Fillet Weld Design & Failure Areas 4

[ColinPearson]

Good Morning Folks!
Sorry for the long post but I have numerous questions here and I'm just starting to get frustrated with myself; I think this should be easy and I'm having more trouble than I'd like to admit.

I have trouble understanding about fillet weld sizing and have scoured the internet looking for really compete examples. I am 'good' at math, but I have to do it in a painstakingly detailed way; I write out all the formulas completely, show all the algebraic manipulations and write all my units that go with every number. When math is done that way, I can see the whole picture and get along fine - it's the shortcuts and skipping written steps that really confuse me. My kiddo is the same way LOL, she will be getting mad at some math homework so I'll take her calculator and have her write down what she knows and tell me what she thinks she is solving for how she thinks she can get there; magically when she takes her time and works through the steps, she always gets it. Anyway, I've read Blodgett with his 'line method' which kind of makes sense, but I don't quite 100% understand. I've read the AISC steel book and their associated examples but a lot of times they leave out multiple steps and/or the example solutions are using a tabulated method (which is fine but I need to know how to do it by hand first).

I need to design a basic shear connection. Something like a PL3/4" that's 12" x 12", fillet welded (both sides) to a pipe column on the "left" side of the clip. The thing it supports is welded to the "right" side of the clip. I'm using 40 kips (vertical down) as the load and length of weld L = 2*11" to account for holding the weld back from each end.

(1) If I were to use Blodgett's 'line method', I can come up with a Fv = 1.82k/in, Fb = 11.9k/in and a resultant of Fr = SQRT(Fv^2 + Fb^2) = 12.0k/in. Then I want to compare that to AISC's required (ASD) strength. My problem there is if I'm using Rfw = 0.6FexxL(w)cos45°/theta (where w = weld size), I get Rfw = w(326.7k/in). Then I would set up the inequality Rfw > Fr ... which is w(326.7k/in) > 12.0k/in ... which leaves you with w > 0.037" and that just seems so unbelievably low that I don't think I'm doing something right. I know there are other minimums to the weld size that won't let me use w = 0.037" = 1/32" but I just want to understand if the math is correct.

(2) If I want to us AISC (ASD) then I would use that same equation Rfw = 0.6FexxL(w)cos45°/theta. I think often I'd be checking a weld on something that we had already existing (such as this PL3/4"x12"x12" clip), so I'd know my 'L', and would try to keep 'w' to a 5/16" max to make it a single pass weld if possible, but within these sorts of guidelines, would I just try various combinations of 'L' and 'w' until I get some allowable strength that is bigger than my actual? That works in a spreadsheet but since the length and leg size of weld show up in the calc for actual stress, then it would be burdensome to iterate by hand. Is there a better way that's more suited to hand calcs? I've seen plenty of stuff that talks about making assumptions of a 1" length and/or a 1" weld leg size, but examples I find are lacking in the sort detail/completeness that I was talking about in my first paragraph so I don't feel confident in what I think I'm learning from them.

(3) Shear yield and rupture ... this one just totally throws me for a loop. Do I need to check yield and rupture for both items that are welded together every time? My understanding is that shear rupture will only happen on a net section (such as through bolt holes), so if I my Agross = Anet can I get by with only checking the yield?

(4) Failure areas - Given the answer from (3), I'm going to be checking the shear yield and/or shear rupture on both the pipe column and the clip itself. For the clip itself, I think the failure area is clear (CLIP THK x CLIP HEIGHT). But for the pipe, it's something like 10' tall and the top of the clip will be about 8' from the base. I can imagine a potential shear failure area of PIPE WALL THK x CLIP HEIGHT (on each side of the clip) but no way can I imagine this thing actually shearing all the way down that 8' to the base of the column. Would my shear area actually be PIPE WALL THK x 8'? It seems so unlikely to have a shear failure that long, but I can't just not check it b/c it sounds silly; I'm trying to show that the failure won't happen with just the math alone.

