--- title: "Torsion of a rectangular hollow thin-wall box" subject: "Mechanics of Solids" chapter: 9 key: "semester_5/mechanics_of_solids:thin-walled-box-torsion-numerical" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/thin-walled-box-torsion-numerical site: BCT Engineering --- # Torsion of a rectangular hollow thin-wall box Mechanics of Solids · Chapter 9: Torsion - Asked in: year not given, 8 marks - Video: [AE 204 Torsion in Thin Walled Closed Sections](https://www.youtube.com/watch?v=LrZ0eIcA89w) — UW Mechanics - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/thin-walled-box-torsion-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-9 ## Question A thin-walled rectangular box section has mean dimensions 100 mm wide and 60 mm deep. The top and bottom walls are 4 mm thick and the two side walls are 6 mm thick. The section is subjected to a torque of 3 kN m. Find the shear flow, the shear stress in each wall, the angle of twist per metre length and the equivalent torsional constant. Take $G = 80$ GPa. Also find the total twist for a length of 1.5 m. ## Short answer Shear flow q = T/(2A) = 250 N/mm. Shear stresses: 62.5 MPa in the 4 mm top/bottom walls, 41.7 MPa in the 6 mm side walls. Angle of twist per metre: 0.0182 rad/m. Equivalent torsional constant J = 2.06×10⁶ mm⁴. Over 1.5 m, total twist = 0.0273 rad (1.57°). ## Answer #### Bredt-Batho theory For a thin-walled closed section, the shear flow $q = \tau t$ is constant around the wall, and the torque is $T = 2Aq$, where $A$ is the area enclosed by the **mid-line** of the wall. ```text 100 mm (mean) +-----------------+ t = 4 mm | | | | 60 mm sides t = 6 mm | | +-----------------+ t = 4 mm ``` #### Data $A = 100\times60 = 6000$ mm$^2$, $T = 3\ \text{kN m} = 3\times10^6$ N mm, $G = 80\times10^3$ MPa. #### Shear flow and stress $$ q = \frac{T}{2A} = \frac{3\times10^6}{2\times6000} = 250\ \text{N/mm} $$ $$ \begin{aligned} \tau_{top,bottom} &= \frac{q}{t} = \frac{250}{4} = 62.5\ \text{MPa} \\ \tau_{sides} &= \frac{250}{6} = 41.67\ \text{MPa} \end{aligned} $$ The maximum shear stress occurs in the thinnest wall (4 mm): $62.5$ MPa. #### Angle of twist per unit length Equating the strain energy to the work done by the torque gives $$ \theta = \frac{T}{4A^2G}\oint\frac{ds}{t} $$ $$ \oint\frac{ds}{t} = 2\left(\frac{100}{4}\right) + 2\left(\frac{60}{6}\right) = 50 + 20 = 70 $$ $$ \theta = \frac{3\times10^6\times70}{4\times(6000)^2\times80\times10^3} = \frac{2.1\times10^8}{1.152\times10^{13}} = 1.823\times10^{-5}\ \text{rad/mm} $$ $$ \theta = 0.01823\ \text{rad/m} = 1.045^\circ\text{/m} $$ #### Equivalent torsional constant $$ J_{eq} = \frac{4A^2}{\oint ds/t} = \frac{4\times 6000^2}{70} = 2.057\times10^{6}\ \text{mm}^4 $$ Check: $\theta = T/(GJ_{eq}) = 3\times10^6/(80\times10^3\times2.057\times10^6) = 1.823\times10^{-5}$ rad/mm. $\checkmark$ #### Total twist for 1.5 m $$ \phi = \theta L = 0.01823\times1.5 = 0.0273\ \text{rad} = 1.57^\circ $$ **Answer:** $q = 250$ N/mm; $\tau = 62.5$ MPa (top and bottom), $41.7$ MPa (sides); $\theta = 0.0182$ rad/m; $J_{eq} = 2.06\times10^6$ mm$^4$; twist over 1.5 m $= 0.0273$ rad ($1.57^\circ$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a thin closed section, the shear flow is A. q = T/(2A) with A enclosed by the mid-line, constant around the wall B. q = T/A, varying with thickness C. q = T t/J D. q = T/(2 perimeter) Answer: A. q = 3e6/(2 x 6000) = 250 N/mm. (Key point: Bredt-Batho shear flow) 2. Which wall has the greatest shear stress? A. The 6 mm side walls, 41.7 MPa B. The 4 mm top and bottom walls, 62.5 MPa C. Both equal 250 MPa D. The 6 mm walls, 62.5 MPa Answer: B. tau = q/t, and q is constant, so the thinnest wall carries the most stress. (Key point: Stress in walls) 3. The twist per unit length uses ∮ds/t, which here is A. 40 B. 70 C. 110 D. 25 Answer: B. 2(100/4) + 2(60/6) = 50 + 20 = 70, giving J = 4A^2/70 = 2.057e6 mm^4. (Key point: Twist and J) 4. Over 1.5 m the total twist is about A. 0.0182 rad B. 0.0273 rad (1.57 degrees) C. 0.0546 rad D. 1.045 degrees Answer: B. theta = 0.01823 rad/m, so phi = 0.01823 x 1.5 = 0.0273 rad. 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