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BTH-1-2014 Shear ?
2

BTH-1-2014 Shear ?

BTH-1-2014 Shear ?

(OP)
Hello folks!
It's been a long day in an already very long week and it's only Monday so maybe some sleep would help, but I have a question about ASME BTH-1-2014 Eq 3-37. It's turned up before on the forum but not really in a way that answers my questions. Really, I think I jsut need a little hand-holding here...

fcr = SQRT( fx^2 - fxfy + fy^2 + 3fv^2) <= Fy / Nd EDIT: original post said "+ fxfy", fixed here

where:
fx = computed normal stress in X direction
fy = computed normal stress in Y direction
fv = computed shear stress

Now... say my XY plane is "flat", and I've got forces Fx, Fy and Fz. Let's say further that all forces act at some moment arm such that there are bending stresses induced by all three forces. Now, I would consider the compression and tension resulting from those bending stresses to be acting normal to the XY plane, but wouldn't you say they acted in the Z direction? I'm confused when the explanation of the variables offered by ASME talks about the normal stresses in both the X and Y directions; do they mean the normal stresses caused by the forces acting along X and Y? I'm looking at BTH-1-2014 Chapter 3-2.5; same chapter for the 2011 version.

Thanks in advance, cheers! (sketch attached)

RE: BTH-1-2014 Shear ?

If I'm not mistaken that the Von Mises general plane stress equation:
Thus it should be your stresses in the Fx, Fx, Fz direction, matching your same coordinate system for your loads.

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

Interesting that you have a + sign before the fxfy term, but I have it listed as a minus sign.

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

(OP)
It's interesting because it was wrong; corrected and thank for the catch.

Bear with me here if you would TehMighty, but I'm not quite up to speed. The subscript on the final term is "v" in BTH-1 ("12" from your source) and they define as "computed shear stress". Obviously the symbols used don't particularly matter, but that to me would seem to include shear sress induce by Fx, Fy and Fz (torsional in my case), is that incorrect?

I'm also still confused because the forces Fx and Fy (in my case) are in the same plane as the shear stress and therefore I'm not following where the "normal" part comes in - I would have thought that normal stress referred to the bending (about the XX and YY in my case).

Any thought? and thanks for taking a minute to respond!

RE: BTH-1-2014 Shear ?

Bear with me as I haven't had my coffee yet and might not explain this correctly.

σ12 (or fv) should be the in-plane shear stress by my understanding. So you have Fx & Fy being the in-plane stresses due to axial forces, and Fxy being the shear stress in the X-Y plane.

Thus, your Fx and Fy should be in the same plane as your shear force (Fxy).

I'm sure there are better references but the one I always end of using (usually because I'm using the FEA feaures) are the help files of RISA and STAAD (when I was still using it) which have nice little graphics showing these stresses:

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

It's worth noting that I'm not an expert in plates and shells and only know enough to squeeze by and be dangerous. ;) I'm sure someone could explain this better than I can.

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

(OP)
I am not even halfway through my first cup, so I hear you!!!

I am with you partially and agree that I've indeed got shear stress due to Fx and Fy that lie in the XY plane. In addition, I've also got torsional shear stress to to a moment/torque about the ZZ axis. The way I'm reading it, the SHEAR stress induce by Fx, Fy and Fz should all be combined in that final "fv" term.

The thing that is really hanging me up is where the "normal" part comes in? This is the equation for analyzing "Combined Normal and Shear Stresses", but as far as I can tell here, there's nothing in Eq 3-37 that's acting normal to the XY plane where my shear stress lives. Now, you did say that I "have Fx & Fy being the in-plane stresses due to axial forces" which I am not quite grasping; I would call Fz and axial force here, but not Fx or Fy. And, obviously, Fx and Fy are normal to each other, but I take this equation as referring to a force/stress acting normal to the plane where the shear lives i.e., as a way to look at the combined effect of bending or axial stresses together with shear.

I hate to beat this horse so badly, and I hope I'm not being a drag here, I'm just not quite there yet. Thank you kindly for your help.

RE: BTH-1-2014 Shear ?

Sorry, by axial forces I meant "plane stresses in the x and y axis". They're all technically axial forces but, yes, they are not your longitudinal axial force.

I think by "normal" they mean stresses normal to the faces of the element being analyzed. So, if we define some discrete element by cutting it out of a larger piece then you have 6 faces on the element (as shown in the pictures above). The Fx, Fy, and Fz stresses are all "normal" to these cut faces.

