Heyo, I'm improving my smallish house and trying to figure out the math involved. I'm replacing my hung ceiling with joists and I'm trying to figure out how thick they should be so I can walk on them and store stuff up there.
Because SEO companies destroyed the internet all I can find is dozens of useless websites like "the spruce" that parrot American code requirements and dimensional lumber.
Dimensional lumber isn't popular in my country and is very expensive. Construction wood in a different shape is cheap. I need to figure out the engineering so I can size my own beams. I'm not an engineer but I can read and do math and this isn't exactly a complicated question, so I think I've got this.
Where I'm at right now is combining two bits of data I've happened across:
1. A 2x6 with a ten foot span can support 760 lbs in the center.
2. the flexural strength wikipedia page (pic related)
Plugging in the numbers for the 2x6 (1.5 inches wide, 5.5 deep, 120 long) into the formula in the picture, I get:
σ(stress) = 3(760lb)(120in)/2(1.5in)(5.5in)^2
which gives me stress = 3015 (not sure what unit this value becomes)
So I'm guessing here, that plugging in differently shaped beams, weights, spans, loads, etc, as long as I keep the stress value under 3015 whatzits I should be safe. My longest span will be only 12 feet. Shear shouldn't matter for something this small right?
Am I doing this right?
Further, can anyone suggest resources for learning structural engineering from zero that don't involve textbooks? Like youtube channels or books that have pictures and are well written?
>Dimensional lumber isn't popular in my country and is very expensive.
And yet you have internet.
So to answer you question in the shortest way possible: deflection. Not stress or strain or load is what will determine the size of your beams.
Ever stood on a branch and bounced up and down? It was holding your weight, the strain you put on it, and even the moment at the tree trunk and it did not fall. It was deflection you were playing with.
>learning structural engineering from zero that don't involve textbooks
Nope. The retard solution is get something rigged up, if it breaks, add more wood and/or a tighter spacing until it stops falling in or wobbling.
Deflection, got it.
>look at charts and err on the side of overbuilding.
That's essentially what I'm doing and it's seeming to work, though it would be nice to have more concrete knowledge.
>type of wood
Here's problem number two, because it's Japanese cedar, which isn't in the English charts, and my Japanese google-fu isn't up to snuff yet.
Actually no, I want to put in a full furnished room up there and use it for storage and chilling out, like picrel.
Floor load is usually specified as psi over the whole area of the floor, not as a point load in the center of the beam. You could pretty easily figure out the stress and deflection with basic mechE knowledge (you seem to already have this). I'm not sure what the limits are, or how they are specified, however. I'm curious if anybody knows where those limits come from. They would definitely depend on the type of wood, bevause span tables are different depending on the type of wood.
In your case, look at some span charts to get an idea of what size lumber you need and then err on the side of overbuilding it.
>so I can walk on them
sounds like you don't plan to install a floor or anything on top of them, and leave it unfinished like picrel (while still having some beams to stand on). Is that right?
I went down this route. Get ahold of some LVL beams that meet your spec. I used them for garage headers, and they're extremely stout.
I get max 273kg for a 2x6 on a 120in span. If you put that load in the middle it will bend 19mm (which is not comfy at all)
The strength of the wood can vary a lot, but it's usually safe to use 17 N/mm2. Even normal pine will be a lot stronger (>80) but the might be defects in the wood.
The elasticity of wood is usually around 10.000 and doesnt vary as much. This is usually also the only thing you need to check for a floor, because it will bend a lot before it breaks.
If you want to calculate this you need to find the "inner moment" (no idea what this is called in English) of the beam (can also be steel/concrete/anything) and calculate the max "moment" of the load (also keep in mind the weight of the beam itself)
I dont think a 2x6 is enough for this span. A 2x8 with 150kg in the center will flex about 4mm which might be acceptable.
If you want to read more about this, just look up the book list for civil/residential engineering in your country for first year university/college. There will be something about basic construction and you can probably torrent the pdf.
Also keep in mind that plywood on top of a 2x6 will create a T-beam which will be stronger than just a 2x6 and a 1x8 will be stronger than a 2x6 (but you might have other issues with that)
Usually, maximum deflection for a beam is L/240. That's 1 inch in the middle of a 20 foot beam.
Thank you so much for this anon, this is exactly what I needed.
Good to know that original bit of data about the 760lbs (345kg) was wrong. I kinda felt it was and overbuilt. My current design is 2.4x18cm beams, spaced at 24cm on center, with plywood screwed onto the top. Math says each of those beams should be stronger than a 2x6, and it's not like I'm parking a car up there so it should be fine.
The 345kg is probably not wrong, it depends on the value you use for the strength of the wood.
There is probably something in your local code about the weight on the floor for calculations. I think usually is something like 175kg/m2, so 2.4x18 spaced 60cm/24inch should be within the maximum deflection of L/240.
If you use 24mm wide "beams" you should also connect the bottom so it cant rotate. It will also look weird and it will be harder to nail/screw the plywood on. It will also be nearly impossible to walk on when there is no plywood attached. 2x8 will make things a lot easier.
You should also check the local prices for your wood. Over here SLS 38x184 is half the price of 50x200 and 22x200 is not even in stock.
2x8s would be easier but they're crazy expensive.
My favorite, cheap, wood shop is closed for the week because it's a long national holiday. They even closed the weekend before and after and the two weekdays this week that aren't national holidays. Their boss is a real bro. Anyway, it's the same reason I'm off and working on the house.
I went to a big box store and the 180x24s are the only thing big enough that's affordable, and waiting for my lumber shop to open negates the purpose of doing this all over the break.
I'm securing them with plywood on top and making hangers out of 2x4s (the only cheap dimensional lumber) because simpson ties don't come that thin. Once the plywood is screwed down I'm gonna use offcuts to space the beams in the center so nothing flexes.
>the "inner moment" (no idea what this is called in English)
waarom de debiele burgereenhedeen broeder
Hello engineer anon, do you know if there is any free software/website where you can make these calculations?
I have been thinking about making a website where you can do this for a few years now, but never really put it online because I'm pretty sure something like that already exists and i just haven't found it yet.
Was thinking about something really basic for diy'ers, so drawing/selecting a profile, selecting the material (wood type/steel classification/concrete) and a basic side profile with loads and supports.
Does anything like that already exist? I regularly see contractors (real life and youtube) make stupid choices because they "feel" something need x beam or x inches of concrete.
> Construction wood in a different shape is cheap. I need to figure out the engineering so I can size my own beams.
Use the American calculator and take the nearest bigger standard size in your country, it just make your own 2x6s.
Calculate all you want but end of the day you want to account for wear, imperfect grain, moisture and whatnot which is exactly what the code tables are for. Also post country
A 2x6 is perfectly adequate for a 12 foot span
Even if you stored your left over bricks in the attic
Structural engineer anon here
Give me some sketches or pics of what youre actually doing
Ah hell yeah. I'm about to go out for a while but I'll be back in a few hours with the sketch
Here you go, let me know if anything doesn't make sense.
Basically you take the length in feet and divide by 2. Round up to the next size board.
Example 20’ deck needs 2x10 at 16” on center. This is always done in a hardwood specifically yellow pine in the south US
The issue is I cannot work with 2x6,2x8,2x10 etc. wood because it's not a standard size here and so is very expensive. While it can be bought, the cost of the project would be about 4 times as much.
Thus I need to know the math behind the American rules so I can redesign to the same level of strength with unusual sizes.
As a point of reference, most houses here are timber framed, made with 105x105 vertical posts and often barely trimmed logs for roof support.
Looks like your roof rafters are supported by the ridge beam. You would have posts every so often
Otherwise sometimes the ceiling joist is part of the roof system for outward thrust. Maybe investigate a bit before you go demo.
I will try to give you the best I can:
Assume 2x6 (1.5in x 5.5in)
L = (length) 10 ft
E = 1600000 psi (Douglas Fir Larch APA No. 2 Grading)
Ix = bh^3/12 (in^4) (mm^4)
Ix = (1.5in)(5.5in)^3/12 = 20.8 in^4
Sx = 1/6bh^2 (in^3) (mm^3)
Sx = (1.5in)(5.5in)^2/6= 7.56in^3
fb = M/Sx (psi) (dunno the metric/mm^2)
Fb = 900 psi (DFL No. 2) (you need to find whatever species you have)
M = wL^2/8 (lb-ft)
w = Qt (plf)
Q = 30 psf (10 psf self weight + misc, 20 psf uninhabitable attic with storage)
t = 16 in o.c.
w = (30 psf)(16"/12") = 40 plf
delta (deflection) = L/240 maximum
delta = 5/384*(wL^4/EI)
Therefore I minimum = 0.00130208333 * w * L^3 / E or 1.38 in^4
M = (40 lb/ft)(10ft)^2/8 = 500 lb-ft
therefore fb = 500*12/7.56 = 793.65 psi < 900 psi, I > Imin
I also can increase that 900 psi allowable * 1.15 (repeatable) * 1.15 (live load) but ignore that because idk what eurocode wants
(Keep in mind I have unit shenanigans here and there like converting to inches)
You should be able to replicate this if you really care but 2x6 Douglas Fir Larch (read literally any 2x6 available here in US) no 2. Grade will work
Dunno the grading in europe but there has to be some standard to define Fb and E. Everything else is math
Saved you $500 for my stamp
>(Keep in mind I have unit shenanigans here and there like converting to inches)
but why are you using decimals and not the superior fractions of an inch?
