real world that is rarely the case, usually there is either load applied uniformly or to points other than at the panel points. In that case the internal forces of the truss are still there, the top chord is still a column in compression. It is also a beam in bending. The more axial compression down the length of the stick, the less bending strength it has. Or, the more the stick is being bent from the side, the easier it is for it to buckle sideways if pushed on from the end. There is a direct interaction between the combined forces and an interaction equation that is used in that situation. If you want to get queasy it is in the NDS under the publications tab at awc.org. The NDS is the building code referenced manual for wood design.

I've designed a few, that doesn't mean you should necessarily listen to me  :D. That equation did make me queasy and I need to revisit it. I went backwards in time to a simpler version that was in an earlier edition of the NDS from the 40's and finally understood it well enough to get that to work. I got the outputs I needed and have not worked on it more but that calc is here;

Untitled (forestryforum.com)

[scoots_36]

Thank you so much for the detailed reply. It really helps me wrap my head around what is going on. 

My joinery skills are elementary at best so your diagrams and pictures made me realize that I may be better off going with post and beam construction using faceplates to reinforce the joints for my first large scale project.

Not necessarily the look I was hoping for but sometimes peace of mind is better than the perfect aesthetic.

[Don P]

Uhh, yeah, on that. This is simply another method of framing, farmers 150 years ago all had the tools and the same construction skills as farmers today. Don't let perfect be the enemy of good, if you can saw replacements this is a good project to learn on. Steel plates do not negate the need to get good uniform wood joints.

There are a number of ways to do this though. This is another way to restrain the rafter thrust without exposing steel at the heeljoint. Calculate the thrust, divide by the fastener single shear value, and run in a few extras. You can also bury a steel rod in that upper face all the way to the opposite side and weld or nut and washer that heel stop. Anyway, that becomes a simple cut to fit. To get that fitted up nicely I did my best then blocked everything up in plane fitted the members together and ran a handsaw through the joint, kerfing it until it seated nicely. Then I plowed a groove for the steel and assembled it

[scoots_36]

Once again thanks for the reply.

The thrust vs shear discussion brings up something I've been wondering for a while...

Would it not be possible to cut a nice clean heeljoint and then send some structural screws through the top chord into the bottom to cover the shear? A 3/8" GRK "RSS" Screw has a shear rating of 3,695 lbs. Here in sunny California with no snow load, that would cover quite a bit of thrust quickly with multiple screws.

I must be missing something because I don't see many people doing it that way. 

[Don P]

Hmm, I see the shear strength of the steel in a 3/8" fastener is 3,695 lbs here;
RSS - Sell Sheet_18-Front Alt (grkfasteners.com)

What you're looking for is the allowable load in wood of your specific gravity in single shear. I suspect that will be in the low hundreds per fastener.
...
...

Here's the meat and taters,
For proprietary structural stuff find the ESR;
ESR-2442 - GRK Fasteners, A Division of Illinois Tool Works, Inc. (icc-es.org)
Scroll to pg6 table 3. look down to the 3/8 x 7.25" and longer. You're shearing parallel to grain there at the heeljoint, Dougfir, SG runs ~.45-.50, ahh sunny, southern dougfir use .45, WRC probably around .35. Assuming green and exposed, I'd multiply by the wet service factor and only use 70% of that dry service allowable. In southern dougfir with a 3/8 x 8" GRK RSS loaded in shear parallel to grain I'm seeing 423lbs x .70 or 296.1 lbs/fastener. Read the rest of the ESR for connection details etc. This is what I design by and am ready to turn over if he asks for papers please, these are the real numbers. You'll notice in reading that, the sales stuff I pulled form the company website says 3695 steel shear strength, the ESR says 1231, No idea what the issue is but the ESR is what matters.

Anyway, there's how to find the allowable shear. Divide the load by the allowable load per fastener to see how many fasteners are needed, then check the spacing specs and see if you can get them in there. There is nothing at all wrong with that approach. Quantify load, provide quantifiable resistance.

[Ljohnsaw]

Bury a healthy piece of allthread, nuts and washers or similar in there.

Don, I'm not an expert but I was told that is not allowed by my engineer.  The wood must rest on a smooth bolt shaft.  In my case, I needed 6" of non-threaded in each timber (so 12").  Since the bigger bolts have 2" of thread, I had to order 14" bolts (3/4" and 1" dia).  Also, the holes must have less than 1/16" or 3/32" (IIRC) of extra space.  So, no over-sizing or drilling crooked and straightening out.  Maybe they figure the allthread will act as a file with heating/cooling expansion/contraction file it's way out the side? :

[Don P]

I was checking in to modify today's faux pas and you found another  :D
Well, I have the NDS open lets see.

I started in the main section on dowel type fasteners in the mechanical connections chapter and got sent to Appendix I where it goes into more detail, and worth reading.
Basically, you can use the full body diameter of the unthreaded portion of a bolt IF the threaded portion occupies less than 1/4 of the length of the hole in the outer ply of the bolted connection. Or maybe an easier way to think about it. At least 3/4 of every member in the bolted connection needs to be bearing on the unthreaded portion of the bolt... if you use the full diameter in your calcs.

