Torsion of thick-walled circular bars

Consider a thick-walled circular bar, with an inner radius $a$ and outer radius $b$, according to the figure below.

Like in the previous section, we study a free-body diagram of a portion of the bar.

For the bar to be in equilibrium the internal and external torques must be equal
$$
\twoheadleftarrow: \quad Q = \Mv
$$

The torque $\Mv$ is the resultant to the shear stress; however, since the bar is thick-walled, we can not assume the shear stress distribution is constant any longer. The figure below illustrates the shear stress distribution, indicating a variation over the cross-section.

So, $\tau$ cannot directly be determined from the definition of the torque $\Mv = \int_A \tau \, r \, {\rm d} A$, as for the thin-walled cross-section. Instead, we must combine compatibility and constitutive relations.

Compatability — shear strains

When a torque $\Mv$ acts on a bar, this produces an angle of rotation/twist $\rot$, varying along the bar. The change in rotation, from one end to the other, is denoted $\Delta \rot$. As shown in the figure below, the torque also produces a shearing deformation $\gamma$ of the bar, as illustrated by the movement of the pink rectangle.

Now, assuming that each cross-section only rotates (no axial displacement and remain their shape), the distance $s$ is expressed in two ways:
$$
s = \shear L = \Delta \rot \, r
$$
Here, we assumed small rotations (angles), such that the displacement $s$ can be calculated as the arch length (angle times radius), this gives $\shear = r \frac{\Delta \rot}{L}$.

By combining Hooke’s law in shear, $\tau=G \, \gamma$, with the compatibility equation above, we obtain for a given angle of twist $\Delta \varphi$ the shear stress distribution $\tau(r)$ as

$$\tau(r) = \frac{G \, r \, \Delta \varphi}{L}$$

showing a linear variation of stress with the radius $r$. The largest stress is, therefore, obtained at the outer surface of the bar.

The main difference between torsion of a thin-walled bar and a thick-walled bar is the stress distribution. For a thick-walled bar, we allow the stress to vary with the radius. However, the internal torque $M_{\rm v}$ must still be the moment resultant of the shear stress:

$$\begin{aligned}
\Mv&=\int_A \tau(r) \, r \, {\mathrm{d}}A =
\braces{\mathrm{d} A =2 \pi r \, \dr} \
&=\frac{G \, \Delta\varphi}{L} \int_a^b r^2 \, 2 \pi r \, \dr =
\frac{G \, \Delta \varphi }{L} \, \Kv
\end{aligned}$$

where $\Kv = \frac{\pi \, (b^4-a^4)}{2}$ is the polar moment of inertia for a thick-walled circular cross-section.

Combining the two equations above, the shear stress due to torsion can be written as:

$$\tau(r) = \frac{\Mv\, r}{\Kv}$$

In the figure below, the shear stress distribution over the cross-section is shown at four positions.

As mentioned above, the largest shear stress for a thick-walled cross-section is obtained for the largest radius, i.e. when $r = b \gives$

$$\tau_{\rm{max}} = \frac{\Mv}{\Kv} b = \frac{\Mv}{W_{\mathrm{t}}}$$

where we have introduced $W_{\mathrm{t}}$ as the section modulus in torsion

$$W_{\mathrm{t}} = \frac{\Kv}{b} = \frac{\pi}{2} \left( b^3 – \frac{a^4}{b} \right)$$

The section modulus $W_{\mathrm{t}}$ can be found tabulated for common cross-sections and is used with the equation above to quickly calculate the largest shear stress given a known torque.

In summary, for an elastic thick-walled circular bar, we have the following relationship between external torque $\Mv$ and change in angle of twist $\Delta \varphi$:

$$\Mv = G \, \Kv \frac{\Delta \varphi}{L}$$

Leave a Reply