= Solution
<stress> describes internal <force> per <area>. A cut with unit normal $\hat{\mathbf n}$ has <traction> $\mathbf t=\boldsymbol\sigma\hat{\mathbf n}$; its normal component is $\hat{\mathbf n}\cdot\boldsymbol\sigma\hat{\mathbf n}$ and its remainder is shear <traction>. <force balance> constrains the <stress> <divergence>, while ordinary local angular-momentum balance makes the <Cauchy stress tensor> symmetric in a medium without body couples.
<strain> describes deformation rather than <force>. For small <displacement> <gradients>,
$$
\epsilon_{ij}=\frac12(\partial_i u_j+\partial_j u_i),\qquad
\frac{\Delta V}{V}=\epsilon_{kk}.
$$
The antisymmetric displacement-gradient part is a rigid rotation to first order and contributes no infinitesimal <strain>. Engineering <shear strain> is $\gamma_{xy}=2\epsilon_{xy}$, so it must not be confused with the tensor component. <stress> has units of <pressure>; <strain> is dimensionless.
An isotropic linear elastic solid obeys
$$
\sigma_{ij}=\lambda\epsilon_{kk}\delta_{ij}+2G\epsilon_{ij},\qquad
w=\tfrac12\lambda(\epsilon_{kk})^2+G\epsilon_{ij}\epsilon_{ij}.
$$
Two independent moduli are required. The <shear modulus> $G$ and <bulk modulus> $K=\lambda+2G/3$ must be positive for stability. <Young's modulus> and Poisson ratio satisfy
$$
E=2G(1+\nu)=\frac{9KG}{3K+G},\qquad
K=\frac{E}{3(1-2\nu)},\qquad -1<\nu<\tfrac12.
$$
For a tensile specimen with free side faces, $\sigma_{xx}=F/A$ gives $\epsilon_{xx}=\sigma_{xx}/E$ and $\epsilon_{yy}=\epsilon_{zz}=-\nu\sigma_{xx}/E$. Pure shear gives $\tau=G\gamma$; hydrostatic loading gives $\Delta V/V=-p/K$. The incompressible limit has $\nu\to1/2$ at finite shear stiffness, not vanishing shear resistance. Crystals can be anisotropic and need a fourth-rank elastic stiffness tensor instead of the isotropic two-constant law.
<elastic deformation> is recoverable on unloading; it need not remain linear at large <strain>. Beyond yield, irreversible rearrangements such as <dislocation> motion cause plastic <strain>, often followed by hardening. Ductile samples can neck before fracture; brittle failure has much less plastic deformation. Engineering <stress> uses initial <area>, while true <stress> uses the changing instantaneous <area>. A decreasing engineering <stress> during <necking> therefore need not mean that local true <stress> decreases. <stress> concentrations, flaws and loading geometry can determine failure before a uniform-material strength is reached. Viscoelastic materials also show rate dependence, creep and <stress relaxation>; their losses cannot be described by a static Hooke law alone.
A tensile test should measure <force> and gauge-length extension independently, correct machine compliance, align the grips, and control <strain> rate and <temperature>. Unloading distinguishes recoverable <strain> from a residual plastic extension. Separate shear or hydrostatic tests determine moduli that one axial measurement does not fix. Compression specimens need <friction> and buckling checks. Work per <volume> is $\int\sigma_{ij}d\epsilon_{ij}$; the recoverable part is <elastic energy>, while a dissipative cycle produces a loss rather than creating <energy>. The original loading/unloading curve below is a chosen constitutive sketch, not measured material data.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-1-stress-strain.png]
{title=Stress and strain}
{description=Small-strain geometry and illustrative unloading.}
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