The EASY WAY to do a Timber Beam Calculation!

[@maarirs12894]

5 Years of architecture study and not one professor explained it this simply. Thank u!

[@RobindeJongh]

That's great to hear, Sriraam

[@rodgerq]

I think this is an issue of study in general. Some tutors will have their own way that their brain manages a problem and so they explain the solutions that way, but for others that doesn't compute so easily and so they struggle to understand.

[@pterodactylx3100]

@@rodgerq Rodger - I agree. It is always useful to use VISUAL explanations wherever possible and Robin does this. Also, I think it is beneficial to make little models out of straws and push them and pull them around to get an idea of where the loads go. You would find that a two-straws thick beam is half again as strong as two single straws side by side. At least I think so after watching the video and doing the calculation in another comment. It is important that the teacher gives the principles involved. This is what Robin does when he ignores all the details (type of wood etc) and gives the basic essentials so we can see what is going on. I would be interested to see his further learning materials.

[@RobindeJongh]

💥Continue your learning journey with BRILLIANT: sovrn.co/16qqwyx

[@staticGenerator4You]

Thanks!

[@TSBoncompte]

I'm literally just getting into this kind of thing (structural calculations) and I must say, it's hard but at least with your vid i feel like i'm getting somewhere. this is therefore excellent!

[@grahamtucker6793]

Thank you so much for your videos that have saved me from setting up some practical tests.

[@shaunleamon6349]

Thanks Robin, great video. You used an example of 2 kN/mK as the land, but can you advise how I work out a specific load on the beam - eg a series of pergola joists? Thanks!

[@morosmith8710]

Hi Robin, any chance you do a video on continuous beam calculations please.

[@dementedmlbb4505]

This is amazing! Could you please explain what happens if we have the same scenario, but we have a small cantilever at the end?

[@arthurappelgren6792]

Why did you not say the allowable stress is Bending Parallel to grain? That way we can see which number on the chart is meant to represent?

[@gatocomic]

niiiice! In Chile we have a tangled method for that, with factors that modified the allowable stress (i imagine that there are the same in professional cases) but this is a good method to do fast numbers! i love it! :) thanks

[@SchoeneLeben]

Thanks for the video! This is really helping me learn this stuff. Do you know any books/texts you would recommend to learn more about these things?

[@AD-rx8wc]

Invaluable mate thankyou, wish I had this when I was revising for my exams years ago!

[@stephenparker6697]

Hi Robin, thanks ever so much for posting this video - it is so clear. I'm a retired director of a civil engineering company and have always had a keen interest in structural engineering. Would you be able to advise or show how the deflection amount on a timber beam is calculated please. Many thanks.

[@RobindeJongh]

Hi Stephen. Thanks! I hope to cover timber beam deflection soon.

[@РинатСафин-б1м]

@@RobindeJongh that would be very nice as well, thank you

[@cooligbeair]

I have a 4”x 6” x 17‘ long beam. And I’m trying to figure out what the maximum load is if there is a post at each end of the 17 feet holding it up.

[@triumfdula8830]

Hi, making a shed with 5 x 30 cm timber beams (30 cm being vertically) would the beams in 5 meters length bend from it's own weight and snow ?

[@chrishaslam1730]

Great video, I am wanting to span 1m on 2 concrete pillars with a steel box section 3x2 as a level guide, on top will be a sip panel as a floor with a 4 x2 insert timber then a sip panel fixed vertically above, the panels are the same width as the span rigid and load bearing, will the steel be effected by the load or will the load be transferred to the pads? Thank you

[@sebronhiob7664]

Sir regarding single point loading in a beam test would you please show me the area of bearing of of a timber in a timber testing.

[@LOL-cs1op]

Hi Robin. Really an useful channel for people who are new to timber design just like me. I wonder if you could do series of full residential house timber design and analysis for 2 Story house with basement unit using Limit state design? It will be a great favor. thank you.

[@RobindeJongh]

Great suggestion! It would have to be a whole series of videos though!

