Unit Stability, torque vs height of CG 3

[BlankCW]

I am struggling with finding a good way to express the stability of a unit. I can look at the torque and say that the torque to keep the unit stable is greater than the torque that wants to tip the unit. But we all know that if the CG is at the top of the unit, it is less stable than a unit with the CG toward the bottom. The height of the CG is not part of the torque calculations.

Torque Calculation See attached PDF.

Moments that keep the unit stable
∑M_st=0=W(X_tp )
∑M_st = 4,800 in-lbs.

Moments that tip the unit
∑M_tt=0=P(h)
∑M_tt = 2,600 in-lbs.

Ratio of Stable Torque to Tipping Torque
M_st/M_tt = 1.84

So, I am wondering if anyone has a good approach to looking at the torque but also consideration for the height of the CG. Maybe there is a two-step approach; first look at torque then look at something else and if both pass you are good to go.

Any thoughts are appreciated.

 

[rb1957]

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[3DDave]

Torque is a really bad way to look at stability.

What is a good way is calculating the amount of energy that is required to tip it over. For the case when the CG is over the pivot that amount of energy is zero.

All other places look at the amount of potential energy that is stored as the item goes from flat on the floor. You didn't include a horizontal offset in that CG. If it was on the opposite side from the pivot it would resist tilting even more, at the same height.

You can resolve torque into energy by multiplying it by the rotation as measured in radians, but then you need to look at the real geometry.

In real geometry parts inside can shift causing the CG to move closer to the pivot, lowering the amount of energy required. For example, if the CG was low because it hangs on a pendulum. That might require more energy than if the CG was high and bolted in place.

 

[BlankCW]

3DDave-

Ok, so we calculate the amount of energy that it takes to tip the unit but what do I do with that information?

The images showing the unit “tipping” over was just to demonstrate how the height of the CG effects the stability, one with the CG toward the top, one at the middle, and one toward the bottom. I didn’t include the height of the CG just because it is not needed in torque calculations. We can consider that there is not “stuff” in the unit to shift the CG as it starts to rotate but I understand your point.

 

[3DDave]

When it goes from flat to just about to tip the CG goes from the initial height you don't care about to the distance from the pivot point to the CG. If the original height is 7 feet and the diagonal distance is 8 feet then that is a gain of 1 foot * weight of potential energy.

No torque involved.

If it weighs 100 pounds and you put in 1 foot*100 pounds of energy into the item, then it will just be at the tipping point. After that the potential energy goes down, causing the item to continue to tilt and either fall back or fall over.

 

[BlankCW]

3DDave-

Ok, so you are suggesting that I compare that amount of energy it takes to tip the unit, when the CG is over the tipping point, to the input force in this case P or 50 lbs?

Is that correct?

 

[3DDave]

Energy cannot be compared to force.

Do you know what potential energy is and how it is measured?

 

[desertfox]

Hi BlankCW

I would specify an angle of tip, so depending on the unit centre of gravity etc then : this unit should not be tipped more than twenty degrees from vertical.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[rb1957]

I thought the question was how much side force can I apply ? The box won't start to rotate until the destabilising moment (Ph) exceeds the stablising moment (Wd) ... whatever height of the CG. no?


"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[HTURKAK]

The stability should be expressed with two checks; ( F.S. could be chosen for )

- Factor of safety against sliding ≥ 1.5

- Factor of safety against overturning ≥ 2.0

-Factor of safety against sliding = ( Sliding resistance of the block )/ (Horizontal force )= W*fs/ P= 200*0.40/50=1.6 (assume the friction coeff. fs=0.4 )

--Factor of safety against overturning = Resisting moment/Overturning moment =200*2/50*4=2.0 O.K.










Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB

 

[Tmoose]

The description of "a unit" is pretty vague.

I'd spend a few minutes thinking about potential for -
-lateral acceleration during shipping or handling in the plant.
-Being picked up or jacked up by one edge. (angular displacement, when the height and position of the CG are forced into play)
-Being bumped by a fork truck or a load suspended and transported by a crane.
-etc.

