At a point on an aircraft aluminum fuselage skin under plane stress, the stress state is $\sigma_{xx} = 80\text{ MPa}$, $\sigma_{yy} = -20\text{ MPa}$, and $\tau_{xy} = 40\text{ MPa}$. The maximum in-plane shear stress $\tau_{\max}$ (in $\text{MPa}$) at this point is closest to:

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### 1. Strategy & Theoretical Principles From Mohr's circle of plane stress, the radius of the circle equals the maximum in-plane shear stress $\tau_{\max}$: $$\tau_{\max} = R = \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}$$ ### 2. Analytical Formulation & Calculation Given: - $\sigma_{xx} = 80\text{ MPa}$ - $\sigma_{yy} = -20\text{ MPa}$ - $\tau_{xy} = 40\text{ MPa}$ 1. Center of Mohr's circle: $$\sigma_{\text{avg}} = \frac{80 + (-20)}{2} = \frac{60}{2} = 30\text{ MPa}$$ 2. Stress difference term: $$\frac{\sigma_{xx} - \sigma_{yy}}{2} = \frac{80 - (-20)}{2} = \frac{100}{2} = 50\text{ MPa}$$ 3. Radius (maximum in-plane shear stress): $$\tau_{\max} = \sqrt{(50)^2 + (40)^2} = \sqrt{2500 + 1600} = \sqrt{4100} \approx 64.031\text{ MPa} \approx 64.0\text{ MPa}$$ ### 3. Distractor Forensics - Forgetting the negative sign on $\sigma_{yy} = -20\text{ MPa}$ gives $\sqrt{30^2 + 40^2} = 50\text{ MPa}$, a frequent error. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Watch signs carefully: $\sigma_{xx} - \sigma_{yy} = 80 - (-20) = 100$. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 1.

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