(5) Final question - I think punching shear might also be involved but I haven't really gotten that far. I'd guess that Id be looking to one side of the neutral axis of the clip (either top/tension or bottom/compression), figuring out the resultant force of that section of the bending moment diagram, determining the stress on the corresponding section of pipe wall and checking that against the allowable shear strength of the pipe material. Is that the general idea?

If you've read this far down, thank you very much and have a GREAT long weekend!!!!!

 

[RPMG]

This isn't a simple shear connection. It is either a bracket with eccentrically loaded welds or a full moment connection. I would recommend consulting a structural engineer. This may be a 5 minute problem, but it isn't a good intro problem to structural engineering.

 

[DrZoidberWoop]

Connection design PE here. It sounds like your problem may be similar to a scenario in Table K1.1 from the AISC manual. I agree with you on the way that the knowledge can sometimes be hidden or scattered throughout different resources over time. You can go about it a few different ways. First of all, the geometry of the situation needs to be clearly described. Depending on your situation, you can most likely size the fillet welds to handle the load with the IR or Elastic method. I've attached some weld design tips I put together a while back. It has a complete formulation of the elastic method from the basic data input stage all the way to the base metal check. It also has tips on other common cases. I hope it helps.

And yes, all relevant limit states need to be checked along the load path. If the required weld size is too large for the base metal of either the plate or pipe to handle, you need to try something else (pjp/cjp, lengthen the plate, etc.). Also be aware that you may have to increase the fillet weld legs to accommodate the outer diameter of the pipe. Be sure your effective throat is still acceptable.

Best of luck to you.

 

[DrZoidberWoop]

I just wanted to add that 40 Kips is a substantial load and not to be taken ahem... lightly. Just to warn you, using ASD, a pair of 11" long 5/16ths fillet welds supporting a 40 kip load that is at the other end of a 12" plate/cantilever-arm can only take about 32 kips (Cweld = 0.589/per vertical leg so about 1.18 from AISC table 8-4 ==> 1.18*5sixteeths*11"/ASD Factor of "2" = 32.4 Kips). And that is without any other reductions, particularly the kind of reductions that should be taken into account when the system is to support movable machinery or something of the like.

If someone around you has experience with this type of thing, please defer to their judgement/calcs on this matter.

 

[RPMG]

In situations like this, over design. Cantilever a beam with flange plates and weld a lug below. Do something that you know will work.

When I hand these problems off to young engineers, I expect errors to be made. Even with the best references and advice, this is a problem that is set up to do wrong. For instance, the problem says simple shear connection, but described as being cantilevered. And Blodgett is from the 60's. His work has made it to AISC. And we haven't even talked about plate buckling, extended single plate connector design, moment on the column, or the attachment to the plate. Also, little welds are being added to a large plate. These are red flags.

 

[Ron]

Without regard to any code or reference criteria, the basic failure of fillet welds occurs primarily in two modes (neglecting prying)....tension and shear. Failure in a properly designed and welded fillet weld will occur in the throat for either condition. The depth of the throat is 0.71 x leg. To make sure this condition exists, it is important that proper welding occurs. The welds should be inspected for proper fusion, undercut, lap at the legs, proper profile and porosity. Inspection should be done by a qualified welding inspector. The welder should be properly certified for all positions encountered.

 

[Settingsun]

Is the 326.7k/in from the first post correct? Seems very high but I don't speak American fluently so might have made a mistake converting to metric.

 

[ColinPearson]

IT department strangled my computer and decided that starting Friday mid-morning that it must upload my entire C: to the cloud so I left it at work. I thought I would get email notices if this thread was answered but I didn't see any all weekend so I thought it got left unanswered. Very happy to see all these replies here this morning, thank you all very much!

@RPMG - thanks for that idea, a piece of beam may very well be what we use.

