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Beam compliance from two support spans separates a cubic span-dependent beam response from a span-independent contribution to the measured mechanical compliance. Assume the same uniform specimen, unchanged machine contribution $C_0$ and a small-deflection central-load model:Here $B$ is beam flexural rigidity and $L_1\ne L_2$. The midspan compliance of a simply supported beam contains the basic numerical fit, while the beam source derives the cubic dependence.^[ref]

An original uncertainty example shows why distinct spans can still be too close. Let the actual model have $B=10\mathrm{N}{\mathrm{m}}^2$, hence $a=1/480{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$. Assign independent compliance standard uncertainties $u(C_1)=u(C_2)=1\mu m/N$, and neglect span uncertainty for this example. Since the slope estimate is linear in the readings,Spans $0.200$ and $0.210\mathrm{m}$ give $u(a)\approx 0.00112{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$, about54percent of the actual coefficient. Using $0.200$ and $0.300\mathrm{m}$ instead gives about3.57percent with the same readout uncertainties. The close-span subtraction strongly amplifies those errors. At the larger relative uncertainty, simply treating $B=1/(48a)$ as approximately linear is unreliable.

This conditioning of two-span beam compliance inference concerns measurement precision, separately from model adequacy. If shear adds a term proportional to $L$, then $C(L)=C_0+a{L}^3+bL$ has three unknown coefficients that two readings cannot identify. Changing contact or fixture compliance can likewise violate the assumed constant $C_0$. Further spans and an independent deflection readout test those assumptions; two fitted points alone do not.

Beam root rotational compliance is the incremental rotation of a beam's attachment relative to its supporting base per increment of applied root torque. For a linear rotational restraint with stiffness $K_{\theta }>0$,The angle is measured in radians, so the compliance can be reported in $rad/(\mathrm{N}\mathrm{m})$. The root's transverse position is held fixed in this model. The rotational spring boundary condition for a beam implements this finite restraint; a clamped boundary condition is the zero-compliance limit. End moments and section rotations are conjugate quantities in beam virtual work.^[ref]

For an original static derivation, apply a transverse tip force $P$ to a beam of length $L$ and constant beam flexural rigidity $B$. The root moment has magnitude $PL$, giving ${\theta }_0=C_{\theta }PL$. In the small-angle approximation, root rotation contributes $L{\theta }_0$ to tip displacement. Bending relative to that rotated attachment gives the usual cantilever contribution, soThis derivation assumes a linear elastic beam, no root translation and an effectively rigid supporting base. The second term is fixture motion rather than a change in material Young's modulus.

For stipulated $L=0.30\mathrm{m}$, $B=6\mathrm{N}{\mathrm{m}}^2$, $C_{\theta }=0.020rad/(\mathrm{N}\mathrm{m})$ and $P=1\mathrm{N}$, the root rotates through $0.006$ radian. Root motion contributes $1.8\mathrm{mm}$, while bending contributes $1.5\mathrm{mm}$. Attributing the entire $3.3\mathrm{mm}$ to a perfectly clamped beam would infer $B=2.73\mathrm{N}{\mathrm{m}}^2$ instead of the stipulated value six.

A proposed measurement uses a known force and lever arm to apply a root moment, with optical markers on the attachment and on the supporting base to measure their relative angle. A separate displacement channel can detect unwanted root translation. Fit the small-load slope of angle against moment; using only the beam-tip displacement would mix attachment rotation with beam bending. Load reversal tests whether one linear compliance adequately describes the fixture or whether backlash and hysteresis require another model.

Bending changes the curvature of an elastic beam, rod or sheet. In the elementary beam model, material on opposite sides of the neutral axis stretches and shortens. Their longitudinal normal stresses can balance to give zero resultant force while still transmitting a nonzero bending moment. A bent ruler illustrates why resistance to changing shape cannot be described by axial stretching alone.

