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I'm thinking about building another trailer. I've built lots of smaller trailers in the past, but this time I'd like to build myself a small tandem axle gooseneck rated for 7,500 lbs.

I am a certified welder, I have a Bachelor's degree in Physics, and I work as a software developer. I have the know how, but I would like some input on material selection.

  • Round Pipe
  • Angle Iron
  • I-Beam
  • Rectangular Tubing

I am currently leaning towards the Rectangular Tubing as the best for support, but I cannot seem to find anything online that confirms this.

After deciding on the best material for the trailer, where would I find good charts to tell me what size and thickness I should use? Obviously, I could go overkill, but I would like to build this one smarter rather than throwing as much iron at it that I have.

Does anyone have any input? Is there a better group to post this in? I was looking for something along the line if Industrial Engineering, but this is all that pulled up.

EDIT:
I was trying to keep this a generic question where someone could tell me something like, "Here is the formula we use, and this is how to use it..." It looks like I won't get that, though.

My heaviest load would be a tractor with a front end loader and a brush cutter on the back with a total weight of 5500 to 6500 lbs. A tandem axle trailer with two (2) 3500-lbs axles can support this load fine. I have selected axles from Southwest Wheel's torsion axle with brakes (the front axle will have brakes, but not the rear).

Trailer length will be 18-foot, and have a gooseneck configuration (it distributes the weight better and pulls smoother than a bumper trailer). For calculations, I'm going to use 7500-lb capacity.

I am looking at the structural data for square tubing using a spec sheet HERE (trying not to advertise another website, but that is where I see data). Page 21 shows data values for various sizes and thicknesses.

There is a line called Bending Factor. For an 18-foot trailer (18 x 12 = 216 inches), 3/8-inch thick 4x2 square tubing shows a Bending Factor of (x=1.03 , y=1.55).

I was using Rogue Fabrication's Calculator yesterday, where I entered the following values: Tube Shape=Square Tubing, Outside Diameter=4-in, Wall Thickness=0.1875-in, Material="Cheap seamed tube", Load=3800-lbs, Tube Length=216-in, and Hazard Factor=1, I got that my material is 1.22 times as strong as the loading conditions.

Next, I tried EasyCalculation's Beam Deflection Calculator, with values of Length=216, Width=2, Height=4, Wall Thickness=0.1875, Force=3750. It shows a deflection of about 100 inches for 2 lengths of rectangular tubing. If I use 4 lengths, that drops the force down to 7500/4=1875 per beam, and deflection down to 50 inches. Those deflection values seem really high. That is more iron than most trailers have.

The old tandem axle trailer I use now only has two (2) lengths of 4-inch angle iron (1/4-thick). It flexes a couple of inches, but not 50 inches. I must be missing something.

How do I calculate the amount of flex a 20-ft length of material would have?

If square tubing is not best, that's fine as long as you let me know what would be better and how you selected that configuration when you comment.

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    $\begingroup$ This question doesn't seem to be within the scope of this site. It is a mix of subjective opinion (every cross-section has its advantages and disadvantages, so you can't define the "best" one without knowing what is precisely meant by that) and of requesting references, neither of which are in scope. $\endgroup$ – Wasabi Sep 21 '15 at 17:12
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    $\begingroup$ The shape of a member affects its area moment of inertia, which affects the loading in the member. You need to define a load, a factor of safety, a material, and then play with dimensions and shape until you find a combination that gets your load*(factor of safety) under the yield strength of the material. I put this in a comment instead of an answer because you haven't defined anything about the trailer except the generic class and that you want a tractor on it. If you want a reference try any "deformable bodies" or "machine design" text. $\endgroup$ – Chuck Sep 21 '15 at 19:33
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    $\begingroup$ We aren't asking to make it as complicated as possible. We're asking to make it as correctly as possible. Figuring out good design is not trivial and should not be taken lightly. $\endgroup$ – Wasabi Sep 22 '15 at 23:11
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    $\begingroup$ You have 3 options - do the math and get the right parts, copy someone else's work and hope they did the math, or guess and hope it works out okay. I mean, what's the worst a 3.5 ton load hitched to a car going 55mph on a populated highway can do? Sarcasm aside, you should probably either take this seriously and do the work or find a different project. $\endgroup$ – Chuck Sep 23 '15 at 0:13
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    $\begingroup$ In our office we have a general rule ... "if something moves, it's mechanical" Maybe try tagging this with Mechanical Engineering? The dynamic loads on the beams and especially the connections would become critical. Just doing a static analysis on the structure would not be enough. $\endgroup$ – NamSandStorm Sep 28 '15 at 12:59
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Here's the formula(s) we use

Beam Bending (available on Wikipedia)

$$EI\frac{d^4\,\delta y}{d\,x^4}=q(x)$$ $$I=\int (y-\bar y)^2 dA$$ $$\bar y= \frac1{A}\int y \;dA$$

$$\sigma_{max} = y_{max}E\frac{d^2\,\delta y}{d\,x^2}_{max}$$

Where $A$ is the cross-sectional area of the beam, $y$ is the position along the axis in the direction of the beam loading, $\delta y$ is the deflection in the direction of loading, $E$ is the modulus of elasticity (search matweb to get a value for your material), and finally $q(x)$ is the load per distance function.

Here's how to use them

For a rectangular tube with height $H$, width $W$ and thickness $t$ we have:

$$\bar y=0$$ $$I=W\int_{-\frac{H}2}^{\frac{H}2}y^2\;dy-(W-2t)\int_{-\frac{H}2+t}^{\frac{H}2-t}y^2\;dy=\frac{H^3W-(H-2t)^3(W-2t)}{12}$$

For $H=4\,in\,,\quad W=2\,in\,,\quad t=.1875\,in$

$$I\approx4.2\,in^4$$

Now there's beam loading, this is likely where you ran into difficulty. First lets take a look at a cantilever beam:

Here there are just two points that are loaded, the support and the tip. Think of a diving board scenario. We'll say the support is at $x=0$ and the load $F$ is at $x=L$ $$q(x)= -\delta(x)F+\delta(x-L)F$$

$$EI\frac{d^4\,\delta y}{d\,x^4}=q(x)$$

$$EI\frac{d^3\,\delta y}{d\,x^3}=\int_{0^-}^xq(x) dx = F$$

This is basically saying that there's a constant shearing stress in the beam the whole way.

$$EI\frac{d^2\,\delta y}{d\,x^2}=\int F dx +C=F(x-L)$$

This expression is for the bending moment in the beam. We know that the free end must have a bending moment of zero so we set the integration constant to accommodate that.

$$\frac{d\,\delta y}{d\,x}=\frac1{EI}\int F(x-L) dx +C=\frac{F}{EI}(\frac12 x^2-Lx)$$

This represents the slope of the deflected beam. Here we know that the slope must be zero at the support so we've set the integration constant accordingly.

$$\delta y=\frac{F}{EI}\int \frac12 x^2-Lx \; dx +C=\frac{F}{EI}(\frac16 x^3-\frac{L}2 x^2)$$

Here we know the deflection is zero at the support so we set eh integration constant accordingly. Now if we just want to look at the deflection at the end we plug in $x=L$

$$\delta y=-\frac{FL^3}{3EI} $$

This corresponds to the equation on last website in your post.

From matweb for medium alloy steel we have $E=30\,000\,ksi$ So plugging in:

$$\delta y= -\frac{3.750\,klb \, (216 in)^3}{3\,30000\,ksi\,4.2\,in^4}\approx -100\,in$$

This is exactly what the online calculator produced. However, if you tried to load a beam like this it would permanently deform. An 18 foot lever arm is really long and will bend the snot out of a 4in thin wall beam with only moderate difficulty. The issue is that a trailer is not a cantilever beam.

so let's take a look at a more reasonable loading scenario. Let's model the axles as a point loads located $40\,in$ and $80\,in$ from the end of the trailer, the $7500 lbf$ load as distributed over the rearward $18\,ft$ and the gooseneck support an additional $5ft$ in front of that.

Now some of our loads aren't known yet, but we can figure out some of them in the process. Some of them we can't though, so let's add an additional constraint. The weight distribution will be split between the axles according to the variable $\alpha$

$$F_{axles}=F_{rear} \frac1{\alpha}=F_{front}\frac1{(1-\alpha)}$$

Now we have:

$$q(x)=-\frac{F}{L}H(L-x)+F_{axels}(\alpha\delta(x-x_{rear})+(1-\alpha)\delta(x-x_{front})+(F-F_{axels})\delta(x-x_{goose})$$

Integrating:

$$EI\frac{d^3\,\delta y}{d\,x^3}=\begin{cases} -\frac{F}{L}x & x\leq x_{rear} \\ -\frac{F}{L}x+F_{axels}\alpha & x_{rear} \lt x\leq x_{front} \\ -\frac{F}{L}x+F_{axels} & x_{front} \lt x\leq L \\ F_{axels}-F & L \leq x \end{cases} $$

Then integrating again:

$$EI\frac{d^2\,\delta y}{d\,x^2}=\begin{cases} -\frac{F}{2L}x^2 & x\leq x_{rear} \\ -\frac{F}{2L}x^2+F_{axels}\alpha (x-x_{rear}) & x_{rear} \leq x\leq x_{front} \\ -\frac{F}{2L}x^2+F_{axels} (x-(1-\alpha)x_{front}-\alpha x_{rear}) & x_{front} \leq x\leq L \\ (F_{axels}-F) (x-x_{goose}) & L \leq x \end{cases} $$

Note that this bending moment must be continuous and both ends must be zero as there is no bending moment applied at the ends (they are free to rotate) This leads to an additional constraint that can be used to find $F_{axles}$

$$F_{axels}=F\frac{x_{goose}-\frac{L}2}{x_{goose}-(1-\alpha)x_{front}-\alpha x_{rear}}$$

However, to keep the expressions shorter lets leave $F_{axels}$ in the expressions.

