Question 10.03: Knowing that column AB (Fig. 10.31) has an effective length ...
Knowing that column AB (Fig. 10.31) has an effective length of 14 \mathrm{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 \times 10^{3} \mathrm{psi}, and the adjusted allowable stress for compression parallel to the grain is \sigma_{C}=1060 psi.
We note that c=0.90 for glued laminated wood columns. We must compute the value of \sigma_{C E}. Using Eq. (10.49) we write
\sigma_{C E}=\frac{0.822 E}{(L / d)^{2}}=\frac{0.822\left(800 \times 10^{3} \mathrm{psi}\right)}{(168 \mathrm{in} . / d)^{2}}=23.299 d^{2} \mathrm{psi}
We then use Eq. (10.48) to express the column stability factor in terms of d, with \left(\sigma_{C E} / \sigma_{C}\right)=\left(23.299 d^{2} / 1.060 \times 10^{3}\right)=21.98 \times 10^{-3} d^{2},
C_{P} =\frac{1+\left(\sigma_{C E} / \sigma_{C}\right)}{2 c}-\sqrt{\left[\frac{1+\left(\sigma_{C E} / \sigma_{C}\right)}{2 c}\right]^{2}-\frac{\sigma_{C E} / \sigma_{C}}{c}} \\ =\frac{1+21.98 \times 10^{-3} d^{2}}{2(0.90)}-\sqrt{\left[\frac{1+21.98 \times 10^{-3} d^{2}}{2(0.90)}\right]^{2}-\frac{21.98 \times 10^{-3} d^{2}}{0.90}}
Since the column must carry 32 \mathrm{kips}, which is equal to \sigma_{C} d^{2}, we use Eq. (10.47) to write
\sigma_{\mathrm{all}}=\frac{32 \mathrm{kips}}{d^{2}}=\sigma_{C} C_{P}=1.060 C_{P}
Solving this equation for \mathrm{C}_{P} and substituting the value obtained into the previous equation, we write
\frac{30.19}{d^{2}}=\frac{1+21.98 \times 10^{-3} d^{2}}{2(0.90)}-\sqrt{\left[\frac{1+21.98 \times 10^{-3} d^{2}}{2(0.90)}\right]^{2}-\frac{21.98 \times 10^{-3} d^{2}}{0.90}}Solving for d by trial and error yields d=6.45 \mathrm{in}.