In order to determine the required bending moment at the rods, you need to consider the entire assembly as a single beam ~16ft long and calculate the bending moment at the threaded rods and at each of their connections from the dead weight of the assemblies. The bending moment in each rod will be the beam moment divided by the number of rods. Since the rods have no connection to each other, they will not behave as a composite section rather individual fixed beams, assuming the both ends of the rod are rigidly fixed. However I wouldn't expect that connection to provide any sort of bending rigidity. Your high stresses are probably correct.

Hi Mark

Are you pivoting off the bottom edge of the lower assembly till it gets vertical? I haven’t done any calcs yet but it doesn’t look like a good idea to lift in the manner you suggest. It would be better to lift using two lifting points say one in the upper assembly and one in the lower assembly, the way you have it at present I would be concerned about the screwed rods failing.

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

MarkJWilson

I became restless and took myself through the pain of calculation. Here is my calculation. Yes, you are correct if you are getting extraordinary stress if you have calculated considering each rod separately. I added crude calculation considering 4 rods together and finding moment of inertia of all rods.




Some points-
-At 0deg, entire beam will be supported and we would not see any of the stress calculated above.
-As we go on lifting till 90deg, we will see range of stress due to change in θ.
-If we assume all rods together support the bending moment then calculation shows lift can be safely done. (Since the 40mm thick plate will apply the BM on all 4 rods simultaneously)
-If all rods will not be supporting collectively (which I doubt), then you have problem which can not be solved by 2 point lifting. May be bigger rods or more lifting points need to be considered.
-Considering the offshore lift will involve dynamic amplification factor (DAF), even if we consider all rods taking the BM together may result in high stress which I have not included since I do not know the DAF. Check that yourself. If that is the case then you may need to resize the rods or provide more lift points/supports to avoid unsafe lift.
-You may need to check the rod thread shear or thread integrity separately.

I hope my calculations are correct and the pain is worth the gain.

does your text agree with your table ?

at 0 deg in the table, the beam is horizontal and max bending.
at 90 degrees, the beam is vertical and in tension

another day in paradise, or is paradise one day closer ?

rb1957

What I mean-
-At 0deg the entire beam will be supported by ground/floor on which it will be resting. Hence I said, not see any of the stress calculated for 0deg. But once we start lift, say 1deg, we can see stress value close to the value calculated for 0deg.

At 90deg, the beam will not have bending moment hence bending stress will be 0. Yes it will be in tension, if its lifted from the ground with both ends in air and no tension otherwise.

MarkJWilson (Mechanical),

A simple hand calculation ( below find ) shows the stresses are not at reasonable level if lifted at from left and and pivoted on right end.

The assumptions are ,
- the wt of four dia mm rods neglected,
- the rods behavior is separate bending elements and fixed at both ends.

If you start to lift from left end pivot around right end,

The lifting force will be max when the rod assembly almost parallel ( at initial ) and the necessary lifting force to start pivoting ;

- Pmax*4953=250*1250+150*4168 ⇒ Pmax= 189 kg ⇒ 1850 N

Max moment at the rods will develop at right end of the rods so ,

Mmax = 1850*2453-150*9.98*1668=2083588 N-mm

Section modulus of single rod Z=Pi*16^3/32=402 mm^3

σ =2083588/(4*402)= 1300 MPa ..


But if you lift from the end of the 1570 mm rod,

Pmax = 250*1250+150*4168/(2500+883)=2719 N

Mmax =2719*(883)-150*9.81*(785+883)=53585 N-mm

and σ =53585/(4*402)= 33 MPa so ,O.K. ⇒ Lifting point should be around 1500 mm from the left end.

Hi Mark

I have my feet in two camps here lol, according to my calculations I agree with NRP99 in that if the four rods act as a composite beam then yes the lift would be safe and on the hand I agree with HTURKAK in that if the rods act independently
then the stresses will be high as his calculations show. Do the rods screw into the spacer plate from both upper and lower side? I presume they do. My gut feeling is that you would get composite action if the spacer plate and the rods are tightly assembled however I wouldn't take that risk if it were my job, I would find a better way of lifting it, why can't it be lifted in two sections and then assembled?

I would also add that I haven't looked at any shear stress on the threaded rods due to the bending action or the tensile loading as the lift progresses and it might actually be the shear stress on the threads where it connects with the lower assembly that becomes the Achilles heel.

