# Trailer Frame Beam Design

IdBdan
TL;DR Summary
Building a trailer. Beam calc's ask for (I) - section calc's give me (Ix) & (Iy)???? Which do I use?
I'm building an aluminum tandem axle (7k total capacity) trailer. Boat (18'/1300#) on the rear and a Ryker Can Am 3 wheel (750#) on the front. It's a 23'-6" x 6'-8" bed. Max live load will be approx 2250#. 900# frame weight. 230# axles.

I'm testing beam sizes with online calculators and need some help.

I'm using a "section properties" calculator to find the Moment of inertia. https://calcresource.com/cross-section-channel.html
I'm using a "cantilevered beam" span calculator for the span distance from back axle to the rear. https://calcresource.com/statics-cantilever-beam.html
And a "simple span" for the distance of front axle to trailer tongue. https://calcresource.com/statics-simple-beam.html

My beam span calculators require Length (L) - Youngs modulus (E) - and Moment of inertia (I). My section properties calculator is giving me Moment of inertia in (Ix) and (Iy) and (Iz). I'm using the (Ix) number in the beam calc box for (I).

My questions: 1. Is using (Ix) correct? Or do I add Ix and Iy together? 2. Do these calculators typically use member self weight in their calc's?

I'm mainly looking for the deflection criteria. I'm trying to keep that under 0.15" (with load). The boats CG is over the axle center so tail bounce should not be a problem. Torsional movement (Iz) is almost non-existent with the cross bracing member size, spacing, and diagonal bracing I've used. The axle system is sliding/adjustable. So I'll be able to tune the tongue weight close to perfection by adjusting the axle location. That should greatly reduce any major lateral instability while towing.

My current choice is an Aluminum Association Channel 8x3.75 - 0.41tf/0.25tw.
Cantilever = 106 in - (69MPa) - MoI 52.04 in^4 - 40# uniform load = (0.101 deflection @ 116")
Simple Span = 144 in - (69MPa) - MoI 52.04 in^4 - 35# uniform load = (0.031 deflection @ 82")

The 32" between axles was ignored. Channel bottom flange is being reinforced with a welded 3/8"x3" sliding axle plate for the center 86" of the total bed length. Both spans are getting that extra moment support and it's not incl in calc. Adds an additional safety factor. The tongue is an A frame that goes under the side channel adding support for another 20% of the front span.

If anyone thinks I'm over designed please let me know why. Aluminum prices are crazy high right now.

Thanks
Dan

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IdBdan

IdBdan
I'm curious why no one can answer my question? Did I do something wrong here?

Gold Member
We’ll, it’s a bit to take in, but personally I didn’t see it until today. 3D loading on a system of beams is a trickier to deal with computationally by hand. There are also dynamic loads also to consider. I’ll see if I can come up with a reasonable approach.

Gold Member
Ignore the torque on the beams from the eccentric loading and the tie down rigging to get a first order approximation of the loading.

From your diagrams this looks to me like:

##R'_t## is the reaction at the tongue divided by 2
##M_t## is the moment from the reaction at the tongue applied to the beam ##M_t = 2 R'_t d##, where ##d## is the perpendicular distance between the tongue and the frame.
##R_a## is the reaction at the axle (solve for ##R_a## in the most forward position)
##P## the weight of the ATV.
##w## is the distributed load from the boat (load per unit length), (with gear etc...)
##W_b## the weight of the beam
##W_c## weight of the crossmembers

By the way, these are all "half loads". (except the weight of the beam ##W_b## and the moment ##M_t##)

You need to solve for the reactions ##R'_t, M_t,R_a## using ##\uparrow \sum F = 0## and ##\circlearrowright \sum M = 0##.

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Mentor
Will you ever tow your boat without the 3-wheeled ATV loaded? If so, the trailer looks like it will be fairly unstable...

Gold Member
Will you ever tow your boat without the 3-wheeled ATV loaded? If so, the trailer looks like it will be fairly unstable...
Oh, that’s a trike…not the motor. Whoops.

berkeman
Mentor
It does kind of look like an outboard motor the way it is drawn...

erobz
Gold Member
After you get the reactions, the graphical approach is to plot Shear/Moment diagrams, with the goal of determining the moment as a function of spatial coordinate ##x## acting along the length of the beam.

