Wind gust duration to overturn a block

[hillbilly63]

TL;DR

How long does a force of H have to be applied to a block to push it beyond it's tipping point?

I'm familiar with statics and calculating the force that will start to tip a block (lets call it H). I'm also familiar with when a block will get to the point of tipping over, rather than sitting back down if the load is removed.

What I'm not familiar with is how to calculate the duration for which H would need to be applied to get the block past the point of tipping over.

I'm assuming an energy based approach is the way to, i.e. the centroid has to be lifted up a certain height in order to get to the tipping point, which can be equated to an increase in potential energy, but I'm not really sure where to go from there.

The image link shows the nomenclature I'm using, as well as force application (H is the total force applied to the vertical face of the block):

Simplifications/Thoughts:
- Assuming block won't slide
- In my application, the force is due to wind. For now, I'm happy to ignore the fact that the area that presents itself to the wind varies as the block tips, as potentially does the drag coefficient.
- Maybe the angle of force should remain perpendicular to the face (force H is due to pressure difference on front/rear surfaces and can only act perpendicular to the surfaces).

 

[erobz]

Probably $$\sum \boldsymbol {\tau} = \frac{d}{dt} \left( I \boldsymbol {\omega} \right)$$ is going to be helpful. Can you try to come up with some ideas about how to apply it? Probably energy too is useful, as you say $H$ is responsible for lifting the center of mass vertically above the corner of the box. I'm sure there is more than one way to skin the cat.

P.S. Is it just me or does

\mathbf{...}

not seem like its "bolding" the symbols $\tau, \omega$?

Probably should move this to physics homework. It will get better coverage there anyhow.

 

[berkeman]

Probably should move this to physics homework.

It doesn't seem schoolwork-like, so it's probably okay here for now.

 

[Lnewqban]

TL;DR Summary: How long does a force of H have to be applied to a block to push it beyond it's tipping point?

... the centroid has to be lifted up a certain height in order to get to the tipping point, which can be equated to an increase in potential energy, but I'm not really sure where to go from there.

That force H induces a moment about the bottom edge of the block that rotates the center of mass.
The weight induces another moment in the opposite direction.

The net moment has to fight the rotational inertia of the block during the completion of the geometrical angle needed to relocate the center of mass directly above the pivot or bottom edge.

The time you look for is based on the magnitude of the achievable angular acceleration.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

 

[tech99]

It seems to me that when the block reaches maximum height with the lowest possible wind speed, it will be in a position of unstable equilibrium. We cannot quantify the increment of additional energy required to start it to tip further; neither can be quantify the duration. In principle, the duration could be very long.

 

[erobz]

It seems to me that when the block reaches maximum height with the lowest possible wind speed, it will be in a position of unstable equilibrium. We cannot quantify the increment of additional energy required to start it to tip further; neither can be quantify the duration. In principle, the duration could be very long.

Seems like a great point. Trouble in paradise. I was imagining it having some non-zero angular velocity at the equilibrium position, but that doesn't have to be true.

EDIT:
But wait, if $H$ is a constant force it's never going to be in equilibrium at that point because the weight is providing no counter torque there?

 

[Baluncore]

If the constant force is sufficient for the upwind edge to begin lifting, then it will accelerate from there towards the equilibrium. That is the minimum constant force required to overturn the block, but in the maximum time, since the initial acceleration is the minimum.

If the constant force terminates, part way to equilibrium, then the block may decelerate to a balance at the point of equilibrium. If either force or duration were greater, the block would be overturned.

If the constant force is greater, then less duration of force is required to reach the equilibrium with velocity ≥ zero.

The computation has become an energy equation, to reach the balance point, with a final velocity of zero.

 

[tech99]

The questions asks for the gust duration for the block to pass the tipping point. A far as I can see this is indeterminate.

 

[Baluncore]

The questions asks for the gust duration for the block to pass the tipping point. A far as I can see this is indeterminate.

The required gust duration, is a function of the constant force, applied during the gust.

 

[jrmichler]

The required gust duration, is a function of the constant force, applied during the gust.

A thought experiment is helpful to understanding how to set up the calculations.

Scenario #1: Assume a constant wind that applies a tipping moment exactly equal to the restoring force of gravity. The weight of the block will be entirely on the bottom right corner, but it will not quite tip. Now assume an infinitesimally small increase in wind force. The block will start to tip with infinitely small angular acceleration, so will take infinitely long time to fully tip.

Scenario #2: Assume infinite wind force. The block will tip in zero time.

Scenario #3: Assume a constant wind force in between Scenario #1 and Scenario #2. The block will tip in finite time. At the start of tipping, the block has a wind force moment of $H * b/2$, and a restoring force moment of $W * a/2$. The wind force tipping moment is greater than the restoring force moment, and the block has a moment of inertia about the tipping point (the right bottom corner), so it has an initial angular acceleration. As the block tips, the restoring moment decreases, so the angular acceleration increases.

TL;DR Summary: How long does a force of H have to be applied to a block to push it beyond it's tipping point?

how to calculate the duration for which H would need to be applied to get the block past the point of tipping over.

