Need advice... I'm designing a floating desktop with dimensions of 1.5" x 65" x 30" oak plywood (~120lbs) with four 1" pipe and flanges connected with 3.5" #14 screws into the middle of 4 studs, with a strip of wood on one side where the desktop will line up with the corner of the wall... the strip of wood will be screwed to the wall studs and bottom of the desktop (strip of wood shown in pic on the end). This will be an office desk with somewhere around 350lb max load (including desktop)
I understand there's many other brackets, legs, and other setups that will be more secure...
The pic is just a mock up of what the view would be from under the desk once the pipes are attached to the wall.
I'm not an engineer, so I'm sure this force diagram is wrong (see attached pic), but can someone tell me if this diagram is a close estimation of the forces at play and also what is the force that will be exerted on the top screw of the mounting plate (I'm considering that only the (4) top screws will carry the load)...please advise
Thanks in advance
Screenshot_20220214-125702_Gallery.jpg
force diag.JPG
Last edited by Maxmil982; 02-14-2022 at 12:58 PM.
Assuming even loading, STD pipe I get the following:
section.png
https://www.engineersedge.com/materi...pes_viewer.htm
stress.png
https://www.engineersedge.com/beam_b...flection_8.htm
The screws to the wall and the flange are your weak point... Use angle (90 degree) support brackets
Tell me and I forget. Teach me and I remember. Involve me and I learn.
The pipes are strong enough. The flanges might be strong enough. But the connection to the wall is where it will fail.
If you're interested, the feature that will cause it to fail is the distance between the bolts in the flange. Think of it this way. The lower bolts will carry the weight of the desk as a shear load, a sliding load. They will "anchor" the desk. The center of weight of the desk will be at some distance out from the wall. The effect will be that the desk will try to "rotate" about the lower bolts in a downward direction. That rotating action caused by the weight of the desk acting at a distance from the center of rotation (the lower bolts) is called a moment load. That downward rotation of the desk will create a very large tension load on the upper bolts. They will pull out or fail allowing the desk to fall.
The reason that the load on the upper bolts is so large is that the distance from them to the lower bolts (the center of rotation) is very short, maybe 3 inches. The more that distance can be increased, the less load the upper bolts will see.
This might help the visual image: the distance from the lower bolts to the center of weight of the 350 pound desk is around 15". This creates a moment load of 15 x 350 = 5,250 in-lb. The strength of the upper bolt connection counter that moment load. It must create an opposite but equal moment, -5,250 in-lb. But the upper bolts are only 3" away from the lower bolts. So create a -5,250 in-lb moment they must withstand a load of: 5,250 in-lb / 3 in = 1,750 lbs! That's why you need to increase the distance between the upper and lower bolts.
So, if you could find some flanges with really big bases that might work. But a simpler way would be to use angle brackets designed for that purpose, as Kelly suggested. The upper and lower bolts in those brackets are much farther apart, thus allowing them to carry greater moment loads.
Originally Posted by jboggs
Many thanks for the detailed write-up. It answered all my questions and gave me a ton of general knowledge, and how to calculate this type of load and supports. I'm abandoning this cantilever idea for the desk because the supports are not even close to being sufficient. I'm going with (4) 19.5" x 13" A brackets for the desk.
Although, I am considering this same style of cantilever support for the matching shelves though. The shelves are smaller at 65"x19"x1.5" and using the same style of pipe and flange supports but 3/4" pipe instead of 1" pipe. The shelf's design load is 200lbs and using your calculations (the distance between the top and bottom screws is 2.33"); the top screw will have to withstand a load of 772 lbs. If you will, can you tell me if I have this right?.... the conservative pull-out strength for the 0.242" diam screw @3" of penetration is 212lbs. Since the total load to be withstanded is 772#, I've divided that by four (for the number of top screws), leaving 193lbs per screw. I know there's not much safety factor in a 212lb capacity, carrying 193lbs, but in reality, the screw can hold a bit more than 212lb, the shelf will weigh a bit less than 200lbs, and I haven't taken into account the ledge that one of the narrow ends of the shelf will rest on (which will carry about a bit of the load, aside from the pipe supports).
