# Replace Aluminum C channel with Steel

1. Apr 2, 2008

### Jim Kopke

Application: Aluminum gooseneck horse trailer incorporates (2) 6 foot long 8 x 2.25 x .25 6062 aluminum C channel beams (sitting long side vertical) from each of the front corners of the box, coming to a point (forming a triangle with the front edge of the box) where they are welded together and rivited to steel plates which suport the (steel) vertical gooseneck post. The beams tranfer the vertical load and braking force from the front of the the box to the gooseneck. For clearance reasons, I wish to cut the bottom 4-5 inches off of each C channel along a 3 foot line at the middle of each 6 foot section, making that 3 foot section 3" high with the 2.25 L on top. To offset the removed material, my thoughts are to reinforce the shortened beam section by bolting steel plate (say 5 feet of 3" x 1"), L or C channel back to back or Box tubing along the top edge of the 8" C channel, spanning the 3 feet where the material had been removed.

My question: Is there anybody out there that can tell me if my assumption that using steel to reduce area is accurate? Or maybe the experience to tell me what steel form/size I need to stay safe and fuctional.....or maybe teach me how to calculate what size and steel shape would roughly possess the strength and integrity of the material removed (or simply match the porperties of the 8" Al C channel)?

If anyone can help me, feel free to eamail with questions. I also have pictures and drawings of what I am trying to do. THANKS EVERYONE (and if you have an electronics questions, I'm your man).

Jim

2. Apr 2, 2008

3. Apr 2, 2008

### TVP45

Are you certain about the 6062?

Although you can do this in principle, the steel replacement will be pretty huge (and heavy) and there's the difficulty of bolting it to the remaining aluminum; if you've just eliminated about 80% of the aluminum's strength, you really don't want to add too many holes.

Send pictures and drawings.

4. Apr 2, 2008

### Jim Kopke

NOTE WEB PAGE HAS BEEN CORRECTED
I have posted several pics on the following webpage.
http://hometown.aol.com/wildwest2a/trailer.html [Broken]

Actual length of material to be removed would only be 2 feet rather than the 3 feet previously stated. Also, I wish to remove 5 inches high, leaving 3 inches. In my crude Photoshop mod, it looks like I'm only leaving an inch or 2.

Last edited by a moderator: May 3, 2017
5. Apr 2, 2008

### Mech_Engineer

Anything is in theory possible, but I have to ask why it is you need that extra few inches of clearance? Why not just extend the bottom of the hitch a bit to gain a few extra inches of clearance at the expense of loading and departure angle?

I suspect that the beam will take a serious degredation in strength if you plan to remove that much material, add a bunch of holes in it, and then bolt some steel plates on. The steel plates will be very weak in bending because of their section geometry, I would instead try to add additional material to the top of the beam within the trailer itself if the bottom of the beam does indeed have to be cut off...

6. Apr 2, 2008

### Jim Kopke

Thanks for the response, Mech. The reason for the additional clearance is to facilitate the back road use of the trailer where we often experience the gooseneck as the high point between a truck moving downhill while the trailer is still moving uphill. One "trailer shop" suggestion was to simply remove the C channel thru the center section and replace it with square tubing, his recommendation being 3" x 4" "heavy wall" aluminum which he claimed to "easily be equal in strength to the C channel, but giving me 5" of additional clearance. Being Al, it could be welded and braced to the ends of the C Channel with aluminum plate. I asked him if he has any engineering data to support his suggestion....and, well......it's a trailer shop. Thanks again for your help. Jim

7. Apr 9, 2008

### Mech_Engineer

There's no way 3"x4" square tube will be as strong as the C-channel you have described which is what's on the trailer right now.

8" x 2.25" C-channel which is .25" thick all around has a moment of inertia of 25.6 in^4, where as 3" x 4" square tube with 3/8" thick walls has a moment of inertia of 9.5 in^4. This basically means the C-channel on the trailer is 2.7 times stronger than the square tube in bending. Even a 3"x4" solid bar of aluminum "only" has a moment of inertia of 16 in^4, still less than the C-channel.

So your worry is clearance over the tailgate of the truck while towing a 5-th wheel horse trailer and going over rough terrain- hills and the like... I have to ask, why not just remove the tailgate when you're going to be towing?

