= Effective beam shear rigidity
{title2=$S=k_sGA$}
\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:
$$
V=S\gamma,\qquad S=k_sGA.
$$
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 $\mathrm{N\,m^2}$. 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.<Applied Mechanics of Solids by A. F. Bower>[§8.7.2, Summary of beam theory, Timoshenko force relation and shear-coefficient paragraph]{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 approximation
$$
\tau(z)=\frac{3V}{2A}\left(1-\frac{4z^2}{h^2}\right),\qquad -h/2\leq z\leq h/2.
$$
It has resultant $\int_A\tau\,dA=V$. Its shear <elastic energy> per unit beam length is
$$
U_s'=\frac1{2G}\int_A\tau^2\,dA
=\frac{3V^2}{5GA}.
$$
Equating 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 $U_s=P^2L/(8S)$ and its contribution to central <displacement> is $\partial U_s/\partial P=PL/(4S)$. Combining it with the bending response gives
$$
\delta=\frac{PL^3}{48B}+\frac{PL}{4S},\qquad
\frac{\delta_{\rm shear}}{\delta_{\rm bending}}=\frac{12B}{SL^2}.
$$
The shear contribution becomes relatively more important as the span decreases.
For stipulated $B=2\,\mathrm{N\,m^2}$, $S=20000\,\mathrm N$, $L=0.10\,\mathrm m$ 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.
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