I have been given the following problem in one of my courses. To say that I'm struggling in the course would be an understatement... I'm just wondering what the difference between shear flow and stress is, wouldn't they be the same in each secion?

Consider the thin-walled section shown below. The section is fixed on one end, and an axial moment (T=12 kNm) is applied to the free end. The section is 6.9 m long. Do the following:

  • Determine the shear flow in each section.
  • What is the maximum shear stress in the section.
  • Determine the angle of rotation of the free end (G = 26 GPa).

enter image description here


2 Answers 2


Shear flow is a quantity which is used to conveniently solve (usually) torsional problems of thin walled beams (it has other applications also). The concept behind it is, that the stress distribution in a wall of a thin-walled beam can be considered constant (while in a circular cross section is proportional to the distance from the shear center.

The relationship between shear flow and shear stress is :

$$q= \tau \cdot t$$


  • $q$ is the shear flow
  • $\tau$ is the shear stress
  • $t$ thickness of the thin walled beam.

In thin walled sections under torsional moment T, the shear flow $q$ can be calculated by:

$$q= \frac{T}{2\cdot A_k}$$


  • T is the torsional moment
  • $A_k$ is the enclosed surface area.

Just to be clear about $A_k$ an example

enter image description here

In the example above $A_k$ is equal to $A_k = b\cdot h$ (not just $2bt + 2ht$).

Regarding the cellular thin walled section (since this seems to me as an exercise I will not give the solution), the idea is That :

  • the algebraic sum of shear flows (coming in: + /coming out: -)at a junction is equal to zero.
  • The Total torsional moment T, is split among the cells.
  • The formula $q = \frac{T}{2A_k}$

With just those three concepts, a system of equations can be derived that can be solved (you can easily find examples on the net - if you show your effort, I would gladly expand this section).


This write-up repeats the things been taught in the class.

enter image description here

Shear flow $f = \tau t$ = constant

For thin wall closed shapes:

$f = \tau t = \dfrac {T}{2A_m}$ and

$\tau = \dfrac {T}{2tA_m}$

In the formulas, $t$ is the wall thickness, $A_m$ is the area bounded by the centerline of the shape. For example, for a round pipe, $A_m = \pi r^2$, and a hollow rectangle, $A_m = bh$, with $r, b, h$ are centerline distance.

The angle of rotation/twist,

$\phi = \dfrac {TL}{GJ}$, in which, $J = \dfrac {4tA_m^2}{L_m}$

$L_m$ is the length of the centerline. For example, $L_m = 2\pi r$ for circular pipe, and $2(b + h)$ for hollow rectangle shape.


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