Then, the elastic curve is given by:

$$EI\frac{d^2y}{dx^2} = M(x) \tag{1}$$

The maximum deflection in the beam is found by setting ##\frac{dy}{dx} = 0##, solving for ##x##, and substituting that into ##y(x)##.

*Note* It's best to use singularity functions ( as opposed to a graphical solution ) for this because of the many discontinuities in loading along this beam. If doing it graphically you would need to solve for a different ##M(x)## between each discontinuity (point load), and solve (1) over each domain, using initial conditions from the last domain. It would be tedious us work to say the least.

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Gold Member
Obviously concerns about the stability of the trailer should be addressed (I don't have much experience with hauling/trailers and what makes them stable/unstable going down the road), but the methods of analysis for the beam won't change.

IdBdan
Ignore the torque on the beams from the eccentric loading and the tie down rigging to get a first order approximation of the loading.

From your diagrams this looks to me like:

View attachment 324006

##R'_t## is the reaction at the tongue divided by 2
##M_t## is the moment from the reaction at the tongue applied to the beam ##M_t = 2 R'_t d##, where ##d## is the perpendicular distance between the tongue and the frame.
##R_a## is the reaction at the axle (solve for ##R_a## in the most forward position)
##P## the weight of the ATV.
##w## is the distributed load from the boat (load per unit length), (with gear etc...)
##W_b## the weight of the beam
##W_c## weight of the crossmembers

By the way, these are all "half loads". (except the weight of the beam ##W_b## and the moment ##M_t##)

You need to solve for the reactions ##R'_t, M_t,R_a## using ##\uparrow \sum F = 0## and ##\circlearrowright \sum M = 0##.
Thank you very much. I apologize that you went to that much effort. I took a shot at completing that calc but it's beyond my abilities. I've called a few local engineers to see if I could pay them to size the beam for me. No seal, just a verbal recommendation with no liability. They were hesitant and I understand why.

Can you tell me if using LX is correct in my question?

And is locating my 2" oc bolt pattern in line with the beam centroid a problem. (see drawing) Would it be better to offset from the centroid?

With personal projects in the past I've just checked span capacities with a uniform loading and added a CYA size up to obtain minimum deflection. I know that's not the best approach but all I have right now. The front span doesn't worry me. It will have an aluminum deck that will lateraly tie all the framing in the first 7'.

It's the tail end cantilever that has my concern. In "half load" the beam will have to carry 40#/LF over 9.25 LF. Cross members only add 5# every 3'. My axle system is sliding. If need be I can slide the axle farther to the rear to compensate.

I don't think moment is going to be an issue when the trailer is loaded. With boat trailers the boat tie downs turn the boat hull into a frame stiffener. And the boat tie down at the bow actually creates a counterweight using the boat as a beam. It's not a signifigant addition but it reduces tail bounce. Moment when it's empty is what I looked at and it's more than acceptable.

I have a 3/8"x3"x 86" axle plate making my bottom flange thickness 7/8" ttl and it extends 46" fore and aft of the axle location. I'm hoping that and the web stiffeners I've added will give it additional strength.

Thanks again! Greatly appreciate the help.

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• TRAILER AXLE.pdf
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IdBdan
Will you ever tow your boat without the 3-wheeled ATV loaded? If so, the trailer looks like it will be fairly unstable...
That was the toughest part of the design. Running multiple scenarios of loading and finding the the right match for a 10 to 15% tongue weight. https://www.engineersedge.com/calculators/trailer-weight-balance.htm

I designed this trailer for my motor home. I want my boats when I travel and I need transportation when I park the motor home.

When the flats boat comes off the trike is loaded from the rear and moves to where the boat is. The bunks for the boat will match the trike wheel base.

When the trike is removed the winch post for the boat slides forward up to 26". 21" is what I calc'd for a 12.5% tongue weight. I'm using airline track with 1" incremental slots.

In odd conditions or different loads my axles can slide 22" forward or backwards from center position in 2" increments. I did that so I can leave the salt water flats boat home on cross country trips. I'll carry my inflatable skiff and the trike. That load requires moving the axle location.

