Shed design: rafters and purlins

[bigmish]

The following is a response to Jim's post on another thread of mine "Skinning the shed: walls (using horizontal nailers)":

To design a purlin rafter system you would start with the purlin. And make it big enough to hold up the load over the span between rafters.

So you'll need to know your snow load, the span between rafters and the spacing between purlins drawn vertically to start.
Using Don_P's calculator figure out the size of the purlin.
Then using that size you can figure the size of the rafter you'll need to support all the purlins.

If you cut away too much of the rafter it will fail under a snow load.

My response:
I'm near Albany in upstate New York, The snow load for the area is 55 psf.

First, figuring the width of the purlins:
the largest area of the roof that needs to be supported by one purlin is approximately 18 ft², times 55 psf for snow load
= 990 lbs live load + 110lbs for dead load (estimate of roof weight) = 1,000 lbs total for roof section
a 4x5 beam of white oak passes all test. Note that I took the depth of the purlin at the joint (5") rather than at the center (6 11/16").

Now, figuring the width of rafters:
including the overhangs the largest area of the roof that needs to supported by one rafter is approximately 111ft², times 55 psf for snow load
= 6105 lbs. live load + 258 lbs. for the purlins (calculated at 83 lbs per purlins) + 157 lbs. for the roof (about 15 2x4s, insulation and interior ceiling)
a 8"x8" beam of white oak passes all test. This is the actual width of the rafter, but not at the joint.
a 8"x3" beam totally fails. This is the remaining wood left at the bottom of the joint when two purlins meet.
a 8"x4.25" beam also totally fails. This beam would be 34 in² which is how much wood there actually in  cross section at the joint (see the last image below)
Do I need to beef the rafters up? if so, do I take the measurement at the joints as the beam depth? Should I include all of the wood in cross section, can I fudge the width and depth to match the actual in² at this point (as I did above to get the 8"x4.25" beam)?

Thanks, Mischa

[Jim_Rogers]

First of all let's discuss how you came up with 18FT² for the load.
How did you get that?

Jim

[Jim_Rogers]

Can you show me a gable view?

[bigmish]

I'm running out the door now but will post ASAP.

In answer to your question: the length from outer edge of one rafter to the other is 7' 5", times the length from the other edge of the top purlin to the peak of the roof which is 2' 5 5/16" = approximately 18 ft².

[Jim_Rogers]

Maybe this will help.



Unless I'm totally wrong on this the load is figured vertically like shown above, not area on the roof plane.

Adjust my figures to make it 7' 5" in length.

Run the numbers again and I'll look at it Monday.

Maybe if farmer77 is following this he can comment if I'm doing this right or not.

Jim Rogers

[Don P]

Yup, you're doing it right Jim.

I'm curious also how the notching is handled. Section 4.4.3 of your '01 edition of the NDS wood handbook says we can't call this a full size beam anymore. My feeling is the section properties need to be refigured for the inverted T shape. Then a beam check based on those properties  ???.

[bigmish]

Jim: below is the requested image. I'll run the numbers ASAP and post them.

DonP: sounds like your saying that by the current handbook, to calculate beam size you use only the dimensions of "section B" in  my image below. That is to say if the rater was 8"x8" and "section B" was 3"x8" the rafter would be considered to be   3"x8", is this correct?   

[Jim_Rogers]

This is where figuring joinery gets very technical, because you're suppose to include both sections.
Bigmish, how are you with excel spread sheets?
I have one that does lots of the math for you.
I could forward that to you, and explain how to use it.

Jim Rogers

[bigmish]

I'm ok with excell; please send!

Thanks Jim, M

[bigmish]

In the mean time I ran the numbers again for the purlins.
I got a slightly diffrent area than you as I used the distance between the mid point between the highest purlin and the next highest purlin, to the ridge:
2' 9" 1/16 (I dropped the 1/16th)
times 7'2" = 19.7 ft² = ~20ft²
times 65 lbs. per ft²
= 1300 lbs.

This leads me to a question: when using the calculator, what should I use for the Span of Beam, 7'2" (the length of the purlin) or 6'10" (the distance between the rafters)?
It passes at the current dimensions for a span of 6'10" but fails deflection at 7'2".

Another question: when calculating the load on the rafters: I should add the weight of the purlins (I calculated 83 lbs. per purlin at the current dimensions), right? That is to say, that weight isn't figured into the 10lbs of dead weight per foot² that we used in the calculation for the purlins, right?

