A Simple Shear Wall

A Wall To Resist Seismic Loading
Wall Information:

Vroof = 2600 pounds.  The weight of wall is 15 psf.  A roof contributes 150 plf to the top
of this wall. The seismic coefficient is 0.183 and we are located in Zone 4.  Height = 12 feet.
The wall is framed from 2x6 Douglas Fir No. 1 except as required.

The Problem:

We want to design a wall with a minimum width to resist the applied force.

Let's start with the easy stuff...

We are going to need compression posts and hold-downs. One at a time...

Compression posts

Now calculate a rough hold-down (tiedown) force by subtracting the resisting moment from the overturning moment and dividing by an approximate distance to the hold-down (generally 6" from opposite end of wall).  Add this load to the vertical loads to calculate a rough compression...

Now calculate the actual hold-down force...

Adjacent Stud Check
The distance "a" (see previous diagrams) indicates the distance to the centroid of compression forces.  If "a" is large, we would check the stud adjacent to the compression post to see if it was overstressed in compression.  In this example, the distance (3*a) is 20.5 inches.  The first adjacent stud is very close to the end of the compression triangle. The total compression is 7450 pounds.  One third of that is 2483 pounds...well under our allowable for such a stud, ok.

Shear and Sheathing
Our unit shear in this wall is 466.4 lbs./ft.  As stated previously, the 1997 UBC will require 3x members at all panel edges. Since the wall is 12'-0" tall, we will also require 3x solid blocking at the 8'-0" elevation. In addition to the end posts, one 3 x 6 stud will be needed in the middle of the wall section.

The thickness of sheathing was not specified in the beginning, so we are left to select that as well as the nailing. Looking at Table 23-II-I-1 on page 2-288 of the 1997 UBC, we see our choices are:

Sill Bolting
If the bottom plate of this shear wall sits directly on top of a foundation stem wall, there is an additional step: bolting the sill down for lateral loads. We want to use bolts that will have at least four diameters of edge distance on the bottom plate so these same bolts can resist perpendicular forces. This means we should use a bolt diameter less than (5.5" / 8) or 5/8".  The Code generally calls for 1/2" bolts so that is what we will use.

Looking at Table 23-III-B-1 on page 2-296, we see that 1/2" diameter bolts do not have a listed value in 2 1/2" thick material.  The value in 1 1/2" material, parallel to grain, is 480 pounds.  The value for 3 1/2" material is 610.  The Code allows interpolation in these instances, therefore the allowable shear is (480 + 610)*(0.5) = 545 pounds.  Because the forces are the result of earthquake loads, we can increase this value by a factor of 4/3 = 726 pounds per bolt.

6 feet * 466.4 = 2798.4 lbs.

(2798.4) / (726) = 3.85 bolts > Use four 1/2" diameter bolts.  Space them at 16" centers near the interior of the wall to leave room for the hold-down bolts at the ends.