IdBdan
It does kind of look like an outboard motor the way it is drawn...
It does now that I look at it. lol

It's a Can Am. Conventional trikes steer one wheel. The Ryker puts the two wheels up front for steering.

Gold Member
Can you tell me if using LX is correct in my question?
If the beam is loaded like:

You are considering bending moments about axis ##{x-x}##, thus ##I_{xx}## is the appropriate moment of inertia to use.

With personal projects in the past I've just checked span capacities with a uniform loading and added a CYA size up to obtain minimum deflection. I know that's not the best approach but all I have right now. The front span doesn't worry me. It will have an aluminum deck that will lateraly tie all the framing in the first 7'.
Applying a "CYA" a.k.a. "Factor of Safety" is indeed good practice, but first we try to get the load approximated as accurately as possible. Sometimes it takes a little effort.
It's the tail end cantilever that has my concern. In "half load" the beam will have to carry 40#/LF over 9.25 LF. Cross members only add 5# every 3'. My axle system is sliding. If need be I can slide the axle farther to the rear to compensate.
So is the weight of your boat 740 lbf?
What is the weight of the Trike?

You state the frame is 900 lbf.

You say the "live load" is 2500 lbf. is that just the total weight of everything?

I'm thinking the weight of the boat is supported by the last 4 cross members, 4 point loads not necessarily evenly distributed. Presumably there is an outboard motor hanging off the back, hence the last support is bearing more load. Just ignoring the small weight of the cross members:

If you can just help me reason out the loadings ( the ##T## from the trike, and the ##F_b##'s from the boat), by providing reasonable weights/distributions/dimensions from the leading edge of the beam, this isn't a terrible computation. You can check it against whatever you had planned.

IdBdan
Thanks for taking your time to do this! I've attached a PDF for a better look at the loading.

My objective was to use rectangular tube. Either 2x5x.25 or 2x6x.25 (6061 T6). The 8" channel was what fit the load numbers I came up with while 'fumbling' my way through this design. It's not a good match for the welded joints at mid span of the web, but it met the load.

Tube gave me a stronger and cleaner weld condition. And it simplified my sliding axle design. I was also using the tube frame as a water reservoir for my brake wash down system. Salt water eats brakes.

The trike loading 'T' is mostly concentrated in the front center of the trike body. So I would say each front wheel is just under a half load. About 285# ea wheel. They are sitting approx 5" in from the beam edge and directly over cross members. The rear wheel isn't carrying any more than 75# and sits directly over a cross member. The area the trike sits on is an extruded aluminum decking that is perpendicular to the cross members. It's clipped to the cross members at the butting edges. And it gets a 1/4" bolt mid span of the cross members in the butt joint of the vertical leg. I'm assuming it's adding to weight distribution with it's lateral attachment?

The rear Fb would be 240# uniform half load with a 120# point half load for the motor?

Then forward of your "Ra" the boat hull would be a 345# uniform half load up to the trike location. I'm not sure how the curve of the hull (not on the frame) works in loading? I'm thinking the area forward of Ra to the point where the hull leaves the frame would carry the full forward load? Because that load falls over the axles I was ignoring any drastic effects in deflection. The 'V' forward bunks don't carry any weight. Just there to align the hull to the winch.

Thanks again! I owe you one.

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Gold Member
How much does the boat weigh without the motor? How much does the motor weigh? You've said 1300 lbf (I assume boat plus motor) in the OP, but in the diagram show 720 lbf (including motor)? Don't cut any loads in half, I'll do all that and figure out the distributions. There is a lot of numbers to juggle and It's getting confusing on who is doing what to the loads.

Gold Member
I think you also need to consider dynamic loading caused by acceleration, braking, hitting kerbs at speed and cornering.

Nik_2213, Lnewqban and erobz
Nik_2213
What sort of damping on those wheels ? As drawn, look like classic 'Indesuspension' with inner and outer square sections set ~45º off. Spacing looks bigger than 'close coupled' so, IIRC, you may need to go up a size / load rating. You must allow for considerable impact loads from potholes, a lot of them torsional.

I've seen a big trailer 'come apart' when A-tongue's welds failed. IIRC, trailer had been sorta 'dancing', rocking from side to side, so may assume metal fatigue. Fortunately, the 'breakaway' cable held, so trailer did not escape, and sturdy 'jockey wheel' kept frame from digging into road...