The necessary force duration is found by integrating from the time the block starts to tip until the center of gravity is vertically over the bottom right corner. As seen from the boundary conditions in Scenarios 1 and 2, that time can be anywhere between zero and infinite.

 

[erobz]

As the block tips, the restoring moment decreases, so the angular acceleration increases.

A confounding factor may be that the torque from $H$ is decreasing with increasing angle of tilt $\theta$. I should probably just do the math, but I'm being unmotivated.

 

[Baluncore]

A confounding factor may be that the torque from H is decreasing with increasing angle of tilt θ.

But during the approach to the tipping point, the area subjected to the gust, will be increasing as the block stands higher on its diagonal, in the wind stream.

Meanwhile, as the leading lower edge starts to lift, the ram air pressure from the gust begins to enter under the block, which increases the overturning torque.

 

[erobz]

But during the approach to the tipping point, the area subjected to the gust, will be increasing as the block stands higher on its diagonal, in the wind stream.

Meanwhile, as the leading lower edge starts to lift, the ram air pressure from the gust begins to enter under the block, which increases the overturning torque.

As far as the actual fluid mechanics go, I completely agree. But not in the constant force (magnitude and direction) simplified model. I was just making the point that there seems to be some not exactly clear "on the surface" variable changes happening with that simplification.

 

[Baluncore]

How long does a force of H have to be applied to a block to push it beyond it's tipping point?

As far as the actual fluid mechanics go, I agree. But not in the constant force (magnitude and direction) simplified model.

In the constant force H, model, your concern over the angle is irrelevant.

In a real fluid dynamics model, the change in slope of the face will be outrun by the increased section and the pressure below the block.

 

[erobz]

In a real fluid dynamics model, the change in slope of the face will be outrun by the increased section and the pressure below the block.

I'm not arguing with you, I said I completely agree. You don't need to double down on that.

 

[erobz]

In the constant force H, model, your concern over the angle is irrelevant.

My emphasis. Are you saying the torque doesn't decrease as angle inclines in the constant force model - meaning I should compute it? My point was that technically it would seem to complicate the argument in post #10, scenario 3 a little bit.

 

[erobz]

Actually I think it is increasing like $\sin ( \theta + \text{const.})$. So no problems there anyhow. I should have just checked it initially.

 

[Baluncore]

My point was that technically it would seem to complicate the argument in post #10, scenario 3 a little bit.

I agree, and we don't need to complicate the argument, once it starts to tip, it will continue, and accelerate faster, while the gust force is applied. We can eliminate all fluid dynamics, and stick to the constant horizontal wind force, H.

The question becomes, to determine the minimum gust duration, for any given constant force H, that will overturn the block.

The first phase is the acceleration due to excess force H at the start of the gust.
The constant force H, will be terminated later, giving the gust duration needed.

The second phase is the inertial deceleration, after removal of H, to almost stop at the tipping point. That can be computed backwards in time and angle, as if the block is falling from halted at the tipping point, back towards the initial position.

The angular rotation from the initial position, to reach the tipping point, is the sum of the angles swept during those two phases.

Both the acceleration and deceleration phases, are integrals, giving angular velocity~angle swept curves for a constant H.

We plot angular velocity against angle swept, with phase one accelerating from the initial origin forward in time, and the phase two deceleration backwards from the tipping point angle. Where the angular velocity curves cross, we can compute the time, which is the minimum gust duration for that given H.

The problem of the infinite time needed to lift off from the origin, or to almost come to a halt at the tipping point, is eliminated by plotting and considering the angular velocity against the angle swept.

 

[jack action]

The thing is that there is no such thing as a constant "wind force" $H$; there's a constant "wind velocity" $v$ though. The equation is then a function of the block angle $\theta$:
$$ H(\theta) h_{cp}(\theta) - mgd_{cg}(\theta) = I\alpha$$
Where:

• $H(\theta) =\frac{1}{2}\rho C_d(\theta)A(\theta)v^2$ is the aerodynamic force due to the wind
• $h_{cp}(\theta)$ is the height of the center of pressure
• $d_{cg}(\theta)$ is the horizontal distance between the center of gravity and the right corner (negative after passing the the tipping point)
• $mg$ is the block weight
• $I$ is the block rotational inertia
• $\alpha$ is the block angular acceleration

Some obvious boundary conditions exist, such as $\alpha \ge 0\ @\ \theta = 0$.

The basic math seems pretty straightforward to me. I don't see where there would be something indeterminate. It is determining all the functions of $\theta$ that is tedious.

 

[erobz]

The thing is that there is no such thing as a constant "wind force" $H$; there's a constant "wind velocity" $v$ though. The equation is then a function of the block angle $\theta$:
$$ H(\theta) h_{cp}(\theta) - mgd_{cg}(\theta) = I\alpha$$
Where:

• $H(\theta) =\frac{1}{2}\rho C_d(\theta)A(\theta)v^2$ is the aerodynamic force due to the wind
• $h_{cp}(\theta)$ is the height of the center of pressure
• $d_{cg}(\theta)$ is the horizontal distance between the center of gravity and the right corner (negative after passing the the tipping point)
• $mg$ is the block weight
• $I$ is the block rotational inertia
• $\alpha$ is the block angular acceleration

Some obvious boundary conditions exist, such as $\alpha \ge 0\ @\ \theta = 0$.

