Timber & Housing Sizing

[jander3]

I've started on a plan for a timber frame shed.  This is a little different as I am more of an organic builder.  Start peeling, cutting, and assembling, then, within 2 or 3 days it becomes clear as to if you are manufacturing a table, an outbuilding, or very labor intensive firewood.  So a plan is a little out of my comfort zone. 

Anyway, I've read a few books on calculating loads and so far I haven't really followed the math.  My building with be 10' x 12'. Below is the general plan.  I will cut three bents for practice, probably not needed.

When I consider a snow load of 30 or 40 pounds and since the building is small, my thoughts are that 6" x 6" red oak or Aspen will be fine.  I will probably use 8" x 8" for practice cutting timbers on the TimberKing 1220, so I'm sure there should be no problem.  However, I have no math to back up my thoughts. 

Looking for the following information:  In the design below what is the load on the tie beams and posts. What size is needed. I think I can figure this using my books.  But I am interested in any thoughts.

The main issue I don't understand is how do I calculate what size shoulder I need for the tie beam.  This looks like the weakest point in this basic frame and I'm not clear on how to figure the load and shoulder size.  Any help would be appreciated.

[Jim_Rogers]

I've started on a plan for a timber frame shed.  ......

  My building with be 10' x 12'. Below is the general plan.  I will cut three bents for practice, probably not needed.

When I consider a snow load of 30 or 40 pounds and since the building is small, my thoughts are that 6" x 6" red oak or Aspen will be fine. 

First of all if using a strong wood like oak for the posts then 6x6 in this size building will be fine. 8x8 will be overkill and make things very heavy.

I won't say 6x6 are ok for tie beam, yet, as I don't know for sure if they will or not.

Basically, that is what you're asking, and want to understand.

Ok, so lets look at your drawing again.

The middle bent will carry the most weight as it has roof over it on both sides. And if we size this middle tie beam to support the load then the two other tie beam won't have any problem supporting them as well. And usually we figure the load on the one that's going to carry the most, first and just size the others the same size to make all the joinery the same, the timber list the same, and it looks nice with all the timbers the same.

If it was a cost issue we may only make the middle one larger and the two outer ones smaller, but I'm going to assume you can make all the timbers the same size, as you have drawn.

So, we now know which timber we're going to figure the size on.
And we do know the dimensions of the frame, and an estimated snow load. We'll figure that you're going to have a heavy winter and we'll use 40 lbs.

Now, that's 40 lbs per square foot of snow load, on the roof. Ok so what's the roof made up of? I'm going to assume boards or planks over rafters, and not a metal roof.
If it were a metal roof then the "load duration" would be less as in the winter when the sun shines on a metal roof and warms it up the snow slides off and it doesn't stay up there as long as an asphalt shingle roof.

But for this example, we'll use a shingle roof. Next we need to add the weight of the roof materials to the load. The snow load is called the "live load" and the roof materials are called the dead load as they are there all the time. And the live load comes and goes depending on the season and weather. So adding the live load and the deal load together we get a "combined" load. Some calculators require separate entries some want one total load entry.

Now we know that the frame foot print is 10' x 12'.

Like this:



Next we'll add the center line of the middle bent and the roof ridge line to this outline:



To figure the amount of roof area that will be supported by the middle king post coming up from the middle bents' tie beam we need to understand that the load is half the distance between each bent, and half the distance to each wall plate. So creating half distance lines and isolating the section of roof we get this:



This shows us that rear of the roof is 5' x 6'  being held up by the center king post, and center tie beam. So, 5x6=30 sq ft.

The center king post and tie beam are holding up 30 sqft of roof.
Standard roofing materials, shingles boards and planks rough out to about 10 lbs per sqft.
That is the dead load. So 40 lbs (live load) plus 10 lbs (dead load)= 50 lbs per sqft "combined load."
Next 50 lbs x 30 sqft = 1500 lbs load on the tie beam.

Now we need to figure if the 6x6 tie beam can support the load and not break in the middle.

Going to the Forestry Forum Tool box we can look at DonP's calculators.

