= Solution
Assume a slender homogeneous <elastic beam> obeying <linear elasticity>, with plane sections remaining plane and no resultant longitudinal <force>. A length element of its <neutral axis> subtending a small <angle> $d\phi$ has length $R\,d\phi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $z\,d\phi$, giving longitudinal <strain> $\varepsilon=z/R$ with signs chosen consistently for tensile and compressed fibres. By <Hooke's law>, its longitudinal <stress> is $\sigma=E\varepsilon=Ez/R$.
The <torque> of these longitudinal <forces> about the <neutral axis> is therefore
$$
B=\int_A z\sigma\,dA=\frac E R\int_Az^2\,dA=\frac{EI}{R},
\qquad
\boxed{R=\frac{EI}{B}.}
$$
Here $E$ is <Young's modulus> and $I$ the <second moment of area>. The displayed relation uses a signed radius and <bending moment>; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative <bending moments> in the later parts, take the deflection $y$ positive downward and the signed <curvature> as $1/R\simeq y''$. Then the small-slope <Euler-Bernoulli beam equation> reads $EIy''=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.
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