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

by acting as a composite beam, I think you mean that the pairs act as couples ?

i think the tension load would stabilise the fasteners, but I wonder about the large displacement between the pairs ... without the tension load, would them tend to displace to reduce the spacing between the pair ? it may be just a small effect.

one thing I think NRP99's analysis misses is displacement. the four fasteners will be much more flexible than the bodies ... so is a straight line a good assumption ?

another day in paradise, or is paradise one day closer ?

hi rb1957

Yes I meant this:- (civil engineering) A structural member composed of two or more dissimilar materials joined together to act as a unit in which the resulting system is stronger than the sum of its parts.

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

Ok my bad. I have done silly mistake in calculating section modulus of rod and the entire calculation needs to be revised for that mistake. Found that after checking the calculation by HTURKAK. Done in haste so may be I need to sit down quite away in sound proof room to achieve the kind of excellence shown by HTURKAK. So here is revised calculation with my approach of assuming simple supported beam with UDL of w/length.



Some points and questions-
-I am getting way high stresses than the HTURKAK, which I do not know why. May be I am doing something terrible which I am not able to pinpoint. Any pointers to look for?
-May be you can use more spacer plates at entire length of rods equally spaced left and right of current plate. This will ensure the middle section becomes stronger and can resist the lifting forces. Just a suggestion.
-Question - How to achieve the excellence of doing simple calculation as shown by HTURKAK? Any suggestions?

Quote (rb1957)

the four fasteners will be much more flexible than the bodies

So the assumption of rods taking all the moment is correct one

Thanks a lot for your help with this everyone, its much appreciated.

It can’t lift using 2 separate lift points as it will be lifted by crane to a 5m plus elevation on the single lifting point at the top end and ran from the side of the vessel. It may be possible to support it though with some temporary rigging while its being lifted from horizontal.

I’m reasonably confident that it can be taken as all 4 rods working as a system. The rods will be screwed into both sides of the plate.

NRP99. Can I ask how ‘Z’ values and ‘I’ values are calculated for all 4 rods as a system? I’m sure what you have calculated is correct, but I am having difficulty following your calculations.

Thanks again for your help.
Mark

MarkJWilson:
There is so much about this problem which we know nothing about, which makes it hardly worth commenting on. You just think…, stand clear, and let him at it, and see what happens. Most often the best learned lessons are from expensive mistakes, by the inexperienced and those with poor engineering judgement. And, their boss’s who allow inexperience (maybe their very own) to do these things. Most of the OP’ers. act like they are guarding national security secrets, so they divulge almost zero info. about their textbook like (not real world) problem. In good part this is because they know/understand so little about their own problem, the big picture, that they don’t even know approx. what info. is needed to start to approach a meaningful discussion about the problem.

Most of us know about how to handle a 16’ long simple beam which weighs about 65 lbs./ft. But, put a spring mechanism at the center of that beam, a radical change in bending and torsional stiffness, and you have a failure mechanism. It appears we have 8” dia., fairly stable beam elements at both ends, about 5.2’ long on the left end and about 8.2’ on the right end. In the middle 2.9’ of this beam, we have 4 – 5/8” round rods, on a 6.5” bolt circle, acting as ligaments btwn. the two solid ends. A 2.9’ long 5/8” round rod is a fairly feeble compression member, doubly so with the reduced bolt circle. These rods are at 12, 3, 6 and 9 o’clock w.r.t. an x-y axis on a cross-section of the beam. The rod connections to the various parts seem kinda dubious without right and left hand threads and uniform tightness and fixity would probably be difficult too. Know, we can start arguing about the stiffnesses of this rod/spring, very flexible, mechanism. It might fail in single rod buckling under compression. Secondary bending and shear my come into play, and it could fail globally by twisting, kinda like you unscrew a jar lid.

Put a strongback on that mother and lift on one end of that strongback, and have a bearing/pivot detail on the other end of the strongback. Lash/connect the strongback to the load at each end of each of the solid parts of the beam. This protects the loaded ends, puts the reaction points on the strongback instead of the load, and eliminates that funny mechanism in the middle from the need of analysis, design and the above arguments/debate.

@NRP99 ...

1) "So the assumption of rods taking all the moment is correct one" ... if you don't mean bolt moment = beam moment /4 ? the fasteners couple carry the beam moment in pairs as couples.

