Torsion of a rectangular hollow thin-wall box

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.

        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 ($1.57^\circ$).