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Mortise shear when in compression #29567 08/29/12 09:52 PM
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borhani Offline OP
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I'm building a small treehouse (really, a tree platform) out of locust. Platform has eight beams, each extending 5 feet radially from tree. Each beam is supported by a strut; see attached figure: [img]http://pxrox.tumblr.com/[/img]

Beam & strut each fit into blocks that are lag-bolted to the tree. Beams have a dovetail mortise/tenon (no peg) to hold them into blocks, and will be tied together horizontally with screwed-in 2x6 flooring that goes all around the tree but extends only 4 feet out (last foot of beam will stick out, for "looks"). Design load (dead + live + snow) on each beam/strut is ~1,200 lbf.

Strut fits into the beam 3 feet out from tree. Angled at 45 degrees, the strut will exert ~1,000 lbf horizontally (outward) and ~1,000 lbf vertically (upward) on the beam (strut thus takes most of the weight load; resulting moments are neutralized by the blocks above & below the beam).

Strut has an oblique tenon (no peg; see figure, in red) to fit into the beam mortise. The mortise bearing shoulder is angled at 22.5 degrees, to take the toe of the oblique tenon. The mortise has ~1 x 5 inches of shear plane to resist the horizontal force. Locust has a max. shear strength parallel to the grain of ~2,500 psi, giving (in theory) the ability to resist ~12,500 lbf of shear. Seems like enough.

But, I am concerned about possible mortise shear-out, in part because a book on timber engineering (Construction en bois: Matériau, technologie et dimensionnement, by Julius Natterer, Jean Luc Sandoz, Martial Rey) suggested 30 (!) shear-resisting nails in the tie of an analogous rafter/tie design (admittedly with ~8x higher loads; see this link: Tenon Example ).

My (aesthetic) goal is minimal hardware...trying to do it the old-fashioned way! Should I be worried?

David

Re: Mortise shear when in compression [Re: borhani] #29568 08/30/12 01:15 AM
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TIMBEAL Offline
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I would not be concerned with the joint where the brace meets the tie. More concern would be the beam to tree.

Re: Mortise shear when in compression [Re: TIMBEAL] #29569 08/30/12 02:35 AM
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borhani Offline OP
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Glad to hear that one of two joints may be good! Would be interested in knowing more of your concerns on the dovetail. Here are some details:

The (irregular) beams are roughly triangular is cross-section, ~4 in. wide across the (slightly crowned) top and ~4 in. deep. The dovetail tenons have a nearly horizontal base that gently slopes upward (angle ~20 degr.) from bottom of beam (AT tree) toward the beam center (as you move away from the tree). Max. depth of cutout always less than one-fourth of beam depth; tenon length is fixed by the supporting block thickness---which holds the intricately shaped mortise---namely 2.5 in. The shear plane thus ends up being ~2.5 x 1.25 in., i.e. ~3 sq. in., giving ~7,000 lbf calc. max. shear load (which is 7x the max. anticipated load).

I plan to relieve the stress in the angular joint at the (non-tree, i.e. outward from tree) end of the tenon by cutting a gentle curve back down to the bottom of the beam. Probably ~3 in. long elliptical curve (not looking forward to this, as it will take a coping saw, and locust is really hard!).

(I had considered a dovetail with vertical sloping sides, but that would mean flipping the beams upside down, putting the broad face on the bottom. Not sure how this would affect beam bending, if at all, but it would give me no good surface on which to attach the decking.)

One other consideration makes me worry (much) less about the shear in the dovetail (which is, after all, going to be in tension to ~1000 lbf): The decking will form a screwed-into-all-beams continuous ring all around the tree. In other words, the deck itself will act to balance the tension loads on one side by the tension on the other, all the way around.

Thoughts?

David

Re: Mortise shear when in compression [Re: borhani] #29570 08/30/12 04:04 AM
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Roger W Nair Offline
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Borhani, there is a great difference between the material strength values you might find on Woodweb or other sources and the approved design values for construction you will find in the National Design Specification {NDS}. For example, in table 4A, base design values for visually graded lumber, shear parallel to grain in red oak is 85 psi and white oak is 110 psi. No value stated for locust. On Woodweb red oak is 1780 psi and white oak is 2000 psi. My guess is that you should use a much lower design thresshold such as the value for white oak but be aware that most every wood species in the NDS is under 100 psi for shear parallel to grain.

Re: Mortise shear when in compression [Re: Roger W Nair] #29571 08/30/12 04:06 AM
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borhani Offline OP
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Thanks, important info. I got my values from Mechanical Properties of Wood (Ch. 4) by David W. Green, Jerrold E. Winandy, and David E. Kretschmann. Why the difference from NDS?

Re: Mortise shear when in compression [Re: borhani] #29572 08/30/12 03:04 PM
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Roger W Nair Offline
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I should add that I referenced the 1991 edition of NDS.

