Torsion constant of a rod. Dimensional derivation

The torsion constant of a circular rod (the torque needed to twist it by a given angle) is easily calculated from the equations of elasticity. Up to a numerical constant, it can also be derived by dimensional analysis, as shown below.

Consider the rod in Figure a), with radius $r$, length $L$ and shear modulus $G$. Its upper end A is clamped. The torque $T$ needed to turn the free end B is linear¹ in the twist angle $\theta$:

$$\begin{equation} T = \kappa \theta \label{twist} \end{equation}$$

Clearly, $\theta$ depends on $r$, $L$ and $G$. Are these the only relevant parameters? How about the bulk modulus $B$, for instance? A non-rigorous way of showing its irrelevance is by considering a material with finite $B$ and $G = 0$ (e.g. a liquid): in this case the torsion constant is clearly zero². We can then write:

$$\begin{equation} \kappa = K \, G^a r^b L^c \label{kappa} \end{equation}$$

The various units are: $[\kappa ] = \text{Nm}/\text{rad} = \text{Nm}$, $[G] = \text{Pa} = \text{Nm}^{-2}$, $[L] = [r] = \text{m}$ and $K$ is a dimensionless prefactor. We can deduce from $\eqref{kappa}$ that $a = 1$ and that $b + c = 3$, but we need more information to determine $b$ and $c$ separately.
[UPDATE: 19/05/2013] In principle, the expression for $\kappa$  $\eqref{kappa}$ could also contain an arbitrary function $f$ of the dimensionless parameter $r/L$, but the discussion below shows that $f(r/L)$ is in fact a constant.

To this end, we make a virtual horizontal cut at the midpoint C (Figure b) and write the torque balance. In A, the clamp exerts a torque $-T$ (to ensure equilibrium of the whole rod). The equilibrium of segments AC and CB requires that they act on each other with torques $T$ and $-T$, respectively. They are identical, so each of them twists by $\theta/2$. In conclusion, for the same torque, a rod half as long twists half as much, so its torsion constant is twice that of the original rod. Hence, $c = -1$ and then $b = 4$, yielding:

$$\begin{equation} \kappa = K \, G \frac{r^4}{L} \label{final} \end{equation}$$

The complete calculation³ yields $K = \pi/2 = 1.57\ldots$, not very far from unity. We can see this as a lucky choice of parameters: working with the diameter $D = 2 r$ instead of the radius would multiply the estimate by 16. On the other hand, $r$ is the more natural elasticity variable, since it is over this distance that the deformation varies (from 0 to its maximum value).

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<sup>1. Rigorously speaking, we can only say that $\theta$ is odd in $T$ (they change sign together), but for small twists the dependence will be dominated by the linear term $\eqref{twist}$. We only consider this linear regime.↩
2. This is an instance of the general principle that liquids cannot transmit static torques. ↩
3. L. D. Landau and E. M. Lifshitz, Theory of Elasticity (volume 7 of the Course of Theoretical Physics), 3rd edition, Elsevier, 1986 (§ 16).↩</sup>