Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 119

grid lines 2 and 3. The intent of placing a limited number of shear walls along grid
lines A and C is to simplify the example and demonstrate the method of calculating the
drag strut forces, not to suggest that this is a reasonable layout. The transfer diaphragm
and the location of its members from the previous example have been maintained. The
diaphragm is broken down into two sections, from grid lines 1 to 2 and from lines 2
to 4. The uniform load of 200 plf is distributed to these sections in proportion to their
individual depths.
Load distribution within the diaphragm (see Fig. 3.20):

Loadsbetween AandBfrom 12 200 25
12

()
5

= = 40 plf

Loadsbetween AandBfrom 24 200 100
1

( )

= = 160 plf

25

Loads between B and C from 2 to 4 = 200 plf

Construction of the basic shear diaphragm (see Fig. 3.21):
Diaphragm 2 between grid lines A and B from 1 to 2:

RR wL

40 35
2

() B

D

700
25

== = = 700 lb v == =

B

28 plf,

B =-28

plf

AB diaph

2

Place the positive and negative sheathing elements and the shears that are transferred
into the boundary elements on the plan.
Diaphragm 1 between grid lines A and C from 2 to 4:

200 15 75 ( (.) + 160 35 32 5

50

( )( .)

R A = = 4090 lb

== -

R
D

4090
100

A

v =- 40 9 . plf

A

160 35 17 5 ( (. ) + 200 15 ( )( 42 5 .)

R C = = 4510 lb

50

R
D

4510
100

v == =+

C

45 1 . plf

C

V B = 4500 - 200 15 () = 1510 lb

V
D

1510
100

v == =+

B

15 1 . plf

B

Place the positive and negative sheathing elements and the shears that are transferred
into the boundary elements on the plan.
The reaction of diaphragm 2 is transferred into the discontinuous strut at grid
line B, which is then distributed into the transfer diaphragm. The disrupted strut
force is in compression, as indicated by the direction of the shears that are transferred
into the strut.