Calculating Section Modulus

[LUGO331]

I want to make a trailer frame using two rectangular steel tubes stacked one on top of the other. I have a copy of the dimensions and properties of various tubing sizes which will give me the Sx of individual sizes but how do I calculate two together? Thank you, Paul

[PinkertonD]

I want to make a trailer frame using two rectangular steel tubes stacked one on top of the other. I have a copy of the dimensions and properties of various tubing sizes which will give me the Sx of individual sizes but how do I calculate two together?

Hi Paul, since you state rectangular rather than square, are you stacking horizontal on horizontal, vertical on vertical or vertical on horizontal (like a "T")?

Unless you are doing the "T" there is not a lot of stiffness to be gained in stacking the tubes so I would just add the figures for both tubes. I would hope that you are designing in a suitable safety factor to accommodate bounce-loading running over rough roads, pot holes and railway crossings etc. Anything gained by stacking the tubes will be a valuable addition.

As JB said in a recent post, "Steel is cheap. Accidents aren't."

[zeke]

My two cents.

If you stack them vertically, like Dave said you add the section moduli .

But if you tie them together by bolting or welding then you can improve the
stiffness significantly.

[LUGO331]

I want to stack two 3 x 8 x .250 and they will be welded together vertically

[PinkertonD]

I want to stack two 3 x 8 x .250 and they will be welded together vertically

3" wide by 16" high is not a good recipe for stable beams. You can do it but the interconnecting structure to prevent twisting will be huge.

Do you have a sketch and/or actual lengths and loads for what you are trying to achieve?

[zeke]

Do you mean that the overall lxh will be

3 across and 16 high?

Or is it 3x8 3x8, side by side?

[PinkertonD]

While I am not a great fan of Wiki "facts," here is a site you may want to read up on before proceeding further. Given the double-3" x 8" x 1/4" per side, I am assuming some quite substantial loads are involved here. That said, it seems employing an Engineer to design this for you is mandatory and not a luxury. In that wiki page, pay attention to the bolded section at the top and the "Safety Factors" at the bottom of the page.

https://en.wikibooks.org/wiki/Structural_Engineering

Just one nervous old fart's opinion.

[LUGO331]

Do you mean that the overall lxh will be

3 across and 16 high?

Or is it 3x8 3x8, side by side?

3 across and 16 high, Paul

[zeke]

Sorry, forgot to follow up.

If the tubes are best case rigidly connected, then the combined I is''

I=2*Ixx+2*A*(h/2)^2

Ixx inertia about the axis thru center for a single tube
A cross sectional area of each tube
h= height of each tube (8")

and the section modulus
S=I/h

But I wouldn't design to this since this is the theoretical case of perfect connection.
Maybe 50% would be OK with a good factor of safety.

[AussomeA]

Interesting that I was just looking into this yesterday, and stumbled onto this forum today. Anyway, I found this link useful:
https://en.wikipedia.org/wiki/Section_modulus

How big of a trailer are you building anyway?? How much weight are you planing on carrying on it?

[Kelly_Bramble]

Interesting that I was just looking into this yesterday, and stumbled onto this forum today. Anyway, I found this link useful:
https://en.wikipedia.org/wiki/Section_modulus

How big of a trailer are you building anyway?? How much weight are you planing on carrying on it?

https://www.engineersedge.com/sectio...ies_menu.shtml

[LUGO331]

The properties of the 8 x 3 x .250 tube are: Ix= 42.4 in4, A= 4.77 in2, Can you help me with plugging these into the formula. Thank you, Paul

[LUGO331]

Thanks Zeke, would you please run the calculation for me, Ixx=35.5, the Area= 4.77 this way I will know that I am understanding correctly. I'm not sure what you mean by the * and ^ symbols. Paul

[zeke]

OK,

I=2*Ixx+2*A*(h/2)^2

I=2*35.5+2*4.77*4^2=71+324.77=396
S=I/h=396/8=49


Oh, the * stands for multiply, the^ stands for exponent.
Say, for example
5^2 means 5 squared or 5*5=25
2^3 is 2*2*2=8

[PinkertonD]

I'm not sure what you mean by the * and ^ symbols.

That right there has to make one nervous.

Given the size material that Paul is asking about I am guessing some significant loads. Who was it that said, "A little knowledge can be dangerous?"

Seriously Paul, I mean no disrespect, but I really think you should get some professional Engineering help with this project.

[LUGO331]

You're funny Pinkerton, here's the dillio: I have been building auto transport trailers for 25 years and I learned from a man who hired me back in 1985 with 50 years in the business, 20 of them building equipment. Early on I found out there are things about the construction of this type of equipment that work and balance as best you can the overall unladen weight with reasonable structural integrity. With that said my calculation skills have long ago fallen off a cliff somewhere in my head but I do know enough to ask what I think is the right question. Traditional trailers run a lower rail and an upper rail seperated by framing and you have a nice strong bridge effect. There is also a style of auto transporter out there that successfully runs a lower rail only design for a span of about 20 feet in front of the axles and I am putting together a new design of this style for my company. Their rail is "C" channel, 1/4" commercial grade steel, and measures 4 wide x 15 tall. The section properties for a 4 x 16 x 5/16 wall has a Sx = 38.5 so my whole thinking is that if I use this as a minimum I will be okay. The reason I like the tube on top of tube is that I can slightly prearch them before welding and this will help the trailer look slightly arched or maybe straight when loaded as oppose to sagging. Zeke's calculation shows that Sx for the two 3 x 8 x 1/4 welded vertically together is 49. That sounds good to me.

[LUGO331]

Thank you Zeke, but I'm not getting part of the equation: 2*4.77*(h/2)^2
2 x 4.77 x 4^2
2 x 4.77 x 16 = 152.64 ?

[zeke]

My bad, you're right!.

That is why I write the formulae first before evaluation
I'm better at formulae .

Gotta watch me

I=71+152.6=224
S=224/8=28

As mentioned before, this is the theoretical maximum S , and in practice you do not use its full value.

By the way, what are you doing with this value. Just to make sure, it is for determining the maximum
stress in the beam from the maximum bending moment. If you don't know it, maybe you should give us an idea
of the beam configuration including its length, support structure and loads.

[LUGO331]

Thanks Zeke, please read the explanation of what I am doing in the response above to Pinkerton. But basically I am using the calculated value to match or exceed what I know is working in the real world which includes all the stresses of over the road conditions and a factor of safety. From what you have shown me I am under my target value of 38.5 so now I can use the formula to change the size of tubing until I reach my goal. Thank you, Paul