|I know there are those out there who will run screaming in horror at the mention of “Math” – so I will keep that to this one page and you can jump past it – in horror or otherwise revulsion. 🙂 – to go directly to Pt V please click this link|
|I know already. Someone is going to point out that “Gee Ed. Just use a metal bridge”. Ummmm. No. I am using wood. That means that I need to span between B2 and B4 and support the section joint at B3 (about 6″ or 24′ full-size). I also need to span between B4 and AR – and support the theoretical joint at B5. The supported section there is not that entire length .. an intermediate bent about where the blue tape is between B5 and AR makes sense but that will still leave a span of around 5″ or 20′ full-size.|
Before I go any further, I’m going to do a little calculations of engine weights.
When I was building my pier I was curious about what weight the narrow gauge engine would have been since that is directly correlated to the size timbers needed for the construction.
I term this process – ‘Pesudo-Engineering’ as I freely admit I know little about this and only amusing myself. In any case in that post I wrote this:
EBT Mikados weighed between 112,00 and 163,000 lbs. A RGW Mike (K36) around 185,000 lbs. Light Mikados built for the Walbash in 1912 weighed 266,000 to 290,000 lbs. Later heavier Mikados at the Walbash weighed around 338,000 lbs.
At the extremes that 112,000 lb EBT Mikado weighed 33% of the heavy Walbash Mike. The RGW K36 at 185,000 lbs weighed 70% of the lightest Walbash model. A little math tells me that to weigh 33% of another it would have been 69% the width, height and length. That RGW K36 at 70% of the weight of the lightest Walbash would have been 89% of the width, height and length.
I’m playing with numbers here but the book “A Treatise on Wooden Trestle Bridges” I used as a source was published in 1897. With that in mind I am going to compare that 112,000 EBT Mikado with the light Walbash Mikado which weighed 266,000 lbs. .. which comes to 42%. A locomotive that is 42% the weight of another .. and for the purposes of this fantasy, reduced proportionally .. would be 75% of it’s full-sized sister in width, length and height.
You might argue that I’m wrong to use proportionally to calculate engine weights – that’s your right. I think I’m pretty close. If you read through the rest of that post I linked I used calculations from “Railroad construction: theory and practice” – where calculations on determining stringer size based on engine weight backs up my numbers I think (at least to a point and that’s ‘good nuff’ for Pesudo-Engineering IMO).
|What about an On18 engine?|
Earlier (Page III) I mentioned that I had calculated the width for an On18 loco.
I found a thread on Trainorders.com where it goes:
Q: What is the width of a narrow gauge passenger car, and what is the width of a standard gauage passenger car?
That gave me enough information then to calculate the width of the car for On18 .. (Pseudo-Engineering again). If I keep the same proportion to the gauge that the Main 2′ used .. then …
3.2 x 18 = 57.6″ or 4’9.6″ .. which I will round off to 57″/4’9″. In O scale that is 1.1875″
The picture shows a 2ft gauge locomotive that I re-sized for 18″ gauge. Note that I came within a inch of the calculated width.
Refer back to he “Engine Weight” bit above. I calculated that a 112,000 lb EBT Mikado weighed 42% of a light Walbash Mikado which weighted 266,000 lbs. That then gives me the numbers to calculate that proportinally the EBT Mikado would have been 75% in width, height and length of the Walbash Mikado.
With the width of a Maine 2′ gauge car of about 6′ 5″ or 77″ .. and my calculated with for my On18 car of 57″ or 4′ 9.6″ .. I can then do this:
57 / 77 = 0.74 and change .. or again .. around 75%.
.75 x .75 x .75 = .42 percent. That just happens to match the .42% weight of the EBT Mikado compared to the Walbash Mikado. Shrug.
So. An On18 Mikado would weigh .. 112,000 x .42 .. or .. 47,000 lbs .. or .. about 24 tons.
No .. I have no intention of an On18 Mikado .. but that puts the upper limits on the weight that my trestle would need to support. Looking at (for me) logical N-scale mechanisms that might be used .. I think 0-4-0, 4-4-0, 0-6-0. I like the 4-4-0 American loco simply because they had the large driving wheels which I think would scale better to an On18 loco.
The 4-4-0 was relatively light .. let’s say 50,000 lbs. That means that my On18 4-4-0 would be about 21,000 lbs or about 11 tons. I’ll keep those numbers in the back of my head .. but look at the timbers needed at say .. 11 tons operating weight but with a maxium at about twice that. I think that is probably good engineering anyway .. figure the load at twice the operational weight for safety purposes makes sense.
In “Railroad construction: theory and practice” pub 1922 it has the following concerning stringers.
“Disregarding all refinements as to actual dimensions, the ordinary maximum loading for standard gauge railroads may be taken as that due to four driving-axles, spaced 5’0″ apart and giving a pressure of 40000 pounds per axle. This should be increased to 54000 pounds per axle (same spacing) for the heaviest traffic.“
[Note: USRA standard designs for 0-8-0 switchers, 4-8-2 Mikado and 4-8-2 Mountain in light versions with an axle load of 54,000 pounds and a heavy version with an axle load of 60,000 pounds. It seems that what was considered light and heavy changed as time passed. The USRA ‘Light Mikado’ is listed at 220,000 pounds. The wheelbase on the USRA 0-8-8 locomotive was 15 ft – which helps explain the earlier statement that a trestle span of 14 ft you can’t get all four drivers on the span – and also with a 15 ft wheelbase there is indeed a 5′ spacing between axles.]