2) I don't a UDL is the correct loading for a lift. UDL for weight and the lift load at the LH end. This makes the beam "difficult" to solve, what's reacting the moment due to the lift and the weight ? do you lift with a strop, a sling attached to both ends, so the lift point is always above the CG ?

another day in paradise, or is paradise one day closer ?

Quote (MarkJWilson)

Can I ask how ‘Z’ values and ‘I’ values are calculated for all 4 rods as a system? I’m sure what you have calculated is correct, but I am having difficulty following your calculations.

Here is what I done- Considered the rods placed as below (which is not ideal placement, I think). Then calculated inertia of 4 rods together with parallel axis theorem about the center of plate.


rb1957

Could you please elaborate your 2nd point. I only understood that UDL is not correct way to consider for this situation. Then I lost somewhere in strop and sling.

instead of My/I, for stresses on the fasteners, I would use M/d (react the moment as act couple) and then /A for stress in fastener. It may be a slightly conservative result compared to yours, but I think the physics "looks" better ... My/I says the four elements are interacting, like with a shear web between them ... but they aren't.

draw a free body of the lift, starting at zero, then at say 30 deg, 90 is trivial (except maybe for were the beam ends up relative to its original position).

if you lift from the LH end, then the beam will translate to be under the lift at 90 deg. How it moves at intermediate states is a question. Does the LH end rise vertically and the other points track as the beam rotates ? What does the free body look like at 0.0001 deg ? If the lift is vertical, and weight is parallel down, how is the offset moment reacted ? It's possible that a "small" horizontal reaction can be applied through the lift point, inclining the lift vector. Remember this beam is a simple three force member (review if needed to understand what this means)

If we lift a beam off the ground, I think we don't apply a vertical force, but an inclined one, thus a horizontal couple reacts the offset moment.

If you lift with a sling, you can move the lift point and keep it close to the CG of the beam (so you're lifting above the original CG position, not the LH end).

clear as mud ?

another day in paradise, or is paradise one day closer ?

Quote (NRP99

Some points and questions-
-I am getting way high stresses than the HTURKAK, which I do not know why. May be I am doing something terrible which I am not able to pinpoint. Any pointers to look for?
...
-Question - How to achieve the excellence of doing simple calculation as shown by HTURKAK? Any suggestions?)

- If the rods are not connected to each other ( to get combined behavior , at least with diagonal elements like truss) , you can not assume the rods will behave as combined and you can not calculate the combined moment of inertia ..
- In this case the four rods will resist as separate elements

- In order to assume group reaction is valid, the rods SHALL BE connected with web elements . The beam theory ( plain sections will remain plain after deformation ) is valid for composite elements when they connected to each other to transfer shear..

Dera MarkJWilson (Mechanical),

If you cannot lift, pivot the assembly from a point say 1500 mm from the left end, you may use a lifting beam. ( temporary wrap the assembly to , say a Channel section ).

Quote (rb1957)

instead of My/I, for stresses on the fasteners, I would use M/d (react the moment as act couple) and then /A for stress in fastener. It may be a slightly conservative result compared to yours, but I think the physics "looks" better ... My/I says the four elements are interacting, like with a shear web between them ... but they aren't.

Understood. I was thinking in that terms of couple force acting on top rod and bottom rod with F=M/83mm. But then I guess, I missed this point in cloud of calculation. What you said is correct- the couple makes physics of problem look more sensible.

On the later part of explanation, Thank you. Got it, almost, I think.

Hi rb1957

I am not understanding how you are approaching this problem but that isn’t to say that you are incorrect, however I have put pen to paper and have laid out my process thoughts and my results are very similar to Those of HTURKAK, I guess it all hinges on how the rods act ie as a unit or individually 👍

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

Here is the second image of my calcs

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

Quote (HTURKAK)

- If the rods are not connected to each other ( to get combined behavior , at least with diagonal elements like truss) , you can not assume the rods will behave as combined and you can not calculate the combined moment of inertia ..
- In this case the four rods will resist as separate elements

Then I am wondering how plate is connected and How the plate transfers the forces to the rods?

Quote (HTURKAK)

- In order to assume group reaction is valid, the rods SHALL BE connected with web elements . The beam theory ( plain sections will remain plain after deformation ) is valid for composite elements when they connected to each other to transfer shear..

