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]

For overturning about polygon side b, try wind from the west with and without F2. Without F2 should govern.

For overturning about polygon side d, try wind from the east without F2.

BA

 

[saintgeorges]

Thank you for the reply BAretired.
But the principle I described in my last two replies is correct or not?

 

[BAretired]

This is not correct:

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

I said this: Wa = (F² + W²)^1/2

Or alternatively, Wa = √(F*F + W*W) (the pythagoras theorem).

Purple arrow was not defined. Did you mean Pink arrow?

Otherwise, I think you have the right idea.

BA

 

[saintgeorges]

Sorry I forgot the squares. Yes pythagoras theorem.
And yes pink arrow (English is my second language, so purple and pink seemed the same to me. Now I see it is not).

Main reason got me confused in this last part was an anchor point for the wind vector. It should be located (that anchor point) on the surface where the wind load acts. Right?

 

[prex]

A fundamental concept in vector theory is that a vector has a line of action, but has not an anchor point. To combine two vectors their lines of action must first of all intersect; if they do not, you need to decompose one of the two in order to have two components that intersect and that you can combine, and you are left with another component that will be considered separately. If the latter is hopefully of little magnitude you could possibly neglect it.
So you correctly determined the wind load (green arrow) as acting perpendicular to the surface of interest and passing through the CoG of the same surface. But now the green arrow and the orange one do not intersect: the simplest way here to combine them is to displace a little the green arrow horizontally to the east so that the two will intersect. Now you slide both arrows to the point of intersection and combine them as explained by BAretired (note that the Pythagoras theorem holds only for vectors at right angles, but this is the case at the moment). The combined vector has its own line of action that you extend down to the floor to determine the condition of stability.
And of course BAretired has a very important point: the most unfavourable condition should be with wind from the west (or, better said, acting perpendicular to side b of the polygon) with or without F2.


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 prex.

But why I can not use this method:

link

to get the resultant (pink arrow) of the sum of orange (selfweight + external vertical load) arrow and green arrow (wind load)?

In that way I will determine the point in which that resultant will act.
I did not know about that vector has a line of action, but has not an anchor point.
But nevertheless we used the centers of gravities until now as an anchor points or points of impact for our vectors. Can we do the same for the pink arrow, by getting it's anchor point/point of impact?
And then calculate it's the direction and intensity from the pytagoras theorem.

?

 

[prex]

Your method holds for finding the CoG of two (or more) masses, but which is the mass associated with the wind action?
Follow my advice above: slide two forces, whose lines of actions intersect, to the intersection point and you get the line of action of the combined force. If they do not intersect try to gently force them to intersect, the approximation obtained could be acceptable.
By calculating the horizontal force as a wind action onto an irregularly shaped surface is splitting hairs: what would you do for the round legs of a table?
There's likely no way to define a rigorous procedure for this calculation, as the loads to check against will never be exactly defined or limited. This reminds me of when my son asked me if a bicycle could be very exactly calculated with FEM, and my answer was: Yes, you can fairly exactly calculate the stresses and deflections, but which are the design loads? Are you going to design for a cyclist of a given maximum weight passing over a hole of given depth and length, or what?
At your place I would fix a somewhat arbitrary amount (5%-20% of the vertical loads) for a horizontal force applied at the table level for a table and at the seat level for a seat and consider this as a design choice against which your pieces of furniture are designed. But of course this won't prevent someone from jumping onto the table and dance on 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

 

[rb1957]

first off, is the horizontal force real (it wasn't mentioned initially), or just something we're imagining to be conservative ?

2nd, why won't the wind affect the structure from the east or west side, were the effect on over-turning will be greater.

 

[saintgeorges]

My apologizes for my late reply. I had some health issues, still do.


@prex:
Thank you.
Do you mind if I disturb you a bit more?
I did not understand that part when you spoke about sliding the forces. Is this what are you talking about:

I slided horizontally the green vector from east to west (?!) until it intersected with the orange vector. You see to the east, but that would mean we are getting away even more from the direction of the orange vector?



quote:
"By calculating the horizontal force as a wind action onto an irregularly shaped surface is splitting hairs: what would you do for the round legs of a table?"

I am using application called Grasshopper to do that. It can get the center of gravity of a surface or volume of free formed or regular shaped surfaces/objects. Therefor I could determine the point where the resultant of that solid load will act.


@rb1957:

BAretired mentioned during discussion that a 5 psi lateral wind load should be included too. I was not aware of this. That is why I did not include it from the beginning.

And yes I will also include this force from all sides of the object with total of four load cases. I used it from the north just to make sure I understand the concept of calculation.

 

[BAretired]

BAretired mentioned during discussion that a 5 psi lateral wind load should be included too.

If you are going to quote someone, please quote accurately. I suggested a lateral wind load of 5 psf, not 5 psi which would be a ridiculous requirement.

A simpler method than this is to consider gravity loads separately from horizontal loads. Calculate overturning moments and stabilizing moments about any edge of the polygon, then determine the safety factor against overturning.



