Cantilever Beam with Flange to Flange Bolt & Weld + FEA

[tKc74]

Hello, working on a pair of transport beams to move a large piece of equipment on a flat deck, beams will cantilever off the ends of the trailer to pick up the loads, think Two Unequal Concentrated Loads Unsymmetrically Placed as a load profile. Working to AS/NZ 3990 standard.

Got existing custom beams from a previous move but they are not stiff nor long enough. Increasing stiffness is straight forward enough, adding a plate to the bottom and RHS sections to the top. This also gets deflection down (Δ/l<250).

For length we'll add some UB I-Beams to the bottom of the existing (see below, UB's overhang vary on each end). The UB's top flange can weld to the existing, but thinking belt and suspenders, I will also add some bolts.

1. Given this load configuration, is it reasonable to assume both weld and bolts can take/share the load given bolts are preloaded? Or will it just be the weld until it fails then the bolts kick in? Not sure I can get weld strong enough on its own as flanges are only so thick.

2. A reasonable way to evaluate the stresses in this weld? Currently evaluating as a line and have the below from my textbook in obtaining a "section modulus" for the weld (f=M/S_w [N/m]), but none are exact to how the load is in relation to the weld. Currently assuming #2 as it appears to be applying the bending along length of weld which is more similar than #3 which is about its leg (if that's right?).

(Based on the assumption above) I found a weld centroidal position and calculated the moment on the weld (as a line) from this point to where the load is applied. Sounds reasonable? Okey

4. Torsional stiffness (from accelerating/deacceleration of load) is handled by 2 chains tied back to the flack deck. I believe this should be sufficient to combat lateral torsional buckling given strong enough chains? Or are stiffeners/other means better?

5. Bolt pattern is calculated by assuming the UB would pivot about the far end of the UB (worst case), confident in this but thought I'd check.

6. I checked the bending and shear stresses at various points across the beam (where beams stack, highest moment, about ends..etc) however the FEA in 2 sections are showing much higher stress than the hand calcs (bending and shear) and not sure why. Looking to add minimal amounts of plates, web doublers, stiffeners as required. From the calcs this should be a stronger section where the 2 beams double up...
Scale set to .9σ[sub]y[/sub] AS 3990. Bending is .6σ[sub]y[/sub] and .4σ[sub]_y[/sub] for shear ((.6σ[sub]y[/sub])[sup]2[/sup]+3*(.4τ[sub]y[/sub])[sup]2[/sup])[sup].5[/sup] = .9σ[sub]y[/sub] ?

#1 In the web of the original beam where it overlaps the UB​

#2 where the edge meets the bottom flat plate (even with a gap between end of UB and plate)​

Apologies if a bit much, any guidance, comments or suggestions is much appreciated. Left out number as much as possible as just looking at getting ideas on how to improve and I'll happily do the leg work in calcs.

Cheers,

 

[desertfox]

Hi tKc74

My first comment would be considering the welding CofG shouldn't it be equally distant from either end of the weld? Your diagram shows it closer to gusset ends of the red outlined welds.
If you are stressing the welds then I am not sure why the fea is showing stress in the beam its welded to. If I were stressing those welds, I wouldn't consider the beam that its welded to, that way it would be conservative because in practice assuming the weld is strong enough some of the load would transfer into the above beam.
The two gusset plates you have welded on I would replace with one central on the beam aligning it with the vertical web of the Ibeam so the load path is through the web and not transferred through the Ibeam flanges.
The addition of bolts? not sure about those because what are you going to preload them to and what external load is actually on the extension piece?
For the bolts to work I would ignore the weld and assume I had to add the extension by bolting alone and see how many bolts,size I required, yes the bolts could work in conjunction with the weld however determining how much the welds or bolts might take is another question.

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

 

[tKc74]

Hi desertfox.

The weld CoG is offset due to the gusset welds, the CoG is the centroid of the UB flange to flange and gusset welds.

The FEA is not modeled with welds, its modeled with the UB bonded to the underside of the existing beam, I figured bonded was sufficient if we went with both Welds and Bolts.
I have also modedled it with just bonding to the edge of the UB where the welds are and with the bolted connections and received similar results. The stress is being taken up by just the upper web it appears?

