= Simply supported beam
{title2=$w=w_{xx}=0\text{ at both supports}$}
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 $0\leq x\leq L$ with transverse <displacement> $w(x)$ and positive <beam flexural rigidity> $B=EI$ has the support <boundary conditions>
$$
\boxed{w(0)=w(L)=0,\qquad w''(0)=w''(L)=0.}
$$
Here $E$ 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 $M=-Bw''$ 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 $w''=-Px/(2B)$. Integrate twice, using $w(0)=0$ and the symmetry condition $w'(L/2)=0$, to obtain
$$
w(x)=\frac{Px(3L^2-4x^2)}{48B}\quad(0\leq x\leq L/2),\qquad
w(x)=w(L-x)\quad(L/2\leq x\leq L).
$$
Thus the <midspan compliance of a simply supported beam> is
$$
\boxed{\frac{w(L/2)}{P}=\frac{L^3}{48EI}.}
$$
The endpoint <slopes> are $w'(0)=PL^2/(16EI)$ and $w'(L)=-PL^2/(16EI)$: they are generally nonzero. This support geometry and central-load formula are used to infer material stiffness from a <three-point bending test>.<Beam Displacements by David Roylance>[Introduction and Multiple integration, Example 1, pp. 1–3; equation 7 for midspan deflection]{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 $\Pi[w]=\tfrac12\int_0^L B(w'')^2\,dx-Pw(L/2)$. 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 contribution
$$
[Bw''\eta'-Bw'''\eta]_0^L.
$$
The terms multiplying $\eta$ vanish because the endpoint <displacements> are prescribed. Arbitrary endpoint values of $\eta'$ then require $Bw''=0$ 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 $PL^3/(192EI)$, 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.
\Image[/beam-support-compliance.png]
{title=Original central-load beam profiles: allowing support rotation gives four times the midspan deflection of clamping both ends.}
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.<Three Point Bend Test Fixtures by Instron>[Principle of Operation and midspan deflectometer description]{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 $I=bh^3/12=4.5\times10^{-11}\,\mathrm{m^4}$ and $EI=9\,\mathrm{N\,m^2}$. 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.
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