web-based beam calculator - RHS's only

Hi. Been at it again. Having learned some PHP - the web-scripting langugage. My beam calculations written in Lisp - done as webpage version.

"Rectangular Hollow Section beam calculations" Best wishes, Rich S

----------------------

I try to do the math and then check it with on-line calculators. A useful result is the deflection at maximum load or yield point, which can be used as a gauge by placing something that high on the beam center and observing its alignment with the two ends when the test or straightening setup is too unstable and risky to approach or put an expensive load cell on. The 3" and

4" channel I found to build my gantry hoist and sawmill tends to twist when near full load.

Check maths - be appreciated. I was exhausted after a through-the-night and could only say it agrees with "The Blue Book" where I spot-checked.

(I was learning a lot of programming with PHP (scripting language for getting the web-server to do calculations and processing for you and put the answers into the webpage being served-up) being so different to the Lisp which is all I have used for years)

I had a warm feeling that that deflection at load would be useful to practical designers.

• I will come back to this *

Usage - strong hint - put in the yield stress as "your stress" and you will get the deflection at the point where any further and the section would yield.

I've done this a couple of times in tests and it is so spot-on right - at least once with Euler-Bernoulli beam (this calculation) and once with Finite Element modelled "structure" (the FEA is doing essentially the same mathematics).

I was so tired then I had to leave as-is and walk-away.

I have heard military training, especially for Sergeants and the the like, deliberately takes you far far far into "impossible" exhaustion so you learn what thoughts do work in that state (unfortunately I found out about the "hallucinatory" world for the first time at a crucial time and messed-up, having no prior experience of it - had I known to trust my "gut instincts" still working I could have produced the good outcome).

The one about coupling the stress and deflection.

So if you put in the yield stress and get the max. load and the deflection for your beam at the point that is reached...

This is what I promised to come back to.

In the function which does that - a bit of algebra, obviously - a lot of variables cross-cancel if/when you run through it on paper first. I was most surprised. That was during that night I couldn't sleep anyway. You end up with a surprisingly small formula which gets the deflection at the "aim stress" the "this is the load which would cause it" output.

Taking three of the big formulae of beams:

M=FL (cantilever beam) M=FL/4 ("simple" beam - it's spanning between two supports)

the massive one which combines material property and geometric property... M=sigma.Z [noting "Z" = I/half-height for a symmetrical section]

and deflection d=F_c.L^3/3EI (cantilever beam) d=F_s.L^3/48EI (simple beam)

the equations have a lot of "substituting into each other".

Hence that which you like - you get the deflection - what you can readily see and measure - at the point beyond which everything goes wrong - that has a backstory which was quite delightful.

In the evening leading into the night I was there with a multi-colour "magic pen" there writing out equations in black, arrowing where they cross-substituted in blue, leaving notes in green and striking though cross-cancelled variables in red.

There is a general function for beam deflection in the code, for debugging during development, but it isn't used in the "working" code which gives the outputs from your inputs.

What you "jumped to noticing you can do" - deflection at the point it all goes wrong - I have tested and seen and you know already that - what happens is not sudden disaster.

My "moment" is recorded here:

was doing the worst I could to a weld - but got to see the Euler-Bernoulli beam calculations giving such accurate predictions that you could actually use them to measure the yield stress of your steel you have in the shop. I did the maths in chalk on a bench-top by the way - I had half and hour with the work held up and had to jump at the opportunity immediately. That's where at the instant you hit that value you are interested in, if you go any further your section which always returned to straight before now has a bend.

The one where I "Finite Element'ed" is here

Both you see - the prediction of deflection at the point of "overload" is accurately predicted. What happens after is what you the individual needs to see at least once. The structure takes more and more bend under progressively increasing resistance to further bend. In a world of "Clipboard Clarence" and "Safety Sam the Management Man" there is an edict and anything beyond what they say is "absolutely disaster". Well it is if you get "fired" for being held to have transgressed "Clipboard Clarence's" edict.

