Complex Diaphragms With Combined Openings and Offsets - The Analysis of Irregular Shaped Structures - page 293

Force at grid line 3 plus 12.5′:

224 2704
2

. + .

F 3125

′ =

2043 − (. ) 12 5 = 201 75 . lb compression

+ .

The shear in the right section changes from a +70.4 plf to a −83.33 plf. The forces of
the positive and negative shear areas are shown above the section.

F 5 =+ 201 75 . − 201 5 . + 282 = 282 5 . lb compression , equals previous calculation

(. 96 44 − 85 16 . )( + 151 99 . − 140 71 . ) ()

F 6 =− 282 25 = 0 lb

2

Therefore the diagram closes.
Force diagrams at grid line B:

F 2 =− 5250 lb chordtension force

(. 82 2 + 207 8 .) ( + 132 2 . + 157 8 .) ()

F 3 =− 5250 + 20 =+ 5 50 lb compression

2

56 04 . + 75 3 .

F 4 =+ 550 − () 10 =+ 106 7 . bcompression

2

100 22 . + 123 33 .

F 5 =+ 106 7 . − () 15 =− 1783 3 . btension

2

(. 96 44 − 25 11 . )( + 151 99 . − 80 66 .

F 6 =− 1783 3 . + () = 0lb 25

2

Therefore the diagram closes.
Force diagrams at grid line D:

207 8 . + 157 8 .

F 3 =− () 20 =− 3656 lb tension

2

56 04 . + 75 3 .

F 4 =− 3656 + () 10 =− 3000 lb tension

2

( 100 22 . + 99 78 . )( + 123 33 . + 76 67 .

F 5 =− 3000 + ()= 0lb 15

2

Therefore the diagram closes.
Force diagrams at grid line E:

99 78 . + 76 67 .

F 5 =− () 15 =− 1323 4 . btension

2

25 11 . + 80 66 .

F 6 =− 1323 4 . + () 25 = 0 lb

2

Therefore the diagram closes.