= Solution
{c}
Because the two <elastic rods> each have thickness $a$, their centrelines are separated by $a$ along the common director $\mathbf d_1$. The bonded geometry is
$$
\mathbf r^+(S^+)=\mathbf r(S)+a\mathbf d_1(S),\qquad S^+=S.
$$
Differentiate with respect to reference <arc length>. The first rod is an <inextensible rod>, and the printed <director kinematics of a rod> give
$$
\frac{d\mathbf r^+}{dS^+}
=\mathbf d_3+a(-u\mathbf d_3)=(1-au)\mathbf d_3.
$$
For the physically admissible orientation-preserving configuration $1-au>0$, the rods have the same unit tangent $\mathbf d_3^+=\mathbf d_3$ and the same planar <material frame of a rod>. Therefore
$$
\boxed{\alpha^+=1-au.}
$$
In particular, the second rod is shortened on the inside of a bend with $u>0$.
The <bending strain of a rod> is defined here by differentiation with respect to reference <arc length>. Since the reference labels match,
$$
\frac{d\mathbf d_3^+}{dS^+}=\frac{d\mathbf d_3}{dS}=u\mathbf d_1
=u^+\mathbf d_1^+,
\qquad\boxed{u^+=u,\quad\beta=1.}
$$
The distinction from spatial <curvature> matters: the second centreline has
$$
\kappa^+=\left|\frac{d\mathbf d_3^+}{ds^+}\right|
=\frac{|u|}{1-au}.
$$
Thus $u/(1-au)$ is its signed spatial <curvature>, rather than the $u^+$ appearing in the given <elastic energy>.
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