I think it goes something like this...



F = M x A

A horizontal acceleration will impart force plus its overturning moment about the base of the cargo.
The Overturning moment OTM = F x h
Where h is the height of nCG over cargo base.
That will tend to rotate the cargo about one of its edges.
It is counteracted by the resisting OTMr = the cargo weight x cargo width/2
OTMr must be greater than OTM to avoid cargo tipover of the cargo on its lifting platform.
You probably want at least a safety factor of 2 there.

At the lifting lugs...
The OTM is counteracted by changing the loads at the lifting points by F= OTM/d,
where d is the distance between lifting points (in the OTM plane).
Rigging load vertical components then become W/N +/- F
Where N = number of lifting lugs.
+ OTMr/d on one side, -OTM/r on the other side.
Assuming 4 equally spaced lugs at the platform corners,
Lugs loads (vertical components) W/N +/- OTM/r/2

Knowing the angle of the rigging cable, you can calculate cable load.
Keep that below your allowable cable tension force.




--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

The following references discuss similar in principle, that may be of interest:

1.
1:08:38:

https://www.fsc.gov.au/blog/ofsc-webinar-lift-plan...

2.
Cl 17.2.4:

https://www.amazon.com/Rigging-Engineering-Basics/...

There is a really good reference from a German insurance group as well, including example calculations.

https://www.tis-gdv.de/tis_e/ls/stabilitaet_von_au...

TBV.  I am not quite sure what you are specifically worried about.  Can you please clarify, and include a diagram.  Assuming 1503-44's diagram is correct, are you worried about the stability of the cargo&platform combined unit, or are you worried that the cargo might move relative to the platform?

If the former, and assuming that your four chains are independently attached to your hook, then Figure 1.1 in that excellent article from ItIsMoving's post correctly states that your overall system is "absolutely stable".

If the latter, and provided that the only movements your system is subjected to are those caused by horizontal movement of the hook, then I think that the "pendulum effect" means that the contact force between your cargo and its platform will always be perpendicular to the platform.

 _{—————————————————————————————————}
^{Engineering mathematician / analyst.  See my profile for more details.}

Thanks for the time sharing answers.

Please, find a diagram illustrating my doubts, the gantry crane lift off the cargo, then travel the cargo some meters to final destination. Beams are not connected between each other.

(Now that we know that you have four, parallel, independent chains...)
A totally unrelated matter is how these chains will share the load.  Since you can never guarantee that all four will be of exactly equal length you should design them each for half the load, not for a quarter of the load.
You probably already know this, but I would be remiss not to mention it.

 _{—————————————————————————————————}
^{Engineering mathematician / analyst.  See my profile for more details.}

Thanks, still the question remains about how to approach stability.

Use a spreader bar arrangement to lift off a single hook from a single crane, and use the guidance in that German publication to simplify things? Or use a spreader bar from one hook on each crane and again use the German publication (it has an example of 2 crane lifts). If done right, spreader bars or other equalising arrangements will also address Denial's comment about sling lengths potentially affecting the load distribution.

Or you could design a custom lifting attachment which fixes solidly into the structure and brings the effective lifting point up to the top (or at least above the COG).

If the picture is correct, add a plug or frame at the top of the hole to capture the chains. This will eliminate the ability to rotate.

If the picture is not correct, perhaps post a correct picture.

I can now see the general arrangement, but I am still unable to visualise the exact possible event you are trying to avoid.  Instead I will give a simplified version of how the cargo will behave when the crane beam accelerates, runs, then decelerates.

Cargo has a mass M. CoG of cargo is a distance L below the crane's hooks.  Initially everything is stationary and the chains are vertical.  Then the crane instantaneously begins to move, with a constant acceleration "a" which it sustains in perpetuity.  Ignore all energy losses, damping, etc.

The motion of the cargo RELATIVE TO THE GANTRY is now simple-harmonic.  Its horizontal acceleration, initially zero, now varies between zero and 2a, at a "natural frequency" of sqrt(g/L) radians/sec (where g is the acceleration due to gravity).  This motion will continue indefinitely for as long as the gantry continues to accelerate (in the absence of any energy losses in the system).  This can be thought of as the response to a sudden "shock".

However the gantry does not accelerate for ever.  At some stage it will stop accelerating and begin coasting along at a constant velocity.  This is another shock.  The effect of this second shock on the cargo's motion depends upon the shock's timing.  It could fully counteract the first shock, or it could be fully additive to the first shock, or anything in between.

But we are not finished yet, because the gantry is still travelling.  It still has to stop moving. Assume it does this by instantaneously beginning to decelerate at "-a", maintaining this deceleration until it is stationary.  We thus have two more "shocks".

So an initial shock, followed by three more shocks of the same magnitude which can each be additive or subtractitive or anything in between.  The worst case (unlikely but possible) is that all are fully additive, which would result in the cargo experiencing a maximum horizontal acceleration of 4×2a = 8a.

[Made an absolutely trivial grammatical correction at 20:55 on 29Sep UTC]


 _{—————————————————————————————————}
^{Engineering mathematician / analyst.  See my profile for more details.}

How far apart are the lifting lugs? If they are too close, instability will be high.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."