The wooden members A and B are to be joined by plywood splice plates that will be fully gl...

There are four parts of glue sharing the \(24 \mathrm{kN}\) as 2 separate parts.

Given average shearing stress in the glue is limited to \(\tau_{\text {ghee }}=800 \mathrm{kPa}\)

Half of the \(24 \mathrm{kN}\) force should be shared by each area part for shearing.

Calculate the cross-sectional area. Let \(A\) be the one of the four areas.

$$ \begin{array}{l} \tau_{\text {ghe }}=\frac{F}{A} \\ 800 \mathrm{kPa}=\frac{12 \mathrm{kN}}{A} \\ A=\frac{12}{800} \\ =0.015 \mathrm{~m}^{2} \mid \frac{10^{6} \mathrm{~mm}^{2}}{1 \mathrm{~m}^{2}} \\ =15000 \mathrm{~mm}^{2} \end{array} $$

Calculate the length of the one glued area. Consider from figure.

$$ \begin{aligned} A &=l \times b \\ l &=\frac{A}{b} \\ &=\frac{15000}{100} \\ &=150 \mathrm{~mm} \end{aligned} $$

Calculate the total length of the glued area. Consider from the figure.

$$ \begin{aligned} L &=2 \times l+8 \\ L &=2 \times 150+8 \\ &=300+8 \\ &=308 \mathrm{~mm} \end{aligned} $$

Therefore, the allowable length of glue is \(308 \mathrm{~mm}\).