= Solution
For a slender <elastic beam>, the <Euler-Bernoulli beam equation> assumes small <strain>, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the <neutral axis> has longitudinal <strain> $\epsilon_{xx}=-y\kappa$, where $\kappa$ is the centreline <curvature>. <Hooke's law> gives $\sigma_{xx}=-Ey\kappa$, with <Young's modulus> $E$.
Choose the internal bending-moment sign so that $M=-\int y\sigma_{xx}\,dA$. Then
$$
\boxed{M=EI_A\kappa,\qquad I_A=\int_Ay^2\,dA.}
$$
Here $I_A$ is the <second moment of area>, with units $\mathrm{m^4}$, not the mass <moment of inertia> of a rotating body. For a uniform material under pure bending, zero axial resultant places the <neutral axis> through the section's <centroid>.
For small slopes, $\kappa\simeq w''(x)$. Combining the moment-curvature relation with transverse <force> and moment balance gives, with a consistent load sign convention,
$$
\boxed{\frac{d^2}{dx^2}\left(EI_A\frac{d^2w}{dx^2}\right)=q(x).}
$$
For constant stiffness this becomes $EI_Aw^{(4)}=q$. The <bending stiffness> $EI_A$ has units $\mathrm{N\,m^2}$. <Boundary conditions> specify displacement and slope at a clamped end, or <bending moment> and shear <force> at a free end.
For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell-x)$, and integration with $w(0)=w'(0)=0$ gives
$$
w(x)=\frac{Fx^2(3\ell-x)}{6EI_A},\qquad
\boxed{w(\ell)=\frac{F\ell^3}{3EI_A}.}
$$
The free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=bh^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic <strains> or short thick beams where shear deformation matters.
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