A thin steel strip of rectangular cross-section has length $L$, width $b = 100\text{ mm}$, and thickness $t = 2\text{ mm}$ ($b \gg t$). The strip is tested under pure torque $T$ in two states: 1. State 1 (Open section): As a flat open strip. 2. State 2 (Closed tube): Rolled into a circular tube of perimeter $b$ (circumference $2\pi R = b$) and seam-welded along the seam. Which of the following statements regarding the torsional behavior of these two sections are **CORRECT**?

(A) The torsional stiffness of the closed circular tube is orders of magnitude greater than that of the open strip ($J_{\text{closed}} \gg J_{\text{open}}$)

(B) For the open strip, the St. Venant torsional constant is $J_{\text{open}} \approx \frac{1}{3} b t^3$

(C) For the closed seam-welded circular tube, the Bredt-Batho torsional constant is $J_{\text{closed}} = \frac{4 A_m^2}{\oint \frac{ds}{t}} = \frac{b^3 t}{4\pi^2}$

(D) Under the same applied torque $T$, the maximum shear stress in the closed tube is significantly higher than in the open strip

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Comparison of Open and Closed Thin-Walled Torsion: 1. **Open Section (St. Venant Torsion):** - Torsion constant: $J_{\text{open}} = \frac{1}{3} b t^3$ - Maximum shear stress: $\tau_{\max,\text{open}} = \frac{T t}{J} = \frac{3 T}{b t^2}$ 2. **Closed Section (Bredt-Batho Torsion):** - Circumference: $2\pi R = b \implies R = \frac{b}{2\pi}$ - Enclosed area: $A_m = \pi R^2 = \pi \left(\frac{b}{2\pi}\right)^2 = \frac{b^2}{4\pi}$ - Torsion constant: $J_{\text{closed}} = \frac{4 A_m^2}{\oint ds / t} = \frac{4 (b^2 / 4\pi)^2}{b / t} = \frac{4 b^4 / (16\pi^2)}{b / t} = \frac{b^3 t}{4\pi^2}$ - Ratio of stiffnesses: $\frac{J_{\text{closed}}}{J_{\text{open}}} = \frac{b^3 t / (4\pi^2)}{b t^3 / 3} = \frac{3}{4\pi^2} \left(\frac{b}{t}\right)^2$ - For $b/t = 100/2 = 50$: $\frac{J_{\text{closed}}}{J_{\text{open}}} = \frac{3}{39.48} \times 2500 \approx 190$ times stiffer! 3. **Shear Stress:** - In the closed tube, $\tau_{\text{closed}} = \frac{T}{2 A_m t} = \frac{T}{2 (b^2 / 4\pi) t} = \frac{2\pi T}{b^2 t}$. - Ratio of stresses: $\frac{\tau_{\text{closed}}}{\tau_{\text{open}}} = \frac{2\pi T / (b^2 t)}{3 T / (b t^2)} = \frac{2\pi}{3} \frac{t}{b} \approx 2.09 \times \frac{2}{100} = 0.042$. - Thus, the shear stress in the closed tube is over $20$ times *lower* than in the open strip. ### 2. Statement Evaluation - **Statement A:** **CORRECT.** Closed tubes are drastically stiffer in torsion. - **Statement B:** **CORRECT.** Standard open strip torsion constant formula. - **Statement C:** **CORRECT.** Exact derivation of closed tube Bredt-Batho torsion constant. - **Statement D:** **INCORRECT.** Shear stress in the closed tube is drastically *lower*, not higher. ### 3. Exam Trap & Standard Reference - **Trap Alert:** Seam-welding a thin slit creates a closed cell, reducing shear stress by orders of magnitude while increasing torsional stiffness enormously. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 18.

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