It depends on the support conditions of your beam and how the load is applied. The centroid of all your sections align, so there is no shear flow in the welds. I think you just need to deal with the local effects at the supports and where the load is applied.

see below.

Canpro,
If the load is being applied to the plate edge, would you have to calculate shear flow to get the load into the channels?

Load is being applied to the plate only. Top and bottom. Lifting device. Creating tension/bending in the plate. I am attaching the channels to the plate to prevent buckling. What I am wondering is, what, if any, loads are being applied to the welds, and if so, how can I quantify them?

The plate sounds really deep for a lifting device. I assume this is to resist the bending stress in the plate? Personally I would go with a conventional lifting beam that you can evaluate with standard methods and weld on gusset plates for lifting eye's.

Is there some reason you have to go with plate / channel design?

Ideem,

Yes we have the channels and plate readily available (on site). So, I'll need to use this configuration.

Follow the procedure below at your own risk:

Weld Design
1) Find properties of the built-up section, I, Q (with respect to the upper flange of the channels).
2) Compute f_v = V*Q/(I*b), in which b = t_p + 2*b_f (of channel).
3) Design the weld for the shear force V_w = f_v*A_flange

See revised, made on 3 May 20 00:36.

Use the built-up sectional properties to derive flexural resistance/capacity, and check deflection. The axial forces (P, T, V) will be resisted by the plate only .

JP20 - I feel your pain

Can you meet the strength requirements with the two C15's? In that case you could consider the welds as transferring the plate load to the channels like a gusset. That isn't the real case since the plate is continuous but if the plate were to buckle you would have enough weld to allow the C15's to pick up the load.

Regarding the loads, it looks like multiple slings on top. If you are using more than one crane it is likely the slings will not all pull in the same direction. I am not sure how much the slings would skew (you will have to estimate that depending on lift). But that would give some out of plane bending which would add to the C15 welds. Fortunately this would be a shear flow problem.

Do you have access to any FEA programs to see what they tell you about the load transfer from plate to C15?

Do you have Blodgetts? I am spit balling here but if you take the distance from the top of C15 to edge of plate and look at the allowable compressive load for supported-free case and then calculate the applied stress assuming section modulus is vertical plate areas outside the C15's - this might give you some idea of plate buckling on the top and bottom edge. If this works how you resolve to the C15 to plate welds I am not sure.

Johns201889, if the centroids align then your Q value in your shear flow calculation goes to zero (Q=a*d, d=0).

JP20, if you're using the channels resisting buckling then I would move them close to the edge of the plate that is in compression - right now your plate edge is about 8" from the channel, not sure on your plate thickness but that seems a bit far. This will also provide a bit more strength in your beam. For sizing the weld, a few things come to mind:

- You need the welds to develop any shear flow that is present
- I suspect the channel will be stitch welded to the plate - you might base the spacing of the stitch on how often the plate needs to be braced
- There might be a better approach to this, but you could base the strength of the weld on resisting 2% of the compressive force in the plate that is tributary to the stitch
- Overall beam stability comes into play, which requires you to develop your weak axis and torsional properties of the beam. Off the top of my head, I'm not sure how much of a connection you require between components to achieve for example the fully effective Iy of the section.

Typically, these kinds of welds end up being minimal. I'd be interested to see some other details here - plate thickness, length of beam, magnitude of load.

Quote (Canpro)

if the centroids align then your Q value in your shear flow calculation goes to zero (Q=a*d, d=0).

Isn't "d" the distance from the centroid of the gross section to the centroid of area consider for shear flow calculations?

CANPRO,

I could be wrong, but isn't Q is the area above the cross section in concern, times the distance measured from centroid of the area to the neutral axis, so the maximum shear flow/stress is at the centroid of the cross section, composite or not?

Whether you are considering the strong axis contribution of these channels to your capacity or not, by welding them to the central plate, you are forcing this built up beam to act as a singular unit. The only way that it won't act as such, is if your welds fail, which is obviously not an allowable scenario. For that reason, you need to design these welds to carry both the strong axis and weak axis load contributions.

In order to determine how much shear will flow into the channels, you need to consider the relative stiffness of the channels to the entire built up section. Take the x axis as being horizontal in your sketch, and the y axis as being vertical. Compare your Ix of the channels to the Ix of the built up beam. Stiffness attracts load, so the channels will carry that proportion of the flexural moment.

Now that you know the proportion of the applied moment being carried by the channels, determine your limit state flexural moment for the entire beam. Now, use that value to determine the flexural moment in the channels. Take this value, divide it by the depth of the channel, and voila you have the force couple required to induce the channel moment. This is the force that your weld must resist for strong-axis loading.

For weak axis design, you can determine the shear flow as normally done for welded wide flange beams, keeping in mind that you have two welds connecting the channels (treated as flanges) to the web. I would personally design this weld to carry enough shear such that the channels can begin to yield at their extreme fibres. This ensures that your weak axis capacity will not be limited by the welds, and that a ductile failure mode can be preserved.

From experience, I anticipate that your strong axis loading will govern the design of these welds. That is, unless the proportions of the plate and channels are out of whack. Keep in mind that your strong axis capacity is going to be limited by buckling of the protruding web above the channels. You can find good references for the treatment of these protruding webs by looking up the buckling behaviour of protruding webs in T-sections.

Quote (Blackstar123)

Isn't "d" the distance from the centroid of the gross section to the centroid of area consider for shear flow calculations?

Correct, and the centroid of the area we are considering (the channel) is the aligned with the centroid of the built-up section, therefore no shear flow.

Quote (retired13)

I could be wrong, but isn't Q is the area above the cross section in concern, times the distance measured from centroid of the area to the neutral axis, so the maximum shear flow/stress is at the centroid of the cross section, composite or not?

My understanding is that it is the entire area of the component (the channel) multiplied by the distance from the component centroid to the centroid of the built-up section.

Think of the shear flow requirement in terms of how you build up your combined section properties. When determining your Ix of the combined section you first just lump together Ix for all individual components - you get to include Ix of the components just for showing up. After that, you apply the parallel axis theorem, which considers the distance from centroid of component to centroid of built-up section (a*d²) - this is where you start to get the most benefit of a built-up section (ie adding flanges to a vertical plate instead of just stacking plates together vertically). But, in order to develop this added benefit in the section (a*d²), you need to ensure the components are adequately fastened together to act as a unit (shear flow, a*d).

In this scenario, since the all of the centroids align, we only get to add our Ix values together, no added benefit from offsetting the channels, and zero shear flow.

Quote (Craig_H)

In order to determine how much shear will flow into the channels, you need to consider the relative stiffness of the channels to the entire built up section. Take the x axis as being horizontal in your sketch, and the y axis as being vertical. Compare your Ix of the channels to the Ix of the built up beam. Stiffness attracts load, so the channels will carry that proportion of the flexural moment.

Now that you know the proportion of the applied moment being carried by the channels, determine your limit state flexural moment for the entire beam. Now, use that value to determine the flexural moment in the channels. Take this value, divide it by the depth of the channel, and voila you have the force couple required to induce the channel moment. This is the force that your weld must resist for strong-axis loading.

I respectfully disagree with this statement. I agree with your first statement regarding the channel taking load in proportion to the stiffness (channel load = Ix channel / (Ix plate + 2*Ix channel). Once you determine the percentage of the load the channel is taking, the weld simply has to deliver that proportion of load to the channel - flexural stress in the channel doesn't come into play, and this is not the same thing as shear flow. In fact, for a simply supported beam with a UDL your maximum shear flow occurs where the flexural stress is zero.

JD20, I didn't see that this is part of your other thread. You had a nice sketch of the beam in that thread that would have been good to include here (I've included it below).



There might be (probably is) a more rational way to approach this, but I would be inclined to take 2% of the your strong axis moment, apply it as a weak axis moment, and design your welds accordingly. As Craig_H pointed out, your channels will take some of the load based on their stiffness, and that load will have to get transferred into the channel along its length and then back into the plate near the support points.