BA

 

[prex]

Well, I suggested to move a bit the green arrow (as you did) in order to get the lines of action of the two vectors to intersect: this operation is incorrect from the standpoint of vector operation, but if the movement amplitude is small, could be acceptable. Otherwise you should decompose first the green arrow into two forces or into a force and a moment, and we would get into a nightmare trying to explain how.
Now that the two forces intersect, slide them along their lines of action to the point of intersection: there you can combine them with the Pythagoras theorem and you get the resultant whose line of action crossing the floor will determine the 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

 

[saintgeorges]

@BAretired:

I owe you an apology. I mixed these two units.
I am using metric unit system, so I could not distinguish the difference between the psi and psf. I used an online conversion and saw the relation between these two is huge. Sorry for that once more.



@pax:

English is not my maternal language, so I did not know the real difference between sliding and moving, related to this problem. Sorry if it sounded like I was misinterpreting your words. But now I understand that "sliding" is related to moving the vector along it's line of action, and "moving" is related to moving the whole line of action (along with vector).

Pax, let me see one more time if I understood you:
I need to move the green vector horizontally (only green vector!) from east to west, like I did in previous post. Then I need to slide both green vector and orange one to the point where the lines of action of orange and green interest, and sum these two vectors with Pythagoras theorem. Like this:

?

I do not intend to repeat what you already said, just to see if I understood it.

 

[rb1957]

why does the wind only act in the Y-driection ? the x-direction would be much more unstable.

is the yellow point the center of mass or the centroid of the Y-face (the center of the airload) ?

is there no load, other than weight, in the z-direction ?

how big is this thing ?

 

[saintgeorges]

This is just an example I made up, not an actual object.
I am trying to understand the concept in order to apply it to future actual objects.
This will probably be some sort of table, let's say around 90cm (around 3 foot) in height, 260 cm in length (about 8,5 feet) and 160cm in width (about 5 feet).

I will apply a wind from the west too. This is just a single load case, with wind from north.

Yellow point represents the intersection point between the lines of action of orange vector (mass of the object) and green vector (resultant of the wind load).
This is what I did according to prex and BAretired guidance:
step1 step2

There is also an external vertical load, acting on that top most surface. Resultant of that load is represented with blue vector. Take a look..
But I did not know how to sum that blue vector with the orange one (mass of object). I tried using this method, but it seems it can be applied only to a two/several vectors representing the masses of two/several objects. Not two vectors one of which represents the mass of an object, and the other representing the resultant of the surface load. Thus I neglected that blue vector, as I do not know how to sum it with the orange one.

 

[saintgeorges]

I made a mistake: yellow vector represents the resultant of the surface load. Not blue!
Here is the image.
Sorry for that.

To bad this forum does not have an edit post function.

 

[rb1957]

in your last image, 'cause both lines of action are inside the ground contact polygon, the loading is stable. but this is the specific result of the specific loading.

BA posted awhile back, determine the overturning moment (restoring or tipping) for each force about a ground contact line.

I think you'll find that we don't like playing "what if". we prefer to have a specific problem, to help you solve.

 

[chicopee]

Why in the world would you want to buy a bench like that? I would not use that for assembly purposes, drawing purposes, fine and rough hand detailing etc... Perhaps for display purposes

 

[saintgeorges]

I am apologizing for the late reply.

@ rb1957:
There were two methods montioned in this post. One method was by - interesecting the lines of action of wind load resultant and objects mass resultant. And then summing those two vectors in the position where their lines of action intersected.

The other one was by calculating the overturning moments for each ground contact line.
But in that method I was confused with Safety factor. I did not know how to calculate it. Or which one should I take? 1.5? 2?
And one of wind resultant components would have to be neglected as it will be parralel to the ground contact line ( I am talking about the far right contact line, the one to the east).


@ chicopee:
This is just an example I made up in 5 minutes. It is not an actual object I am going to build. Just something that will help me understand the concept, and then use it on my future projects.

 

[BAretired]

The other one was by calculating the overturning moments for each ground contact line.
But in that method I was confused with Safety factor. I did not know how to calculate it. Or which one should I take? 1.5? 2?
And one of wind resultant components would have to be neglected as it will be parralel to the ground contact line ( I am talking about the far right contact line, the one to the east).

Let F1 be the unit weight. Suppose that F1 acts at distance 'a' inside and normal to the contact line in question.

Let F2 be the superimposed load. Suppose F2 acts at distance 'b' outside and normal to the contact line in question.

Let H be the sum of horizontal forces acting normal to the contact line in question at a distance 'h' above the ground.

For that particular contact line:
Stabilizing moment SM = F1*a
Overturning moment OM = F2*b + H*h
Factor of safety FOS = SM/OM

Whether you choose a minimum safety factor of 1.5 or 2.0 is a matter of engineering judgment. There is no code specifying a safety factor for furniture (or if there is, I am not aware of it).




BA

 

[saintgeorges]

BAretired, are we neglecting one of Wind load resultants components (Wp)?

F1 - resultant of the self weigh of object (dead load)
F2 - resultant of outer, external surface load
W - resultant of wind load
Wn - component of W, normal to overturning axis
Wp - component of W parallel to overturning axis


Also, did I draw the "h" distance correctly? Is that "h" you are talking about?