For a homogeneous slender elastic beam in small-deflection pure bending, take $x$ along the beam and $y$ perpendicular to its centroidal neutral axis. Let $\kappa$ be the signed centreline curvature. The plane-section kinematics and Hooke's law giveHere $E$ is Young's modulus, taken constant over the section. Choosing the positive bending moment as $M=-{\int }_{A}y{\sigma }_{xx}dA$ yieldsThus the second moment of area $I$, rather than just the cross-sectional area, controls bending stress. The beam flexural rigidity is $EI$. Zero axial resultant requires ${\int }_{A}ydA=0$, locating the neutral axis at the section centroid under these assumptions.^[ref]

A four-point bending test makes the constant-moment region experimentally accessible. Two outer supports carry a specimen while two inner contacts apply the load. The Illinois teaching laboratory uses adjustable inner and outer spans on an Instron 4400 frame and records load and deflection electronically; it also tests notched specimens to examine stress concentration.^[ref]

For an original symmetric loading example, place the outer supports at $x=0,L$ and apply total downward force $P$ equally at $x=a,L-a$, with $0<a<L/2$. Each support reaction is $P/2$. Between the inner contacts the transverse shear force is zero and the bending moment is constant, $M=Pa/2$. This central interval realizes the ideal beam conditions for pure bending away from contact disturbances. A three-point bending test instead has its maximum moment at the single central loading contact.

Choose $L=160\mathrm{mm}$, $a=40\mathrm{mm}$, $P=40\mathrm{N}$, rectangular width $b=10\mathrm{mm}$ and thickness $h=4\mathrm{mm}$. Then $I=b{h}^3/12=53.333\mathrm{m}{\mathrm{m}}^4$ and $M=0.800\mathrm{N}\mathrm{m}$. With stipulated $E=70\mathrm{GPa}$, the model givesThese independently chosen numbers describe an assumed elastic specimen, rather than measurements of a particular material.

The same relation can infer a flexural modulus from a measured bending moment and curvature, provided the geometry and beam assumptions are adequate. Added axial force contributes a uniform longitudinal normal stress and can shift the zero-stress line. Short beams require attention to transverse shear strain; large rotations require a geometrically nonlinear description. The small-deflection Euler-Bernoulli beam equation and the more general Euler elastica cover different levels of that modeling.

Conditioning of two-span beam compliance inference describes how perturbations in two measured compliances and support spans affect the inferred bending coefficient. In the model$C_0$ represents an unchanged series contribution and $B$ is beam flexural rigidity. The cubic beam term follows from the midspan compliance of a simply supported beam.^[ref] The subtraction is sensitive when the spans are close, or when the beam-response difference is small compared with readout errors.

Let $D=L_2^3-L_1^3\ne 0$ and temporarily treat the spans as exact. The sensitivity coefficients are $\partial \hat{a}/\partial C_1=-1/D$ and $\partial \hat{a}/\partial C_2=1/D$. Consequently,This is the standard measurement uncertainty propagation rule applied to a linear combination; the covariance term matters when the readings share errors.^[ref]

For an original numerical comparison, stipulate $B=10\mathrm{N}{\mathrm{m}}^2$, so $a=1/480{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$, and independent standard uncertainties $u(C_i)=10^{-6}m/N$. Spans $0.200$ and $0.210\mathrm{m}$ give $D=0.001261{\mathrm{m}}^3$ and $u(\hat{a})=0.001122{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$, about 54 percent of $a$. Increasing the second span to $0.300\mathrm{m}$ gives $D=0.019{\mathrm{m}}^3$ and $u(\hat{a})=0.00007443{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$, about 3.57 percent. The wider spacing makes the same compliance readout precision much more informative.

Span uncertainty enters through additional sensitivity coefficients:For uncorrelated span readings, their first-order contribution to $u^2(\hat{a})$ is $9{\hat{a}}^2[L_1^4{u}^2(L_1)+L_2^4{u}^2(L_2)]/D^2$, before any compliance/span correlations are included. An identical additive readout offset cancels from $C_2-C_1$, whereas a changing offset does not.

