= Solution
At distance $y$ from the <neutral axis>, bending <strain> has magnitude $y/R$ and the corresponding <stress> is $Ey/R$. Its moment is therefore
$$
\boxed{B=\int_A y\frac{Ey}{R}\,dA=\frac{EI}{R},\qquad I=\int_Ay^2\,dA.}
$$
The <second moment of area> is about the neutral bending axis and has units of <length> to the fourth power. This assumes a slender linear-elastic beam, small <strains> and plane sections remaining plane, not necessarily small deflections.
For horizontal compression $F>0$, moment balance across a short arc gives $dB/ds=-F\,dy/ds=-F\sin\theta$. Since $B=EI\theta'$, the exact-angle <Euler elastica> equation and its first integral are
$$
\boxed{EI\theta''+F\sin\theta=0,\qquad
\frac{EI}{2}\theta'^2-F\cos\theta=C.}
$$
The first integral follows by multiplying by $\theta'$ and integrating; no small-angle replacement has been made.
For tension set $F=-T$. The straight far-field ends give $C=T$, so on the increasing-angle <tensile elastica loop> branch
$$
\theta'=2\sqrt{\frac{T}{EI}}\sin\frac\theta2,\qquad
\ln\tan\frac\theta4=s\sqrt{\frac{T}{EI}}.
$$
Choosing $s=0$ at $\theta=\pi$ gives
$$
\boxed{\theta(s)=4\arctan e^{\alpha s},\qquad\alpha=\sqrt{T/(EI)}.}
$$
The terminal <angles> are $0$ and $2\pi$, the same physical tangent direction. At the top, $\theta'(0)=2\alpha$ and the <bending moment> is $2EI\alpha$. The horizontal tensile <force> has lever arm $h$, so
$$
\boxed{Th=2EI\alpha,\qquad h=2\sqrt{EI/T}.}
$$
Integrating $(x',y')=(\cos\theta,\sin\theta)$ independently gives $x=s-2\tanh(\alpha s)/\alpha$, $y=2\operatorname{sech}(\alpha s)/\alpha$, confirming the height. This ideal loop neglects self-contact and thickness at its crossing.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-1-elastica-loop.png]
{title=Original analytic tensile-elastica shape and its tangent-angle and curvature profiles}
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