×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Checking for overturn

Checking for overturn

Checking for overturn

(OP)
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

RE: Checking for overturn

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

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

(OP)
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.

RE: Checking for overturn

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.

RE: Checking for overturn

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

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

(OP)
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).

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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

?

RE: Checking for overturn

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

RE: Checking for overturn

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

BA

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

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.

www.PeirceEngineering.com

RE: Checking for overturn

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

BA

RE: Checking for overturn

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?

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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.

RE: Checking for overturn

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

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

(OP)
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)

?

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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
?

RE: Checking for overturn

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?

RE: Checking for overturn

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

RE: Checking for overturn

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

www.PeirceEngineering.com

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

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.

RE: Checking for overturn

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) = 14o. The resultant has a value of 103# and slopes at an angle of 14o 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

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

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).

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

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

RE: Checking for overturn

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

RE: Checking for overturn

This is not correct:

Quote (saintgeorges)

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

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

(OP)
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.

?

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

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.

RE: Checking for overturn

(OP)
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.

RE: Checking for overturn

Quote (saintgeorges)

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

RE: Checking for overturn

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
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Checking for overturn

(OP)
@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.

RE: Checking for overturn

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 ?

RE: Checking for overturn

(OP)
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.

RE: Checking for overturn

(OP)
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.

RE: Checking for overturn

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.

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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.

RE: Checking for overturn

Quote (saintgeorges)

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

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

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.

RE: Checking for overturn

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

RE: Checking for overturn

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, ...

RE: Checking for overturn

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

RE: Checking for overturn

(OP)
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?

RE: Checking for overturn

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.

RE: Checking for overturn

(OP)
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.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources