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The neutral axis is the line in the cross-section of an elastic beam where longitudinal bending strain vanishes. For a homogeneous beam in pure bending, zero resultant longitudinal force gives ${\int }_{A}zdA=0$, so the neutral axis passes through the centroid. Fibres on its opposite sides are stretched and compressed. The second moment of area about this axis controls the relation between bending moment and curvature.

In units of elastohydrodynamic penetration length and inverse forcing frequency, $y_t=-y_{xxxx}$. Prescribing $y(0,t)=y_0\cos t$, zero bending moment $y_{xx}(0,t)=0$, and decay at infinity givesThe two components have phase velocities $-1/S$ and $1/C$. Their unequal spatial attenuation produces a nonreciprocal shape cycle even though the endpoint actuation is reciprocal.

Overhung idler loading occurs when an idler gear's mesh resultant acts outside the axial span of its shaft supports. The overhang creates a lever arm, so the nearer mechanical bearing can carry a larger transverse load than the mesh resultant and the farther support can have the opposite reaction direction. This geometry is separate from whether the idler's two transmitted torques cancel.^[ref]

Place ideal radial supports A and B at $z=0$ and $z=L$. An upward mesh resultant $P$ at $z=L+e$, with overhang $e>0$, gives static equilibriumHenceFor $P=200\mathrm{N}$, $L=100\mathrm{mm}$ and $e=30\mathrm{mm}$, A acts upward with60N while B acts downward with260N. Their signed sum balances the applied200N. The larger reaction at B is a lever effect, not an additional external mesh force.

The overhung shaft bending moment immediately next to the near support has magnitude $Pe=6\mathrm{N}\mathrm{m}$ in this ideal model. That bending moment can exist while the net shaft-axis torque on the steady idler is zero. Doubling overhang at fixed load doubles the near-support bending moment and changes the reaction magnitudes to120 and320N.

Idler bearing load sharing must therefore include the gear's axial position, not just its pitch radius and speed ratio. Loads in independent transverse directions can be balanced separately and then combined as vectors. A support model with more constraints requires compatible deformations as well as force balance. Shaft bending and support flexibility can also alter the gear face load distribution, linking the overhang geometry to tooth loading.

An overhung shaft bending moment is the internal bending moment produced when a transverse load acts beyond the shaft's support span. For a single end load $P$ at overhang distance $e$, its magnitude immediately next to the nearer support isin the ideal two-support shaft model. The relevant lever arm is the distance beyond that support, rather than the distance between supports. Manufacturer bearing-load guidance distinguishes this arrangement from a load located between bearings.^[ref]

Place ideal supports A and B at axial positions $z=0$ and $z=L$, and apply an upward load $P$ at $z=L+e$. With upward reactions positive, static equilibrium givesChoosing the internal moment sign consistently with the left-hand force sums, the bending moment along the shaft isIt is continuous at B, reaches $Pe$ there and vanishes at the free end. An ideal bearing need not apply a concentrated couple for the shaft's internal moment at its location to be nonzero: the shaft continues beyond that support.

For original values $P=200\mathrm{N}$, $L=100\mathrm{mm}$ and $e=30\mathrm{mm}$, the reactions are $+60$ and $-260\mathrm{N}$ and the peak moment is $6\mathrm{N}\mathrm{m}$. Doubling $e$ at fixed $P$ doubles that moment. Changing $L$ while keeping $P,e$ fixed changes the reactions but leaves $Pe$ unchanged.

For a uniform solid circular shaft in simple elastic bending, overhung shaft bending stress follows from ${\sigma }_{\max }=Mc/I$ and $I=\pi d^4/64$:At diameter $d=20\mathrm{mm}$, the example gives about $7.64\mathrm{MPa}$. This excludes stress concentrations and other loads; a shaft's allowable load requires more than this nominal estimate. Bending can remain even when opposing gear torques give zero net shaft-axis torque, because transverse-force moments and axial twisting are different components of the mechanical loading.

Overhung shaft bending stress is the longitudinal normal stress caused by the bending moment from a transverse load beyond the shaft's supports. It differs from twisting due to shaft-axis torque. In the ideal two-support model, the largest overhung shaft bending moment has magnitude $M=Pe$, where $P$ is the end load and $e$ the overhang.^[ref]

For a homogeneous linear elastic beam, plane sections in the beam model give longitudinal strain $ϵ(y)=\kappa y$ relative to the neutral axis. With Young's modulus $E$,where $I$ is the second moment of area. Thus $\sigma (y)=My/I$. For a solid circular shaft of diameter $d$, $I=\pi d^4/64$ and the outermost distance is $d/2$, givingAn original example uses $P=200\mathrm{N}$, $e=30\mathrm{mm}$ and $d=20\mathrm{mm}$. Then $M=6000\mathrm{N}\mathrm{mm}$ and ${\sigma }_{\max }\simeq 7.64\mathrm{MPa}$. Reducing diameter to $16\mathrm{mm}$ multiplies nominal stress by $(20/16{)}^3\simeq 1.953$, raising it to about $14.9\mathrm{MPa}$. The cubic diameter dependence makes a diameter change different from an equal percentage change in overhang.

