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Contents | 1. Introduction to structural design | 2. Loads |

Introduction to wood | Material properties | Sectional properties | Design approaches | Construction systems | Tension elements |

The reduction in allowable compressive stress, *F _{c}* , to account for buckling is accomplished by multiplying

"Idealized *C*_{P}" = (safety factor)/*F*_{c}

(3.9)

In practice, given the pattern of column failure represented in Figure 1.56, the Euler equation must
be modified to account for crushing and non-elastic behavior, especially at low slenderness ratios.
The column stability factor, *C _{P}*, does just that and more, replacing σ

(3.10)

In Equation 3.10, *A* = [1 + (*F _{cE}* /

A full description of *C _{P}* can be found in Appendix Table A-3.4, along with other adjustments to the allowable compressive stress.

For non-pin-ended columns, the unbraced length, *l _{e}*, is multiplied by an effective length coefficient (Appendix Table A-1.2) to account for the change in critical buckling stress resulting from more
or less restraint at the column ends.

** Problem definition.** Check the capacity (allowable load) of a 10 × 10 Douglas Fir-Larch Select Structural column 8.5 ft high, used indoors, supporting live load (

*L* = 40 kips; *L _{r}* = 20 kips;

** Solution overview.** Find relevant material properties and adjustment factors; compute adjusted allowable
stress; find capacity by multiplying cross-sectional area by adjusted allowable stress; compare
capacity to governing load combination.

*Problem solution*

**1.** From Tables A-3.3 and A-3.9, find material properties *F _{c}* and

**2.** Find adjustment factors for *F _{c}*:

a. From Appendix Table A-3.4 Part *B*, *C _{M}* = 1.0.

b. From Appendix Table A-3.4 Part *A*, *C _{F}* = 1.0.

c. Find load duration factor, *C _{D}*, and the governing load combination: Two load combinations from Appendix Table A-2.7 (for allowable stress design) should be considered:

(*D* + *L*)/*C _{D}* = (50 + 40)/1.0 = 90.0.

Then, looking at *D* + 0.75*L* + 0.75(*L _{r}* or

*D* + 0.75*L* + 0.75(*S*)/*C _{D}* = (50 + 30 + 15)/1.15 = 82.61

or

*D* + 0.75*L* + 0.75(*L _{r}*)/

Dead plus live load (*D* + *LD*) governs, so *C _{D}* = 1.00, and the load used to design (or analyze) the column is (

d. From Appendix Table A-3.4, find the column stability factor, *C _{P}* (to account for buckling):

From Appendix Table A-3.9, find *E*'* _{min}* =

*E _{min}* = 580,000 psi;

*l _{e}* = 8.5 ft = 102 in.

d = 9.5 in.

*F _{cE}* = 0.822

*F _{c}** =

*c* = 0.8 for sawn lumber.

*A* = [1 + (*F _{cE}* /

*B* = (*F _{cE}* /

*C _{P}* =

**3.** Compute adjusted allowable stress in compression: from step 2, *F _{c}** = 1150 psi and

**4.** Find capacity, *P* = *F _{c}*' ×

**5.** Check capacity: since the capacity of 96.7 kips ≥ governing load combination of 90 kips, the
column is OK.

The value of *C _{P}* = 0.934 indicates that buckling has reduced the column's allowable compressive stress to 93.4% of its "crushing" strength.

** Problem definition.** Find the lightest cross section
for a wood column (Douglas Fir-Larch Select
Structural) that is 8.5 ft high, used indoors, on
the second floor of the 3-story building shown
in Figure 3.17, supporting live load (

Figure 3.17: Framing plan and building section for Example 3.4

*L* = 40 psf; *L _{r}* = 20 psf;

** Solution overview.** Find relevant material properties and adjustment factors (assuming a provisional
value for

*Problem solution*

**1.** Using Appendix Tables A-3.3 and A-3.9, find material properties *F _{c}* and

**2.** Find adjustment factors for *F _{c}*, except for

a. From Appendix Table A-3.4 Part *B*, *C _{M}* = 1.0.

b. From Appendix Table A-3.4 Part *A*, *C _{F}* = 1.0 (assuming that "dimension lumber" will not be used).

c. From Appendix Tables A-3.10 and A-2.7, *C _{D}* depends on which load combination proves to
be critical. To find

(*D* + *L*)/*C _{D}* = [25(600) + 27.2(300)]/1.0 = 23,160 lb

Then, looking at *D* + 0.75*L* + 0.75(*L _{r}* or

(D + .75*L* +.75*S*)/*C _{D}* = [25(600) +.75(27.2)(300) +.75(30)(300)]/1.15 = 24,235 lb

or

(*D* + .75*L* +.75*L _{r}*)/

The first case of the second load combination governs (using dead, live, and snow load), so *C _{D}* = 1.15, and the load used to design the column is (

25(600) + 0.75(27.2)(300) + 0.75(30)(300) = 27,870 lb.

