Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 136

Main diaphragm section from C to D, uniform load:

120 60
2

() 3600

60

== = 3600 lb = = 60 plf, = - 60 plf

RR v v

2 3 2 3

Determination of the diaphragm net shears (see Fig. 3.31):

Location (Grid Lines) Net Shears

1 from A to B +115.58 plf

3 from A to B -79.5 plf

1 from B to C v = +115.5-(231) = -115.5 plf

2 from B to C (left) v = +40.5-(231) = -190.5 plf

2 from B to C (right) v = +40.5 + (57.75) = +98.25 plf

3 from B to C v = -79.5 + (57.75) = -21.75 plf

Determination of transverse collector/chord forces (see Fig. 3.32): The chord force along grid
line A from 1 to 2, starting at grid line 1, is

( vvx + ) ( 115 5405 15

2

. + .)()

F =

1 2

= = 1170 lb compression

2

2

The diaphragm shear at grid lines 2 and 3 changes from +40.5 plf to -79.5 plf, respectively.
The forces along that interval can be broken down into positive and negative forces by
calculating the areas of the similar triangles shown in the figure. The maximum positive
force in the chord is equal to F 2 plus the positive triangular area.

F max = 1170 + 410 = 1580 lb compression

This force is equal to the force of the negative triangular area. Therefore the force
diagram closes.
The transfer diaphragm chord force along grid line B, starting at grid line 1, is equal to
From 1 to 2,

( 115 5 . + 115 5 05 190 5

2

.) (. + + .) ()

F 2 = 15 = 3465 lb tension

B

From 2 to 3,

(. 98 25 - 40 5 95 21 75

2

.) (. + - .

F 2 = () 60 = 3465 lb tension

B

Therefore the diagram closes.
The transfer diaphragm chord force along grid line C, starting at grid line 1, is
equal to

From 1 to 2,

115 5 . + 190 5 .

F 2 = () 15 = 2295 lb compression

B

2