8. Apr 9, 2008

### Jim Kopke

Ah....a man with the math! Thank you, thank you thank you! To answer your last question first, if the trailer in not square behind the truck (usually the case), the sides of the truck box become the point of interference. OK, given the areas of concern per your response, I've rethought this and put together a couple sketches to illustrate a new and improved idea (I hope). If you would, please take a minute and check out the new sketches and let me know your thoughts. In the new design (posted on the url shown previously), all is aluminum and all searms are heli-arced. Thanks again so much for your help. Jim

9. Apr 9, 2008

### Mech_Engineer

So the really critical metric you can use to help evaluate the effectiveness of any modification is the http://en.wikipedia.org/wiki/Second_moment_of_area" [Broken], sometimes called "moment of inertia" but not to be confused with an object's mass-moment of inertia. Most standard cross-sections will have a solved equation you can use to help you compare how "strong" different geometries will be in bending. Since the C-channel currently on the trailer is the performance metric you're aiming for, you'll want to try and get close to that (about 26 in^4).

For 3"x3" hollow tube with .25" walls for example, the second moment of the area can be calculated using the following equation (from Roark's Formulas for Stress and Strain, seventh edition):

$$I_{1} = \frac{b*d^3-b_{i}*d_{i}^3}{12}$$

Where b is the outside dimension on the "short side" and d is the dimension on the "tall side," and bi and di are the correspoding inside dimensions.

Using this equation you can see the area moment of inertia for one 3"x3" .25" wall tube would be 3.5 in^4. Putting two side by side would effectively put the two in series, so the moment of inertia would double to about 7 in^4, still far short of your goal... Obviously I-beams and C-beams are very strong in bending.

It's important to note that increasing the "tall side" dimension "d" by just one inch to 4 inches increases the moment of inertia by 100% to 7 in^4. This is because the area moment of inertia is a function of d to the third power...

Last edited by a moderator: May 3, 2017
10. Apr 9, 2008

### Mech_Engineer

Here's a suggestion- what if you used an I-beam or C-beam with a much wider flange but lower height?

An I-beam that is 4.75" tall (outside dimension), 6" wide at the flanges, has 0.375" thick flanges, and a 0.75" thick web has the same moment of inertia as the current c-channels on the trailer- 25.6 in^4.

Some equations for standard cross-sections-

http://www.efunda.com/math/areas/SquareIBeam.cfm

in the case of the linked I-beam, you will be interested in the moment of inertia about the xc axis, which is its strongest direction.

Similarly, a rectangular tube that is 4.5" tall, 6" wide, and has half-inch walls will provide a little bit better moment of inertia than required- about 27 in^4. A tube that is 4" tall would have to be 8" wide with 0.5" walls.

Last edited: Apr 9, 2008
11. Apr 9, 2008

### Jim Kopke

THANKS AGAIN! Being ingorant of the properties here, I passed over the shorter I beam idea assuming that the tubing would provide more efficient support. Such an I-beam would also be much easier to fabricate with less potential for weld failures. I'll post here in the next couple days after I familiarize myself with the math, try a couple examples and apply it to a practical design. Thanks also for the efunda link. Knowledge is a wonderful thing! Jim

12. Apr 10, 2008

### Jim Kopke

OK, please check me on this. Using bd^3 - h^3 (b-t), I find that an I-beam 6"wide, 4"tall, with .75" flanges and a 1.0" web gives me 25.489 in^4.

If my math is correct, this would work perfectly well. At each end of the I-Beam, I'd weld a 8 x 6 x .5 plate to interface (welded) with the existing 8" C channel.

Am I on the right track here? Thanks again for your help. Jim

13. Apr 10, 2008

### Mech_Engineer

Yup, that looks correct to me. An I-beam with 4" overall height, 6" width, 0.75" flanges and 1" web, has a moment of inertia of 25.49 in^4.

However, I wouldn't bother trying to keep the C-channel at all, I would try to replace as much of it as possible (as in cut out the c-channel and replace with the new I-beam) to minimize the number of welded joints, which will be weaker than a solid beam. Cutting the C-channel in the middle and trying to graft in the i-beam will add a lot of welds and a lot of stress concentration points.