CANPRO,

Would you mind to take a look on the linked file (pay attention to p.3). Thanks. Link

• https://files.engineering.com/getfile.aspx?folder=5a02e078-8b1d-4dcb-8ed6-3

retired13, none of those examples show the component in question crossing the NA. The illustration looked familiar, it’s the same as my mechanics of materials book (Hibbeler), but I have an additional figure. See below, figure (b).

CANPRO,

Interesting. How about case (c). Actually, I don't think about component, but treat all parts togather as a single unit, then decide the location needs to be investigated/connected. Simple check is that, for a rectangular shape, the maximum shear is at the mid-section, where A_top & A_bot are maximum. Anyway, it just provided for your information.

Now I realize (b) demonstrates the case with the centroid of the leg is offset from the NA of the built-up section, thus, Q=A'*y'. If centroid coincident with NA, we can't use take this short cut, or this method, then we need to use (c), if the location shown is the interest. But does not mean shear flow is zero. Find an example for square built-up section, you will know what I mean.

In seeking to further understand this quote:

Quote (CANPRO)

Once you determine the percentage of the load the channel is taking, the weld simply has to deliver that proportion of load to the channel - flexural stress in the channel doesn't come into play, and this is not the same thing as shear flow. In fact, for a simply supported beam with a UDL your maximum shear flow occurs where the flexural stress is zero.

I am curious what "load" you see being transfered between the central web plate and the side channels? To me, in order to ensure continuity of the cross-section, the flexural stress where the web and channel meet must be the same. Hence, the welds must be able to engage tension and compression in the channel flanges, thereby inducing a continuity of flexural stress. The welds cannot transfer much vertical shear due to the obvious geometry of the channels.

I think we are in agreement that shear flow is zero in strong axis flexure, and due to the low magnitude of weak axis loads, will likely not govern there either.

Grag H,

Bending produces compression and tension stresses in the cross section, and the stress is zero at the NA. IMO, I don't consider compressive and tensile stress (flexural stresses) are the same as shear flow (stress). There is shear flow at the junction of the center plate and the channels, as well as flexural stress. The weld needs to satisfy both.

There is shear flow in strong axis bending isn't there? If it were a fully solid composite section wouldn't the shear flow sort of look like what's shown in my sketch? I think you would need to use the shear flow equation from the top protruding plate web to the welded joint. As shown on the sketch, you'll find H with the shear flow equation. I would then separate that H load into the two welds and remaining plate web based on the stiffnesses, like previously mentioned in this thread.

johns20188,

Very good presentation. Other than shear stress at the weld, also check flexural stress at the same location. IMO, the stresses are additive (fw = √fb²+fv²).

I understand computing the shear flow at top channels that results from the max shear (V) resulting from the shear-moment diagram. Got that, we’re good.

However, I don’t understand how to apply the moment to the weld in terms of k/in.

Do I take max moment and divide by the distance from the top (or bottom) to the top (or bottom) of the channel? That would give me a kip value, so I would need to divide by another length. Would I divide by the length of beam or what?

The beam is 40’ long. Max shear is 60k. Max moment is 100 ft-k.

M*y/I = fb (ksi = k/in^{2 =} k/in/in; M in kips-in, y in inch, I in in⁴.

And I take my “y” as the distance from centroid to top (or bottom) of channel?

K/in/inch of weld:
Ok say we have 10 k/in/inch of weld. I’m having a hard time correlating that. Because that’s what I was trying to do Friday and it didn’t seem right. Can you explain further please?

JP20,

As I understand it (please correct me if I'm wrong), the shear flow result, H, is derived from bending stresses as you can see from the calc in the sketch above. And H is the logitudinal shear that needs to be resisted, resulting from the bending stresses, and is in units of force/unit length.

In additional to that, I believe you need to account for vertical shear in your weld strength.

1) Yes.
2) k/in/in of channel flange width, bf. Weld demand for bending is fb*bf, the unit turns out k/in. (sorry, made mistake before)

Craig, I think we agree on this part –All the centroids align, there is no shear flow in the weld. Since this is the case, you could look at this as (3) separate elements taking some portion of the total load. The channels in this case are not part of the load delivery or the support of the combined section – so if they’re carrying any vertical load from one point to another, the load must enter and exit the channel.

For arguments sake, lets say Ix plate is 2x Ix channel, so total combined Ix of the section is 4x Ix channel. Therefore each channel carries ¼ of the load, which must enter the channel somewhere near where the load is applied, similarly, the load must exit the channel near the supports.

Quote (Craig_H)

The welds cannot transfer much vertical shear due to the obvious geometry of the channels.

As I was thinking about how to reply to you, I gave this more thought – I don’t think the channel is fully effective. The vertical shear has to transfer in and out of the channel through the flanges (bending) and is kind of a soft load transfer – you’d need a full FEM model of the beam to confirm, but I suspect only the flange of the channel are doing much. See my proposed effective section in my sketch – for the loading/support conditions, this is what I would consider effective in strong axis bending, I would take the entire cross section as effective for weak axis section properties.

Quote (Craig_H)

To me, in order to ensure continuity of the cross-section, the flexural stress where the web and channel meet must be the same

But continuity of the cross section is shear flow. You just said you agree shear flow is zero – the only purpose of the welds at that point is to make sure the vertical load is delivered to the components in proportion to the stiffness. If the channel and plate were the exact same depth and the load was bearing on the top of the plate/channel, they wouldn’t need to be welded together at all to share the load – and the stress in the channel and adjacent plate would still be the same – no weld required.

Quote (Johns20188)

There is shear flow in strong axis bending isn't there?

There absolutely is not. Whoever gave you the star for this has a fundamental misunderstanding of shear flow, or didn’t look at your post close enough. The work you show is for calculating the shear flow in the plate just above the channel, not in the weld between the channel and the plate.



Retired13, respectfully, you are fundamentally wrong and you’re giving JP20 incorrect advice. The bending moment in the beam does not come into play when calculating shear flow. Shear flow is not about how much bending stress is in the section, it is about how fast the bending stress in the section is being developed. Shear flow is about how much we’re adding to the bending stress, not how much is already there.

JP20, forget about the maximum moment when you’re checking shear flow. You need the maximum change in moment (location of max shear) – this is where your shear flow is greatest. You seemed to have made it clear that you understood that a few posts ago, but then you went back down the rabbit hole about bending stress with retired13 again.

CANPRO,

Not to argue. The junction of the plate and the channel is suffering two separate actions occurred simultaneously - bending and shear.
Actually, the channels in OP's beam feel very little of the shear force, rather, the majority force comes from the push (compression), or pull (tension) due to bending stress on the channels. But still need to be designed for both forces conservatively. I think OP can judge by his own from this point on. Therefore I rest my case.

JP20,

Please ignore my suggestion on weld design for shear (made on 1 May 20 19:51). The steps shall be revised to read as follows:

Weld Design for shear
1) Find properties of the built-up section, I, Q (with respect to the area above the upper flange of the channels).
2) Compute fv = V*Q/(I*b), in which b = tp + 2*bf (of channel).
3) Design the weld for the shear force fv*Ab_flange

Then fv should be combined with fb as discussed above.

Quote (CANPRO)

There absolutely is not. Whoever gave you the star for this has a fundamental misunderstanding of shear flow, or didn’t look at your post close enough. The work you show is for calculating the shear flow in the plate just above the channel, not in the weld between the channel and the plate

.

What is the difference between the shear flow calculated in my sketch and the shear flow calculated in your book's figure 7-15 c and b?

I’m now a little confused. It seems 3+ have differing opinions on how the load is transferred.

The plate itself is fine to take the entire bending moment (1.25”x36” plate) so I’m not too worried about the bending moment in the steel.

I specifically want to know what the welds are going to take load-wise.

I presume it will take shear flow from the max shear and the shear flow due to the max moment. This has to be completed tomorrow and built Monday. Need clarifications stat lol.

My username says (Structural) but really I’m construction (took my PE exam in construction). And work for construction company. Maybe if I would have took the Structural exam I would know but I didn’t.

The channels will feel stress only when bend to shape, that needs to conform to the deflection of the plate they are connected to. Nearly 100% shear force will be resisted by the middle plate. You shall have a structural engineer review your design. Hope it is not too late.

I think it will work with 5/8” fillet on both channels, but I want to be super confident. These conflicting statements are concerning.

It is a confusing matter, as this type of built-up shape is rarely seen in practice. I've not, and don't like it.

I don’t want to rehash all of the comments above, but a 5/8” weld will take several passes. I always used 5/16” as the maximum single pass weld. Saves time and money.

gjc

Hi JP20
I was wondering why you cannot redesign the lifting beam to a more convenient shape which would allow an easier analysis of the welds? For instance my sketch attached

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

• https://files.engineering.com/getfile.aspx?folder=fe6d0eb4-5b1d-4baf-8fa9-0

The bending moment in the built-up/composite beam does not come into play when calculating shear flow. To suggest otherwise is fundamentally incorrect. Jp20, read CANPRO's posts again.

The longitudinal stress in the plate and the channel either side of the weld is constant under flexure (strain is linear with depth). There is no differential stress going on that suggests a longitudinal force is being transferred along the weld due to bending in this type of symmetric arrangement.

Agent, my turn to be wrong or misunderstanding you. Apply your argument to a typical 3-plate I-girder where there is shear flow in the welds to create the composite section. The stress at the top of the web is the same as at the bottom of the top flange. It's not the stresses at a single section; it's difference in stresses at sections separated by a short (infinitesimal) distance along the beam that result in the weld stress.

I think shear lag means the central plate acts as the web for half of the flange width of the channels (or maybe more given the load application). It's better placed geometrically than the channel webs.

Edit: This forum need a serif font so "I girder" looks like an I girder!

I read through this morning and as I understand JP20 situation max M=100 kip-ft, plate is 36x1.25 and C15 channels (unknown size) and length 40'. I saw a comment the plate would take the load. I decided to check this using AISC section F11 (shockingly straightforward for AISC). I come up with allowable bending 804 kip-in - vs 1,200 kip-in applied. I assume unbraced length 40'. I could be wrong on the calc so please verify for yourself. Note I am using AISC ASD method with Cb=1.

I would strongly encourage you to revisit the assumption the plate will take the load.

If the plate will take the load than the C15's are your suspenders and belt - so the weld size is not as critical. However if the plate will not take the load than the C15's are critical and weld size is critical. As several people have pointed out this is a non-standard section and opinions are going to vary on the right approach. Personally I am more concerned about overall weak axis buckling (whole thing folds up like a potato chip) and buckling of the outside edges in compression.

ldeem,

The channels are C15x50 and the plate is considered continuously braced because the weld will be continuous and the flanges of the channels (turned around) will be resisting any out of plane bending that may occur due to the low axial load induced from the top two corner attachment points. The max moment is actually more near 80ft-k .

There will not be any loads applied in the weak axis direction.

JP20

I think there are several ways to look at this and each method has its merits. You will probably need to consider each method and take the most conservative approach.

So back to the stability/bracing - The two C15's will form a box section which will be good in torsion so will resist the LTB of the plate (twisting of the section). However I am not sure how you would go about proving the two C15's will provide the needed bracing to assume the plate is fully braced. An alternative would be to consider the two C15's as a box section and check that for the full 40' unbraced length. AISC-05 notes long box sections can fail in LTB but they don't give equations since deflection would normally control. That is true for buildings but your case is unique so if you go this path you should check LTB of the box shape. There are lots of long span box sections used in crane girders so this would be a good place to start. CMAA or AIST Technical Report 6

There have been several threads on stability bracing over the last few months which may be worthwhile to read up on.

How would the section ever see twisting action?

JP20, part of the requirements for stability bracing is that it resists the section twisting. You see this when looking at stability bracing points on a long beam. A full depth stiffener with a brace at the mid-point can be considered a brace point for LTB because it keeps the beam flanges from twisting. This is my understanding however I am not an expert in stability bracing requirements.

In your case for the plate only to carry the load you have to assume some amount of lateral stability bracing. In other words taking the full 40' length does not work. So the question is do the two C15's provide the lateral stability bracing to justify using some length less than 40'. In a way I can see that working because the two C15's form a box and therefore at torsionaly rigid, On the other hand I have never designed a beam that needed stability bracing and that I used members welded along the length of the beam like this. In my experience the closest I have come to something like this is having two trusses parallel to each other and I assume the chords are braced at each point where there are cross connecting members.

If it were me and because I don't know if the two C15's provide enough stability bracing I would look for another analysis method (or change the plate so it will work). The closest I can think of are crane box girders and do the analysis using the two C15's as a box girder spanning 40'. If this works it leads to the question of how to weld the C15's to the plate and satisfy the analysis assumptions made. Also you need to satisfy the load transfer from the plate to the box girder.

You have a really tricky problem on your hands here. It's good you are reaching out for opinions on this form but that is no substitute for having a peer review. Hopefully you have someone in your office or can hire a consultant familiar with your company to do a peer review of your calculations and drawings. You might also want to consider doing a weight test on this to be sure it will work.

Quote (Agent666)

The bending moment in the built-up/composite beam does not come into play when calculating shear flow. To suggest otherwise is fundamentally incorrect. Jp20, read CANPRO's posts again.

The longitudinal stress in the plate and the channel either side of the weld is constant under flexure (strain is linear with depth). There is no differential stress going on that suggests a longitudinal force is being transferred along the weld due to bending in this type of symmetric arrangement.

Longitudinal shear/shear flow IS the result of differential bendinging stress over an infinitesimally small length along the longitudinal length of the beam. The bending tensile and compressive stresses on the cross section cause there to be internal longitudinal shear stress to resist them (otherwise how does the beam section stay together?). Shear flow is the sum of the longitudinal shear stresses resulting from the bending stresses. It doesn't matter that there's no differential bending stress along the flange, there's still shear flow since there is still bending stresses (there's still shear flow in the flange of an I beam even though the flange bending stress is constant along the flange).

Weldment is a 3D element, thus subjects to stresses in all three orthogonal directions. The weld design stress/force is the vector sum of all stresses/forces.

Can someone break down how you convert a moment to longitudinal shear?

Quote (JP20)

I want to be super confident.

Rightly so. For repeat use of a lifting device, someone has suggested "peer review" and "load test", that are essential for your success.

Agent666, thank you for being another voice of reason in this thread.

Quote (Johns20188)

What is the difference between the shear flow calculated in my sketch and the shear flow calculated in your book's figure 7-15 c and b?

We're talking about the shear flow between the channel and the plate - your calculations were for shear flow between two portions of the plate...completely different plane, you're at the wrong location.

Quote (Johns20188)

It doesn't matter that there's no differential bending stress along the flange, there's still shear flow since there is still bending stresses (there's still shear flow in the flange of an I beam even though the flange bending stress is constant along the flange).

I agree with this, but we're not talking about a beam section made of 3 plates with offset centroids, we're talking a plate and 2 channels all aligned. It makes a difference.

Say for example you had two beams - one is solid 6"x6" steel and the other is a collection of 6 plates 1"x6" stacked vertically side by side but not welded together. If the plates are loaded on the top edge and bearing on the bottom edge, both beams will have the exact same strong axis bending properties and the exact same stress profile, there is no weld required for the loose collection of plates to behave the same as the solid square beam (because their centroids all align). If the loose plates were stacked horizontally (offset centroids), then you would now require a weld between layers of plate to achieve the same properties.

