Question 6.8: A Z-section is subjected to bending moment M = 3 kN·m at an ......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize:
Properties of the cross section: Use the results of Example D-7 in Section D.8 of Appendix D. Moments of inertia and orientation of principal centroidal axes are
\quad\quad\quad\quad  I_z = 32.6 \times 10^6 ~ mm^4 \,\,\, I_Y = 2.4 \times 10^6 ~ mm^4
\quad\quad\quad\quad \theta_{p1} = 19.2° \quad\quad\quad\quad \theta_{p1} = (19.2)\frac{\pi}{180}radians
Coordinates (y, z) of points A, B, D, D’, E, and E’ for the rotated principal coordinate axes (labeled y and z in Fig. 6-32) are
\quad\quad\quad\quad \theta=-20\left({\frac{\pi}{180}}\right)\mathrm{radians}
\quad\quad\quad\quad y_A = \frac{h}{2} \cos (θ_{p1}) + {\Bigg\lgroup} b – \frac{t}{2}{\Bigg\rgroup}\sin (θ_{p1}) \quad y_A = 121.569 ~~ mm \\ \quad\quad\quad\quad y_B = – y_A \quad\quad\quad\quad\quad\quad\quad y_B = -121.569~~ mm \\\quad\quad\quad\quad  y_D = \frac{h}{2} \cos (θ_{p1}) – \frac{t}{2} \sin (θ_{p1}) \quad\quad\quad\quad y_D = 91.971~~ mm \\ \quad\quad\quad\quad y_D {^\prime} = \frac{h}{2} \cos (θ_{p1})\quad\quad\quad\quad\quad\quad\quad y_D {^\prime} = 94.438~~mm \\ \quad\quad\quad\quad y_E {^\prime} = -y_D{^\prime} \quad\quad\quad\quad\quad\quad\quad y_E {^\prime} = -94.438~~mm \\ \quad\quad\quad\quad y_E = -y_D \quad\quad\quad\quad\quad\quad\quad y_E = 91.971~~mm \\ \quad\quad\quad\quad z_A = {\Bigg\lgroup} b – \frac{t}{2}{\Bigg\rgroup} \cos (θ_{p1}) – \frac{h}{2}\sin (θ_{p1}) \quad z_A = 45.024~~ mm \\ \quad\quad\quad\quad  z_B = -z_A \quad\quad\quad\quad\quad\quad\quad z_B = -45.024 ~~mm \\ \quad\quad\quad\quad z_D = \frac{-h}{2} \sin (θ_{p1}) – \frac{t}{2} \cos (θ_{p1}) \quad\quad\quad\quad z_D = -39.969~~ mm \\ \quad\quad\quad\quad z_D {^\prime} = \frac{-h}{2} \sin (θ_{p1})\quad\quad\quad\quad\quad\quad\quad z_D {^\prime} = -32.887~~mm \\ \quad\quad\quad\quad z_E{^\prime} = -z_D{^\prime} \quad\quad\quad\quad\quad\quad\quad z_E{^\prime} = 32.887 ~~mm \\ \quad\quad\quad\quad z_ E = -z_D \quad\quad\quad\quad\quad\quad\quad z_E = 39.969 ~~mm
Bending moments (kN·m) M = 3 kN·m:
\quad\quad\quad\quad M_{y}=M\ \sin(\theta)\qquad M_{y}=-1.026\ \mathrm{kN}\cdot\mathrm{m} \\ \quad\quad\quad\quad\quad\ M_{z}=M\ \cos(\theta)\qquad M_{z}= 2.819\ \mathrm{kN}\cdot\mathrm{m}
3. Analyze:
Bending stresses at A, B, D, and E (see plots of normal stresses in Fig. 6-32b):
\quad\quad\quad\quad\sigma_{_A}={\frac{M_{y}z_{A}}{I_{y}}}\,-\,{\frac{M_{z}y_{A}}{I_{z}}}\,=\,-\,19.249\,-\,10.51\,3\,=\,-29.8\,\mathrm{MPa} \\ \quad\quad\quad\quad\sigma_{_B}={\frac{M_{y}z_{B}}{I_{y}}}\,-\,{\frac{M_{z}y_{B}}{I_{z}}}\,=\,\,19.249\,+\,10.51\,3\,=\,29.8\,\mathrm{MPa} \\ \quad\quad\quad\quad\sigma_{_D}={\frac{M_{y}z_{D}}{I_{y}}}\,-\,{\frac{M_{z}y_{D}}{I_{z}}}\,=\,\,17.088\,-\,7.953\,=\,9.14\,\mathrm{MPa} \\ \quad\quad\quad\quad\sigma_{_D}{^\prime}={\frac{M_{y}z_{D}{^\prime}}{I_{y}}}\,-\,{\frac{M_{z}y_{D}{^\prime}}{I_{z}}}\,=\,\,14.06\,-\,8.167\,=\,5.89 \,\mathrm{MPa} = -\sigma_{_E}{^\prime} \\ \quad\quad\quad\quad \sigma_{_E}={\frac{M_{y}z_{E}}{I_{y}}}\,-\,{\frac{M_{z}y_{_E}}{I_{z}}}\,=\,\,-17.088\,+\,7.953\,=\,-9.14 \,\mathrm{MPa}
Location of neutral axis:
\begin{array}{c}{{\quad\quad\quad\quad \tan(\beta)=\frac{I_{z}}{I_{y}}\,\tan(\theta)}}\\ \quad\quad\quad\quad {{\beta=-78.6^{\circ}}}\end{array}
4. Finalize: An alternate approach is to use the generalized bending theory for nonprincipal centroidal axes in the cross section that are parallel to the sides of the cross section. The y and z axes in Fig. 6-33 are now aligned with the web and flanges. From Example D-7, the moments and product of inertia are I_z = 29.29 × 10^6 mm^4, I_y = 5.667 × 10^6 mm^4, and ~I_{yz} = 9.366 × 10^6 mm^4.
(Note that I_{yz} is positive for this orientation of y and z axes.)
The y and z coordinates of point A, for example, are now easily computed in this system of axes as y_A = h/2 = 100 ~mm, z_A = b – t/2 = 82.5 ~mm. Applied moment M is at angle θ + θ_{p1} = 39.2° to the z axis, so the components of moment M along y and z axes are
\quad\quad\quad\quad M_{y}=-(3\mathrm{~kN}\cdot\mathrm{m})\mathrm{sin}( θ + θ_{p1})=-1.896\mathrm{~kN}\cdot\mathrm{m} \\ \quad\quad\quad\quad M_{z} = (3\mathrm{~kN}\cdot\mathrm{m})\mathrm{cos}( θ + θ_{p1})=2.325\mathrm{~kN}\cdot\mathrm{m}
Entering the numerical values given here in the general flexure formula (Eq. 6-47) gives the same compressive stress value as that reported previously:

\quad\quad\quad\quad \sigma_{_A}\,=\,{\frac{(M_{y}I_{z}\,+\,M_{z}I_{y z})z_{_A}\,-\,(M_{z}I_{y}\,+\,M_{y}I_{y z})y_{A}}{I_{y}I_{z}-I_{y z}^{2}}}=-29.8\ \mathrm{MPa}
Use of y and z coordinates of other points of interest on the cross section will give the normal stresses at B, D, and E listed in step 3.