Pivoting Cantilevered Worktable Support Material

I'm making a worktable that can only be supported in the center. I will need access to both the area above and below on each end. I won't be able to use legs or ceiling suspensions toward the ends of the table, therefore, it will need to have two cantilevered ends stretching out from the center.


My primary question concerns the material it would be made from. I'm wide open on what material to use. I would like to keep cost and weight down, but I also need something not too difficult to work with. I know a lot of folks use aluminum for this type of thing, but I'm concerned whether it will hold up well at the hinge/stop location (explained below) and I think I will have more trouble getting someone to weld it. I decided to go with steel square tubing just to have something to start with, but I could easily be swayed to another material if it will meet my needs.

I’m estimating the load to be [B]500lbs[/B]. It will be moveable and will sometimes rest on the extremity of one or the other cantilevered ends.

I will need to be able to tilt the table a few degrees off level, so the table will be supported by a pivot/brake mechanism in the center. It can be pivoted when the brake is released, but will normally have the brake applied and that will prevent movement about the pivot point.

The table is [B]80" x 60"[/B] and will be supported in the center along the short cross section, which means that there will be two [B]40"[/B] cantilevers placed opposite of each other. It will be comprised of 3 pieces. A center section that is [B]16"[/B] wide and two hinged sections that are [B]32"[/B] each.

[CENTER][CENTER][B]32 + 16 + 32 = 80

[ATTACH]2469[/ATTACH] [ATTACH]2470[/ATTACH] [ATTACH]2471[/ATTACH]
[ATTACH]2472[/ATTACH] [ATTACH]2473[/ATTACH]
[/B]
[/CENTER]
[/CENTER]
The cantilevered ends need to be able to fold up (on hinges) into a stowed position when not in use to provide maximum space in a multi-use area. (Due to space restrictions, I may need to have one side folded up and still be able to work on the other side. So, the method that is used to secure the stowed end piece needs to be clear of the work surface.)

[CENTER][ATTACH]2474[/ATTACH] [ATTACH]2475[/ATTACH] [ATTACH]2476[/ATTACH]

[/CENTER]

[ATTACH]2477[/ATTACH] [ATTACH]2478[/ATTACH] [ATTACH]2479[/ATTACH] [ATTACH]2480[/ATTACH]
I'm also considering splitting the table down the middle into two halves that would be bolted together in order to make it more manageable for relocation. This adds even more complexity to an already complex problem, but I'm afraid the weight of the assembled unit will be too much to easily handle when moving.

[ATTACH]2481[/ATTACH]

I have tried to educate myself about beam calculations and I have a formula that gives me information about the moment at the fixed center.

[B]Moment = 500lbs (40") = 20,000 lb/in = 1667 lb/ft[/B]

But now I don't know what to do with that. I checked some websites for steel suppliers and they don't appear to offer the kind of specs on their products that I would need to utilize that equation.

The problems I am trying to solve are:


[LIST=1][*][B]Material type, size and amount to adequately support the load[/B][*][B]A brake mechanism that will hold firmly and also allow the table to rotate a few degrees[/B][*][B]A hinge/stop that will hold the table completely flat when in the non-stowed position[/B][*][B]A way to prop each hinged section in the stowed position (independent from the other)[/B][*][B]A way to attach the two halves together if I decide the weight is too much to leave in one section[/B][/LIST]

I confess that I'm in over my head. I'm not even sure what discipline is most appropriate to ask these questions of. I've looked around on this forum and there appears to be some very knowledgeable people here. If nothing else, maybe someone can point me in the direction of who better to ask. If I can get at least some of these questions answered I will be very happy.

Doing an analysis without any engineering training or education will be challenging. I do suggest that you determine the risk associated to yourself, end user and employer before proceeding.

See: [URL]https://www.engineersedge.com/Technology_Engineering/are_you_doing_the_engineering_yourself_15406.htm[/URL]

First thing you should know is that it it easier to over-engineer a structure for strength then it is to engineer it just right. Over engineering for strength is usually costly but safe. So, you want the apparatus to very much stronger than required - we call this the Factor of Safety.

First, I suggest you look for and find similar-too designs to get ideals. I suggest that you search the internet for Rotating weld fixtures (images.) You’ll likely use a pillow block bearing and a pin-hole on a disk to lock out rotation or some adjustment screw lock mechanism.


This webpage overviews the general approach to structural analysis.

[URL]https://www.engineersedge.com/beam_bending/structural_beam_stress_and_deflection_13746.htm[/URL]

What you need to learn and know:

Area moment of inertia: - this is the cross sectional properties of a structure shape.
[URL]https://www.engineersedge.com/section_properties_menu.shtml[/URL]

Strength characteristics of engineering materials (steel and Aluminum etc.) Yield strength is an important number you want your factor of safety such that maximum loading on your frame is way below this for the steel or aluminum you use. See:
[URL]https://www.engineersedge.com/manufacturing_spec/properties_of_metals_strength.htm[/URL]

Loading configuration on your rotating frame. A full analysis of the frame is likely out of reach however, you can identify the highest stress location and do an analysis there.

