Question: For redhoss

Tue 7 Sep 2010 - 6:32 pm UTC

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ANSWERED on Tue 12 Jan 2010 - 10:24 pm UTC by redhoss

Question: For redhoss

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Asked by mbutler315 on Tue 12 Jan 2010 - 7:07 pm UTC:

Redhoss- you helped me last week with a question about 4x2 steel tubing
used in a custom trailer application. You mentioned your son builds
trailers. I have a design question regarding the trailer. Heres what I
know: The ideal set-up of a trailer would have 60% of its weight placed in
front of the axle, 40% to the rear. This ratio gives a 10-15% tongue
weight. My trailer is 8 ft long (not including the tongue which is about
48").The axle is 35" from the rear,so as it sits I am assuming the ratio is
64/36. I will be adding a 2500 lb piece of equipment to the trailer.The
piece is perfectly round with 54" diameter. Here's what I don't know: Can I
place this equipment 2-3" behind the axle?(best case scenario for me). If
not, can I safely place it directly over the axle 50/50 (2nd best scenario
for me) or do I absolutely have to place it in front of the axle?(worst
case scenario for me). I plan on using a scale to check the tongue weight.
However, once I begin installing the equipment it really can't be moved. I
need to keep the tongue weight around 400lbs. The finished weight of the
trailer should total about 4000lbs.
I'm hoping you and your son can give some guidance. Thank you- Mark

Uclue Researcher Answer by Researcher redhoss on Tue 12 Jan 2010 - 10:24 pm UTC:

Hello again Mark, my son agonizes over this exact problem on many of the
trailers he builds. If we use your numbers, it is easy to come up with an
answer.

Summing the moments about the axle(unloaded):

R(tongue weight) x 109 = 4000 x 13
R = 477#

This falls within your 10/15% number.

Summing the moments about the axle(loaded with the load centered 3" behind
the axle):

R x 109 = 4000 x 13 - 2500 x 3
R = 408#

Assuming your numbers are accurate, this is probably pretty close to what
you will see. Is this what you are looking for. Please ask for a
clarification if you have any questions.

Request for clarification by mbutler315 on Tue 12 Jan 2010 - 10:38 pm UTC:

So with the load centered about 3" behind the axle, the tongue weight will
be 408lbs or about 10% of the total weight. Correct? Can you explain the
numbers in your equation so if I have to adjust I can recalculate? Thank
you-Mark

Uclue Researcher Answer clarification by Researcher redhoss on Wed 13 Jan 2010 - 3:35 pm UTC:

Draw yourself a sketch of a side view of the trailer. Put a pivot mark at
the axle location, an R at the hitch, an arrow labeled 4000# pointing down,
and an arrow labeled 2500#. Locate all these forces as you have described
and calculate their distances from the axle pivot point. Then you will see
that:

109 = 48 + 96 - 35

13 = 48 - 35

3 is obvious from your sketch

Now look at the direction the loads try to rotate the trailer about the
axle pivot. I think you will then understand. This is a pretty rough
approximation and you could refine the numbers using weights that you
calculate. Obviously I have left out the weight of the tongue. The
principle of summation of moments is a very useful tool. The sketch you
drew is called a free body diagram. You are now an engineer.

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