= Beam shear correction coefficient
{title2=$k_s$}
The \b[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> $\gamma$,
$$
S=k_sGA,\qquad V=S\gamma,
$$
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.<Applied Mechanics of Solids by A. F. Bower>[§8.7.2, Timoshenko force relation and following shear-coefficient paragraph]{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 are
$$
U_s'=\frac1{2G}\int_A\tau^2\,dA=\frac{V^2}{2k_sGA}.
$$
Thus
$$
k_s=\frac{V^2}{A\int_A\tau^2\,dA}.
$$
The <Cauchy-Schwarz inequality> gives $V^2\leq A\int_A\tau^2\,dA$, hence $k_s\leq1$ 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 $G=80\,\mathrm{GPa}$, $A=100\,\mathrm{mm^2}$ and $V=1000\,\mathrm N$. Then $S=6.667\times10^6\,\mathrm N$ and $\gamma=1.5\times10^{-4}$ radians. Omitting the correction gives $1.25\times10^{-4}$, 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 $1.875\times10^{-4}$. 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 $PL/(4k_sGA)$, whereas bending varies as $PL^3/(48B)$; a span comparison can help separate them.
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