[@LOL-cs1op]

@@RobindeJongh Of course it will and it will be a great content which will benefit many like me. A practical approach how they design at offices for construction. Looking forward to hear from you. I hope you get the time to help us. Thank you for replying me. ✌️

[@richardparsons7012]

Thank you for sharing. I'm struggling to get a clear answer from my assessor re beams with multipul members. Ie- 3 45x195 bolted together to make a beam 135x185. Is the figure I pop in for Z simply the total section dimension, or do I have to make considerations given the combined elements? Really appreciate the videos, you're a bg help.

[@danielintegrity7050]

I am practicing to learn staad to find max moment, etc., And make excel spreadsheet to check simple design. This is a great video. Ty

[@chrise202]

Load per meter - What if all the load is applied on the center only? Eg: 500kg at midspan of a 10m beam, eg at 5m. Would you consider the load per meter be 500/10=50kg/meter ? From a bearing capacity does it make a difference if the load is applied evenly vs only at center point, or maybe at 1/3 of the lenght?

[@RobindeJongh]

Hi. The formula for bending moment changes depending on where the load is. Try checking online for a list of the formulas.

[@cliveh3330]

Hi Robin. Good work for spreading the word. Can this example be used for an unrestrained beam that is not a floor or rooof joist fixed along its upper edge. Am trying to get to grips with timber design and see your allowable bending stress for C24 timber is 7.5N/mm2 wheras Eurocode 5 starts off with an allowable stress (in bending) of 24 kN/mm2 which is then reduced by numerous factors but still seems to end up a lot higher than the 7.5. Is it a case that not all the Euroocode factors have been applied or BS 338 provides a much more conservative design.

[@RobindeJongh]

Hi Clive. WIth Eurocode you have to also factor your loads, so that effectively removes any benefit over British Standards.

[@MrMac1986]

Great video, coming in handy for my 2nd year of CivEng. What would the difference be if its a mid-span point load?

[@RobindeJongh]

The equations for bending moment and deflection would be different.

[@MrMac1986]

@@RobindeJongh I see, do you have any resources on how to calculate such a scenario? Would be most useful for my assignment

[@facutinmoldova]

noise from that pen killing me arhhhhhh

[@LifewithLewy]

Gooooood me too

[@arfski]

@@LifewithLewy Call it ASMR and people love it and the view count will shoot up.

[@thoeunhor8655]

Wow 🤩 really lol

[@arvindgadri6643]

I have a 2x4 stud wall which was has been constructed on site using 400mm centres. The stud wall is the inner leaf to a single brick exterior wall and the new stud is supporting the load from a new roof. Dead load of 1.81 kn/m2. Nothing built above as it is a bungalow. Can I use this calculation to see if the stud wall can support the load from the roof?

[@RobindeJongh]

Hi Arvind. For vertical loads on stud walls you would be better with this tutorial: kzbin.info/www/bejne/bXi1iaBvaK6Geck

[@samuelthedude]

In the US, I'm assuming we look at column on for Fb Bending Force in PSI?

[@honeyobenza3420]

determine the axial column load capacity of a wooden column, having a size of 10" x 10", and a height of 6 ft, using Yakal, 80% grade

[@arfski]

0.4*fcu*Ac + 0.75*Asc*fyA

[@RobindeJongh]

That's only for concrete.

[@RobindeJongh]

See this video: kzbin.info/www/bejne/bXi1iaBvaK6Geck

[@deborahskeete]

Please can you share your knowledge on vaulted ceilings

[@RobindeJongh]

Hi - try this one: kzbin.info/www/bejne/i3mTfoeer7Sel6c

[@arthurappelgren6792]

Where did you get the numbers for the C24 Timber? Is that a chart? Okay, I got that from your downloadable chart. What species is the wood in this demo chart?

[@omorigojira6194]

C24 is a standardised strength, its graded to meet those criteria as a minimum afaik

[@73Ferret]

Brilliant, clear explanation - thank you! I’m looking at putting a 200l horizontal immersion tank in a loft - span is 3m, so am I right in thinking that if I put doubled up and bolted together 50x100 c24’s in under each support that will be ample??

[@RobindeJongh]

All I can say is that a water tank is heavy, and if it is above people's heads, you might be best looking for a structural engineer to be on the safe side.