A measurement check can place opposed axial strain gauges at one shaft station. In a simple axial-plus-bending model, readings ${ϵ}_{+}=44$ and ${ϵ}_{-}=-32$ microstrain give axial strain $({ϵ}_{+}+{ϵ}_{-})/2=6$ microstrain and bending strain $({ϵ}_{+}-{ϵ}_{-})/2=38$ microstrain. This separates two contributions that a single gauge would mix.

The nominal formula excludes local geometric concentrations, multiaxial loading and yielding. It is not a fatigue-life prediction. Under a stationary bending direction, a rotating material point can experience alternating stress, motivating rotating shaft bending fatigue even when the external load is steady. A permissible load requires the actual geometry, material and load history.

The tension term $T{w}_{xx}$ is the transverse restoring force from a nearly straight, tensioned rod. The bending term $-EI{w}_{xxxx}$ is the force from Euler-Bernoulli flexure, where $E$ is Young's modulus and $I$ the second moment of area. The right side is transverse inertia. In the displayed one-dimensional equation $\rho$ denotes linear mass density; if it denotes volumetric density instead, the inertial coefficient must include cross-sectional area.

For a Fourier mode, $w_{xx}=-k^{2}w$, $w_{xxxx}=k^{4}w$ and $w_{tt}=-{\omega }^{2}w$, givingFor $k>0$ the phase speed is $c_p=\omega /k=\sqrt{T/\rho +EI{k}^2/\rho }$, strictly increasing with $k$ when $EI>0$ and $T\ge 0$. Thus shorter wavelengths propagate faster in phase. The group velocity $c_g=(T+2EI{k}^2)/\sqrt{\rho (T+EI{k}^2)}$ is also increasing, so the statement is consistent with wave-packet transport.

At a simple support the displacement and bending moment vanish: $w=0$ and $EI{w}_{xx}=0$ at $x=0,L$. The eigenmodes are $\sin (n\pi x/L)$ for $n=1,2,\dots$. Setting $T=0$ in the dispersion relation therefore yieldsThe integer is positive for a nonzero mode; $n=0$ gives the identically zero sine function, and negative integers repeat the same modes. Bending stiffness makes the frequencies quadratic in mode number, unlike a purely tensioned string.

Use the same first variation on the half-plane, with no restrictions on the edge traces of $\eta$. Write $M=w_{xx}+\nu w_{yy}$. Two $x$ integrations of the $M{\eta }_{xx}$ term produce the boundary term $-M{\eta }_x+M_x\eta$ at $x=0$. The mixed term, after one integration in $x$ and one along the edge in $y$, contributes $2(1-\nu )w_{xyy}\eta$. All other boundary terms vanish at infinity. HenceThe bulk equation makes its area term zero. Stationarity therefore gives exactlyThe two edge traces can be prescribed independently: use $\eta (x,y)=\chi (x)[f(y)+xg(y)]$ with $\chi =1$ near zero and compactly supported smooth $f,g$. The fundamental lemma on the edge then givesThey are zero bending moment and zero effective transverse shear at a free straight edge. The second includes the edge derivative of the twisting moment; omitting the mixed contribution would give the wrong coefficient.

Work to first order in the slope $w^{\prime }$ and take $B$ constant. The elastic beam is slender, its cross-sections remain plane and normal to its centreline, transverse shear deformation is neglected, and the loading is quasi-static. There are no distributed couples. In a non-contact interval the distributed force is $-p{\mathbf{e}}_z$ and there is no axial distributed force. These are the assumptions of the small-deformation Euler-Bernoulli beam equation, with an optional prescribed tension retained in transverse force balance.

