1. Mar 12, 2014

### stevewindmill

Hello everyone :)

I have a problem which I would love help with, very briefly I am having a model windmill built as an exhibition stand. As it's height is just over 3 metres I am being asked some pretty tough questions by the health and safety team at the show.

Specifically "what weight is required to stop the windmill overturning".

It's mainly built of honeycomb cardboard, but with a wood frame underpinning. I have calculated its weight, footprint, and have most measurements available to me, but the physics behind their question baffles me.

Facts I know:
Main structure 3.36m high
Total height including sails 4.5m high
Footprint of base 3.31sq/m
structure tapers and at a height of 3.3m the 'footprint' is 1.69sq/m
Total weight of honeycomb cardboard external shell 37kg
Timber internal structure 59kg
motor 5kg at a height of 3m
total weight of 101kg
43kg below a height of 1m
will add a weight at each corner of the base but need science behind how much.

We've talked about 20kg in each corner, adding 80kg to the total weight, but this is just based on guesswork sadly.

I'm assuming that the question will actually need more specifics - what weight is required to stop the windmill overturning is very vague. I'd like to be able to say that the weight of an average-above average person applied at a specific height of 1.5m to the side of the structure, or similar, would definitely not topple the windmill (with calculations to back it up). Input is appreciated here also - it's unlikely that someone could apply their entire weight with a sideways force at this height???

I'm really stressing out over this, the company designing and building the stand have been very helpful but when it comes down to this level of physics, we're all just staring blankly and scratching our heads.

I have lots of measurements and details here so if there's anything glaringly obvious that would help calculations just let me know.

2. Mar 12, 2014

### Travis_King

Wind blows against the windmill, this causes a force. That force (maybe taken to be at hub of the blades) is at a distance away from the mounting point (the base), this creates a moment. This moment must, in your case, be resisted by some weight at the edges of the base. Basically, they want you to tell them the expected wind load moment at the base, and then relate how much weight is needed at the edge of the base to prevent the mill from toppling.

3. Mar 12, 2014

### AlephZero

I think you need some professional advice (from somebody who has indemnity insurance if the advice turns out to be wrong!) about this. If structure of that size and weight falls over for some reason, it could easily injure somebody.

You have to consider the worst possible case. Suppose an evacuation alarm goes off, and 100 people are trying to get past your windmill to the exit. Wanna bet that none of them will bump into it, or get pushed against it, and knock it over?

4. Mar 13, 2014

### stevewindmill

There is no wind as we're inside a building, so i'll have to use the possibility of impact from a person to be the 'moment'

I have been looking at professional advice and hopefully will get somewhere soon,
In the mean time I'm trying to put together any calculations I can to get across to the organisers, as I am past deadline for submission.

It is worth mentioning that in the plans we have a base which extends out 600mm from ground height to 400mm high, with a protective fence which will then extend to 650mm high, making it fairly difficult for anyone to actually reach the structure itself. Anyone falling or being pushed would most likely either bounce off the fence or break through the fence and end up laid down on the 400mm high 'shelf' created by this base.

I've looked at the calculation
F(mass) * L(mass) = F(applied) * height
> therefore: F(applied) = F(mass)*L(mass) / height where 'L' = lever arm
giving the total weight of the entire stand as F(mass),
the 600mm distance from ground to edge of windmill as L (lever)
and the theoretical height of impact as 'height'.

From what I gather, this calculation gives a result based on the structure having a completely even weight distribution (which it is not, it is extremely bottom heavy) therefore we can assume that the result is absolute worst case answer and the structure will actually require more force to topple.

Using the calculation I found that with no additional weight in the base, the structure would require a force of 41.74kg applied at a height of 1.5m.
With 80kg added the structure would require 74.81kg of applied force at that height.

This assumes the following:
The structure would not slide at all, and any horizontal force could ONLY result in tipping (unlikely since it will not be attached to the floor)
The honeycomb board which makes up the side panels would hold the force and not break. This product is very strong but I do think that if 70kg of force were applied at a given point on the wall it would simply crease and break through, leaving the wooden structure supporting.

To put those numbers of theoretical applied force into perspective I have been experimenting with a scale against a wall. I've tried and I can't put more than 40kg against the wall.

If anyone has comments or ideas i'm really keen to hear them.

Thanks again
Steve

5. Mar 13, 2014

### Travis_King

fyi, OSHA's guardrail guidelines dictate that a guardrail must:
Where the top edge is defined to be:
This is an estimate of the person's full weight being pressed against the top rail. If you are expected to meet something similar to this criteria, then in order to safely withstand collisions, it should be able to resist at least 200 lb at a point roughly at the average person's back/shoulder level (maybe around 60 inches).

Since it's cardboard, isn't designed to be a fall protection system, and has a fence around it, I doubt that's a realistic design criteria. I would ask the health and safety team what scenarios they are trying to prevent, and then work with them to ensure that your display meets those requirement. You have the masses of the respective parts of your windmill, and you have the dimensions of the base (so you know the pivot point of where it would tip).

Calculate the location of the center of mass for the whole structure including the base; the mass of the structure is acting on that point as a moment about the tipping point (keeping it grounded). Calculate the force required at the approximated 60 inches that would counteract that moment.

6. Mar 13, 2014

### AlephZero

It has 69kg of wooden structure inside the cardboard (plus other hard objects like the motor).

But from the later post, it seems like the real issue is designing a guard rail that meets the specifications, not modifying the windmill itself.

I bet you can apply a lot more than 40 kg as a dymamic load. About twice your body weight would be a reasonable guess. But don't injure yourself measuring this experimentally!