[URL]https://www.engineersedge.com/beam_calc_menu.shtml[/URL]

Chances are you're going to want a weldment (weld frame) using rectangular or square steel. Design around standard shape sizes and materials, like A36 carbon steel. To analyze a single weld intersection, see:
[URL]https://www.engineersedge.com/weld_design_menu.shtml[/URL]

Lastly and most important get a design review before building from pragmatic and qualified . Don’t confuse personal preference with actual good engineering design practice.

This will likely get edited by myself several times BTW.

You said you calculated the moment created by a 500# load at the corner 40" from the center (1667 ft-lb) but didn't know what to do with it. Well, first thing, it's "ft-lb", not "lb/ft". There's a huge difference in meaning between those two terms.

That number is going to affect LOTS of things. For example, based on the images above, it looks like the whole table will tip over.

2nd thing - It also means that your "brakes" at the rotation axis will have to withstand a 1667 ft-lb load. That is a LOT! Picture hanging 3/4 of a ton (1500 lb) from the end of a 1 foot long stick.

3rd thing - that means your folding hinges will also see a tremendous shear load when flat. You already know you're going to have to do something to keep those joints flat. If you were just planning on inserting some shim material at the bottom, read the next paragraph.

Let's talk about the hinges. The 500 lb load at the table edge will be 31.5" from the hinge centerline. That gives us a moment of 31.5 x 500 = 15,750 in-lb. There are 8 hinges on a side so we'll divide that load by 8 (although that isn't really realistic). 15,750 / 8 = 1969 in-lb at each hinge. If your frame is made from 1.5" tube, that means that the bottom of EACH joint will see a compressive force of 1969/1.5 = 1,312 lb. Like I said. That's a LOT!

If there is any way for you to start small and progress up from that I would recommend it. For example try the same load on a table half that size and see what happens. Just remember this - each failure means you're one step closer to success.

Kelly Bramble
There is definitely some similarity between Rotating weld fixtures and my project. I will look over these links.

jboggs
lol, you are right about the table tipping over. I didn’t put a lot of thought into the support since I don’t know yet what I’m supporting. The final product will likely end up being bolted to the floor. I’m going to try not to be intimidated by the numbers you just tossed out there.

Thank you both for your thoughtful replies!

I have been going through the information and trying to put together the necessary info to use one of the calculators.

[URL="https://www.engineersedge.com/beam_bending/calculators_protected/beam_deflection_9.htm"]https://www.engineersedge.com/beam_bending/calculators_protected/beam_deflection_9.htm

[/URL]So far I have:
Modulus of E 29,000,000
Moment I=wl 20,000
Load W 500
Distance l 40
Distance x
Distance z

But I'm not sure what x and z represent. Can anyone help clarify that please?

[ATTACH=CONFIG]2483[/ATTACH]

The calculator delivers two stress calculation points, one at the support and one at a distance x from the support.

Thus, there are two stress "s" calculations and two deflections "y". One at l distance and one at x distance from the support.

Z=section modulus of the cross-section of the beam = I ÷ distance from neutral axis to extreme fiber.

The equations are linked on the webpage.

[URL]https://www.engineersedge.com/beam_bending/beam_bending9.htm[/URL]



[QUOTE=reg_vermont;16777]I have been going through the information and trying to put together the necessary info to use one of the calculators.

[URL="https://www.engineersedge.com/beam_bending/calculators_protected/beam_deflection_9.htm"]https://www.engineersedge.com/beam_bending/calculators_protected/beam_deflection_9.htm

[/URL]So far I have:
Modulus of E 29,000,000
Moment I=wl 20,000
Load W 500
Distance l 40
Distance x
Distance z

But I'm not sure what x and z represent. Can anyone help clarify that please?

[ATTACH=CONFIG]2483[/ATTACH][/QUOTE]

ok, I have some numbers

Mod of El (lbs/in^2) = 29,000,000
Moment I=wl (in^4) = 20,000
Load W (lbs) = 500
Distance l (in) = 40
Distance x (in) = 32
Distance z (in) = .5

Stress Hinge (lbs/in^2) = .1000
Stress Support (lbs/in^2) = .5000
Deflection x (in) = .000
Deflection l (in) = .00002
Section Modulus (in^3) = 40000.000

[ATTACH=CONFIG]2484[/ATTACH]

I don't know what to make of these numbers except that I must have done something wrong.
The deflection seems way too small.
Even if these are correct, I'm not sure what this tells me about how many sticks of 1" tubing to use.
Any elaboration is welcome.

It's a wide - wide plate.. 500 lbs distributed over that area at 29,000,000 Modulus of elasticity.

I'm guessing the plate is 1" thick"?

Seems reasonable to me with the given numbers.

Thanks everyone for your help.

I'm throwing in the towel.

I originally thought I would save time and expense by doing some theoretical calculations rather than going through trial and error.

However, it has turned out to be that the math is so complicated that it is taking me longer to figure this out theoretically than it would for me to just buy some tubing and hang a sand bag off the end and watch what happens.

I wish you all well!