Hope this was a nice follow-up chat.

The one about the maths cross-cancelling - as you'll also be commending to anyone "on the path", you can only see that by having some sheets of paper, preferably a multicolour pen, and running through it yourself.

I woke up this morning feeling great having slept like a log from early, after yesterday was "different" having not slept the night before.

Best wishes, Rich Smith

Deriving the expression for deflection at that already-calculated Force:

Caution here is that on-paper often put "=" when mean "==" "=" is "assign" or "are equal at this one specific instance" where "==" means "always equal to". What is derived has no generality - only useful in this case (?) A stopped clock is correct twice a day. What I derive is only right at one of those "coincidence points"...

Taking cantilever beam case.

M=FL M=sigma.Z [noting "Z" = I/half-height for a symmetrical section] d=F_c.L^3/3EI

M=FL -> F=M/L into d=F_c.L^3/3EI d=(M/L)L^3/3EI=ML^3/3EIL=ML^2/3EI

M=sigma.Z -> M=2.sigma.I/H into ML^2/3EI

(2.sigma.I/H)L^2/3EI=2.sigma.IL^2/3EIH=2.sigma.L^2/3EH

d is abtained for cantilever beam by this algebra in my program

2.sigma.L^2/3EH Astonishingly simple!

When/if do "simple beam" remember M=FL/4

Nifty stuff Rich👍

If you haven't found this already... you might like Marv's bunch of little calculator programs:

Maybe give you some more ideas and/or another way to double check some of your calculations.

Nifty stuff Rich👍

If you haven't found this already... you might like Marv's bunch of little calculator programs:

Maybe give you some more ideas and/or another way to double check some of your calculations.

Leon Fisk Grand Rapids MI

----------------------

I like spreadsheets for cutting mathematically definable shapes like spheres. By putting the constants, multipliers and step size into fixed cells and referencing them in the formulas it can be adjusted for roughing and finishing cuts with steps as small as your patience allows.

It generates a pair of columns with the part's dimensions for checking, and another pair with the machine settings to produce them, infeed and carriage travel.

Cross-check

detail provided of test, and has both Euler-Bernoulli beam and Finite Element Analysis modelling - which agree and the physical measurement agrees]

"100x50by7.8, of "S355" (nominal 355MPa)" "The end supports for the test beam with test weld are set at 600mm gap between these supports"

So:

RHS Height

100e-3

RHS Width

50e-3

RHS Wall-thickness

7.8e-3

RHS beam length

0.6

RHS material yield stress

355e6

RHS material Young's modulus

210e9

My new online beam calculator for RHS's agrees with "The Second Moment of Area, "I", of a square-cornered rectangle of the overall dimensions of the RHS = 2.443e-6 m^4" "Section Modulus, "Z" = 2I/h = 4.886e-5 m^3"

"The central load causing the selected stress is

115644.490402Newtons" (/ 115644.490402 9.81 1e3) ;; 11.78842919490316 That predicted is 11.8Tonnes-force The beam did not return to straightness after 11.6Tonnes of applied force. Very very close prediction [Western European steel "just" makes the yield stress - they have such accurate control that the manufacturers can and do do that - gives a lovely steel to work with in the workshop - punches, drills, bends in 3-point rolls, welds, etc. exactly as you'd wish] The slightly lesser force to bend real<->predicted could be due to the rounded corners of the real section slightly reducing the Second Moment of Area. The predicted elastic bending at 355MPa yield 1.014286e-03 m (Nov 2020 programme) and "At that maximum stress, the deflection at the middle of the beam is 0.00101428571429metres (1.01428571429mm)" (online calculator) agree.

So that carefully checked work of Nov. 2020 with two different theoretical treatments and physical engineering tests and this new online beam calculator agree.

Both the 2020 mathematical work and the online calculator are based on right-angled-corner sections (slight deviation from actual sections with rounded corners).

Would be glad of any checks anyone else does.