My question is am I right it considering the total bending moment (772lbs) that needs to be countered, will be approximately divided by four for the four screws?
Good job! Yes, you are right, generally.
You never mentioned what material the screws were going into. Steel, wood, drywall, concrete? That, and the length of engagement, will be the determining factors in their "pull out strength".
I have not double checked your math but it sounds right. My main concern is the lack of any safety factor at all. One of the incontrovertible laws of engineering is that the real world NEVER operates exactly as we expect it to. When your shelf has fallen on the floor, and maybe hurt someone in the process, the excuse that "my calculations showed that it should have worked" doesn't mean much. When you operate that close to failure, very minor things can have a major impact. You don't want your shelf to fall on the floor when someone slams a door nearby, or when your cat jumps up on it, or when your neighbor's rambunctious 6 year old tries to hang from it, or because you over-torqued a couple of the screws thus slightly stripping them, or when any number of other unforeseen circumstances arise. "Play it safe" is not just a familiar phrase. I wouldn't be satisfied with anything less than a 2:1 safety factor, and sometimes more than that, much more!
Thanks again for the response. I'm starting up the project again, now that the shelves are finished. To answer your question, the screws are going into standard 2x4 framing studs. I used the formula found here: https://tinyurl.com/9h92t3d3 for calculating the allowable withdrawal force of the screw. I'm using a heavy duty construction screw with a 0.3" shank and 2" thread penetration length. Using 0.36 specific gravity for the wood, the allowable withdrawal force came out to be 222lbs (the screw will need to resist 193lbs)...
I read your reply about using an adequate safety factor, and I think the 222lbs has a hefty safety factor baked in, because on that site it says that a general "allowable" withdrawal force is less than the maximum force....also the equation they are using for the force is:
F = 2850lbs*(specific gravity of the wood)^2*D*L
...which is a different equation than the one here on Engineer's Edge
https://www.engineersedge.com/calcul...ance_15385.htm
which is:
F = 15,700lbs*(specific gravity of the wood)^2*D*L... which seems to be the maximum force instead of allowable force, which came out to be 1,200lbs (which would be plenty to have enough safety factor)
If you get a chance can you look over the details here and let me know if you think I am on the right track with these equations?
Looking at both webpages - there IS a source shown on Engineers Edge but not on the reference webpage (unknown source).
Backing up a little Specific Gravity = (density of wood) / ( density of water )
Reading the text below the force shown on Engineers Edge is defined as the "Maximum" and the text on the reference website uses the words "allowable" - allowable according to who?
So the 2,850 vs 15,700 are constants likely reflecting FOS as you stated. Which would suggest 5.5 for the reference webpage.
Looking at the references (bottom of webpage) on Engineers Edge the formulas check out
reference.png
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Ok thanks for the info! I see you're in Bold Springs; I live in a little west in Gwinnett Co
There's a previous post where I used jboggs' equations on a narrower shelf, than the original desktop, with a 200lb design load instead of 350lb. It came out to be approximately 200lbs needed to be resisted at each screw. Do you think this cantilever support system is adequate considering the 5.5 factor of safety on the withdrawal strength of the screws?
Also to counter deflection and the bottom of the metal flange crushing the drywall, I'll be placing nuts (the same width/length of the drywall) between the stud and the bottom hole of the flange.
I know this setup is probably a bit unorthodox, but I am really opposed to using traditional brackets and interfering with my minimalist vision for these floating shelves.
Thanks again for your input
Last edited by Maxmil982; 03-20-2022 at 01:09 PM.
"Cantilever" think pry bar at the support reaction end - you'll need to do the math at that support to determine the loading the fasteners will likely see.
here: https://www.engineersedge.com/beam_b...m_bending9.htm
As with any engineering material the math is perfect world which is why we have FOS. Your materials can vary, screws and assembly orientation induced loading and assembly. Be generous with wood if there's cyclic loading.
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Thanks for all your help. Here's a few pics of the shelf in action.... The plants only ended up weighing about 70lbs (which was much less than the design load). The shelf's been up for a few weeks and has held up just fine with no noticeable defelction and some capacity to spare.
Thanks again20220423_174451.jpg20220423_174459.jpg
Looking good!
Thanks... and thanks for all the help and formulas
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