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Beam span and general structure information for post and beam pavilion

Started by Eric_M68, January 23, 2021, 09:34:57 PM

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Eric_M68

Hello everyone,
 
I'm planning on building a pavilion next summer and I have a few questions regarding the structure and others points.

I'm a hobbyist woodworker and I have some knowledge about home renovation but not at all about structure (span, load, etc.)

I live in southern Quebec where there is snow but not that much (10" on a month). But there's rarely more than 3-4 inches of snow covering the shed at all times.

I based the following information on online plans I found, videos I watched (which I found very educating).

Timber available to me is pine or hemlock (not sure which specific species)

The pavilion will be either 14'x14' or 12'x14', the longest length being front and back.
The posts are 6"x6", 7' high (my first plan used 8'x'8').
All beams, king posts are 6"x10".
The curved knee braces are made from 4"x10". They 'span' 2 feet in each direction.
Rafters are 2"x8", 6'6" long, 12" on center, and overhang is about 8". Roof angle is 25 deg (5.6:12 pitch).
The roof will be covered with asphalt shingles.

Not sure what to use between the rafters and the shingles.
1"x4" (or 6") strapping covering the whole area.
1"x4" (or 6") strapping spaced 12" with a 3/4" plywood.
Just 3/4" plywood.

Now, the beams... I've read a lot of posts, with a lot of information but most of it I just don't understand. It's the technical terminology that loses me (being French speaking doesn't help). Tried using the calculators but I just don't know what everything means. Sorry.

Are the 6"x10" enough? 6"x12", or maybe 4"x"12" would be better but, because of their height, I think it would look out of place.

Do the knee braces offer enough support to make the span smaller?

Let me know if more information is needed.

I can upload my sketchup file if need be.

Thanks in advance for the information...



 



 



 

LarryG

Very nice drawings. What software program are you using. Sorry I can't be any help I'm in the learning stage myself. 
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Don P

Your English is far better than my French  :D
If you can post the file in sketchup 2016 it might help with explaining. I'm assuming the species are either eastern white pine or eastern hemlock? @SwampDonkey
The biggest unknown I see right now is snow load. We are designing for the heaviest snow load in 50-100 years rather than average. See if you can find the design snow load for your location, google that term and your local jurisdiction and see if you can find it. From there we'll go slow and see if it makes sense, often the math and terminology "clicks" on our own project.


Here we go;
SCAN0002.pdf (fore-engineering.ca)
and another;
jabacus - Snow Load Calculator

Braces do not reduce span. One old adage to remember is "load goes to stiffness" then think about where the true stiffness is in that area, it's really the post.

SwampDonkey

Probably eastern white pine, eastern hemlock. Although, red pine and jack pine range there, of the two I think you can find jack pine in the hills between Quebec city and Chicoutimi.
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Eric_M68

Quote from: LarryG on January 24, 2021, 06:22:08 AM
Very nice drawings. What software program are you using. Sorry I can't be any help I'm in the learning stage myself.
Thanks. This was made using Sketchup.

Eric_M68

Quote from: Don P on January 24, 2021, 07:34:36 AM
Your English is far better than my French  :D
If you can post the file in sketchup 2016 it might help with explaining. I'm assuming the species are either eastern white pine or eastern hemlock? @SwampDonkey
The biggest unknown I see right now is snow load. We are designing for the heaviest snow load in 50-100 years rather than average. See if you can find the design snow load for your location, google that term and your local jurisdiction and see if you can find it. From there we'll go slow and see if it makes sense, often the math and terminology "clicks" on our own project.


Here we go;
SCAN0002.pdf (fore-engineering.ca)
and another;
jabacus - Snow Load Calculator

Braces do not reduce span. One old adage to remember is "load goes to stiffness" then think about where the true stiffness is in that area, it's really the post.
I cannot, for the life of me, find the button that allows me to upload anything other than pictures to my library. I figure I need to put it in the depository, so I read the instructions there but it doesn't work. I did post a question regarding this issue there as well...

As for the snow load, I check both links you posted and here's what came out... The city right next to mine is Beloeil, Qc. 


 


 

Eric_M68

Another thing I wanted to mention...

There's a company that sells "cheap" pavilions and pergolas (Yardistry) and those are being sold here. The beams on a structure that's 12x14 are made out of 2 2x6 with a metal plate sandwiched between them. And they don't even span the whole length. How can those be sold here? How can those support anything over a few inches of snow.



