# Steel selection for building a trailer

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.

• 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.
– Wasabi
Sep 21, 2015 at 17:12
• 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. Sep 21, 2015 at 19:33
• 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.
– Wasabi
Sep 22, 2015 at 23:11
• 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. Sep 23, 2015 at 0:13
• 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. Sep 28, 2015 at 12:59

# 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.

• +1 for the spectacular answer. I hope @jp2code realizes the effort involved in making "just a trailer". Sep 30, 2015 at 15:36
• @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. Sep 30, 2015 at 17:15
• @jp2code it's this or guesswork. Sep 30, 2015 at 17:57
• @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.
– Rick
Sep 30, 2015 at 19:04
• 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. Dec 18, 2015 at 2:22

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:

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).

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.

• 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 Jan 4, 2016 at 22:28
• 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. Jan 4, 2016 at 22:38

If this is your first time working with structural steel sections, or simply looking for accurate data on their mechanical properties, find the official "Handbook of Steel Construction" for your region. Here in Canada it is "Canadian Institute of Steel Construction (CISC): Handbook of Steel Construction" and in the USA it is the "American Institute of Steel Construction (AISC): Steel Construction Manual". I am not sure about other countries but they pretty much all follow the same title format and are either called "The Handbook of Steel Construction" or "Steel Construction Manual". It should be fairly easy to find the official version for your region if you search for that.

As someone who stumbled across this thread trying to research these same questions, I know how hard it can be to find reliable answers. So I cannot stress this enough. GET YOURSELF A COPY!! This book will literally answer every question you have and I really wish I would have found it sooner.

Cheers, eh.

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.

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.

If it was me I would use I beam for the entire frame and gooseneck I built one that I pulled with my 2500 HD silverado and used 4 x4 x 30# I beam it was 25ft with a 3.5 ft dovetail and I hauled my pulling truck on it... And my other one is a gooseneck as well but it is made with 12 x 14 # high-tensile steel and I have with the dovetail included a total of 40ft I have 6ft of dovetail on it that raises and lowers on large hydraulic cylinders under it but I haul heavy equipment on it I haul everything from my 12 to 15 ton dozer to my tracked cat skid steer and I've even had my 15 to 20 ton excavator on it but it is built to haul a GVW of 30 to 35 ton it has 4 tandem dual Tire Axel's under it with one set of overload tags and it's all air bag suspension and I pull it with my 3500 HD diesel 4 x 4 Chevy so I hope this post helps you or anyone with A question on building yourself a gooseneck trailer and also when you weld it should really do a 3 pass a root a hot pass and finally a half inch wide cap sorry I'm. IN THE PIPELINERS UNION we double and triple up all welds hope it helps.

• Welcome to engineering.SE! If you could add some punctuation to your answer, it would benefit a lot, as of now, it is still difficult to read. Jul 30, 2020 at 13:15
• Use capital letters in the right places too. This is site policy. See Write to the best of your ability. Jul 30, 2020 at 15:57
• I don't like your answer, Ronnie. If you use high tensile steel, then you should not be welding on it because that will ruin the structural integrity of the metal at the weld. Next, my question stated that I am only pulling 7500 lbs, so your answer talking about a 20 ton excavator and trailers built to haul 30 to 35 tons is just not applicable here. Jul 30, 2020 at 17:28

I’m not a structural engineer but I do understand materials and statics. My degree includes engineering, physics, other sciences and math. Long story short I worked as a civil engineer.

The first problem no one mentioned is the tandem axles being under sized. I’m you have to take the weight of the trailer and add it to your tractor and implements. That trailer is pushing 2500# I’m guessing. Ad the 6500# of your tractor and equipment and you’ve already exceeded the 2 3.5 ton axles capacity. The axles should be 5 ton each. Funny none of the engineers noticed that piece of information. 😂 BTW I am “beefing up” a much smaller trailer to haul my subcompact tractor without the implements just to get it serviced by the dealer. It’s a great \$1,000 trailer for hauling stuff to the dump or maybe a hay ride one day. My trailer can handle 1550# and my tractor weighs 1499# dry. That’s pushing it really close. All I’m looking for is about 200# more capacity. Otherwise I might have to remove my rear tires and put bicycle tires on just for the ride. 🙃