To balance out the wall, the tension force resisted by the hold-down must be paired with a compression force at the opposite edge. Proper selection of the compression member involves two checks: one for the strength of the member and one for where that member meets the remainder of the construction.

 Simple Columns When determining the allowable load on a simple column it is important to know its slenderness ratio - its length relative to its least width dimension (L/D2).  The lower the (L/D2) ratio, the stronger the column.

 Braced Columns In shear walls we can expect that the sheathing material will be attached to one or both faces of the end post/column.  This will generally act to brace the member against buckling in the weak (D2) direction.  Thus the largest L/D ratio may indeed be (L1/D1). Where panel sheathing is not used, calculate the L/D ratio based on the spacing of the intermediate blocking, i.e. L2/D2.

Beginning with the 1997 Uniform Building Code, the design of wood columns and allowable wood stresses are referred to the ANSI/NFoPA NDS-1991 National Design Specification for Wood Construction, published by the American Forest & Paper Association. Here is how it works:

The NDS formulas seek to establish a Column Stability Factor, (Cp), for a column of a given dimension.  This is a fancy way of arriving at a value between 0 and 1 by which the maximum axial compression for a given grade of wood is multiplied in order to create an allowable column load.

1. When a column is supported throughout its length to prevent lateral displacement in all directions, Cp = 1.0.

2. Calculate the maximum L/D ratio for the column using the diagram above.  Use (L1/D1) or (L2/D2) whichever is larger.
Caution: The slenderness ratio may not exceed 50 except that during construction it may not exceed 75.

3. The column stability factor shall be calculated as follows:

A = (1/2c)*(1 + Fce/Fc)

B = (1/c)*(Fce/Fc)

Cp = A - SQRT( A*A - B)

in which...

Fc = tabulated compression design value multiplied by all applicable adjustment factors except Cp (see below).

Fce = (Kce* E)/SQR(L/D Ratio)

Kce = 0.3 for visually graded lumber and machine evaluated lumber (MEL)

c = 0.8 for sawn lumber

c = 0.9 for glued laminated timber

Fc can be found by referencing the NDS Supplement. Wood values based on grade are listed for various species along with adjustment factors based on the size of member under consideration. The table for Douglas-Fir Larch is presented below for your reference.

 TABLE 4A - BASE DESIGN VALUES FOR VISUALLY GRADED DIMENSION LUMBER DOUGLAS-FIR LARCH Design values in pounds per square inch (psi) Grading Rules Agency Species and Commercial Grade Size Classification Bending Fb Tension parallel to grain Ft Shear parallel to grain Fv Compression perp. to grain Fc-perp. Compression parallel to grain Fc Modulus of Elasticity E Select Structural 2" to 4" thick 2" & wider 1,450 1,000 95 625 1,700 1,900,000 WCLIB WWPA No.1 & Better 1,150 775 95 625 1,500 1,800,000 No.1 1,000 675 95 625 1,450 1,700,000 No. 2 875 575 95 625 1,300 1,600,000 No. 3 500 325 95 625 750 1,400,000 Stud 675 450 95 625 825 1,400,000 Construction 2" to 4" thick 2" to 4" wide 1,000 650 95 625 1,600 1,500,000 Standard 550 375 95 625 1,350 1,400,000 Utility 275 175 95 625 875 1,300,000

Tabulated bending, tension, and compression parallel to grain design values for dimension lumber 2" to 4" thick shall be multiplied by the following size factors:
 SIZE FACTOR, CF Fb Ft Fc Grades Width Thickness 2" & 3" 4" Select Structural, No. 1 & Better, No. 1, No. 2, No. 3 2", 3" & 4" 1.5 1.5 1.5 1.15 5" 1.4 1.4 1.4 1.1 6" 1.3 1.3 1.3 1.1 8" 1.2 1.3 1.2 1.05 10" 1.1 1.2 1.1 1.0 12" 1.0 1.1 1.0 1.0 14" & wider 0.9 1.0 0.9 0.9 Stud 2", 3" & 4" 1.1 1.1 1.1 1.05 5" & 6" 1.0 1.0 1.0 1.0 Construction & Standard 2", 3" & 4" 1.0 1.0 1.0 1.0 Utility 4" 1.0 1.0 1.0 1.0 2" & 3" 0.4 n/a 0.4 0.6

And now an example for a 2 x 4, No. 1 D.F., 8'-0" tall, with sheathing bracing the member continuously against buckling in the weak direction...

L/D = 96" / 3.5" = 27.43 (< 50, ok)

Fc = 1.15 * 1450 = 1667 psi

Fce = (0.30* 1,700,000)/SQR(27.43) = 677.82

A = (1/1.6)*(1 + Fce/Fc) = 0.879

B = (1/0.8)*(Fce/Fc) = 0.508

Cp = A - SQRT( A*A - B) = 0.364

Allowable unit load = (Cp) * (Fc) = 606 psi (< Fc-perp. = 625 psi, ok)

Total Load not to exceed (1.5) * (3.5) * (606) = 3,181 pounds.

We've covered the basic forces, overturning, hold-downs and compression posts used in shear walls, but there are some additional constraints to be considered before we can wrap up the design portion of this presentation.