This isn't my house in the pic but it looks exactly the same. Don't worry, I'm not demoing anything except for the hung ceiling, which is visible under the insulation in the pic and is about 1/8th of an inch thick. The actual structure of the house will remain.
Thank you for the writeup on how to do this calculation. I'll try to do the math with my setup now:
Have a few questions on units
What do lx, Sx, M, w, Q and I stand for? I want to look them up.
Let me see if I've got this right:
L= (length) 11.8ft
E = modulus of elasticity = 914000-1420000psi (Japanese Cedar) = Call it 1000000
lx = bh^3/12 (in^4)
lx = width x height^3/12
lx = (0.95 x 7.1^3)/12 = 28.3 in^4
Sx = bh^2/6 (in^3)
Sx = (0.95 x 7.1^2)/6 (in^3)
Sx = 7.98 (in^3)
Fb = bending psi = 625
(couldn't find for Japanese cedar but Alaskan cedar has similar E values so probably has similar Fb values)
M = wL^2/8 (lb-ft)
w = Qt (pounds per linear foot)
Q = 30psf
t = 9.5in oc
w = (30psf)(9.5"/12") 23.75 plf
delta (deflection) = L/240 maximum = 11.8/240 = 0.05 ft
delta = 5/384*(wL^4/EI)
(I'm assuming the fraction is a constant)
delta = 5/384*(23.75*11.8^4/1000000*I) (by the way who is the fucker who made uppercase i and lowercase L the same fucking symbol I'm gonna choke him out)
delta = 5/384*(23.75*11.8^4/1000000*I)
delta = 0.013*(460460/1000000*I)
0.05*I = 0.006
I think I lost you here, my number doesn't make sense. Can you explain the part from delta a little bit more?
Whatever the case, with as far as I could get I can tell that the floor beams are about 25% stronger than 2x6s, and that with the spacing the load per beam is about half as much, so it looks like it'll be just fine.
Thank you! There's very few cases where a research paper writing style is anything but an impediment to understanding. I'll get a hold of this book.
i also had a course on beams but to see you use inches and psi instead of mm and pascals i honestly want to gauge my eyes out
I can use either man, see the metric units in my plan design? I used inches because he did and I didn't want to mess around with changing units.
Ix = moment of Inertia
Sx = Elastic Section Modulus
w= is my representation if a uniform load on a simply supported beam, which is what you have, other cinfigurations may have no w at all and so deflection and M would be differently calculated
M = flexural moment, integral of shear diagram, which is an integral of the actual loading
Q = area load in lbs/sq ft, imagine 40 lbs in every single ft x ft across your entire attic, it's structurally sound up to that per code. It also is checked for servicability hence the L/240 which is emperically derived.
Most of this calc assumes you will have sheathing of some sort so I guess I forgot that bit
Let me check the calcs
E = 1000000 psi
I = 28.3 in4
Sx = 7.98
w = 23.75 plf = 1.98 lb/in
Delta = 5*(1.98 lb/in)(11.8ft*12in)^4/384/1000000/28.3 = .366 in, or roughly 3/8" sag fully loaded
Allowable was .59 inches so youre good there
M = 1.98 lb/in*(11.8*12)^2/8 = 4963 lb-in
Fb = 4963/7.98 = 622 psi, and call it good because you do have allowable factors that can increase bearing stress only by 32%
Your depth is not great but your spacing makes up for it and we're being conservative
To everybody else I get that imperial measurements are dumb but the US studies both (in STEM) and then in practice its all imperial because I am talking to real people (tradesmen, architects, etc) and the average person isnt versed in both. If I make anything less than a 1/4" on drawings they either call it cunt hairs or tell me their tape doesnt go down that small.
OP sounds like youre fine
>Further, can anyone suggest resources for learning structural engineering?
Yes, I can. It's conversational, basically written for non-engineers. I found the steamboat chapter hilarious, probably because fuck yeah 'murrika.
Here it is on archive.org
Thanks anon, I really appreciate your trouble here. I'll post some pictures of the build if this thread is still up tomorrow night.