There is the "Contact Form" on my website if you want to raise a private query. Seen from "root" of weldsmith.co.uk - is

Both the 2020 mathematical work and the online calculator are based on right-angled-corner sections (slight deviation from actual sections with rounded corners).

---------------

If the corner is truly a quarter circle the area reduction is pi/4 times the wall thickness.

Wouldn't it be a bit more complicated than that on how much metal is where? The "square" corner is the material thickness too. Then there is a more complicated effect on assigning where the metal is to what it contributes to the Second Moment of Area?

In the context that

• the radius sections have at the corners / edges is a bit variable anyway

• the wall thickness variation of the sections would make a bigger difference than accurately representing a rounded corner compared to a "square" corner ???

Nett - my impression is leave as right-angled corners so at least the user knows where they stand. The simplest model.

I am missing here some simple and obvious logic I clearly haven't grasped?

There's yet another argument to "leave as-is". You are going to be selecting promising sections from "The Blue Book". Which has properties which are best known by empirical testing done by experts. Particularly the "Class" of the beam. Will it bend - but be heavy. Or be light - but prone to buckling. Only the designer of this structure in question can know how to navigate that. Obviously - "my" "simplistic" beam calculation cannot. But the low "cost" of running calculations with my "web server-side scripted" "beam calculator" means you can develop your ideas - and record a trail by copy-and-pasting the output into a "project file".

When you've found the apparent candidate - do anew the calculations using "The Blue Book"'s tabulated

• Second Moment of Area ("I")

• Section Modulus ("Z")

• Moment capacity (in N.m)

• the statement of the section Class ("1" to "4")

• etc. for the chosen solution and take that as "THE design case".

You get straight into the F=M/L (cantilever) and F=4M/L ("simple beam") starting from the tabulated values - where the human can readily see if the arithmetic looks about right (the early stages with Second Moment of Area take a series of steps where any one non-obvious error can give wrong output which is difficult to intuitively spot (reason to write functions to do those calculations - so what you did in the best of times and tested with mathematic values which give obvious mental-arithmetic-obvious outputs - and spot-checked "realistic" section values - serves you machine-faultlessly when you are fraught, tired and distracted))

You must have experience of such "navigation" as a skilled knowlegeable project driver...

Regards, Rich Smith

<snip>

You ever read?

Quite the slog but I don't regret suffering it...

Your kinda discussions always bring it to mind😉

Should I take that as a compliment?!?

A long time I read Pursig's "Zen and the Art of Motorcycle Maintenance".

I used to tweak and repair motorcycles. Mechanical skill probably featured in my "unusual" Doctoral research in science. My abilities propelled me into a lot of problems - which then became my "story" and shaped my personality.

My Doctoral years were an almost hallucinatory experience. I only got that because I "realised" my mind does interesting things if I let it. All the interesting, crazy and wild things happened when I departed on my path of being a dedicated career person.

Once at tea-break (we were in the UK) we looked around at each other and spontaneously broke out laughing - we were all there palid, strained, shaking hands, drawing on cigarettes like a vacuum cleaner, etc, etc, etc. Such is life's experience.

So - I can assure you I am fine myself with your conjecture.

Best wishes, Rich Smith

<snip>

Cool! and I'm not surprised :) Would have been helpful had I read it earlier in life, maybe...

<snip>

You ever read?

Quite the slog but I don't regret suffering it...

Your kinda discussions always bring it to mind😉

Wouldn't it be a bit more complicated than that on how much metal is where? The "square" corner is the material thickness too. Then there is a more complicated effect on assigning where the metal is to what it contributes to the Second Moment of Area?

In the context that

• the radius sections have at the corners / edges is a bit variable anyway

• the wall thickness variation of the sections would make a bigger difference than accurately representing a rounded corner compared to a "square" corner ???

Nett - my impression is leave as right-angled corners so at least the user knows where they stand. The simplest model.

