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

 

[CANPRO]

only difference is now you can tip in a couple of difference directions. Each force will produce an overturning moment on the x and y axis. Should be able to find an example in a statics text book

 

[prex]

What you miss is that overturning is not around a point, but around a line. This line must be chosen to be the most unfavourable one; in your sketch it should join your red point to the opposite corner of the other support wall.
Then you take the moments with respect to the line.

prex
[URL unfurl="true"]http://www.xcalcs.com[/url] : Online engineering calculations
[URL unfurl="true"]http://www.megamag.it[/url] : Magnetic brakes and launchers for fun rides
[URL unfurl="true"]http://www.levitans.com[/url] : Air bearing pads

 

[saintgeorges]

Thank you for the replies, both of you.

@prex:
Is this the axis(line) you are talking about:


Until now I always took the point with the larges compressional stress on soil/ground, as my overturn reference point. This was related to 2d objects (extruded in third direction). But now I understand that it was in fact not a point, but a line. For example this point "A" is not a point, but a line:

Right?

So in this case, the overturn line (axis) I am looking for should look like this:

?
But I am not sure that opposite corner of the other support wall, is the place where the second largest compression stress on ground will appear?!
Now I am confused. Should I get my overturn line/axis by simply connecting two points with the largest compression stress on the ground, or not?




@caneit:
Is this what you are talking about, when you mentioned x and y axis:

?

Then moments from F1x and F2x are not dangerous as they act against the overturn reference point.
But moments caused by F1y and F2y are dangerous. But in this case both F1y and F2y forces are acting in the same direction, not opposite in comparison to the point "A".
So what should I do?

Can those principles from the statics text book be applied to objects like chairs and tables? Because of the lifting forces that will appear on some parts of supports, and lifting foundations is not desirable in structures. But in here I do not have foundations, just table/chair legs?!


Thank you.

 

[frv]

prex,

rXd in three directions must equal zero. Not sure why it would matter whether you define a line or not. You can take the sum about any point you wish.

 

[BAretired]

Consider each force and its distance 'n' normal to the overturn axis. If the force lies outside the axis, n is positive. If it is inside the axis, n is negative. If it is on the line, n = 0.

If Σ F*n is positive the unit will overturn. If it is negative, it will not. If it is zero, it will be borderline, i.e. the least additional positive moment will cause it to overturn.

BA

 

[prex]

Right, saintgeorges, that is the line I was thinking of.
To make such an analysis, the best is to introduce the concept of the polygon circumscribed to the support points (this approach being valid of course for a rigid structure, not for a mechanism). This is a polygon that includes all the points of contact with the floor and is composed of lines that may become each one in turn an overturning line. By calculating the moment of the external forces wrt each one of the sides of the polygon you'll be able to determine which one might become first the hinge for an overturning.
As far as the soil pressure is concerned, this is a separate matter. For a given overturning line, you'll have to determine the total vertical component of the loads and then how this can be distributed on the points of contact with the floor available along the line under analysis. This distribution will in turn depend on the position of the vertical resultant of the loads and also on the stiffness of the structure if the points of contact are more than two.

prex
[URL unfurl="true"]http://www.xcalcs.com[/url] : Online engineering calculations
[URL unfurl="true"]http://www.megamag.it[/url] : Magnetic brakes and launchers for fun rides
[URL unfurl="true"]http://www.levitans.com[/url] : Air bearing pads

 

[saintgeorges]

Thank you for the replies.

@pres: Is the center of that circle you are mentioning the center of the gravity of object? Because I can determine the center of the gravity with FEA applications, which I already did in here - for both self weight of the object (blue arrow on images) and external surface load (yellow arrow).

Another problem emerges: I have four line which I got from the supports. And I can not circumscribe an irregular shaped quadrilateral into circle. So I approximated quadrilateral into triangle:

?

Well it seems that both center of gravity force and surface load force, are on the same side of the overturn axis, and inside the triangle. Which means the object is safe when it comes to overturning?
Or am I wrong?

 

[saintgeorges]

About the soil part: the problem I have is with chairs and tables. So no soil is included, just solid ground bellow chairs and tables. I mentioned the soil issue, because all of this overturning checking, I copied from the overturning of a support wall (because I do not know any other method).

 

[prex]

Well..., I mentioned a polygon, not a circle!
Draw the convex polygon (normally a quadrilateral, possibly a triangle) that encompasses all the support points: if the CoG of the self weight falls inside the polygon, the object is stable under its own weight. If the resultant of the external loads (plus weight) points inside the polygon, the object is stable under the external loads.
And of course it will be more stable when the point where those resultants fall on the floor is far from the sides of the polygon (but inside it).

prex
[URL unfurl="true"]http://www.xcalcs.com[/url] : Online engineering calculations
[URL unfurl="true"]http://www.megamag.it[/url] : Magnetic brakes and launchers for fun rides
[URL unfurl="true"]http://www.levitans.com[/url] : Air bearing pads

 

[BAretired]

I agree with prex. If the c.g. of all gravity loads falls inside the polygon connecting the floor supports, the furniture will be stable under gravity loading. A horizontal force applied above the floor, however, could render it unstable under combined loading.

