2
$\begingroup$

In Eurocode, two equations for loads are given for STR limit state:

$$1.15K_{FI}G_{kj,sup}+0.9G_{kj,inf}+1.5K_{FI}Q_{k,1}+1.5K_{FI}\sum_{i>1}\psi_{0,i}Q_{k,i}$$

$$1.35K_{FI}G_{kj,sup}+0.9G_{kj,inf}$$

Designer is required to calculate using both, and the design value of the load comes from the equation that gives the least favourable result. But what is the actual difference of these equations, why do we need two? I can see that the bottom equation (6.10a) only takes into account permanent (dead) loads (letter $G$) and the one above considers temporary (live) loads as well (letter $Q$).

Why don't we need to consider the live loads in one of the equations? My lecture material says that the lower equation is rarely the deciding one, but it must always be checked. I've read my lecture materials and the Eurocode section on this, but I can't really get any good explanation of this. If somebody could explain the difference between the two equations and why we need them both I would be very grateful!

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ which version of EN1990 you got the equations from? Could you explain the kj and sup, inf subscripts in your post? $\endgroup$ – NMech Oct 1 '20 at 12:03
1
$\begingroup$

I am not sure which version of Eurocode you have this (I have access to an older version circa 2002), and 6.10a and b have a slightly different form. In any case I will answer and I will later update my answer.

Equation A: Dominant secondary actions and combination of transients loads

First of all: equation a is not only one equation. It represents all possible combination of temporary actions on a structure (could be wind, snow, fire, earthquake etc).

Since Qs are temporary loads, it is highly unlikely that all of them will occur together. So, what the Eurocode suggests, is to assume that any point in time only one of the actions will be dominant, while all the other are multiplied by a reduction factor (see $\psi_{0,i}$). So if you got 4 temporary actions, you need to perform 4 times that calculation.

The Dead weights remain constant. You need to notice that, for $G_{kj, sup}$ the coefficient is slightly less in eq. a (set to 1.15) compared to equation B.

Equation B: Dominant dead loads

Equation B represents only dead loads, but it has a slightly higher coefficient for $G_{kj, sup}$. In this scenario, the dominant load is the dead load, and no temporary loads are considered.

**Update:**I found a reference online regarding the finnish SFS-EN. What happens is that $G_{sup}$ is the upper design value, and $G_{inf}$ is the lower design value, which can be used if for example you have a building with maximum occupant capacity or completely empty. The two values create a type of average (leaning towards the upper limit, in order to make the calculation more direct.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is my local Eurocode variant, SFS-EN 1990. The sup subscript indicates the highest statistical value of something, while inf indicates the lowest possible value (at least this is how I understand it). $\endgroup$ – S. Rotos Oct 1 '20 at 14:43
  • $\begingroup$ Regarding the combination of transient loads: if we have 4 of them, do I need to choose each as the dominant one once, with every combination of minor load? For example if I had wind, snow and load from people, would I choose wind as the dominant one, multiply snow and people load by $\psi_{0,i}$, then choose snow as the dominant one and multiply people load and wind by the reduction factor, and finally choose people load as the dominant and reduce wind and snow? $\endgroup$ – S. Rotos Oct 1 '20 at 15:23
  • $\begingroup$ Exactly that is the procedure. Mind you it gets more interesting with temporary actions because you might have to consider different directions. So you might have to check the wind from the north, or assume that all the snow somehow is stored on one half of the floor etc... It can get messy. $\endgroup$ – NMech Oct 1 '20 at 17:11
  • $\begingroup$ Okay, thank you for your help! I understand it better now. $\endgroup$ – S. Rotos Oct 5 '20 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.