Question 6.7: A channel section UPN 220 is subjected to a bending moment M......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize:
Properties of the cross section: The centroid C is located on the axis of symmetry (the z axis) at a distance
\quad\quad\quad\quad c = 2.14 ~cm
from the back of the channel (Fig. 6-31)⁴.The y and z axes are principal centroidal axes with moments of inertia:
\quad\quad I_{y}=197\,{\mathrm{cm}}^{4}\quad I_{z}=2690\,\,{\mathrm{cm}}^{4}
Also, the coordinates of points A, B, D, and E are

Bending moments: The bending moments about the y and z axes (Fig. 6-31a) are
M_{y}=M\sin\ \theta=(2\operatorname{kN}\cdot\operatorname{m})(\sin\ 10^{\circ})=0.347 \operatorname{kN}\cdot\operatorname{m} \\  M_{z}=M\sin\ \theta=(2\operatorname{kN}\cdot\operatorname{m})(\cos\ 10^{\circ})=1.97 \operatorname{kN}\cdot\operatorname{m}
3. Analyze:
Bending stresses: Now calculate the stress at point A from Eq. (6-39):

\sigma_{A}\,=\,\frac{M_{y}z_{A}}{I_{y}}\, – \frac{M_{z}y_{A}}{I_{z}} \\ \quad = {\frac{(0.347\mathrm{~kN}\cdot\mathrm{m})(-0.0586\;\mathrm{m})}{1.97\times10^{-6}~\mathrm{m}^{4}}} – {\frac{(1.97\mathrm{~kN}\cdot\mathrm{m})(0.110\;\mathrm{m})}{2.69\times10^{-5}~\mathrm{m}^{4}}} \\ \quad = -10.32~ MPa – 8.06~ MPa = -18.38 ~MPa
By a similar calculation, the stress at point B is
\sigma_{B}\,=\,\frac{M_{y}z_{B}}{I_{y}}\,  –  \frac{M_{z}y_{B}}{I_{z}} \\ \quad = {\frac{(0.347\mathrm{~kN}\cdot\mathrm{m})(0.0214\;\mathrm{m})}{1.97\times10^{-6}~\mathrm{m}^{4}}}  –  {\frac{(1.97\mathrm{~kN}\cdot\mathrm{m})(-0.110\;\mathrm{m})}{2.69\times10^{-5}~\mathrm{m}^{4}}} \\ \quad = 3.77 ~MPa + 8.06 ~MPa = 11.83~ MPa
These stresses are the maximum compressive and tensile stresses in the beam.
The normal stresses at points D and E also can be computed using the procedure shown. Thus,
\quad\quad \sigma_{D}=-4.29\,\mathrm{MPa},\,\,\,\,\sigma_{E}=-2.27\,\mathrm{MPa}
The normal stresses acting on the cross section are shown in Fig. 6-31b.
Neutral axis: The angle β that locates the neutral axis [Eq. (6-41)] is found as
\tan\;\beta= \frac{y}{z} = \frac{I_{z}}{I_{y}}\;\tan\;\theta=\frac{2690\;\mathrm{cm}^{4}}{197\,\mathrm{cm}^{4}}\,\mathrm{tan}\;10^{\circ}=2.408\;\;\;\beta=67.4^{\circ}
The neutral axis nn is shown in Fig. 6-31, and note that points A and B are located at the farthest distances from the neutral axis, thus confirming that σ_A  ~ and ~ σ_B are the largest stresses in the beam.
4. Finalize: In this example, the angle β between the z axis and the neutral axis is much larger than the angle θ (Fig. 6-31) because the ratio I_z /I_y is large.
The angle β varies from 0 to 67.4° as the angle θ varies from 0 to 10°. As discussed previously in Example 6-5 of Section 6.4, beams with large I_z /I_y ratios are very sensitive to the direction of loading. Thus, beams of this kind should be provided with lateral support to prevent excessive lateral deflections.