[@TotuBine]

Was it ample?

[@christianblake3997]

@@TotuBine No he’s still in Hospital after the water tank fell on him 🤣🤣🤣

[@pterodactylx3100]

@@TotuBine It would make an interesting project on an engineering course to make these calcs then put a tank on a beam and add water until the beam snaps, then see how close 'real life' is to the theory. I suspect there is a wide variation as a lot depends on the given strength of the wood, but the wood dries out, or gets damp.

[@73Ferret]

@@TotuBine well yes it must be cos I'm still ok!

[@danielshirey8024]

awesome lesson! Thank you

[@Seeker64]

As an American General Contractor and Timber Framer, how much of the formula changes when working in feet and lbs.?

[@Seeker64]

Does the bottom of the formula, dividing by 8 change?

[@RobindeJongh]

The moment formula stays the same no matter what units you use. So in the US you will want the resulting units of moments to be lbf.in, and will input w in lbf/in and L in inches.

[@Giuseppe_coachgius]

Does it makes sense within a floor system to individually calculate beams based on specific loads a ting for each beam (floor load width of each beam)? Would I obtain a combination of joists sections that would specifically suit what is going on in specific parts of the floor system?

[@avgjoe-cz7cb]

Live load/Dead load ? Can I span 12' with a 2x8 at 16" OC?

[@honedanny28]

Trying to work out load for a 6x6 cedar beam 4.2m length with .75kn per m2

[@pawelmacpoof5687]

Is there considerations for knots and imperfections for wood, a job i am applying for need basic wood design. Talking about that is there any uk standards like there are for road design?

[@RobindeJongh]

Hi. Timber is graded by how free of knots it is and its strength. The grade (C16, C24 etc) takes all that into account. The UK codes for timber design are BS EN 1995 or BS 5268.

[@pawelmacpoof5687]

@@RobindeJongh cool thank you.

[@balabyekubojohnchristom241]

How are you professor? Iam Balabyekubo I.J.C I have problem sir and asking for solutions. In floor system consists of one way spanning slabs with beams, there are side beams which don't carry slabs that span in short directions meaning they don't carry slab loads. They just provide restraint. How can you get loads to necessitate the design of that side beam which don't carry slab? Thanks.

[@morosmith8710]

Thanks for your content love it, however isn't section modulus Iy = bd^3/12

[@RobindeJongh]

The video uses elastic modulus. For a deflection calculation use second moment of area, which is what you mean I think.

[@morosmith8710]

thanks @@RobindeJongh

[@ronhorton21]

Okay, that left me completly crosseyed.

[@lhrete9878]

Can we apply this formula for steel except Z.

[@discostuart1981]

Great vid is there a separate video for the deflection ?

[@RobindeJongh]

Hi - lots of people have asked for this so I hope to get a video done soon!

[@discostuart1981]

@@RobindeJongh great news just been working out a timber Ridge beam.... Found this really useful

[@Joe14579abc]

Have a building 20ft wide by 60 ft long what size of floor trusses can i use without anything in the middle, just straight, foundation on each side will hold the load.

[@RobindeJongh]

Hi John. Drop me a note here and I'll see if I can help: forms.gle/CUy6G2nBxjooekUz6

[@plummetplum]

Hi Robin, I've been looking at a size of floor joist beam @ 400mm centres for my loft conversion to span 3.4m. I'm trying to minimise the joist depth. According to the the span tables I can use a 147x75 beam but according to your calc this beam isn't suitable. I take it there are more calcs needed to account for spacing etc? Do you have a video on that? Thanks.

[@loganswift8865]

To Account for spacing you must first calculate the tributary area of the beam at the given spacing, then calculate line load based on the tributary area. I would look at how you calculated your load. Beam tables are always built with a factor of safety and overestimate the sizing

[@plummetplum]

@Logan Swift Thanks, I bought Robin's spreadsheet, which does the calcs and factors in spacing, so all good.

[@Ftseinvestor]

Hi, is the calculation method the same for a roof rafter as it is for this example? Thanks

[@RobindeJongh]

Yes it is. You may be able to take a higher value for timber strength due to load sharing between members, and you need to account for the sloping member if it is a pitched roof.

