A friend who had served as, I think, a machinist’s mate, on a nuclear aircraft carrier once told me that when the ship is going at full speed, the drive shaft for each propeller develops about a half a revolution of twist from one end to the other because the torque is so great - and this on a shaft about 6 feet in diameter. I gather that the shaft is also pretty long, though obviously no longer than the ship itself.
Can anybody tell me if it’s true?
>Carriers of the Nimitz class have four steam tubine driven propeller shafts turning massive props and powered by up to nine nuclear reactors. Along with this official description of the drive system comes an official top speed of “+33 knots”, meaning that is as fast as the navy will admit to them being able to travel. That number is on the low end. I heard from an engineer serving on the Nimitz back in college that under maximum acceleration – say, from a near dead stop in Puget Sound to full ahead – those lengthy shafts deliver so much torque to the props that they twist one and one half revolutions difference from the head of the shaft to the prop.
[from http://peristaltor.livejournal.com/17602.html]
Well, gee, that didn’t take long to find. And it says one and a half, not just one half!
But is it true?
If you have a cite for the diameter of the shaft, whether it’s hollow or not (and if so, the inside diameter), the length of the shaft, and the peak torque it undergoes, it shouldn’t be hard to calculate the amount it twists.
From Wikipedia, I see that a Nimitz-class ship has 2 reactors and 4 shafts total, with each reactor producing 140,000 shaft horsepower (104MW)(another page says 260,000 total, FTR). Well, I can say that the shaft of an 800MW turbine, something I see all the time, is certainly not more than 2 feet in diameter, perhaps only 18 inches or so, so I’m really doubting the 6-foot diameter shaft quoted. Now power isn’t torque, of course, but I’m wondering just how much torque is applied to those shafts. If we knew the speed in rpm the screws turn, we could figure it out by the standard hp-torque equation.
In short, I’m also really doubting the “turns 1 and a half times” quote too. But I’m not claiming it’s wrong without doing calcs myself.
Based on my Navy experience (admittedly destroyer, not carrier) it would be something in the neighborhood of 200 to 300 rpm.
And, from my experience aboard a Destroyer, the shafts are hollow. And on a Destroyer 36" in diameter.
I agree with your skepticism about the 6 ft shaft, having had some experience (way back when) with large power plant turbines. I also had a bit of skepticism about the 1/2 turn or 1-1/2 turns of twist on the shaft. I just had to look this up a bit more.
The Mechanical Engineers’ Handbook (Wiley, 98) gives an equation for torque and torsional deformation, which is the technical description of “twist.” They give:
(theta) = TL/KG, where T = twisting moment; L = length; G = modulus of rigidity; and K is determined by the geometry of the shaft.
For a solid cylindrical shaft, K = ( (pi)d^4 )/32. So doing all the algebra,
T = ( (pi)G(theta)d^4 )/(32L)
I found a value (EngineeringToolBox.com)of G for nickel steel of 76 GPa, or 11.02 x 10^6 lb/sq in, so doing all the arithmetic gives a value of T (assuming I did that all correctly) of just over 52 million ft-lbs of torque. Assuming 60 rpm, that’s 52 million ft-lb/sec, or about 70.4 MW of power. These are for a 150 ft shaft (a WAG based on the size of the ship) and pi radians of deformation (1/2 turn).
So (if I’ve calculated everything right and used the right formulas) it seems reasonable to see such a tremendous twisting of the shaft.
If you’re really interested in more, here’s a YouTube video of a turning ship shaft from the Maine Maritime Academy.
I saw “Super Carrier”, about the construction of one of the newest carriers in the fleet (GHW Bush I think) on one of the Discovery channels recently. My memory is a little fuzzy, but they did a bit on the drive shaft assembly, and I want to say it was about 24-30" in diameter, and it was definitely hollow. The wall thickness on the shaft was around 3". It certainly wasn’t 6’ in diameter.
Let me try going through this from the other side.
So let me work backwards to get to theta.
K = ((pi)*d^4)/32
If d is 6 feet, then
K = (3.14159 * 6ft^4)/32
K = 127.23 ft^4
So we have theta = (52,000,000 ft*lbf * 150ft) / (127.23 ft^4 * 1586.9 x 10^6 lb/ft^2)
(note I converted G to a foot-basis)
theta = (7.852x10^9 ft^2*lbf) / (2.019x10^11 ft^2 * lbf)
the units should cancel out, and I get
theta = 0.0389 rad (2.22 degrees of turn)
If two people say they’re Jesus, one of them must be wrong…what assumption did you use for the shaft diameter?
OK, let me try with some other assumptions.
Wiki gives 2 reactors at 140,000 shaft horsepower each, and 4 shafts. So let’s say 70,000 shaft horsepower per shaft. Now, let’s also take the assumption of Xema and say that the screws turn at 200 rpm (low-end of his range, meaning more torque on the shaft). And let’s take what Bad Samaritan and Rhubarb said, roughly, and assume a shaft of 36 inches in diameter, 3 inch wall thickness.
