Torsion of a thin open channel section

Method

For a thin open section, the section behaves like a set of narrow rectangles. Each strip has the same twist per unit length $\theta$, and carries a torque in proportion to its $J_i = b_it_i^3/3$ (membrane analogy).

$$J = \sum \frac{b_it_i^3}{3},\qquad \theta = \frac{T}{GJ},\qquad \tau_i = \frac{T\,t_i}{J},\qquad T_i = T\,\frac{J_i}{J}$$

Torsion constant

$$\begin{aligned} J_{web} &= \frac{100\times6^3}{3} = \frac{21600}{3} = 7200\ \text{mm}^4 \\ J_{flange} &= \frac{60\times8^3}{3} = \frac{30720}{3} = 10240\ \text{mm}^4\ \text{(each)} \\ J &= 7200 + 2(10240) = 27680\ \text{mm}^4 \end{aligned}$$

Shear stress

$T = 200\ \text{N m} = 2\times10^5$ N mm.

$$\begin{aligned} \tau_{web} &= \frac{Tt_w}{J} = \frac{2\times10^5\times6}{27680} = 43.35\ \text{MPa} \\ \tau_{flange} &= \frac{2\times10^5\times8}{27680} = 57.80\ \text{MPa} \end{aligned}$$

The maximum shear stress, $57.8$ MPa, occurs at the long edges of the thicker flanges.

Twist

$$\theta = \frac{T}{GJ} = \frac{2\times10^5}{80\times10^3\times27680} = 9.03\times10^{-5}\ \text{rad/mm} = 0.0903\ \text{rad/m} = 5.17^\circ\text{/m}$$

Torque shared by strips

(The two flanges together carry 74.0 percent.)

The calculation neglects the stress concentration at the re-entrant corners and the extra stiffness at junctions; a correction factor (about 1.1 to 1.3 for rolled sections) is used for rolled channels and angles.

Answer: $\tau_{web} = 43.4$ MPa, $\tau_{flange,max} = 57.8$ MPa, $\theta = 0.0903$ rad/m ($5.17^\circ$/m); torque shared: web $52.0$ N m, each flange $74.0$ N m.