Wood Column Stability Factor Example #1

Fig. 1: Sawn lumber columns loaded from the overhead level

Column Stability Factor Explained:

The column stability factor is to ensure that weak-axis buckling or torsional buckling does not occur over non-laterally supported lengths. A full breakdown of the process can be found here.

Example #1:

Find the allowable compressive strength of a 2x6 Douglas Fir, which spans 10'-0" in the strong axis and is braced every 2'-6" in the weak axis. Assume you are using construction grade "Douglas Fir-Larch"

Assumptions:

• All of your Adjustment Factors are unity except for your Duration Factor Which will be set at two months (C_D = 1.15).
• Assume a pinned-pinned connection (K_e = 1.0)

Step #1:

Verify the wood is capable of spanning the necessary distance.

Note: Since the column is supported intermittently in two directions, both directions must be checked concurrently to see what is the governing case.

Weak Axis
l_e / d = 30" / 1.5" = 20 > 50 <okay>

Strong Axis
l_e / d = 120" / 5.5" 21.8 > 50 <okay>

Note: Since K_e = 1.0, l_e = l

Step #2:

Start by solving for the critical buckeling design value, F_cE :

Weak Axis
F_{cE} = {{0.822(555,000)}\over {\left(20\right)^2}} = 1130 psi

Strong Axis
F_{cE} = {{0.822(555,000)}\over {\left(21.8\right)^2}} = 950 psi

Note: E'_min = E_min * C_M * C_t * C_i * C_T


where:

Step #3:

Solve for F_c*:
F_c* = F_c * C_D * C_M * C_t * C_F* C_i = 1650 psi *1.15*1.0*1.0*1.0*1.0 = 1898 psi

where:

Step #4:

Solve for F_cE / F_c*

Weak Axis
F_cE / F_c* = 1130 / 1898 psi = 0.595

Strong Axis
F_cE / F_c* = 950 / 1898 psi = 0.501

Step #5:

Solve for:

Weak Axis
C_{P} = {{1+0.595}\over {2*0.8}} - \sqrt{\left[{{1+(0.595)}\over {2*0.8}}\right]^2-{{0.595}\over 0.8}} = 0.497

Strong Axis
C_{P} = {{1+(0.501)}\over {2*0.8}} - \sqrt{\left[{{1+(0.501)}\over {2*0.8}}\right]^2-{{0.501}\over 0.8}} = 0.434

Step #6:

Multiply C_P by F_c* to obtain your allowable compression force (F_c')

Weak Axis
F_c' = 1898 psi * 0.497 = 943 psi

Strong Axis
F_c' = 1898 psi * 0.434 = 823 psi (Governs)

Note: In this example, strong axis bending would govern over weak axis.

References:

1. American Forest and Paper Association, "National Design Specification for Wood Construction", 2005
   • http://www.awc.org/pdf/WDF14-4-2005NDSarticle.pdf