Checking for overturn

[saintgeorges]

This question is not strictly related to Mechanical Engineering, it's more Industrial Design I guess, but maybe some of you know the answer.

I got puzzled for the last few days with an overturn check, and decided to ask for help online.


I used to do an analysis for for overturning effect of chairs and tables, by hand based on 2d support walls overturn check from the structural engineering. This is an example of bench:

("A" point represents the reference point for overturning)
Overturn condition:
F1 * l1 ≤ 1.5 * F2 *l2


But what should I do when the objects are not planar (planar but extruded in third dimension like this bench)?
Here is an example of a 3d irregular shaped object:

It is not possible to check for overturn this kind of irregular shaped object by hand. Because forces no longer lie in the same plane:

Some other combinations:

Can anyone help me with this issue?
How to check objects for overturn, when their overturn moments do not lie in the same plane?

Thank you for the reply.

P.S.

here is .3ds file of this object:

http://www.mediafire...qxx7gqp1ndt9s6x

 

[rb1957]

what is the red line on the right of the polygon (in the top view) ? (it doesn't look like the ground contact).

sorry but this is easy ... calculate the moments about the critical contact line for each force. there are ways to simplify this for simple loads. there are ways to demonstrate stability for simple loads (that don't work well with complicated loads). it's easy to determine the component of force causing a moment about a line, and hence the moment the load causes about the line.

safety factor is also an easy concept. it is a factor that can be applied to all loads to create a critical loading. say you have a bolt with a shear allowable of 1000. if you have two loads applied (in x- and y-directions) 500 lbs each, the resultant is 707 lbs and the safety factor is 1.414.

The other way of calculating a safety factor is to apply it against the allowable; what factor reduces the allowable to equal the applied. this makes sense in the example, and also in your first loading (when it seemed you had one fixed stable moment and one variable moment) but later loadings don't readily allow this type of SF calc. so you usae the previous suggestion.

 

[BAretired]

saintgeorges,
I thought I had responded to your last post, but either it has disappeared or I did not post it correctly before.

We are not neglecting anything. Wp is a vector parallel to the overturning axis. As such, it does not contribute to overturning in this case, but it would contribute if we were considering a different overturning axis.

A more critical condition would be to consider wind normal to the axis under consideration.

rb1957,
The red line on the right is a ground contact line. If you look at earlier sketches, you will see an additional diagonal support not visible on the plan view.

BA

 

[rb1957]

yes, i see that it is a line between two contact points (back on 15th Sept, 06:34).

my suggestion, too late now, is to examine how to solve problems with simple examples first.
understanding how something tips over is one thing, understndin the calcs is another, see how to apply forces another ...

once you understand the basics (moments about a tipping line) then you should be able to see how to apply this principle to other loads.

i think a key thing to appreciate is to ask yourself how can this thing fail ? how can it tip over ? what loads would it sensibly see to make it tip ?
this will lead you to consider wind loads acting in the worse direction, to apply distributed loads an a conservative manner, ...

 

[rb1957]

and in all this we haven't mentioned "is the structure (and the support) strong enough ?"

 

[saintgeorges]

Thank you for the reply rb1957.

All four red lines in top view are contact lines. I just bolded them in order to be seen more clearly.

I understood you about the moments calculation. But as W force in not perpendicular to the far right red axis, I need to decompose it into two forces - Wp and Wn. Where Wp is parallel to far right red axis, which means we neglect Wp. Right? We only include Wn * h + F2 * b into Overturning moment?

 

[rb1957]

yes, althought i'd quibble with your wording, we're not neglecting Wp, it's moment contribution is zero (neglecting implies that it's contribution is small, insignificant) it's fully correct to say the moment of W force about the tipping line is Wn*h.

and since we're playing "what if ...", if the wind was applied at 90 degs (against the other sides) it would be way from significant.

 

[saintgeorges]

Understood.

Thank you rb1957, BAretired, pax, I owe a couple of beers to all of you, for the help and patience you had.
I wish you all the best.