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Stability Calculations

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DanielMeidan

Mechanical
Jan 7, 2015
6
Hi everybody,
I designed a single post table consists of the following:
1 - Base plate with 4 glides.
2 - Post connected rigidly to the base plate with connection plate at the top.
3 - Table top plate (round).

This table is tested for stability by placing a weight off center.
The weight location around the center at the most unstable location.
ONLY vertical forces exist in this stability test (NO HORIZONTAL FORCES).
The loading force is known, marked as L.
Stability is measured as the ratio of required force to tip over the table, to the loading load L.
The requirement is to exceed the load L by minimum of 25%.

My question is:
Does stability depends in the height of the the table top from the floor.
Attached a PDF picture of the tested table.
According to my calculations height is not the factor, however test results given to me claim that the higher the table the lower the stability.
Can someone provide me with the static equations of the stability.
Thank for support

Daniel Meidan P.Eng.
 
 http://files.engineering.com/getfile.aspx?folder=99da9a63-ef18-4fb6-8f64-d9cfe6d91fe8&file=STABILITY_TEST_BASE_GRB22H28.pdf
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agree, height should not be a factor (presuming that it's enough to prevent the table from touching the floor).

stability would be critical if the pressure from the table base onto the floor went to zero. this should be enough for a P.Eng to figure out the details.

another day in paradise, or is paradise one day closer ?
 
In theory all forces are vertical and height doesn't matter.

In reality, if your test arrangement is not perfectly level, if deflections are sufficiently large under test loads, there will be horizontal reactions at the base.
 
I agree with the test results. Depends, of course, on the diameter of the center column. The higher the table the more stability would depend on how much the stand will buckle. In other words, you would have to add the deflection "δ" of the column to the eccentricity of the load.
δ = ε*(1 - cos(p*L))/cos(p*L) where ε = eccentricity, L = column length and p = (P/E*I)0.5
 
DanielMeidan,

Have you tried drawing a 2D orthogonal free body diagram? We all were taught this for a reason.

Have you considered the dynamic case in which your mass decelerates to a halt when it makes contact with the table top?

--
JHG
 
thx for cleaning up the threads !

could you repost your test results ... this is a purely static problem, right?

it sounds odd that for two tables identical except for height, that the taller one is more unstable (unless the table top is rotating significantly.

another day in paradise, or is paradise one day closer ?
 
When you have moved the center of gravity of the system to a distance from the central axis greater than the radius of the bottom plate your table will tip over. As mentioned by Occupant a taller table will tip over sooner because the central column and bottom plate will bend more under the same load causing your center of gravity to move outward and your table falls over under a lower load.
 
Stability test
Thanks everybody.
This is a simple static case, no moving parts and the table is horizontal.
From the equations that I used the height is not a factor.
As for test results, I get tip over or not, no numbers.
Please your comments.
Thnx

Daniel Meidan P.Eng.
 
 http://files.engineering.com/getfile.aspx?folder=13b2693d-4345-4852-860e-317408c06319&file=STABILITY_TEST_FIG_04.pdf
is G the weight of the table or the total weight (table+load) ?

you have the weight of the table (Wt) acting on the table center, the load (L) at some distance X, and the radius of the baseplate,(R). the CG of the two weights (table + load) should be inside the baseplate ...

CG = LX/(Wt+L) < R

another day in paradise, or is paradise one day closer ?
 
Hi RB1957
The weight of the table without the weight is G, and it acts along the table center line.
The weight is located at the distance R from the Table center line.
The distance X is the fulcrum which is a tangent line between two glides.
When the moment that the table provides (G * X) is greater then the load L applies L*(R-X) the table is stable.
I calculated the required load to tip over and the ratio to the test load is the stability.
In other words the over capacity that the table weight provides to overcome the eccentric load.
Hope I provided the full picture,
Please comment.
Thank you.

Daniel Meidan P.Eng.
 
rb1957 is right in calculating the balance point of the table. And any deflection in the center post or at the center post to baseplate connection will add to X. Resulting in tipping over of the table.
 
then it's LX/(L+G) < R ...

where's the CG of the two weights ? one weight(G) is at zero, the other (L) is at X ... G*0+L*X = (L+G)*cg

the only way i see height entering into it is if the floor (that the table is standing on) isn't horizontal.

another day in paradise, or is paradise one day closer ?
 
?? Isn't the weight cantilevered off the post? When you do the free body diagram, you should get a diagonal connection from the base to the weight. The weight decomposes into two vectors, one of which is perpendicular to the diagonal connection. That force vector x lever arm is torque, so a taller post is a longer arm, so more torque. The taller the post, the less stable it is. That's why top-heavy things are unstable.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
isn't the FBD the two loads and the reaction at the CG ?

thinking about it, the imbalance moment, Lx, is reacted by a "bending" stress at the baseplate, peak = (Lx)*R/(pi*R^4/4)
and when this equals the deadweight pressure, (G+L)/(pi*R^2), so it should start to tip when ...
4Lx/R = (G+L) ... Lx/(G+L) = R/4 ... that's a bit of a surprise ... if G = 3*L, then x = R ...

higher tables are less stable when you apply horizontal loads

another day in paradise, or is paradise one day closer ?
 
Hello rb1957 and IRstuff

There are two ways to calculate this stability case.
They both and in the same equation where the height is not a factor in the equations.
Please see the attachment.

From all the discussions today I may come to the conclusion that the test was not performed in the same way and details for all heights.

Thanks for the support

Daniel Meidan

Daniel Meidan P.Eng.
 
 http://files.engineering.com/getfile.aspx?folder=bf16adae-ab7b-49d1-8f96-a08c1bafa3bd&file=Stability_calculations_equations.pdf
Top heavy items are only unstable when the center of gravity to the base contact point is not a vertical line.

This is a simple problem of moments. Sum of the moments equal zero to balance the table on the edge of the base. Using the edge of the base as the zero distance.

Weight of table * Radius of base - Load * (X-radius) = 0
where X-Radius is the overhang of the load in relationship to the base radius and X>Radius for instability.

or

G*R-L(x-R)=0
for balance G*R=L(x-R) Looking at the equation, if X<R then stable, if X=R it is stable.

Solving for R to get the Radius of the base as a function of the load and distance from the central support.

R=LX/(G+L)

To fall over G*R-L(x-R)<0

and G*R < L(x-R)
R < L*X/(G+L)
 
I don't understand your calc ... I was using x the position of the load L (or F) for your R and my R is the radius of the base.

based on my "bending stress for imbalance moment" = dead weight "pressure" ... (Lx)*R/(pi*R^4/4) = (G+L)/(pi*R^2) ...
I realize this is the onset of tipping, that you can use a partial disc with a linearly varying pressure to react the dead weight and the imbalance moment (basically having the resultant act at the CG).

I think it'll tip before Lx/(G+L) = R


another day in paradise, or is paradise one day closer ?
 
Hi rb1957

Here are some clarifications
1 - The table will tip over around the line tangent between two glides.
This is the pivot line or the fulcrum.
This is the point A shown in the sketch.
The distance from this line to the center of the table is X.

2 - The total weight of the table G (without the test load L), applies a moment of M1=G*X

3 - The test load L is located in the distance of R from the table center.
The distance of the test load L to the pivot line or fulcrum is R-X
The test load L applies a moment of M2=L*(R-X)

The table will start to tip over when M1=M2
If M1>M2 then the table will not tip over.
In this case, the center of gravity of the table with the test load, is between the table center line, and the pivot line.
I hope this explains my calculations.
The point of this discussion was to get other professional opinion that the height does not affect stability.
Thanks.





Daniel Meidan P.Eng.
 
You are assuming all the weight is at G, which ought not be the case. The restoring moment, that keeps the table from tipping over should be acting from the center of mass, which is where the height has to appear.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
You might want to look at thread559-154454

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
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