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Governing equations

These are the field identities behind the entries in the library: how deformation is measured, how forces are carried, how mass and momentum are conserved, and how those statements become a discrete system on a mesh.

Continuum mechanics

Motion, stress, and conservation, before any choice of element or solver.

Kinematics

Deformation gradient

F = ∂x / ∂X

A material line element in the reference placement is mapped to the current placement. Finite-strain measures are built from this tensor. Translation, rotation, stretch, and shear are shown as a homogeneous motion of a grid.

Kinematics

Jacobian

J = det F = dv / dV

Local volume change. Mass balance in the reference placement is ρ₀ = ρ J. Incompressibility is J = 1, which is the constraint on the principal stretches of rubber.

Kinematics

Green–Lagrange strain

E = ½ (Fᵀ F − I)

Strain on the reference configuration. When displacement gradients are small, it reduces to the infinitesimal strain. A pure rotation leaves E at zero and does not leave ε at zero.

Kinematics

Infinitesimal strain

ε = ½ (∇u + (∇u)ᵀ)

The symmetric gradient of displacement. This is the strain used in linear elasticity and in most introductory finite element courses.

Kinematics

Polar decomposition

F = R U = V R

A unique rotation R and symmetric stretches U, V. Strain is built from U or from C = U², so a rigid rotation does not count as stretch. Stretch then rotate, or rotate then stretch — the two paths meet.

Kinematics

Almansi strain

e = ½ (I − b⁻¹)

Finite strain on the current placement, with b = F Fᵀ. Like Green–Lagrange, it vanishes in a rigid motion.

Kinematics

Velocity gradient

L = Ḟ F⁻¹ = D + W

The symmetric part D is stretching; the skew part W is spin. This is the rate that is conjugate to Kirchhoff stress. D, W, and L on a material cross.

Stress

First Piola–Kirchhoff

P = J σ F⁻ᵀ

Nominal stress: force on a current area, measured per reference area. Work-conjugate to Ḟ. The same force on two areas.

Stress

Second Piola–Kirchhoff

S = J F⁻¹ σ F⁻ᵀ

Stress on the reference placement, work-conjugate to Green–Lagrange strain. Hyperelastic laws are usually written as S = ∂W/∂E.

Stress

von Mises equivalent

σ_vm = √(3/2 σ′ : σ′)

Magnitude of the stress deviator, scaled to match uniaxial tension. J₂ plasticity yields when this reaches the current yield stress. The cylinder in principal-stress space.

Stress

Cauchy traction

t = σ n

Force per current area on a surface with outward normal n. Without couple stresses, σ is symmetric. On a deforming body that traction rides with the current surface. Turn the cut and the same σ produces a different t.

Balance

Conservation of mass

∂ρ/∂t + ∇ · (ρ v) = 0

Local mass balance. If the motion is incompressible, this is equivalent to a divergence-free velocity field.

Balance

Linear momentum

∇ · σ + ρ b = ρ a

The divergence of stress, together with body force, balances inertia. In a static problem the right-hand side is zero. On a supported body the reactions on the supported area close the global balance.

Constitutive

Isotropic Hooke’s law

σ = λ tr(ε) I + 2μ ε

Linear elasticity in the isotropic case, written with the Lamé parameters. The compact form is σ = ℂ : ε. Uniaxial tension produces lateral strain −ν ε.

Constitutive

Neo-Hookean stored energy

W = (μ/2)(I₁ − 3) − μ ln J + (λ/2)(ln J)²

The simplest isotropic hyperelastic law. Incompressibility drops the volumetric terms and enforces J = 1 with a pressure. Principal stretches at J = 1.

Balance

Angular momentum

σ = σᵀ

Without couple stresses, Cauchy’s second law is symmetry of σ. It is an algebraic restriction, not a differential one.

Fluids

Incompressible Navier–Stokes

ρ (∂v/∂t + v · ∇v) = −∇p + μ ∇²v + ρ b

The same momentum balance, closed by Newton’s law of viscosity, together with ∇ · v = 0.

Finite elements

The weak form of momentum, and the linear system obtained after interpolation and assembly.

Weak form

Principle of virtual work

∫ σ : ∇δu dV = ∫ t · δu dA + ∫ ρ b · δu dV

A weak statement of linear momentum. Finite elements replace the trial and test displacements with shape-function expansions. Stationarity of the potential is the same statement when a stored energy exists.

Interpolation

Isoparametric map

x(ξ) = Σ N_a(ξ) x_a,   u(ξ) = Σ N_a(ξ) a_a

The same shape functions describe geometry and the unknown field. Derivatives in physical space use J = ∂x/∂ξ. Parent square to warped quad.

Discrete system

Element stiffness

Kᵉ = ∫ Bᵀ ℂ B dV

The strain–displacement matrix B holds the symmetric gradients of the shape functions. Assembly scatters Kᵉ into the global sparse matrix.

Discrete system

Linear elasticity on a mesh

K a = f

After interpolation and assembly, the unknown nodal coefficients satisfy a stiffness system whose right-hand side carries the applied loads.

Quadrature

Gauss rule

∫ f dV ≈ Σ_q w_q f(ξ_q) det J(ξ_q)

Constitutive evaluations happen only at the quadrature points. Full versus reduced integration on a bilinear quadrilateral.

Mixed methods

Incompressibility

∫ δp (∇ · u) dV = 0

A displacement-only space cannot represent ∇ · u = 0 without locking. Pressure is interpolated independently, and the pair must satisfy an inf-sup condition.

Numerical methods

Nonlinear problems are reduced to a sequence of linear solves; dynamics add a choice of integrator.

Nonlinear solve

Newton–Raphson

KT Δu = −R(u)

A residual R(u) = 0 is linearized at the current guess. The tangent KT is the Jacobian of that residual. Consistent linearisation is what makes the convergence quadratic.

Stability

Euler load

Pcr = π² EI / L²

The linear eigenvalue at which a hinged–hinged column loses uniqueness of the straight solution. Geometric stiffness, not a change in E, is what makes the tangent singular. An imperfect strut.

Path following

Arc-length constraint

‖Δa‖² + ψ² Δλ² = Δs²

The load factor λ is an unknown. The extra equation lets the iteration pass a limit point where load control would stall. Riks and Crisfield.

Dynamics

Semi-discrete momentum

M ä + C ȧ + R(a) = f

Mass, damping, and a (possibly nonlinear) internal force. Explicit methods lump M; implicit methods form a tangent at each step.

Dynamics

Newmark

a_{n+1} = a_n + Δt v_n + Δt² [(½−β) ä_n + β ä_{n+1}]

Average acceleration (β = ¼, γ = ½) is unconditionally stable for linear problems. Central difference is the explicit member of the same family.

Contact

Complementarity

g ≤ 0,   p ≥ 0,   p g = 0

Gap, pressure, and the requirement that one of them vanish. Friction adds a tangential law and a nonsymmetric tangent.