 

[JStephen]

If the unit is perfectly rigid until it tips over, it won't matter where the CG is.
If there is some flexibility in the structure or in the base, you can do a p-delta analysis, and that will incorproate that difference. Basically, calculating the deflection, then calculating the additional moment due to that deflection.
If the tipping is due to wind or seismic, you can consider the properties in either calculation as well.

 

[BlankCW]

3DDave-

Yup I understand potential energy, but I don’t understand what you are suggesting. Are you suggesting that I calculate the difference in potential energy from the stationary position to when the CG is over the tipping point? Lets say that I do that, I am missing what that information would tell me about the stability of the unit.

Desertfox-

So I could specify an angle of tip based on the input force and then create rules that relate to the height of the CG. For example, if the CG is toward the top on the unit, that specified angle needs to be less than 10 degrees but if the CG is toward the bottom maybe that specified angle needs to be less than 20 degrees.

Is that what you are getting at?

Rb1957-

No I was not asking about lateral force.

Hturkak-

So that is how I approach the issue generally but I skip the Slip v Tip and just looked at the tip portion but base the input lateral force as what it would take to slip (f_s=μ_s N). What you are expressing, I think, shows the issue at hand. If we take the example that I had at the top our ratio (safety factor) is 1.84:1 just under your 2:1 factor of safety. So how do we fix that, we have two options; we can increase the width of the base or add ballast to lower the center of gravity. If we can’t increase the size of the unit our only option is to add ballast. If like in my example the unit is symmetrical, we can all the weight we want but because there is a relationship between the input lateral force and the weight of the unit that safety factor ratio doesn’t change. We inherently understand that a lower CG is more stable than a top heavy unit.

So my question is how can we express that?

Tmoose-

The term is meant to be vague because it is more of a “whiteboard” question then solve this specific problem. I am not sure those use cases are relevant to the question. Really those are just input forces applied at different locations which would definitely have an adverse effect on the unit.


JStephen-

I agree that when looking at only moments the CG height doesn’t matter and only proximity to the tipping point matters. Interesting thought, looking at the deflection in the base and including that moment. I am not sure the height of the CG has any influence in those calculations.

 

[desertfox]

Hi BlankCW

The answer to your question is yes, to my way of thinking is that somebody handling/hoisting the unit will have a physical reference for manipulating the unit safely.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[3DDave]

You compared force to energy, so I have some doubts.

Still, the more energy required to get to the tipping point the more stable the item is.

Did you notice the difference in height when the CG is lower is far greater than when the CG is higher, so the energy required is greater when the CG is lower?

Your "angle of tip" vs. force is the amount of force required to initially lift the item. The force at every point after that is lower because the amount of potential energy to rotate it at every angle increment is lower. But the force depends on the direction and location the force is applied. If the force is directly beneath the CG and lifts up it is highest. If it is on a crowbar it can be far lower.

The only thing that doesn't change is the change in potential energy as the item rotates, that's the only measure of stability one needs.

 

[BlankCW]

Desertfox-

Thanks for your feedback, I will keep exploring that approach.


3DDave-

Ignoring your misguided sarcasm because it is unproductive.

I agree that the more energy required to get the unit to the tipping point the more stable the unit is. How do you quantify that in a way that says this unit is stable or unstable. In torque there is a comparison between the “tipping torque” and “resisting torque”. Bring it back to my original question I think heigh of the CG should play a part in the stability of the unit and it doesn’t if you just look at moments.

For this example, the input force or applied force is horizontal.

 

[MintJulep]

In naval architecture the stability of a ship is described by its "righting moment curve" or "static stability curve".

The same concept works for describing the resistance of a thing to tipping.

Because the diagram is moment vs angle it's effectively an energy approach, as suggested by 3DDave.

Because it uses a moment, it takes away "what height are you pushing at" question. The CG height defines the shape of the curve.

With a curve you know everything there is to know. At any angle of tip you know how much torque is trying to return it to the original position. You know what angle the righting moment goes to zero and the thing tips over.

@BlankCW
I think the struggle you are having is that you are looking for "an answer". I.e. Stable is greater than some number. Unstable is less than some number.
That isn't going to happen. The best you get is a comparative analysis. "A is more stable than B", or "Stability is increased by this change".

 

[BlankCW]

MintJulep-

I think you are right in your assessment I am looking for “an answer” and that might just be misguided.

Thanks for your advice.