Quote (JP20)

My username says (Structural) but really I’m construction (took my PE exam in construction). And work for construction company. Maybe if I would have took the Structural exam I would know but I didn’t.

I want to be super confident.

This might not be the advice you want, but its the advice you need - you're in over your head with this one. And there is nothing wrong with admitting that. You aren't going to learn this enough to be confident in your design by Monday. And frankly, you've been given questionable advice here so if this is your last resort I think its time to ask for help (in person). The use of continuous 5/8" fillet welds is far more than you need to fully engage the channels - if you don't have an automated welding process you'll probably have the rest of the week to figure this out while they make 4 - 40' long 5/8" fillet welds. With that said, I'm happy to help you sort through some of the technical issues.

Quote (JP20)

Can someone break down how you convert a moment to longitudinal shear?

This right here makes me want to give up - I think it is has been made clear in this thread that you don't need the moment to calculate longitudinal shear. The moment does not develop longitudinal shear - it is the change in moment (shear) that creates longitudinal shear. Johns20188 explained it best when he said "Longitudinal shear/shear flow IS the result of differential bendinging stress over an infinitesimally small length along the longitudinal length of the beam" - it is the change in bending stress from point along the beam length to another. That is the change in bending moment, which is shear - it is not the bending moment.

Based on your plate size and channel size, the channels are only taking about 6% of the load each so they're really only helping with lateral stability properties - I think you know that. How are you checking the unbraced strength of this section? You have a very flimsy compression flange (just the bit of vertical plate) and I doubt it has much capacity over 40' unbraced length. You need to make sure your built-up section actually works before you worry about how to weld it together.

I am going to put a horizontally-orientated plate (between holes) on either side of the flat plate that protrudes above the channels.

It doesn’t seem as if one person on this thread can answer the needed questions tbh.

JP20, what question has gone unanswered? I'm not confident you know what questions you should be asking. Shear flow calculation is straight-forward and can be found in your college mechanics of materials book.

If you're welding additional plates near the top of the beam your channels are doing even less. How are you checking the unbraced strength of this beam? It is a mono-symmetric section and that check isn't exactly straight-forward.

Canpro, I’m not and I don’t know how to tackle that. Do you know how?

JP20, where are you practicing? CSA S16-14 (canadian steel code) has a nice procedure for mono-symmetric sections (cl 13.6), but I believe this only applies to I or T shape sections - there is good reason for this, you're going to want a horizontal compression flange to get you any sort of strength of 40'. I'm sure AISC has a similar procedure.

Your sketch in the other thread showed your slings inclined which will induce significant compression in your beam, has that been considered?

USA. Yes they induce 12.5k max in axial compression. I checked this incorporating area of section and it yielded a KSI below the allowed (computed from the formula provided by BTH)

Quote (steveh49)

Agent, my turn to be wrong or misunderstanding you. Apply your argument to a typical 3-plate I-girder where there is shear flow in the welds to create the composite section. The stress at the top of the web is the same as at the bottom of the top flange. It's not the stresses at a single section; it's difference in stresses at sections separated by a short (infinitesimal) distance along the beam that result in the weld stress.

I think shear lag means the central plate acts as the web for half of the flange width of the channels (or maybe more given the load application). It's better placed geometrically than the channel webs.

Edit: This forum need a serif font so "I girder" looks like an I girder!

I agree it's the difference in moment between two points that causes the shear flow. Another way of expressing the change in moment along the length is by saying V ∝ dM/dx......... i.e. the change in moment is proportional to the shear. We know shear flow is proportional to the shear flow. Nothing to do with longitudinal stresses derived from bending in a beam as some people were suggesting.

With an I girder, it fundamentally a different beast to what is being proposed here. The compression force in the top flange (say) has to transfer across the weld to resolve with the tension force in the bottom flange. For the symmetrical section about the centroid proposed, this is not the case. Load in channel can resolve through the channel web and doesn't necessarily need to go through the plate. There are effectively two sections placed beside one another constrained to the same deformation. I believe as others have said that if the centroids are all aligned for the different bits of steel. Then there is no shear flow.

I did some more thinking on this while I was sleeping which can be dangerous. What you have got is three bits of steel constrained to the same deformation with centroids aligned. With all the load being introduced via the plate to the system. One way to think about this scenario is that the loads are shared based on relative stiffness, work out the relative stiffness and you know the loads going between each member, size the welds for this vertical force. You have the added complication of getting the loads out at the ends of the channels.

It's fundamentally no different than making a flitch plate out of timber/steel/timber plies. Provided that the centroids are aligned, there is no shear flow. It is simply some members with dissimilar stiffness sharing a common load. Once you know the stiffness and how the loads are transferring between members and introduced to the members you can design fixings between the plies to carry the proportion of the load required to each member and follow the loadpath required for equilibrium

If you had to build some sort of model to represent what is going on regarding the vertical load transfer perhaps you could do something like this. With pin ended links constraining the two members to the same deformation under whatever loading you were applying. Load in these links is going to give you an indication of the vertical load transfer between the elements. This vertical load transfer is what you should be designing the welds for in my view.

Agent666, I mentioned the concept of treating it as 3 separate sections and splitting the load based on the the stiffness - but you’re transferring the vertical shear in the channel web to and from the plate via flange bending, so I wasn’t sure how effective this would be, feels like that load transfer might be a bit soft to say its fully effective. In terms of the model you proposed, your links wouldn’t be rigid (channel flange bending) and the load would only be applied to the plate. The pin support on the channel would be a vertical roller...only the plate is supported/loaded.

I am wondering if introducing 1/2” horizontal stiffeners will effectively brace the “top flange) which is the flat plate.

Also wondering if I could solve a lot of my problems by moving the plate down to be flush with the top flanges of the channels therefore I have a good compression flange. Then I could weld three eyelets on top of the plate. Although I would then have an excess amount of plate below. But that’s in tension so not much to worry about I suppose. Am i thinking right?

If I took this approach would this eliminate my unbraced length dilemma?

JP20,

After further review, I've to admit I made a mistake to include bending in the consideration. The weld should be designed for the shear flow at the flange-plate junction. Sorry having to misled you, and waste your precious time.

Quote (JP20)

It doesn’t seem as if one person on this thread can answer the needed questions tbh.

Yeah, we're killing you here. Sometimes we get so bunged up in fancy theoretical fun that we forget that we're dealing with a real engineer trying to solve a real problem in an expedient way.

The sketches below show how I'd tackle this if were a real problem crossing my desk. Go forth and be profitable my friend. Expend your serious mental horsepower on some problems more worthy of it.

Quote (KootK)

Yeah, we're killing you here. Sometimes we get so bunged up in fancy theoretical fun that we forget that we're dealing with a real engineer trying to solve a real problem in an expedient way.

I might have gotten slightly side tracked with trying to explain some basic shear flow concepts, but I've been trying to give JD20 some sound and practical advice.

Quote (KootK)

Expend your serious mental horsepower on some problems more worthy of it

As someone who expends a fair amount of mental horsepower designing these types of devices to be both safe and efficient, I have to say I take some offense with this. This design is not as straight forward as it seems. This beam has a significant unbraced length at 40', and the section is built up and mono-symmetric which complicates the LTB check - and it is a lifting device so you bet your ass it will see its design loads.

Quote (KootK (from the PDF markup))

If you want to save a little money on this, I guarantee that one channel will suffice

KootK, I have a mountain of internet-respect for you, but you are adding to a pile of bad and now dangerous advice. Have you seen the proportions of the plates and channels he is using? Plate is 1.25"x36" and the channels are C15x50. You can check my math, but I get a radius of gyration of about 2" in the weak axis. Unbraced length is 40', so kl/r is about 240...this thing can't even function as a proper compression member over 40' with two channels and you're advocating for a single channel. OP has stated he hasn't checked for LTB and doesn't know how - this isn't the time for OP to just solve this in an expedient way.

