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The beam shear angle is the small-angle difference between centreline slope and material-section beam cross-section rotation:With transverse coordinate $z$ and kinematics $u_x=-z\theta (x)$, $u_z=w(x)$, the engineering shear strain is ${\partial }_z{u}_x+{\partial }_x{u}_z=\gamma$. The Euler-Bernoulli beam equation imposes $\gamma =0$; a shear-deformable model permits $w$ and $\theta$ to vary independently. The primary beam account distinguishes these kinematics and relates transverse force to the resulting shear measure.^[ref]

For an original local example, suppose shape measurements give $w_x=0.015$, while section markers give $\theta =0.010$ radians in the same sign convention. Then $\gamma =0.005$, about $0.286$ degrees. Setting the section rotation equal to the observed slope would erase this measured discrepancy by assumption.

In a simple linear shear model, the transverse force resultant is $V=S\gamma$, with effective beam shear rigidity $S=k_{s}GA$. Here $G$ is the shear modulus, $A$ the material area, and $k_s$ a section correction coefficient. Stipulating $G=1.0\mathrm{GPa}$, $A=20\mathrm{m}{\mathrm{m}}^2$ and $k_s=5/6$ gives $S=16667\mathrm{N}$ and $V\approx 83.3\mathrm{N}$. The correction coefficient accounts for using a reduced section model; it is not universally fixed at $5/6$.

A physical test must measure both the centreline shape and section orientation, for example by tracking markers during loading. Shape alone determines $w_x$ but cannot independently determine $\gamma$ without section-rotation information or an additional mechanical model. The comparison also requires compatible spatial resolution and angular measurement calibration.

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.