Is the plate not playing role of connecting member of all rods? I considered that rigid plate(compared to rod stiffness of course) connected to 4 rods and moment applied to rigid plate about horizontal central axis. I am wondering whether this assumption is totally wrong.

If I use F=M/d as rb1957 said and then for stress in rod F/A applies. So M=9199578.38Nmm for 0deg and D=166mm, F=55420N and then A=201.0619mm^2 gives stress in rod=275MPa which is again different approach than you. But still plate plays role of transferring the moment as "couple" of forces. I do not see why plate plays no role in either of situation.

"assumed pivot" (at the RH end) is an assumption ... the lifting point moves laterally (not saying it can't, up you'd think the lift would be up and the beam RH end would slide laterally to under the lift.

can you make a free body out of your diagram (first post) ? I can't see a moment restraint balancing the offset moment between the lift and the weight ??
Would the lift cause a up reaction (the end of the beam would push against the ground), so that the lift force is a portion of the weight (I'd thought it would be all the weight),
and now the weights are reacted by the lift and the RH end reaction. As I write this it sounds so obvious !!? (probably I confused myself thinking the lift = the weight)
then no lateral couple ... (sigh)

I would've assumed that the "kgs" were "kgf" (force not mass)

another day in paradise, or is paradise one day closer ?

Quote (NRP99
Quote (HTURKAK)
....

Then I am wondering how plate is connected and How the plate transfers the forces to the rods?
Quote (HTURKAK)
- In order to assume group reaction is valid, the rods SHALL BE connected with web elements . The beam theory ( plain sections will remain plain after deformation ) is valid for composite elements when they connected to each other to transfer shear..

Is the plate not playing role of connecting member of all rods? I considered that rigid plate(compared to rod stiffness of course) connected to 4 rods and moment applied to rigid plate about horizontal central axis. I am wondering whether this assumption is totally wrong.

......)

DEAR NPR99,

- THE PLATE DOES NOT TRANSFER ANY FORCE..THE ASSUMPTION ,' THE RIGID PLATE DISTRIBUTES THE BM TO THE FOUR RODS ' IS NOT CORRECT AS LONG AS SHEAR EXISTS..

- IN ORDER TO ASSUME THE FOUR RODS BEHAVIOR WILL BE COMBINED BEHAVIOR AND THE MOMENT OF INERTIA WOULD BE THE INERTIA OF THE ASSEMBLY ( AS PER UR CALCULATION ), THE BEAM ASSEMBLY SHALL SUBJECT TO PURE BENDING AND THE INDIVIDUAL BEAMS ( IN THIS CASE FOUR RODS ) SHALL BE RIGID CONNECTED , AND THE SECTION AT THE TOP AND BOTTOM ASSEMBLY CONNECTIONS SHALL ROTATE RIGID..

IN ORDER TO CLARIFY AND JUST FOR SHARING THE KNOWLEDGE , I COPY AND PASTED A RELEVANT WORKED EXAMPLE FROM THE BOOK (Advanced Stress and Stability Analysis BY V. I. Feodosiev )











I HOPE THIS RESPOND ANSWERS YOUR QUESTIONS AND DOUBTS..

GOOD LUCK..

Hi rb1957

Agreed that the pivot at the right hand end is a big assumption because like you say the whole assembly will at some point will slide which is part of the reason I only considered a 10 degree rotation. Now from my calcs above if I assume a couple as you suggest then the moment is 2100000Nmm which I then divide by the 117.63 dimension which yields 17890.6N, if I divide this by two and then find the rod stress using the rod area I get 44.5 N/mm^2 which is pretty close to what I got originally assuming the rods act collectively (50.3N/mm^2). Now another question if we rotate the rod arrangement by ninety degrees so that it looks like the diagram that NRP99 posted then we have two rods on the neutral axis and two rods at extremity of the beam ie
top and bottom so how would you share the couple now?. If I check the stresses in the rods at top and bottom of the beam with this arrangement it is 1.3 times greater than the 50.3N/mm^2 I got previously. If I have misunderstood what you are saying it might be worth you posting your calculations so I might understand better 👍

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

Thanks again for the amazing support you have given on this. It’s been incredibly helpful and educational. And thanks for providing such detailed calculations that even I can just about follow.

@Desertfox… one thing I can’t follow is how you have calculated the bending moments per your diagram? I’ve been staring at it and playing with my calculator for half an hour but can’t figure out how 2100 Nmm x 10^3 has been reached. Please can you clarify?

