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Force distribution of a tilted 20 ton Ring?

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Skyba
#1
Sep25-13, 08:50 AM
P: 7
Hi dear PF'ers,

I have to develop a tank lifting system similar to the picture : http://www.astanks.com/Lifting%20method.JPG. The lifting would be distributed evenly over a dozen of chain hoists pulled by motors.

Each chain hoist has a maximal load of 5 metric tons. Because the motors may not lift evenly, the tank might tilt a bit and the force distribution would not be equal.

The tanks that are to be lifted vary in diameter and height. The weight goes from 10 to 40 metric tons.

I would like to define maximal circular interval between each jack as well as the maximal tilt angle of the tank, as a function of tank dimension, so that each jack carries no more than 2.5t (a security factor of 2 seems appropriate). I do not know where to start... Should I use a simulation program ? Is wind negligible ? Any gut feeling on how the force distribution compares to the horizontal situation ?

Thank you for help !

Basile
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Baluncore
#2
Sep26-13, 08:26 AM
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Are you increasing the capacity of the tank by cutting it from it's floor, lifting it and inserting a new, possibly thicker section wall?

If you lift from the base the maximum weight will be transferred to the lowest side, so you cannot balance the weight distribution using equality of jack load. Does the tank still contain any fluid? If so that will run to the lowest point and aggravate the situation.

If there is no floor or flange at the bottom of the wall then the tank it will not remain circular if there is any imbalance between the jacks. You need to be able to take the weight on all jacks and then advance them all in stages based on height of lift, while monitoring for excessive load.

Wind will be a problem. Your circle of jacks need to have some lateral stability. Guy wires to the tank would be a liability as they would convert horizontal wind pressure into vertical forces on the jacks.

Are you restricted to manually controlled adjustment or do you have a computer based control system that can read height from laser gauges and load limit detectors ?
Skyba
#3
Sep26-13, 10:49 AM
P: 7
Hi Baluncore,
Thank you for your answer.

Are you increasing the capacity of the tank by cutting it from it's floor, lifting it and inserting a new, possibly thicker section wall?
Exactly. We are actually building new tanks, so they don't contain any fluid.

weight will be transferred to the lowest side
Yes, the question is, how much ?

If there is no floor or flange at the bottom of the wall then the tank it will not remain circular if there is any imbalance between the jacks
That is a very good point. The jacks look like this : http://img844.imageshack.us/img844/8128/0h7o.jpg, and they all pressed from inside to the outside of the wall. If the jack spacing is small enough, we should be able to limit the tank's deformation.

Until today, the tank was lifted completely manuallly (one person on each jack) and an overcharge on a hoist could be easily felt and corrected. We would like to automate the process by pulling every chain hoist with synchronised motors. In order to do so, I need to estimate the maximum tilting angle and wind so that no jack overloads.
We have a possibility of incinometer sensoring. But we would like to evaluate the necessity of such a feedback before investing in it. It may turn out that the maximum tilting angle can be seen by the naked eye and/or that the motor speed difference to achieve such an angle is unlikely.

Anyway we would choose a closed loop only in last resort, because the implementation is too complex and the system needs to be easily set up / changed / manipulated by an operator. For now, the inclinometer is thought to stop the system and the operator would then manually correct the position of each sensor by deactivating some motors.

But If we use an inclinometer, we have to set the maximum acceptable tilt... Any idea on how to calculate that ?

Here is some additional info on the tank, if it helps anyone :
The tank is made of A36 steel and the bottom rings are often 9mm thick.
Diameter is around 8 to 15m, height 6 to 10m, weight 10 to 40 tons. There are some other, additional structures that need to be lifted with the rings, like reinforcements and circular metal stairs.

SteamKing
#4
Sep26-13, 11:21 AM
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Force distribution of a tilted 20 ton Ring?

It's all very nice to chit-chat this problem on a forum, but you should have your jacking arrangements reviewed by a competent mechanical or structural engineer. There are too many things which can go wrong trying to do this type of procedure by the 'seat of the pants'. At the very least, the structure of the portion of the tank being lifted needs to be inspected and analyzed to determine if it is strong enough to withstand the lift without damaged.
Baluncore
#5
Sep26-13, 01:04 PM
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Because the jacks surround the tank there will always be a pair of jack stays to bear the wind pressure. The two diagonal stays on the jacks need to be strong enough to stop the wind moving the tank wall. There need to be enough jacks to spread the wind pressure around the wall and so prevent buckling of the flexible wall. Jack stay design will come down to your region's maximum recorded wind gusts and the tank's aerodynamic profile.

Each jack will need some way to monitor safe load. That must stop them before they break something. It could be self contained within each jack. If each had an adjustable threshold then you could correct for tilt by adjusting the load by stopping some motors.

