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

 

[BAretired]

If F is the resultant of F1 and F2, then if F is inside the polygon, you will not have overturning. However, if F2 is outside the polygon, you will not know the value of the safety factor against overturning unless you calculate it with the above equation.

It is also correct to say that if F2 is inside the polygon, there is no overturning moment, so it is not necessary to check the above equation.

I think that prex was also suggesting that if there is any doubt about the precise magnitude or position of F2, more conservative assumptions may be warranted.

BA

 

[prex]

Thanks BA, ditto!
saintgeorges, by "If the resultant of the external loads (plus weight) points inside the polygon" I meant "If the resultant of the external loads (plus weight) is pointing inside the polygon": points is here a verb, not a plural noun! Note that this statement is true independently of the actual direction of the forces (being them vertical or not).

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]

I see. My bad sorry.
But again my question will be more or less the same:

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

?

 

[prex]

Yes, but the value of SF is up to you, 1.5 may be small, may be high, it depends on your confidence on value and position of external loads. If you were sure that you have an envelope for your loads (this is normally the case when calculating a structure), then a value of 1.1 could be sufficient.

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]

Consider a hypothetical case where F1 and F2 are gravity loads occurring on one edge of the polygon. The resultant of F1 and F2 will also occur on the same edge of polygon.

Although theoretically there is no overturning moment, such a situation is unacceptable because a gentle breeze could blow it over. The resultant of all forces must point to a position sufficiently inside any edge of the polygon to prevent overturning.

BA

 

[saintgeorges]

Here is the resultant of self weight and external load I got (orange circle):

It seems that this approach is better as I do not know which safety factor should I use in the upper equation (F1 * a ≥ F2 * b * SF).
In this approach I just need to determine whether resultant of F1 and F2 is outside of polygon or on it's edges, which automatically means that overturn will occur
?

 

[rb1957]

once you're inside the polygon it's stable, no matter the SF you apply to the load.

what you might try is ... what if the load isn't completely uniform? what if the load peaks towards the edge?

 

[BAretired]

It must also be recognized that the structure illustrated above may become unstable by virtue of its narrow base dimension normal to the critical rotation axis.

A horizontal force applied above the floor would change the direction of the resultant force. If it changes it enough to intersect the floor outside the base polygon, the structure will overturn.

Most building codes specify a minimum lateral design pressure of 5 psf on interior partitions to account for the possibility of air movement. To assume zero horizontal pressure is not a realistic assumption.

BA

 

[PEinc]

I thought this was a chair design. Why would we consider building codes?

http://www.PeirceEngineering.com

 

[BAretired]

Is it a chair or a table? I thought it was a table. In any case, PEinc, you are correct in suggesting that building codes do not govern the design of furniture. They do, however indicate the possibility of lateral forces acting on interior elements. Accordingly it might be prudent to consider a minimum lateral wind pressure acting on the chair or table.

BA

 

[saintgeorges]

This is not an actual object. I made it up. So it can be a chair or a table. I just want to understand the principles, for further calculation on actual objects (either chairs or tables).

About the furniture codes: My cousin works in a furniture factory.
They use the same codes structural engineers use for the wood. Only difference was in intensity of loads.
The problem is that all the shapes of the furniture were very strict, not so irregular formed. So there was no need for checking on overturn.


I did not now about those 5 psf. I though I only need to apply vertical load. Thank you for the information BAretired.

But if I am going to apply the horizontal force too, than the approach with finding whether resultants of loads act inside of the polygon is not possible. As horizontal force will act in the plane which is perpendicular to my polygon plane, there for I can not use it in order to check wether it is inside or outside of polygon:

So that returns us back to previous approach with moments: F1 * a ≥ F2 * b * SF

Right?

 

[paddingtongreen]

Can you be sure that the loads will land simultaneously? otherwise...

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

Wrong! You can consider the horizontal force. If a horizontal force H is applied at the top of the furniture item and the sum of gravity loads is F, the resultant is at an angle to the vertical of tan^-1(H/F).

