How many load combinations do I have

$n$ = the number of ‘dead’ loadgroups (for example: self weight, dead loads)

$k$ = the number of ‘life’ loadgroups (for example: life load A, wind, snow, temperature)

The total number of load combinations in ULS FC according to EN 1990 [–] equ. 6.10 will be:

$2^n \cdot (k \cdot 2^(^k^-^1^) + 1)$

Example:

$n$ = 3, $k$ = 5, total number of load combinations ULS FC according to EN 1990 [–] is:

$2^n \cdot (k \cdot 2^(^k^-^1^) + 1)=2^3 \cdot (5 \cdot 2^(^5^-^1^) + 1)=648$

Remarks:

This formula assumes all load groups have a $\Psi_0$ factor different from 0.
This formula does not included the effect of:
- Linked load groups
  Linking 2 dead loads, will divide the number of load combinations by 2. Linking 3 dead loads, will divide the number of load combinations by 2*2, and so on.
- Incompatible load groups
- Sub load cases
- Crane loads
  Each position of the crane load will be translated into a sub load case behind the scene.
For the EN 1990 [NL] the total number must be multiplied by 2. This national annex uses equations 6.10a and 6.10b, instead of equation 6.10.