I cannot strictly account for difference between the material lab tests to failure and approved design values but consider this: the lab test evaluates what the sample can withstand right now under a set of specific conditions and design values give authoritative assurance on performance without degrade over decades under much wider conditions. In a way the tests tell us that wood is strong and design values tell us that time is the leveler.

Re: Mortise shear when in compression [Re: Roger W Nair] #29573 08/31/12 04:29 AM
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borhani Offline OP
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After much reading, I have to admit I still don't understand the discrepancy. Perhaps we're referring to different shear values, or applying them to different (and perhaps inappropriate) situations?

From what I can tell, "Fv" = the NDS "Design Value for Shear Parallel to Grain"; Fv seems to be used to determine whether the "actual" (parallel, horizontal) shear stress within a horizontal beam is allowable, or not. You would think that Fv should equal the "Shear Parallel to Grain", yet Fv is about 10- or 20-fold lower, on the order of 100 psi.

By contrast, the tabulated values for "Shear Parallel to Grain", for Black Locust, are 1,760 psi [green] and 2,480 psi [12% moisture content]. These values do not vary more than about 2- or 3-fold, with locust being one of the highest, and some cedars & firs being the lowest, ~600 psi.

The *actual* beam shear stress, "fv", is calculated according to a standard formula. In a rectangular beam, for instance, fv is based on the (vertical) shear force P (e.g., half the load if the beam is supported only at the ends) and the beam breadth & height, b & h: fv = 3P/2bh. fv less than Fv is OK; fv greater than > Fv is bad, signalling possible failure.

I have not been able to find exactly how the low NDS Design Fv values are determined or estimated. One document suggested that Fv = 40 + 266*G, where G is the wood specific gravity. Another, from the USDA Forest Service, says:
Quote:
Shear design values for solid-sawn structural members
are currently derived from small clear, straight-grain
specimens (ASTM 1995a). ... Because of the
placement of the member within a structure and the
local climate, the occurrence and degree of splitting are
varied and unpredictable. Published shear design values
(AFPA 1991b) account for this uncertainty by
assuming a worst case scenario---a beam that has a
lengthwise split at the neutral axis
.
(My locust has essentially no checking; it's solid like a rock.)

Does anyone know how the Design Fv is really determined, or why it is so much lower than the measured (literally, by shearing wood blocks) Shear Parallel to Grain value?

Or are Roger and I just mixing apples & oranges, and the shear I originally was worried about (e.g., a mortise giving way) is not the same shear as what the NDS is talking about (because it's assuming worst-case-scenario beams)?

Thanks,
David

Re: Mortise shear when in compression [Re: borhani] #29574 08/31/12 02:22 PM
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Roger W Nair Offline
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Yes, we might be mixing apples and oranges, how about a bunch of bananas? The first banana is I referred to the 1991 NDS values, unfortunately values for shear parallel to grain have been doubled since 1991. Link to 2005 NDS Supplement

http://www.awc.org/pdf/2005-NDS-Supplement.pdf

Sorry about that. Still leaves an order of magnatude of difference between values established in the lab and the approved design values.

For clarity, my first post related to the shear plane calc in your second post. It is established practice to use NDS Fv value times shear plane area to evaluate joint strength for dovetails.

Re: Mortise shear when in compression [Re: Roger W Nair] #29575 08/31/12 05:10 PM
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borhani Offline OP
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Awesome, thanks Roger. Locust seems much like Oak or Maple in many respects, so an Fv value of ~200 psi seems reasonable to use.

The joint I described in my second post is indeed much like a dovetail. Based on the above Fv value, I do need the restraining effect of the decking---the dovetail itself could only hold ~600 lbf (about half of what is needed).

The mortise in my first post has a shear area of ~5 sq. in. if I include only the *bottom* plane of the mortise in the area calculation. Including the sides as well gets me up to ~12 sq. in., for ~2400 lbf, about twice what I need. Is it OK to include bottom & sides of a mortise in the shear area calculation?

Thanks again for your valuable input!
David

Re: Mortise shear when in compression [Re: borhani] #29576 08/31/12 09:48 PM
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Roger W Nair Offline
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David, I honestly am confused by your first and last post. When framers speak of shear planes, we are concerned with the supporting grain beyond the joint, aka relish. When joints apply force close to and towards the beams end and when tenons are expected to be in tension, then end distance and relish become a matter of concern. So with a strut close to the end, we take tenon depth times end distance times NDS Fv value times 2 equals the force the joint can safely resolve. We multiply by 2 because there will be two shear planes of equal area. For pegs we take center line of peg to end distance times tenon thickness times Fv times number of pegs times 2 [two shear planes per peg} equals force limit. In the case of your strut to beam joint, you have ample end distance.

In the case of the French manual, the drawing seems to me to be of a major rafter landing with a scant tenon with a nailed cleat supporting the load, not applicable to your case.

Last edited by Roger W Nair; 08/31/12 09:50 PM.
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