Quote (JP20)

If I took this approach would this eliminate my unbraced length dilemma?

It helps improve the unbraced strength of the beam, but you still need to check it.

Edit: couldn’t sleep because I realized I messed up my Iy calc, and subsequent radius of gyration. Kl/r is closer to 200, but I stand by what I said.

Yup, I took op's sketch to be somewhat to scale and, therefore, the beam nowhere close to 40' long. Nobody gets it right 100% of the time. I think that the proportions that I assumed are pretty clear from my sketches. I don't see the interwebs breaking over this.

KootK, I think you get it right enough times...but you know how it is on e-tips, how often do you a reasonably proportioned sketch?

I'm not worried about the interwebs breaking - I'm worried about OP's spreader beam folding like a pretzel and hurting somebody.

JP20, I double checked the slenderness ratio of your beam and I get ry = 2.32" and a kL/r = 206 > 200 which means this beam does not have any reliable compression resistance. You've stated that you didn't check this long unbraced beam for its unbraced strength, and it is apparent you didn't take the unbraced length into account when checking the compression.

Whatever bending/compression capacities you have calculated for this beam are severely overestimated - your beam design is not safe

You are missing or misunderstanding far too many concepts here to get a quick answer from e-tips. You either need to talk to your local senior engineer and get help, or if there is no other engineer at your company, its time to admit this is out of your area expertise and get outside help.

Quote (Agent666)

Load in channel can resolve through the channel web and doesn't necessarily need to go through the plate.

I suppose you could say the same for a rectangular hollow section. How does it decide whether to use its left web or its right web?

Is a weld's stiffness so low that channel's web has more control over the flange tip than the plate that is welded directly to the tip? I wouldn't have thought so, unless the weld is undersized.

Quote (Agent666)

It's fundamentally no different than making a flitch plate out of timber/steel/timber plies.

Flitch beams are usually made from stocky elements in the direction that matters. This channel-plate section has the slender channel flanges. I think it will make a noticeable difference (shear lag and section distortion).

I'm interested in Canpro's effective section from 2 May @ 23:46: it amounts to an I-beam with the web extending above and below the flanges. That doesn't seem consistent with the case for no shear flow in the welds.

steveh49, if you take just the channel flanges as effective then there is shear flow in the welds. I don't know if you read through this whole thing, but there are bigger problems here than shear flow.

Quote (JP20)

Can someone break down how you convert a moment to longitudinal shear?

Quote (CANPRO)

This right here makes me want to give up - I think it is has been made clear in this thread that you don't need the moment to calculate longitudinal shear.

Different people think and understand differently. Whilst there is another way of calculating this, working with longitudinal stresses may be easier to understand. The explanation in the image below works in terms of longitudinal stresses right up until the end, when V=dM/dx is substituted. I've used this to help explain this to other engineers in the past when they just couldn't grasp the V*Q/(I*t) method. Then they see the physical meaning of Q.

I have read through and it sounds like the scheme is dead in the water in terms of lifting a load. So we're just left with the theoretical shear flow argument.

I concur with CANPRO 's slenderness calculation. kl/r>200 even though compressive loads are low it is a show stopper in my opinion.

The more I read here the more concerned I get. JP20 if this is for construction lifts I would toss the idea of building up a section from available materials as being technically infeasible and go rent a proper lift beam from someone like LGH. If you are insistent on fabricating something in the field toss this C15 idea and design a proper plate girder where design methods are straight forward and well documented.

Rigging failures are all to common in construction. Loads are hard to estimate when you consider dynamic effects as things move, high wind blowing your load around, equipment jerking due to some mechanical problem, outriggers settling, etc. etc. This is no place for a design that just meets code calculated strength.

With that being said, what do you suggest to make the beam fully-braced?

ldeem, what did you use for "K" and "A" in your calculations?

Are you suggesting that getting kl/r < 200 will fix the buckling issue?

JP20 k=1, I used my built up section calculator to find ry, length 40'

I am saying kl/r means you need to change direction. In my opinion the plate and C15 configuration is hard to analyze (as you can see by the length of this thread) so change your direction to a configuration that you are comfortable analyzing.

Quote (steveh49)

Different people think and understand differently

I agree 100% and I love to see someone approach a problem differently than I do so I can learn. However, to suggest that you can take the moment value at any given location and turn it into a longitudinal shear is fundamentally wrong - there is no other way to look at it. I've mentioned this example 3 times now in this thread - a simply supported beam with a uniform load has its maximum shear flow demand at the supports where there is zero bending stress.

steveh49, I get that you understand this and that you're suggesting to use the bending stress to help rationalize shear flow, and from this thread it is apparent that a lot of engineers take that route. But clearly that is not effective because some of engineers responding to this thread have a fundamental misunderstanding of shear flow.

Quote (JP20)

With that being said, what do you suggest to make the beam fully-braced?

You don't make the beam fully braced. You design for the unbraced length to ensure it has the proper capacity. Here are some key items you need to review:

• Buckling of compression members
• Unbraced strength of bending members
• Lateral torsional buckling
• members subject to axial and bending stress (beam columns)

I think we've gotten quite caught up in semantics around shear flow here. Trying to put that aside, and in light of the OP's beam dimensions now being clear, I will provide some input on the design of lifting devices such as this one. I have designed several and analyzed them for more load cases than I can count. Generally, here is how I have approached these multi-hole lifting beams:
- It is often easiest to have a central plate provide the vertical shear capacity, and have it protrude above and below the lateral bracing portions of the cross-section so that you can drill holes and use those portions as lugs. That is perfectly vanilla.
- These devices are susceptible to both local and global buckling. The protruding, unbraced webs often limit the allowable stress due to their tendency to buckle locally. Because your top slings are not vertical, your lifting beam is exposed to combined compression and flexure, making it susceptible to global buckling (KL/r stuff). There's a double whammy with the flexure as those angled top slings both impose compression and additional flexure due to their eccentricity relative to the beam neutral axis.
- Due to the susceptibility to global buckling, I have always designed these devices to be quite laterally stable. Use closed circular or rectangular sections to prevent LTB sensitivity. Make these portions significant so that the beam is stable, and you can also utilize them for additional flexural strength. Your central plate should not constitute the majority of your Ix. A typical cross-section (with lug holes shown) is shown below:

- I take the weak axis design quite seriously. There's potential for a suspended load to shift or induce slight lateral loads on the beam, which we know is susceptible to global buckling. I have generally designed the weak axis stiffening elements to be able to yield without splice connections or welds failing.

OP, if you haven't written us off here, you might want to reconsider your choice of lateral stiffening, and choose something that maximizes the geometry of that stiffening (such as shown in the image above).

Craig_H, I think you're giving JP20 excellent advice on the design procedure.

Quote (Craig_H)

I think we've gotten quite caught up in semantics around shear flow here

I honestly feel like I'm stuck in the twilight zone right now. The debate about shear flow was far from semantics. There was a serious misunderstanding going on here, and people were giving JP20 advice that was fundamentally incorrect. I can't bring myself to re-hash this any more, its all in the thread above.

CANPRO, perhaps semantics was the wrong way to say that. I think it was a very worthwhile discussion, but perhaps would be best served by a dedicated thread. I feel a bit bad continuing to discuss the intricate details of the shear flow, but then again, it's essentially what the OP asked. On that train of thought, heck why don't I close out my own curiosity:
- I agree with you regarding the "effective cross section". The flanges of the channel are not going to transfer vertical shear into the channel webs, at least not much of it. The effective cross-section essentially looks like a WF section with protruding webs. I am including your sketch below for everyone's ease of access. In that case, and using your linked shear flow document, the shear flow would be given as shown below:






- Following that, the approach that I suggested early on of resolving the flexural moment in the channel into tension and compression loads that are carried by the weld seems invalid. I'll admit that deep in my gut I still feel that there's efficacy to that approach. I back-derived the approach, and it yields very near similar results. My approach almost matches the generic shear flow equation, but includes a term to account for the flexural stiffness of the flange which I believe is ignored by the general approach to shear flow. Bottom line, I think we agree, and would arrive at similar results through different approaches. Your approach is definitely more textbook.

