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Above is the definition of fundamental basic wind velocity from Eurocode 1. According to its instructions, wind loads on building are supposed to be calculated based on this velocity. It says that this velocity needs to have an annual risk of being exceeded of 2 %. But isn't this quite a risk? If we design our building's extreme wind loads according to this value, doesn't that mean our building has 2 % chance of failing? To me that seems quite high.

Also according to Eurocode, in reliability class RC2, we should have a reliability index of at least 4.7, which approximately corresponds to probability of failure less than $10^{-6}$. This is much lower than 2%.

NMech
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S. Rotos
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    " If we design our building's extreme wind loads according to this value, doesn't that mean our building has 2 % chance of failing?" No. Suppose the design velocity is 25 m/s. Do you think a velocity of 25.01 m/s is *100% certain* to cause failure? – alephzero Apr 17 '21 at 13:13
  • @alephzero Well, maybe not 100%, but for that wind velocity we cannot say anything about the reliability anymore. If we only have one velocity to work with, which has 2% probability to be exceeded, then we know only that our design has 2% reliability. The point is, we don't know what the probability of 25.01 m/s wind has, and so we can't say anything about the reliability anymore, right? – S. Rotos Apr 17 '21 at 13:32
  • @S.Rotos I think you are confusing the reliability index in this. There is a very good example in this set of [slides](https://eurocodes.jrc.ec.europa.eu/doc/WS2008/EN1990_5_Vrouwenvelder.pdf) that explains the concept of reliability class. In order to calculate it you need to know the mean value and the standard deviation of the resistance of a structure and the load. – NMech Apr 17 '21 at 14:23
  • @NMech I have actually read those slides. So here is what I understand: the reliability index is directly related to the probability of failure, which is calculated using the probability distributions of the load and the resistance. But to get the distribution of the load (here the wind load) you would need a more complete knowledge about the distribution of wind speeds than one value that has a probability of being exceeded of 2%, which in turn is something Eurocode does not give. Do you have more knowledge as to what kind of wind distribution Eurocode assumes? – S. Rotos Apr 17 '21 at 15:09
  • @S.Rotos No I don't have any knowledge on the wind distribution that is assumed in Eurocodes (It should be a Weibull and not a gaussian distribution which is very skewed). However, even if someone knew the distribution (which is *very* specific to the location), you'd have to incorporate that into the deviation of the total actions on the structure. So, the point, is that you can't directly relate the 2% probability to the probability of the Reliability class. – NMech Apr 17 '21 at 15:19
  • The 2% is the probability of the 10 min average wind occuring in a year, and the other (reliability) is related - at least IMHO- to the standard deviation of the load that would be exhibited at that wind speed. So you are looking at the std. deviation of the load at e.g. 30[m/s] (added to the other actions). – NMech Apr 17 '21 at 15:26
  • How much safety factor is built in to the design code for a wind loading of X? It is very unlikely to be 1.00. If it is 2 for example, you would expect NO damage for a wind speed 1.4 times the design speed, since wind loads scale as speed squared. If you want to find the "correct" risk factor, you need to take *all* the relevant facts into account, not just look at one "2% per year" number and jump to the wrong conclusion. – alephzero Apr 17 '21 at 15:44
  • Practically speaking, we are not designing the structures for wind load alone, as mentioned by NMech's response. Every category of design loads has the chance to be exceeded, so that's why there are load factors, load combinations, and strength reduction factors, that all are intended for combating the uncertainties anticipated for materials and loadings. After all factors, the risk may go down to one out of thousands or even lower. – r13 Apr 17 '21 at 16:13
  • @S.Rotos, regarding the variability of the load, if you look at Eurocode 1 part 4, section 4.4 Wind turbulence, there is an indication about how the actual peak load is calculated based on the variability (its sort of coming back to me, but I have to see work I've done about 10 years ago to tell you more). (Although only flashes are coming back) I seem to remember is that the basic peak load could be multiplied several times over, based on orography and other parameters. – NMech Apr 18 '21 at 16:40

2 Answers2

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If you take a read more carefully the Eurocode, it says "2% annually".

That should be translated that in a full year, there is only a 2% chance to get to that wind level. So one simple way to think about this is that there is one chance in 50 years for that failure to occur (which is considered a reasonable scenario.

The more correct mathematical way is that, if there is a chance of 2% in a year to exceed that wind, then the annual chance is 98%. Since, the chance in a year is statistically independent from the next year, then the chance of not exceed the wind speed for 10 years would be equal to $0.98^n$. So, the chance that the wind would be exceeded (in a 10 min average) once after 10 years is equal to $1-0.98^{10}=18\% (approx 20\%)$, after 20 years it increases to $33\%$, and after 35 years there is approximately a $50\%$ to get a 10-min average gust.