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

Your trouble here stems from attempting to use the 2D Von Mises failure criterion on a 3D stress problem. Google "Von Mises 3D" and you'll get the idea.

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.

RE: BTH-1-2014 Shear ?

KootK has it, and that's what I was missing. I didn't read your problem statement fully and was trying to explain only your equation as written. If you have 3D stresses then this gets a lot more complex and is probably beyond me. Short answer is you cannot use the equation as written.

My understanding is you will have to revert to the full Von Mises yield criterion equation:

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

(OP)
KootK, I actually have, and it's a reminder of some basics that I need to brush up on for sure!

That said, unless an object was ONLY loaded in (say) the X and Y directions, then I don't see a way that I would have shear stress AND stress normal to that plane. I would think that one would need some axial (tension/compression) or bending moments about those XX and YY planes to induce shear AND normal stresses. That said, the equation is just sitting there to use without a huge amount of clarification. If you look at my sketch, if I were to constrain Fz to lie ALONG the Z axis thereby eliminating the moment about ZZ-axis and the resulting torsional shear, do you feel like that equation 3-37 would more aptly apply to that situation? How would that Fz ever end up having an affect on the equations results?

I feel like maybe the issue is a bit wider than the intended application of the equation; I'm trying to understand what I should do to correctly apply the intent of the check in a situation where the given equation may not exactly apply to. Thanks again!

RE: BTH-1-2014 Shear ?

(OP)
Kind of what I was thinking that y'all were driving at TehMighty and KootK. How have you approached something like this in your work when the prescribed equation doesn't really fit the situation, but it is what it in the code? I know that equations are not magic bullets and they only apply for given sets of criteria, within limited ranges, etc, but I'm not sure how to proceed when it seems liek the prescribed equation doesn't describe the situation accurately.

RE: BTH-1-2014 Shear ?

My gut feel here is that the effect of Fy will be negligible and can be neglected, taking you back to the 2D equation applied to the XZ plane. Of course, you're relieving advice from a guy who doesn't know your loads, proportions, or boundary conditions. So proceed with caution.

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.

RE: BTH-1-2014 Shear ?

If I'm fairly clear on the intent of the original equation (in this case a Von Mises yield criteria) then I usually adjust it as I feel is appropriate. If I'm not clear than it's off to eng-tips, senior engineers, and/or the governing body of whatever specification I'm referring to.

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

(OP)
KootK: Let's say that my XY plane is "flat" ( I guess ALL planes are flat!) and it represents some structure or other thing that a lifting lug would be welded to. For ease of imagination, let's say our XY is overhead. That means that Fz would be vertical and as you'd guess, that would be the direction of loading due to the weight suspended from this lug. I'm working on a super-spreadsheet to design and/or analyze existing lugs so I'd like the ability to have the load apply anywhere along a 180deg path i.e., 100% along +Y to 50%/50% between Y and Z (45 deg), 100% Fz (90deg) and so on until all the load is in the -Y direction. In addition to that, let's consider 'sidepull' to act in the X direction, so I'm including a max out-of-plane loading angle of something like 5deg-10deg. As you can see, I've got a real interest in the 3D problem; even the small amount of sidepull is of interest because a) it happens in real life all the time, and; b) it's against the weak axis and therefore worth being mindful of. Hope that clears up what my intent is.

TehMighty: Thanks for your input. I'm actually drafting an interpretation request to ASME BTH about the question now.

Thanks folks!

RE: BTH-1-2014 Shear ?

Can you just ignore some of the plate so that there is no torsion? Just neglect enough of the plate on the left side so that your force is at the centroid

RE: BTH-1-2014 Shear ?

(OP)
Possibly, yes, at least when we have design control, but I would like the option to leave it in the spreadsheet. In practice, I've seen a lot of lugs that have the hole significantly eccentric w.r.t. the lug centroid. Say, for instances where the weld needs to be pretty long, then you can either have a much taller lug and taper it up from the base to the pin hole, OR, you can keep the lug fairly shallow in height but offset the hole way to one side so you still have appropriate clearance for rigging hardware to attach. In a straight pull situation, that is fine, but if you're a few degrees off straight and a foot or so off of the centroid, I'd like to leave the torsion in there.

RE: BTH-1-2014 Shear ?

