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Beam cross-section rotation is the signed change in orientation of a material section in the bending plane. The Euler-Bernoulli beam equation assumes that sections remain plane and perpendicular to the deformed centreline. With a consistent angle convention, the section rotation therefore agrees with the centreline tangent angle; for a graph $w(x)$,The last step is a small-angle approximation, with the angle measured in radians. This kinematic relation underlies the axial bending strain, rather than being an extra condition imposed by every support.^[ref]

For an original geometric profile, take $w(x)=a{x}^2$ with $a=0.10{\mathrm{m}}^{-1}$. At $x=0.50\mathrm{m}$, the slope is $0.10$, while the exact tangent angle is $\arctan (0.10)\approx 0.09967$ radians. Replacing it by $0.10$ overestimates the angle by about 0.33 percent. This is a chosen shape illustrating the approximation, not a measured equilibrium profile.

The beam shear angle records a possible difference between centreline slope and section rotation in a model that permits transverse shear strain. The Euler-Bernoulli kinematics set that difference to zero; a short or shear-compliant specimen can require a different model. A simply supported beam leaves its end section rotation free, whereas a clamped boundary condition prescribes it. Neither zero transverse displacement alone nor zero endpoint bending moment requires zero rotation.

Bending-stiffness bias in the vibroscope area estimator occurs when bending resistance raises a strand's frequency but its area is inferred with a flexible-string model. In the independently derived simply supported beam model in vibroscope area estimator, let $B=EI$ be the bending rigidity, $T>0$ the tension and $L$ the length. The dimensionless ratio and area error areHere $A$ is the actual area in the model's measurement configuration. The parent article derives this relation; no circular-area assumption is needed for the ratio once $B$ is specified.

For an original example, choose $r=0.25$ and $A=80\mu {\mathrm{m}}^2$. The uncorrected estimate is $64\mu {\mathrm{m}}^2$, twenty percent below the model area. Doubling only $L$, with the same $B$, $T$ and material configuration, divides $r$ by four. The new ratio is $1/1.0625=16/17$, giving $\hat{A}\approx 75.29\mu {\mathrm{m}}^2$, about 5.88 percent low. A length-dependent discrepancy is therefore a useful diagnostic to investigate, rather than evidence that the material area itself changed.

The boundary conditions are essential. Simply supported endpoints impose zero displacement and zero bending moment. The companion experimental theory instead treats fully clamped endpoints, with zero displacement and slope, and derives its own frequency and deformation corrections.^[ref] The displayed correction cannot be transferred unchanged to that apparatus.

Two-mode separation of tension and bending stiffness supplies another check within the same simply supported model. A model correction for bending bias also remains distinct from the standard input uncertainties treated in vibroscope area uncertainty.

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.

Idler bearing load sharing determines how an idler support reaction is divided among the shaft's supporting bearing elements. Zero net torque on an ideal steady idler gear does not remove its mesh-induced transverse force. The bearing loads additionally depend on axial positions and the supported-shaft arrangement. Manufacturer guidance separates a radial load between two bearings from an overhung load and treats gear loads as applications of those cases.^[ref]

Model the shaft as a simply supported beam with supports A and B at axial coordinates $z=0$ and $z=L$. Let a gear resultant $P$ act upward at $z=a$, and take upward reactions as positive. Static equilibrium requiressoFor a200N mesh resultant at $a=40\mathrm{mm}$ between supports100mm apart, the downward reactions are120N and80N. Equal division occurs only for the centered load in this symmetric ideal support model.

For overhung idler loading at $a=130\mathrm{mm}$ with the same support span and resultant, $R_A=+60\mathrm{N}$ and $R_B=-260\mathrm{N}$. The farther support's force reverses direction, and the nearer bearing carries more than the applied resultant. This does not violate force balance: $60-260+200=0$. If a particular support cannot provide the required reaction direction, the assumed support model must change.

For loads in two transverse directions, solve each component balance before forming the resultant at each bearing. Helical gears add gear axial force; allocating that thrust requires the actual bearing constraints and cannot be inferred from the transverse lever rule. More than two effective supports generally require deformation compatibility as well as equilibrium. These calculations estimate support forces, rather than directly determining bearing life or internal rolling-element contact loads.

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 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 three-point flexural test loads a specimen through one central contact while two separated supports carry the reactions. In a small-deflection simply supported beam model, an incremental central force and the resulting midspan displacement determine the midspan compliance of a simply supported beam. With a known second moment of area, that response yields a flexural modulus.^[ref]

The apparatus must distinguish specimen deflection from machine travel. The manufacturer's fixture description uses adjustable lower anvils and a central loading member; a direct midspan plunger can transmit deflection to an extensometer. A load cell supplies the force record.^[ref] The crosshead displacement can also contain deformation of the frame and contacts.

For an original readout example, stipulate span $L=0.120\mathrm{m}$, width $b=12\mathrm{mm}$ and thickness $h=4\mathrm{mm}$, so $I=6.4\times 10^{-11}{\mathrm{m}}^4$. Suppose a linear loading interval gives direct midspan compliance $C_s=0.020mm/N$, while machine travel gives $C_m=0.025mm/N$. Applying the same beam relation to the two readouts givesThe apparent modulus is twenty percent lower. The numbers are chosen to isolate the displacement-channel effect; they do not characterize a particular fixture. Inferring a material Young's modulus further assumes the beam's linear, homogeneous bending model is adequate. Contact indentation, support motion and transverse shear strain can otherwise alter the measured compliance. The parent simply supported beam supplies the deflection derivation and support-comparison figure.

Two-mode separation of tension and bending stiffness uses two identified transverse eigenmodes to separate coefficients that a single resonance cannot determine independently. For the uniform simply supported beam model derived in vibroscope area estimator,Thus plotting $f_n^2/n^2$ against $n^2$ gives an intercept associated with tension and a slope associated with bending. Frequencies alone identify these ratios; determining dimensional $T$ and $B$ also requires the remaining scale quantities.

An original exact example takes $f_1=800\mathrm{Hz}$ and $f_2=\sqrt{3328000}\mathrm{Hz}\approx 1824.28\mathrm{Hz}$. The two transformed values are $640000$ and $832000\mathrm{H}{\mathrm{z}}^2$ at $n^2=1$ and $4$. Their difference givesHence $r=b/a=1/9$. Using the first mode as if $b=0$ would use $640000$ instead of $576000$ for the tension coefficient; the corresponding area correction is a factor $10/9$ under this model.

Two observations fit two free coefficients, so that fit alone does not validate the model. A third identified mode should satisfy $f_3^2=9(a+9b)=10368000\mathrm{H}{\mathrm{z}}^2$, or $f_3\approx 3219.94\mathrm{Hz}$, in this example. Repeated measurements and another mode can test consistency.

For $T>0$ and $B\ge 0$, the same model requires $2\le f_2/f_1<4$. A ratio outside that range contradicts at least one assumption or the mode assignment. Actual clamped grips require the clamped-end frequency model described in the companion theory.^[ref] Do not label arbitrary neighbouring resonance peaks as modes one and two without additional evidence.