I think you dont understand the problem!
I give you a output example:

A snubber at node 350 is skewed -27,3 degrees about globel Z.
The dyn.case has following results:

Restraint Forces (global system)

Node FX FY FZ

350 1096 0 566
527 272
1 X/3M 1 X/3M

With this forces the snubber load can be calculated
Fres.= SQRT ( FX*FX + FZ*FZ).
The force at the snubber is Fres.= 1223,5.

Now look in the output local element forces:

Node Fa Fb Fc

350 1247 310 510 Element 340 to 350
840 73 241
2 X(2) 2 X(2) 5 Y(6)

350 1233 831 476 Element 350 to 360
829 221 178
2 X(2) 7 X(2) 2 X(2)

When i assume the sum of Fb is the snubber load then i get 310+831 = 1141 !
This value is not Fres. = 1223,5.
Resume:The local forces can not be a base for calculation restraint forces in dynamic case or dyn+static combination case. Look the restraint forces have the biggest part from
1 X/3M (missing mass)and the local forces come from 2 X(2) and 7 X(2) !!
The calculation for the local restraint load at the snubber with the globel force is
very simple. But if you have restraint in the space with 3 skewed axis then it is not fast and simple to calculate from hand.

In the staic load case the local forces are correct for looking the restraint load !
I think in dynamic load case not !!