Timber Frames Calculation. 01. The Basics.

xacob
6 min readNov 2, 2020
Image: Author.

The mission of a structural frame is to bear the loads it has to support during its planned lifetime.

Think about the chair you are sitting on right now. The chair reacts to your weight (load) with the same force in the opposite direction. That reaction is the pressure you are feeling in your butt. In a timber frame, the system works the same: It reacts to the load with the same intensity in the opposite direction. It’s Newton’s principle of action and reaction. It means, the more you load it, the more a frame reacts to that weight. The structure and the chair can do that but not for every load because everything has a strength limit. From there the structure breaks apart or is deformed to a point that makes it not useful anymore. Think about your chair again. Intuitively, you know it would be probably crashed under the weight of an elephant.

A chair is not that important, but in construction, architects and engineers put tons of material over the head and under the feet of humans. We better make sure we don’t reach the breaking point on any part of the structural frame.

How to calculate Timber Frames.

You need to follow four steps:

  1. Design a structure you think will work.
  2. Calculate the loads.
  3. Calculate how much of those loads get to every point of the frame.
  4. Calculate the strength of the weakest point of your frame.

Compare the load on that point (E) with the strength or resistance of the wood there (R).

If loads are the same or lower than strength, E ≤ R, your system works.

If loads are bigger than resistance, E > R, then it doesn’t work. It breaks.

In case b. you have to start over again with a new design.

Easy, isn’t it? Well, just in theory. Unfortunately in real life loads vary a lot, because people and nature are variable and complicated. An office building can be full of people during the day and empty at night. A house works reversely. We can put more furniture or less in our buildings. A football stadium or a parking lot has to support more weight than a house or an office. Wind changes in direction and intensity and snow can add a significant load to the structure some times of the year. If we go to less probable loads, we can mention earthquakes, gas explosions or plane crashes. When you design a building, you have a whole range of loads, with more or less probability to happen.

That means, loads aren’t a value but a probability with normal distribution:

Now, from that diagram, what point would you take as useful value for your load? Some of you would take the average, Emed. That’s an idea, but, in a hard winter, with a lot of snow on your roof and the building full of people celebrating a big Christmas party, your Emed load would be below the real load.

If you take Emed, you will be covering 50% of the possible combinations of loads, and that’s extremely risky. There is a consensus between engineers, and it says that you have to take a load that covers 95% of all load posibilities. This way, there’s just a 5% probability for the loads on your building to go above your chosen load. We call that load “Charasteristik Load” and note it as Ek.

I can read your mind. You are thinking “5% probability is still a risk I don’t want to take” and you are right. 5% risk means that you could go above it eighteen days per year. Still too much. To solve this, (Ek), the basis value, will be multiplied with a security coeficient called gamma (γ). With this operation, you will get (Ed), the “design” load, and that is what you use to make the calculations. Now the probability of the real loads to surpass your design loads has become really low.

γ x Ek = Ed

Let’s do an example to see how much security we have: For a house, engineers and architects take 200kg/m2 of the variable load over your floor (2KN/m2). This value takes into account people and furniture. To reach that load, a 90m2 apartment has to be crowded with 3 people every squared meter. 270 people!!! That would be the wildest party ever. This is your Ek, but that’s not crazy enough. Engineers apply the gamma coeficient (γ) to make that number bigger. In this case, they use γ=1.5.

γ x Ek = Ed Ed=1,5 x 200 = 300 kg/m2 (3 KN/m2)

For a 90m2 appartment, engineers, and hopefully, you too soon, calculate that you could put between 360 and 450 people there. Four to five people per square meter. Almost an impossible event.

Remember: You take a rare big load for your building (95% probabilities or less to be reached) and then you make it bigger to be sure that the probabilities of reaching them are extremely low.

R Calculation

Good! We have converted a normal distribution of loads into a value Ed that we can compare with the strength of the building (R).

BUT!

As you can imagine, resistance or strenght R is not a value, but a standard distribution. We can use the same thought process to get to our design value of R, which we will call Rd.

In this case, as a standpoint, we don’t take the average of the resistance, Rmed, either. We will take a value for strength or resistance that apperars only on the 5% of wood. If we build a floor with 100 beams, just 5 of them will be under that value, the characteristic value of the resistance (Rk)

Now we put more security, and for the design of the structure, we make the characteristic resistance lower with a gamma coeficient (γ).

Rk/γ = Rd.

Let’s see this that with an example:

You can calculate the tension (N/mm2) at which the wood breaks apart. If you do that a lot of times, you get a lot of values for R. In the illustration below, every black point is a test, and it is placed depending on the obtained resistance. Look that there are a lot of tests that come around 50N/mm2 and that the distribution of the results follows a normal distribution. We don’t take Rmed (50N/mm2). That would be too risky. We take the point where 95% of the tests are above Rk. In the illustration below, the wood characteristic resistance is 24 N/mm2.

And after this, we apply the gamma coefficient (γ). For normal wood, it yoult be γ=1.7

Rk/γ = Rd. 24/1.7 = 14N/mm2.

Look at that. We have wood beams that reach more than 75N/mm2, and the average R is 50 N/mm2. But at the end of the day, we take just 14 N/mm2!!!!

It means, that we are using normally around 28% of the total strength of the wood. On the other hand, the real loads are normally much lower than the design loads, around 50%, that’s why buildings are so safe.

Here you can see the whole picture, the relationship between E and R, and the real and calculated security.

https://en.wikipedia.org/wiki/Probabilistic_design

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