Here's the formula(s) we use

Beam Bending (available on Wikipedia)

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

$$sigma_{max} = y_{max}Efrac{d^2,delta y}{d,x^2}_{max}$$

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:

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

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

$$Iapprox4.2,in^4$$

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 $x=L$ $$q(x)= -delta(x)F+delta(x-L)F$$

$$EIfrac{d^4,delta y}{d,x^4}=q(x)$$

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

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

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

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:

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

Integrating:

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

Then integrating again:

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

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 $F_{axles}$

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

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

Now the slope will be:

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

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.

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).

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.

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.

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.

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.