[Don P]

I can answer a little on the calc. Span is correctly the distance from face of support to face of support + 1/2 of each required end bearing. (clear span+ the number in the bottom left box normally)

I set up the pass/fail on deflection to check for floor joists, the most restrictive requirement. If you look at the calc there is an allowable maximum roof deflection (3rd box down under deflection), check that manually against the deflection output (top box). The top box should show a lower number than the 3rd box. A roof can deflect more than a floor, I've chosen a fairly conservative roof deflection at 1/240 span. Many buildings are designed with 1/180 span allowable roof deflection. A bouncy roof is of less concern than a bouncy floor. These are judgement calls, I tried to stay conservative.

Your dead load can be assumed, that 10-15 psf we're always kicking around, or it can be accurately figured by totalling up the actual weights of all the materials up there including the self weight of the beam.

Jim, I'd like a copy too if you don't mind. I'm still trying to calculate the new section properties longhand, and I'm slow.

My calcs do not allow manual entry of section properties at present, they are doing that internally for simple rectangles (your width and depth inputs). For complex shapes its on to the spreadsheet or longhand. The calcs can be rewritten to allow such inputs pretty easily if you know the section properties. I did this once to spec some steel angle, there are tables of published section properties for common steel shapes. Needless to say every wood shape is custom and it was getting complex fast, so I kept it simple to cover the most common useage, a rectangle.

I'll sit on my hands now, this is getting good.

[bigmish]

Just to confirm I understand you right Don P:
* the purlin is 86" end-to-end
* the pockets are 3" each
* therefor the distance from the face of one rafter to the other is 80"
* so the span to use is 83" (80" + 1/2(3"+3"))
Correct?

Also, how does one figure "required end bearing", what is the "the number in the bottom left box" (what box, where)?

Thanks, M

[Jim_Rogers]

Bigmish:
About length of timbers.
We were taught at the one week course I took that a structural engineer is going to figure that the piece (purlin in this case) will join the other piece (rafter in this case) and that the length will be half way into the second piece on each end.
The spread sheet I use figures one half the width of the second piece.
I believe that we should use the actual length of the piece not just the unsupported length, but I will follow Don_P methods if it comes to making a choice.

Now, I'd like to address the space under the purlin at the rafter, to bigmish.
This space is only holding up half the load, as this purlin has two ends, so when you figure the load at that point, reduce the total load by half and run the numbers, your original rafter may now pass.

I'm working on an explanation to the spread sheet and will forward that to you soon.

Jim Rogers

[srt]

Sorry I can't offer anything intelligent on the question asked.  However, I have an interest in this timber framing thing, and have been reading past posts about the programs used to do the drawings.  BigMish, what program are you using?  Thanks,  srt

[Don P]

For span, there's nothing wrong with using the full length of the timber, conservative is good.
Here is what I was remembering
NDS 3.2.1
Span of Bending Members
For simple, continuous and cantilevered bending members, the span shall be taken as the distance from face to face of supports, plus 1/2 the required bearing length at each end.

In your example Mischa you're mostly right, your "required" end bearing is probably less than the 3".
The box titled "Minimum Bearing Each End (sq")" on this calc;
https://forestryforum.com/members/donp/beamclc06b.htm
Gives the required bearing in square inches. With that and your joint width you would then have a length to add to the clear span. The figuring there is kind of circular, you might need to run a couple of times to get it all to work. To be honest if a beam were getting that close you're probably pushing too hard. That's just how to figure the bare minimum span. Yup, full length is ok 

I've gotten a look at the spreadsheet, that looks like the ticket. Thanks Jim

One thing I saw written today that is always worth remembering since we're all learning our math. Doing calculations is math, an engineer knows when and how to use the right math.

[bigmish]

In your example Mischa you're mostly right, your "required" end bearing is probably less than the 3".
The box titled "Minimum Bearing Each End (sq")" on this calc;
https://forestryforum.com/members/donp/beamclc06b.htm
Gives the required bearing in square inches. With that and your joint width you would then have a length to add to the clear span. The figuring there is kind of circular, you might need to run a couple of times to get it all to work. To be honest if a beam were getting that close you're probably pushing too hard. That's just how to figure the bare minimum span. Yup, full length is ok 

Maybe I'll use this calc, might just go full span, is easy and safe.

I've gotten a look at the spreadsheet, that looks like the ticket. Thanks Jim

Agreed.

BigMish, what program are you using?  Thanks,  srt

It's called SketchUp. It's very good and very free.

Now, I'd like to address the space under the purlin at the rafter, to bigmish.
This space is only holding up half the load, as this purlin has two ends, so when you figure the load at that point, reduce the total load by half and run the numbers, your original rafter may now pass.

