Determining Axle Load On Trailer Assembly

So here's the rundown. I work for a machinery moving company and we occasionally haul objects well in excess of 150 tons. We occasionally
get asked what our axle loads will be and just recently found out the method we had been using isn't very accurate. There is software out there that will do this but it's very expensive. So what I'm looking for is if anyone can tell me how to do the following or at least point me in the right direction as I've tried treating things as an assembly of simple beams but the numbers just don't look right. We have access to measure anything we need with the exception of weights which are either measured previously or given from the manufacturer.
The basic assembly is we have a semi-tractor, it's weight, the lengths between steer & 1st drive axle, 1st & 2nd drive, 2nd & 3rd drive axles, and the fifth wheel position (adjustable) in relation to any of those wheels.
Behind that is a dolly, with all wheel to wheel measurements, weight, and wheel to kingpin & wheel to fifth wheel measurements. It has 4 axles.
Behind that is a trailer which we know the weight, kingpin to wheels, and wheel to wheel dimensions.
On the trailer is a load, which is roughly a uniformly distributed box. We know it's overall length and distance from center of gravity to the kingpin, and consequently, the wheels.
Is there a simplified formula, or a process that I can go through to determine this? I don't have any engineering background but I understand some of the basic concepts. I tried figuring it out using a simple beam uniformly loaded partially distributed but that didnt seem right nor did it work out at all when i transfered that force to the dolly calculations. Any help would be greatly appreciated!
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Dear joshya...:
On Jun 10, 12:30 pm, snipped-for-privacy@gmail.com wrote: ...

A picture would help...

Sum the moments formed by each load or axle, using the "fifth wheel" of each segment as zero for the coordinate system. A moment is another name for a torque, and is a force times a distance, *with a sign* that indicates a direction. All the moments must sum to zero, or the segment would start spinning.
Choose a side of the assembly (say the US driver's side), and use your right hand (say) to assgn signs. If the thumb is located at the fifth wheel, and each force component is aligned with your fingers, the direction you thumb must face makes the sign positive or negative.
If the axles are close together, (or are far from teh fifth wheel) they each can be assumed to provide the same lifting force.
5th LOAD X1 X2 | dL | | dX |
0 = - dL * weight_LOAD + dx * force_BOTH_Xles.
Anyway, rather than wave my hands any further... http://en.wikipedia.org/wiki/Free_body_diagram http://204.68.195.102/reports/tswstudy/Vol3-Chapter5.pdf http://trailer-bodybuilders.com/mag/trucks_weight_distribution_math /
David A. Smith
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This is all good information, and I'll be posting with a link to the diagram we use of the trailer, but the issue I seem to be having is that, as you'll see in the diagram, some of the wheels lie directly under the 5th wheel position so their distance along x is 0, which makes some of these equations a little wonky as it's certainly not ttaking the entire load or no load.
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Here's a copy of the diagram we use to write down the information. http://www.hennes.us/_documents/70%20Trailer.pdf
It has most of the neccesary measurements. The only thing it doesn't show is the kingpin & 5th wheel locations, which are as follows:
On the truck the 5th wheel is 6" in front of the middle drive axle. On the dolly the 5th wheel is 10" behind the first axle.
Just to throw some numbers at this, it was figured using the old inaccurate method that 180,000# load should be putting 31,500# on each trailer axle, 22,500# on each dolly axle, 21,000# on each tractor drive axle, and 15,000# on the steer axle. Now I understand that not all the axles in each group will take exactly the same amount as each one is slightly further away, but a simple force * distance would result in the steer axle, being the farthest from the load, taking the most weight, which cannot be the case as it's designed to be the structurally weakest axle. Thoughts?
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On Jun 11, 6:57 am, snipped-for-privacy@gmail.com wrote:

http://www.hennes.us/_documents/70%20Trailer.pdf
You will draw *three* separate free body diagrams. * One for the truck, with the downward force from the dolly on the fifth wheel on the truck. * One for the dolly, with the downward force from the trailer on the dolly's fifth wheel. * Last for the trailer proper. You can assume each axle's springs are uniformly compressed (hence evenly loaded), since the trailer will stay largely horizontal.
Solve the trailer first.
I would not assume the dolly stays fully horizontal, so each axle will be slightly differntially compressed. But that does give you a simplification, that the dolly's structure is non-deformable, so from the pivot point each axle futher away form the truck's fifth wheel will carry porportionally more or less load.
The truck loading will be much more complex.
You might be money ahead to find a transportation professional to run these calcs for you? Consider the penalties if damage is incurred, and if the penalties are all paid by "John Hennes Trucking", because they did not do due diligence...
David A. Smith
(PS. I am NOT such a professional.)
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On Tue, 10 Jun 2008 12:30:12 -0700 (PDT), snipped-for-privacy@gmail.com wrote:

Draw a straight line, with wheel positions marked to scale, and the load position CogG marked at the scale position.
Then if the trailer is static, the forces acting clockwise equal the forces acting anti clockwise if their positions are taken into account.
Take as the pivot, any point - say the end of the tail. The load is its weight times its distance from the end. The wheel support is last axle force times its distance from the tail end PLUS the next to last axle force times its distance from the tail end PLUS ... The front axle force times its distance from the tail end.
You have too many unknown forces at the axles to get a definitive answer immediately for axle forces, but you could start by (wrongly) supposing that all four (or however many) axles provide the same support force. You CAN work out that force. Then do the same again, ignoring the force from the two axles furthest from the load. This gives you a low limit and a high limit on each axle, which will get you towards the ball park.
Brian W
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This doesn't seem to work out quite right either as the way a trailer is designed is such that all 4 axles on the back of the main trailer should be taking a *relatively* similar load, the dolly's axles should also be taking a similar load amongst themselves but quite a bit less than the trailer axles, same for the tractors drive axles being less than the dolly and much less than the trailer, and finally the tractor's steer axle should have the lowest of all. If I go about it your way, the steer axle should be taking the most load as it's furthest from the physical load...
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On Wed, 11 Jun 2008 06:27:42 -0700 (PDT), snipped-for-privacy@gmail.com wrote:

Hmmm...you don't quite get the idea of opposing moments balancing out: think of a long lever with a weight near the far end, so the near end needs a low force to lift the heavy far weight. Anyway, I ran a calculation on your diagram on the back of an envelope, just for the trailer wheels, and the numbers you have for the sample load look close enough: (these are the 'incorrect' numbers as you describe them) for the 180k lb load plus the 123 k lb trailer deadweight, I got 30.8 k lb. average per axle, leaving 129 k lb on the trailer fifth wheel
Brian W
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On Wed, 11 Jun 2008 20:19:14 -0500, Brian Whatcott

Make that: 180 k lb load and 72 k lb trailer weight....
Brian W
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Dear Brian Whatcott:
wrote:

I wonder if a deformable tralier frame would make the difference? Seems to me it would only add / enhance differential loading to the individual axles. The resultant at the fifth wheel "should not" change...
David A. Smith
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On Thu, 12 Jun 2008 06:23:25 -0700, "N:dlzc D:aol T:com \(dlzc\)"
///

Some double axles have that kind of load leveling built in, with a swinging link on the adjacent leaf spring ends supporting the axles. Going over a bump with one wheel high and the other low, they still both get a more equal share of the load than without the swinging link. You may have come across a 'swingle-tree' that connects two horses hauling a cart: one horse can step forward a little, but the other horse still gets to pull its share. On a fancier scale, I recall visiting an arcraft facility where an airframe was being fatigue tested. The attachment points along the wings were connected to a beam two at a time, and these two beams were connected to a collector beam two at a time,and so on, making a kind of giant swingle tree For a trailer with pneumatic axle springs, load-sharing is a matter of connecting all the bladders together [in principle] so they are guaranteed to provide similar forces, though they may ride high or low over bumps.
Brian W
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Yea I redid things your way and I think I understand what you were saying now. It turns out after some digging and getting the original guy who did the numbers to come in that he was getting accurate numbers, he was just going about getting them in a very round-about way. His method doesn't make a whole lot of sense step by step but in the end he's determining the right values so they still worked out to the same resultants. Thanks for the help.
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