A practical span comparison holds the specimen, contacts, loading position and displacement channel as consistent as possible. Close spans amplify measurement noise; simply choosing distant spans can instead expose a failure of the assumed model. Shear produces a term proportional to $L$, and changing contacts can change $C_0$. A third span tests those assumptions. Finally, inversion to $B$ is strongly nonlinear near $a=0$: the 54-percent case cannot be adequately summarized by a small symmetric relative uncertainty on $B$.

Effective beam shear rigidity is the coefficient $S$ relating a transverse shear-force resultant $V$ to the beam shear angle $\gamma$ in a reduced linear beam model:Here $G$ is shear modulus, $A$ is material area and $k_s$ is a beam shear correction coefficient. The coefficient $S$ has units of force, whereas beam flexural rigidity has units $\mathrm{N}{\mathrm{m}}^2$. A shear-deformable beam permits section rotation to differ from centreline slope; Bower gives the corresponding force relation and explains the correction coefficient's dependence on section geometry and the approximation used.^[ref]

An original energy calculation explains one familiar correction. For a homogeneous rectangle of width $b$, thickness $h$ and area $A=bh$, adopt the classical parabolic shear-stress approximationIt has resultant ${\int }_A\tau dA=V$. Its shear elastic energy per unit beam length isEquating this to the reduced-model energy $V^2/(2S)$ gives $S=5GA/6$. Thus $k_s=5/6$ follows from this particular energy-matching approximation; it is not a universal material constant or an exact coefficient for every beam theory.

The effect on a three-point bending test can be derived from that reduced energy. For a simply supported beam under central force $P$, the magnitude of shear force is $P/2$ on each half. Hence $U_s=P^{2}L/(8S)$ and its contribution to central displacement is $\partial U_s/\partial P=PL/(4S)$. Combining it with the bending response givesThe shear contribution becomes relatively more important as the span decreases.

For stipulated $B=2\mathrm{N}{\mathrm{m}}^2$, $S=20000\mathrm{N}$, $L=0.10\mathrm{m}$ and $P=10\mathrm{N}$, the two contributions are $0.1042$ and $0.0125\mathrm{mm}$. Shear adds twelve percent of the bending deflection. A span series with a load cell and direct midspan displacement readout can distinguish cubic bending dependence from linear shear dependence, provided contact and fixture contributions are also controlled. Inferring $G$ from the response additionally requires the section geometry and correction model.

The midspan compliance of a simply supported beam is the incremental central deflection per incremental central load, with the support geometry fixed. For a uniform, small-deflection Euler-Bernoulli beam equation model without axial preload,Its units are length per force. The simply supported beam contains the derivation; the experimental use is to infer beam flexural rigidity or, with known geometry, Young's modulus.^[ref]

An original two-span example separates beam compliance from an assumed constant machine contribution. Suppose the same uniform specimen is tested over spans $0.20$ and $0.30\mathrm{m}$. A machine-travel readout gives $0.060$ and $0.155mm/N$. Assumewhere $C_0$ is unchanged between the tests. Subtracting the readings givesHence $B=1/(48k)\approx 4.167\mathrm{N}{\mathrm{m}}^2$ and $C_0=0.020mm/N$. The inferred specimen compliances are $0.040$ and $0.135mm/N$, with ratio $3.375=(0.30/0.20{)}^3$. The raw readings have a different ratio because they include $C_0$.

This beam compliance from two support spans relies on the unchanged-series-compliance assumption. Changing supports, contact indentation or shear contributions can invalidate it; two points fit its two coefficients without independently validating the model. A third span can test the cubic dependence. An incremental slope also differs from simply dividing a total displacement, including an initial offset, by the current force.

The moment-free natural boundary condition for a beam is the zero-end-moment condition obtained when an endpoint can rotate freely and no external couple is applied there. For positive beam flexural rigidity $B$, it is $B{w}_{xx}=0$. This differs from prescribing the endpoint beam cross-section rotation.

