Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 123

the shear value of 39.6 plf on the opposite side of the collector. The force in this collector
is equal to

F = (39.6-4.6)20 = 700 lb compression

This is equal and opposite to the direction of the strut between lines 1 and 2, and therefore
it closes the force diagram. The direction of the net transfer forces into the strut and
collector indicates that the members are in compression.
The strut at grid line C has a uniform shear of 69.6 plf from grid line 2 to 3 and a shear
of 45.1 plf from grid line 3 to the start of shear wall SW1. The shear transferred into the
boundary members along that line is in the negative direction and shows that the strut
is in compression. The force in the strut at line 3 is

F = -69.6(20) = -1392 lb compression

The force in the strut at the start of shear wall SW1 is

F = -1392-45.1(80-15) = -4323.5 lb compression

The net shear at the shear wall is equal to +288.3 plf, acting in the opposite direction.
Therefore the force at the end of the wall at line 4 is

F = -4323.5 + 288.3(15) =0 lb
Therefore the diagram closes.
Next, examine the transverse collectors at grid lines 2 and 3. The collectors at these
locations also serve as the chords for the transfer diaphragm. The resulting net shears and
their direction transferred into the transverse collectors at lines 2 and 3 can be referenced
in the figure. Observe that the transfer shears applied at 2A are 51.4 plf on the right of
the collector and 28 plf on the left. The shears act in opposite directions, which requires
them to be subtracted from each other. It can be seen from the diagram that the net
applied transfer shear is a uniform 23.4 plf compression from grid line A to B. Therefore
the force in the collector at grid line 2B is

F 2 = 23 435 .( ) = 819 lb compression

AB

On the opposite side of line B, the collector force is

39 6696
2

. + .

15 819

F 2 = () = lb compression acting i nopposite direction

BC

At grid line 3, the net applied transfer shear above line B is a uniform 10.5 plf, and
below line B the shears are a uniform 24.5 plf.

F = 10 535 .( ) = 368 lb tension

3 AB

F = 24 515 .( ) = 368 l bbtension acting in opposite derection

3 BC

All the force diagrams close and receive tension and compression forces in
accordance with the anticipated diaphragm movement. Therefore the analysis has
been verified. ▲