 

Don P

In the reply window click on the tab that says "Attach documents and other options" when that opens you'll see "Attach:" and under it click "choose file" use that to find the skp file on your computer and then click "open" and that should load it here. you can put it in this thread or in the plans repository, either one is fine.

The beam in your last post is called a flitch plate beam. Steel has high bending strength and stiffness, the wood provides thickness to take care of buckling in a thin piece. That said, that looks like a pretty miserable beam. Just because someone does something doesn't mean it needs to be repeated.

Thanks for the snow load info, it looks like we'll use 50 psf snow load for design. That is called the "Live Load". We'll add to that the weight of the roof materials themselves, known as the "Dead Load", usually 10 psf (Pounds per square foot) is adequate. So our total design load will be 60 psf.

I'll stop here for now and see if you can get a sketch loaded.

Eric_M68

Well, the attach file section wasn't there before... Maybe there's a minimum number of posts before this gets activated.
In any case, here's my Sketchup file.

And thanks for the load terminology explanation... baby steps ;)

Don P

I usually start at the top, collecting loads as I work my way down through the building, so let's start with the ridge beam. Live load is measured horizontally (snow falls vertically and the numbers we have are for level surfaces). Dead load is technically measured along the slope but we use an average dead load number that is safe to combine with the live load on the level, so I usually combine them... don't worry if that was a little too much info right now, watch how I do the math here.



 
Half of each rafter is supported by the lower plates, half of each rafter is supported by the ridge. So in this case our "tributary area" uniformly (evenly) loading the ridge is about 7' wide x 14' long, 84 square feet.

The design load we decided on yesterday is 60 pounds per square foot.
84 square feet x 60 pounds per square foot =5040 lbs on the ridge beam.

We decided the dead load is 10 psf, multiply that by 84 sf = 840 lbs DL (dead load)

Span is the distance between ridge beam supports plus a couple of inches.
I would call the span here 156"



 

I think you have enough to play with the calc here;
Design for Bending (forestryforum.com)
Let me know what you come up with.
Come back with any questions or if you get stuck.

Don P

Hmm, something's not clicking. Maybe I shouldn't have made the snow yellow.

Try this pic, it shows the load on the ridgebeam, maybe a little differently;


 
These are the inputs to type in to the calc;
Total Load: 5040
Dead Load: 840
Span: 156
Width: 6
Depth:10
Species: #1 Eastern Hemlock B&S
Click "Show Result"
You should get passes in bending, deflection and shear.

Try Eastern White Pine, #2B&S and I think you'll need to upsize the width and depth.

LarryG

Thanks Don P this helps me out a lot to understand it also. Nice detail on loads
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Eric_M68

Thanks a lot Don.

The "yellow snow" made sense to me ;) Your post really helped me understand how to calculate loads. Much appreciated.

I did select the same species as you and got good results. Just got to be careful which grade I select. Speaking of which, I'm guessing P+T means pressure treated but, what does B+S mean?

So, I'm guessing that since all my beams are the same size, and that the ridge beam is the one supporting the most weight, I should be ok. Right?


Don P

Oh we're just warming up  :D.  For the plates if you want to optimize size use the same calc but your trib area will be about 4' x 14'.


 

There is another gorilla in the room. Each tie beam has half of the ridge beam load landing in the center of it, a "midspan point load". A concentrated load at midspan produces twice the bending moment in the beam compared to the same load evenly distributed along the beam.

What do you do if you are walking on a pond and hear the ice crack? Spread your feet wide or lay down. Standing with your feet together is a concentrated point load, spreading the load out lowers the bending stress. Our feet are together here, the ridge posts are a concentrated load on the ties.

Our ridge beam load was 5040 lbs so each support post is carrying 2520 lbs and delivering that load to the center of the tie beam.


 
The correct calc for that situation is here;
Midspan Point Loaded Simple Beam (forestryforum.com)
In that calc you will have to manually enter the design values for Fiberstress, Modulus of Elasticity and Shear. The easy way to find those is from the previous calc but those design values are also listed here;
Maximum Allowable Fiberstress in Bending Values (forestryforum.com)

P&T: Posts and Timbers, square-ish sections, most often posts but can be used as beams. They are less efficient as beams and have slightly lower design values.

B&S:Beams and Stringers, members that are 2" or more different in width and depth, 6x8, 8x12, etc. These are generally used as beams and have slightly higher design values.

