Trailer Loading and CQ - Hypothetical Discussion

[Beltzington]

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]

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]

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]

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

 

[Beltzington]

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]

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]

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]

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]

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]

Some math wiz please figure this out! :laughing: