Question 16.3: Knowing that column AB (Fig. 16.25) has an effective length ...

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.