We will select the "beam and column" calculators section.

In that section we will select "single beam point load mid span" as this calculator will basically do what we want and help us understand what size beam will work.

The first line we have to enter is the load on the beam in pounds. We just figured it as 1500 lbs. We enter that, and look at the next line.
It asks us for the width of the span in inches. As this is the width of the building which is 10' converted to inches we enter 120.
Next it asks for the width of the beam and the depth of the beam in inches, so we enter 6 and 6.
Now it asks us for the values of the wood being used.
In order to find that we need to select one from the attached list.
Opening the list we look at red oak. And we see two sections: Beams and stringers, posts and timbers. And as our timber is a 6x6 it doesn't qualify to be a beam or stringer.
(If a timber is deeper then it is wide by more then 2" then it is assumed that is going to be used as a beam and therefore becomes a piece of lumber from the "beams and stringer" section.)

So we have identified which section, we're going to use as "posts and timbers" next we need to understand or select the grade. The grade required for timber frame structures is usually grade 2 or better. So we'll use grade 2 values of 725, 1.0 and 155 for the three values we need in DonP's calculator.
Enter these values and press "show results".....



and we see several sections we need to review and understand.

Let's look at them in the next post....

[Jim_Rogers]

Let's look at the last section on the right first.

This section is called "shear" :



This means will the fibers of the timber shear between each other going from end to end of the timber. When you bend a magazine from end to end the faces of the pages will slide a little so that the magazine can bend. This is shear, and when the fibers shear then they don't have the combined strength of the timber.
We see that the needed amount is 7.25 and we got 36 so this passes, for sure.

Next lets look at the middle section, called deflection:



What this tells us is that this tie beam is going to deflect or bend down under this load .5 of an inch or 1/2" from end to end.
The maximum allowed for a floor system to deflect is .33 inches and this exceeds it. But as this is a shed or out building and won't have any sheet rock ceiling to crack we most likely could let this go. And we see in a roof it would just pass.

So lets, finally, look at the first section. It is called "bending":



The calculator figures the load as a term called a "moment".  And the size of the timber needed to support or resist that moment is 62.06 and we have only 36 so this size timber fails in bending.
Usually if we can solve the timber sizing for bending, it will automatically solve it for the other two needed sections of the calculator.

So a 6x6 grade 2 red oak timber will not support this roof.
Yes, there maybe some load transfered over the center ridge plate to the other king posts and down them to the other tie beams, but I wouldn't build it with a 6x6.

As we have everything entered into the calculator we can change one value and hit show results again. Just change the 6 to an 8 in depth and press show results:



Now we see that the beam needs 62.06 and we have 64 so it passes.

This tells us that a 6x8 tie beam will hold up that weight and it will deflect less than 1/4" and it will not fail in shear.

Jim Rogers

[Jim_Rogers]

Now we need to look at the ends of the tie beam and see what area is needed to hold this up at the post.

So, in review, we have a 6x6 post and a 6x8 tie beam joining it.

Like this:



And when we pull the tie beam out we see the joint like this:



And to understand the area of the mortise and housing that will support the load we see that area as this:



It is shaded in blue.

To understand what I have done, I'll tell you that I created a through tenon and mortise with a 1 1/2" thick tenon, and a 1/2" housing. This is a standard tie beam to post connection. You use a through tenon so that the tenon is long enough for the peg hole spacing requirements. And the housing is a standard 1/2" under the actual size of the timber, standard rule. The tenon width (1 1/2") is because of the rule that tenons are 1/4 the beam thickness (6" / .25 = 1.5").

So we get the shaded area. Which is 1 1/2" wide by 5 1/2" or 8.25 square inches. And the housing area is 1/2" x 6" or 3 square inches. So the total area in blue is 8.25 plus 3 or 11.25 square inches.
This 11.25 square inches will be supporting half the load figured above in the previous stories.
That is, in review, 1500 lbs divided by 2 or 750 lbs. Each end of the tie beam holds up half the load.

Now in the post the load is being held up by the end grain of the post.

