Question 6.6: A wood beam AB of rectangular cross section serving as a roo......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize:
Loads and bending moments: The uniform load q acting in the vertical direction can be resolved into components in the y and z directions (Fig. 6-23a):
\quad\quad q_{y}=q\ \cos\ \alpha\ \ \ q_{z}=q\ \sin\alpha \quad\quad\quad (6 – 25 a,b)
The maximum bending moments occur at the midpoint of the beam and are found from the general formula M = qL^2 /8; hence,
M_{y}\,=\,\frac{q_{z}L^{2}}{8}\,=\,\frac{q L^{2}\;\;\sin\;\alpha}{8}\qquad M_{z}\,=\,\frac{q_{y}L^{2}}{8}\,=\,\frac{q L^{2}\;\;\cos\;\alpha}{8}\quad\quad\quad (6 – 26 a,b)
Both of these moments are positive because their vectors are in the positive directions of the y and z axes (Fig. 6-23b).
Moments of inertia: The moments of inertia of the cross-sectional area with respect to the y and z axes are
\quad\quad\quad I_{y}={\frac{h b^{3}}{12}}\quad I_{z}={\frac{b h^{3}}{12}}\quad \quad\quad (6-27 a,b)
Bending stresses: The stresses at the mid-section of the beam are obtained from Eq. (6-19) with the bending moments given by Eqs. (6-26) and the moments of inertia given by Eqs. (6-27):

The stress at any point in the cross section can be obtained from this equation by substituting the coordinates y and z of the point.
From the orientation of the cross section and the directions of the loads and bending moments (Fig. 6-23), it is apparent that the maximum compressive stress occurs at point D (where  y = h/2  and  z = -b/2) and the maximum tensile stress occurs at point E (where  y = -h/2   and   z = b/2). Substitute these coordinates into Eq. (6-28) and then simplify to obtain expressions for the maximum and minimum stresses in the beam:
\sigma_{E}=-\sigma_{D}\,=\,\frac{3q L^{2}}{4b h}\!\left(\frac{\sin\,\alpha}{b}\,+\,\frac{\cos\,\alpha}{h}\right)\quad\quad\quad (6-29)
3. Analyze:
Numerical values: The maximum tensile and compressive stresses are calculated from the preceding equation by substituting the given data:
\quad q = 3.0 ~kN/m ~~ L = 1.6 ~m ~~ b = 100 ~mm ~~ h = 150 ~mm ~~ \alpha = 26.57°
The results are
\quad\quad \sigma_{E}=-\sigma_{D}=4.01\;\mathrm{MPa}
Neutral axis: In addition to finding the stresses in the beam, it is often useful to locate the neutral axis. The equation of this line is obtained by setting the stress [Eq. (6-28)] equal to zero:
\quad\quad {\frac{\sin~\alpha}{b^{2}}}z-\,{\frac{\cos~\alpha}{h^{2}}}\mathcal{Y} = 0\quad\quad\quad\quad (6-30)
The neutral axis is shown in Fig. 6-23b as line nn. The angle β from the z axis to the neutral axis is obtained from Eq. (6-30) as
\quad\quad \tan~\beta =\frac{\mathcal{Y}}{z} =\frac{h^{2}}{b^{2}}~\tan~\alpha\quad\quad\quad (6-31)
Substituting numerical values gives
\tan~\beta=\frac{h^{2}}{b^{2}}~\tan~\alpha=\frac{(150~\mathrm{mm})^{2}}{(100~\mathrm{mm})^{2}}(\tan~26.57^{\circ})=1.125~~~\beta=48.4^{\circ}
Since the angle β is not equal to the angle α, the neutral axis is inclined to the plane of loading (which is vertical).
4. Finalize: From the orientation of the neutral axis (Fig. 6-23b), note that points D and E are the farthest from the neutral axis, thus confirming the assumption that the maximum stresses occur at those points. The part of the beam above and to the right of the neutral axis is in compression, and the part to the left and below the neutral axis is in tension.