Good call on the gusset, might add that one and keep the other two as well.

The bolts I would figure to pre-load to 75% yield (Gr. 8.8 ~M20's currently). I can assume no weld when sizing the bolts and just add as many as required, however if I went the other way and assumed no bolts when specing the weld I don't believe I can get it there on its own and that's why I wanted to confirm the above approximation in weld section modulus (along with everything else).

Cheers,

 

[desertfox]

Hi tKc74

I would ignore the gusset welds and just stress the other welds outlined in red, if you are just stressing the welds the horizontal beam above the lower one shouldn’t be included in the calculation in my opinion because that means lower and upper beams are essentially one piece and that also assumes the weld is strong enough or isn’t present as you have stated, so can you post the hand calculation for just the welds? What loads are placed on this device, if you can provide some dimensions for the lengths of beams and weld size I can do a rough hand calculation.

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

 

[HS_PA_EIT]

Just some quick thoughts:
1.) accounting for combined strength of weld and bolts is difficult- figuring how much stress goes where/strain compatibility. In the AISC 360-2015 spec, it's not allowed except in the design of shear connections on a common faying surface with strain compatibility considered.

2.) The arrangement is creating a fairly complex load path, hence some of the confusion with including the gusset welds etc.

3.) Lateral torsional buckling is complex for cantilevered beams. If it is a problem, it will be worse with these open profiles, maybe consider using closed sections/box beam.

4.) Pay very close attention to the bonded condition between parts in the model and simplify where you can.

5.) Consider diagonal stiffener plates in the high stress panel zone.

6.) keep weld access in mind.

I hope these might help a bit.

 

[WARose]

1. Given this load configuration, is it reasonable to assume both weld and bolts can take/share the load given bolts are preloaded? Or will it just be the weld until it fails then the bolts kick in? Not sure I can get weld strong enough on its own as flanges are only so thick.

You never assume that....you assume which ever one is stiffer will take all the load. (Typically the weld.) If you need the bolts for erection purposes....you could design each to take 100% of the load.

2. A reasonable way to evaluate the stresses in this weld? Currently evaluating as a line and have the below from my textbook in obtaining a "section modulus" for the weld (f=M/S_w [N/m]), but none are exact to how the load is in relation to the weld. Currently assuming #2 as it appears to be applying the bending along length of weld which is more similar than #3 which is about its leg (if that's right?).

You need to get Blodgett's book on weldments. It can guide you through stuff like this. I'm not certain as to what is welded and what is not so I cannot advise (at this point) what your [weld] section modulus is.

4. Torsional stiffness (from accelerating/deacceleration of load) is handled by 2 chains tied back to the flack deck. I believe this should be sufficient to combat lateral torsional buckling given strong enough chains? Or are stiffeners/other means better?

I'm not 100% sure I follow the situation.....but if you are saying you are operating under the assumption that the (suspended) chains are providing LTB restraint for any beams.....that assumption needs to be investigated. There are a number of resources that can do that. See this thread for a few:

https://www.eng-tips.com/viewthread.cfm?qid=446703

6. I checked the bending and shear stresses at various points across the beam (where beams stack, highest moment, about ends..etc) however the FEA in 2 sections are showing much higher stress than the hand calcs (bending and shear) and not sure why. Looking to add minimal amounts of plates, web doublers, stiffeners as required.

You sure some of that isn't due to stress concentrations? When designing steel by FEA plate models.....that happens. You can design steel on paper and have it pass via code.....then model it with FEA and get high stresses that don't pass. If there is a way to do it by hand....do it. Otherwise you will wind up possibly overdesigning things or chasing your tail on concentrations.

