Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 121

Applying the remaining shear on the sheathing element shows that the transfer
diaphragm shear between grid lines B and C is positive.

700 15
50

()

V A = = 210 lb

V
D

210
20

A
TD

v = = = 10 5 . plf

A

The direction of the reaction is also acting to the left. Applying the remaining shears
on the sheathing element shows that the transfer diaphragm shear between grid lines
A and B is negative.
Determination of diaphragm net shears (see Fig. 3.21):

Location (Grid Lines) Net Shears

A from 1 to 2 -28 plf

A from 3 to 4 -40.9 plf

B from 1 to 2 +28 plf

A from 2 to 3 v = -40.9-10.5 = -51.4 plf

B from 2 to 3 (above line) v = +15.1-10.5 = +4.6 plf

B from 2 to 3 (below line) v = +15.1 + 24.5 = +39.6 plf

C from 2 to 3 v = +45.1 + 24.5 =+69.6 plf

C from 3 to 4 +45.1 plf

Determination of net shears at shear walls (see Fig. 3.21): Always observe the direction of
the shear transferred into the boundary members.

wL
L

200 50
210

()
()

v = = =+ 500 plf here L = 10 ft

SWA SW A

2

SWA

v diaph =-51 4 . plffrompreviouscalculation

v = v - v =- =+ 500 51 4 . 448 6 . plf

net SWA diaph

200 50
215

()
()

v = = + 333 3 . plf here L = 15 f t

SWC SWC

v diaph =-45 1 . plf

v net = 333 3451 . - . =+ 288 3 . plf

An enlarged plan of the shears applied between grid lines B and C from 2 to 3 is
shown at the lower right of the figure.
Determination of longitudinal and transverse strut/chord forces (see Fig. 3.22): Starting with
the longitudinal strut along grid line A from 1 to 2, the strut force at line 2 is equal to

F = -28(25) = -700 lb compression, which is the same as at grid line 2B

The net shear in the transfer diaphragm between grid line 2 and the start of SW2 is
-51.4 plf. Continuing 5 ft into the transfer diaphragm to the start of shear wall SW2, the
strut force becomes

F = -700-51.4(5) = -957 lb compression