Right, good point. This weekend I hope to have some time to run the numbers. I'll let you know what comes of it.

.M

[srt]

Mish,  Thanks for the reply.  I took a couple autocad courses at the local community college a few years ago, but just couldn't justify the $$ for the program.  I'll check into Sketchup!  srt

[Raphael]

I have AutoCAD 2000 and a copy of TimberCAD installed into it but haven't had the time to learn how to make it work.
Unfortunately SketchUp won't run on my iron.  :(

[Don P]

This thread has been on my mind this week, I didn't want it to slip into the archives without some comments. If I understand what full scale testing is showing, capital IF, the section to use in design of a center third notched rafter or beam is the remaining vertical rectangle. In this case it would be a 2x8.

Again if I'm understanding right, if the purlins are mortised into the rafters with a mortise in the neutral axis of the rafter (mid depth) then the rafter retains 95% of its 8x8 rectangular section properties.

The plans I typically build from stack joists or purlins on top of beams with blocking between. Not high art but it preserves the full dimension of the beam and stays within the letter of the present code reference. The TF Guild is working on testing to expand understanding of that area.

[Max sawdust]

Keep going guys.  Purlins are an engineering beast that beat me ;)  (Retreated to rafters)

I will conquer Purlins during the project I will start designing this winter (and this thread helps!)

max

[bigmish]

the section to use in design of a center third notched rafter or beam is the remaining vertical rectangle. In this case it would be a 2x8.

OK, I understand your meaning and it makes sense.

Again if I'm understanding right, if the purlins are mortised into the rafters with a mortise in the neutral axis of the rafter (mid depth) then the rafter retains 95% of its 8x8 rectangular section properties.

Now you're losing me! Firstly, what do you mean by "neutral axis"? What source or formula did you use to get the "95%". Finally, if it has 95% of the strength, what is point of the figuring to get the "2x8" above?

The plans I typically build from stack joists or purlins on top of beams with blocking between. Not high art but it preserves the full dimension of the beam and stays within the letter of the present code reference. The TF Guild is working on testing to expand understanding of that area.

Hmmmm, interesting. You must use hardware to attach the purlins to the rafter, right? What/where could I obtain? Can you explain the blocking: what's it for, where do you place, do you have any photos or diagrams?

Thanks, M

[Jim_Rogers]

bigmish:
The neutral axis of a timber is the center where the fibers are not being pulled apart or pushed together, when the timber is under load.
Jim Rogers

[Don P]

The neutral axis..
Think about a beam being pushed down on by a load on top. The top face is in compression, the fibers trying to become shorter. The bottom edge is trying to become longer, the fibers are being pulled apart in tension. If you mentally take flat slices through the beam from each edge towards the center, the forces of compression and tension diminish until at some point near the middle there is a flat, horizontal slice that is neither in compression or tension, the neutral axis.

A mortise aligned along the neutral axis, that breaks neither bottom nor top edges, has far less impact on bending strength than a notch in the compression or tension face. A tension face notch is the most strength reducing.

A tusk tenon has some reinforcement above the tenon to help with vertical shear yet preserves the compression edge, of the beam. I think the 95% is probably from empirical calculation not testing yet.

Think about an I beam, steel is expensive and heavy so they have removed all unnecessary parts of the beam. That removed area around the vertical web is where a mortise will do much less damage than if material is removed from one of the flanges.

On the last job we had some 6x16 glulam beams supporting sections of the second floor. The timber joists were set on top with pieces of the same height blocking installed between each joist as it was set. The blocking can be narrower or wider material than the beam to made a visual step on the beam. This was all drilled and lagged together and down to the glulam. The only real downside is it can make things too deep sometimes.

[Jim_Rogers]

here are a couple of examples of a tusk tenons

[Jim_Rogers]

At the eastern conference of the timber framers guild, during the pre-conference workshop on engineering a frame, there was a discussion of notching timbers or using tusk tenons.
The comment was made by the speaker, a professor of engineering at a university, that if you use a tusk tenon and have the mortise in the tie beam near or at the neutral axis the load carrying capacity of the beam, the tie beam, was reduce to about 95%.
This is just an estimate or an assumption. You really should do the math to prove this, is what he told me when I questioned him further about it.
This 95% is what Don_P is referring to....

When you take a rafter and cut in the top edge and make a drop in purlin, you reduce the load carrying capacity of this rafter to the rectangle left, not including the section under the notch. So in bigmish's design it makes the rafter a 2x8, not an 8x8.

This is why tusk tenons are better than drop in purlins. But they are harder to cut, and even harder to assemble. Not impossible, just harder.

Jim Rogers