Question 16.CA.3: Knowing that column AB (Fig. 16.26) has an effective length ...
Knowing that column AB (Fig. 16.26) has an effective length of 14 ft and 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 × 10³ psi, and the adjusted allowable stress for compression parallel to the grain is \sigma_C = 1060 psi.
Note that c = 0.90 for glued laminated wood columns.
Computing the value of \sigma_{CE}, using Eq. (10.35), gives
\sigma_{CE}=\frac{0.822 E}{(L / d)^2}=\frac{0.822\left(800 \times 10^3 psi\right)}{(168 in. / d)^2}=23.299 d^2 psiEquation (16.34) is used 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,
\begin{aligned} 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}}\end{aligned}Because the column must carry 32 kips, Eq. (16.33) gives
\sigma_{\text {all }}=\sigma_C C_P (16.33)
\sigma_{\text {all }}=\frac{32 \text{ kips}}{d^2}=\sigma_C C_P=1.060 C_PSolving this equation for C_P and substituting the value into the previous equation, we obtain
\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 in