Now the slope will be:

$$\frac{d\,\delta y}{d\,x}=\frac1{EI}\begin{cases} -\frac{F}{6L}x^3+C_1 & x\leq x_{rear} \\ -\frac{F}{6L}x^3+F_{axels}\alpha \frac12 ( x-x_{rear})^2+C_1 & x_{rear} \leq x\leq x_{front} \\ -\frac{F}{6L}x^3+F_{axels}(\alpha \frac12 (x-x_{rear})^2+(1-\alpha)\frac12(x-x_{front})^2)+C_1 & x_{front} \leq x\leq L \\ (F_{axels}-F)\frac12 (x-x_{goose})^2 +C_2 & L \leq x \end{cases} $$

And at this point I moved over to a numerical solution. I integrated again and found values for all the constants such that both the slope and displacement were continuous and the displacement at the goose and the rear axle were zero. The resulting deflection had a maximum at about 2 inches. But I used the full load and I should have used half the load giving 1 inch. That sounds about right to me.

Note that the peak bending moment is $9\, kN \,m$ which when multiplied by half our height and divided by our area moment give a peak stress of $38 ksi$ this is about 13% the yield strength of the medium alloy steel on matweb. You mihgt hink this would be sufficient then however, this is only for a static trailer, not one moving and bumping around.

The acceleration forces on a trailer could easily triple the load over short periods. Additionally, the bumps in the road will cycle the loading making it not the yield strength you want to look at but the fatigue strength at the appropriate number of cycles you'd like the trailer to last. The fatigue strength may be as low as 10% of the yield strength, so I would want a minimum load factor of about 30 (3/10%), then add a factor of safety of 2 and your beams need to be about 60 times stronger than would be required to just meet your yield stress in a static load scenario. In short, I would go with bigger beams.

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    $\begingroup$ +1 for the spectacular answer. I hope @jp2code realizes the effort involved in making "just a trailer". $\endgroup$ – Chuck Sep 30 '15 at 15:36
  • $\begingroup$ @Chuck, I doubt any gooseneck trailer manufacturers have used these calculations. This is a spectacular answer that I may eventually accept, but I would like to know how manufacturers determine what size material they need when building a trailer for a given load range. $\endgroup$ – jp2code Sep 30 '15 at 17:15
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    $\begingroup$ @jp2code it's this or guesswork. $\endgroup$ – Chuck Sep 30 '15 at 17:57
  • $\begingroup$ @jp2code Like most problems, once you solve the problem once you can make a tool to re-do all the calculations when your numbers change. So, no they don't go through this for every trailer design. They made a tool to do it for them. Then they probably verify their design with an FEA analysis. I doubt any gooseneck trailer manufactures have used less than this level of detail calculations, it's just likely to be embedded in a tool similar to the online tools you found. $\endgroup$ – Rick Sep 30 '15 at 19:04
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    $\begingroup$ I'm not sure that this really adds anything useful to real world design. I once designed a 6 lane road bridge without integrating anything at all. I think that it's still standing. Engineer's don't integrate. $\endgroup$ – Paul Uszak Dec 18 '15 at 2:22
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Here is some additional information and lengthy discussion on trailer design criteria. There is even a white paper in the thread on loading and safety factors that should be used:

Loads for Trailer Design

There are a lot of other threads on that site providing good information regarding trailer design.

For what it's worth, I'd start my structure design with some type of rectangular steel section in mind for a trailer. They are regularly available and "easy" to work with(cut, drill, weld, mount other components, etc).

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For the structure of a trailer rectangular section tube is likely to be the most efficient compromise between strength stiffness and ease of design and manufacture. Round tube is a bit stronger weight for weight but much more difficult to assemble and join accurately, simply because rectangular tube has convenient flat surfaces.

As already mentioned thing like this aren't designed by calculus in the real world and by far your best bet is to copy an existing design as failures in this sort of application tend to occur when you get unexpected load concentrations rather than considering the design as an approximated beam so unless you have access to FEA software then paper calculations are a bit pointless.

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  • $\begingroup$ I had hoped that one of the engineers on this site could have said, "It is best to use X for {something} pounds". In the end, I just guestimated: i.imgur.com/mkOJrhS.jpg $\endgroup$ – jp2code Jan 4 '16 at 22:28
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    $\begingroup$ The problem is that the actual load on the trailer is a small part of the overall problem what I will say is that for a 3000 kg load on an A frame about 3m long 100mm x 50mm rectangular box section (3mm wal thickness) is the right sort of ballpark to give you a comfortable factor of safety. $\endgroup$ – Chris Johns Jan 4 '16 at 22:38
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The easy answer is don't design - cheat. Go and look for a trailer similar to what you're after. Photograph it and measure all the bits. (Don't act like you're trying to nick it). Similar sections will do, but I'd err on larger sizes.

Now your issues will be:-

  1. Welding the joints. I'm not sure what type of welding you're certified in, but fillet welding 10mm steel is not the same as tacking on car wings.
  2. Brakes. You need to make sure that they work. How will you test them? Just the fact that the trailer doesn't roll down your drive doesn't mean they're working.
  3. In England, if you took this onto a public highway, it would need a test certificate.

I hate the whole Health & Stupidity nonsense, but do not underestimate the responsibility you will be assuming if you drive this down a road at speed.

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