 

[tKc74]

@desertfox: So your saying as soon as the lower beam reaches the upper one you can ignore the lower beam and just consider bending about the upper (i.e, don't combined the two section modulus, just use the upper?) If so, I'm not sure I follow that justification in not using the stacked modulus. I have attached a sketch with a load.
Upper beam is a custom unit, Section Modulus = 4517511mm^3
Lower beam is a UB610x101, Section Modulus = 2515269mm^3
Combined, Section Modulus = 5723368mm^3 (using weaker bottom, top is 1.202*10[sup]7[/sup]mm^3)

Fillet welds are 12mm, both sides of the flange. Currently I am using #2 in the third photo of my original post to get a section modulus then calculate load on the line (f=M/S_w [N/m]) per unit length, again unsure if geometry applies to my scenario but its the closest I have found, really hoping someone can answer that, or know a directly applicable section modulus formula for this load configuration.
M=rxF, F=350kN, r is the distance of overhang + distance to centroid of weld (1365mm + 285.6mm (including gussets)

@HS_PA_EIT (:
[ol 1]
[li]Figured as much, I should design each to handle 100% load then?[/li]
[li]How does the load path get more complicated with the gussets? I figured it was creating a smoother path for the load?[/li]
[li]I've considered a box section. Can I add plate from outside edge to outside edge of the flange? Or is it better to be between the flanges?[/li]

[/ol]

5. I have used diagonal stiffeners and it doesn't make much of a difference.​

@WARose (:
[ul]
[li]Similar to HS_PA_EIT, ill try to design each for 100%, tho this will be difficult with the weld, but could be achievable with additional gusset.[/li]
[li]Had a look thru a version of Blodgett's, but could find anything to match my geometry, if you're unclear of the geometry see the first photo in this post.[/li]
[li]Ill have a look at this when I can get into its details[/li]
[li]That's what I thought, but even when going with 4 elements across the thinnest web, I am still getting the same high stress's, I would like to know WHY the FEA is wrong, not just assume it is[/li]
[/ul]

 

[WARose]

Had a look thru a version of Blodgett's, but could find anything to match my geometry, if you're unclear of the geometry see the first photo in this post.

If I follow that right (i.e. it is a continuous weld from a beam to that plate), the right [weld] section modulus for the weld is #2 in your OP (i.e. d[sup]2[/sup]/3). And the moment arm would be the cantilever + half the length of the weld. (I came out with 1897mm.) Considering just the 350 kN load, in bending only, I came out with the weld forces being 1.75 kN/mm. (Or 9.92 kips/inch.)

By the way, your pic shows some kind of sandwich plate between the upper and lower beams....but the elevation does not.

That's what I thought, but even when going with 4 elements across the thinnest web, I am still getting the same high stress's, I would like to know WHY the FEA is wrong, not just assume it is

It's hard for me to say more than I have (based on what I see in this thread).

 

[Agent666]

You should share what you did by hand, you say FEA disagrees with hand analysis. Based on what's presented we don't know the other side of the story here.

Keep it simple, to me all support load from the bottom cantilever tries to go through gusset, so place central gusset and design/check as such.

Fundamentally you have a cantilever cantilevering under another cantilever. Draw the moment and shear diagram on the assumption of supports transferring loads at the ends of the overlap in each member, this is is a good starting point to understand how the loads want to transfer.

 

[desertfox]

Hi tKc74

I attach my calculation for the weld stress due to bending, however the stress is far to high to be safe. I will do the bolt calculation later today, however I don’t think your design will work as it stands. In my earlier post I was saying if you are just stressing the weld then ignore the upper beam.

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

 

[JLNJ]

It seems to me that a 1 meter long weld loaded as sketch by desert fox might just "unzip" and not evenly distribute out the forces as you are hoping. The member need to be very, very stiff to spread the load to full length of the weld. To my eye, it just doesn't look right.

 

[desertfox]

Hi tKc74

Had a look at the bolts but there are no positional dimensions on the sketch, so I placed the fourth row of bolts a 1M from the left hand end of the upper beam and then placed three more rows at 40mm centres going from right to left from the bolts at the 1m distance.
The position and the size of the bolts are unlikely to work, the bolt pair at the 1m position will see 120KN load due to the offset of the external load and when added to the proportion of the direct load each bolt sees then the bolt will exceed the proof load for an M20 8.8 bolt.
I cannot see how you can combine the welds and bolting to hold because you would need to know the stiffness of the welded components and of the bolted joint to understand how much load they would see in practice.
According to my calculations neither the weld or bolts can hold 100% of the 350kN, it needs a redesign of the proposed modification in my opinion.

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

 

[WARose]

It seems to me that a 1 meter long weld loaded as sketch by desert fox might just "unzip" and not evenly distribute out the forces as you are hoping. The member need to be very, very stiff to spread the load to full length of the weld. To my eye, it just doesn't look right.

That's a good point JLNJ. And even considering that length of weld....as per my (and desertfox's) calcs....it doesn't make it.

 

[Agent666]

I agree the calculation is only really valid if the same stiffness of the underlying system applies right around the weld on both sides of the weld. As detailed the load wants to concentrate on the webs of the beam at the tip of the top beam (i.e stiffest part). Someone earlier on referenced Blodget, he has a great list of principles for weld design, and one of them if I recall is that the load naturally wants to go to the stiffest location. It's logic that you don't need FEA to demonstrate, because it's just the way things want to work.

As I noted previously, if you draw the moment and shear diagrams of two members stacked like this with some vertical links between the beams at the ends of top and bottom beams it is fairly easy to convince yourself that most of the load is reacted at the ends.

So provide the connection there for the loads being transferred.

Keep it simple? Simple is likely cheapest.

 

[tKc74]

Hi Team, thanks for the responses, this community is amazing, much appreciated.

@desertfox, why have you used the section modulus #3 from my third photo in the OP? I'm using #2 and so did WARose [d^2/3, as opposed to d*b]

First, just focusing on the weld taking 100% load:

Lengthening the overlap of the beams is not much of an option, top beam is existing and bottom would interfere with other components.
Weld size can increase, but is limited to the thickness of the UB610, this would mean a 14mm leg max.

Original design I had 2 gussets but was mentioned earlier to ignore, need to ask why now? This would, in my opinion, increase the overall weld length AND shift the weld centroid closer to the external load creating a smaller moment on the weld. This also gives the most amount of weld where it is needed (at the gussets being tripled up, which is at the base of the overhang)

I would ignore the gusset welds and just stress the other welds outlined in red

If 3 gussets with 200mm long legs (cant make longer again due to interference with other components, plus a centered gusset would better transfer load from web to web) with 14mm fillet, I get the following.

M=rxF => (1367mm+304.5mm)*350kN = 584.3 kN*m <---------------------- Overhang distance + distance to weld centroid including gussets
S=d[sup]2[/sup]/3 => (1065mm)[sup]2[/sup]/3+3*(200mm)[sup]2[/sup]/3= 418075.0 mm[sup]2[/sup] <----------------- This is the section modulus of the flange to flange weld plus 3x 200mm long gussets.
t=14mm * 2[sup].5[/sup]= 9.9mm <------------------------------------------------------------- New weld throat thickness

σ_weld=M/(S*t) = 141.2 mPa

Given G300 steel with an Ultimate Tensile of 430 mPa, the electrode will be 430mPa, allowable stress in a fillet weld is 1/3 UT according to AS 3990 9.8.2(b)

σ_allow1/3)*430 mPa = 143.3 mPa

Gets it over the compliance line, yes?


Bolts in taking 100% load:

These are easier to change. I'm able to
[ol 1]
[li]Include more bolts[/li]
[li]Increase bolt size[/li]
[li]Increase grade (12.9 if need be)[/li]
[/ol]
Or any combination of the above. Preference is to keep to 4 bolts and increase grade, but open to comment.

I've revised my calc and have gotten the below load on the furthest bolt, is this equation correct? 4th and furthest bolt placed 1M from the right of the bottom beam, spaced 65mm apart. Four rows of bolts with 2 bolts in each row.

Agent666 I've made a bending moment diagram, but it assumes the overall beam is one continuous section. It would look like your sum diagram.

What you're proposing is interesting. Is that a flat plate protruding down from the upper beam on the back and similar from the bottom beam on the front with bolted connections to each other? What is this achieving exactly, apologies if I am unclear, but drew what I thought you were trying to show.

Thanks in advance.