----------------------

The test is how much closer to the book value that simplified correction brings you. I agree that it may not be worth the added complexity unless the improvement is significant, but I'd rather under than overestimate yield strength. Usually I measure the actual dimensions of what I bought for the CAD model and it's almost always been near the low end of the tolerance.

Hello all

Wrote following as example of a beam calculation in a project.

Following on from creating the "beam calculator"

"Rectangular Hollow Section beam calculations" webpage.

I have a feeling beam calculations are often like this. Finding your way along. Same as for the other case I made a webpage for

"Cantilever drilling platform - analysis using beam calculations"

------------------------------------------------

I didn't give any idea why you might want a beam calculator when interested in mines...

Take for example Wheal Round (a hypothetical case).

It's accessed from high ground and water flows from adits draining mines in the area. So there's a good chance a lot of the workings remain dry.

Problem - the main shaft is blocked by a boulder 30m down which is estimated to be 3 Tonnes.

The promise of many levels and interesting finds in Wheal Round has a project to open the main shaft. Resin-anchor in an eye-bolt and hoist the boulder out.

The beam would span 5m across the top bank around the shaft.

Wheal Round is in that high place with a small path as only access, so that beam needs to be very portable.

It's lifting a boulder, there's no-one underneath anything and the hoist can be worked from a safe distance, so the consequence is just about zero of the beam totally collapsing by local plastic buckling.

So we can look to even a "Class 4" section (it could locally buckle and instantly collapse on overload).

Considering all, we'd like a factor of 3 safety on the beam load bear

- mainly for it if boulder jams and the load on the beam with hoisting increases.

So that's 9 Tonnes point capacity at the middle.

Box-section - take as being S355 steel - 355MPa = 355e6Pa yield.

We'll get an estimate were to start looking for candidate sections in "The Blue Book"

Pardon that I use "Lisp" in-line to calculate things. Using the "emacs" text-processor, I can "fire-up" its interpreter on Lisp expressions and it will insert its answer into the file.

M=FL/4 for a "simple beam" (end supported; central load)

M = ...

(/ (* (* 9 1e3 9.81) ;; 88290.0 ;; N ;; (gravity - 9.81N/kg) 5 ;; m length ) 4e0 ;; ) ;; 110362.5 ;; N.m 1e3) ;; 110.3625 ;; kNm

Going straight to "The Blue Book" for hot-finished Rectangular Hollow Section

looking down the "Mc,y,Rd" column plausible sections start at around 250x150x5mm - with a weight of

30.4kg/m

For squares - Square Hollow Sections

sections start at around 200x200x6.3mm - with a weight of

38kg/m

Going with 250x150x5mm in my beam calculator

"

Input values you provided

Variable Value (in familiar units) Height 250e-3 (250mm) Width / breadth 150e-3 (150mm) Wall thickness 5e-3 (5mm) Beam length 5 (m) Maximum stress 355e6 (N mm^2) Young's Modulus (E) 210e9 (N mm^2)

Beam intrinsic sectional properties ... Derived variable Derived value (in familiar units) Second moment of area: 3.403250e-5 (3403.25cm^4) Section modulus: 2.722600e-4 (272.26cm^3) Moment: 96652.3 (N.m)

Beam extrinsic properties ... "simple beam"

The central load causing the selected stress is

77321.84Newtons

At that maximum stress, the deflection at the middle of the beam is

0.0281746031746metres (28.1746031746mm) "

My calculation and "The Blue Book" agree - For "I" the Second Moment of Area is 3.403250e-5m^4 = 3403.25cm^4 (my) to 3360cm^4 (Blue Book).

For what it's worth general geometric page

"Z" as 269cm^4 (Blue Book) for "my" 272.26cm^3 - same.

We expect "my" "I" and "Z" to be a bit higher because "my" section is a rectangular-cornered hypothetical sections, whereas the real section has rounded corners, putting a bit less metal for from the neutral plane where it would have counted the most. But "my" calculation is quick to use to step through the logic processes of selecting "candidate" beams.

