Question 10.CA.3: Knowing that column AB (Fig. 10.32) has an effective length ......
Knowing that column AB (Fig. 10.32) 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_{C E} , using Eq. (10.51), gives
\sigma_{C E}=\frac{0.822 E}{(L / d)^2} (10.51)
=\frac{0.822\left(800 \times 10^3 psi \right)}{(168 in . / d)^2}=23.299 d^2 psi
Equation (10.50) 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
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}} (10.50)
=\frac{1+21.98 \times 10^{-3} d^2}{2(0.90)}-\sqrt{\left[\frac{\left.1+21.98 \times 10^{-3} d^2\right]^2}{2(0.90)}-\frac{21.98 \times 10^{-3} d^2}{0.90}\right.}
Since the column must carry 32 kips, Eq. (10.49) gives
\sigma_{\text {all }}=\sigma_C C_P (10.49)
\sigma_{\text {all }}=\frac{32 kips }{d^2}=\sigma_C C_P=1.060 C_P
Solving 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.