Otherwise, use the root diameter when looking at the tables or doing calcs.
If we are using that allthread as a real shear fail safe in my heeljoint it needs investigating as such, assume the chunk of wood in red on my heeljoint drawing has sheared off and the threaded rod is what restrains it. I can't remember but I think there was around 7 kips of thrust in that heel. Root of that 1/2" allthread was probably 3/8... and I'm below the chart, bummer  :-[. Using full diameter smooth 5/8" would be good for about 1,000lbs. A 1" dia would be good for around 2 kips. I need 7, we're going to run out of wood before that can really do the job. I'd still do it, just give me three steps towards the door, and it holds the joint together in uplift, but excellent point. When you use the root diameter you take on more damage (net section loss in the wood) and less strength (the small diameter of the steel) than if you use smooth and can use all the steel and all the hole.

Oversize on a bolt hole should be 1/32 to 1/16" oversize in the NDS. In something like a truss, account for that drift before it locks up.

Ha, drift must have been in the back of my mind. If you call a smooth stick of steel a drift, you can drill up to 1/16" undersize, drive the drift in (no splitting allowed) and use the full diameter for shear. The joint isn't going to move about. Uplift is low.

You can probably chase that rabbit down more in the wood handbook.

For my faux pas on today's post, there is potentially an angle of grain reduction/adjustment that might be applied to the GRK shear strength. The nice thing is most of these companies have good tech departments. If going that route I'd draw and dimension what I wanted to do and run it by them.

In the end what we're seeing is the thrust is probably going to exceed fastener capacity on all but the smallest truss going that route. You are probably right in the gray.  For that type of simple sawn and butted joinery as it sizes and loads up I think it'll take something more like split rings or shear plates and bolts, John's properly sized bolts!, to get there.

[Rgdsolution]

Hey I see you guys digging deep on this stuff and that's what I keep doing but my sense keeps telling me to trust the grey areas even if I don't like thinking that way usually. Let me ask you this. When you say 7 kips thrust on the end of that beam I think we'll geez if that wood grain happens to be nice and straight then who knows but to me that's like the difference between dead weight and a rope that's got friction on it. You know how you can use your finger to hold up more weight than your arm can hold if it's wrapped around something once or twice- I picture a metal piece or even a couple of through bolts right after the heel cut as like the finger holding a knot from unraveling. Would you agree? I have a similar maybe simpler version. I have 5x7 rafters over a 8" wide plate and it has 4" of overhang on the inside of plate. I feel confident it would be good but to ensure the rafter doesn't split down the middle I want to cut a little square out of the top and put a rafter clip or just two 1/4" screws right through to the plate. I kind of wonder what the actual propensity it has to split is but I feel like it's kind of like the finger on the knot and so if their before it splits it should be perfectly fine. Have you ever seen anything still fail even after being overprotected like that??

[Rgdsolution]

I meant a square out of the tip for the rafter clip 

[Don P]

Or within it. there were/are a number of cast "shoes" and other methods of shear enhancement (which I guess is really what we are talking about). The most common method nowadays is probably split rings or shear plates but spike grids, those gang nails plates, etc are all up to the same thing. Looking at old church and industrial work shows some neat solutions that are kind of out of our box, just make sure they, and then you, understand what is going on.

[Rgdsolution]

I was just reading in Steve chapels book and came across a shear note: 'by far the most common cause of beam joinery failure is due to horizontal shear stress' I thought since that was the topic here yesterday I'd share that. Though I figure it could happen I didn't realize it is prevalent. Maybe another knot in the rope could work better than a finger. I will say sometimes It seems crazy how much extra figuring goes into stuff that looks full proof in a picture but better safe than sorry I guess. 

[Don P]

 :D Remember full proof is half strength.

That sounds about right. You've seen here we can usually talk through bending strength pretty quickly and get that into the right range. Connections are the most common failure point and shear is probably the most common mode of failure there. Something like that heel shearing, the relish of a tenon beyond the peg tearing out in double shear... and one we saw just the other day that kind of frosts my cookies. Someone posted asking about the strength of a structural screw. The numbnuts in sales widely published the ultimate strength of the steel. The allowable, useable strength of the steel is 1/3 of that. But the steel in a wood joint probably isn't the limiting factor. The allowable shear strength of a wood joint using that screw is around 1/10th of what the sales nuts are publishing.

Methods of shear enhancement in joints usually involves spreading the stress over a larger area. Using many nails in a joint rather than one bolt might not look as pretty but it is often stronger and more resilient because it spreads the load over a wider area that the wood can handle. Also if one nail fails the joint is probably still quite strong. If the relish behind one bolt fails, you're probably not redundant enough to say the joint is serviceable, in fact it may be painfully obvious. The gang nail plates on lightweight metal connected trusses is the same thinking. All those little teeth are transferring around 80 lbs per square inch across the joint. Well within the allowable stress on the wood. Split rings and shear plates are doing the same thing. Instead of the bolt bearing on the wood with just the diameter of its shank, the plates or rings increase the effective diameter and surface area in contact with wood to several inches diameter and the bolt shank is bearing on steel or iron.

In old work when there is a horizontal shear failure underway I've seen structural screws driven across the member to reinforce and hold the piece together.