Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 140

From A to B,

F 1 =- 115 520 .( ) =- 2310 lb compression

B

From B to C,

F 1 = 115 520 .( ) = 2310 lb compression opposite ( direction)

B

The force diagram closes.
The strut force along grid line 2 is

From B to C,

98 25 . + 190 5 .

F 2 =- () 20 =- 5775 lb compression

C

2

The reaction of diaphragm 2 = 3600 lb; therefore the total shear to grid line 2 is

R 2 = 5775 + 3600 = 9375 lb

See the alternate check in the figure.

9375
20

v = v = = 187 5 . plf

SW 1 SW 2

+ 30

v =- = v v 187 560 . - = 127 5 . plfatshear walls

net SW iaph

At the end of SW1,

F =- 5775 + 127 520 .( ) = - 3225 lbcompression

At the start of SW2,

F =- 3225 - 60 10 () = - 3825 lb compression

At grid line D,

F D =- 3825 + 127 5300 .( ) = lb

Therefore the diagram closes.
Looking back at the direction of the transferred shears in Figs. 3.31 and 3.32, it can be
seen that for the transverse members, the chord at grid line A is in compression because
that section of the diaphragm was assumed to span from grid line 1 to 3. The transfer
diaphragm was designed to act as a propped cantilever beam. The top chord at grid line B
is in tension, and the bottom chord at grid line C is in compression. For the longitudinal
direction, all the struts are in compression. All the forces applied to the collectors, chords,
and struts act as expected and have confirmed the analysis. ▲

A problem frequently arises when sloped roofs are involved is that shown in Fig. 3.34.
The different depths of the diaphragm cause a vertical and horizontal offset at grid line 2B.
Transfer of the discontinuous chord forces at this area can be very complicated because of
the horizontal and vertical offset occurring at the same location. The diaphragm framing