Trailer Loading and CQ - Hypothetical Discussion

Beltzington · Jan 20, 2015

My cold must be effecting the brain cells, I think I have the basic gest but the teeter-totter effect is messing with me.

Assuming a 10,000lbs equally dispersed fixed load, a 40-ft. trailer and a single axle. If the axle was located at 24' it would be bearing 9,000lbs the hitch would have 1,000lbs. Feel free to check my math.

What is bothering me is the suspended 16' load, in a static situation I am guessing it would stay suspended. But how much external force would it take to change this, running over a curb at 20mph. How about the angle of the load, say driving up a steep hill?

Keep in mind this is hypothetical.

Diamondpilot · Jan 20, 2015

Beltzington said:
My cold must be effecting the brain cells, I think I have the basic gest but the teeter-totter effect is messing with me.

Assuming a 10,000lbs equally dispersed fixed load, a 40-ft. trailer and a single axle. If the axle was located at 24' it would be bearing 9,000lbs the hitch would have 1,000lbs. Feel free to check my math.

What is bothering me is the suspended 16' load, in a static situation I am guessing it would stay suspended. But how much external force would it take to change this, running over a curb at 20mph. How about the angle of the load, say driving up a steep hill?

Keep in mind this is hypothetical.

You are going to have to ask someone way smarter than me....lol

Chris

MossRoad · Jan 20, 2015

Beltzington said:
My cold must be effecting the brain cells, I think I have the basic gest but the teeter-totter effect is messing with me.

Assuming a 10,000lbs equally dispersed fixed load, a 40-ft. trailer and a single axle. If the axle was located at 24' it would be bearing 9,000lbs the hitch would have 1,000lbs. Feel free to check my math.

What is bothering me is the suspended 16' load, in a static situation I am guessing it would stay suspended. But how much external force would it take to change this, running over a curb at 20mph. How about the angle of the load, say driving up a steep hill?

Keep in mind this is hypothetical.

You said equally dispersed.... so the load is 40' long? I meant that as a question....

tcartwri · Jan 20, 2015

The answer is; load the trailer such that the truck sits level front to back. And don't go over the speed limit.

Beltzington · Jan 20, 2015

MossRoad said:
You said equally dispersed.... so the load is 40' long? I meant that as a question....

Lets say the 40' trailer is the only load.

CobyRupert · Jan 20, 2015

Beltzington said:
What is bothering me is the suspended 16' load, in a static situation I am guessing it would stay suspended. But how much external force would it take to change this, running over a curb at 20mph. How about the angle of the load, say driving up a steep hill?

Keep in mind this is hypothetical.

You're talking about the 16' behind the single axle right? How much external force would it take to change this? Change this to what?
In general, I'd guess the amount of weight on the rear axle will change if driving up a hill, as a function (a sine function I believe) of the angle.

Beltzington · Jan 20, 2015

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.

MossRoad · Jan 20, 2015

If you have 10,000# trailer 40' long with an axle at 24'....
Every 4' of length has 10% of the load.
So, 6000# on one side of the axle and 4000# on the other side.

I don't know the mathematical formula to figure out how much weight will be on the tongue end. Do you?

MossRoad · Jan 20, 2015

I'm going to guess 1667 pound tongue weight. Why? Well, I'm just guessing, that's why.

The formula I found is Fe = Fl dl / de
Where:
Fe is the effort force.
Fl is the load force.
dl = distance from load force to fulcrum.
de = distance from effort force to fulcrum

We know de = 24'
We know the 4000# on the back cancels out the 4000# on the front, so we have 2000# to deal with.
So let's say the 2000 extra pounds is centered in the last 8' of the front. Probably not right, but I'm just guessing. So, 20' from the axle.

Fe = (2000 x 20')/24'
1667 = (2000 x 20')/24'

MossRoad · Jan 20, 2015

Some math wiz please figure this out! :laughing:

MossRoad · Jan 20, 2015

Here's a link to the formulas...
Levers

dodge man · Jan 20, 2015

Sounds like problems I worked in college. Just the trailer sitting still is a static problem. I think your getting into "moments" with this problem, meaning the weight farther away fron the axle has more effect than the weight close to the axle. I can't answer your question though. Then you get into the dynamics of the load when it moves, a entire new set of problems.

dodge man · Jan 20, 2015

I'm not sure the lever formula is100% valid since you don't have a point load but a load of the trailer over its entire length.

timswi · Jan 20, 2015

If it were me, I'd just hire a small crane. I did that with a 50' beam when I built my house and the crane barely even grunted so I'm certain it could handle a trailer frame of similar length..The owner said what he was doing for me was nothing for the truck.. Much safer.

MossRoad · Jan 20, 2015

dodge man said:
I'm not sure the lever formula is100% valid since you don't have a point load but a load of the trailer over its entire length.

I agree. I figured 4000# hanging 16' off of each side would cancel each other out. So would the 16' on each side carrying that 4000#. So, I came up with 2000# in a length 8' long starting 16' from the axle and ending 24' from the axle, then took the middle of that, which is 20', since I don't know any better! :laughing:

MossRoad · Jan 20, 2015

Trying to find a calculator to calculate tongue weight.... pretty much everyone tells you to use a tongue scale.... no kiddin? There's got to be a mathematical formula for it somewhere.

Beltzington · Jan 20, 2015

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

timswi · Jan 20, 2015

Beltzington said:
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

very interesting math/physics question though.

Beltzington · Jan 20, 2015

Here is the proprosed bridge location 16' will just make level on the other side.
View attachment 407770

CobyRupert · Jan 21, 2015

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: