Question 1.7: A compressive force of 3 kN acts on the inclined wooden comp...
A compressive force of 3 kN acts on the inclined wooden component, which is joined with the other parts as shown in Figure 1.23. Find out the normal stresses on the areas of contact defined by MN and NR and also the shear stress (average) along horizontal plane defined by PNR. All dimensions are in mm.
Let us draw the free-body diagram of the inclined component [Figure 1.24(a)].
From Figure 1.24(a), it is clear that
\begin{aligned} F_{ MN } &=3 \times \sin \theta \\ &=3 \times \frac{3}{5} \quad\left\lgroup \text { as } \tan \theta=\frac{3}{4} \right\rgroup \\ &=\frac{9}{5} kN =1800 N \end{aligned}
Similarly,
F_{ NR }=3 \times \cos \theta=3 \times \frac{4}{5} kN =2400 N
As F_{ MN } is the horizontal force, it will cause a shearing force at the interface, defined by KPN [Figure 1.24 (b)]. From consideration of equilibrium, this shearing force is 1800 N.
The area of the interface concerned = (75 × 40) mm² = 3000 mm². Therefore, average shear stress = 1800/3000 = 0.6 N/mm² = 0.6 MPa.