[@Ftseinvestor]

@@RobindeJongh Thanks for the reply its much appreciated, How would I account for the sloping member of say a pitched roof between 15 - 22 degrees.

[@Seeker64]

Where did you get the Allowable stress of 7.5 n/MM2? Is this in a book? Is this the same as F sub b in bending values from a say the Timber Construction manual?

[@RobindeJongh]

The allowable stress of 7.5 N/mm2 comes from the UK timber code BS5268, and I have chosen a common timber grade here of C24. Your Fb does appear to be broadly similar, thought the material factors of safety used to arrive at that figure may be different.

[@sheepsy90]

@@RobindeJongh Hey, can you look at this resource and explain how to derive the 7.5 N /mm2 - As far as i can see the conversion from MPa in N7mm2 would be 7.4 kN /mm2 which sounds wrong.

[@samozabijaka]

I also wonder why its recomennded 7.5 where the limit stress for C24 pine & spruce is .... 24 N/mm2, or am I looking at wrong specs ?

[@arfski]

@@samozabijaka Table 8 of BS5268 part 2 gives C24 Bending parallel to grain as 7.5 N/mm2 so yes you are, possible you're looking at the Compression perpendicular to grain figures?

[@RobertSimmonds-o6q]

You glossed over pretty quickly where that 7.5 factor comes from. Is it from a British Standard?

[@RobindeJongh]

7.5N/mm2 is the value taken from the british standard for bending strength.

[@osuosubi3379]

Thanks very much

[@discostuart1981]

Hi Robin ever fancy doing a ridge beam video. I have calculated a beam with my understanding of half the roof load sitting on the ridge beam and 25% sitting on each wall plate. Our building inspector says that this is incorrect and it should be 2/3s to 3/4s of roof load acting on the beam.

[@RobindeJongh]

Thanks for the suggestion - I'll add it to the list. A ridge beam would generally take half the roof load, but for this you will need to take into account the pitch (slope) of the roof. See my course here for full worked examples of loading: structural.thinkific.com/courses/steel-beam-calculations

[@DB1956]

So for single storey shed with an opening of 180cm in a load bearing wall will I get away with doubling up on 2 by 6s or should I get 2 by 8s (por indeed 2 by 7's- the space between the two walls is about 12 feet.. We never get more than a couple of inches of snow here.

[@RobindeJongh]

Send me some more info and I'll see if I can help: forms.gle/fSMzoXMJPrpzVctv6

[@loganhogan953]

I had to stop watching the video because the noise of the pen scraping onto the paper was giving me goosebumps

[@RobindeJongh]

Thanks for the feedback. I’m moving to pen tablet in future videos so will try to improve this.

[@loganhogan953]

@@RobindeJongh you're all good my brother you're a smart man

[@awesomedave8484]

@@loganhogan953 some people pay for apps that make this happen... what its called called ASMR.

[@DiscoFang]

Came to the comments to say this. Felt like my ears were being sucked inwards by my brain.

[@DiscoFang]

@@awesomedave8484 This is ASMR only in the same way fingernails on a chalkboard is.

[@alexandruagapie7977]

The calculation formulas are correct, but the bending moment should be considered at the middle of beam`s length as it is supported at both ends. So L=2/2=1 resulting in BM=0.25 kNm.

[@RobindeJongh]

Thanks for watching. The video shows the mid-span moment for a simply supported beam with a udl load, which is wL^2 /8

[@MrKumar-eo2ix]

But its all about udl beam

[@rabakasultana3065]

Thanks

[@tlrieder1]

Thanks I was having a real hard time finding as simple an explanation as this. Now I just got to find the variables, are they labeled the same in imperial?

[@arfski]

Imperial? You're either old or American, the world has moved on!

[@MrNewman457]

absoulte hero, thanks for explaining this!

