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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.

The beam shear correction coefficient $k_s$ adjusts a reduced beam model's shear response for the nonuniform shear stress over a material section. With shear modulus $G$, area $A$, transverse force resultant $V$ and generalized beam shear angle $\gamma$,where $S$ is effective beam shear rigidity. Bower's primary account calls this the Timoshenko shear coefficient and explains that its value depends on geometry, Poisson's ratio and how the approximate theory is matched to the underlying solid.^[ref]

An original energy-matching derivation makes the correction precise for one approximation. Assume uniform $G$ and a chosen shear-stress field $\tau$ with resultant $V={\int }_A\tau dA$. The shear elastic energy per unit beam length and its reduced representation areThusThe Cauchy-Schwarz inequality gives $V^2\le A{\int }_A{\tau }^{2}dA$, hence $k_s\le 1$ for this definition. Equality requires uniform stress. This bound belongs to the stated energy-matching definition rather than every convention called a shear correction coefficient.

For the parabolic rectangular stress approximation used in effective beam shear rigidity, the squared-stress integral is $6V^2/(5A)$, giving $k_s=5/6$. Choose $G=80\mathrm{GPa}$, $A=100\mathrm{m}{\mathrm{m}}^2$ and $V=1000\mathrm{N}$. Then $S=6.667\times 10^6\mathrm{N}$ and $\gamma =1.5\times 10^{-4}$ radians. Omitting the correction gives $1.25\times 10^{-4}$, underestimating this generalized shear angle by one sixth.

The generalized angle also differs from the peak local shear strain. Under the same parabolic approximation, peak stress is $3V/(2A)$, giving peak strain $1.875\times 10^{-4}$. The values describe different reductions of the same chosen stress field. Identifying $k_s$ from a bending test therefore requires the section model and displacement readout, not merely a material modulus. For a three-point bending test, the shear deflection contribution varies as $PL/(4k_{s}GA)$, whereas bending varies as $P{L}^3/(48B)$; a span comparison can help separate them.

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.