The simply supported beam gives the full principle of virtual work derivation. At its right support, the admissible virtual displacement $\eta$ obeys $\eta (L)=0$, but ${\eta }^{\prime }(L)$ is arbitrary. The remaining bending boundary contribution is $B(L)w^{\prime \prime }(L){\eta }^{\prime }(L)$. Its vanishing for every permitted end rotation requires $B(L)w^{\prime \prime }(L)=0$. Beam virtual work pairs end moments with rotation variations and forces with displacement variations.^[ref]

An original modified boundary example shows why the missing external couple matters. Define a signed right-end couple $m_L$ by its external virtual work $m_L{\eta }^{\prime }(L)$. The boundary coefficient then becomes $B{w}^{\prime \prime }-m_L$, so the free-rotation condition is $B(L)w^{\prime \prime }(L)=m_L$. With $B=2\mathrm{N}{\mathrm{m}}^2$ and $m_L=0.010\mathrm{N}\mathrm{m}$, the endpoint second derivative is $0.005{\mathrm{m}}^{-1}$, not zero. This sign definition fixes the couple convention explicitly.

Clamping instead sets ${\eta }^{\prime }(L)=0$; that restriction permits an unknown reaction moment rather than forcing it to vanish. Finite rotational restraint leads to a rotational spring boundary condition for a beam. These alternatives distinguish a prescribed motion from its conjugate natural load condition.

A rotational spring boundary condition for a beam lets an endpoint rotate while imposing a restoring torque proportional to that rotation relative to its support. Let $K_{\theta }>0$ be the spring stiffness, with zero rest angle. Its stored elastic energy is $K_{\theta }{\theta }^2/2$, where the small beam cross-section rotation is $\theta =w_x$.

At a right endpoint whose transverse position is fixed, add this energy and the applied-couple potential $-m_L\theta (L)$ to the bending energy. The coefficient of the free rotation variation isHere $B=EI$ is beam flexural rigidity, and $m_L$ is defined by external virtual work $m_L\delta \theta (L)$. End moment and rotation are conjugate variables in the beam principle of virtual work.^[ref] The spring term and the following example are independently derived. At the left endpoint, the bending boundary term has the opposite sign: with no applied end couple, $-B(0)w^{\prime \prime }(0)+K_{\theta }\theta (0)=0$.

For an original fixture model, hold the left endpoint position fixed, restrain its rotation by the spring, and apply a downward end force $P$ at distance $L$. The root moment magnitude is $PL$, giving root rotation $PL/K_{\theta }$. Bending relative to that rotated root contributes $P{L}^3/(3B)$ to the tip displacement; rigid root rotation contributes $P{L}^2/K_{\theta }$. ThusWith $P=1\mathrm{N}$, $L=0.20\mathrm{m}$, $B=2\mathrm{N}{\mathrm{m}}^2$ and $K_{\theta }=20\mathrm{N}m/rad$, these contributions are $1.333$ and $2.000\mathrm{mm}$, giving $3.333\mathrm{mm}$ total. Root rotation is $0.010$ radian, within the stipulated small-rotation model.

As $K_{\theta }$ increases without bound, the root approaches a clamped boundary condition. At zero stiffness this particular end-loaded arrangement becomes a rotational mechanism; it has no finite static deflection for nonzero $PL$. A small-couple fixture check can estimate $K_{\theta }$ from measured moment and angle. The resulting beam root rotational compliance is distinct from deformation of the beam material.

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.

A uniform elastic beam with endpoint conditions $y=y_{xx}=0$ has normal modes $\sin (n\pi x/L)$ for positive integers $n$. Substitute $y=X{e}^{-i\omega t}$ into $EI{y}_{xxxx}+{\rho }_{\ell }{y}_{tt}=0$ and impose the four endpoint conditions to obtain the displayed spectrum. Here ${\rho }_{\ell }$ is mass per unit length, $I$ is a second moment of area, and $EI$ is beam flexural rigidity. frequencies proportional to $n^2$ distinguish flexural waves from the nondispersive ideal stretched string.