[@tkoster4950]

So when calculating the max load for c24 in the dimension 50x200, (inverse function) this would be : W is 5 KN/m, ((5*2^2/8)=2.5). For a beam of 2 meter the max load is 10Kn, being 1000kg, the weight of a small car, or 1000 liters of water. That seems an awfull weight to strain the beam, even if its supports are only 2 meters apart. Is this correct, or do I make a calculation error ?

[@RobindeJongh]

5kN/m seems very high. I'd suggest first working out the load, then as a seperate calculation work out the bending moment, then thirdly the section size needed.

[@pterodactylx3100]

T Koster did the same calcs as I did above in another comment. His 10kn is two lots of 5kn/m, two lots as there are two metres in his beam. His 5kn is 2.5 times the 2kn that Robin started off with, as the load to carry. But maybe this is the load at which the beam actually snaps, and if the weight of the car is all in the middle. I wonder in real life if structures are made double or ten times max expected load. You cannot make a loft roof then expect to put a heavy library of books up there. I assume most people understand this but some people will not. I assume that the first thing to happen would be a sagging roof and splits in the ceiling, as even though one beam is starting to bend at this point, the others nearby would take on some of the strain. It shows what a responsible job this is compared with say an accountant where the errors can just be put right by changing the numbers and no serious outcomes.

[@cbalano]

What is the bending moment if you have a 2 KN force right in the middle of the beam (applied over a negligible length, not applied over the 2m of the beam)?

[@RobindeJongh]

If we take it as a point load in the centre of the span, the formula is PL/4, with P being the load and L the span length.

[@cbalano]

@@RobindeJongh Thank you, good sir!

[@tdahli]

Why do you only square the 200 (d) and not the f*b*d ?

[@arthurappelgren6792]

it looks like you have written, the product of (bxd) squared and then dividing that by 6 and the multiplyinng it by the 7.5? But if I do it that way I get 4,500,000,000. Still not even close. But how oyu enter it is different than you have written.

[@TravisInCanada1]

When you say, lets say there are 2 Kn/M Where did you come up with 2? I understand that number could be 3 or 4 or whatever but what I am asking is how to determine that number in order to use it with the rest of the equation?

[@RobindeJongh]

Hi Travis. Take a look at this loadings video which may help: kzbin.info/www/bejne/g4fNpo17fJKVhpY However, bear in mind that for timber design we do not factor the loads.

[@pterodactylx3100]

Very interesting. Unfortunately I went to a 'posh' school (at the time the parents did not pay extra as it was Direct Grant in those days) and it was beneath them to talk about practical courses when all that mattered was how many students got to Oxford & Cambridge (I was not one for them). I only mention this to describe just how low the standard of career advice in schools was a few decades ago & from what young people at school tell me, it still is low or non-existent. It was the same in my father's time and he was born in 1929 - he wanted to do art & poster design but the school made him study Latin. Schools should be obliged to have people from industry in different careers come into school, or at least videos shown to them explaining various careers and courses. Teachers are not generally practically-oriented people by their nature and cannot be trusted with this task. From your video I learned that the beam's weakness is proportional to the square of the length, so one twice as long is 4 x as weak, or 1/4 as strong. Also interesting is the way the strength is proportional to the square of the depth. This means that you are better having a 4" beam rather than two 2" beams, even though the amount of wood is the same. The bending strength of two is Constant + Constant but the bending strength of one twice as thick is Constant squared, so the strength ratio is (C sqd /2C) = C x C / 2C = C/2 = half again as strong for the same weight of wood. Another point is that the load is 1kn/m and the beam in your calculation gives 2.5 x needed strength. If one kn is 16 stone then 2.5 kn is 40 stone or 4 average people. Does this mean the beam can take 4 average people standing in 1m of the MIDDLE of the beam provided there is little weight either side of them? But it cannot take these 4 people (40 stone) IN ADDITION to any further significant weight on other parts of the beam?

[@holdurhorse9149]

How about the deflection check ?

[@RobindeJongh]

Hi. Here's my deflection check video: kzbin.info/www/bejne/iXWlepmKZs6qq8U

[@davidreid2301]

Why does the depth of a beam not come into this BM equation? Surely that changes the bending moment, no?