Thanks. Let me throw this out there... thinking about changing the section to the following so the compression will, without a doubt, be transferred to a wide flange:

Replace the channels with stiffer angles might help. Also, why not use standard WT section!?

Craig_H, I agree with your last post. I understand your gut feeling on this, shear flow comes from the change in bending stress so it isn't a stretch to associate the presence of bending stress with shear flow. On the surface, it is very intuitive to follow the approach that was incorrectly presented in this thread (using bending moment to get shear flow).

I think what you were alluding to in your statement about flange stiffness (and I think retired13 was trying to make this point earlier) is that if you look at the weld as part of the cross-section, it does indeed see stress from bending. But with shear flow, we're looking at the weld stress developed from two components trying to move relative to each other - this is the critical stress when sizing the weld, this is the load that will cause a failure plane within the weld itself (traditional weld design, weld shear at effective throat).

JP20, the channels can't easily be positioned that way - you're going to have a decent size fillet weld between your two 1.25" plates and the channel won't fit there unless you notch the top flange. If you really need to make this work with the material you have on hand, I believe this is your best shot - turn your channels into flanges and use the 1.25" plate as the web. The channels give your compression flange great Iy properties. And I think this gets us back to a shear flow condition that we can all agree on. Please review the list of topics from my previous post - I feel as though you're still missing key concepts required to make this a safe design.

Quote (JP20)

My username says (Structural) but really I’m construction (took my PE exam in construction). And work for construction company. Maybe if I would have took the Structural exam I would know but I didn’t.

Don't feel badly, this is actually a very complex problem for most structural engineering specialists. The quite effective, yet difficult to predict nature of the LTB rotational restraint at the supports and load application points of these types of beams makes evaluation a difficult challenge. That said, it's not as though you're the first engineer to attempt this. Below are the papers that I have on the subject. Member WArose turned me on to the first and it's quite good.

Quote (JP20)

Thanks. Let me throw this out there... thinking about changing the section to the following so the compression will, without a doubt, be transferred to a wide flange:

I would give some serious consideration to something like Craig's recommendation, repeated below. Not that it necessarily has to be ac circular section etc.

While it may not optimize your material costs to the nuts, one very simple way to shed the complexity involved with this stuff is to design yourself a cross section that would work for both strength and stiffness even if it did roll over in lateral torsional buckling. Flip the thing sideways and make it work that way as a beam column.
Obviously, that makes for a section with a lot of weak axis meat to it. But, then, you've got a fairly simple and reliable solution that you can execute without too much fuss.

One problem with your current section is that, with your three point support condition, sometimes the compression flange will be the bottom flange. In that case, you'll have your edge stiffening on the wrong side.

KootK, at the risk of being labelled difficult, I'm going to be disagreeable in this thread again. I know you can handle some disagreement and healthy debate, so here we go:

Both of the references you listed are for I-beams, and every reference I've seen for LTB buckling checks are for an I-section or at least something with a distinct compression flange. With the section proposed by Craig and re-posted by you, what would you consider to be effective compression flange? A portion of the round member near the top? I think it is important to have a compression flange with some healthy Iy properties near to the extreme compression fiber and I don't think you get that here - and I believe this is the basis of the references you've presented. If you took a standard I section, kept the same overall depth, and moved the compression flange down 20% of the beam depth you surely wouldn't get the same LTB resistance out of the flange. And to me this makes sense - any time LTB comes up here you drive home the point that the beam has to deflect sideways and rotate about a point in space. If you lower your compression flange, it gets closer to the point of rotation and becomes less effective in resisting this rotation. This is why I proposed using the channels as an I-section.

Quote (CANPRO)

Both of the references you listed are for I-beams

Yup, but I would argue that:

1) The general principles still largely apply and are worth knowing about and;

2) Depending on the direction OP takes, treating his real cross section as an equivalent wide flange might prove attractive.

Quote (CANPRO)

With the section proposed by Craig and re-posted by you, what would you consider to be effective compression flange? A portion of the round member near the top?

That's what I had in mind, yes.

Quote (CANPRO)

I think it is important to have a compression flange with some healthy Iy properties near to the extreme compression fiber and I don't think you get that here .

The approach that I recommended would treat the flat plate as entirely sacrificial except for local transfer of loads in and out of the system. As such, I've little concern for the Iy properties near the extreme compression fiber.

That said, I voiced a similar concern in OP's previous thread on this topic.

Quote (KootK)

..your critical failure mode is likely to be out of plane plate buckling of the portions of the flat plate extended above and below the channels.

If the design assumes that the plate is to be fully utilized in flexure, I imagine that one would check this failure mode via b/t ratios or something similar. With plate extensions 1.25" thick and 8"+/- long, I don't see this being much of a problem.

I agree the latest solution by CANPRO is more functional/desirable, but then why not use a standard WF, or I, as suggested way back to the beginning of this thread!?

By the way, johns20188 has opened a thread devoted to the discussion of designing this type of built-up members. For interested parties, please jump in and provide your opinions, and not to side track the JP20's attention.

My thoughts from the file uploaded earlier

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

Getting back to the original question, at ultimate load, each channel is stressed to Fy, compression tension on top, tension compression on the bottom. The total force in the top and bottom of the channel is Fy*A/2. That has to be developed by a weld of length L/2. If we assume uniform distribution from end to centre, the weld must carry a force per unit length of Fy*A/L where A is the area of channel and L is the span of the built-up section.

BA

Quote (BAretired)

Getting back to the original question, at ultimate load, each channel is stressed to Fy, compression tension on top, tension compression on the bottom. The total force in the top and bottom of the channel is Fy*A/2. That has to be developed by a weld of length L/2. If we assume uniform distribution from end to centre, the weld must carry a force per unit length of Fy*A/L where A is the area of channel and L is the span of the built-up section.

I disagree with a couple parts of this statement. First, if you look at the proportions of this beam, there is no chance that the channel flanges get to see Fy before this fails - the beam is far too flimsy to reach yield before buckling.

Further, it has been thoroughly discussed here that the development of bending stress in the channel flanges does not come from the welds - centroids of all sections align, Q = 0, no shear flow in the welds. Any bending stress developed in the flanges of these channels would come from shear flow between channel web and flange.

Kootk, I get what you were suggesting now - the bit of plate above the circular section is essentially just a continuous lug. I'd be comfortable using that bit of plate for checking the cross-section strength (healthy b/t ratio as you suggest), but for LTB checks I would assume that bit of plate doesn't exist and that top of beam = top of circular section.

BA's method is the practical method of design, the stresses create a couple on the channels, and force the channels to separate from the center plate, thus require weld to glue the pieces together.

Quote (CANPRO)

I disagree with a couple parts of this statement. First, if you look at the proportions of this beam, there is no chance that the channel flanges get to see Fy before this fails - the beam is far too flimsy to reach yield before buckling.

Who cares? I don't know how the size of the channel was determined, but the capacity of a section in flexure, built up or otherwise, is developed when everything above and below the N.A. is stressed to Fy, one in compression and one in tension. In order to get the force into the channel, the only connection you've got is the weld to the plate at the top and bottom of the channel.

How long is the beam? Do we know? Check it out. The OP needs answers quickly and you can't do any better than weld to fully develop the channels in bending.

BA

The beam is 40' long, with 3 slings on top and multiple lift loads below. The channel wasn't sized, but an attempt to prevent buckling.

Well, if the channel size is C380x50 (C15x33.9) and the span is 20'-0" or say 6000mm, the weld must be good for 6430*300/6000 = 320N/mm which would require an intermittent weld of 6mm (not worth getting excited about).