Thanks Again
Mark

Hi Mark

I took moments at the point loads, so to find the bending moment at the point where the 250kg acts,I took moments to the left of that point so :- F * cos10 * (785+2919) - 150*9.81 * cos 10 * 2919 = bending moment where the 250kg acts. Similarly to find the bending moment where the 150kg acts I take moments to the left of that position so : F * cos 10 * 785 and that’s how I got the original two values, the bending moment diagram between the two points is a straight line and so I drew a bending moment diagram to scale and measured the value from the diagram. Hope this helps

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

Mark, what sort of experience do you have ? The calc was well laid out (unless you're saying df dropped a clanger).

The clac shows that the RH end in supplying the rest of the weight reaction ..
loads weight is (150+250)*9.81 = 4000 N (I think we get here, to forces in N, whether the bodies mass or weight is given in kg)
Lift is 1857 N, so RH reaction is 2143 N

another day in paradise, or is paradise one day closer ?

Hi Mark

Having read through HTURKAK's last post I believe the rods will only act as individual beams and not act together as I originally thought, in which case I think the lift is doomed in its present state.

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

Quote (desertfox
....Having read through HTURKAK's last post I believe the rods will only act as individual beams and not act together as I originally thought, in which case I think the lift is doomed in its present state.)

Eventually i am glad to be understood and a pink star for this respond..

doomed is a pretty big word !

for 10deg, df calculates BM = 2.1E6, but HTURKAK calcs 9E6 ? (Nmm)

At zero deg, (max BM) the lift is approx 2000 N (the RH end reaction about 2500 N)
the moment is approx 2000*2500-1500*1300 = 5E6-2E6 = 3E6 Nmm (close to df's)
if the spacing of the active pair of fasteners is 166mm, then the couple is 18E3 N
if the bolt is 16mm dia, then stress is 18E3/200mm2 = 90 MPa (which looks low, but then I don't work in metric)

18kN is about 4000 lbs (2 tons) ... but a 1/4" (aerospace) bolt has an allowable of 4600 lbs, so 16mm is ok ??

another day in paradise, or is paradise one day closer ?

Hi rb1957

I think even using the moment couple it assumes that the rods work as one unit.

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

HTURKAK

I still feel that my assumption not totally wrong if not completely accurate. The behavior of rods will be in between the theoretical attached rigid plate and no plate like the combination of figure 237 and figure 36 of your post.
This is not just thought. I actually ran a simulation today (thought was not getting out of my head as a hard core analyst). I modelled Tool as single solid part considering rods "firmly" connected to the plate and both bodies. Adjusted mass of both bodies as given in the sketch by OP. Overall mass = 0.415tonne due to weight of plate and rods. Then fixed the lift point and only vertical point constraint of 5deg rotated tool at the other end where it just touches the ground. Applied only gravity load with elastic properties.

The von Mises stress at the bottom rod (my previous post image of plate with rods) at 2054mm from LH is 106MPa peak and 90MPa peak at 2453mm from LH. So what is assumed by me and rb1957 that the plate plays role in the transferring the forces is correct. Further this proves what rb1957 is saying that physically the plate will impart axial forces on the rods as couple. But there is something more to it.

From LH total mass supported = 4071.2N, then vertical reactions at close to 0deg, RL=2035.58N RH=2035.58N. Then 2035.58*2054-(150*9.81*(2054-1570/2))=2.314E6Nmm. Then M/d=2.314E6/166=13.938kN. Then the axial stress in the rod = 13.938kN/201.06mm^2=69.33MPa which is not close to 106MPa? rb1957 already shown the calculation for 90MPa at 2453 from LH.

But there are other things in the results. The rods at neutral axis not playing any role much but the stress is not entirely zero, either. As opposed to theoretical consideration of 90deg and 270deg rod-neutral axis rods will not be playing any role. Same is said about the rod bending. The 484mm length part rods are bending which is why the results are not matching with 69MPa. There is clearly bending stresses in the all 4 rods of 484 length. The behavior is in between as I said above. For the 369mm length rods, the bending component is less but it is still "bending" "collectively". May be the rods are "firmly" connected that is why this may happen, you say. But I think even if, lets say, rod and plates or bodies are not attached "firmly" (there is some gap between the rod, plate and bodies as gap in rods and holes) after some relative play, rods are forced to dance together.

Thanks very much for clarifying Desertfox, much appreciated.

