Diaphragms With End Horizontal Offsets - The Analysis of Irregular Shaped Structures - page 126

Loads from A to B:

wD
D

200 25
125

()

From 12 to =

12
14

-

= = 40 plf

-

wD
D

200 100
125

( )

From 24 to =

24
14

-

= = 160 plf

-

Loads from B to C:

From 2 to 4 =200 plf

Determination of shear wall shears: The force is distributed to each shear wall in accordance
with its length.

L SW

++ =++=

8815 31 ft

1 SW 3 SW 4

LR
L

V

8 5000
31

( ) 1290 3

8

.

SW 1 BC ,

SW
SW

1
1

V = = = 1290 3 . lb v = = = + 161 3 . plf

∑

SW 1 SW 1

L

SW

5000
10

V = 5000 lb v = =+ 500 plf

SW 2 SW 2

VV = = 1290 3 . lb v = v = 161 3 . plf

SW 3 SW 1 SW 3 SW 1

15 5000
31

( ) 2419 4

15

.

V = = 2419 4 . lb v = = + 161 .. 3plf

SW 4 SW 4

Construction of the basic shear diagram (see Fig. 3.24):
Diaphragm section between grid lines A and B from 1 to 2:

RR wL
AB

40 35
2

()

== = = 700 lb

2

R
D

700
25

v =

B

= = 28 plf, R = - 28 plf

diaph A

12

Diaphragm section between grid lines A and C from 2 to 4:

200 15 75 ( (.) + 160 35 ( )( 42 5 .)

= -

4090
100

R = = 4090 lb v =- 40 9 . plf

A

50

dia ph

160 35 17 5 ( (. ) + 200 15 ( )( 42 5 .) 4510

100

R = = 4510 lb v = = + 45 1 . plf

C

50

di aph

1510
100

V = 4510 - 200 15 () = 1510 lb v = =+ 15 1 .plf

B diaph

Determination of the force transferred into the transfer diaphragm (see Fig. 3.24): The force
transferred into the transfer diaphragm can be determined by summing the shears along
grid line B from 1 to 2. The net shear along the length of shear wall SW1 is equal to the
shear wall shear minus the diaphragm shear.

v net =+ 161.3 -=+ 28 133.29 plfatshear wall