= Solution
Set $\ell=\sqrt{EI/F}$. The <Euler elastica> equation becomes $\theta''=\ell^{-2}\sin\theta$. Multiplying by $\theta'$ gives the <first integral>
$$
\frac{EI}{2}(\theta')^2+F\cos\theta=C.
$$
Straight asymptotic ends require $\theta'\to0$ and a horizontal tangent, so $C=F$. This is the <separatrix> of a <simple pendulum> with its angle shifted by $\pi$. A nonconstant solution moves from $\theta=0$ to $2\pi$, with the same physical tangent at both ends, taking infinite <arc length> to approach either limit. One explicit representative is
$$
\boxed{\theta(s)=4\arctan e^{s/\ell}.}
$$
Using $\cos\theta=1-2\operatorname{sech}^2(s/\ell)$ and $\sin\theta=-2\operatorname{sech}(s/\ell)\tanh(s/\ell)$ and integrating $r'=(\cos\theta,\sin\theta)$ gives the <tensile elastica loop>
$$
\boxed{x(s)=s-2\ell\tanh(s/\ell),\qquad y(s)=2\ell\operatorname{sech}(s/\ell).}
$$
This solution is nontrivial, tends to horizontal straight tails, and has finite excess <bending energy>. It is also necessarily self-intersecting. Indeed $y$ is even and $x$ is odd. For $z>0$, let $f(z)=z-2\tanh z$. Near zero $f(z)<0$, while $f(2)>0$, so there is a root $z_*\in(0,2)$. At the two distinct parameters $s=\pm\ell z_*$, both points have $x=0$ and the same $y>0$. Numerically $z_*\simeq1.91501$.
Any nonconstant smooth solution with the same straight asymptotic conditions has this <first integral>. Between consecutive horizontal orientations, separating variables gives the displayed solution up to translation, reflection, and reversal. It cannot pass through a straight equilibrium at a finite parameter, by uniqueness of the <ordinary differential equation>. Thus all such nontrivial branches have the same loop crossing. This is an ideal zero-thickness rod solution; a physical rod with self-contact excluded cannot realize the crossing without adding contact forces.
\Image[/past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2011-tensile-elastica-loop.png]
{title=The tensile elastica loop and its self-intersection}
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