RB1957 no I certainly wasn’t questioning what Desertfox had written. I just wasn’t understanding it. My background is mechanical (more pressure containing) but I seldom carry out calculations and haven’t done any beam type calculations since college which was more than a couple of decades ago! 😊 Without a worked example I was really struggling with this. Thanks so much for providing such comprehensive support which was well beyond any expectations. You have definitely helped my own understanding (plus shaken a few cobwebs loose) and I have a much better idea now on how to carry out bending stress calculations in future.

I’ve taken into account everything that’s been said. I’ll work on the principal that it should be taken as an individual rod rather than all rods acting as a system. I think there may be scope to increase the plate size enough to put new larger diameter rods. I’ll also look to change the material to something of a higher yield strength. It means getting new parts designed and manufactured which will be very tight based on the deadline but better that than the interface failing. I’ll also discuss with our offshore service tech’s to see if we can find a means of supporting it as its lifted from horizontal.

Thanks again for all your support. It has very much been appreciated.
Have a great weekend everyone.
Mark

I think you're missing something Mark. I think we're (me and HTURKAK, are saying we "think" the bolt stress is very low. I believe the bolts will react the moment as a couple, producing low stresses in the 16mm bolts. The bolts can work as a pair, there is no change in endload along the bolt. Fine tune the calc to include the small moment that would develop along the bolt length (between the plates) as the bolts (all 4) carry the shear.

Understand you're doing something your experience has made unfamiliar. I expect you hit the books, and are using us as sounding board.

I think the structure you have will work ok (based on the small amount of info we have). I would look into displacements as I expect the bolts are a lot more flexible than the plates. This will be a tricky calc, assuming the plates are rigid will make it easier

another day in paradise, or is paradise one day closer ?

Quote (NRP99
HTURKAK

I still feel that my assumption not totally wrong if not completely accurate. The behavior of rods will be in between the theoretical attached rigid plate and no plate like the combination of figure 237 and figure 36 of your post.
This is not just thought. I actually ran a simulation today (thought was not getting out of my head as a hard core analyst). I modelled Tool as single solid part considering rods "firmly" connected to the plate and both bodies. Adjusted mass of both bodies as given in the sketch by OP. Overall mass = 0.415tonne due to weight of plate and rods. Then fixed the lift point and only vertical point constraint of 5deg rotated tool at the other end where it just touches the ground. Applied only gravity load with elastic properties.)

It is true that the behavior of rods in this case, will be 'in between ' .. However, engineered calculations shall be based on MINIMUM ENSURED , GUARANTEED level.

I am not sure what kind of simulation you have performed but in this case , the buckling of compression members (top chord), second order effects , shear will affect the behavior.

If you look to the picture Fig. 237 of may previous post, the two elements are touching , so having single curvature ..In this case, the top, bottom chords would have different curvatures.

Regarding the moments and shear developing at rod assembly, i reviewed my hand calculation and pasted below;



I would like to see ,If you can post your simulation model ..

Here are images from simulation.
Model-RP1 is constrained in all DoF but RX free. Other side vertex touching the ground is fixed in Y only. Gravity applied in -Y. Tool body rotated by 5 deg wrt horizontal


Displacement


Reactions


Stress


Different results in another software but overall behavior is same as above.

be careful with von Mises stress ... it hides the -ve stress.

is your FEA supporting my "back of a fag packet" calc ... then 16mm fasteners have a high margin ?

another day in paradise, or is paradise one day closer ?

Very interesting but sadly I haven't got access to software to run the analysis but looking at HTURKAK's post 20th January and specifically at FIG 237 it does show if you clamp the ends rigidly for two beams albeit they are touching you can achieve combined action. I would agree with the sentiment that the four rods would act between the two conditions of acting singularly or combined. I think the arrangement would work better if the rods were positioned as per my sketch on the 20th January rather than having two rods on the neutral axis of the beam. Also the analysis does show that the stresses are similar to assumption of taking rb1957's assertion of considering a couple but it doesn't account for the lower stress on the rods positioned on the neutral axis bearing in mind that all the rods should see the same shear force of 385N. If I assume that the bending moment is shared equally over the four rods without regard to there position within the tool assembly but each rod taking a quarter of the bending moment and acting as a beam individually then I get a bending stress of 81.59N/mm^2 which would support the lift being okay.
At the end of the day its how comfortable the OP is with lifting on the rods and how close to real life the stress analysis as been modelled in terms of restraint etc (no disrespect intended to those who have run the analysis).
One point I have to raise though is how do screw four rods into a single plate central and two 200mm bar diameters on either side without having to use nuts any where, what i am saying is you can't assemble it in my eyes if the central plate and the two 200mm diameter bars just have tapped holes or I am i missing something?

