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Shearwalls calculates the total diaphragm design force Fpx on each floor in each direction from equations 12.10-1,-2, and -3 in 12.10.1.1, and determines the transfer forces from discontinuous shear lines on the levels above, factored by the overstrength factor Ω0 that are to be added to this force.
Shearwalls does not model or design diaphragms, but in the Seismic Information table it outputs both Fpx and the total force of Fpx plus the transfer forces, to aid you in your own diaphragm design calculations. The diaphragm design force is also used to determine the shear line force used for drag strut calculations and in-plane anchorage.
Note that this diaphragm force is a minimum force according to 12.10.1.1; if a force derived from structural analysis using the Equivalent Lateral Force procedure in 12.8 is greater than that derived from the diaphragm force, it should be used for diaphragm design and is used by Shearwalls for drag strut calculations.
The loads and forces derived from the diaphragm design force can be viewed in Plan View in place of the shear wall design loads and forces derived from ASCE 7 12.8 by checking a Show/Hide menu item when in the Loads and Forces action. A Show/Hide menu item in Elevation view also allows you to view the forces derived from the Diaphragm Design force to be used for drag strut and anchorage design.
Redundancy Factor for Transfer Forces
Note that the he redundancy factor r is not applied to transfer forces from discontinuous upper shear lines as per items 5 and 6 in the list of conditions in 12.4.3.1 for which ρ = 1, i.e. the design of elements to which Ω0 is applied. This is in seeming contradiction with the Commentary C12.10.1.1 which says that the redundancy factor applies to diaphragm transfer forces, but in a communication ASCE clarified that the Commentary applied to typically smaller transfer forces due to changes in lateral stiffness rather than those due to Irregularity Type 4 Offsets to which Ω0 is applied.