 

[desertfox]

Hi tKc74

Sorry I took my Z from another ref book and read the wrong line it should have been the value you stated.
Now to answer the questions, I said to ignore the gusset welds because we only have a reference Z for two welds a certain distance apart and the gusset welds were at a different width and not connected to the flange welds. However even if you could add the gusset welds and reduce the centroid of the weld nearer to the load it won’t help you, because the shear stress in the weld will increase on the two long vertical legs ( the 1m long welds ) on the opposite side of the centroid ie 1065-304.5= 760.5, so the far left side or compressive side of the weld if you want to call it that would have a different stress compared to the tensile side of the weld where the gussets are positioned. In reality by adding the gussets you have two Z values; one for the two vertical legs plus the gussets above the centroid and Z value below the centroid. So I don’t think your weld calculation is valid and in my view it is to close to the allowable limit given that I would not be confident on the way the calculation as been done.

Regarding the bolts the 173KN you have calculated exceeds the Proof load for an 8.8 bolt so you would need either a 10.9 or a 12.9 grade bolt.
However you would need to preload these bolts to at least 1.5 * 173.5KN load you have calculated to allow for errors in torquing up bolts, in doing this though you would now exceed the proof load for both 10.9 and 12.9 grade bolts so I doubt this would work. The real limitation on the bolting might not be the bolt material grade but the yield stress of the material being clamped under the bolt head, very often you can obtain a bolt material for the preload you require but when the components are assembled the desired preload cannot be obtained because the material being clamped yields under the load of the bolt way before the bolt is torqued to its desired value.




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

 

[tKc74]

Hi,

Desertfox I completely understand what you mean, adding the gussets by simple superposition is not the way to go about it. Analogous to an unsymmetrical I-beam, you do infact have different section modulus to the left and to the right of the centroid.

The engineer inside of me needed to know what these values where however, I found this resource. Hobert - Welding Formulas and Tables.

https://files.engineering.com/getfi...file=Hobert_-_Welding_Formulas_and_Tables.pdf

On P.17 (P.10 in PDF) it gives the formulas to derive any weld shape. Then on P.19 (P.11 in PDF) under 9.3 Example 2 it gives a bit of a worked example of something similar. Following these formulas from these pages I get:

I[sub]y[/sub]=∑I[sub]y[sub]o[/sub][/sub]+∑l*x[sup]2[/sup]
I[sub]y[sub]o[/sub][/sub] = Moment of inertia of each element about its own c.g
x = distance from c.g. of weld patter to c.g of an element
l = length of element

and

S=I[sub]y[/sub]/x
x = distance from c.g. of weld pattern to desired points of weld.

(beams actually 1063mm...not 1065mm apologies)
S[sub]y_left[/sub] = 671747 mm[sup]2[/sup]
S[sub]y_left[/sub] = 1012760 mm[sup]2[/sup]

I then checked it with this program I found here on eng-tips:

https://www.eng-tips.com/viewthread.cfm?qid=443549

https://github.com/buddyd16/Structural-Engineering/tree/Python3_migration

This produced the same values as I got from my hand calcs:

So back to the stress equation:

σ[sub]left[/sub]=M/(S[sub]y_left[/sub]*t[sub]weld[/sub]) <---------------------This is the stress left of the weld centroid. Compressive
σ[sub]right[/sub]=M/(S[sub]y_right[/sub]*t[sub]weld[/sub]) <-------------------This is the stress right of the weld centroid. Tensile


I assume the load would look something like this then?

If so, for the r in M=rxF, I get as the following:

σ[sub]left[/sub]=(350 kN * (1063mm-303.659mm))/(671747 mm[sup]2[/sup]*9.9mm) = 40 mPa <-------------------Taking the r equal to the end of the weld from the centroid. See (my poor) graph in image above.
σ[sub]right[/sub]=(350 kN * (303.659+200mm))/(1012760 mm[sup]2[/sup]*9.9mm) = 26.35 mPa <-------------------Same as above, but to the other end of the weld. See (my poor) graph in image above.

Not confident on that r value (stress seems low), hoping to get some feed back tho.


