Static vs rolling friction wheel loads 2

[WBell]

I am modeling a capstan (winch) equipment support that pulls three (3) rail cars with a wheel load of 22.5 kips each. Assuming the first two cars are in motion, rolling friction applies to the eight (8) wheels (4 per car) of the first two cars and static friction applies to the four (4) wheels of the last car.

Calculating the resistance to rolling using the equation given in Machinery's Handbook 29th Edition (p. 167) the cars in motion create a horizontal force, F_r = (W x fr) / radius. For the given cars, with a wheel radius of 4.08 inches, and mu = 0.02, the rolling resistance, F_r = 22.5 kips x 0.02 / 4.08 inches = 0.11 kips /wheel.

The force required to overcome static friction, F_s = 22.5 kips x 0.80 = 18 kips / wheel (using a static coefficient of friction = 0.80 for steel on steel).

The total force acting on the capstan chain is = (8 wheels x 0.11 kips/ wheel) + (4 wheels x 18 kips / wheel) = 72.9 kips, which is making the anchorage for the equipment challenging. I would appreciate any comments on the approach and the magnitude of the pull force exerted by the capstan.

 

[Ron]

With the static wheels, you are not trying to overcome staic friction but another source of rolling friction. You are not trying to slide those static wheels but you are trying to cause them to roll.

 

[WBell]

Thank you for the response. I have since found several sources that discuss rail car wheel to rail friction and the consensus seems to be that the rolling resistance of the stationary car is 2 - 2.5 times that of the cars in motion. The value calculated in this way is orders of magnitude less than the force calculated using the static coefficient of friction for the last car. For the given configuration, which is three cars with a combined total weight of 270 kips, to set the cars in motion, using a rolling friction coefficient of, say 0.0065 inches, a multiplier of 2.5 for the initial rolling friction coefficient, and a wheel radius of 4.08 inches, it would only take a pulling force of 650 lbs to start all of the cars in motion.

 

[BridgeSmith]

I could be mistaken, but I would think that inertia would be a bigger obstacle than friction in this circumstance. The force needed for acceleration of a 270 kip load, even if the acceleration is small, I would expect to be more than 650 lb.

 

[Ron]

HR10...I agree. 650 seams awefully light.

Also...what type of rail car has a radius of 4.08"? Are you off by a factor of 10 on everything?

 

[dhengr]

Wayne Bell:
Rolling friction with a steel wheel and a steel rail is kinda a misnomer. Rolling resistance is probably a better phrase for the situation, but friction does keep the wheel rolling instead of sliding. There is starting and rolling friction in the wheel or axle bearings also. The wheel starting problem is highly dependent of the wheel diameter and the wheel load. Take a look at the Hertz bearing stress problem in any good Strength of Materials or Theory of Elasticity textbook, you are literally pulling the wheel up out of a small indentation (cylindrical depression) in the rail. The railroads, AAR and wheel and bearing manufacturers have done a lot of work on this subject. Of course, their interest is generally high mileage, high load, high speed, long life, min. maintenance, etc.

What kind of railcars are these, mine cars or in-plant transfer cars of some sort. Your 8” whl. dia. and the 22.5k load is a fairly high load on a fairly small whl. RR’s usually think in terms of load ranges like: 27.5k/33” dia. whl. 32.9k/36” whl. and 39.4k/38” dia. whl. But, I’ve designed around whl. dia’s. and loads like your's before on in-plant, heavy hauling and mining equip. You also have to account for any incline in the track and this may be as small as some track settlement/deflection under load. Your 73k pulling force seems much too high, and your .65k seems much too small to me. Where did you get your 2 or 2.5 times pulling force numbers from? I can’t seem to locate the couple texts I’d look at on this issue, and it’s been years since I really worked with it. In starting a heavy train, RR engineers will often back into a long train, as much as they can to put slack in as many of the couplers as possible. Then, when they start forward they are starting each car individually, moving it several inches before they start to pull on the next car, in line, from a stop.

