Open Front and Cantilever Diaphragms - The Analysis of Irregular Shaped Structures - page 226

lb shearonleft side of grid line2

VRw
L

=- ′ = () 20 1500 - 200 20 () =- 2500

2 1

= -

2500
60

v =- . 41 67 lf diaphragmunitshear lefttsideofgrid line2

2 L

∑

M 1 = 0 summing momentsabout grid line 1

200 30 15
20

( ()

RV
2

== = 4500 lb shearatgridline2

2

VRV
R

=- = 4500 - 2500 = 2000 lb shearonright sideeofgrid line2

2 2 2 L

2000
60

v = =+ . 33 33 lf diaphragmunitshear righttsideofgrid line2

2 R

lb shearatgridline3

VV w
R

=- ′ = () 10 2000 - 200 20 () = 0

3 2

Construction of the collector force diagram at grid line 1 (see Fig. 6.14):
Net shear at shear walls at grid line 1:

R
L

1500
34

∑ 1

44 11 .

v = = =+ plf

SW

SW

v net = 44 125 911 . - =+ . plfshear wall shearminus diaphragmshear

Starting at grid line A:

F = 19 11 18 . () = 344 11 . lb tensionatend of SW 1

lb tensionatstart of SW2

F = 344 11 . - 25 10 ( ) =+ 94 11 .

lbtension at endofSW2 F = 94 11 . + 19 11 12 . () = 323 44 .

lb compression at startof SW3

F = 323 44 . - 25 16 ( ) =- 76 56 .

lb at grid lineD

F =- 76 56 . + 19 11 40 . () =

Construction of the collector force diagram at grid line 2 (see Fig. 6.14):
Net shear at shear wall SW4:

R
L

4500
12

v =

2

= =+ 375 plf

SW

SW 4

vV v
net

= - =- - 375 41 67 . 33 33 . =+ 300 plf

SWdiaph 4

Starting at grid line A:

1650 lb compression at startofSW4

F =- (. 41 67 + 33 33 22 . () = -

F =- 1650 + 300 12 () = 1950 lb tensionatend of SW 4

lb at grid lineD F = 1950 - 75 26 () = 0

Therefore the diagram closes. ▲