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At distance $y$ from the neutral axis, bending strain has magnitude $y/R$ and the corresponding stress is $Ey/R$. Its moment is thereforeThe 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 $dB/ds=-Fdy/ds=-F\sin \theta$. Since $B=EI{\theta }^{\prime }$, the exact-angle Euler elastica equation and its first integral areThe first integral follows by multiplying by ${\theta }^{\prime }$ 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 branchChoosing $s=0$ at $\theta =\pi$ givesThe terminal angles are $0$ and $2\pi$, the same physical tangent direction. At the top, ${\theta }^{\prime }(0)=2\alpha$ and the bending moment is $2EI\alpha$. The horizontal tensile force has lever arm $h$, soIntegrating $(x^{\prime },y^{\prime })=(\cos \theta ,\sin \theta )$ independently gives $x=s-2\tanh (\alpha s)/\alpha$, $y=2\mathrm{sech}(\alpha s)/\alpha$, confirming the height. This ideal loop neglects self-contact and thickness at its crossing.