I wrote this last night, it might be too much information or it might help someone, who knows  :D

I've maybe thought of another way to explain bending moment in beams. I think most people understand the basics of a torque wrench, we use it to put a calibrated amount of rotational force on a bolt and measure that force in foot-pounds. The same rotational force is what is being figured in the bending equations in the beam calcs.

If we have a lever 1 foot long and hang a 1 pound weight on the end of it, the fixed end wants to rotate with, as well as the lever experiences, 1 foot pound of bending force. The moment, that desire to rotate, is 1 foot pound.

If the lever was 2' long, 2 foot pounds. If the lever was 1 foot long and the load was 2 pounds, again 2 foot pounds. Moment is a function of load and span.

Imagine if the ridge beam above were rigidly fixed at one end, like a diving board. If the beam extended out 13 feet (156") and the load on the end was our 5040 lbs, the bending moment would be 13 feet x 5040 lbs, or 65,520 ft-lbs.

A simply supported beam, that is, a beam sitting on a pair of sawhorses, in other words the ends are free to rotate, is essentially 2 torque wrenches with the handles welded together. If you load it, it still wants to rotate, the left side wants to rotate clockwise, the right side counterclockwise. If you spread the load out evenly along the beam, the rotational force in this configuration is 1/8 of the moment produced by the cantilever described above. 65,520 ft-lbs/8 = 8190 ft-lbs

When you plug the numbers into the beam calc and notice the max moment in the bending section, there it is 8190 ft-lbs.

If that helped explain things we can visit this again when we get to the tie beam, a concentrated point load at midspan.

Don P

Oh, more on that too much information  ::)
Notice the maximum bending moment when you check the tie beam. The span of the tie and ridge is the same. The point load on the tie beam is half the load that was on the ridge beam but because it is concentrated at midspan the bending stress on each is identical.

Sod saw

Having read thru this series of lessons, I am impressed by the explanation and sketches.  They look as though my Dad (architect) was still hand drawing when he quit practice at age 95.  Do not ever stop making sketches, they help to "see things" that computer programs may overlook.

Here in NY State a building permit would probably be required in most townships for a building such as the one proposed and even though your math may produce a solid structure, , , an architect or engineer would be required to provide stamped drawings.

I envy you your project and wish you well.  Are you going to cut your pieces from your own trees?    Please post some photos as you progress thru your build.  

Oh, one last thing.  Don't make it too small. They tend to shrink as time goes on when more and more friends stop by to visit.
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Eric_M68

Allright... Here goes. That's a lot to take in. Frankly, I had to read your posts a few times to get it, for the most part.

For the lower plates, there's no problem since the tributary area is a little more than half the width (4/7) of the one for the ridge beam and I intend to use the same size beams everywhere. 

For the tie beam, and if I read you right, I should be ok since the point load supports only half the total load, even though its concentrated at mid point and poduces twice the force . Am I reading this right? It evens out.

What's the horizontal shear, though? This is where I see major differences between the two (half of the beam load to be exact).

Thanks again for taking the time to explain all this.



 

Eric_M68

Quote from: Sod saw on January 29, 2021, 12:35:28 PM
Having read thru this series of lessons, I am impressed by the explanation and sketches.  They look as though my Dad (architect) was still hand drawing when he quit practice at age 95.  Do not ever stop making sketches, they help to "see things" that computer programs may overlook.

Here in NY State a building permit would probably be required in most townships for a building such as the one proposed and even though your math may produce a solid structure, , , an architect or engineer would be required to provide stamped drawings.

I envy you your project and wish you well.  Are you going to cut your pieces from your own trees?    Please post some photos as you progress thru your build.  

Oh, one last thing.  Don't make it too small. They tend to shrink as time goes on when more and more friends stop by to visit.
Haha.. Thanks. I'm kind of limited in size since I already have some paver stone there. But, I rarely "entertain" that many people ;)
The posts and beams will be coming from a local saw mill. I don't have the patience nor the time to start learning yet another trade lol

Don P

QuoteFor the tie beam, and if I read you right, I should be ok since the point load supports only half the total load, even though its concentrated at mid point and produces twice the force . Am I reading this right? It evens out.
Yes, you got it. In this instance it produces the same bending moment so you are fine. I wanted to take advantage of the situation to show the difference in the two loading conditions, it seemed to be the perfect example. The tie with only half the load concentrated at midpoint is experiencing the same bending force as the ridge beam with double the load evenly distributed along its length.

