Built Up Wood Beam - Beam Stability Factor 4

[codySTR]

Given a 3-ply sawn lumber bending member (face width oriented for strong axis bending), what width value, "b", would you use when evaluating the Beam Stability Factor, CL, in NDS 2015 3.3.3?

For example: Let's say you have a 3-ply 2x10 bending member. Ignore the possibility that you may have continuous lateral restraint or that you may have something preventing rotation at the member's ends (so CL doesn't default to 1.0). Do you evaluate the CL factor using b=1.5" for one 2x10 and multiply your design capacity by 3? Or rather, do you take b=4.5" and determine the bending capacity of the overall 3-2x10?

Thanks, team.
-Cody

 

[desertfox]

Hi There
I might have missed something but if I assume 'b' is 1.5 and work out the bending capacity of a single 2x10 and multiply by three, I arrive at he same answer as using 'b' as 4.5 and working the three beams has one.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[tony1851]

Don't know what your code allows, but won't there be an increase in bending capacity of about 10% or so by using the three beams?

 

[oldestguy]

Then there is the builder using several 10 ft. planks for a built up beam measuring 18 feet long, with a few joints near the middle.

 

[Jerehmy]

You evaluate it as one beam, b=4.5”, as long as theyre properly fastened together.

Desert fox - im not sure thats true. You get a greatly reduced CL for a single 2x10 than you would for all 3 together. CL reduces the allowable stress. The 3-ply will have a higher allowable stress than three single plies due to reduced LTB.

You also get the 15% repetitive member factor for 3-ply members and greater but thats separate from CL.

 

[desertfox]

Hi Jerehmy

I calculated the 'I' value of the beam based on a single beam with a width of 1.5" x 10" and multipling by 3 and then on a beam having a width of 4.5" x 10" using the standard formula b*d^3/12 and found no difference, I don't have access to the standard mentioned so I guess I could easily have misunderstood the OP post.

regards

desertfox

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[BAretired]

The problem is that a 1.5" wide beam tends to buckle whereas a 4.5" beam has less tendency to do that provided the laminations are sufficiently tied together to be considered a solid member. On a short span, both beams have the same capacity.

BA

 

[KootK]

b = 4.5" will make a big difference, and it's an interesting question. For b = 4.5" to apply, the compressed zone of the beam needs to be laminated across plies such that the plies don't just travel together but actually act compositely for weak axis bending. Luckily, it generally doesn't take much strength or stiffness to get this done for stability applications. I've never attempted to check the capacity of the fasteners to do this job and probably never will so long as I'm using typical fastening arrangements which would include max fastener spacing on the order of 2 x beam depth and some mechanism for preventing fastener draw out (screw / nail clinching).

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[Lomarandil]

Actually, I'm not sure about that KootK. One of my previous mentors apparently studied the ability of typical (dowel) fasteners to make wood members composite as part of some undergraduate research.

I can't find a thesis or anything to back up my memory of the discussion, but he mentioned that there was generally too much slop and crushing at the fasteners to develop full composite action. Instead, you got an intermediate state -- not fully composite, but not just 3x individual capacity either.

I'd imagine that in the real world, it falls into the category of "usually works, but for different reasons"

----
The name is a long story -- just call me Lo.

 

[KootK]

he mentioned that there was generally too much slop and crushing at the fasteners to develop full composite action. Instead, you got an intermediate state -- not fully composite, but not just 3x individual capacity either.

Sure, it's an intermediate state with any built up composite section no matter the material. Heck, it's really an intermediate state even in solid sections once the effect of shear deformation is considered. But sure, if somebody's put pen to paper on this, then I'll happily defer to it for the global strength and/or stiffness case. I think that it's worth noting that fastener demand for stability concerns are an order of magnitude less than they are for global strength and stiffness however. If our 2% steel numbers are to be believed, perhaps closer to two orders of magnitude. Find that thesis man!

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[codySTR]

Thanks for the discussion, everyone.

Desertfox: For purposes of bending stress checks using basic mechanics, you're correct in your approach. However, the National Design Specification for Wood (NDS) requires application of several adjustment factors. One of these is called the "Beam Stability Factor" (CL), which is dependent upon member depth-to-breadth ratio as well as some other considerations like support conditions, compression edge restraint, etc.. I think the CL factor ultimately boils down to mechanics with a lot of curve-fit equations to cover up some real-life behavior.