@DrZoiderWoop - Thank you very much for the attachment, I'm going through that now. Could you please help me understand where the "Cweld = 0.589/per vertical leg so about 1.18" comes from? Since I was assuming a=11" and the eccentricity of the load was 12", (e = a*l = 12", therefore a=1.09) I interpolated AISC Table 8-3 for "a" between 1.0<1.09<1.2 (for k=0) and I get C=1.19 ... clearly this is pretty much the same number as your 1.18, but since I'm not sure where Cweld=0.589 came from (which you appear to have doubled to get a=1.18), I'm not sure if this is just a coincidence that our numbers match or if I'm doing something fundamentally different than what your describing? One further question: is Table 8-3 saying to use EITHER the tabulated value of "C" or the Cmin given at the top, whichever is smaller?

@Ron - agreed. We've got qualified welders and ample QC to ensure the welding is performed correctly. I appreciate your nod to the practical matters of actually getting the real-life physical weld made.

 

[DrZoidberWoop]

@ColinPearson. Our numbers match because we used roughly identical methods. The AISC book is set up to account for 2 vertical weld lines, while the IR method code I wrote calculates for 1, but can be multiplied by 2 to account for a pair of vertical weld lines.

And Cmin is not the variable you should be interested in because you already have the geometric constraints under control. You know the load and the maximum length of your welds. The critical information you need is fillet weld size (Dmin). In this case, Dmin = (ASD Factor of 2 * Sqrt(Shear Load^2 + Axial Load^2))/(Weld Length * Cweld) = (2*40 Kips)/(11" * 1.18) = 6.16/16ths. The weld required to develop the plate, per AISC page 10-102, is (5/8)*(plate thickness)= 0.47" or 1/2" inch. So in this case, the plate base metal will not control. However, we need to look at the pipe. If it's A53 Grade B (Fy = 35 KSI, Fu=60 KSI), your maximum permissible fillet leg size will be Dmaxpipe = the minimum of either:

1) (0.6*Fu of the pipe * actual thickness of the pipe wall)/(ASD Shear Rupture Factor of 2* 0.928 Kips/in/16th)
or
2) (0.6*Fy of the pipe * actual thickness of the pipe wall)/(ASD Shear Yielding Factor of 1.5* 0.928 Kips/in/16th)

If the pipe can handle it, you should select the 1/2" fillet weld to develop the shear plate. If the pipe wall is too thin to handle the welds, then you have to apply some engineering judgement because you will have to either resize the weld/plate length or apply welds that will rupture before the ductile failure of the plate, which can be dangerous in certain scenarios. All of this is relevant only to the design of the welds. You still have to check all the remaining limit states for the plate and pipe.

 

[ColinPearson]

@DrZoiderWoop - Thank you VERY much. If you've got the time and inclination I would very much to continue this conversation. I've had a response typed up and have been changing it over and over since about 10 minutes you responded as I've been making more and more sense out of the whole thing. I've got to leave work to go to a school function of my daughter's but would leave to finalize my response and post back tomorrow. Thanks again!!!

 

[ColinPearson]

@DrZoiderWoop:
Firstly, and as before, thank you kindly for your assistance. I have no problem doing the legwork to learn more on my own, but sometimes it's easy to get stuck without a little guidance. I very much appreciate yours here.

Given the values for "Cweld" obtained with your method and/or by interpolating AISC Table 8-4, I follow your 6.16/16ths for "Dmin". I also understand what you're saying about not needing to worry about the "Cmin" formula as given above the Table 8-4 since I already know the load and geometry of the problem.
QUESTION (A): I guess that if I did NOT know all the particulars (say if I knew the geometry and weld size and wanted to determine how much load the weld group could support), I would I ever have to check that my combination of "Pa", "D" and "L" all conspired to ensure I had a C >= Cmin ... or is it just that “Cmin” is the “C” value for k = 0? That seems to make sense to me b/c dividing by a smaller number would give you a larger “Dmin” required so a quick check with “Cmin” would give you a more conservative weld size.

