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Shear Design for Wood Beam with Biaxial Bending 2

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MJC6125

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Apr 9, 2017
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If you are designing a wood beam with bending in two directions (strong axis and weak axis), how come you only need to check the shear stresses in each axis independently? Is there no sort of resultant or combined shear stress that develops on the beam cross section face? Seems like this should be a basic question, but I'm having trouble wrapping my head around it.
 
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I have checked EC-5 if any recommendation for the combined shear stress due to bi-axial loading but only biaxial bending cases defined. You did not mention if the section rectangular or not. If the section is rectangular , the combined shear stress will be maximum at the intersect of the x–x axis and y–y axis .

It will be conservative to combine the applied shear forces about the x–x and y–y axes and limit this total to the shear capacity

My opinion..




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Think about the actual shear distribution in a rectangular member (parabolic?) and think about how a shear in both axes might combine in terms of stresses. Only a very small region might have the maximums combining from each axes in my opinion.

 
It would be interesting to know how shear is resisted in wood with Tau xy and Tau xz components... a question for CWC or AITC, perhaps?

I sent this on to the CWC...

We are looking at a right hand coordinate system with longitudinal axis of beam along the x-axis. The top of the beam is the positive y-face, and the front of the beam is the positive z-face.

A question has arisen about the handling of wood shear stresses for a beam member loaded in both the y direction and the z direction. Is there an interaction diagram that can be used to confirm the combined effect of Tau xy and Tau xz, the respective shear stresses?

thanks, Dik Coates, P.Eng.(Man)



-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 
I would sum the vector shear stresses and compare to the allowable. Alternatively, you could calculate the shear flow Q for a rotated rectangle and obtain your demand from NDS 2018 3.4-1, fv=VQ/Ib. I would think you should get the same answer. Technically I think AWC could argue this equation and the description "The actual shear stress parallel to grain induced... shall be calculated as follows" covers biaxial shear stress, since they don't specify about which axis shear needs to be checked, although AWC needs to rewrite this section for clarity (would it kill them to throw some torsion provisions into this mix too? jeez who's steering this ship?)

Agent, technically only a single point would experience maximums from both directions, the center of a rectangular beam. I'm not sure this makes me feel any more comfortable though. I would prefer not to fail any part of the cross-section.

Interesting question. I've often wondered the same thing about similar topics but never bothered to ask. We spent a great deal of time studying principal axes and stress transformations in school and it's not clear the codes have taken these into account. I assume the reason why bending and shear aren't combined into one check is because the maximum stresses are located at different parts of the cross-section? And even if the transformed stress at other points of the cross-section exceed both the maximum shear stress and maximum bending stress, it can't be by much, and probably within the limits of our factors of safety, right? Right.

But a couple stresses that do have maximums at the same point would be axial + shear, on which NDS is mute. Kinda seems like NDS is behind AISC and ACI in the modern age of hyper-codification, but maybe that's a good thing? I mean, the human race is currently drowning in over-regulation on more than one front.
 
Wouldn't a square root of the sum of the squares apply here? These shear stresses aren't directly additive so a Von Mises or something like that would apply would it not? Failing that, I would suspect that shear1/allow + shear2/allow <=1 would be a sufficient check. It's an esoteric question. Curious what "official" response we will get.
 
I don't know... it depends on the testing results (wood being orthotropic). When I get a response from the CWC, I'll respond asking how the interaction with direct stress (tension or compression) and if it's the same as it currently is (an oversight).

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 

was planning to...

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 

...might not work... see my response to lex. I don't know how the properties of engineered wood and if they can be considered as isotropic.

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 
Mike Mike - I'm not convinced axial loads produce shear stress in the same sense. Sure, you get diagonal tension and that's how concrete test cylinders fail, but wood has better tension/compression ratios than concrete. I get your point but it took me a minute. I don't think that failure mode has manifested in clear wood testing. Perhaps that's why it's not in the code. Now if you had torsion and bending, those shear stresses would potentially add/subtract. But the torsion would be "point" maximum (sand heap analogy) and the horizontal shear from bending would be planar (parabola with maximum at the middle of the cross section).
 
MJC6125 said:
Yes, I was thinking a rectangular section like a 2x8.

For a high aspect ratio member like that, I imagine that you probably max out your weak axis flexural capacity long before you induce meaningful weak axis shear stress.

Obviously, this doesn't answer your interesting question regarding the code intent for general cross sections. For that, I'm liking Agent's non-coincident peak stress thing. The only thing that nags at me there is that I know that grading rules on checking allow member to be so severely checked that that the center of the cross section is practically all one has left.

I have another, unsubstantiated hypothesis. I suspect that most real world shear failures actually start as flexural tension failures originating at imperfections, similar to how it works with concrete. In this respect, your biaxial flexural check would kind of shield you from biaxial shear failure.

 
dik, I think weak and strong axis can be considered isotropic for the purposes of calculating shear stresses. In reality the shear strength in the tangential direction is typically higher than in the radial direction, but we don't know what part of the log the 2x8 was cut from, so my understanding is the published allowable shear stress values are based on the lower of the two directions.

lex pat, that's a good point and I'm probably missing something here, but to dig into it a bit, let's say the center of mjc's 2x8 is subject to 100psi tension and 100psi shear. At a 32 degree rotation, this transforms into 0 shear and 162psi tension (check my math, haven't done mohr's circles in a while). Worse yet, the tension is no longer oriented parallel to grain. As to the sand heap analogy, are you referring to the analogy first presented by Nadai at a meeting of the German Society of Applied Mathematics and Mechanics in Marburg, Germany in 1923? I agree torsion and bending interact, and can't be separated, from a mechanics of materials standpoint.

kootk, interesting hypothesis, does it apply to deep beams?
 
IIRC, wood's shear strength parallel to the grain (which is what has been traditionally listed in NDS tables) is weaker than what it is perpendicular to the grain. So for the bi-axial bending case, generally what I have seen people do is take the parallel to the grain allowable as being good for both directions.....and combine their utilization with a simple interaction. [I.e. (fv[sub]1[/sub]/Fv)+(fv[sub]2[/sub]/Fv)]
 

It's not the dimensions, it's the material... wood has substantially different properties depending on the axis... I just don't know what the effect is. I'd normally ask my 'Wood Guy' but he's no longer with us...

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 
in the aerospace world, we would use an square root sum of squares interaction, with different allowable strengths in the two directions: MS + 1 = 1/ sqrt( (taux/Fsx)^2 + (tauy/Fsy)^2 )
 
Heard back...

"The combined effect of shear stress is out of the scope of CSA O86:19, so we cannot provide a straightforward answer.

A possible solution would be to combine linearly the shear ratio of both directions, similar to what is done for two-way bending members. An example for two-way bending members is given at p.136 of our Wood Design Manual 2020 (see image below).

However, the current draft for the next Eurocode cycle is set to use the square value of each term when dealing with shear in both directions, which would be less conservative.

This is a subject that will most likely be evaluated in the next cycle of CSA O86. We are also considering adding guidance on this subject in future editions of the Wood Design Manual.

Should you have any other questions, do not hesitate to reach out."

That was a lot of help... The sketch sent had to do with combining moments... Less conservative? but correct?

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik
 
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