Choose the internal section resultant on the right face of a segment aswhere $N$ is the transverse shear force, and take the section bending moment as $M{\mathbf{e}}_y$. Longitudinal and transverse force balance on a segment of length $dx$ give $T^{\prime }=0$ and $(T{w}^{\prime }+N{)}^{\prime }=p$. Balancing the torques about its left end gives $M^{\prime }=N$: the torque from the transverse section resultant is $-Ndx{\mathbf{e}}_y$. ThereforeWith this positive-$y$ bending moment convention, the linear relation between bending moment and curvature is $M=-B{w}^{\prime \prime }$. Thus $N=-B{w}^{\prime \prime \prime }$, and substitution givesThe tension need not be small compared with the transverse restoring terms, but products arising solely from finite deformation, such as the change of axial extension proportional to $(w^{\prime }{)}^2$, are outside this linear model. A separate axial constitutive constraint would be required to determine an unknown tension from fixed end positions.

In an open contact interval, the upward distributed normal contact force is $R$. At this order the smooth obstacle supplies no axial force; its vertical resultant changes the downward load to $p-R$. HenceFor nonadhesive unilateral beam contact, the obstacle can push but cannot pull, so $R\ge 0$. The accompanying constraints are $w\ge f$ and $(w-f)R=0$. A negative $R$ would require adhesion and would invalidate the proposed contact region. If $f$ is four times differentiable, the distributed normal contact force inside an open contact interval is $R=B{f}^{\prime \prime \prime \prime }-T{f}^{\prime \prime }+p$.

The jump conditions express a connected elastic beam with no concentrated horizontal force or applied couple. Continuity of tension, $[T]=0$, follows from horizontal force balance. Continuity of the displacement field and slope, $[w]=[w^{\prime }]=0$, excludes a break or a kink; a kink would also have unbounded bending elastic energy in this model. Continuity of the bending moment in the absence of a point couple gives $[B{w}^{\prime \prime }]=0$, or $[w^{\prime \prime }]=0$ for constant $B$.

The only allowed discontinuous section resultant is the transverse shear force. Integrating vertical force balance through an upward point force $R_i$ gives $[N]=-R_i$. Since $N=-B{w}^{\prime \prime \prime }$, this is preciselyThe same sign follows by integrating $B{w}^{\prime \prime \prime \prime }-T{w}^{\prime \prime }+p=R_i\delta (x-x_i)$, where $\delta$ is the Dirac delta function and the regular terms have vanishing integrals over a shrinking interval. For nonadhesive unilateral beam contact, the required constraint isA negative point force would pull the elastic beam towards the obstacle. At an endpoint where $R_i=0$, the third derivative is continuous as well.

Let the contact interval be $[-s,s]$, where $w=-H$. Its slope and curvature vanish. Continuity of the displacement field, slope and bending moment at $x=s$ givesOn the free interval $s<x<L$, the cubic polynomial satisfying these three conditions is $w=-H+C(x-s{)}^3$. Put $d=L-s$. The end boundary conditions becomeConsequentlyThis requires $\alpha \ge 3H/L$, with $s=0$ at the transition. The full displacement field isIt satisfies nonpenetration. In the open contact interval the distributed normal contact force is $R=B{w}^{\prime \prime \prime \prime }=0$, since the obstacle is flat and $p=T=0$. The normal contact force is concentrated at the two endpoints. At $x=s$, $w^{\prime \prime \prime }$ jumps from zero to $6H/d^3$; at $x=-s$, it jumps from $-6H/d^3$ to zero. Both upward point forces are thereforeAt $s=0$ they merge into the single midpoint point force of the preceding regime. These formulae belong to the small-slope model; interpreting arbitrarily large $\alpha$ as a finite physical rotation would exceed its assumptions.

Measure $x$ from the clamped end and let downward deflection $y$ be positive. In the small-slope elastic beam model, the bending moment has magnitude $M(x)=F(l-x)$ and $EI{y}^{\prime \prime }=M(x)$. Integrating with $y(0)=y^{\prime }(0)=0$ givesFor the rectangular cross-section, $I=w{h}^3/12=6.75\times 10^{-7}{\mathrm{m}}^4$, and thereforeRequiring $\delta \le 0.2\mathrm{m}$ with $F=1079.1\mathrm{N}$ yieldsThis calculation neglects beam self-weight as stipulated, and also transverse shear deformation and geometric nonlinearities. The end slope at the limit is $3\delta /(2l)=0.10$, so the small-slope approximation is reasonable for the intended estimate. Meeting a stiffness specification alone does not establish safe strength or fatigue life.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

Zero applied end force and the force balance ${\mathbf{n}}^{\prime }=0$ imply $\mathbf{n}(S)=0$ everywhere. The moment balance then reduces to ${\mathbf{m}}^{\prime }=0$.

