Math question

0.065 wall mild steel tubing 8.5" long and 2.25" diameter. If the tube was welded on one end to a plate, how much force can the tube take before bending if a force was placed on the other end perpendicular to the length?

Thanks engineers!

 

Thanks...you got me in the right direction and I found these calculators

http://www.botlanta.org/converters/dale-calc/bending.html

 

The calc is 'ok', but keep the resultant stress below the Proportional Limit and be sure to factor in the material's temp as well. The enclosed diag is for a plain jane A36 structural steel. You should lookup the stress v strain characteristics of your particular steel. A high chrome moly (for example) will react quite different when compared to an A36.
Rgds,
Vincenzo
PS: In the following graph, you can see part of the reason why F-car headers deform: 2/3rds+ of the strength is gone at higher temps...
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Ahhhhh.........bit of a problem there, FB ----

I was originally going to post up some helpful, DIY, formulas you could use to run the numbers for yourself, but I deleted it after I re-read your post a bit more carefully ---- as there is a problem. You also have the same problem in using that online calculator.

The problem is that the standard equations for elastic beam deflections / stresses theory (and that calculator is using them as well) only, and specifically, apply to "linear, prismatic beams" ---- in other words, they only work if the beam is long, slender, and of uniform x-section.

Your design is of a fixed-end beam that does not meet those criteria because basically it is too short relative its width ("I" value, to be more specific). So, its deflection and maximum load capability need to be calculated using the principles for "Short Beam Theory" --- and those are rather tedious, to say the least. Also, the geometry of that tube (beam) is such that, buckling may very well be your failure mode --- particularly depending on how it is actually loaded --- well before the maximum yield stress is reached due to bending and flexure stresses.

So, really the analysis for buckling failure modes needs to be run as well. All of this can, of course, be done (short beam stress & buckling analyses), but not from any quick set of formula calcs. Honestly, it is tedious enough that we don't even do it much these days. Instead, it is far quicker and more accurate to build a solid model (CAD) and run a numerical FEM analysis of such a design.

Of course, you don't care about any of that Engineering nonsense ---- you just want to know what load your beam can carry ! And, frankly I would strongly suggest that if you need to know the answer with a fair amount of accuracy (and maybe you don't ? --- it depends on how critical your application is), your best way would be to build a reasonably similar "test" beam, duplicate the scheme in which it will be mounted & loaded, and simply apply a progressive (and measurable) amount of load until failure.

Sorry FB that I can't provide you with a cleaner, easier answer --- like, hey, here you go....it'll hold XXX lbs.

 

Finnerty - good points to consider. Where is Shigley when you need him!

However, won't a 'short beam' (with no buckling) take a higher bending load with less deflection than a slender beam as long as the stress is below the proportional limit?

I agree with you that buckling collapse of the tube is the likely ultimate failure mode. Adding an end cap to the tube would help. Regardless, I'd bet that keeping the stress appropriately low per the slender rod calc's will likely yield a safe result.

I'd be more worried about weld failure at the end plate.

I need to dig up Shigley... it has been too many years.

Rgds,
Vincenzo

 

Thanks gents. No one is getting held to anything. I generally empirically over build everything. I'm trying not to do that and build what I need with a factor of safety. I'm more interested in concepts than absolutes not being an engineer. So the concept of long beam bending stress and the short beam buckling all make foggy sense to me getting clearer as I search those topics. The answer to my question is really to just cut down my time spent experimenting to get me in the ballpark. Because yes Finnerty that is my general MO to build stuff and then destory it until I get it right. It's a racer's mentality...OK built it light until it breaks then strenthen what broke until you find the next weak link.

 

Yes --- given the same applied load, the shorter beam will experience a lower maximum bending moment, therefore lower peak flexure stress.

But, given the same amount of deflection, the shorter beam has less capacity to absorb the strain energy, therefore it will be weaker.

So, it will depend upon whether the design is "max load-driven" or "max deflection-driven".

That would be a very good thing to do. That would help dramatically reduce the possibility of local buckling at the end, where the load will be applied ---- also, it will significantly add resistance to buckling along the full span of the tube as well.

FB ---- Will your application allow you to cap / close the free end of the tube ?

Indeed, the weakest point will be near the margin of the weld --- where the parent material's strength has been degraded from the welding. Welds themselves are strong (if done properly), but failure typically occurs immediately adjacent to the weld. Also, in this application, that location is also going to see the peak bending moment / stress ---- so, it's a double-whammy there --- if the tube does not buckle somewhwere else first, that will be the weak spot.

FB ---- Any chance you can add a reinforcement collar at your welded end ? Or, can you use a piece of butted tube (thicker wall section at the end) ?