The force at yield the beam calculator gives: (/ 77321.84 9.81 1e3) ;; 7.881940876656472 ;; Tonnes-force We might seek a bit higher; were it not for seeing the weight of plausible sections - see later.

Continuing...

So the excess load carrying capacity is still okay - in the region of what we wanted.

Using my beam calculator alone now for other sections seen in "The Blue Book"

If that section fell over on its side, presenting 150x250, it would still support the boulder, with 5.95 Tonnes capacity. Not as it would be recommended to allow that to happen, but there wouldn't be the buckled beam, chain-block and boulder dropping back down the mineshaft.

300x100x5mm looks plausible too. 300x100x5 if it fell over - and it is tall and narrow - would give 41932.6N -> (/ 41932.6 9.81 1e3) ;; 4.274475025484199 Tonnes so it would likely be "game over" or best keep well clear and lower the boulder back from whence it came. So if contemplated 300x100x5 - which is 30.4kg/m - would have to bring some strong "keep upright" support with "feet" - likely a steel fabrication weighing a few kg each end.

These two - 250x150x5mm and 300x100x5mm - are looking-in at about

30.5kg/m, and the beam to span 5m is 6m long.

(* 30.5 6) ;; 183.0 ;; kg

Thinking of the narrow path - if roped the steel beam so it just lofts off the ground when a transverse pole at each end is lifted by two people each side - so eight people lifting total, that's (/ (* 30.5 6) 8) ;; 22.875 ;; kg per person - doable. But for a pole long enough to get 4 people on the pole, it would be too long (wide) to fit between the trees and bushes and through gates on the narrow path. Mind's-eye sees that as not a great option.

Path is quite straight, so look to a "tail wheel" at the back - some compliant pneumatic tyre from eg. a trailer rolling along supporting the back of the beam - and have a forward-sticking-out pole with four people on it lifting and pulling.

Other lifting strategies examined:

• Longitudinal pole with four people at each end? Well to be roped with the beam just lofting above the ground, the pole has to be 6m + about 2 x 2.5m = 11m, which is some pole - has its own weight.

• Pole attached to each end and carry at shoulder-height - NO! Never where if say someone tripped and everyone fell over there is a significant risk of someone getting hurt.

"Tail wheel" - could have additional people pulling on ropes for the steep bits (not pushing from behind - not exposed to being "run-over" if all goes horribly wrong.

So - is doable. Go with whichever of those sections

250x150x5mm 300x100x5mm or even 200x200x6.3mm (38kg/m - more grunting and perspiration) is found to be available on a blagging / cadging mission...

The reason for the beam calculator is exploring options and scenarios easily, at negligible effort to the person. A person can with care get through one beam calculation with its several steps. For several beam scenarios a person tends to get overwhelmed and lose their way.

There has been "freestyle" use of parts of the beam calculations.

There has been use of "The Blue Book".

However, the "beam calculator" made for a good portfolio of calculations presenting the case for the option suggested.

I didn't give any idea why you might want a beam calculator when interested in mines...

Take for example Wheal Round (a hypothetical case).

It's accessed from high ground and water flows from adits draining mines in the area. So there's a good chance a lot of the workings remain dry.

Problem - the main shaft is blocked by a boulder 30m down which is estimated to be 3 Tonnes.

--------------------

That's interesting, and quite similar to my built-up 16' (4.88m) gantry beam which I designed to lift 1.5 tons of log or boulder and move it about 3~4 meters after I've hand-carried all the parts uphill to the site by myself. No part weighs over 25kg when completely disassembled, that's why I used four channel sections instead of the obvious single WF beam that I couldn't move or store under cover. It worked fine to load a 2100 Lb log onto my sawmill, though I had to lower the log at the middle to step the central shear legs over it.

The supports are two tripods of 2" x 10' pipe, plus shear legs at the center splice for loads over a ton. A boat trailer winch u-bolted onto one leg of each tripod lifts the beam and levels it on sloping ground, i.e most of my property. Then chains suspend the beam and its load.