Regarding your options 1) and 2) ... I believe you have basically taken AISC14th's expression (8-2b) Rn/(ASD Fillet Weld Factor Of 2.0) = 0.928DL and set it equal to the expressions in AISC14th Section J4 for shear rupture and shear yielding, respectively. I was trying to figure how you got down to the simplified version in your post, so I began by setting Eq(J2-3) equal to both Eq(J4-3) and Eq(J4-4). I solved both versions and get the same exact numbers that your already-algebraically-simplified versions give me. Per your suggestion, I would then select the smaller of those as my "Dmax" as governed by the pipe thickness. That makes sense as it seems to essentially set the strength to be governed by the pipe material (w.r.t. shear yield and shear rupture) and then you solve for the corresponding weld size. As it turns out, with a 6" S/40 pipe, the "Dmax" is larger than would be allowed by the wall thickness.
QUESTION (B): The loading and geometry are "kinda" known for this problem b/c I'm trying to use what we have on site. We have a small variety of pipe and plate material so thicknesses are generally known, but I could do something like change the height of the bracket. In this case, would you suggest I start with the "Dmax" as calculated above, then go back to Table 8-4 and alter the length of the weld group (which in turn alters "a" which requires interpolating an updated "C" the tabulated values) until Dmin(Table8-4) <= Dmaxpipe? That would make sense to me b/c the “Dmaxpipe” does not depend on the length/shape of the weld group, only material properties and wall thickness.

QUESTION (C): Before I get to this question, I should say that I've revised my load to 20 kip as it's going to work out better to have two supports which will equally share the original 40 kip load. Given that bit of information, and the fact that Dmax = 0.246" = 4/16ths, I'm inclined to think changing up my material selections to make this work. If I use PL1/2" for the bracket then per AISC 10-102, my D = (5/8)*(1/2") = 0.3125" = 5/16". If I switch to some HSS6x6x3/8”, then even using the lowest grade A500 mat’l, my Dmax = 5/16”. Finally, I’ve added a second support so given my new load of P = 20 kips, my new Dmin = 3/16” per Table 8-4. I think I could safely choose w = 1/4” and satisfy all of those requirements. Does that sound reasonable? Then of course I will check the bracket itself and the pipe itself as well.

QUESTION (D): If the particulars in my previous paragraph are correct, it’s fortunate that the material selections we have on hand appear to work out nicely. In this case, it’s the end of a job and we’d like to use as much leftover material as possible though if need be we can certainly order whatever we need. In some instances, however, we don’t have many options and are forced to use the material on hand. Could you elaborate a bit on what you alluded to at the end of your last post? If there are no obstructions, I can see where one would often just resize the plate. On the other hand, the option is to use a weld that is strong enough on its own but does not allow the full strength of the plate to be realized. In this particular job situation, the load is very well known; would you tend to worry more about a weld rupture failure vs. ductile plate failure in a situation where the loads we less well known?

 

[DrZoidberWoop]

A) If you know the geometry (L) and load distance, you know the "C" value. "C" is independent of load magnitude or weld size (within reason). The Cmin equation would start with the weld size/length and end up determining a maximum "a" in the ex=aL equation. There's nothing else of value you could really extrapolate there.
A.1) Yes. Your use of the Equations J4 are correct. However, when you say "As it turns out, with a 6" S/40 pipe, the "Dmax" is larger than would be allowed by the wall thickness" I don't understand because the only independent variable is the pipe wall thickness.
B) If you've selected your plate size and the required weld, you could identify a minimum pipe thickness by using AISC Eq. 9-2.
C) If you have enough workable flat on your HSS, I'd default to a 5/16" fillet weld just to be conservative. I'm not reading too deeply into your geometry, but you need to get this calculation package checked by a local PE.
D) Unless you are designing a booby-trap, you should always prefer to have a ductile failure over a brittle failure. It's the difference between 1) a heavy object sagging down slightly and 2) having that object snap off and fall on your head. Go the route of a ductile failure.

FYI, this problem shouldn't take a qualified individual more than a half hour to design and draft. If you are in a pinch for time, just ask a qualified individual near you. Engineers love to feel needed and as if they're the smartest person in the room. Best of luck to you.

 

[ColinPearson]

@DrZoiderwoop - I misspoke a bit in that last one, the D I was talking about was as prescribed by the (5/8)t ... that was the weld size that was larger than the pipe wall was thick. Thank you very kindly for your help and I hope you have a great weekend!