In practice, whenever I've had to do something like this I would resort to a FEA computer model to calculate the Von Mises stresses rather than try to build a spreadsheet that covers all the bases. Generally it ends up taking a little more time in total but the added flexibility of a 3D FEA program is quite useful and generally easier to review as well.

Professional and Structural Engineer (ME, NH, MA)
American Concrete Industries
www.americanconcrete.com

RE: BTH-1-2014 Shear ?

(OP)
Well, I've got Inventor and a pile of books about it's use and an additional book and online lecture series I bought specific to the FEA capabilities, so that's one thing I'm learning right now. I very well may just leave that stress as a user-entered value to be determined in an FEA. Good idea.

RE: BTH-1-2014 Shear ?

Working out the demand side of the 3D Von Mises check is certainly doable. Personally, I wish that codes wouldn't micro manage the design process by specifying stuff like this explicitly. Based on what I know of your system, the real issues will be effective flexural/buckling width, bi-axial flexural stress, and cantilevered plate buckling. I doubt that shear is even worth checking except, perhaps, locally at the hole as bolt tear out etc.

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.

RE: BTH-1-2014 Shear ?

(OP)
Thanks for the visual! Appreciate all of y'all's help today.

RE: BTH-1-2014 Shear ?

(OP)
I've actually submitted two separate inquiries to ASME regarding BTH-1; I'll post back and let you folks know what they say.

RE: BTH-1-2014 Shear ?

Colin,

What exactly are you trying to design? The lug, or the thing that the lug attaches to? When I designed padeyes in the past, our company decided that the item being lifted wasn't covered under BTH-1. The item being lifted would still be checked, but only to whatever code to which it was designed - for me, this meant checking a lot of J4 checks in AISC 360.

Now then, for the padeye itself, it was basically major axis bending, minor axis bending, and tension in combination (eqn 3-35). Weld at the base gets sized for the bending(s), tension, and the shear(s). I'd only drop in to eqn 3-37 if I did a FEA of the lug as plate elements, and I can't imagine analyzing a lug with FEA.

Just my .02

RE: BTH-1-2014 Shear ?

(OP)
winelandv - Just looking at the lug/padeye itself. I agree with you that the thing it attaches to is not covered by BTH-1 other than IMO you could make a case that the section 3-3.4 on Welded Connections would require a little checking of the thing that gets the lug. J4 is good stuff for the thing that gets the lug as you say.

Also agree that 3-35 must be checked, but we have seen many lugs that get loaded at quite shallow angles with respect to their base. That is, if the lug is made out of b" (long) x t" (thick) plate that is h" (tall), then the angle from the "bt" plane is often fairly low; say less than 45 degrees is not uncommon. Therefore, there is a significant amount of shear, but at the loading is arranged, there is also bending about both the major axis (about the axis through the "t" dimension in my example), bending about the minor axis(here, the axis through the "b" dimension), and some axial force (along the axis parallel with the "h" dimensions). To cover all bases, there are also times when there is torsion about the "h" axis as well. I know that torsion is never good, and maybe those are some shallow angles to be loading something, but that is how I see them on numerous jobs so I'd like to be able to correctly address the calcs. Anyway, the presence of that worth-investigating amount of shear along with at least bending and axial (though not always torsion) got me looking at Eq.(3-37) for the lug plate itself; do you think I'm off base here? Thanks for your input, have a good eve!

RE: BTH-1-2014 Shear ?

Less than 45 degrees? Your lifters are really putting the shackles through their paces. Also torsion... yikes. I see why you're struggling for a BTH-1 solution here - it was written for the majority of cases for lifters and lifter attachments. Usually, loads stay in one plane and rotate around the pin. Sometimes, you get the out of plane loading during uprighting. Torsion, though, I've never seen it addressed, and I'm struggling to come up with a scenario for it. If you need to get all these forces through your lug, though, may I suggest lengthening the lug, and adding four triangular stiffeners (both sides of the hole, both sides of the lug). Larger I for the out of plane bending, and can resolve the torsion into a force couple into those stiffeners
|
----|----
|<------hole assumed in center of lug
----|----
|
(lug plan view)

On the other hand, while possible to try and design for the way things are being done, maybe some push back the other way is needed, saying, "we can go to these limits, but not beyond."

RE: BTH-1-2014 Shear ?