Quote Quote by Skyba
But If we use an inclinometer, we have to set the maximum acceptable tilt... Any idea on how to calculate that ?
Any detectable tilt should be corrected as soon as it is detected. Tilt needs to be measured in at least two dimensions. I believe that the measurement of tilt would be unnecessary if height could be controlled.
Skyba
#6
Sep26-13, 03:14 PM
P: 7
Guys thanks for the heads-up, but my problem is much simpler than that =)

Because fortunatly, I do not have to re-design the tanks nor the jacks. These have already been tested on the field for years with a nominal load of 2,5 tons. I simply have to "upgrade" the lifting from manual to motor-driven. The only difference being that when it was manual, one could feel the load directly on each chain hoist. Some of the lifters would slow down / accelerate to compensate for the lift.

My starting point is to have a dozen of exactly same motors to do the job. Let me put my question in an other way :
If we assume that one motor would have a max. speed difference of 1% to any other, is it safe to assume that the resulting tilt of arctan(DeltaH/Radius)=arctan(0,01*2m/6m)=0,2 degrees would not have any significant impact on the load distribution, thus being well within the security factor of 2 ?

I am just trying to see that I dont overrate the tilt problem !
Baluncore
#7
Sep26-13, 06:27 PM
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Tilt will only be a problem with tall tanks that have a small diameter. It is a cosine effect. A few degrees is not a problem, but because of the instability, any tilt will immediately result in positive feedback which will amplify the tilt. For that reason you need to prevent or control the tilt as soon as it is detectable.

It will be difficult to find load insensitive synchronous electric motors. Ideally your motors would have a shaft tachometer and speed regulator that guarantees they all advance at the same rate. They may also need a maximum current limit to prevent overloading with an alarm to stop a lift when it occurs. How are you going to detect overload of any one jack if you have no human feedback ?

If you model the entire mass of the tank as being concentrated at the centre of the top of the tank then you will get a realistic estimate of the distribution of jack load.
Skyba
#8
Sep27-13, 05:24 PM
P: 7
Thank you Baluncore. I will make some simulations and tell you my plan as soon as I have one ^^.
Skyba
#9
Oct1-13, 04:22 PM
P: 7
Well, as it turned out on a simulation, the tilt of the tank does not change much. For an inclination of 1%, on a tank of 20t, 8m diameter, 8m height and 7 jacks, the forces vary from 3.12e+004 N for the most charged jack to 3.01e+004 N for the least loaded. For 5% inclination, (which is huge!) the forces are respectively 3.35e+004 N and 2.8e+004 N. The model I tested is one of the highest tanks we would use with respect to the diameter. I attached the report images to this post.

I wonder thus if we need an inclinometer any more. Even 1 can be seen by the naked eye. And if the motors have an oversize factor of 3 to 5 (does not make any big price difference at this low HP), we dont have a risk of positive feedback either. The system would be stable.

The only risk is if one motor goes "out of (horizontal) line". Since the tank is quite rigid any acceleration would take on all the force. What about adding a rope segment between the chain hoists and the tank, for the elasticity to compensate any overweight ?
Attached Thumbnails
Flip Simulation - 1 degree in positive x axis.png   Flip Simulation - 5 degrees to the right.png  
Baluncore
#10
Oct1-13, 08:16 PM
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Looks good.
It should be possible for all other jacks to carry the unbalanced load if one jack fails catastrophically. That may require more than a factor of two in some situations.

I think you could benefit from some form of load sensor on each jack. That could be a simple rubber block in compression that deforms sufficiently at say 2000kg to trip a micro-switch. The trip would stop that jack motor(s) and sound a general alarm. The rubber would provide some flexibility and load sharing during normal operation. It would also provide a visible indication of load.
Another solution could be a very short hydraulic cylinder with a pressure gauge and overpressure switch. There would be no flow of fluid, just pressure monitoring. A small gas bladder in the compression side would give some load balancing between the jacks.


If you lift with hydraulic cylinder jacks from the lower edge, the tank will tend to tilt as, (without metering), pressure will decide flow volume. But if you were to lift from the upper edge it would remain stable. It is a reality that you must lift from below. But you still have that cosine advantage, if you can keep it true there should be no problem.
Skyba
#11
Oct2-13, 06:30 PM
P: 7
Balcunore, cheers again for your time
That could be a simple rubber block in compression that deforms sufficiently at say 2000kg to trip a micro-switch.
I am having a hard time finding any load cell that simple. Most of them are analog and cost more than $150. I'll look further into it, and if you have a suggestion on the name, i'll find it rather soon ^^

Basile
rollingstein
#12
Oct2-13, 10:30 PM
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Quote Quote by Skyba View Post
Balcunore, cheers again for your time


I am having a hard time finding any load cell that simple. Most of them are analog and cost more than $150. I'll look further into it, and if you have a suggestion on the name, i'll find it rather soon ^^

Basile
For the sort of tank lifting project you describe isn't $150 per cell reasonably cheap? :)
Baluncore
#13
Oct2-13, 10:30 PM
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I would make my own. This is the first I found as a commercial product. There must be many others.
http://www.mackayrubber.com.au/products/snubber/ Download the pdf for data with the graph.
Skyba
#14
Oct3-13, 08:35 AM
P: 7
Wow ! This is just EXACTLY what I was looking for. Awesome.
It'll help even out the load if one motor is ahead. I'll buy a bunch of those :D
There is no switching mechanism though..


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