BA

 

[saintgeorges]

My bad prex, you are right. You did not mention the circle. Sorry for that.
So this is what you are talking about:

(I bolded the quadrilateral sides in the Top view, in order to have better view)

Blue vector represents resultant of self weight, and yellow vector resultant of external load. So it seems that this object is stable, right?



But what if the resultant of external loads is outside of polygon? Does that automatically mean that regardless of the intensity of that resultant and intensity of resultant of self weight and it's normal distances from the particular quadrilateral side/axis - the object is not safe?
An example:

Does this automatically mean that this object is not safe for overturning?

Or is there a condition:
F1 * a ≥ F2 * b * SF

where SF should be some safety factor, let's say: 1.5

?

 

[rb1957]

the resultant is the resultant of whatever forces you consider important, include weight if you think it's important; since weight will always be inside the contact polygon (ie alsways stabilising) it should be conservative to ignore it.

the resultant of your forces A & B will be on a line between A & B. if it is inside the polygon you're stable. if you want to graphically include your SF, reduce the size of the polygon.

going back to your original overturning result ...
"Overturn condition:
F1 * l1 ≤ 1.5 * F2 *l2"

by geometry overturning occurs when F1*l1 = F2*l2, no?
but your expression says (for the same F2) a larger F1 (or l1) is stable, yes?
which sounds wrong to me, i'd've thought it was F1*l1 < F2*l2/SF ... this gives you a smaller F1 for a given F2

 

[BAretired]

The self weight of the table can be considered a counterweight, so the equation:
F1 * a ≥ F2 * b * SF
seems reasonable.

BA

 

[prex]

Self weight should always be included, as it will be present all the time. The stability condition will consider both the self weight and the external loads, as you propose.
However the safety factor should be evaluated considering other factors: the confidence on the fact that the external loads are really maximums (in value and position) is important. If the value is known, but the position is less certain, a way of assuming a safety factor would be to place the resultant on the far edge of the table, but other strategies are possible.

prex
[URL unfurl="true"]http://www.xcalcs.com[/url] : Online engineering calculations
[URL unfurl="true"]http://www.megamag.it[/url] : Magnetic brakes and launchers for fun rides
[URL unfurl="true"]http://www.levitans.com[/url] : Air bearing pads

 

[PEinc]

Ignoring the weight of the chair, Mot = F1 x L1 and Mr = F2 x L2

If SF > 1.5, Mr > 1.5 x Mot Therefore, F2 x L2 > 1.5 x (F1 x L1), which is not what the original post indicates.

http://www.PeirceEngineering.com

 

[BAretired]

Not so.
Ignoring the weight of the chair, Mot = F2 x L2 and Mr = F1 x L1


BA

 

[rb1957]

i agree with Pierce (and BA). BA is solving for SF = 1 ... when the overturning moment = the restoring moment ... F1*l1 = F2*l2.

if we want to apply a SF then it should be making the situation more stable, yes?
assuming F2 fixed and F1 varible then the safety factor should reduce the applied load (F1) ... F1*l1 = F2*l2/SF
or the SF can be applied to F2, increasing it ... F2*l2 = SF*F1*l1

no?

 

[BAretired]

F2 is the applied load. F1 is the dead weight.
The overturning moment is F2*b (or F2*L2)
The restoring (or stabilizing) moment is F1*a (or F1*L1).

To provide a safety factor,
F1*a ≥ F2*b*FS
OR
F1*L1 ≥ F2*L2*FS

Sheesh!


BA

 

[saintgeorges]

I would like to thank you for all the replies.

The reason for all this confusion is my fault, as I mixed the colors of the resultants and markings on first object(chair) and on second object.
So these are the correct markings and colors for both objects:

F1 - resultant of the self weight (dead load)
F2 - resultant of the external load
a - normal distance from F1 to reference overturn axis
b - normal distance from F2 to reference overturn axis
red point A - reference axis for overturning
red lines - reference axes for overturning


I guess now it is clear that:

F1 * a ≥ F2 * b * SF

where SF (Safety factor) should probably be ≥ 1.5 ?


But I am not sure I understood prex when he said:
"If the resultant of the external loads (plus weight) points inside the polygon, the object is stable under the external loads."

Does this mean, that if the resultant of the external loads points are inside the polygon, then I do not need to check for the overturn.
But when it is outside of the polygon (like in this case) then I need to check for an overturn using above equation (F1 * a ≥ F2 * b * SF)

?


Thank you.