(OP)
An example where you'd get some torsion and have such shallow angles would be lugs located radially on a hemispherical head being used to lift vertical pressure vessel; no uprighting necessary as the vessel would remain vertical. In these cases, let's say 6-8 lugs spaced equal (angular) distances around the head, the angle of slings w.r.t. horizontal may be about 60 degrees or better, but due to the curvature of the head where they are welded, the angle w.r.t. the base of the lug is considerably less. So, the rigging is not necessarily being aggressively loaded, but the lug itself sees a pretty flat angle. More on the torsion part in a bit.

Maybe I can ignore torsion, but until I had my calcs running smoothly and had an opportunity to see the actual numbers, I was hesitant to do so. I've got two situations that are in mind:
1) a lug welded to some piece of structure used to pick up "stuff" - generally much lighter duty. These are the ones where significant weak axis bending is to be expected b/c boilermakers are boilermakers and they will make it happen, engineer's nerves be damned. In all of these cases, I would put the lug hole located axially, so there is biaxial bending and tension, but never the torsion. Simpler case although I did have the question about normal+shear and Eq.(3-37)that I'm working through.

2) Lugs welded to the head of a vessel as described above. If the length of lug req'd (due to length of weld req'd) is fairly long, we would slide the hole to one end to allow clearance for the shackle/turnbuckle without having to make the lug excessively tall. Be amaze by my keyboard artwork!!! The single quote marks and periods are just for spacing, I couldn't get it quite right but I think they convey the message.
' __________
' |..O...........| <--------- lug with hole to one end
' |...............|
~~~~~~~~~~~ <--- weld to vessel

...instead of...

''''''''''' ____
'''''''''''/..O..\
'''''''''/.........\ <--------- lug with hole at lug CL but tall enough to allow enough taper to provide adequate clearance for shackle
'''''''/............\
'''''/...............\
~~~~~~~~~~ <--- weld to vessel


If all the rigging came to a single point, then even with this longitudinal offset of the hole w.r.t. the "LUG_WIDTH x LUG_LENGTH" axis, then there would be no torsion. However, since the rigging does not come to one exact single point, then some over the lugs are loaded across the minor axis to a relatively small degree; this is the torsion I was trying to capture. winelandv, it sounds as if you are familiar with the subject but if anyone is not, one reason the rigging would not come to a single point is that the crane hook would often be a "sister" or "duplex" or "ram's horn" - a pair of hooks with the distance between CLs of the hooks being something like 18"-30", more or less depending on size. Heck, even with a single hook, I've seen situations where some of the lugs could be loaded out of plane by 10 degrees or more. To illustrate a bit further, say we had 6 equally-spaced lugs arranged concentrically and aligned radially; the lugs at 0* and 180* may be loaded in major axis only, but the lugs at 60*, 120*, 240* and 300* would all have some minor axis loading as well, due either to the rigging collecting at one or the other of two hooks, or even just due to the multiple slings having to lay next to each other and not being able to a=occupy the same physical space.

I like the idea of the stiffeners, and definitely will look at putting that into my spreadsheet. Thanks again for your time and thoughts.

RE: BTH-1-2014 Shear ?

Ok, I'm keeping up with you now.

Sounds like for case 1 you're good to go.

Case 2 - I say that there is no torsion at each lug. Thinking through the FBD for each lug individually, there's still just one resultant force that's going through (ideally) the center of the shackle pin - thus no twist. Now, if there's too much slop in the shackle in the hole (pin length > lug thickness), the shackle could shift to one side moving the resultant off of the "vertical" centerline. We prevented this by welding cheek plates to the lug so that width of the lug + cheek plate was at least 90% of the pin length. You probably know this, though. But that's my reasoning behind no torsion, or at least, small enough to be ignored compared to the bi-axial moments and tension.

RE: BTH-1-2014 Shear ?

(OP)
Maybe I'm getting the old panty-bunch over nothing. I should have just provided a better picture to begin with , see the attachment and you can why I'm thinking about torsion. My life would definitely get one step simpler if I decide it's OK to ignore!

You can see the lugs at 120* and 300* (* = degrees) are loaded straight on, but 0*, 60*, 180* and 240* are all loaded at about 11* w.r.t. their longitudinal axis. The hole is well offset like my upper "sketch" above. Maybe I'm worrying about nothing? I'm just worried that 11* or so created a moment about the axis that would be basically normal to the vessel surface; would you think the effect is small enough to discount?

Thanks!

RE: BTH-1-2014 Shear ?

Well...

