Tipping force for Tank Stand

[kos1]

Hello good people of Eng-tips,

I need your help with a unique problem I have encountered. I am designing a tank stand for our shop and I need to calculate how much force would be needed to push the tank over the stand. See attached picture for reference, I can provide additional dims if needed. Thanks in advance for any help/tip you might provide.

Thank you
FR

 

[dijonnaise]

Where are you getting stuck in solving this problem?

 

[EdStainless]

Draw the FBD.

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P.E. Metallurgy, consulting work welcomed

 

[ChorasDen]

Is a lateral force the only concern here? What about a tangential force 'rolling' the cylinder off the mounts?

 

[kos1]

In reality this is how the tank would be tipped over:

This is what i think the calculation would look like, but not 100% confident:

 

[dik]

Is the lateral force from wind?... if not, I'd consider a small nominal lateral load, in case someone left the door open. Is this an area with seismic events? There may be some tank standards that stipulate a minimal horizontal force.

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[dik]

The work out would be the weight of the tank times the height change in tank support.

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[WARose]

Assuming we are talking about a safety factor of 1.5.....i came out with 5697 N being the tipping force.

In actuality, I think sliding is probably more of a concern (unless you've decided the support will somehow stop that). Running with a coefficient of friction of 0.2 (and a sf of 1.5) I got the resistive force at 1306 N.

 

[BAretired]

I believe you are correct with F = W.a/h. Of course, F would be the pushover value, so you would want to stay a little below that, say 0.6F.