Regarding the fastener requirements to reach composite behavior in bending: I'm not sure how I'd go about really calculating the requirements and I'm not sure I'd want to, either (like KootK). But to just say minimum fastening is sufficient seems a little sketchy: the built-up compression member fastening requirements per NDS2015 Section 15.3 get pretty intricate to the point where I'd be concerned that workers won't really achieve the proper nailing/details/construction. The caveat, though, is that's specifically for column-type compression members and we're discussing bending... But this CL factor does carry over into the combined bending and compression checks.

I think for most applications the difference in capacity isn't going to be a game changer. I'll probably determine capacity based on individual ply capacity, multiplied by "n" number of plies as well as the repetitive member adjustment factor mentioned by Jerehmy (thanks! forgot about that little guy).

 

[NorthCivil]

Off topic but somewhat relevant. I've heard that the fire resistance rating of a multi ply wood beam is only as good as it's thinnest member. Something about the members twisting and allowing flames to creep in the cracks

 

[Canuck65]

I use WoodWorks for a lot of my wood design. Some time ago when analyzing a multiply wood beam without lateral restraint I was finding that the (Canadian) KL (Lateral Stability) factor was being calculated on the 1.5" dimension for a 3 ply beam. Therefore the bending capacities were quite a bit lower. I looked into this further and found that there is documented research from testing which confirms that even if you connect the plies really well, you cannot achieve similar test results as if a full dimension member. Therefore, as a result of the research, basing the KL factor on the single ply dimension is what has prevailed. I cannot put my finger on the description/research at the moment, but if have time will look for it and post. I will admit the first time I came across this it was a bit of a surprise.

 

[KootK]

If it's good enough for Woodworks, it's good enough for me. Thanks for sharing that Canuck.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[codySTR]

Thanks for sharing, Canuck65! I'd like to see the research you're referring to if you manage to find it.

 

[TehMightyEngineer]

Also intrigued if you can find the research for this.

Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries

https://www.facebook.com/AmericanConcrete/

 

[KootK]

I vote for cody having to call Woodworks and chase it down. Who doesn't like a homework assignment from a stranger? This, combined with Lo's mentor's work, raises other potentially ugly concerns for me. It's not uncommon for designers to use built up 2X headers as weak axis wind girts in wall systems. You know, if they consider wind load at all to begin with.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[jayrod12]

My typical detail for large windows has a sill plate above and below the headers to act as the lateral resisting element. The one I always see ignored outright is the sill at the bottom of large windows, it always ends up being a single member unless you specify it otherwise which I'm yet to see anyone but myself do.

 

[Canuck65]

A little bit more information.

1. In Woodworks, the default setting is to calculate the KL (Lateral Stability) factor based upon a single ply, for multiply beams that are laterally unsupported. If your engineering judgement dictates otherwise, there is a menu option to toggle this assumption off.

2. Excerpt from WoodWorks help file for multiply beams: "6. For lateral stability calculations for the lateral stability factor KL in CSA O86 7.5.6.4.4, Sizer offers you the a Design Setting choice of whether to use the full member width or the single ply width as b in the expression for the slenderness ratio and in the limiting ratios for which the stability factor can be assumed to be 1.0. Research has shown that nailed and bolted beams have at most 30% composite action effect in terms of resisting torsional buckling, so. It is recommended to use single ply width unless adhesives are used to laminate the members together."

3. Keep in mind that this discussion mostly disappears if you have lateral support, which is more often the case.

4. There are provisions in CSA-O86-14, cl. 7.5.6.3 and cl. 7.5.6.4 that are pertinent to the discussion as they provide ways to consider a multiply beam as a single unit for calculating the lateral stability factor.

More digging needed to identify the source research material.

 

[KootK]

My typical detail for large windows has a sill plate above and below the headers to act as the lateral resisting element.

I do that too when it works. I was just reviewing a competitors detail yesterday where there was much ado about how lamination was accomplished. The plies had to be separated by packing etc rather than tight together. Clearly gunning for composite stiffness.

Research has shown that nailed and bolted beams have at most 30% composite action effect in terms of resisting torsional buckling

30%? How disappointing. I was in the wood design industry before I got my formal engineering education. One of the first neat things we hit in mechanics of materials was how to fasten a wood box beam together based on VQ/It etc. I remember thinking that I couldn't wait to apply it in the wild. I'd not have been so enthusiastic if I'd known it was a bunch of hooey.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.