If we're staying within the realms of beam theory, we can go with this approach which is valid for any material that exhibits linear-elastic behaviour before yield:
The curvature of a beam is related to the applied moment and it's flexural stiffness:
$$
\begin{align}
\kappa = \frac{M}{EI}
\end{align}
$$
where, for a cicular rod, $I = \frac{\pi d^4}{64}$. The curvature is also equal to the inverse of the bent radius, giving the following:
$$
\begin{align}
R = \frac{1}{\kappa} = \frac{EI}{M}
\end{align}
$$
For a rod of circular cross-section, the stress at the extreme fibre is related to the applied bending moment and the elastic section modulus:
$$
\begin{align}
\sigma = \frac{M}{Z}
\end{align}
$$
where for a circular rod, $Z = \frac{\pi d^3}{32}$. This gives:
$$
\begin{align}
\therefore M = \frac{\sigma \pi d^3}{32}
\end{align}
$$
Therefore, if a given material has a certain yield strength, $\sigma = f_y$, the minimum radius can be determined through the substitution of the above equations:
$$
\begin{align}
R_{min} = E \, \frac{\pi d^4}{64} \frac{32}{f_y \pi d^3}
\end{align}
$$
$$
\begin{align}
\boxed{\therefore R_{min} = \frac{E d}{2 f_y}}
\end{align}
$$
where $E$ is the elastic modulus of the material, $d$ is the diameter of the rod and $f_y$ is the yield strength of the material.
Refer to this solution to question 6a on page 8 for a similar solution for a rectangular section. Note that the end result is the same for both a rectangle and circle as both have a distance of $D/2$ from the neutral axis to the extreme fibre.