 

[WBell]

Thank you all for a good discussion. I have always felt that more information will help clarify a problem and work towards a successful solution. I will respond in outline form to some of the comments/suggestions.
1. The railcars are used in a battery recycling center and hold two (2) crucibles and weigh 40,000 kg (88.2 kips) per the vendor specifications (Engitec Technologies). I have attached a picture of the cars showing the configuration of the cars. The cars pull up under a furnace where lead from batteries is melted down and then poured into crucibles mounted on the cars. The wheel radius is given as 4.08" in the equipment specifications.
2. Since the original post, I have acquired a copy of the loading specification which gives the equipment loads to two (2) surface mounted plates. The cars are pulled back and forth on ASCE 60 gage rails and the loading specification gives the loading to the plates as follows:
a) Lateral load = 11 Short Tonnes (22 kips) to each plate, for a total lateral load = 44 kips;
b) Vertical load on front plate = 24 Short Tonnes (48 kips) downward (compression);
c) Vertical load on back plate = 22 Short Tonnes (44 kips) upward (tension)
3. These numbers confirm a subsequent analysis I performed with RISA 3D. I modeled the system with a node assigned the mass of the equipment (capstan winch) and applied a lateral force = 44 kips to get the anchor bolt reactions that hold the plates onto a mat foundation. I have asked the vendor to confirm the vertical tension force of 44 kips, but I feel comfortable with that load because of the good agreement confirmed by the analysis. I tied the node representing the equipment CG to the mat foundation with "spiders" - rigid, massless links.
4. The friction question is the area I'm most uncertain of; depending on which reference you use, there are several values given for rail-wheel steel. I started with Machinery's Handbook 29th Ed., which gives a coefficient of static friction of 0.8 for clean steel-steel and 0.16 for lubricated steel-steel. The recommended coefficient of rolling resistance is given as 0.02 inches for iron (no value given for steel?) to be used in the equation for calculating the resistance to rolling = W x f / r (p. 167).
5. I worked backwards to obtain a resistance to rolling force of 45.3 kips using a wheel load, W = 22.5 kips, a coefficient of static friction = 0.5, and a coefficient of rolling friction = 0.0065. The assumption is that the first two cars are in motion and exert a rolling resistance in combination with the last car starting to move and exerting a static friction force.
6. The more I think about what is happening, I think it is probably a problem of bearing friction with the axles. The vendor's engineers will be looking back at the original load calculations, but in the meantime, the contractor would like to proceed with a foundation design for the equipment. I am struggling with the tensile load on the rear mounting plate and am trying to convince the owner/contractor that a simple spread footing will not work without a relatively large footprint. I would like to put in some drilled shafts and mount the equipment on a pad tied into the shaft.

 

[dhengr]

Wayne Bell:
For some reason I was thinking you were primarily involved in the design of the railcars. The info. you finally gave is very good, quite informative and complete, and really starts to paint a good picture of your engineering problem. More info., sketches and photos are almost always better. But now, alas, it appears you are actually involved in the design of the capstan foundations, so you are short on good info. again. The spec. sheet, arrangement dwg., found. layout dwg. with track location would all be really helpful now. I wouldn’t argue much with the capstan manuf’ers. force numbers, if the P.O. told them correctly what to provide, and you say your calcs. kinda confirm these numbers. I would put my own FoS on the numbers for my uncertainty about those numbers and soils conditions, etc. But, I am confused about the front and back pls. Are there two of each, at the corners of the capstan base; or one of each, front and back, and each pl. spans btwn. the two front corners of the base, or the back corners of the base. I don’t know how big this base is. Whatever the contractor likes (or not) the back ftg. has to be heavy enough (plus a FoS) to hold the base down, and as you suggest, the tension seems the most troublesome. Drilled and filled, maybe cased, piers are pretty clean and practical compared to digging for a spread ftg. in a plant and under tracks. An improvement, maybe…, sizes are for concept only; make yourself a found. base, W24’s, 5-6’ apart, 20’ long, and parallel with the rail, with some cross framing. Attach the capstan base atop this found. base, at its front. Now the tension reaction at the found. happens with an extra 10-15’ of lever arm, and is much smaller. Put the controls up there, with a chair, call it an in-plant patio base, and they’ll love it.

 

[BridgeSmith]

Being unfamiliar with capstan system you're proposing to use, I don't know whether there is the capability to provide gradual acceleration. I hope so, because any sort of 'hard start' scenario will put tremendous force on all the components in the system.