Question 6.5: A 4-m long cantilever beam (Fig. 6-21a) is constructed from ......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Maximum bending stresses when the load is aligned with the y axis.
1, 2. Conceptualize, Categorize: If the beam and load are in perfect alignment, the z axis is the neutral axis, and the maximum stresses in the beam (at the support) are obtained from the flexure formula:
\quad\quad\quad \sigma_{\mathrm{max}}\,=\,{\frac{M y}{I_{z}}}\,=\,{\frac{P L(h/2)}{I_{z}}}
in which M_z = -M = -PL ~and  ~M_y = 0 ~ so  ~ M = PL ~ is the bending moment at the support, h is the height of the beam, and I_z is the moment of inertia about the z axis.
3. Analyze: Substituting numerical values gives
\quad\quad \sigma_{\mathrm{max}}={\frac{(45\ \mathrm{kN})(4000\ m m)(250\ \mathrm{mm})}{68740\ \mathrm{cm}^{4}}}=65.5\ \mathrm{MPa}
4. Finalize: This stress is tensile at the top of the beam and compressive at the bottom of the beam.
Part (b): Maximum bending stresses when the load is inclined to the y axis.
1, 2. Conceptualize, Categorize: Now assume that the beam has a small inclination (Fig. 6-21b), so that the angle between the y axis and the load is α = 1° .
The components of the load P are P cos α in the negative y direction and P sin α in the positive z direction.
3. Analyze: The bending moments at the support are

The angle β giving the orientation of the neutral axis nn (Fig. 6-21b) is obtained from Eq. (6-21):

\tan\beta=\frac{y}{z}=\frac{M_{y}I_{z}}{M_{z}I_{y}}=\frac{(-3.14\;\mathrm{kN}\cdot\mathrm{m})(68740\;\mathrm{cm}^{4})}{(-180\;\mathrm{kN}\cdot\mathrm{m})(2480\;\mathrm{cm}^{4})}=0.878 \quad \beta = 25.8°
This calculation shows that the neutral axis is inclined at an angle of 25.8° from the z axis even though the plane of the load is inclined only 1° from the y axis. The sensitivity of the position of the neutral axis to the angle of the load is a consequence of the large I_z /I_y ratio.
From the position of the neutral axis (Fig. 6-21b), note that the maximum stresses in the beam occur at points A and B, which are located at the farthest distances from the neutral axis. The coordinates of point A are
\quad\quad z_{A}=-92.5\,\mathrm{mm}\ \ \ y_{A}=250\,\mathrm{mm}
Therefore, the tensile stress at point A [see Eq. (6-19)] is

\sigma_{x}\,=\,\frac{M_{y} {z}{}}{I_{y}}\,-\,\frac{M_{z}{y}{}}{I_{z}} \quad \quad (6-19)\\
\sigma_{A}\,=\,\frac{M_{y} {z}_{\!A}}{I_{y}}\,-\,\frac{M_{z}{y}_{A}}{I_{_z}}
=\;\frac{(-3.14\;\mathrm{kN}\cdot\mathrm{m})(-92.5\;\mathrm{mm})}{2480\;\mathrm{cm}^{4}}\;-\;\frac{(-180\;\mathrm{kN}\cdot\mathrm{m})(250\;\mathrm{mm})}{68740\;\mathrm{cm}^{4}}
\quad\quad = 11.7 ~MPa + 65.5 ~MPa  =  77.2 ~MPa
The stress at B has the same magnitude but is a compressive stress:
\quad\quad \sigma_{B} = – 77.2 ~MPa
4. Finalize: These stresses are 18% larger than the stress σ_{max} = 65.5 ~ MPa for the same beam with a perfectly aligned load. Furthermore, the inclined load produces a lateral deflection in the z direction, whereas the perfectly aligned load does not.
This example shows that beams with I_z much larger than I_y may develop large stresses if the beam or its loads deviate even a small amount from their planned alignment. Therefore, such beams should be used with caution, because they are highly susceptible to overstress and to lateral (that is, sideways) bending and buckling. The remedy is to provide adequate lateral support for the beam, thereby preventing sideways bending. For instance, wood floor joists in buildings are supported laterally by installing bridging or blocking between the joists.