[@RobindeJongh]

Bending moment is a function of the length of the beam and the amount of load applied. Bending strength is a function of the depth and breadth of the beam, which is worked in to the value of "Z".

[@davidreid2301]

@@RobindeJongh Thank you! 25 years ago was a long time! The cogs are a tad rusty now, too! Cheers!

[@padraicfeehily6732]

excellent.

[@Ammy007ification]

I love your hand writing 💗 awesome.

[@MelissaMiller-ym4nq]

ok well, now my brain just hurts.

[@ioananghel715]

Nice to watch u!

[@jamesderoc6717]

clear as mud

[@mradamhowell]

Is this method transferable to steel beams?

[@RobindeJongh]

This method is specific to timber beams.

[@freespeech878]

Well thanks you tube fit ruining a perfect example my covering up the working out of the formula

[@masterman280]

Shouldn’t L be 1m instead of 2?

[@RobindeJongh]

Hi. See 2 mins in where the beam is shown as 2m long, L=2m

[@moi984]

Nice! Looks like a Dutch name. Hi from Amsterda

[@ZoomStranger]

As a dyscalculic, this is arabic to me but thanks for the demo of your skills.

[@ajaikumar44]

Nice explanation 😀, but Im afraid if the moment of inertia formula is to be crosschecked. The formula for area moment of inertia is bh^3/12 Thank you

[@RobindeJongh]

Hi. I didn't cover moment of inertia in the video (see 3:50). The formula is for section modulus, which is correct.

[@philwell76]

4 squared is 8?

[@jaybay2233]

16

[@rok1917]

this video maybe good for a simple example but really bad for people who want to calculate their own beams. For Instance the fc,m for C24 is not 7,5 and also the person doesnt canlculate the beam on buckling witch is normally the defining calculation with timber.

[@RobindeJongh]

In british standards, the allowable stress is 7.5N/mm2, which incorporates a safety factor. If you design to limit state methods then that figure wouldn't be correct. Also in BS you don't work out the buckling directly, but limit the depth to breadth ratio. Hope that helps 👍

[@TheJohnson861]

I wish I wasn’t thick

[@rodgerq]

Same here. I'm just intelligent enough to know how dumb I am and it's really not a good place to be 🤣

[@christianblake3997]

@@rodgerq just read your comment today 3 months on and it had me rolling in stitches🤣🤣🤣

[@gambart2002]

I can't listen to the sound of that pen. 😕

[@RobindeJongh]

You're not alone!

[@arthurappelgren6792]

So I multiplied 7.5 x 50 x 200=75,000 then 75,000 x 75,000 =5,625,000.000 to square it then 5,625,000,000/6=937,500,000. How did you come up with 2,500,000? I am not even close. I don’t have a fancy calculator v]but for simple multiplication and Addition I thought a straight forward one woudl work. What is the math you are using? What did I do wrong? So using a scientific calculator I entered this 7.5 x 50 x 200(squared button) I got this. 40,000. dividing that by 6 equals 6,666.666667 which to me seems right when oyu divide 40 by 6 you 6.6667 What gives. Does math change from your world to mine? Really confused. I am hoping to write a spread sheet to simplifiy the calculations, but I wanted to make sure I understand the steps. What did I miss?

[@RobindeJongh]

Hi Arthur. Just do the sum like I did on the calculator, and when you get to the squared, multiply by 200 again instead. Hit the equals button, then divide by 6. See if that works for you.

[@jaybay2233]

Do 200² first (40,000) times that by 50, then times by 7.5. You get 15,000,000. Divide that by 6 you get 2,500,000

[@CADstruction]

get a new pen:P

[@chriswest1996]

Yikes! Never use 1 or 2 for a sample calculation!

[@RobindeJongh]

Chris - you're right. What was I thinking!

[@YouThinkAboutThis]

Easy my arse

[@larryc1964]

Prefer imperial measurements. This does not register with me

[@RobindeJongh]

Thanks Larry. It would be nice to be able to reach the US audience with some imperial calculations but having looked into it all the formulas are so drastically different, I fear it would greatly complicate it.

[@coolhluke8089]

You lost me??????

[@RobindeJongh]

Which bit?