Structural Stability of Free Standing Objects

I am currently embarking on a project to build some freestanding tubular framed structures assembled from galvanized steel and/or aluminium scaffold tubes. I was wondering if you knew of any simple formulas which could be use to approximately work out the structural stability of an object for a given dimension and weight; or if there is an easy to use software package with this functionality.

The data of most interest to me is that relating to preventing a structure from tilting or sliding. For example in the case of a single length of tube which will be raised perhaps 200mm from the ground and will be subject to significant forces coming from a sideward angle rather than straight above. It would be nice to be able to work out on paper roughly how heavy it would need to be and / or how wide the supporting legs will have to be in order to stop it sliding along the floor or tipping onto its side.

Obviously prototyping & trial/error will produce satisfactory results in the end but I'd rather start from more than just an educated guess to minimise wasted time in the workshop + wasted materials.

Anyone got any clever suggestions or tips?

It seems like you want a cross between truss structures and beam bending.

I've just had an exam on it so if you have a design and some dimensions I can help.

It sounds like more a question of 'will it fall over?' rather than 'will this structure collapse/yield/snap' Is that correct?
If the question is: 'will it slide?' then the answer is challenging as there are so many unknowns. If the structure is on grass then the 'coefficient of friction' (grip) could be really high if the tubes are pushed into the ground or really low if the feet are flat and smooth. if the feet are flat and smooth I would guess the friction is about 0.2 (i.e. if all the download on the structure is 10kg (structure weight plus load on it) then it will only take 2kg of sideways load to make it slide). If the feet dig into the ground then the friction coefficient could well be 2, so the same structure would require 20kg to make it 'slide'.
Stability is an easier issue to estimate. The simple answer is: resolve moments around the contact points (http://en.wikipedia.org/wiki/Moment_(physics)).

Mark you are thinking along the right lines. I've been building robots long enough to know how to over engineer something to brunel standards without fear of it collapsing. The issue is about sliding and tipping.

Floor surface will be a typical sports hall so varnished/polished wood or the modern polymer equivalent

P.S. that wikipedia link isnt working for me

http://en.wikipedia.org/wiki/Moment_(physics)

The second ) messed the link up.

Ok, so now the challenge is estimating the friction coefficient and the size and direction of the side loads. If you can choose the 'foot' material, use something soft/grippy like rubber, then you'll get a coefficient of friction of around 1. So if side load is less than weight plus down load then the structure won't slide. If you can't choose or must use a hard foot material then you'll need to do tests to see what the friction will be.
For stability, resolving moments will tell you if its stable, provided you can estimate the side load. My explanation of moments in this context is:
Consider every load/force that stops or causes the structure to topple. Take that load and multiply it by the perpendicular distance from the contact point you think the structure will fall over/rock onto to the force of interest to calculate a 'moment'. The sum of all restoring moments must be more than the sum of all toppling moments.

So if the structure is symmetrical and 20cm wide and weighs 10kg then its 'restoring moment' will be 100kg-cm whether its pushed over to the left or the right (the weight acts in the middle so the distance from the weight force to the contact point is 10cm). This means that a horizontal load acting at 20cm high needs to be less than 5kg or the structure will topple.Â

Does that make sense/help?

Hi Mark,

Thanks for the in depth response. It does help but I possibly only half understand all of what has been said. I am one of those people that you could call mathematically challenged, so anything involving numbers, letters or formulae needs to be explained in very simple steps with no assumptions of prior knowledge!

I do find things much easer to grasp if they are applied to something in the real world so perhaps you wouldnt mind explaining your theory applied to this photograph. Its not exactly what i am building, but is of a similar style of construction and used in a similar way. - For the sake of argument lets assume the structure in the photo is 4m long, 1m high, with 1m wide 'feet' and weighs a total of 100kg.

http://visual.merriam-webster.com/im...lance-beam.jpg

Is there an easy/inexpensive way I could set about measuring the side loads and likewise the coefficient of friction?

I have no idea how much force i would exert upon a given area if i took a decent run up and threw myself at it but this would be a useful value to have should I be able to find a reliable/accurate way of recording it.

Also do you know how the distribution of weight factors into our calculations. Again using the above photo as an example, would it make a difference if the beam was solid with all the weight at the top or if the beam was lightweight and hollow with all the weight in the feet? If this would make a difference how would this be accounted for in the calculations?

I've had a few thoughts on how to get a handle on your questions. I'll outline my thoughts, if you think they're worth further effort then we can refine the methods/go into more detail.
For determining the friction do you have access to a similar piece of equipment?
If you can get access to something with a similar weight and foot design then you can work out the friction coefficient. The scientific way would be to weigh the item, then measure how much force it takes to push it along the ground (should be on a representative surface). This could be done with some bathroom scales but it'll be difficult.
The alternative is to tie a piece of string/rope to the equipment and see at what point it transitions from sliding to toppling. If you tie the rope at the base of the equipment and pull horizontally then, provided you can pull with enough force, the equipment should slide. If you tie the string to the top of the equipment then it will probably topple. There should be a transition point like this:
https://lh4.googleusercontent.com/-9.../stability.jpg
At the point of transition from toppling to sliding the friction force will be as high as it can be, and will equal the pull of the string. crucially, with this method you don't need to measure the actual force, just where the string is attached and how wide the equipment is
If the equipment is symmetrical then you€™ll know where the weight force line acts. The reaction force from the foot must act through the intersection of the weight force and the string force. Once you know the angle the reaction force takes then you can work out the ratio of the normal reaction to the friction force. This could be done simply with measurements with a tape measure.
The ratio is horizontal distance from end of foot to centre of mass divided by height to string. This ratio will tell how much side load will cause any given piece of equipment to slide, if you know that equipment€™s weight.
I'll describe the effect of height of the centre of mass in the next post.

As you might be able to see from the previous illustration, the height of the centre of mass doesn't affect how easily it is to initially tip the equipment.
It does affect how easy it is to cause the equipment to totally fall over though
Taking two extremes with the centre of mass (CoM) at the base and at the top you can see the effect on the angle of no return:
https://lh5.googleusercontent.com/-p...0/toppling.jpg

Mathematically you can equate this to the energy to topple the equipment. In the first case, from level to toppling over, the CoM lifts just 0.2m. In the second case the CoM lifts 0.5m. The energy is given by height x gravity x mass. So 0.5m x 10N/kg x 100kg gives 500 Joules for the second case. If you relate this to a high jumper (1.5m x 10N/kg x 80kg = 1200J) then 500 Joules isn€™t much compared to what an athlete can put out in one jump, but it is quite high.

The other way to work out what force you can push with is to think about what you could push in a leg press. This is typically several times your own weight. Here's how I'd resolve moments once I'd estimated the force:
https://lh4.googleusercontent.com/-a...00/moments.jpg
So the 200kg assumed force is drawn on at a realistic angle. X is the perpendicular distance between the line of the applied force and the foot the equipment will topple onto.For stability 200kg x 'X'metres must be less than 100kg x 0.5metres. So for this particular case, x must be less than 0.25m

Mark,

Thanks so much, this is a great help!