Trailer Loading and CQ - Hypothetical Discussion

[MossRoad]

Thanks for all the input, I worked my numbers out from center toward the ends, 4' = 1000lbs. If the axle is carrying 9k at 24' then all that is left for the "tongue" weight is 1000lbs. At this point I figured there has to be a formula. I do need to know the axle weight and the tongue weight to insure they are within my tractors lifting ability and the axle rating. What really concerns me is the effect on the balance during the move, I guess I could use some light rope to secure to the FEL for some control but if things really go south the rope would break and the 10K bridge can go where it wants without me and the tractor attached. Then I will call the crane company

You've got 6000 pounds in front of the axle and 4000 pounds behind the axle. Yes, the axle is carrying most of the weight. And the 16' behind the axle with 4000# on it balances out the first 16' and 4000# in front of the axle. But you have an additional 2000# that starts 16' in front of the axle and goes 8 more feet towards the front of the trailer.

So now you basically have a 2000# weight hanging 8' in on a 24' long lever. Its going to take 2/3 more force than half the weight. So, 1000# + 667# = 1667# of tongue weight. I think.... :laughing:

Man, where are the people that can do this math for us???? :confused3:

 

[MossRoad]

Ok, now I understand your question.
Tell me where I go wrong:
32' feet, representing 8000 lbs are in balance. (10000lbs x 32/40)

That leaves 2000 lbs and 8' of trailer, the center of this weight being 4' from the hitch, and 20' from the axle.
Said differently this is 2000 lbs at a spot that is 1/5 & 4/5 of the span. So 4/5th (1600lbs) on the tongue and 1/5th ( 400lbs) + the 8000lbs (in balance)= 8400lbs on the axle. :confused3:

That looks close to me, too.

 

[George2615]

Why would anyone consider a 40' trailer with a single axle. Must be one heck of an axle. It would be like riding a see-saw. A 40'er should have 3 axles, 2 minimum.


My bad, somehow I skipped over the OP's reason in post 7

 

[aczlan]

Why would anyone consider a 40' trailer with a single axle. Must be one heck of an axle. It would be like riding a see-saw. A 40'er should have 3 axles, 2 minimum.

This is why:

Ok, this is not entirely a hypothetical discussion. I want to use a 40' flat rack container to bridge a deep stream on my property. For the money it can't be beat but there is the whole matter of getting it in place.
Once the "bridge" is delivered my thought is I will weld a temporary axle at 24' that will leave 1000lbs on the other end I can easily lift with my loader. I will then be able to back the "bridge" into the stream and the 16' foot suspended behind the axle will reach the other side. I can remove the axle and use wench and hydraulic jacks to set the bridge. I am concerned with controlling the load with a 6,000 pound tractor and do not want to be around if the "bridge" decides teeter in the other direction.

Aaron Z

 

[RedNeckRacin]

The answer is based on the uniform loading of 250lb/ft to get 1667 lb at the hitch and 8333 lb on the axle. :thumbsup:

The forces that you can get into during motion are well out of my engineering experience but I would say that if you are anywhere close to the weight of the trailer being moved i.e. if your tractor doesn't weigh close to 2/3 the weight on the flat, or at least as much as the trailer on a hill, I'd find some creative solution to moving it such as a crane that was mentioned, or someone with some bigger equipment. The last thing you want to do is get beyond your tractor capabilities and have that trailer dragging you somewhere you don't want to be. Is there anyway to put this thing on a dolly to get a second axle under it until you get close and then fix the axle and push the trailer across it so that you aren't fighting another factor?

The equation incase anyone is curious:

Hitch=(W/(2L))*(L^2-a^2)=((250lb/ft)/(2*24ft))*(24ft^2-16ft^2)=1667lbs

 

[aczlan]

It might help with control on the hill is if you got an axle with brakes (like a mobile home axle) and run power to the axle.

Aaron Z

 

[Beltzington]

The answer is based on the uniform loading of 250lb/ft to get 1667 lb at the hitch and 8333 lb on the axle. :thumbsup:

The equation incase anyone is curious:

Hitch=(W/(2L))*(L^2-a^2)=((250lb/ft)/(2*24ft))*(24ft^2-16ft^2)=1667lbs

Thanks RedNeckRacin I knew I was over simplifying the weight distribution, looks like MossRoad and CobyRupert where much closer than I.

Maybe I should have titled this post "Hold My Beer and Watch This." Actually once the ground dries up the roll-off should be able to drop the container 50' from where it needs to be so brakes and tandem axles not really needed. In theory I can install the axle through the pallet pocket on the container, pick the 1667lbs end up with my loader and push it straight back into the ditch. The real fun will be what happens when the 10k rolls into the ditch.

 

[Sodo]

For better visualization heres a pic I found on google images searching "railcar bridge"


There's a way to do this without hiring a crane, and have some fun too. I would worry a little about the single axle sinking into the ground you might do all this better on logs. You have a 6000 lb tractor, but?.do you have logs? Or can you get some?

Here's an idea - skid a couple big logs across the stream and set them wider than the container. Then have them crap the container onto a fulcrum log crosswise. Set another log then drag the container over the logs, leapfrogging fulcrum logs until you get to the stream. If you can keep it near the balance point you can lift one end with the tractor. It either slides over the fulcrum log, or with luck, rolls the log, and doesn't roll on down into the stream. Cantilever it over the rollers on the two stringer logs then drag it across. You could have it halfway across just on the fulcrums.

Pic doesn't show it but the log stringers are 10 feet apart and your roller logs are 15 feet long. In the end the bridge is sitting there with a log stringer about a foot on either side.

Your stringer logs have to be sized to support about 2500lbs (each) in the center so about 12" log could do it. More roller logs is better because it distributes the load. If this option is even feasible we could work on the sizing.

 

[CobyRupert]

The answer is based on the uniform loading of 250lb/ft to get 1667 lb at the hitch and 8333 lb on the axle. :thumbsup:

Redneck is correct. And MossRoad. I just asked a Professional Structural Engineer licensed in 4 or 5 states. (Hopefully, he knows what he's talking about )

 

[zzvyb6]

The "bridge" weighs 10,000 lbs and is 40 ft long. Assuming that an axle plus wheels and/or tires (whatever) can hold up that much weight, the solution really amounts to where the axle should be placed in order to manage a reasonable tongue load. (That means: what's the hitch capacity of your towing apparatus (truck, tractor, loader, etc.)?

So, taking moments about an axle placed some distance X from the center of the 'bridge' (its uniformly built, so the c.g. is midway), means that the 10,000 lb bridge weight times the distance from the c.g. to the axle location (X), is balanced by the acceptable hitch load times its distance from hitch to the axle (which is X + 1/2 the length of the 'bridge'). So, if your max hitch load is 1000 lbs and 1/2 the bridge length is 20 ft, then (10,000*X) = 1000*(20 +X) and solve for X. In this case, the axle should be placed 2.2 ft behind the midpoint of the bridge to get a 1000 lb tongue load.

Do the math. I'd be careful about how much your soil will support with that much wheel load. The more hitch load you can take, the less the axle load will be. If you are backing it down hill, the gravity load (from the angle) may exceed the traction of your tow vehicle tires and it will slide downward out of control. The higher the hitch load, the less weight will be on a truck's front tires and that's bad for steering and braking if you are backing up this bridge-trailer downhill.