Understanding Structural Physics and Steel Tubing for Utility Trailer Design

[pityocamptes]

Hi, all. New here. Currently I'm a computer science student so I have some background in physics. I do not know very much about mechanical/structural physics or the equations associated with them. Currently, I am trying to build a small utility trailer and have questions about the steel tubing size that is appropriate for the load I would like to carry. I have downloaded some beam structure software, which calculates shear and the maximum bending moment.

I do have some questions on how to interprete the graphs associated with shear and max bend moment. If anyone can direct me to a tutorial or simply explain these graphs and what they mean I would appreciate it (vert and diag lines intersecting the x axis).

I also need some info on how to determine what size and gauge steel to use. I know what my max load is that I want the trailer to carry (and have an idea of the size of the tube) but I am having a hard time figuring out how to determine what gauge steel to use. I could go with oversized steel but I really do not want to overbuild it as weight issues might become a problem. Any help would be greatly appreciated.

 

[FredGarvin]

If you can describe the loading of what your situation is (free body diagram) that would help us a lot.

As far as shear and bending moment diagrams, that all falls under pure bending of beams.

http://www2.umist.ac.uk/construction/intranet/teaching/ul222/exp/sfbmdex.htm
http://em-ntserver.unl.edu/NEGAHBAN/Em325/10a-shear-and-bending-moment/Shear%20stress%20in%20beams.htm
http://lowery.tamu.edu/cven305/305/HowToShearandMoment.htm

 

[pityocamptes]

Thanks FredG! I was able to figure out the section modulus and max bend moment with the steel tubing I would like to use.

Here are some problems maybe you or someone else can help me with. My trailer is 36" X 48". I would like to use 2x2x1/8 steel tube (mild steel with a yield of 36,000 psi). I would like the trailer to hold a max of 800#. When I did my calcs I basically (just to see how the steel size would hold) figured a straight 36" piece being supported at both ends with a point load in the center. I did the same for the 48" piece (front to back). I did not take into account intersections. I also used a saftety factor of .55. So for the 48" piece I came up with a max bend moment of 757# and for the 36" piece (width) I came up with a max bend moment of 1010#.

I realize that these numbers will change considering the design. My question is how will the design, shown below, affect the outcome? Will my max bend moments be higher since the tubes will not in practice have a center point load? Any help appreciated. Thanks.

Attached is a rough pic of my trailer design. The lines in black are the frame and the lines in red are a sub frame/rails fastened beneath the black lines frame.

 

[FredGarvin]

In reality, as long as you have good welds at the junctions, your current analysis is more or less a worse case scenario. The presence of joints will not really hinder you in this case. They should help distribute the load (if I understand your drawwing) and thus reduce induced stresses.

Now that you have your max bending moments, did you go back and use that to back out your desired moment of inertia and thus the proper sizing of the steel box tubing?

 

[pityocamptes]

In reality, as long as you have good welds at the junctions, your current analysis is more or less a worse case scenario. The presence of joints will not really hinder you in this case. They should help distribute the load (if I understand your drawwing) and thus reduce induced stresses.

Now that you have your max bending moments, did you go back and use that to back out your desired moment of inertia and thus the proper sizing of the steel box tubing?

Thanks Fred. So basically the beam will support a higher poundage if the load is distributed, and the numbers I came up with are a worse case scenario, as you mentioned, for a max point load? Since I'm new at the structural physics "thing" maybe you can shed some light on what the Moment of Inertia is used for (is this a case where the trailer, with load, might fall into a pot hole and the force on the beam has just increased?) and how I would go about using it (formula). Thanks again.

 

[FredGarvin]

The moment of inertia is a mathematical description of how mass is distributed about some reference axis. In your case, a higher moment of inertia usually means a stiffer beam. For example, take a 2x10 floor joist. if you lay it flat, the board will have a certain moment of inertia and it will, intuitively, deflect quite a bit. Now take that same board and put it on it's side. It's the same board, but the moment of inertia will become quite a bit larger (due to the way it is calculated for a rectangular cross section) and the board will not deflect nearly as much.

 

[pityocamptes]

The moment of inertia is a mathematical description of how mass is distributed about some reference axis. In your case, a higher moment of inertia usually means a stiffer beam. For example, take a 2x10 floor joist. if you lay it flat, the board will have a certain moment of inertia and it will, intuitively, deflect quite a bit. Now take that same board and put it on it's side. It's the same board, but the moment of inertia will become quite a bit larger (due to the way it is calculated for a rectangular cross section) and the board will not deflect nearly as much.

OK. Thanks, that makes sense to me. Will calculate moment of inertia and post back for some more suggestions. Thanks again.

 

[pityocamptes]

OK, here is what I have:

2 x 2 x .120 square tubing ->


M = maximum bending moment = 9091 in-lb.
M = .55 (safety factor) * 30,000 (psi) * .551 (section mod) = 9091 in-lb


y = distance from the neutral axis of the cross section to outer edge of beam = 1 inch

I = moment of inertia of cross section; for square = .551 in^4

Maximum Bending Stress = M y / I = (9091 in-lb)*(1 in)/(.551 in4) = 16,499 lb/in^2

Max Deflection = yMax = FL^3/(48EI) = (800 lbs * (36 in)^3 ) / (48 * (30*10^6) * .551 in^4) = .047 "
I'm not sure if this formula is for tubing or solid beam?



Ok, I've done the math, hopefully right Can anyone verify? Also, what numbers do I need to be aware of and how they affect my design/load/tube size? Do I need to calculate anything else? Thanks again!

 

[pityocamptes]

Bump - to see if anyone might have had a chance to review?