FWIW, may I suggest provision for two (2) spare trailer wheels plus a pair for your trike-ish runabout ??

Tom.G
IdBdan
How much does the boat weigh without the motor? How much does the motor weigh? You've said 1300 lbf (I assume boat plus motor) in the OP, but in the diagram show 720 lbf (including motor)? Don't cut any loads in half, I'll do all that and figure out the distributions. There is a lot of numbers to juggle and It's getting confusing on who is doing what to the loads.
Sorry for the confusion. You had mentioned half load in a prior post and I thought you were treating the frame 'per side'.

The (down arrows) 720 / 690 / 650 figures are the weights and locations I used to calculate the axle location and the tongue weight. I thought they would apply.

I've made some changes since my OP to the Ryker platform material. I changed to an aluminum decking and added a battery to the bow/front of the boat. So my max weight has changed slightly.

Boat
Dry hull weight = 740#
Motor weight = 239#
Accy's / batteries/ fuel/ etc = 411#
Total = 1390#

Front Platform
Ryker = 650#
Aluminum Deck = 110#
Total = 760#

TTL = 2150

Cross members = 2x4x.25 rectangular tube - 6061 T6 round corner (3.3# / LF)

Frame weight estimated total remains the same 900#.

Thanks again for all this help.

IdBdan
What sort of damping on those wheels ? As drawn, look like classic 'Indesuspension' with inner and outer square sections set ~45º off. Spacing looks bigger than 'close coupled' so, IIRC, you may need to go up a size / load rating. You must allow for considerable impact loads from potholes, a lot of them torsional.

I've seen a big trailer 'come apart' when A-tongue's welds failed. IIRC, trailer had been sorta 'dancing', rocking from side to side, so may assume metal fatigue. Fortunately, the 'breakaway' cable held, so trailer did not escape, and sturdy 'jockey wheel' kept frame from digging into road...

FWIW, may I suggest provision for two (2) spare trailer wheels plus a pair for your trike-ish runabout ??
I'm using tandem 3.5k torsion axles at 22.5 degree down. They are mounted directly to the frame and better than leaf springs when it comes to reducing flex and stiffening the frame. They eliminate bounce typically caused by standard leaf spring axles. They are also independent of each other and ensure one wheel is always on the ground as the other is lifted from extreme bumps. And the tire reaction to a bump is not directly vertical. It's rotational and to the rear.

Trailers fail because people think they can load them any way they want. Tongue weight to load weight is essential if you want it to track properly at variable speeds.

One spare tire mounts to the tongue on an operable axle hub. It's 5 bolted. If there's a break away, like you said, in lieu of relying on the jack stand wheel the spare hits the ground. The main axles also have disc brakes that automatically apply with a break away.

I couldn't change a tire on the Ryker if I wanted to. It takes special tools and torque spec's to change a front tire. And the rear tire is a direct drive system. It has an automatic transmission. Bad tires on that bike are a tow truck call.

Gold Member
I think you also need to consider dynamic loading caused by acceleration, braking, hitting kerbs at speed and cornering.

It would not be a bad thing to consider regulatory safety in this case. If this is in the United States, the federal Department of Transportation has some things to say about trailer design, materials of construction, braking requirements, hitch designs, tow vehicle capability, and signal needs.

I expect states to have similar concerns. If this will be a licensed/tagged vehicle for regular use, you may want to talk to an insurance professional about whether or not it (or its contents) will be insurable without professional design and approved capacity rating.

The manufacturing of trailers has a great deal of liability associated with it--People crash. They also overload things.

If you have an original design, you are taking on both the design and manufacturing liability. Designing, and, more importantly, properly welding aluminum trailers so that they don't fail, fatigue, etc. is worth considering.

tech99, berkeman, erobz and 1 other person
Gold Member
I have no expertise in structural engineering and or trailer design and the dynamic loads ( @tech99 post #17 points out) they might experience under normal operation, and this is not a validation of the overall trailer design in any way. This kind of structure and 3d loading is much better suited for FEA software. You know how to draft, so perhaps just get a license and learn to the basics of that software would be better comprehensive analysis. As others have said, you'll have to ensure your design meets regulations (federal and/or state) and whether or not the trailor and the load can be insured given self design/manufacture. See @ChemAir response #21.

