The EASY WAY to do a Timber Beam Calculation! 

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

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

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!

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

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

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

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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!

absoulte hero, thanks for explaining this!

awesome lesson! Thank you

Thanks very much

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

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?

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.

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?

Nice to watch u!

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.

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?

excellent.

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.

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 ?

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

Thanks

Nice! Looks like a Dutch name. Hi from Amsterda

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??

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.

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.

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?

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.

I love your hand writing 💗 awesome.

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.

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

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.

Great vid is there a separate video for the deflection ?

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

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?

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.

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

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)?

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?

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

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.

ok well, now my brain just hurts.

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

Please can you share your knowledge on vaulted ceilings

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?

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.

Can we apply this formula for steel except Z.

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?

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

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?

Okay, that left me completly crosseyed.

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 ?

noise from that pen killing me arhhhhhh

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?

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

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?

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.

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

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

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.

Is this method transferable to steel beams?

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

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

How about the deflection check ?

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

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

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

Shouldn’t L be 1m instead of 2?

clear as mud

I wish I wasn’t thick

4 squared is 8?

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.

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

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?

get a new pen:P

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

Easy my arse

Prefer imperial measurements. This does not register with me

You lost me??????