**3.** Select cross section by trial. The stability factor, *C _{P}*, cannot be determined directly, since it depends upon the cross-sectional dimensions of the column which haven't yet been found. Design therefore turns into an iterative process, repeatedly making and testing assumptions about
the column's stability until the tests (i.e., column analyses) confirm the assumptions. To begin
the iterative process:

a. Assume a value for *C _{P}*, for example,

b. Compute *F _{c}** =

c. Compute *F _{c}*' =

d. Compute the provisional required cross-sectional area, *A _{req}*:

*A _{req}* = axial load / stress = 27,870/1058 = 26.3 in

*Trial 1:*

**1.** From Appendix Table A-3.12, select trial cross section based on provisional required area of
26.3 in^{2}: A 6 × 6 has an area of 30.25 in^{2}, but since the provisional required area of 26.3 in^{2} was based on an assumption about the column's stability (*C _{P}* = 0.8), it is not immediately clear whether the choice is correct: what we must enter into at this point is the first step of an iterative process. We start by checking the 6 × 6 for its actual capacity and comparing this capacity to the applied load. This process is identical to the timber column analysis method illustrated in Example 3.3.

**2.** From Appendix Table A-3.4, find the actual column stability factor, *C _{P}*, for the 6 × 6 column:

From Appendix Table A-3.9, find *E*'* _{min}* = the adjusted minimum modulus of elasticity =

*l _{e}* = 8.5 ft = 102 in.

*d* = 5.5 in.

*F _{cE}* = 0.822

*F _{c}** =

*c* = 0.8 for sawn lumber.

*A* = [1 + (*F _{cE}* /

*B* = (*F _{cE}* /

*C _{P}* =

**3.** Compute the adjusted allowable stress in compression:

*F _{c}*' =

**4.** Find capacity, *P* = *F _{c}*' ×

**5.** Check capacity: the capacity of 28,405 lb is greater than the actual load of 27,870 lb. In other
words, analysis shows that the 6 × 6 column is acceptable. If the capacity of a 6 × 6 column were insufficient, we would try the next largest size, i.e., a 6 × 8; and then an 8 × 8, etc. until a cross section was found with adequate capacity. In this case, however, even though the 6 × 6 is acceptable, it is possible that a smaller column size will also work, for two reasons: first, the next
smaller size (a 4 × 6) falls under the dimension lumber size classification, which has a higher
allowable compressive stress than what was assumed for posts and timbers. Second, allowable stresses for dimension lumber generally increase as the cross-sectional area gets smaller, due to the size factor adjustment. For these reasons, we now check a 4 × 6 column.

*Trial 2:*

**1.** From Appendix Table A-3.12, a 4 × 6 has an area of 19.25 in^{2}.

**2.** From Appendix Table A-3.4, find the actual column stability factor, *C _{P}*, for the 4 × 6 column:

From Appendix Table A-3.9, find *E*'* _{min}* = the adjusted minimum modulus of elasticity =

*l _{e}* = 8.5 ft = 102 in.

*d* = 3.5 in.;

*F _{cE}* = 0.822

*F _{c}** =

*c* = 0.8 for sawn lumber.

*A* = [1 + (*F _{cE}* /

*B* = (*F _{cE}* /

*C _{P}* =

**3.** Compute the adjusted allowable stress in compression:

*F _{c}*' =

**4.** Find capacity, P = *F _{c}*' ×

*A* = 19.25 in^{2}; then, *P* = 615.9(19.25) = 11,856 lb.

**5.** Check capacity: the capacity of 11,856 lb is less than the actual load of 27,870 lb. Therefore, the 4 × 6 column is *not* OK: select the 6 × 6 column from Trial 1.

© 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.