Complex Diaphragms With Combined Openings and Offsets - The Analysis of Irregular Shaped Structures - page 320

Force diagrams at grid line 2, starting at grid line A:

(. 93 84 −+ 84 )( . 51 84 − 42 ) ()

=− 10 =− 98 4 . ompression

F A lb c

. 2

2

( 42 − 38 72 . )( . + 87 28 − 84 ) ()

F B =− 98 4 . + 30 = 0 lb

2

Therefore the diagram closes.
Force diagrams at grid line 3, starting at grid line A: The shear on the sheathing
elements from grid line A +10′ to B, at 3 and 5, changes from negative to positive values.
The forces along grid lines 3 and 5 due to this change are calculated by similar triangles
and are shown at the bottom left of the figure.

(. 93 84 −+ 21 )( . 51 84 + 21 ) ()

F A = 10 = 728 4 . lb te nsion

. 2

2

F max = 728 4 . + 178 48 . = 906 88 . lb tension

The shears are additive because of the direction of the shear.

F B = 906 88 . − 906 88 . = 0 lb from similar-triangle forces calculated

Therefore the diagram closes.
Force diagrams at grid line 5, starting at grid line A:

(. 91 88 −+ 21 )( . 49 88 + 21 ) ()

F A = 10 = 708 8 . lb co mpression

. 2

2

F max = 728 4 . + 184 52 . = 893 3 . lb compression

The shears are additive because of the direction of the shear.

F B = 893 3 . − 893 30 . = lb from similar-triangle f orcescalculated

Therefore the diagram closes.
Force diagrams at grid line 4, starting at grid line B:

(. 91 88 −+− 84 )( 84 76 8 .) ()

F D =− 40 =− 286 8 . lb com pression

2

49 68 . + 768 .

F E =− 286 8 . + () 10 = 0 lb

2

Therefore the diagram closes.
Force diagrams at grid line 5 from B to E, starting at grid line B:

( 105 − 91 17 . )( . + 76 863 − ) ()

F D =− 40 =− 552 lb comp ression

2

( 63 − 768 . )( + 105 − 49 68 .

F E =− 552 + () 10 = 0 lb

2