Calculating the Specific strength for the rod

Dywyddyr
Ok, let’s run very simple calculations

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So, we have got the electro motor and the rod. Radius of the rod is 1 meter and length-1 000 meters. The volume of such is 3140 cubic meters. Assuming that the rod is made of Carbon nanotubes (their density varies, but let’s take 0.1 tons/m^3) its mass will be 314 000 kg. And the mass of the sphere is 1 000 kg (negligible to rod’s mass). Tensile strength of the Carbon nanotubes is 3 600 MPa (written here: http://en.wikipedia.org/wiki/Ultimate_tensile_strength ) and this must be multiplied to rod’s cross-section’s area (3.14 square meters)-we will receive 11 304 000 000. As for rotational velocity-let’s say that the rod begins rotating and when completing this revolution (that is 360 degrees) the speed of the rod’s end (and hence sphere’s speed) will be 5 000 m/sec……then what?

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Rotation creates g-forces. The faster you spin something the more it will "weigh".
Check here, use the relevant values, and calculate the final effective weight of the mass.

 

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As I know the acceleration/g-load for the circular motion is calculated by means if following formula:

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So, the acceleration on the rod’s end (on sphere) when the rod makes one complete revolution is equal to:

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25 000 m/sec^2, that is 25000/9.8 m/sec^2 2551 g.
So, we have value g-load, then what?

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So if the mass is subjected to ~2,500 g does that not suggest that that it will "weigh" ~2,5000 times more than usual?
Hence what was, for example, 1 tonne will now effectively be 2,551 tonnes.

 

Yes, the sphere's weight (not mass) will be 2551 times more

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But the way, do you know what happened to my thread "How to rotate stretched cable"? Nobody answered after my last post, did I write something very stupid?

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No.

It's more probably a case of "we've gone as far as we can at the moment towards an answer".
Some threads move slowly, some don't.
That particular one is waiting for a flash of inspiration

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I see, perhaps nobody has asked this kind of question before, so it needs a lot of think

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So, we know the g-load of that sphere when it moves at the speed of 5 000 m/sec-2551 g. Then what?

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Er, so now you know the load applied to the cable...

Post 58.

 

Dywyddyr
So, we know applied load (that 2551 g) and 11 304 000 000 (as I wrote is post 61 tensile strength of the Carbon nanotubes (3 600 MPa) was multiplied to rod’s cross-section’s area), now should we compare them?

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Yes. Once you know the effective weight of the mass then it's simply "is the cable strong enough to hold it?"
Compare the mass with the (tensile strength multiplied by area). And don't forget the factor safety!

Re-reading your post: Applied load is NOT 2,551 g but 2,551 x the mass of the, uh, mass on the end of the cable.
If it's a two tonne mass then the load is ~5,000 tonnes but if it's a 10 tonne mass then the load is ~25,000 tonnes.

 

Dywyddyr

So, I used the SI system during calculation, therefore the received values should be correct and we see that 11 304 000 000 is more than 2551 g, so this rod will not be broken, right?

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You mentioned it on the previous page

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what is this? How safety factor can be estimated?

Something new….you mean that 11 304 000 000?

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NO!!!!

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Again: the EFFECTIVE LOAD (g x the mass) is compared with the (tensile strength x area).

A factor of safety is not estimated: it is assigned. Start here. Basically it is a reduction in "working strength" to allow for out-of-ordinary conditions. If anything occurs to your motor, for example, and the rotation increases then the effective load will increase. The total load (in normal use) must ALWAYS be low enough so that a certain overload is catered for without failure.
If the cable is, for example, capable of taking 1 tonne then you should make sure that under any reasonable failure of the system to work as planned (motor speed up, external impulse, etc etc) then the cable will still hold. Therefore a competent engineer (or even an incompetent engineer who wishes to remain employed) will make sure that the mass on the end is LESS than than 1 tonne. A factor of safety of 2, for example, will allow the mass to double without failure: i.e. a normal load of 500 kg.

Effective weight = actual mass weight x g load. If your mass on the end of the cable is 1 tonne and it's under 2,551 g of acceleration then the effective load is 2,551 tonnes.

