IEC60909 Fault Analysis

You are right. In 4.5.2.4 it is stated:

For unbalanced short-circuit currents, the flux decay in the generator is not taken into account, and equations (66) to (68) apply.

That means the factor μ for near-to-generator and q for near to induction motor it may be calculated only for symmetrical three phase short-circuits. For all other unbalanced short circuits [as per 2.3.2 Application of symmetrical components]

it is not applicable.

So the effect of offset current is considered negligible?

Or perhaps the method to calculate the thermal effect has enough built in unknowns that the introduction of another calculated variable will not cause the result to be exceed the range of the probable calculation result, if excluding the offset data.

What this standard says here is to neglect the DECAY of the generator flux. That means to take it entirely untouched [q=1,μ=1].μ and q are always less than unit.

That's not what OP is stating, so their original assumption is wrong? You apparently are stating that the asymmetrical portion of the fault current is used in calculating the energy applied to the switchboard?

If we take Ik"=20 kA and R/X=0 then the Ipmax=2*sqrt(2)*Ik" =56.57 kA[Ip cannot be more]

Actually, if we take Ip at 1 cycle after the three-phase fault start[1/50=0.02 sec] for 0.02 sec

q=0.84+0.26*e^(-0.26*IkG"/IrG)

Let's say IkG"=20 kA for IrG=1kA q=0.84+0.26*EXP(-0.26*20)=0.8415 Ip=0.8415*2*sqrt(2)*20=47.6 kA

but Ib=0.8415*20=16.83 kA rms

Now, if we neglect the generator flux decay q=1 and Ip=56.57 kA and Ib=20 kA

Other than symmetrical breaking current rating, switchboard also has rating for making current. Making current takes care of asymmetrical current which is 2.6 to 2.8 times. So while selecting the switchboards take care of both symmetrical breaking and asymmetrical making current. For example while selecting the switchboards breaker, take care for breaker symmetrical breaking as well as making current.