Centre of Mass of Moving Equipment

[Aytacoglu]

Hi All,

I am trying to carry out a calculation for a transportation arrangement of an equipment in a plant room.

The equipment we have is a transformer and it is quite heavy (~11 tonnes).

The proposal from the contractor is to use skates to undertake this. They are proposing to put 2no. skates at the back of the equipment, and 1no. skate at the front, so that someone can divert the machine and bring it to its final position.

My concern is on how to calculate the load using this method. Obviously this like a 3-point system where you have 2no. skates at the back and 1no. bigger skate at the front. I am looking to calculate the force that will be exerted by each skate on the floor. It seems like the distribution is not going to be even due to the arrangement and we can't assume 11 tonnes equally distributed 11/3 = 3.67 tonnes each skate.

My question is how to work this out and how to get the Centre of Gravity so that I can accurately calculate the distribution between different skates?

I have attached an image of the arrangement for reference.

It would be great if someone can assist with this.

Kind Regards,

 

[desertfox]

Hi Aytacogulu

In order to do this you would need to know where the centre of gravity is currently or have a reasonable estimate of its position, do you have that information?

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

 

[Aytacoglu]

Hi Desrtfox,

No I do not know the centre of gravity and that's what I am looking to calculate using the position of these skates. I do have the exact position of the skates.

 

[SWComposites]

Assuming the cg of the transformer is in the center of it, just conservatively assume the front single skate sees 50% of the total load.

 

[Enable]

Isn't this little guy the CG (pretty close to center but somewhat offset towards the front)?

 

[desertfox]

Sadly without the centre of gravity of the device you wouldn’t be able to calculate the loads on the skates. However I notice that there is a red and black circle within the drawing you show which looks like the symbol for the centre of gravity unless you have marked that on?
If I make the assumption that the centre of gravity is in the centre of the 2456 dimension then the 11tons divides by 2 and would equal 5.5 tons back and front, so now looking at the two skates at the back where the 1021 dimension is, then the 5.5 tons to the rear would each have 2.75 tons on each rear skid.
That’s making the assumption on the centre of gravity being dead central in length and width of the device, I have made that assumption based on looking at the drawing that everything seems symmetrical in side and front elevations, if that isn’t the case then the skids will have different loadings. Best I can do given the information.

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

 

[human909]

IMO play it safe. Forget COM games, use 4 skates and make conservative assumptions.

I recently did this with a 17T machine across an elevated steel deck flooring. For stability 4 skates were used and it was conservatively assumed that half the load 8.5T could end up on one skate.

Additional floor joists were added at 12mm steel plate was added everywhere that would be used as a rolling area for the skates. Rather than an operator assisted trolley, chain blocks were used to move the item. Sure it was slower but it was effective. The whole operation of removing one 11T item and replacing it with a 17T item was expected to take two days but we got it done in one. Which was great considering we needed to pay for a 250T crane to site.