 

I ran the specified design thru my FEA program, not super high end that Dave has access to I'm sure but is still a good approximation to real world testing.

Material is A36 carbon steel, the concern is the weld as the program assumes best case uniformity etc...

Hopefully the jpegs are readable, I have doc files as well but the pics are easier to read for layman.

This is calculated with a remote load of 1200lbf at 200mm from the weld plate end.
First pic is FOS or factor of safety, the overall is 5.6 and higher
2nd is the Von Mises stress, it peaks near the weld at 206,692,632; yield is 250,000,000.

ignore the deformation as it's greatly exaggerated for visual purposes, maximum displacement is 5.522e-001mm
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Here's the buckling study, as noted the deformation will occur near the weld. The deformation is greatly exaggerated for visual purposes, max deformation is 4.136e-001mm. for the math challenged that's 0.4136mm or 0.0162".
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Nice job Scott ----

If FB wants to supply you with more detail about the free end condition ---- such as what may be mounted to it and the nature of the applied load configuration.

Perhaps you would be kind enough to run your study again to provide a more refined result ??


BTW, what is your s/w tool there ? ....the output looks a bit like it came from COSMOS-xpress (Solid Works) ?

 

It's SW11 Pro.
I've used the flow analyses on the manifold project and real world testing cam within .5% of the calcs.

If FBB wants to provide a bit more I can run some prelim calcs for him. In the above the plate is the assumed fixed body, however as you know it's attachment will also need to be calculated, esp if it's not the size I used 1ft sq.

these were static and buckling tests, there is still the possibility of frequency or harmonics causing failure that would need to be run too. I don't have the parameters so I couldn't factor that.

 

Classic Euler third order differential equation for column buckling.

Assuming fixed base and pinned top and A36 carbon steel modulus of 29ksi

about 3000 pounds.

i pulled this all from wikipedia.


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someone check my math as i was in teh bathroom at the waldorf astoria..

 

Review the equation and I think you'll eventually find the error to the values given in the math. No help given as one only learns through their own research.

 

forget the math errors, its the wrong equation! Euler is for column, this should be a beam equation. goes to show you that you should be doing math in a bathroom.

 

try again, 930 pounds round down to 800 to be safe
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goes to show how moving a bit further out reduces the load that can be supported. I ran the clacs at 200mm or 7.8" vs 8.5"

I'll re-run them at 8.5" in a bit and report the findings for buckling.

ohh and Sledge, way to go keeping it 'ol school. I've gotten lazy relying on the sw to handle a majority of number crunching. Although in my defense I have been spending time brushing up on linear algebra, use or loose it - man is that a painful truth!

 

I dont practice structural enginering anymore, so i dont have any of the cool new tools. Just relying on basic statics and mechanics of materials

I am old school, i dont even have facebook.

 

I don't either, but then again I'm just paranoid happens when you've worked for the DOD.

 

i went to the sort of highschool you dont want people who went there finding you

how much is that software you are using? just curious.

 

Hey you guys are awesome! I did not know this little thread had gotten so far. Thanks all you smart guys for chimming in. For some reason this thread did not refresh on the user CP so I never saw any updates to it. So my bad for not replying. Anyway my reason for the quiry is to add this 8.5" tube and weld it to my bumper hidden behind fascia on my racecar. The goal of this tube was to use it as a jack point to lift both front wheels at the same time. That was no problem. The question was basically how thick do I need to make it to also use as a tow point to winch up my trailer. I know I can just weld a mass of metal and everything is happy but I'm trying to minimize weight and not be such a caveman with all of my building. I was going to cap the end and build the capped end with a 2.5" ring that extends out so I can hook up there. Another idea to stop the buckling but stay light using thin wall tube is to weld a 1.5" x 8.5" tube and then weld a 2.25" x8.5" over that like a tube in a tube. Cap weld the end and that should be very strong syas my pea brain. But I have no clue how strong.

I thank you guys for teaching me some stuff like the long beam short beam theory ideas. It all make empirical sense. I love those FEA. I see them alot when the engineers chime in on our rollcage building threads. It is nice to put numbers and sound math theory to what you think should happen.

 

All the information above is for static loading, and in the case you mention above there will be a bit of dynamic load, which introduces more force. Without getting yourself into myriad of calculations, i would just size the tube for static load, and bump it up all couple of wall thickness sizes to be safe.

Geno

 

not to mention the attachment to the frame. I'd say possible ribs or half cone, I'll run a simulation a bit later for dynamic loading and failure on them. not knowing the layout in the front makes this harder, however all cars have a cross member in the front that will support the weight of the car for lifting. Is this more for ease of lifting?

Are there any regulations against having essentially a jousting pole jutting out that can run the risk of puncturing both vehicles in the event of an accident?