A beam cross-section with centroidal second moments of area $I_{xx} = 4.0 \times 10^{-5}\text{ m}^4$, $I_{yy} = 1.0 \times 10^{-5}\text{ m}^4$, and product moment of area $I_{xy} = 0$ is subjected to a pure bending moment $M$ inclined at an angle $\theta = 45^\circ$ to the $x$-axis ($M_x = M \cos 45^\circ$, $M_y = M \sin 45^\circ$). The inclination angle $\alpha$ of the neutral axis to the $x$-axis is given by $\tan \alpha = \frac{I_{xx}}{I_{yy}} \tan \theta$. The value of $\alpha$ (in degrees) is closest to:

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### 1. Strategy & Theoretical Principles In unsymmetric bending when coordinate axes are principal ($I_{xy} = 0$): The direct bending stress at any point $(x, y)$ is: $$\sigma_z = \frac{M_x}{I_{xx}} y + \frac{M_y}{I_{yy}} x$$ Along the neutral axis, the bending stress vanishes ($\sigma_z = 0$): $$\frac{M_x}{I_{xx}} y + \frac{M_y}{I_{yy}} x = 0 \implies y = -\left(\frac{M_y I_{xx}}{M_x I_{yy}}\right) x$$ The inclination angle $\alpha$ of the neutral axis satisfies: $$\tan \alpha = \frac{I_{xx}}{I_{yy}} \tan \theta$$ where $\theta$ is the angle of the applied moment vector to the $x$-axis. ### 2. Analytical Formulation & Calculation Given: - $I_{xx} = 4.0 \times 10^{-5}\text{ m}^4$ - $I_{yy} = 1.0 \times 10^{-5}\text{ m}^4 \implies \frac{I_{xx}}{I_{yy}} = 4.0$ - $\theta = 45^\circ \implies \tan(45^\circ) = 1.0$ $$\tan \alpha = 4.0 \times 1.0 = 4.0$$ $$\alpha = \arctan(4.0) \approx 75.96^\circ \approx 76.0^\circ$$ ### 3. Distractor Forensics - Candidate assuming neutral axis is perpendicular to moment vector gets $\alpha = 45^\circ + 90^\circ = 135^\circ$ (valid only if $I_{xx} = I_{yy}$). ### 4. Exam Trap & Standard Reference - **Trap Alert:** In asymmetric bending, the neutral axis is generally **not** perpendicular to the applied bending moment vector. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 16.

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