In the tie beam the load is being held up by the side grain of the tenon and the area of the beam that rests on the housing. This is perpendicular to the grain.

In order to see if these areas are enough to support this load we have to first find the values of red oak parallel to the grain and values of red oak perpendicular to the grain.

This values are known and are published in books that designers and engineers use. These values are in a book called the NDS, Nation design specifications.

I have one of these books and use it all the time. It has many tables of values in it.
We need to use the table for the right size timber. The timber size is larger than a 5x5 so we use the table labeled "4D" for timbers 5x5" and larger.
Next we find that sections for red oak.
And we find the line for red oak grade #2 post and timbers again. This time we use the values in the column for "Compression perpendicular to the grain" F(with a subscript c and a symbol that looks like an upside down "T") and it reads 820. And that is 820 lbs per square inch. So this means the tie beam will not fail as we have 11.25 inches to support 750 lbs.
Next we look at the column "Compression parallel with the grain" F(subscript c) and we see the value of 350. We have 11.25 sq in x 350 = 3937.5 lbs of support and we only have 750 lbs to support so that area is ok, even with some shrinkage.

To to answer your question of the size of the mortise or housing to support the timber, the above design will work ok and it should support it fine.

Of course that is without any floor load on the tie beams....

Jim Rogers

[jander3]

Jim,

Thank you for your time on this.  I owe you.   Your explanation is clear and concise.  I didn't think it was rocket science.  Well, actually, I am a nuclear engineer and it was driving me nuts.  I scoured "Build a Classic Timber Frame House" and the tables and engineering information in Chappell's "A Timber Framer's Workshop." The formulas were straight forward but I couldn't figure out how to apply them.  One read through your posts above and it is clear.   I do plan on exploring the formula's that support Don's calculations some more to satisfy my curiosity.  

Your explanation shows me why the tie beams in most of the books and drawings I have are a bit larger in size.

One last question:

If I eliminated the middle bent in the above design, would the load on one king post (on the gable) still be base on 30 sq ft (1/2 the load since there is a post on each end)?  Or do I have to take a different load because the posts are on the end.  I'm thinking the post load would be the same, but the size of the ridge would need to increase to support spanning 12' instead of 6'.


Jon

PS

If you were to write a book or something, this type of detail on loads sure would help a guy figure things out.

[Dave Shepard]

I didn't read all of Jim's posts, maybe later, so if this is already covered, just ignore me. :D I find that in these small frames, joinery requirements often trump the engineering. What I mean by that is that you often have to use a much larger timber just to accommodate all the mortises and tenons in a particular spot. In my last project, I had a brace on opposite sides of a post. Had I used a 6x6 and housed both braces, I would have been limited to 2.5" brace tenons. The oak 6x6 would have been more than adequate, but I went with a 7x7 to accept the braces.

[jfl]

One last question:

If I eliminated the middle bent in the above design, would the load on one king post (on the gable) still be base on 30 sq ft (1/2 the load since there is a post on each end)?  Or do I have to take a different load because the posts are on the end.  I'm thinking the post load would be the same, but the size of the ridge would need to increase to support spanning 12' instead of 6'.


Jon

Hi,

Yes the load of the 2 external bent will be the same as calculated above by Jim.  Instead of having 1/2 of the central load on the central bent and 1/4 of the load on each external bent, you get 1/2 of the load on each bent, so you get the same calculation for the tie beam.  However the ridge beam now has to take the load from 5x12x50=3000. Basically, the load that was on the central king post in Jim's calculation is now on each of the end bent.  Where is that load coming from? From the ridge beam! I'd say that you have to look at the load like something that "flows" from top to bottom, sometimes dividing itself between members, sometime joining (adding) together on a downward member. 

So if you enter 3000 lbs, 144 inch span, 8x6 beam size an same wood design values as Jim used into Don's Calc Uniformly Loaded Simple Beam calculator, you'll see that section modulus required is 74.482, so a 8x6 does not do the job.

8x8 would work (85.333 section modulus).

jf

[jander3]

Thank you jf, with your input, I think I have it now.