What may be worth getting excited about, however, is the question of buckling. This needs to be properly checked to determine whether or not the proposed section is adequate for the rated capacity of the lifting beam.

BA

Quote (BAretired)

Who cares? I don't know how the size of the channel was determined, but the capacity of a section in flexure, built up or otherwise, is developed when everything above and below the N.A. is stressed to Fy, one in compression and one in tension.

Presumably, whoever is designing the beam would care if the failure mode is by section yield or buckling.

Quote (BAretired)

How long is the beam? Do we know? Check it out.

This thread is up to 90 replies now. And it is a continuation of another thread. I'm sorry if I'm being a bit cranky right now, but if you read through the thread you would have seen the section sizes and beam length, and you would have seen the long discussion regarding the load in the welds. BA, I have a lot of respect for you and I've learned a lot from you over the years, I mean this with all due respect.

Quote (BAretired)

The total force in the top and bottom of the channel is Fy*A/2. That has to be developed by a weld of length L/2

Quote (retired13)

BA's method is the practical method of design, the stresses create a couple on the channels, and force the channels to separate from the center plate, thus require weld to glue the pieces together

I can't believe I'm still here saying this. The bending stress in the flange of the channel does not come from the weld. This is fundamentally incorrect. The centroids of all components align with the neutral axis of the built-up section, there is no shear flow in the weld. There is no tendency for the interface between the channel and the plate to slip relative to each other in the longitudinal direction. Therefore the weld cannot possibly be responsible for building up bending stress in the channel flange. All of these other methods might sound rational but they are just work-arounds for doing it proper - and the proper way isn't hard! Its a college level shear flow problem. If OP was to build this beam as proposed, there would be some stress in the welds (as has been discussed here and in Johns20188's new thread) but it isn't shear flow.

CANPRO and BA,

You are invited to the thread opened by johns20188, so more people can join the discussion without browse through and read the lengthy responses.

Three slings creates a problem of indeterminacy which may be okay, but I prefer a determinate system. Two slings, I believe would be better. If the beam is forty feet long, the pickup points would not be at the ends, but say a' from each end leaving a span of b' in the middle. Then 2a+b = 40, so if a = 8, b = 24'. I don't know, but I think that's about right.

Cantilever moment = 8^2/2 = 32w; Span positive moment = 24^2/8 -32 = 40w

Then check for buckling and modify a and b as required. If that doesn't work, beef up the section using whatever material is available on site.

BA

I decided to just use a W16x77 with lugs welded on top and bottom flanges in line with the webs. Kl/r for weak axis is 196. However this is a lot of determining factors in BTH manual that I still have to check. I did check welds earlier today and they are good being 8” long on both sides of lug - 1/2” fillet.

This beam is going to be rated for 25 Tons of lifting.

The part I hate most is that no matter what you do to this type of lifting device, you cannot laterally brace it. If someone knows of a way to, let me know! Otherwise I’ll be using BTH manual to design as 40’ unbraced.

I'm with CANPRO et al on:

1) The shear flow theory that applies (or doesn't apply) here and;

2) the value of a correct theoretical understanding even if a gross, overkill approach would do.

I'll tell my story the the long way let folks pick and choose which bits they might light to challenge.

THE TWO COMPONENTS OF COMPOSITE FLEXURAL FASTENING

3) All components of the cross section must be induced, by the fastening, to assume the same curvature along their independent axes. This is usually the low demand aspect of fastening demand.

4) Where the joining of the components would increase the moment of inertia of the assembly above the sum of the separate components, the fastening must also axially extend and axially compress the component parts uniformly. This is usually the high demand part of the fastening demand as outright elongating or shortening a piece of steel really takes some doing.

THE SPECIAL CASE OF ALL COMPONENT NEUTRAL AXES BEING COINCIDENT

When this is the case, #4 is unnecessary and #3 is all that applies with respect to the fastening. This should make intuitive sense because, really, if your moment of inertia isn't going to get pushed any higher than SUM(Ix_components), why should you have to pump a ton of capacity into making that happen? You shouldn't, of course.

HOW TO FASTEN WHEN THE COMPONENT CENTROIDS ARE COINCIDENT?

It doesn't take much really. And there are two options. In step by step fashion:

1) Based on the ratio Ix_component / Ix_assembly, figure out how much load goes into each competent part. This will create a condition wherein all component parts have the same centroidal curvatures along their lengths and share a common cross section strain profile (this last bit is really what the crux of composite flexural action is).

2) Having determined the load on each component, the shear and moment diagrams for those components become determinate. They are what they are and they need to be internally consistent.

3) The problem now becomes one of imposing the requisite shear and moment diagrams from #2 on the component being fastened given that it may be that no external load is applied to the component. There are two fundamental approaches:

a) Transfer increments of shear from the assembly to the component to replicate the shear diagram that the component should have. This will result in the desired moment diagram falling into place. In this case, the welds would be designed for vertical loads only of a magnitude equal to the distributed load going to the component divided by two, applied top and bottom. This is often how we do it in wood and the fastening demand is very light except, sometimes, at reaction points, cutoffs etc where large amounts of concentrated load must be moved in or out of the component locally.

b) Transfer increments of moment from the assembly to the component to component to replicate the moment diagram that the component should have. This will result in the desired shear diagram falling into place. In this case, the top and bottom weld lines would be designed for horizontal loads only of a magnitude equal to [V/h], varying along the length of the beam [V = shear; h = distance between weld lines]. This is the method that I would favor for this application given that the path for vertical shear transfer is made a bit flexible as a result of it needing to pass through the channel flanges and induce transverse bending there.

One or the other of 3a or 3b will usually make more sense than the other based on the connection style etc. Where it's not obvious, I'll design for the worst of both cases. It's usually a pittance either way.

One could certainly be forgiven for getting the impression that #3a and #3b are both versions of "shear flow" (especially #3b). Both methods do move load around via weld shear after all. For me, the difference is that neither method attempts to impose a uniform axial strain on a component of the section which I take to be the thing that defines a classic "shear flow" problem. I suppose that another way to look at it is to say that the coincident centroids case is really just a special case of the shear flow problem in which it is possible, but not necessary, to transfer load into the component members via horizontal shear.

Quote (JP20)

The part I hate most is that no matter what you do to this type of lifting device, you cannot laterally brace it. If someone knows of a way to, let me know!

Whatever would you be bracing the beam to? It will be floating around arbitrarily in space. At best, you can "brace" the slender parts of your compressed elements to attached, less slender parts. A better term for that is probably stiffening rather than bracing though.

Quote (CANPRO)

This thread is up to 90 replies now. And it is a continuation of another thread. I'm sorry if I'm being a bit cranky right now, but if you read through the thread you would have seen the section sizes and beam length, and you would have seen the long discussion regarding the load in the welds. BA, I have a lot of respect for you and I've learned a lot from you over the years, I mean this with all due respect.

I apologize for joining the discussion so late. It is true, I didn't read through the thread and again, I apologize for that. Thank you for the compliment and I can say that I have a lot of respect for your contributions to this forum.

EDIT: Please ignore all the red notations. CANPRO is absolutely correct.

Quote (canpro black, ba red)

I can't believe I'm still here saying this. Neither can I.

The bending stress in the flange of the channel does not come from the weld.
It really does. Without the weld, the channel would feel no bending stress while the plate would elongate or shorten at the elevation of the top and bottom of the channel.

This is fundamentally incorrect. The centroids of all components align with the neutral axis of the built-up section, there is no shear flow in the weld. There is no tendency for the interface between the channel and the plate to slip relative to each other in the longitudinal direction. Therefore the weld cannot possibly be responsible for building up bending stress in the channel flange. There is no other way to get flexural stress into the channels.

