Naked Science Forum
Non Life Sciences => Technology => Topic started by: circustavern on 04/07/2019 16:29:48

i am installing a gym and want to calculate whether the concrete floor will be distorted by continuous dropping of the weights.
2x20kg circular weights, dropped from a height of 1m.
Next is to work out how to calculate/test what the distortion might be without actually damaging the floor. I know of companies that provide load testing but this wouldn't be the same as the impact from dropping...?
Any help gratefully received.

Without getting into SR and GR the force depends on the ratio of the dropping distance to the stopping distance.
If a one Kg weight falls for one meter and stops within one mm the force exerted is 1000Kg.
That's why you get hurt if you fall from a window or have a car crash

So you need to know the stopping distance in order to calculate the stopping distance?!! Tricky problem this. The smaller the distance for deceleration, the larger is the force, so if the weight stops in 0.5 millimeters the 1Kg becomes 2,000Kg or 19620 Newtons. Also remember that the suggested 1mm stopping distance is shared between the weight and the floor. Probably the steel weight will deform (and spring back) more than the concrete. One good point is that the force is distributed over an area, so if the contact area between your weight and the floor is say 10 square millimeters the actual floor load is 1/10 of that of a 1 millimeter area. The gym that I use has thick rubber mats in areas were weights are present, these would be a help.

So you need to know the stopping distance in order to calculate the stopping distance?
No.
the actual floor load is 1/10 of that of a 1 millimeter area.
No.

Any small attempt to explain my mistakes would be gratefully received.

OK, given that the posts before yours had explained how you could calculate the forces if you knew the stopping distance, how did you conclude that " you need to know the stopping distance in order to calculate the stopping distance"?
Nobody had said anything about how " to calculate the stopping distance".

if the contact area between your weight and the floor is say 10 square millimeters the actual floor load is 1/10 of that of a 1 millimeter area.
The load on the floor is (shockingly) the load on the floor.

OK, given that the posts before yours had explained how you could calculate the forces if you knew the stopping distance, how did you conclude that " you need to know the stopping distance in order to calculate the stopping distance"?Nobody had said anything about how " to calculate the stopping distance".
Thanks for your reply. Please bear with me, my brain (never very bright) came into existence during the war, it's now even more feeble!
I understand that given the stopping distance you can calculate the force required to decelerate the falling weight. Syphrum gave an example where the stopping distance is one mm, but surely this distance depends on the characteristics of the materials absorbing the impact? The force on a concrete floor would be much greater than on say a rubber floor, because the stopping distance (on rubber) would be much greater. Perhaps I am asking, "how do you know the stopping distance?"

if the contact area between your weight and the floor is say 10 square millimeters the actual floor load is 1/10 of that of a 1 millimeter area.The load on the floor is (shockingly) the load on the floor.
Again, thanks for your reply. Perhaps I should have said the "load per unit area is 1/10...." I seem to remember, when working in design of warehouse vehicles, that floor loadings were based on this sort of measurement.

OK, given that the posts before yours had explained how you could calculate the forces if you knew the stopping distance, how did you conclude that " you need to know the stopping distance in order to calculate the stopping distance"?Nobody had said anything about how " to calculate the stopping distance".
Thanks for your reply. Please bear with me, my brain (never very bright) came into existence during the war, it's now even more feeble!
I understand that given the stopping distance you can calculate the force required to decelerate the falling weight. Syphrum gave an example where the stopping distance is one mm, but surely this distance depends on the characteristics of the materials absorbing the impact? The force on a concrete floor would be much greater than on say a rubber floor, because the stopping distance (on rubber) would be much greater. Perhaps I am asking, "how do you know the stopping distance?"
Well, yes.
That's the point of the rubber matting they use in gyms.
And if it is a few mm thick then it must mean that the stopping distance is a few mm.
Steel landing on concrete would give a smaller distance and a larger force.
Again, thanks for your reply. Perhaps I should have said the "load per unit area is 1/10...." I seem to remember, when working in design of warehouse vehicles, that floor loadings were based on this sort of measurement.
Floor loadings are complicated.
The lab where I work has a reinforced concrete floor and it's rated for something like 3 tonnes per square metre. (If I remember rightly).
But, I am 75 Kg or so, and I don't go through the floor, even though, as I walk, most of my weight is carried on a few square cm of my toes.
Say its a strip under my shoes an inch by 4 inches that holds up my weight as I walk.
That's about 1/400 of a square metre, so it should hold 1/400 of 3 tonnes
3000 Kg/ 400 gives 7.5 Kg and I'm about 10 times that.
Ther are, in effect, two different "maximum" floor loads.
There's one where the floor supports fail, and there's one where the material fails locally.