Choose an untwisted straight configuration: all three directors are constant, ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$, and $\mathbf{u}=0$. The constitutive equation giveswhich is constant in space and therefore satisfies the moment balance. The required end bending moment is consequentlyIts sign opposes the positive intrinsic curvature of an elastic filament. If the initial position and frame from part (ii) are retained, this solution is simply $\mathbf{r}(S)=S{\mathbf{e}}_3$ with ${\mathbf{d}}_i={\mathbf{e}}_i$; the clamp supplies the opposite reaction moment at the other end.

A freely swimming body has neither an imposed end couple nor an imposed end transverse force. Its end bending moment is therefore zero, and the transverse force $S=G_x$ is also zero at both ends. These are the four free-end bending boundary conditions; they do not require the end displacement or slope to vanish.

Put $R=m_b{h}_{tt}+\mathcal{D}(m\mathcal{D}h)$, using the physically consistent sign. Integrating $G_{xx}=R$ with the two leading-end conditions givesThe two trailing-end conditions consequently requireThese free-end compatibility for a swimming beam conditions are total transverse force and torque balances. They explain why an arbitrary plausible prescribed $h$ does not suffice: two integration constants can satisfy only two end conditions unless the forcing already satisfies both compatibility integrals.

The missing freedom is the swimming recoil correction. An observed or prescribed deformation $h_d$ relative to the body must be supplemented by a lateral translation $Y(t)$ and a small yaw $\Theta (t)$:To identify $Y$ with the actual centre of mass displacement, first remove the body-mass-weighted mean from $h_d$. One may also remove its body-mass-weighted first moment about $x_c$ to define the yaw reference uniquely. This changes the decomposition of the observed motion, not the physical displacement.

For $\xi =x-x_c$, the recoil contribution to the forcing isInsert $R=R_d+R_{\mathrm{rec}}$ into the two compatibility integrals, using $1$ and $\xi$ as weights. This gives two coupled ordinary differential equations for the two recoil motions. Their principal acceleration matrix isFor a nondegenerate positive mass distribution it is a positive-definite matrix: its quadratic form at $(a,b)$ is ${\int }_0^L(m_b+m)(a+b\xi {)}^{2}dx>0$ unless $a=b=0$. Thus initial translation and yaw determine the recoil through these equations, and the corrected $h$ supplies the two compatibility conditions. The four free-end conditions determine the bending moment together with two rigid recoil motions; they do not overdetermine a physically complete swimming motion.

Choose the internal transverse force $S(x,t)$ so that the net internal transverse force on a slice is $[S(x+dx,t)-S(x,t)]$. Its linear momentum balance is thenThe coordinate $x$ labels body slices, so their lateral material acceleration is $h_{tt}$; the fluid transport operator $\mathcal{D}$ must not be substituted for this body acceleration.

Let $G$ be the total internal bending moment, including both passive elastic and active muscular contributions. With the compatible moment orientation, the net couple on the slice is $(G_x-S)dx$. The cross-sectional rotary inertia is smaller than the translational inertia by the square of the transverse-to-longitudinal length ratio, and is neglected in this slender-beam approximation. Slice angular momentum balance therefore gives $S=G_x$. Eliminating $S$ from the slice linear momentum balance provesThis equation is the balance law for an active bending beam for a swimming fish; a constitutive relation for the passive bending moment or the muscle actuation is not needed to derive it. With the force-on-body convention established above it becomesAs a check, constant $m$ and rigid lateral acceleration give the effective inertia $m_b+m$. Combining the printed positive force in part (i) with the minus sign in the beam equation would instead give $m_b-m$. That is the same sign defect, rather than a negative hydrodynamic added mass. If the positive quantity in part (i) is called $F_y$ as a force on the fluid, the beam equation must contain $+F_y$.

The overdamped filament bending equation is linear, so sinusoidal forcing has a sinusoidal steady response at the same frequency. Write $h=h_0\mathrm{Re}[e^{i\omega t}F(\eta )]$ with $\eta =x/\ell$. Substitution givesUsing the elastohydrodynamic penetration length from the previous part reduces this ordinary differential equation toThe boundary conditions become $F(0)=1$, $F^{\prime \prime }(0)=0$, and decay of $F$ and its derivatives at infinity. The derivative condition expresses zero bending moment at the driven end, rather than a clamped slope. This separation of variables constructs the periodic response; an unspecified initial shape can add a transient solution of the homogeneous evolution problem.