Raising the boulder isn't enough; you need to move it off the hole. A modified gantry trolley might work on your rectangular tubing but in my experience doesn't add much resistance to the beam bowing sideways and twisting.

When assembled the ~100kg beam can be moved with two fixed caster wheels bolted to a short plank under one end, and a trailer coupler on the other, coupled to a trailer dolly, home-made, like this: I can muscle it uphill with a block and tackle (or my tractor, if it runs) and let it downhill with abseiling gear. A similar long and narrow trailer for logs and cut beams is manageable by hand with a 300kg load.

It doesn't corner well, a second coupler and dolly for the tail end might help if you are actually asked to build your sample problem. In a pinch a long handled shovel can lever up a pretty hefty load and swivel to move it sideways.

Path is quite straight, so look to a "tail wheel" at the back - some compliant pneumatic tyre from eg. a trailer rolling along supporting the back of the beam - and have a forward-sticking-out pole with four people on it lifting and pulling.

-----------------

Trailer tongue jacks are heavy-duty swivel casters with load lifting capability.

one has a short URL, they are much cheaper elsewhere.

A single-wheel one can be upgraded for soft ground with replacement wheels on either side, on a longer axle. It won't track well and needs a steering + towing handle, one of mine is the handle from a broken snow shovel with a fork of steel flat stock that connects to the extended axle, another is a pipe tee with slots milled in the ends for the flat stock. The long pipe handle screws on to move by hand, unscrews for storage or to tow the shop crane it's on behind the tractor. The rather high mounting plate isn't a problem if attached to the base of a gantry hoist or mast of a crane but it makes them inconvenient for a low riding lift platform unless you can weld.

I have the shop crane, a trailer and a log splitter that all weigh around

500 Lbs and can be moved by hand on level ground without excessive effort. As a test I've pulled them uphill without the garden tractor by using a block and tackle (faster) or a lever chain hoist (safer). A chain fall is very slow and doesn't work well horizontally. First I tried cheap cable pullers but was disappointed with how quickly they wore out and failed despite being greased. The steel is very soft. A vehicle-mounted winch was best, assuming you can drive to the hilltop, but then you could tow with it instead. the axles and bearings helps somewhat, if you have a lathe. On mine the ratchets deformed when manufacturing slop allowed them to misalign.

Sounds like a large tyre piano moving dolly would be suitable for the task. I was impressed by how easily the piano movers removed my mother's piano with one and how well it coped with multiple 6" - 7" steps.

Sounds like a large tyre piano moving dolly would be suitable for the task. I was impressed by how easily the piano movers removed my mother's piano with one and how well it coped with multiple 6" - 7" steps.

------------------- They appear to use hand truck wheels on 5/8" axles. I bought a batch of used solid tire wheels with needle bearings for 3/4" axles that have served well for loads as large as an 1100 Lb boulder but I haven't found anything similar for sale new.

Yes you are right...

"Wheal Round" is imaginary, of course, but yes you would need to move the boulder sideways away from the shaft.

I kept the "Wheal Round" case as short as possible and about beams.

It proves the minimum strength beam to lift the boulder is the maximum weight beam you could move.

[you could make a truss structure from much smaller lighter sections to span the mine-shaft (the dimensions are about what you find in Cornwall, where the mining areas are littered with such shafts) - but, again, I didn't want to go away from beams - so ignoring all this...]

So the safety case is that no-one approaches the lift and its equipment during the lift. Given this...

I took it that, lifted above the shaft, you could drag a couple of beams across the shaft with ropes, and onto the beams you land the boulder.

You've already tied a "tag-line" to the strop suspending the (chain?)block from the beam...

Load off it, you pull the strop a quarter-metre (10 inches) along the beam then lift the boulder again, which will drift the boulder sideways by that quarter-metre. Repeat, repeat, repeat, ... until the boulder is on the berm. Maybe more steel sections as a slide-way and use a "Tirfor" to haul the boulder up and over the berm and bring it down to level ground.

???