Standard Shear Walls - The Analysis of Irregular Shaped Structures - page 329

3

8 vh

EAb
vh
G t

where = bending deflection

= shear deflection

075 . he

n =

nail slip contribution

hd

e

= hold down slip

b

v = unit wall shear, plf
h = wall height, ft
b = wall width, ft
E = modulus of elasticity of studs/chords, psi
d a = hold-down slip from manufacturer, in
e n = nail slip per IBC Table 2305.2.2(1), in
G t = panel rigidity through thickness (pli) per IBC Table 2305.2.2(2)
A = area of boundary elements, in 2
As an alternate, the deflection can be calculated using the three-term equation provided
in Section 4.3.2 in the 2008 Special Design Provisions for Wind and Seismic (SDPWS-08).

h ∆

8 vh

EAb

3

vh

δ SW = + +

a

SDPWS-08 Eq. 4.3-1

1000 G b

a

where ∆ a = vertical elongation of wall anchorage system (including fastener slip, device
elongation, rod elongation, etc.).

Example 9.1: Single-Story Shear Wall Analysis
The one-story structure shown in Fig. 9.1 is 50 ft long by 24 ft wide with 12 ft high walls.
The roof consists of 4 : 12 double-pitched trusses spaced at 24″ o.c. A 4 ft shear wall with
a drag strut occurs at grid line 2, and a full-length shear wall occurs at grid line 1. Wind
loads control the design. The structure is located in the 90 mph zone, wind exposure C.
The uniformly distributed wind load to the roof diaphragm has been calculated to be
129 plf in the end-zone pressure area and 86.4 plf in the interior zone. The total factored
shear applied to the top of the 4′ and 24′ end walls is 2385 lb. The resulting sum of the
wall and tributary roof dead loads is equal to 135 plf. The dead load reaction from the
strut or header applied to the 4 ft wall is 150 lb. The analysis of the shear walls shown
in Figs. 9.1 and 9.2 is as follows:
4 ft shear wall, (see Fig. 9.2):

12
4

AR . . ==< 335 . OK if wood structural panels are used ( blocked )

V = 2385 lb diaphragmrea ctiontogrid line ncluding effectsof 2 end-

zone wind pressure.

2385
4

v SW = = 596 3 . plf