Write $\omega =kc$, $q=k(1+i)$, and $E=e^{qL}$. The local resistive-force theory force per unit length on the filament iswhere $\mathbf{t}=(X_s,Y_s)$ is the unit tangent and $\mathbf{v}=(X_t-U,Y_t-V)$ the material velocity relative to the fluid. Inextensibility gives $X_s=(1-Y_s^2{)}^{1/2}=1-Y_s^2/2+\cdots$. With $X(0,t)=0$, $X_t=-{\int }_0^s{Y}_s{Y}_{st}d{s}^{\prime }+\cdots$ is second order and has zero time average for a periodic prescribed stroke. The actual small-slope condition is $k|H|e^{kL}\ll 1$ for fixed $kL$; $kH\ll 1$ alone is not uniform when $kL$ grows.

At first order, the transverse force is $f_y=-\mu K_N(Y_t-V)$. Adding the spherical-head drag force $\mu K_{N}L\delta V$ and imposing zero total force givesEquivalently $V(t)=\mathrm{Re}[-i\omega H(E-1)e^{-i\omega t}/(qL(1+\delta ))]$. This form is useful for the subsequent time averages.

At second order, the axial resistive-force theory force isAdding the head drag force gives the head-drag correction to planar flagellar propulsion:since $\overline{X_t}=0$. Directly, $Y_s=kH{e}^{ks}(\cos \vartheta -\sin \vartheta )$ and $Y_t=kcH{e}^{ks}\sin \vartheta$, with $\vartheta =k(s-ct)$. HenceAlso ${\int }_0^L{Y}_{s}ds=Y(L,t)-Y(0,t)$, whose complex amplitude is $H(E-1)$. Using $\overline{\mathrm{Re}(a{e}^{-i\omega t})\mathrm{Re}(b{e}^{-i\omega t})}=\mathrm{Re}(a\overline{b})/2$, together with $\mathrm{Re}(-i\omega /q)=-c/2$, givesAs $|E-1{|}^2=e^{2kL}-2e^{kL}\cos kL+1$, the requested mean speed follows:Anisotropic resistance is essential: $\gamma =1$ removes propulsion at this order. The mean motion is opposite to the direction of wave propagation when the bracket is positive, since the head's axial velocity is $-U$.

For the basal bending moment of a planar flagellum, take positive moment to be the positive out-of-plane torque exerted on the distal filament at its base. The leading hydrodynamic torque on that filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$, so balance requiresAn explicit unsimplified time-dependent answer isThe first term follows from ${\int }_0^{L}s{e}^{qs}ds=[E(qL-1)+1]/q^2$. This is the total internal bending moment required by the hydrodynamic load. It equals the contractile moment when passive elasticity is neglected. If a specified passive moment $M_{\mathrm{pass}}$ also acts, the required active moment is $M_0-M_{\mathrm{pass}}$; no elastic modulus is supplied here.

The prescribed base coordinate needs a geometrical qualification. Literally, a coordinate measured from the attachment point must have $Y(0,t)=0$, whereas the printed waveform has $Y(0,t)=H\cos \omega t$. The preceding calculation reproduces the printed speed by treating that waveform as the prescribed coordinate, including its moving transverse offset. If the intention is instead a fixed attachment at the origin, replace it by $\tilde{Y}=H[e^{ks}\cos k(s-ct)-\cos \omega t]$. ThenThese follow by replacing $Y_t$ with $Y_t-Y_t(0,t)$ in the same force balance. The basal moment likewise uses ${\tilde{Y}}_t$ and $\tilde{V}$. Thus the printed kinematics and its claim that the origin is the actual attachment cannot both be imposed without this correction.

An Euler strut is a slender initially straight elastic column under axial compression. For a uniform bending stiffness $EI$ and a small transverse displacement $y(x)$, the second-order potential energy under a conservative compressive load $P$ isThe bending term is positive; shortening releases load potential. Stationarity yields $EI{y}^{(4)}+P{y}^{\prime \prime }=0$. Pinned ends impose $y(0)=y(L)=0$ and zero end bending moments, $y^{\prime \prime }(0)=y^{\prime \prime }(L)=0$. Substitution of $y=A\sin (n\pi x/L)$ givesThe straight configuration loses stability first in the longest admissible bending mode. The second moment of area $I=\int z^{2}dA$ is geometric, not a mass moment of inertia. Boundary conditions matter: a fixed-free column has effective length $2L$, while fixed-fixed ends have effective length $L/2$, changing the threshold by factors $1/4$ and $4$. Slenderness, elastic material response, alignment and the weak bending direction must be checked. Imperfections produce bending before the ideal threshold, and a short column can yield or crush first. Linear buckling theory does not predict an unlimited deflection or a complete postbuckling curve.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

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.