# How is any kind of structural reliability achieved if the probability of annual wind speed exceedance is 0.02? 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%.

• " 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? Apr 17 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? Apr 17 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 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. Apr 17 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? Apr 17 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. Apr 17 at 15:19

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.

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).

• 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.. Apr 17 at 14:00
• @S.Rotos I was adding the "correct" mathematical way to look into probability. Hopefully that would make more sense now Apr 17 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). Apr 17 at 14:16
• @NMech good job, removing my comments :) Apr 17 at 14:18
• @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. Apr 17 at 15:51