Gust factor

Also keep in mind that this is a gust of wind is measured in an average of 10 minutes (there are approximately 50000 10-minutes is year).

When you take the average of a 10 minutes, then you know that somewhere in the 10-minute there are higher values. Usually, they measure an average of 3s for a gust, and even then the wind speed (depending on a multitude of factor, surface roughness, height from ground, location, surrounding topology ), the gust factor to convert a 10 min value to a 3 second value is in the order of 1.5 to 2. So a 10 minute average gust of 100 kph, probably has within it a gust wind of about 150 to 200kph (which is insanely high), since the load load quadruples if you double the wind speed.

accidental actions

Additionally, the wind (as snow and fire) are considered chance(/accidental I am not sure about the translations) actions, and as such they get combined with the dead loads and the other actions.

If I remember correctly (from the top of my head - this is for illustration purposes), if determine a load of $P_D$ for dead load, $P_W$ for wind and $P_S$ for snow you'd need to perform checks for the following combinations (for a class of structures):

  1. Only dead loads: $150\% P_D$
  2. Only dead loads and wind: $100\% P_D + 100\% P_W$
  3. Only dead loads and snow: $100\% P_D + 100\% P_S$
  4. Only dead loads wind as a primary action and snow as secondary: $100\% P_D + 90\% P_W + 20\%P_W$
  5. Only dead loads wind as a primary snow and wind as secondary: $100\% P_D + 90\% P_S + 20\% P_W$

The above percentages are for illustration purposes only (I'll try to find the correct combinations but it will take me a while).

NMech
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  • Hmm, I still am missing something. If we consider one year, we then have the probability of 2 % of exceeding the fundamental basic wind velocity. Is that not the probability overall? Or is the total probability in 50 years something different? Maybe I'm missing something about probability in general.. – S. Rotos Apr 17 '21 at 14:00
  • @S.Rotos I was adding the "correct" mathematical way to look into probability. Hopefully that would make more sense now – NMech Apr 17 '21 at 14:11
  • @SolarMike you are right on both accounts a) there are slighthly less than 100k (87600 to be exact) 10 min intervals in a year (I missed a zero and the 10min, thanks for noticing), and b) hospitals or other critical structures (eg dams) don't apply to the Eurocode rules. Also, another reason, that they didn't design to Eurocode standards in New Orleans is that US doesn't have to (Europe technically also doesn't have to because Eurocode is considered a standard and not legislation). – NMech Apr 17 '21 at 14:16
  • @NMech good job, removing my comments :) – Solar Mike Apr 17 '21 at 14:18
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    @S.Rotos "Maybe I'm missing something about probability in general." You are missing the fact that you can't treat one source of load as a "reason for failure" in isolation from everything else. – alephzero Apr 17 '21 at 15:51
  • @alephzero Consider a scenario where we are designing some structural element that is subject to wind load alone, for example a stiffening brace. This kind of elements exist, that are not meant to carry other types of load. We design that element to withstand a wind load calculated using the basic wind that has 2 % chance of being exceeded. This means that there is 2% chance that the brace experiences a load that is higher than what the brace can withstand. How does this not meant there is a 2% chance of the brace failing? What else is there to consider? – S. Rotos Apr 18 '21 at 13:37
  • I found some research that says the extreme value of the wind speed is Gumbel distributed. I see that these distributions go into quite low probabilities after 0.02.. maybe it is true that when we apply a safety factor, the failure load probability indeed does go down few orders of magnitude. I will ask a follow up question on this and return to this later. – S. Rotos Apr 18 '21 at 14:44
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Buildings are typically designed with a 50-year lifespan. So a 2% yearly chance of passing the limit makes intuitive sense.

But you're right: that'd basically mean we expect the structure to collapse sometime during those 50 years.

So that load with a 2% yearly chance of being surpassed makes sense... but only as a starting point.

The thing to remember is that you aren't getting that wind load, calculating its internal stresses and then checking to see if your members resist that load.

After all, there's a bunch of other safety factors involved in that calculation. If you're using LRFD design procedures, then there are two factors: one for the applied load and one for the structure's strength. The factors themselves depend on the code you're using.

In Brazil, the relevant code is usually NBR 8681, which gives a safety factor of 1.4 for wind loads (when they're the primary load in a combination). For the structural strength, the code depends on the material. For steel, it's 1.35.

So, using these coefficients, you start out with a wind load $W$ representing a 2% chance. But you then effectively design your structure for a load of $1.4 \cdot 1.35 W = 1.89W$, almost double the initial value.

And given that the distribution of wind speeds is roughly normal (well, more Weibull, but close enough) and that we start at the tail of the distribution (2%), getting a load twice as large as that throws us deep into the tail, with a much lower chance of happening, basically ever.

Wasabi
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