Using the calculations that the Narrow-Gauge locomotive weighs 42% of the Standard-Gauge locomotive we get a pressure of 16800 pounds per axle. If I then take my 42% calculation between the On18 and the Maine 2′ gauge locomotive I calculate that my On18 would be 17.5% of the Standard Gauge locomotives which would give an axle load of 7,080 pounds per axle.
They point out that “.. the span of trestles is always small, is generally 14 feet, and is never greater then 18 feet except when supported by knee-braces. The greatest load that will ever come on any one span will be the concentrated loading of the drivers of a very heavy locomotive. With spans of 14 feet or less it is impossible for even the four pairs of drivers to be on the same span at once.”
I guess I’m slow but I finally understood what the author was getting to here. Since you can’t have all four drivers on a single span at one time then that portion that IS on the single span has to be calculated.
Getting back to the 4-4-0. Let’s say .. I donno .. 7′ wheelbase on 60″ drivers. That should be somewhere in the area I think. In in N-scale that would be about a .525″ wheelbase and .375″ dia drivers. Converting to 1:48 and we get a 25″ wheelbase with 18″ drivers. (note: Just playing with numbers here people – don’t get upset.)
What tells me though is that the entire weight of the locomotive would be on that span between the bents so I will use that earlier calculation of normal weight of 11 tons and maximum of 24 tons for the stringers etc.
Note: They said that .. the span of trestles is always small, is generally 14 feet, and is never greater then 18 feet except when supported by knee-braces.“. My widest span is 24′. That’s six foot past their “never greater then 18 feet” .. true .. but we are looking at a load that is in the order of 18% of what they were supporting.
We then get a chart for ‘Stresses on various spans …’ for spans of 10 to 18 ft. Just below the chart it refers back to the earlier 54000 pounds per axle for heavier traffic .. where it states ..
“ … there will be no appreciable error in assuming the corresponding values, for a load of 54000 lbs. per axle, to be 54/40 of those given in the above tabulation.“
Ok. Then for my 7,0000 lbs. per axle load for my narrow-gauge then the corresponding value would be 7/40 or 17.5% .. of course. 🙂
From the chart, with a span of 12-ft. the load on one cap will be 49,440 lbs. For our On18 locomotive that will convert to 8,652 lbs. If the stringers and cap are made of long-leave yellow pine (why not), the allowable value, according to Table XXI, for “compression across the grain” is 260 pounds per square inch; this will require 34 square inches of surface. If the cap is 9″ wide (a place to start – but I previously used 12″ wide caps for my pier for On30. Seems reasonable to use something lighter for the On18), this will require a width of 3.7 inches.
What this is saying is that the stringer needs to be 3.7 inches wide minimum sitting on a 9 inch wide cap to keep from crushing the wood fibers. This the minimum and a place to start.
From the chart showing ‘Stresses on various spans …” we find that (for Standard-Gauge) and a Span of 12 feet, the Max. movement, ft. lbs is 82,600. Multiplying that times 17.5% for On18 we get 14,278 Max. movement, ft. lbs.
For rectangular beams. Moment = 1/6*R’bh2. Using R’ the safe value is 1300 lbs. per square inch.
14,278 x 12 = 1/6 x 1300 x 3.7 x h2
From which h = 14.6 in.
A stringer needs to be 3.7″ x 14.6″
I used what I had on hand which was 0.100″ x 0.25″ .. which translates to 4.8″ x 12″
Just for fun I re-did the numbers using those stringers ..
14,278 x 12 = 1/6 x 1300 x 4.8 x h2
This time h = 12.8 in.
Cool. Double the stringers of course. If one breaks .. the other one should still support the load as it is only .8″ short. Since in 1:48 this is only 0.016″ .. I can live with that.
The chart “Stresses on various spans” only ran to 18 ft. I put the numbers into Excel and extended to 24 ft. That gave me a Max. movement, ft. lbs for 24 ft of 226,464. Multiplying that by 17.5% to convert to our On18 loco gives us a Max movement of 39,631.2 ft. lbs.
Now we run the numbers for that 24 ft. span.
39,632 x 12 = 1/6 x 1300 x 4.8 x h2 (using that same stringer width to start)
I get about a 21.4″ stringer .. or about .445″ O scale.
.. that’s all well and good except I have nothing that size (or close) in my strip-wood stock.
Let’s see. I have some 0.180″ x 0.260″ which would be .. ummm ..8.64″ x 12-1/2″ .. ok ..
39,632 x 12 = 1/6 x 1300 x 8.64 x h2
This time solving for h I get – 16″. Drat. Deeper is better.
Oh oh. I have a piece 0.180″ x 0.330″. That translates to a 8.64″ x 15.84″ beam. Cool. That will work.
|End of Math|
|Ok. Those in the grip of Mathphobia can go to the next page now! 🙂|