Tipping Over a Table: How Much Overhang?

[kevelev]

TL;DR Summary

Designing a table and need to know how much overhang for certain weight, or how much weight for certain overhang.

I'm designing a table and need to know how much overhang I can have. Assuming the table is completely rigid, is symmetric, and has weight W acting on the CG, how much overhang x can I have? For instance, I know that zero overhang could have L = ∞ (if completely rigid) as the load wouldn't be outside (i.e. to the right of) the tipping point TP. I also know that as L increases, the CG will shift to the right, meaning the distance a will decrease. So the tipping point will be when a = 0. Is there an equation that incorporates the changing tipping point?

 

[berkeman]

Welcome to PF.

Why do you want an overhang on this table? Is it for artistic content, or some practical reason? Can you show us some sketches of your work so far?

 

[kevelev]

 

[kevelev]

Couldn't get the image to work at first. Overhang is for artistic reasons yes.

 

[berkeman]

Well, it also leaves room for chairs and such when people are sitting at the table. Can you show us your FBD for the table to get the equations involved in the Tipping Point calculation?

 

[kevelev]

I posted the image with the known forces.

 

[berkeman]

Well, can you post the equation for the sum of Moments, and say what it takes to avoid tipping it over with the extra downward edge force?

 

[hutchphd]

The equation where it doesn't tip is given by $$Wa\gt Lx $$ Or if you prefer $$L \lt W\frac a x$$. These both say that the "lefty" torque in the picture must be bigger than the "righty" torque about the obvious tipping point. That make sense?
EDIT: Sorry I stepped on @berkeman and gave it away...

 

[kevelev]

The equation where it doesn't tip is given by $$Wa\gt Lx $$ Or if you prefer $$L \lt W\frac a x$$. These both say that the "lefty" torque in the picture must be bigger than the "righty" torque about the obvious tipping point. That make sense?
EDIT: Sorry I stepped on @berkeman and gave it away...

Seems there was an issue with the formatting of the equation...I'm not sure what it's saying.

 

[hutchphd]

I will hand this to Mr @berkeman

 

[berkeman]

No, no, you are doing fine.

 

[kevelev]

If we assume the table was massless and only forces W and L were acting on it (as well as a reaction force at the tipping point), then we could say Wa = Lx and then solving, then max overhang x = Wa/L. The problem is that the table isn't massless so if a 200 lb person sits on the end, then the new CG of the table + the person will shift to the right.

 

[berkeman]

Seems there was an issue with the formatting of the equation...I'm not sure what it's saying.

It's the sum of Moments equation that I was alluding to. It doesn't make sense?

 

[kevelev]

No I got the sum of the moments. It was just hard to read with all the dollar signs and strange formatting.

 

[berkeman]

If we assume the table was massless and only forces W and L were acting on it (as well as a reaction force at the tipping point), then we could say Wa = Lx and then solving, then max overhang x = Wa/L. The problem is that the table isn't massless so if a 200 lb person sits on the end, then the new CG of the table + the person will shift to the right.

Well, the drawing makes it look like the table has plenty of mass. And yes, whatever situation you envision (distributions of loads), that will modify your FBD and sum of Moments.

 

[berkeman]

No I got the sum of the moments. It was just hard to read with all the dollar signs and strange formatting.

Oh, got it. Yeah, the PF software uses something called "lazy LaTeX" rendering, which doesn't render it until you refresh your browser once. After that it should be rendered correctly (at least for that particular thread).

 

[berkeman]

(BTW, there is a useful "LaTeX Guide" in the lower left of the Edit window in case you would like to learn how to post math at discussion forums)

 

[kevelev]

I have a CAD model of the table. With a known density, it calculates the weight of the table and center of mass. I also created a CAD model of a large lead weight and placed that on the edge of the table. That made a new CG and I was making sure that the point of the new CG was still within the edges of the base of the table. This is one way of doing it, but I wanted to have a formulaic approach rather than one by trial and error of varying lead weights and overhangs.

 

[kevelev]

Long story short, the table width is 48", the base is 36". That's a 6" overhang on each side. The table weight is about 350 lbs. My trial and error approach shows that with my design I can have about 1200 lbs. at the edge before it tips.

 

[berkeman]

This is one way of doing it, but I wanted to have a formulaic approach rather than one by trial and error of varying lead weights and overhangs.

Yeah, the formulaic approach is just to use the sum of Moments equation.

 

[kevelev]

So the size and weight of the table are a factor for it's stability in this case.

 

[berkeman]

Yes, absolutely!

 

[kevelev]

*huge factor...

 

[hutchphd]

I thought you meant that TP was the "edge" of base of the table. You want that (call it the "base edge" to be right of the "balance point" of the force moments (torques). I am assuming W is the weight of the table and L is the Load (person's weight). Thats what the equations say.

 

[kevelev]

*huge factor...