A wide thin sheet of width $W$ under uniform tensile stress $\sigma_\infty$ contains a small circular hole of radius $a$ ($a \ll W$). According to Kirsch's classical elasticity solution, the maximum tensile stress concentration occurs at the edge of the hole on the diameter perpendicular to the loading direction. The theoretical elastic stress concentration factor $K_t = \frac{\sigma_{\max}}{\sigma_\infty}$ is:

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### 1. Strategy & Theoretical Principles From Kirsch's elasticity solution for a circular hole of radius $a$ in an infinite plate under uniaxial tensile stress $\sigma_\infty$ along the $x$-axis: $$\sigma_\theta(r, \theta) = \frac{\sigma_\infty}{2} \left(1 + \frac{a^2}{r^2}\right) - \frac{\sigma_\infty}{2} \left(1 + 3\frac{a^4}{r^4}\right) \cos(2\theta)$$ ### 2. Analytical Formulation & Calculation At the boundary of the hole ($r = a$): $$\sigma_\theta(a, \theta) = \sigma_\infty (1 - 2 \cos(2\theta))$$ - At points on the diameter parallel to loading ($\theta = 0^\circ, 180^\circ$): $\sigma_\theta = \sigma_\infty(1 - 2) = -\sigma_\infty$ (compressive stress of magnitude $\sigma_\infty$). - At points on the diameter perpendicular to loading ($\theta = 90^\circ, 270^\circ$): $\cos(180^\circ) = -1$, giving: $$\sigma_\theta = \sigma_\infty [1 - 2(-1)] = 3 \sigma_\infty$$ Therefore, the elastic stress concentration factor is: $$K_t = \frac{\sigma_{\max}}{\sigma_\infty} = 3.0$$ ### 3. Distractor Forensics - Candidate evaluating elliptical hole formula $K_t = 1 + 2(a/b)$ for $a=b$ directly confirms $1 + 2(1) = 3.0$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Circular hole in uniaxial tension produces $K_t = 3.0$ and a transverse compression of $-1.0 \sigma_\infty$. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 1.

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