Any small attempt to explain my mistakes would be gratefully received.
OK....
So you need to know the stopping distance in order to calculate the stopping distance?!!
The question was about computing force. Measuring/computing stopping distance is one way to begin to approach this. Nobody suggested needed to know X in order to calculate X.
The stopping distance is misleading. The floor is going to distort some distance, and the object hitting it will also distort. You seem aware of this in your comment below.
The smaller the distance for deceleration, the larger is the force, so if the weight stops in 0.5 millimeters the 1Kg becomes 2,000Kg or 19620 Newtons.
For a point mass decelerating uniformly over half a mm, 19620 Newtons under Earth gravity, yes, but deceleration is not uniform over that distance. Force will vary, and thus hit a max value probably somewhere near the bottom of the distortion on the floor, and that max value may be above 19620 Newtons, or possibly less depending on what exactly the 0.5 mm measures.
Also remember that the suggested 1mm stopping distance is shared between the weight and the floor. Probably the steel weight will deform (and spring back) more than the concrete.
Yes, the object will continue to change its center of gravity even beyond the point where the floor reaches its maximum distortion. The stopping distance is thus not a nice clean value like 0.5mm. It's a dynamic process.
Is the 0.5 mm a measure of max floor distortion, or is it a measure of the distance traveled by the center of gravity of the falling object from the moment of first contact to the moment of reaching the lowest point? If the former, the max force is possibly less than 20k Newtons, and if the latter, it is definitely more.
One good point is that the force is distributed over an area, so if the contact area between your weight and the floor is say 10 square millimeters the actual floor load is 1/10 of that of a 1 millimeter area.
The change of area does not change the load (force) at all. The area is pretty irrelevant to a computation of force and has more to do with pressure. The question did not mention pressure (which is measured Pascals, not Newtons), even if pressure is more likely to damage a floor than is force.
Like the force computation, the max pressure computation cannot be computed by force/area. That would only be the average pressure, not the max pressure reached presumably near the middle given a round mass hitting a flat floor.
The gym that I use has thick rubber mats in areas were weights are present, these would be a help.
That would serve to reduce both force and pressure, yes. It reduces the former by increasing stopping distance, and it reduces the latter by increase of contact area.
I understand that given the stopping distance you can calculate the force required to decelerate the falling weight. Syphrum gave an example where the stopping distance is one mm, but surely this distance depends on the characteristics of the materials absorbing the impact?
Sure, a more spongy material will have a longer stopping distance, so a 1 Kg nerf ball dropped from a meter will exert far less force on the floor.
"how do you know the stopping distance?"
You don't. You might be able to compute it if you were a materials engineer, but I'm not one of those. Best to just measure it.

Floor loadings are complicated.
The lab where I work has a reinforced concrete floor and it's rated for something like 3 tonnes per square metre. (If I remember rightly).
But, I am 75 Kg or so, and I don't go through the floor, even though, as I walk, most of my weight is carried on a few square cm of my toes.
They had airplane cabin floors rated (local) for 100 tons per square meter, but these women would go on with perhaps 50kg mass and heels that were one sq cm which works out to half a million pascals, about 2.5 times the rate on the deck. They'd punch holes in them. They had to ban the shoes on some airlines.
Say its a strip under my shoes an inch by 4 inches that holds up my weight as I walk.
That's about 1/400 of a square metre, so it should hold 1/400 of 3 tonnes
3000 Kg/ 400 gives 7.5 Kg and I'm about 10 times that.
As you point out below, your 3 tons/meter is more of a larger scale average. The airplane obviously isn't going anywhere if actually loaded to 1% of that rated average pressure.
Ther are, in effect, two different "maximum" floor loads.
There's one where the floor supports fail, and there's one where the material fails locally.
Indeed, and there can be even more ratings, all at different scales.

It's also interesting to note that the 3 tonnes per square meter is roughly a third of the load exerted by the air pressure.
Fortunately, there's air on both sides.

Thanks to everyone for your detailed replies. I shall sit down and read everything in detail!

I would think if you were worried about the floor the thickness of the concrete, and if it was reinforced with rebar or concrete might be needed to take into consideration.

If I slip over and fall on my gluteus maximus I drop more than 40 Kg from more than waist height to the floor.
If the floor can't cope with that loading then it's a monumentally poor design.
We don't need to address details like rebar if the floor fails when someone falls over.

This work is what one would call, a "tour de force".
I think about and work with biological systems all the time and I was still blown away.

The gym that I use has thick rubber mats in areas were weights are present, these would be a help.