Now IIRC, to find the twist on a hollow shaft, we look at the outside and inside diameters.
K = ((pi)*(do^4 - di^4))/32
If do is 3 feet, and di is 2.5 feet (assuming 3 inch wall thickness), then
K = ((3.14159) * (3ft^4 - 2.5ft^4)) /32
K = 4.117 ft^4
And torque will be = (70,000 hp 5252)/200 rpm = 1,838,200 ftlbf
(note - this is much less torque than earlier assumed!)
So we have theta = (1,838,200 ft*lbf * 150ft) / (4.117 ft^4 * 1586.9 x 10^6 lb/ft^2)
(note I converted G to a foot-basis)
theta = (2.757x10^8 ft^2*lbf) / (6.534x10^9 ft^2 * lbf)
the units should cancel out, and I get
theta = 0.0422 rad (2.40 degrees of turn)
Am I the one who is doing something wrong here?
Somewhere I’ve gotten the idea that hollow shafts are stronger than solid, maybe from that show mentioned above.
The shafts on my ship, an LPD, were maybe 24" in diameter. Some guys used to wrap their arms and legs around them and take a ride when they were turning slowly.
Peace,
mangeorge
Within limits (that is, setting aside Euler and other buckling), and with equal mass of material per cross section, a hollow shaft will generally be stronger than a solid one due to the mechanics associated with the geometry. If the two shafts you’re comparing are exactly the same outer diameter, then the solid one, having more material overall per cross section, will be stronger.
Also, depending upon the geometry of course, a hollow shaft can potentially be stronger than a solid one of greater overall material per cross section. Again, it all depends on the geometry. And you always have to watch out for buckling.
IANA engineer, so I’m gonna tackle the issue from a qualitative aspect.
The ship is about 1000 ft long. A typical layout would have the drive turbines about 2/3rds of the way aft & the props at ~ 90% aft. So shaft length ought to be ~25% of ship length or 250 ft.
The FOAF of the OP said 0.5 revolutions of twist, ie 180 degrees. The FOAF of poster #2 said 1.5 twists, or 540 degrees.
That would represent a twist of ~0.8 to 2.2 degrees per foot of shaft length. We have another WAGged shaft length up-thread of 150 feet, which would further increase the twist to ~1.2 to ~3.6 degrees per foot.
As I said, IANA engineer, but I am a mechanic and all around handy dude & that amount of twist would produce buckling in any material I’m familiar with.
Una’s numbers, ie about 1/100th the amount stated by the FOAFs, smells a lot better to me.
Whoa! Yeah, exactly. Freakin’ idiots. 
I thougt kids were smarter nowadays!
Oops - forgot to list all the assumptions. I used a 2-ft diameter shaft. If I re-run the calculations using a 6-ft diameter shaft, I get the same theta as you did with the 52 million ft-lb torque.
I also get the same value of theta that you did for the hollow shaft (3 ft OD, 2.5 ft ID, 70,000 hp), so we’re either both right or both wrong - it’s just a matter of the assumptions.
My calculations don’t consider any sort of failure - the shaft could well have exceeded its limit with that much power being transmitted down its length, not to mention any additional stresses that would be put on the shaft if there were any upsets.
Personally, I think the lower shaft HP and lower twist are much more likely. I was surprised to see that using the initial assumptions would give as much twist to the shaft as it did.
My son is getting to retire after 20 years on the USS Carl Vinson and USS Eisenhower as a nuclear mechanic, and his response to all the questions is below:
The assumption for a 6’ diameter shaft is about right. I’ve never measured it, but my arms are just shorter than the diameter.
The max rpm we can go is 171. That limit is there to prevent shaft damage. Also, we have limited throttle rates to prevent breaking the shaft.
The horsepower rating is correct. 70k HP per main engine. 2 per plant, 1 per shaft.
It’s almost impossible to measure the amount of twist on the shaft, but speaking to some of the 10 pound heads (engineers) that work for RPPY (Reactor Plant Planning Yard), a good estimate is about a full revolution of twist.
Bar torsion is used instead of springs in some cars.
Cast Iron is brittle stuff, filled with carbon , because the smelting process was first just mixing iron ore into coke and fanning it with gallows… And the casting process leaves the grains too large … and full of defects.
But good steel is very much better than cast iron, its very elastic. This means it can strain a lot, and yet return back to shape. Also, it takes a lot of force to strain it (steep slope in hookes law part of the curve), so it can take a lot of a force and then return back to shape…
So, don’t go and test your iron or cheap steel (fencing wire, or $1 a set screwdrivers) for this property, but good steel is very elastic like that.
There is a similar twisting effect under high torque for nuclear-powered submarine shafts, which are also hollow. The shaft diameter is more like 2-3 feet, though.
Macabre! Or do you mean “bellows?”