Question 16.3: Knowing that column AB (Fig. 16.25) has an effective length ...
Knowing that column AB (Fig. 16.25) has an effective length of 14 ft, and that it must safely carry a 32-kip load, design the column using a square glued laminated cross section. The adjusted modulus of elasticity for the wood is E = 800 × 10^{³} psi, and the adjusted allowable stress for compression parallel to the grain is σ_{C} = 1060 psi.
We note that c = 0.90 for glued laminated wood columns. We must compute the value of σ_{CE}. Using Eq. (16.33) we write
σ_{CE} =\frac{0.822E}{(L/d)^{2}} = \frac{0.822(800 × 10^{3} psi)}{(168 in./d)^{2}} = 23.299d^{2} psi
We then use Eq. (16.32) to express the column stability factor in terms of d, with (σ_{CE} /σ_{C}) = (23.299d^{2}/1.060 × 10^{3}) = 21.98 × 10^{-3} d^{2},
C_{P} = \frac{1 + (σ_{CE} /σ_{C})}{2c} – \sqrt{\left[\frac{1 + (σ_{CE} /σ_{C})}{2c}\right]^{2} – \frac{σ_{CE} /σ_{C}}{c}}
= \frac{1 + 21.98 × 10^{-3} d^{2}}{2(0.90)} – \sqrt{\left[\frac{1 + 21.98 × 10^{-3} d^{2}}{2(0.90)}\right]^{2} – \frac{21.98 × 10^{-3} d^{2}}{0.90} }
Since the column must carry 32 kips, which is equal to σ_{C}d^{2}, we use Eq.(16.31) to write
σ_{all} = \frac{32 kips}{d^{2}} = σ_{C}C_{P} = 1.060C_{P}
Solving this equation for C_{P} and substituting the value obtained into the previous equation, we write
\frac{30.19}{d^{2}} = \frac{1 + 21.98 × 10^{-3} d^{2}}{2(0.90)} – \sqrt{ \left[ \frac{1 + 21.98 × 10^{-3} d^{2}}{2(0.90)} \right]^{2} – \frac{21.98 × 10^{-3} d^{2}}{0.90} }
Solving for d by trial and error yields d = 6.45 in.