For example, suppose F = 100# and H = 25#. Then tan^-1(H/F) = tan^-1(0.25) = 14^o. The resultant has a value of 103# and slopes at an angle of 14^o from vertical. Since the wind can blow in any direction, you would assume it blows normal to the axis of rotation.

If the height of furniture is h, your orange circle moves by a distance 0.25h. If we assume h = 29" (the height of my desk), your orange circle moves 7.25" normal to the axis of rotation. If it ends up outside the polygon, the furniture will overturn.

It will be more critical without the live load F2 because F2 is inside the polygon and contributes to the stabilizing moment.

BA

 

[prex]

As implied by BAretired's (correct) way of reasoning, if you calculate the resultant of all the loads (in the most unfavourable combination, as explained by BAretired), then the condition of stability is that the resultant vector should cross the floor inside the polygon. The safety factor could be also included by saying that the landing point should be a given distance far from the sides of the polygon: this way of reasoning would allow to automatically exclude those cases where the polygon is too narrow to insure sufficient stability.

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

 

[rb1957]

BA's completely correct in suggesting that adding a horizontal load will increase the overturning moment (by H*h, H = horizontal force, h = height). conservatively H acts to increase the overturning moment ... it could act to reduce it (really trying to tip the structure the opposite way).

now instead of just looking at a plan view of the loading and seeing if the resultant acts within the contact polygon (1st one to make a parrot comment has to buy a round in the pub), not you have to look at the ground plane and project the line of action of the resultant ('cause now we're dealing with a load vector with two components).

 

[saintgeorges]

Thank you for the replies.

But I am not sure I understand.
Is this the force (Wa) you are talking about:

(check this link for larger image)

?

"Wa" creates the moment around which of these four red overturn axes? a, or b?

 

[prex]

Well, it seems that your are mixing up things a bit, now.
1)Assuming your orange force F is where you represent it (right view) and W is an horizontal load also acting there, then Wa is the resultant of the two and, as it points outside the polygon, the object will overturn, and it will turn around axis b.
2)However your F orange load is also depicted elsewhere, inside the object. If this other position is the correct one, then your vector sum in the right view is completely incorrect. It is difficult to explain here all the basics of vector operations, you should consult a basic book on statics.

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]

Wa is the resultant of F and W. F is fixed in direction, namely vertically down. W (or H as I called it previously) lies in a horizontal plane but can be in any direction depending on which way the wind is blowing. The magnitude of Wa is (F² + W²)^1/2.

The resultant force Wa creates moment about every side of the polygon a, b, c and d but the one which we have been considering until now is side b because that was thought to be critical for overturning.

If resultant Wa intersects the floor outside any side of the polygon with wind blowing normal to that side, the furniture will overturn about that side.

Sides a and c do not appear likely to be a concern assuming your piece is drawn to scale but overturning appears possible about either of sides b or d.

BA

 

[saintgeorges]

Thank you.

I think I am finally starting to understand.
I took a wind coming from the north part of the object.
Please correct me if I am wrong:

(larger image link)

Orange arrow (F) represents sum of vectors selfweight (F1) + vertical external load (F2)
Green arrow (W) represents resultant vector of horizontal wind load
Pink arrow (V) represents sum of vectors of selfweight + vertical external load (orange arrow) and horizontal wind force (green arrow).

For an anchor point of green vector I took the center of gravity of the surface on which the wind load acts.
I got an anchor point for Pink vector, the same way I got it for Orange one, using this method:

link

Intesity of the Pink vector is based on what BAretired said: (orange vector + green vector)1/2.

Intersection point between the support plane (plane where these 4 read overturn axis lie) and direction line of the Purple vector, shows that Purple vector "acts" inside the polygon. Which means the object is stable for this direction of wind.

Right or am I wrong again?

 

[saintgeorges]

If above image seems not clear, I recorded a video with different angles on the model:

http://www.youtube.com/watch?v=Ib8jRLh5jR8&fmt=22