That being said, here is what I get:

#Note#: The beam is actually statically indeterminant due to the dual axle system. That would require more computational analysis that I'd like to go through. I've combined the tandem axle reactions into a single reaction at the central location between axles.

Here are the parameters I used: I gave a dynamic load factor of 120%

Here are the resulting Shear/Moment/Deflection Diagrams:

Here are the equations that produced them:

\begin{align} V(x) = R_t - \lambda x - P_{atv}\left< x-a \right>^0 - P_{atv}\left< x-b\right>^0 + \tag*{} \\ R_a \left< x-d\right>^0 - w \left< x-e\right>^1 -\frac{P_{motor}}{2} \left< x-f\right>^0 \tag{shear} \end{align}

\begin{align}M(x) =M+ R_tx - \frac{1}{2} \lambda x^2 - P_{atv}\left< x-a \right>^1 - P_{atv}\left< x-b\right>^1 + \tag*{} \\ R_a \left< x-d\right>^1 - \frac{1}{2}w \left< x-e\right>^2 \tag{Moment} \end{align}

\begin{align} y(x) = \frac{1}{EI} \left( \frac{1}{2}Mx^2 + \frac{1}{6}R_tx^3 - \frac{1}{24} \lambda x^4 - \frac{1}{6}P_{atv}\left< x-a \right>^3 - \frac{1}{6}P_{atv}\left< x-b\right>^3 +\tag*{} \right. \\ \left. \frac{1}{6}R_a \left< x-d\right>^3 - \frac{1}{24}w \left< x-e\right>^4 + C_1 x \right) \tag{Deflection} \end{align}

The deflection was solved with initial conditions ##x=0,y=0##, ##x=d, y=0##. Which yields for ##C_1##

$$C_1 = \frac{1}{d} \frac{P_{atv}}{6} \left( (d-a)^3 + (d-b)^3 \right) + \frac{1}{24}\lambda d^3-\frac{1}{6}R_td^2-\frac{1}{2}Md$$

Here are the reaction equations:

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IdBdan
Thanks! Can't tell you how much I appreciate your time on this. And I agree that anything you're helping me with is totally and unquestionably hypothetical.

One last question. In your 'parameters', is the M.o.i. (46.13) what I need to exceed or is that the M.o.i of the member you tested? The reason I ask is because your deflection graph show's -1.4" at the tail end? Wouldn't that be considered excessive? I've been trying to stay under 0.25". Am I being to strict with that requirement?

Thanks again.

Gold Member
One last question.
No need for it to be the last question.
In your 'parameters', is the M.o.i. (46.13) what I need to exceed or is that the M.o.i of the member you tested?
The M.o.I is for the C channel that you specified( 8x 3.25x.41x.25). I just rough calculated it though, if you have an exact value from the manufacturer, we can use that. I didn't notice that you had given 52 in^4 in the OP, is that the manufacturer spec?
The reason I ask is because your deflection graph show's -1.4" at the tail end? Wouldn't that be considered excessive?
It's hard to tell. The normal stresses in the beam are not very high ( S.F. ##\approx## 5 on material yield stress ignoring forces from the second axle, and combined torsional loading effects present). The stresses are what determine the cycles until fatigue failure for the beam...so having the average stress low is a good start.

As for the deflection, I would try to tame it by boxing the beam with some flat plate in a good location. My gut is telling me about a few feet on either side of the axles would help ( if you could imagine taking your finger and pushing the curve down toward zero close to the 150 inch range it seems like you could do some damage to the tail deflection issue), but I haven't computed anything (its tedious business to compute with the moment of inertial changing along the beam).

It's likely that the assembled frame is stiffer than just the beams. How much...going to have to consult FEA on that. The issue with not trying to stiffen the beams is if it is deflecting under that load, then whatever is tied to it is going to act like a stiffener i.e. your boat in this case. So, while it may not deflect as much, that trades off with the stress of the load being carried partially by the boat. I don't know if that is a good trade?

Gold Member
## \text{M.o.I} = 50 ~\rm{in^4}##