= Solution
<Elasticity> describes recoverable deformation. For small displacements $\mathbf u(\mathbf x)$, the <infinitesimal strain tensor> is the symmetric part of the displacement gradient,
$$
\boxed{\epsilon_{ij}=\frac12(\partial_i u_j+\partial_j u_i).}
$$
Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is $\gamma_{xy}=2\epsilon_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.
The <stress tensor> specifies the <force> per unit area on each oriented plane: $\mathbf t(\widehat{\mathbf n})=\boldsymbol\sigma\widehat{\mathbf n}$. Positive normal stress is tensile in the convention used here; tangential traction gives <shear stress>. In the absence of internal couples, angular-momentum balance makes the <stress tensor> symmetric. Force balance gives $\nabla\cdot\boldsymbol\sigma+\mathbf f=0$ in static equilibrium, where $\mathbf f$ is body <force> per unit volume.
<Linear elasticity> relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:
$$
\boxed{\sigma_{ij}=\lambda\epsilon_{kk}\delta_{ij}+2G\epsilon_{ij}.}
$$
Here $G$ is the <shear modulus> and $\lambda$ is the other <Lamé parameter>. For uniaxial loading, <Young's modulus> is $E=\sigma_{xx}/\epsilon_{xx}$, with lateral contraction described by <Poisson's ratio> $\nu=-\epsilon_{yy}/\epsilon_{xx}$. The <bulk modulus> is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants are
$$
\boxed{E=2G(1+\nu)=3K(1-2\nu),\qquad K=\lambda+\frac{2G}{3}.}
$$
Stable isotropic elastic <strain energy density> requires $G>0$ and $K>0$, implying $-1<\nu<1/2$. The incompressible limit is approached as $\nu\to1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.
The elastic <energy density> is $w=\boldsymbol\sigma:\boldsymbol\epsilon/2$. In the isotropic linear model this becomes $\lambda(\operatorname{tr}\boldsymbol\epsilon)^2/2+G\boldsymbol\epsilon:\boldsymbol\epsilon$. The quadratic form explains why reversible loading stores <energy> and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.
For a slender beam described by the <Euler-Bernoulli beam equation>, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,
$$
\boxed{M=EI_A\kappa,\qquad I_A=\int_A y^2\,dA.}
$$
$I_A$ is a <second moment of area>, with units of length to the fourth power; it is not a mass <moment of inertia>. For a rectangular section of width $b$ and thickness $h$, $I_A=bh^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w''$, and the <Euler-Bernoulli beam equation> is $EI_Aw''''=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F\ell^3/(3EI_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.
In <torsion of a solid circular shaft>, a cross-section rotated by angle $\theta(x)$ has shear strain $r\theta'(x)$. Therefore
$$
\tau(r)=Gr\theta',\qquad \boxed{\mathcal T=GJ\theta',\quad J=\int_A r^2\,dA=\frac{\pi a^4}{2}.}
$$
Here $J$ is the <polar second moment of area>. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal T\ell/(GJ)$, and its largest shear stress is $\mathcal T a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.
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