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Help With Bending Stress Calculation 5

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MarkJ_

Mechanical
Jul 8, 2021
9
Hello,
I was hoping to get some advice on how to calculate the strength of a new lifting interface. I have 2 existing tools that need to be joined together rigidly for lifting offshore. There are hydraulic lines between the 2 assemblies and access is also needed at the lifting interface for making these up. For this reason and because of space constraints I am looking to use 4 x 16mm OD threaded rods with a support plate in between.

The top assembly with a single lifting point on its end face is rigid and weighs 150Kg. The bottom assembly is also rigid and weighs 250Kg. I have attached a sketch which shows the arrangement. The support plate and lower support rods (359mm long) are already existing which is why the support plate isn’t in the middle.

So the complete assembly will be lifted from horizontal to vertical using the single lifting point on the top assembly.

I was hoping someone could offer some advice on how best to calculate the bending stresses in the rods?

I tried doing this as a simply supported beam assuming the supports are where the rods attach to the top and bottom assemblies. I took second moment of area as being the formula for solid bar and calculated the bending moment halfway between the support rods and support plate. The stresses are very high and I’m pretty sure my calculation is incorrect.

Calculations are not my strong point. I would appreciate any advice?
Thanks
Mark
 
 https://files.engineering.com/getfile.aspx?folder=ee499a3c-12d7-42ed-9591-51af0658f0aa&file=Tool_Lifting_Sketch.pdf
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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.
Capture-1_r3n1ic.jpg


Capture-2_xadlqc.jpg


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. [smile]
 
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.[bigsmile]

“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.

Capture-2_rxz5fu.jpg


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?

rb1957 said:
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 ?
 
MarkJWilson said:
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.
Capture_3_ravff7.jpg


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 ?
 

- 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 ).

 
rb1957 said:
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.[smile]
Capture_5_ah9z2i.jpg
 
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 👍
B2F5BA98-A715-4B2E-90BF-F4A5D0E99692_vttov6.jpg



“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
E640058B-1E70-4F29-957E-06B4D0691524_dk8g12.jpg


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