Quote (CANPRO)

All of these other methods might sound rational but they are just work-arounds for doing it proper - and the proper way isn't hard! Its a college level shear flow problem. If OP was to build this beam as proposed, there would be some stress in the welds (as has been discussed here and in Johns20188's new thread) but it isn't shear flow.

I believe it is, but I think we are both repeating ourselves.

CANPRO is correct. Please ignore my comments in red.


BA

Quote (JP20)

I decided to just use a W16x77 with lugs welded on top and bottom flanges in line with the webs. Kl/r for weak axis is 196. However this is a lot of determining factors in BTH manual that I still have to check. I did check welds earlier today and they are good being 8” long on both sides of lug - 1/2” fillet.

This beam is going to be rated for 25 Tons of lifting.

The part I hate most is that no matter what you do to this type of lifting device, you cannot laterally brace it. If someone knows of a way to, let me know! Otherwise I’ll be using BTH manual to design as 40’ unbraced.

Using a WF is definitely a good idea.
I believe you should use only two slings, not three.
I believe you should pick up substantially closer than 40' (possibly 8', 24', 8').
It's not possible to brace it, but the beam must be capable of handling the compression for an unsupported length of 24' (or whatever central span you decide upon).

BA

BA, well then with no hard feelings, let the debate continue. Although this might be better served for the other thread. But to quickly answer back to your statement "It really does. Without the weld, the channel would feel no bending stress" - the bending stress in the channel flange would result from shear flow between the web of the channel and the flange. The channel and plate could be joined at the neutral axis only and you'd still get the same bending stress in the channel - as the plate deforms, the channel is forced to deform with it and the load is shared in proportion to the bending stiffness, and bending stress still develops in the channel flange due to shear flow between the web and flange.

JP20, I like BA's idea of moving the lift points in - also reduces our hook height requirement on the crane. Although, your unbraced length in bending could still be 40' depending on load placement (entire beam could be in negative bending).

Quote (CANPRO)

BA, well then with no hard feelings, let the debate continue.

Certainly there are no hard feelings. Why would there be? And certainly, the debate should continue when time permits.

EDIT: There is no need...I concede (sigh) that I was wrong and that you were correct.

Regarding your other point, I assumed that load placement would be more or less uniform along the bottom of the beam. If it is possible to place large loads on the cantilever and little or nothing on the span, the geometry of the lift points might change. Or perhaps there could be alternate snatch points when the load is placed near the extremities.

BA

BA, you should read the other thread started by Johns20188.

I feel sorry for your heavily edited posts - turns out I was wrong. There is shear flow in the welds if it forms a multi-cell closed structure. Others were right here, but for the wrong reasons - but I think that highlights that validity of your original method. It wasn't technically correct, but through "brute force" you ensured there was a valid load path. I didn't necessarily disagree with this approach, I believe it will almost always produce a conservative design - I was just disagreeing with the technical accuracy of the process.

CANPRO,
I haven't read the other thread by Johns20188. I will try to get around to it in due course.

The reason I conceded was that, after thinking about it for a few moments (something I should have done earlier), I reasoned that, even if you bolted the channels to the plate with bolts adequate to transfer the channels share of the applied load, based on stiffness, you wouldn't need a weld on the flange tips (you may need it for weak axis bending or prevention of buckling, but not for strong axis bending).

Maybe there are still aspects of the problem which I am missing, but my synapses were definitely misfiring when I responded to your earlier objections.

BA

BA, you're going to enjoy this other thread. The bending stress in the section would be the same with or without the weld. The weld is not required for the section to act compositely (all components have common N.A.) - however, in this case, since we create two closed cells in a single section, you develop shear flow in the weld. It doesn't change the bending stress, but the shear flow does exist - and at that point the simple VQ/IT method fails us. The part that really caused confusion is the multi-cell cross-section. If this was (2) channels turned toe-to-toe and and welded to form a box section, there would be no shear flow in the weld.

BA,

I think we are in line on that shear force is taken by the center plate only, the channels are only the riders. But, the center plate will bend under load, and force the channels to have the same displacement and deflection due to stress and strain compatibility, and weld in the connections is required to make the parts stick together to be compatible.

Quote (JP20)

I decided to just use a W16x77 with lugs welded on top and bottom flanges in line with the webs. Kl/r for weak axis is 196. However this is a lot of determining factors in BTH manual that I still have to check. I did check welds earlier today and they are good being 8” long on both sides of lug - 1/2” fillet.

The choice of WF section will depend on load and span. You will probably need to comply with OSHA requirements for load factors. I am not familiar with them offhand, but I suspect they are higher than normal load factors for beam design for a building.

Quote (JP20)

This beam is going to be rated for 25 Tons of lifting.

If the span is 40' and the load is 25 tons or 50 kips, the maximum moment is 50*40/8 = 250ft-k.
But you told us something different in an earlier post:

Quote (JP20)

The max moment is actually more near 80ft-k .

Quote (JP20)

The part I hate most is that no matter what you do to this type of lifting device, you cannot laterally brace it. If someone knows of a way to, let me know! Otherwise I’ll be using BTH manual to design as 40’ unbraced.

If you use a span of 40', you will require a pretty hefty beam. Why don't you consider reducing the span by having a cantilever at each end? Not only do you reduce the span, but you apply a negative moment at each end which directly reduces the positive moment in the central span.

In a way, you are laterally bracing the beam at the pickup points by applying the load below the beam, preventing rotation about the beam axis. You could brace the top flange between pickup points with a top channel or plate, but it is probably more economical to select a WF which can span the central span without additional bracing.

The cables, chains or slings make an angle with the beam. They put moment and compression into the central span which needs to be taken into account.

Finally, if you feel uncertain about the design, I would strongly recommend that you retain an engineer familiar with this type of beam.

BA

CANPRO,

I still have not read the other thread, but I assure you, I will.

retired13,

We have been all around the mulberry bush on that topic. It is no longer relevant to the OP as he has decided on a different shape for a beam. That does not mean we should not discuss it, but I for one am not up for that discussion at the present time. First, I would like to read the other thread.

BA

JP20,

I think you know the design criteria well, but rusty in the real design. Your 40' beam length is a problem, not only in design, but you will require ample space, and equipment that capable of handle this operation. I suggest to sort through internally to gather ideas, and refine the plan and method to carry out the task. Also, you'll be definite benefitted from engaging a structural engineer, who is familiar with the nature of your business and the task on hand. My last suggestion, for lateral stability concern, a strong box beam may work, but not for present length.

If total load plus beam weight = W
then W/2 is vertical load at each pickup point.
w = W/40
F = W/2sinA (sling force)
P_span = W/2tanA (compression in span)

M_cant = wa²/2
M_midspan = w(2b)²/8 - wa²/2 = w(b² - a²)/2

If a = 8', b = 12'
M_cant = 32w
M_midspan = 40w


If a = 10', b = 10'
M_cant = 50w
M_midspan = 0



BA

Wow! After reading through this thread, I can't believe how horribly awry the discussion got at points. In my opinion, EVERYTHING Canpro said in this thread is spot on and was really one of the only consistent voices of reason from the beginning. I applaud him for sticking with it. It's too bad that so many others muddled the conversation with really bad advice and halfhearted responses. Sadly, much of the bad advice came from several rather well respected individuals on this forum. Of course, I am coming into this conversation after the fact, so it's easy for me to criticize with the advantage of hindsight, however, I'm really not sure why it took the OP so long to realize that the original design was trash and to just go with a rolled wide flange with welded lifting lugs.

STrctPono,

The OP said the best, "I think it will work with 5/8” fillet on both channels, but I want to be super confident. These conflicting statements are concerning."

We all been struggle through the rare/weird shape the OP attempted to use. Pretty much the posts after that comment were focus on checking each others understanding of matters, along the way, I think some good points have been raised, that potentially can benefit people who has patient to read all posts through, and me included, a beneficiary of this lengthy discussion. You shall visit another thread to see the finale of these arguments. Link