How to find a bend curve in 2D? I am looking for a solution on how to find the (approxmiate) shape when bending a rigid-flex circuit board. Please see the abstract sketch below. I have two solid objects ($A$, $B$) which are connected by a thin and flexible but non-stretchable strip of material.
Given that I know
(i) the position snf rotation of the fixed parts $A$ and $B$ in 2D space and
(ii) the length $d$ and thickness $w$ of the flexible connecting strip
how can I calculate the shape of the flexible strip when it is bent?
Any help would be highly appreciated, approximate solutions (which may assume $w=0$) are fine.

 A: Admittedly I'm not an expert in this, but I can have a crack at a toy model. Let's assume that the material is elastic and naturally "wants" to be straight. Let the 2D path of the flex PCB follow a curve $\boldsymbol{c}(s)$ parametrised by arclength, i.e. satisfying $|c'(s)| =1$ everywhere as $0<s<L$.
Calculus of Variations
A natural way to do implement what you want is to build a curvature-penalising energy functional
$$ H[\gamma(s)] = \frac{k}{2}\int dl F(\kappa(s)) =\frac{k}{2}\int dl F\left( \sqrt{x''(s) + y''(s)}\right)$$ where the function $F$ accounts for the possibly nonlinear relationship between stress and strain. An ideal elastic material has $F(x)=x$, which is a poor approximation in this regime. The boundary conditions are given by the position and orientation of the start and end points - $x(0), y(0), x(L), y(L), x'(0), y'(0), x'(L), y'(L)$ are all fixed. Without loss of generality you can choose coordinates such that $x(0)=y(0)=y'(0)=0, x'(0)=1$.
To minimise $H$, you simply compute the funcitonal derivative:
$$\frac{\delta H}{\delta x_i} = 0$$
$$\Rightarrow k\left(\frac{F'(\kappa)}{\kappa^2}x_i'' \right)''=0$$
By integrating twice, you end up with
$$x_i''(s)k F'(\kappa(s)) = \kappa(s)^2[A_i s + B_i]$$
$x_i, i=1,2$ is a stand in for $x,y$ components.
Even in the 'simple' elastic case, this is a bit of a mess. Let's use the fact that it's parameterised by arclength to write $x' = \cos \theta(s), y' = \sin \theta(s)$. Remarkably, this simplifies life a lot (divide the x equation by the y equation to get here)
$$ \theta(s) = \cot^{-1}\left(-\frac{A_y s + B_y}{A_x s + B_x}\right)$$
or, in a singularity free (but still exact!) formulation,
$$ -(A_y s + B_y)\sin(\theta(s)) = (A_x s + B_x) \cos(\theta(s))$$
The difficulty comes with determining the constants $A_i, B_i$.
The actual path can be computed using $\boldsymbol{c}(s)= \int_0^s d\tau (\cos(\theta(\tau)), \sin(\theta(\tau))) + \boldsymbol{x}_0$.
Note that there are only three free parameters: a uniform rescaling does not alter the fraction's value.
Boundary conditions give us the rest. $\theta$ has the physical interpretation as the angle of the tangent vector relative to the starting point. We need to start and finish at the right angle, so we have the linear constraints
$$\cos(\theta_0) B_x = -\sin(\theta_0)B_y$$
$$\cos(\theta_L)(A_x L + B_x) = -\sin(\theta_L)(A_y L + B_y)$$
Hang on, we seem to have an issue- we've only used two of the four boundary conditions, and we only have one parameter left. This is because knowing the starting angle, finishing angle and material length L already constrains us to a 1D curve in the plane - knowing the x coordinate alone is enough to uniquely specify the flex PCB length. Imposing this condition on $\int_0^L ds \sin(\theta(s))$, or the same with $\cos$ finishes the job.
Unfortunately, I've been a bit cavalier with domain choice and assuming things aren't zero. In particular, this solution is only valid on the $\theta$ domain $0<\theta<\pi$, so these calculations can't account for 'S shaped' buckled curves. (In principle this can be accommodated by defining the curve piecewise)
More Practically
The integration to get explicit $x(s),y(s)$ can in fact be done analytically if you restrict to $0<\theta<\pi/2$, when $\cos(\theta) = \frac{1}{\sqrt{1+\tan^2\theta}}$
$$ x(s) = \int_0^s dt \cos(\theta(t)) = \int_0^s dt \frac{A_y t + B_y}{\sqrt{(A_x t + B_x)^2 + (A_y t + B_y)^2}} = A_y f(s) + A_x g(s)$$
$$ y(s) = \int_0^s dt \cos(\theta(t)) = \int_0^s dt \frac{A_x t + B_x}{\sqrt{(A_x t + B_x)^2 + (A_y t + B_y)^2}} = A_x f(s) -A_y g(s)$$
where
$$f(s) = \frac{1}{(A_x^2 + A_y^2)}\sqrt{B_x^2 B_y^2 + (B_x^2 B_y^2 + 2A_y B_y + 2A_x B_x) s + (A_x^2 + A_y^2)s^2}$$
$$g(s) = (A_x B_y - A_y B_x) \mathrm{arctanh}\left[\frac{A_x B_x + (A_x^2 + A_y^2)s + A_x B_x + A_y B_y}{(A_x^2 + A_y^2)(B_x^2 B_y^2 + 2( A_x B_x + A_y B_y)s + (A_x^2 + A_y^2)s^2)}\right]$$
Similar solutions can be recovered in other quadrants, differing by a handful of minus signs. Parameters can then be fixed using the earlier equations.
A: The curvature is proportional to the local bending moment.
According to the picture, it seems that the strip is firmly held by the parts A and B, in a way that it can not slide. In that case, the situation is pure bending and the curve is an arc of circle. See http://emweb.unl.edu/NEGAHBAN/Em325/11-Bending/Bending.htm
The situation is different if a strip is firmly held in A (by a vise for example), and we bend it by pushing the other end with a hand. In this case, the bending moment at the hand is zero (as the strip is free to slide there) while it is maximum at the vise. The curvature is not constant and the form is not simple, except for small displacement (angle $\alpha \approx 180^\circ)$
