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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.

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