Question regarding the extreme wind velocity distribution in Eurocode - Engineering Stack Exchange most recent 30 from engineering.stackexchange.com 2022-01-20T17:42:48Z https://engineering.stackexchange.com/feeds/question/42666 https://creativecommons.org/licenses/by-sa/4.0/rdf https://engineering.stackexchange.com/q/42666 0 Question regarding the extreme wind velocity distribution in Eurocode S. Rotos https://engineering.stackexchange.com/users/13372 2021-04-18T15:07:36Z 2021-04-18T16:36:57Z <p><a href="https://i.stack.imgur.com/YKYw2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YKYw2.png" alt="enter image description here" /></a></p> <p>Above is the definition of the probability factor from Eurocode 1-4. It is used in context of calculating the extreme wind load on a building. Ignoring some other factors, it is used as follows:</p> <p><span class="math-container">$$v_b = v_{b,0} * c_{prob}$$</span></p> <p>where <span class="math-container">$v_{b,0}$</span> is the extreme wind velocity with less than 0.02 probability of being exceeded. The factor <span class="math-container">$c_{prob}$</span> is used to reduce the extreme wind velocity in cases when considering time periods shorter than the 50 year-return period (0.02 probability).</p> <p>I assume the formula comes from Gumbel distribution:</p> <p><span class="math-container">$$p = e^{-e^{-\frac{v-\mu}{\beta}}}$$</span></p> <p>which can be solved for <span class="math-container">$v$</span>:</p> <p><span class="math-container">$$v = \mu-\beta \ln(-\ln(p))$$</span></p> <p>where <span class="math-container">$p$</span> is the probability of encountering extreme wind <span class="math-container">$v$</span>.</p> <p>Taking the ratio <span class="math-container">$\frac{v}{v_{b,0}}$</span> and considering that <span class="math-container">$v_{b,0}$</span> is defined to have to probability of being exceeded of 0.02, we get:</p> <p><span class="math-container">$$\frac{v}{v_{b,0}}=\frac{\mu-\beta \ln(-\ln(1-p))}{\mu-\beta \ln(-\ln(0.98))}$$</span></p> <p>Defining <span class="math-container">$K = \frac{\beta}{\mu}$</span>, we get:</p> <p><span class="math-container">$$\frac{v}{v_{b,0}} = c_{prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))}$$</span></p> <p>Finally to my question: <strong>Where does the exponent <em>n</em> come from in the Eurocode formula?</strong> As I have derived to formula in the picture, there is no need for the exponent. Is my derivation flawed? Is there something I don't know?</p> <p>EDIT: Only reason I assumed Gumbel distribution was that I was able to get a similar formula using it. There seems to be nothing in Eurocodes that goes into detail as to what distribution is actually used.</p> https://engineering.stackexchange.com/questions/42666/-/42667#42667 0 Answer by NMech for Question regarding the extreme wind velocity distribution in Eurocode NMech https://engineering.stackexchange.com/users/19906 2021-04-18T16:36:57Z 2021-04-18T16:36:57Z <p>Your derivation seems absolutely reasonable, as does your assumption about the Gumbel Distribution, with one minor detail.</p> <p>IMHO, the reason why the exponent <span class="math-container">$n$</span> exists, is that the Eurocode does not really care about the velocity but about wind pressures (ultimately about loads on the structure). So the idea, is that you are trying to capture the extreme value of the wind pressure (though the wind speed).</p> <p>So all the analysis you carried out is not about <span class="math-container">$\frac{v}{v_{b,0}}$</span> but about wind pressures <span class="math-container">$\frac{q}{q_{b,0}}$</span>. So, instead its:</p> <p><span class="math-container">$$\frac{q}{q_{b,0}}= c_{\color{red}{q},prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))}$$</span></p> <p>However, because the wind pressure <span class="math-container">$q$</span> is proportional to the square of the wind velocity so:</p> <p><span class="math-container">$$\frac{q}{q_{b,0}} = \frac{v^2}{v_{b,0}^2}=\left(\frac{v}{v_{b,0}}\right)^2$$</span></p> <p><span class="math-container">$$\left(\frac{q}{q_{b,0}} \right)^{0.5} = \frac{v}{v_{b,0}}$$</span></p> <p><span class="math-container">$$\left(c_{q,prob} \right)^{0.5} = \frac{v}{v_{b,0}}$$</span> <span class="math-container">$$\left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5} = \frac{v}{v_{b,0}}$$</span></p> <p>However, because <span class="math-container">$c_{prob}$</span> is a factor that is applied to velocity:</p> <p><span class="math-container">$$\frac{v}{v_{b,0}} = \left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5}$$</span></p>