QuoteWhat's the horizontal shear, though? This is where I see major differences between the two (half of the beam load to be exact).
Horizontal shear, take a thick phone book and bend it, watch what the pages do as you bend it. They slide past one another as you bend, that is essentially horizontal shear. Generally in a shorter, heavily loaded beam this becomes the "control" where as beams get longer deflection (sag), or bending (tearing of the "extreme", or bottom fibers) controls design.

If a beam does horizontally shear it typically rips down the centerline of its length, becoming two, thinner, beams stacked on top of one another. It then sags tremendously and if the force is high enough then tears out in bending, not good!

The overall equations for maximum shear for both of your conditions is essentially identical BUT, the tie beam is carrying half the load. That is why you are seeing half the maximum shear force in the tie beam. We should probably talk about this more, the shear force along the length of the beam is different as you move away from the centerline but the maximum shear, what we are most concerned with is the same, more later if you want.

Here's some pics I have stored here from when we talked about horizontal shear a few years ago.

First a stack of thin strips set up as a simply supported beam (simply supported: the ends are not rigidly retrained but are free to rotate, a typical wood beam)




Uniformly loaded, sorta :D, hey they were handy.
 



Notice the center has not displaced horizontally. This is a key concept, shear typically passes through zero at the point of maximum bending moment.
 



Maximum shear is at the ends, notice the horizontal displacement.
 

 

That last part is also important. The allowable shear check does take into account allowable checks or splits BUT, those only occur when the timber is fully seasoned, you won't see them in a green or partially dry timber. If you are getting close on shear in the calcs pay careful attention to the ends.

This is getting long and I think I just dropped a handful on you, better stop here for now to see if you have questions, but if you want to delve deeper just holler, actually the next step might help clarify this some, or turn it all into mud :D.

Eric_M68

Love the pics you used as an example. Really shows how the shear effect can be problematic. I never thought this could happen with a beam. The load must be really heavy for this to happen.


QuoteIf you are getting close on shear in the calcs pay careful attention to the ends.

What do you mean when you say to pay attention to the ends? The way they are attached to it supporting beam or post?



Quote ...actually the next step might help clarify this some,

The next step? There's a next step? I thought I was ready to start building now ;)

When I work on furniture, the wood needs to be dried carefully in order for the pieces to stay straight after assembly. Does the same thing apply when building outdoor structures? The mill where I called last summer to get some prices told me the wood would still be "green" (not sure what the term would be in english).

Also, I was thinking of using long lag screws to tie all this together (found those at Lowes: FastenMaster 50 #0 to x 10-in Ecoat Hex-Drive Structural Wood Screw) Would those be strong enough? Considering the way my beams interlock each other...

Don P

Yes, shear usually controls with short, heavily loaded beams.

These two beam diagrams show the equations behind the calcs but notice the shear diagrams. With the uniformly loaded beam the magnitude of the shear force increases towards each end where the internal shear force with a center point loaded beam is constant within the timber from the load point out. 

Also follow the line down through the diagrams and notice that where shear passes through zero is also the point of maximum bending moment.


 

 

My comment about watching the ends is that if shear is the "control" then look carefully at the end for checks, splits, joinery with sharp re-entrant corners (joist notches!)... things that tend to propagate a shear. 

I prefer dry wood but often as not am pressed by the same issues as everyone else just be aware things will be more likely to not fit as well later, bearing surfaces change shape or contact green wood is half as strong as dry, all the usual problems with using green wood.

I can't see the original drawing from this post window but will take a look at the sketchup, better post this before the computer eats it.

Don P

I think this area needs more thought;


 
The tie beam has a sharp cornered notch with the bottom edge of the notch unsupported at the post. There's that re-entrant notch at a point of high shear we were talking about. I'd be tempted to insert the bottom edge into the plate beam a bit there to support the bottom edge.

The screw through that joint is not getting into the post. If a screw is holding the roof to the end grain of the post I'd look at the roof as having about 15 psf of lift in high wind and check the withdrawal from end grain connection capacity of the connector.

The braces have no connection yet, this is an area where I am not comfortable. Open structures without bracing walls rarely pencil out. Just some thoughts. Nails and bolts tend to be inadequate if you do the math.  A mortise and tenon or at least a notch to capture the toe's of the braces to help lock them firmly in place. If you use green material shrinkage is a factor. Provision for tightening packing wedges would be one possible solution to keeping the braces snug. Shorter braces tend to jack the beams they are attached to up if they are asked to provide serious racking resistance, the longer the braces are, the larger that triangle is, the better.

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