A laminated wooden beam made of three planks of 100mm X 50mm glued together to form the solid section of 100mm X 150mm. The allowable shear stress in the glue joint is 0.35N/mm2. If the beam is 2m long and simply supported at the ends find the safe point load the beam can carry at midspan.

Let the safe point load at mid span i.e \frac{2}{2}=1m be W (N)

Now, moment of inertia of section about neutral axis

$I=\frac{100*(50+50+50)^3}{12}=2.8125*10^7 mm^4$

Maximum shear force F=\frac{W}{2}

Now, shear stress at the glue joint=$q_{1-1}=\frac{F a bar y}{Ib}$

where $y_{top}=\frac{50*100*\frac{50}{2}+50*100*(50+\frac{50}{2})+50*100*(50+50+\frac{50}{2})}{50*100+50*100+50*100}$

$\bar y_{top}=75mm$

$q_{1-1}=0.35 N/mm^2$ at glue joint (given)

And b=100mm and $Q=50*100 m^2$ is area above (1)-(1)

$\bar y=75-\frac{50}{2}=50mm$

$q_{1-1}=0.35=F*(\frac{50*100*50}{2.8125*10^7*100}$

F=3937.5N

i.e W=3937.5*2

W=7875 N

The safe point load the beam can carry at mid span is 7875 N