Torsion of thin-walled circular bars

Shear stress — equilibrium

Consider a thin-walled circular bar loaded by an external torque $\Mv$, according to the figure below. The bar has the thickness $h$ and mean radius $\bar{r}$.

Make a section through the bar and draw a free-body diagram, of the left part, as shown below.

By considering moment equilibrium, one sees that the internal torque $\Mv$ must balance the external torque $Q$ i.e.

$$\twoheadleftarrow: \quad Q – \Mv=0$$

The internal torque $\Mv$ is the (moment) resultant to a shear stress distribution $\tau$, around the cross-section, as illustrated below.

The torque $\Mv$ is a so-called sectional force, just like the normal force $N$, but note that $\Mv$ is not a force but a moment.


Since the cross-section is thin-walled ($h$ is small compared to $\bar{r}$), we can assume that the shear stress distribution is constant over the cross-section (in the radial direction). For the same reason, we also assume that the radius is constant, $r=\bar{r}$. The internal torque is then obtained by integration as:

$$\Mv = \int_A \tau \, r \, {\rm d} A =
\tau \, \bar{r} \, A =
\tau \, \bar{r} \, 2 \pi \, \bar{r} \, h$$

where $A$ is the cross-sectional area calculated as “thickness times circumference”.

The shear stress becomes

$$\tau=\frac{\Mv}{2 \pi \, \bar{r} \, h}$$

and is illustrated below (at four positions). Note the constant magnitude.

In the next section, we continue with studying thick-walled cylindrical cross-sections.

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