Bolts:
Okay the pre-load required is high, but to what you are saying with the bolts, isn't it common to pre-load until .75-.9 yield, so for a Gr8.8 M20 Bolt that would be:
σ[sub]y[/sub]=.75*600mPa*245mm^2 = 110.3 kN (conservative at .75, but would be much more at .9)

This evenly distributed over a M20 washer (very conservative, d[sub]o[/sub]=37mm, d[sub]i[/sub]=21mm ) would impact the following stress onto the clamping material:
110.3 kN/((37mm/2)[sup]2[/sup]-(21mm/2)[sup]2[/sup])= 475.2 MPa
So this would yield/embed common G300 plate...I'm most certainly missing something here cuz this happens all the time?
Came across an article here on it, but still not convinced:

https://www.assemblymag.com/articles/93778-analyzing-bolted-joints-for-clamp-load-and-joint-stress

“If the women don't find you handsome, they should at least find you handy.” - Red Green

 

[desertfox]

Hi tKc74

just check your maths below according to the original diagrams the bending moment should be 350kN*(1367+304.6)
the figures in your last post suggest the moment from the 350KN is 350KN*(1063-303.659)see below:-

σleft=(350 kN * (1063mm-303.659mm))/(671747 mm2*9.9mm) = 40 mPa <-------------------Taking the r equal to the end of the weld from the centroid. See (my poor) graph in image above.
σright=(350 kN * (303.659+200mm))/(1012760 mm2*9.9mm) = 26.35 mPa <-------------------Same as above, but to the other end of the weld. See (my poor) graph in image above.

I think your moment is wrong which explains the low stress values.

The bending moment previously was given in an earlier post and if you use the moment below the welds are way over stressed:-

M=rxF => (1367mm+304.5mm)*350kN = 584.3 kN*m <---------------------- Overhang distance + distance to weld centroid including gussets

The bolts are commonly tightened to 0.75 or .9 of there proof stress but one should always consider the material being clamped for the reasons given in my earlier post, your bolt calculation and I agree shows an external force of 173KN which falls within the proof load of a grade 10.9 or 12.9 bolt however due to errors in torquing up the bolt and to ensure the joint doesn't open on loading you would need a preload of about 260KN so putting it outside the proof load of the 10.9 or 12.9 grade bolt.
Now at that sort of preload you may well find that the material being clamped as yielded and the joint will not be able to function as intended.


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

 

[tKc74]

Hi, yes okay the moment I think should be that original value. Was looking at it thru my load diagram I posted in my last post, 3rd photo. But I'll go with the original moment as per below:

M=rxF => (1367mm+304.5mm)*350kN = 584.3 kN*m <---------------------- Overhang distance + distance to weld centroid including gussets

However, i'm really after is a peer review of the section modulus I found in my last post and if its derivation was valid and could be used. See my last post for a clear explanation of how I arrived to these new section modulus's.

Sy_left = 671747 mm[sup]2[/sup]
Sy_right = 1012760 mm[sup]2[/sup]

Given a Sy[sub]left[/sub] (compressive) = 671747 mm[sup]2[/sup] the stress would still be under the compliance limit:

σ[sub]left[/sub]=((1367mm + 304.5mm) * 350kN)/(671747mm[sup]2[/sup] * 9.9mm) = 87.39 MPa <------------- [sub]left[/sub] is left of centroid, compressive, the right (tensile) section modulus is much higher so dont need to look at.


Bolts:

Okay plate may yield... Given that the clamped material will yield, what is the best way to avoid this while achieving the pre-load required, options include any variation or combination of below:
[ol 1]
[li]Include more bolts[/li]
[li]Increase bolt size[/li]
[li]Increase grade (up to 12.9 if need be)[/li]
[/ol]

Surely some combination of 1,2 or 3 can be used to take 100% load if need be?

Cheers,

“If the women don't find you handsome, they should at least find you handy.” - Red Green

 

[desertfox]

Hi tKc74

You can increase the number of bolts and or the size however I would recommend a n increase in bolts rather than size.
I will look closely at the section modulus derivation and post later, normally the section modus a s units of m^3 and not m^2 as in the posting but I think that is not accounting for the throat area of the weld, so you might be correct.

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