Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 148

At grid line 3,

= -

506 7 . - 446 7 .

15 7150 5

F 3 () =- . lb tension

D

2

At diaph. c.l.,

020
2

+

F CL =- 7150 5 . - ( ) 5 =- 7200 5 . lb tension

At grid line 4,

060
2

+

F 4 =- 7200 5 . + ( 15 ) =- 6750 lb tension

D

At grid line 5,

480 + 420

F 5 =- 6750 - () 15 =- 13 500 , lb tension

D

2

At grid line 6,

300 + 600

F 6 =- 13500 + () 30 = 0 lb tension

D

2

Therefore the diagram closes to zero.
Determination of transverse strut/chord forces (see Fig. 3.39):
Forces at grid line 1, starting at line A:

V SW = 12 000 , lb

1

12 000
16

,

v SW = = 750 plf

1

v diaph = 333 3 . plf

v net =- =+ 750 333 3 . 416 7 . plf

At the end of SW1,

F = 416 716 .( ) = 6667 2 . lb

Strutforce F = 333 320 .( ) = 6667 2 . lb tension

The force diagram closes.
Forces along grid line 2, starting at line A: At grid line 2 (above C),

F 2 = ( 111 1859 36 . + .)() = 7092 lb tension

C

At grid line 2 (above C),

F 2 = 506 714 .( ) = 7093 8 . lb tension

C