The basic math seems pretty straightforward to me. I don't see where there would be something indeterminate. It is determining all the functions of $\theta$ that is tedious.

I think we were just trying to get @hillbilly63 to engage with the simplest model that doesn't have obvious holes in it. Once they chime in we can see just how far they are willing to take it?

 

[DrClaude]

P.S. Is it just me or does

\mathbf{...}

not seem like its "bolding" the symbols $\tau, \omega$?

Standard $\LaTeX$ has no bold version of the greek letters.

 

[erobz]

Standard $\LaTeX$ has no bold version of the greek letters.

Thanks! I'm sure I've wasted time in the past trying to make them bold...

 

[DrClaude]

Thanks! I'm sure I've wasted time in the past trying to make them bold...

There is a package you can use in your $\LaTeX$, nut it is not part of MathJax so won't work on PF.
https://ctan.org/pkg/bm?lang=en

 

[renormalize]

Standard $\LaTeX$ has no bold version of the greek letters.

Huh??
$\alpha,\beta,\gamma,\text{... vs. }\boldsymbol{\alpha},\boldsymbol{\beta},\boldsymbol{\gamma},\text{...}$
($\alpha,\beta,\gamma,\text{... vs. }\boldsymbol{\alpha},\boldsymbol{\beta},\boldsymbol{\gamma},\text{...}$)

 

[erobz]

Huh??
$\alpha,\beta,\gamma,\text{... vs. }\boldsymbol{\alpha},\boldsymbol{\beta},\boldsymbol{\gamma},\text{...}$
($\alpha,\beta,\gamma,\text{... vs. }\boldsymbol{\alpha},\boldsymbol{\beta},\boldsymbol{\gamma},\text{...}$)

Nice! So they just require a different command.

 

[hillbilly63]

Hello all, many thanks for your replies. I will take some time over the Christmas break to explore and understand the suggestions and come back with my questions.

@berkeman: this isn't a homework problem, I'm a structural engineer and this is a real life problem! We typically design the structures based on simple static equilibrium and a FoS, effectively designing against any tipping. I'm interested in understanding what the real resistance is to total failure, given that peak gust speeds only occur for short durations.

 

[tech99]

In my job designing radio towers, for design purposes we would take the wind force from the 3 second once-in-50 year wind. The wind speed was slightly tapered down the structure. We made the foundations able to withstand the resulting uplift with a safety factor of 1.5. We also allowed for a certain weight of soil to assist the weight of concrete, this soil being a cone around the block. The steel members were designed to comply with the working stress in the code at that time, the main issue being compressive strength. I do not know how to deal with the question of dynamics posed by the question, but in practice I think that once the structure moves it has failed.

 

[Baluncore]

... , but in practice I think that once the structure moves it has failed.

That is true for a buried structure, since once the anchors have begun to fail, it becomes easier to continue the movement.

Once cracks appear in the soil around an anchor block, air can enter quickly under that block. Atmospheric pressure then appears under the block, and so it can lift more rapidly during a gust.

Where the anchor is set in deep wet soil, the anchor block has buoyancy that reduces its effective mass, but air cannot immediately get under the block. It will then be a strong, steady wind, that will lift the anchor through a wet fluid soil.

Anchors rely on a conical sucker of damp clay, deep in the soil, to survive gusts of wind. Failures of anchors are more likely to occur during a drought, when the soil dries out, and a wind gust dislodges the anchor.

The same is true of trees. Look at the roots of a fallen tree, notice how the roots form the shape of a sucker, like an upside down dinner plate, that seals against the damp clay of the deeper soil horizon. Trees are blown over during windstorms, especially during unusually bad droughts, when the sucker fails to seal in the dry cracked soil.

Take a look at these roots, to see the middle of the area of the "sucker" that sticks the tree to the ground. Then imagine what happens if the top of the fallen tree is cut off, while there is still tension in the remaining roots. The ABC are careful not to mention any cutting in the article, but a later picture shows the upright stump.
https://www.abc.net.au/news/2019-03...r-finding-son-crushed-by-fallen-tree/10937070

 

[Lnewqban]

... Trees are blown over during windstorms, especially during unusually bad droughts, when the sucker fails to seal in the dry cracked soil.

Even when air can't flow underneath the roots, Florida sandy soil is known for liquefying in floods and removing much of any tree's resistance to tipping moment for any tree.

https://www.accuweather.com/en/weather-news/why-trees-topple-in-high-winds/348991

Specially during hurricanes, there is previous period of intense rain before the eye of the hurricane arrives.
The soil then becomes saturated quickly and following hurricane-force winds can do much damage to big trees of foliage and tall palm trees.

The ABC are careful not to mention any cutting in the article, but a later picture shows the upright stump.

https://www.abc.net.au/news/2019-03...r-finding-son-crushed-by-fallen-tree/10937070

That is a sad story.