And I'm sure the answer is "always been done this way", but I don't like the turnbuckle being used like that. I haven't opened my Crosby in a few years, but I have no memory of an allowable out of plane loading on a turnbuckle. Any chance of slapping a shackle on the lug, and using a super short, high capacity nylon to connect the shackle to the turnbuckle?

Another thought, any chance of getting the lug turned so as to line up with the rigging?

RE: BTH-1-2014 Shear ?

(OP)
Yep, thought about both of those. I ended up using a shackle in between the lug and the turnbuckle b/c I didn't like it either. We also used cheek plates such that the shackle was bearing on about 85% of its pin width. That eliminated the lousy loading no the TB but left the loading on the lug still a question in my mind. I had thought about cutting the lugs differently so that they all aim directly where they 'should', but decided it wasn't worth the increased fabrication cost, layout/fit-up time, or the possibility of using the wrong lug in the wrong place. Based on previous experience and the homework that had been done (once we fixed the TB loading with extra shackles), I had no concern about the lug, rigging or vessel itself; I just want to improve my calcs. All in all, it flew well, but I made a mental note to revisit the issue when between projects and I had time to tune up my calcs. It's kind of a pet project that I've been working on in my spare time but I'm enjoying the exercise.

For the record, Crosby states (in caps) that, "TURNBUCKLES RECOMMENDED FOR STRAIGHT OR IN-LINE PULL ONLY".

And, again, thanks for the input.

RE: BTH-1-2014 Shear ?

(OP)
If anyone is interested...

QUESTION TO ASME (WITH RESPONSES):
For members that are subject to combined normal and shear stresses, Equation (3-37) limits the critical stress to fcr <= Fy/Nd. If the member is loaded in the X direction (in such a way as to induce bending, average shear and torsion), Y direction (in such a way as to induce bending and average shear) and Z direction (in such a way to induce axial stress and bending), could the Committee please clarify the following:
1) Is "fv" a combination of all shear stress due to Fx, Fy and Fz, whether an average shear stress or shear stress due to torsion?
YES
2) Is "fx" a stress normal to the XY plane, induced by the bending moment of Fx about axis YY?
YES
3) Is "fy" a stress normal to the XY plane, induced by the bending moment of Fy about axis XX?
YES
4) Does Equation 3-37 correctly apply to a situation where there are loads in three directions?
YES
5) If the answer to 4) is "no", then could the Committee please provide a recommended form of an appropriate equation (such as a full 3D version of a Von Mises yield criteria) that is consistent with the nomenclature used in BTH-1?
NOT ANSWERED SINCE 4) WAS "YES"

More to follow when I get their response about some question w.r.t. the weld design in these cases.

RE: BTH-1-2014 Shear ?

Hmmm...

A few thoughts:

1) Try it using average shear stress with the max bending stresses and tension stress. If it checks out, no worries.

2) If it doesn't check out, then I would refine the analysis to use the actual shear stress VQ/It. This would probably entail checking the critical stress at tenth (or hundredth) points in both axis (x and y) which just jumped the complexity of the spreadsheet.

3) Or just hand wave it away by noting that your max normal and shear stresses don't occur at the same location - which is why we don't do a combined axial/moment/shear check in beams (at least beams without torsion).

Hopefully this post is helpful. smile

RE: BTH-1-2014 Shear ?

(OP)
The answers from ASME led me to believe that the fx and fy were stresses in "native" or "obvious" Height direction ("Z" axis), and that they were due to about the "native" or "obvious" Length x Width directions ("X" and "Y" axes). I hadn't checked the Commentary in BTH because in the middle of writing this sheet, I switched from 2011 to 2014 which moved the Commentary to the back, and, I suppose therefore out of mind. Despite what my informal answers from ASME appeared to say, I found:

"Equation (3-37) is the Energy of Distortion Theory relationship between normal and shear stresses (Shigley and Mischke, 2001). The allowable critical stress is the material yield stress divided by the applicable design factor, Nd. For the purpose of this requirement, the directions x andy are mutually perpendicular orientations of normal stresses, notx-axis and y-axis bending stresses."

It does appear that KootK and TehMighty were correct (again!). Had they used some other subscript, I probably wouldn't have stumbled so badly on this one, but less than 3" away on the same page, there are multiple other topics of discussion where "x" and "y" are talking about the "native" axes of the thing you're designing. Perhaps a poor choice of subscripts, but I still wish I had snapped to a little (lot) quicker. Thanks KootK, TehMighty and winelandv for your time and valuable input!!!

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