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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.

A simply supported beam is prevented from moving transversely at its two supports while its ends remain free to rotate. In the small-deflection Euler-Bernoulli beam equation, a uniform span $0\le x\le L$ with transverse displacement $w(x)$ and positive beam flexural rigidity $B=EI$ has the support boundary conditionsHere $E$ is Young's modulus and $I$ is the second moment of area about the bending axis. The second pair expresses zero applied endpoint bending moment, since $M=-B{w}^{\prime \prime }$ in the convention used here. It does not set the endpoint beam cross-section rotation or slope to 0. A clamped boundary condition instead constrains both displacement and slope.

For an original derivation, apply a downward central point force $P$ and take $w$ positive downward. Assume small elastic deflections, constant $B$ and no axial preload. Symmetry and force balance give an upward reaction $P/2$ at each support. On the left half, the bending moment is $M(x)=Px/2$, so $w^{\prime \prime }=-Px/(2B)$. Integrate twice, using $w(0)=0$ and the symmetry condition $w^{\prime }(L/2)=0$, to obtainThus the midspan compliance of a simply supported beam isThe endpoint slopes are $w^{\prime }(0)=P{L}^2/(16EI)$ and $w^{\prime }(L)=-P{L}^2/(16EI)$: they are generally nonzero. This support geometry and central-load formula are used to infer material stiffness from a three-point bending test.^[ref]

The moment-free natural boundary condition for a beam also follows from the principle of virtual work. For a central load, the total potential energy is $\Pi [w]=\frac{1}{2}{\int }_0^{L}B(w^{\prime \prime }{)}^{2}dx-Pw(L/2)$. An admissible virtual displacement $\eta$ satisfies $\eta (0)=\eta (L)=0$, while its endpoint first derivatives remain free. Two integrations by parts produce the boundary contributionThe terms multiplying $\eta$ vanish because the endpoint displacements are prescribed. Arbitrary endpoint values of ${\eta }^{\prime }$ then require $B{w}^{\prime \prime }=0$ there. The free end rotation and zero end moment are conjugate parts of the same variational boundary condition.

Clamping both ends changes the central-load deflection to $P{L}^3/(192EI)$, one quarter of the simply supported value for the same $P,L,E,I$. It introduces endpoint reaction moments; it does not merely rename the same support. The clamped-beam Green function provides the corresponding fourth-order boundary-value solution. The support conditions likewise determine the simply supported beam normal modes, so a static stiffness measurement and a vibration measurement must use the same mechanical support model when they are compared.

A three-point bending test places a specimen on two separated supports and applies a central force. A load cell measures force, and a midspan displacement readout measures response. A direct plunger and extensometer can measure specimen deflection separately from machine crosshead displacement, as in the published Instron fixture arrangement.^[ref] Fitting a proportional small-load segment estimates $E=L^3/(48IC)$, where $C$ is the measured incremental midspan compliance. Record support spacing, load position and cross-section dimensions with that estimate.

For stipulated $L=0.50\mathrm{m}$, rectangular width $b=20\mathrm{mm}$, thickness $h=3\mathrm{mm}$ and $E=200\mathrm{GPa}$, the second moment of area is $I=b{h}^3/12=4.5\times 10^{-11}{\mathrm{m}}^4$ and $EI=9\mathrm{N}{\mathrm{m}}^2$. An incremental $1\mathrm{N}$ central load gives $0.289\mathrm{mm}$ incremental deflection under the simply supported model, versus $0.0723\mathrm{mm}$ for the clamped model. Incremental readings remove an unchanged baseline deflection in this linear model; support settling, contact indentation and large rotations require separate interpretation.