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

I don't think the plate is a significant feature of this design. The shear is distributed between the four fasteners, the moment between two (or four if your clock the load 45 degrees). I guess the plate is reducing the moment due to shear, with the plate fixing the fasteners mid-span, sure that'll make the bending due to shear smaller, but shear at the middle should be very small ... the lift is about 1/2 the weight, and the LH body is about 1/2 the weight (so I wouldn't've thought the bending due to shear/4 was anything to worry about).

Still, df's practical questions are still there ... maybe the OP didn't show the nuts (to save us worrying about them, and getting distracted) ?

another day in paradise, or is paradise one day closer ?

I wasn’t worried about the shear reducing the bending moment because the shear force in the region is constant and the intersection of the rods does not coincide with the maximum bending moment of the tool were the shear force would be zero however it’s appears to me that whether you assume the rods as couples resisting the bending moment or sharing the moment over four rods the figures for stress are within 20-25% of each other. The more important question now lies in how that assembly is achieved because screwed rods into the plate or the opposing end is liable to be a more rigid connection than that retaining with nuts, which will need a clearance hole in the central plate or one of the other ends, so in practice our rigid rods screwed in at both ends cannot happen and the restraining of the beam ends to achieve some kind of combined action during bending is not looking as good as before.

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

I guess we all are shooting in the dark. Unless we know the "Tool" really, its difficult to "predict" the behavior of lifting. And besides exact assembly information, there is DAF for offshore lift which will make the lift more risky. The stress we see in those pictures may get magnified based on what the DAF is.

Don't know the about OP's lift, but certainly my brain was lifted with so much insights from members. Thank you - HTURKAK, rb1957, desertfox and MarkJWilson.

Hi NRP99

It certainly is a good thread and certainly got my brain thinking, I hope the OP comes back to tell us about how it is assembled.
I would like to thank all for making it a very interesting thread.👍😀

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

Desertfox/NRP99/rb1957/HTURKAK thank you so much for your help with this.

And thanks for your patience with me plus really helping develop my knowledge on this topic.

It has been very much appreciated.
Mark

Hi Mark

Can you tell us about how you assemble those rods?

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

Absolutely Desertfox.

Top assembly (150KG) 4 x 16mm rods are screwed into M12 tapped holes on 166mm PCD.

Other end of these rods sit in 17mm diameter counterbored holes that are 10mm deep in the top face of the support plate. This end of the rods is tapped M12 and are fixed in place through the bottom side of the plate with 4 x M12 fasteners.

I maybe didn’t make this clear but the 4 x rods between the support plate and bottom assembly are separate from the ones between the top assembly and support plate. These 4 rods are on the same 166mm PCD equally spaced in between the 4 rods that attach to the top assembly. They are assembled the same way but the 17mm counterbore is on the underside of the plate and they are fastened through the top again with 4 x M12 fasteners.

These 4 x 16mm rods are again screwed into M12 tapped holes on the flange plate on the bottom assembly (250KG).

It needed to be done this way because there isn’t much room and I’m avoiding many existing features on both the top and bottom assemblies. There are hydraulic lines, air lines as well as control data cables that run through the centre of the tool.

The other thing to mention is that the lower rods (that fit into the 250KG bottom assembly) and support plate were designed much earlier and are already existing. The requirement to fit the top assembly came along much later and it needs to be a set distance from the bottom assembly. This is why the support plate isn’t bang in the centre between the 2 assemblies and the rods aren’t of equal lengths.

Thanks again
Mark

Hi Mark

Thanks for updating us on how the rods are assembled, I couldn’t see how it was possible from the original tool sketch but it’s clear now👍

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

the important detail there is the "M12 tapped hole" ... so the 16mm fasteners are effectively 12mm ,,,

another day in paradise, or is paradise one day closer ?

Very true rb1957 that’s what I was saying about how the detail of fastening the rods was achieved I thought the screwed rod was all M16 however it depends how much bending the screw or bolts actually carry if any but it is significant.

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