---
title: "Torsion of a thin open channel section"
subject: "Mechanics of Solids"
chapter: 9
key: "semester_5/mechanics_of_solids:open-channel-torsion-numerical"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/open-channel-torsion-numerical
site: BCT Engineering
---

# Torsion of a thin open channel section

Mechanics of Solids · Chapter 9: Torsion

- Asked in: year not given, 6 marks
- Video: [MAE 3323 - Torsion of open and closed sections](https://www.youtube.com/watch?v=PKm_8QXvBRY) — Ronald Delahoussaye
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/open-channel-torsion-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-9

## Question

A thin-walled open channel is made of a web 100 mm $\times$ 6 mm and two flanges each 60 mm $\times$ 8 mm (the dimensions are length $\times$ thickness of each strip). It is subjected to a torque of 200 N m. Using the narrow rectangle approximation find the maximum shear stress in the web and in the flanges, the angle of twist per metre, and the torque shared by each strip. Take $G = 80$ GPa.

## Short answer

Using narrow rectangle approximation, J = Σ b_i t_i^3/3 = 27680 mm^4. Maximum shear stress: 43.4 MPa in web, 57.8 MPa in flanges. Twist per metre: 0.0903 rad/m (5.17°/m). Torque: web 52.0 N m, each flange 74.0 N m.

## Answer

#### 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

| Strip | $J_i$ (mm$^4$) | Torque $T_i$ (N m) | Share |
|---|---|---|---|
| Web | 7200 | 52.0 | 26.0 % |
| Each flange | 10240 | 74.0 | 37.0 % |
| Total | 27680 | 200 | 100 % |

(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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the open channel, J is
   A. Sum of b t^3/3 of each strip = 27680 mm^4
   B. Sum of b t = 1320
   C. 4A^2/∮ds/t
   D. b t^3/12 of the web only
   Answer: A. 7200 + 2(10240) = 27680 mm^4. (Key point: Torsion constant)

2. Maximum shear stress occurs
   A. In the web, 43.4 MPa
   B. In the thicker flanges, 57.8 MPa, since tau = T t/J
   C. At the corners only
   D. Equally in all strips
   Answer: B. tau_i = T t_i / J, so the thickest strip (8 mm) has the largest stress. (Key point: Maximum stress)

3. Torque in each strip is shared in proportion to
   A. Its area
   B. Its Ji = b t^3/3
   C. Its length
   D. Its thickness
   Answer: B. All strips twist equally, so Ti = T Ji/J: web 52 N m, each flange 74 N m. (Key point: Torque shares)

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