 

Dywyddyr

Sorry, when rushing I apparently confused this. So-2 251 000 kilograms versus 11 304 000 000, the latter (that is the rod) wins

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So, in other words the honest engineer should always take into consideration some unforeseen circumstances and calculate everything so that there to be some “reserve” for increasing the mass (or rotational speed of the rod) of the sphere if everything is going perfectly. It looks like the automobile that can ride at let’s say at the speed of 300 km/h but the careful driver rides at less one-200 km/h to save car and its resources

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And how much should be this factor of safety? To what value is it equal generally in engineering world? 2 would be enough?

So, it is increased weight due to acceleration, that one that flyers and astronauts face during flight, right?

 

Yes, if the end-mass is nominally 1 tonne. (Note - 2,551,000 - you put a 2 in place of a 5).

More or less.

Not quite, since the car will be designed with a safety factor and if the manual says 300 km/hr then the vehicle will be capable of withstanding speeds exceeding that. (Whether the engine can get it to those speeds is beside the point).

Ah, now here's where the problems start. It depends on usage, customer requirements and practice. 2 is a reasonable factor (e.g. in normal use it runs at half of the maximum it can withstand), but are there any circumstances where this can be exceeded? - bearing in mind that if the cable fails then you're sending a 1 tonne mass off into space at 5 km/ sec -

http://www.projectrho.com/rocket/spacegunconvent.php
Your 1 tonne weight becomes huge potential bomb.
You have to weigh up the possibility of the cable failing against the cost of lawsuits and loss of life.
But since your result gives a very large margin then you should be okay. One thing to check (if this were going into production) would be creep. If something is under stress or tension (even at values below the nominal yield) for extended periods of time it can eventually deform and fail at values significantly LOWER than nominal.

Yep.

Oh, and another consideration would be the strength of the attachment of the cable to the spindle. After all that's subject to the mass of the rotating cable AND the mass of the rotating end weight. And BOTH are undergoing acceleration from the rotation.

 

Dywyddyr

I can imagine what will be if that sphere rotates at the speed of 15-20 km/sec

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So, as I see factor of safety cannot be determined early theoretically, but it needs some practical experiments and experiences also

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As for Creep, it does not matter since “In materials science, creep is the tendency of a solid material to slowly move or deform permanently under the influence of stresses”. So, as I see it generally occurs very slowly, if the rod rotates during a lot of time, but “my” rod will rotate only for several seconds….

By the way, where is the atmosphere’s resistance? We discussed the strength of the material and applied load and comparing these two factors would be enough in vacuum but on the Earth we have got quite dense atmosphere and it will resists rapid motion, so apparently the rod should have more strength, right?

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Oh okay. I assumed that it was going to be a constant rotation, but if it's not then creep isn't worth considering.

Not really. An atmosphere mostly means that you will have to put more power into rotating the mass to compensate for atmospheric drag.
If you really want to think about what operating in an atmosphere will do take a look at kinetic heating

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But, again, if the operation is short-lived then that's not worth consideration either.

 

Dywyddyr

So, it is matter of simply spending more energy….

As I know this online-calculator here: http://www.grc.nasa.gov/WWW/BGH/stagtmp.html enables to calculate the temperature of the moving/rotating body in the Earth’s atmosphere

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Yes, I know this. Moreover, I have read that during the nuclear tests under Project Orion http://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion) the steel plates covered with Graphite stayed intact after the explosion and these plates were placed at nine meters away from epicenter. This seemed to me incredible but at the web-site (non-English) there was given good explanation: the very high temperature (20 million degrees by Celsius or so) lasts only several seconds and therefore much heat is not generated….

 

Yep, but, for a short-lived rotation at least, the difference is probably negligible compared to the overall requirement of rotating the mass itself.

Nice find. I could have used that a year or so back when we had a question of atmospheric heating. I'll book mark it. Thanks.

 

Dywyddyr

You are welcome

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The calculator the link of which you gave me in post 4 is very valuable, but is there the one that would calculate the deflection (and everything other) for the rod that would NOT have free, unsupported ends? Imagine that we need to deliver long rod in space by means of Space Elevator like this:

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Here rod’s both ends are held, by Space Elevator’s cabin and by train, but the rod will be deflected and will have shape of arc and not straight line, so can I calculate this deflection?

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Um, post 2:

I.e. what you have so far is a special case of a supported beam, normally they ARE supported at both ends.
However, SS Beam calcs (that's Simply Supported, by the way, not a Nazi unit) usually also look at horizontal situations.
You may have to start digging, Giecks' would be a good reference.
Or maybe there's an add-on which will do the work for you in your graphics (CAD?) package?