Statics and Equilibrium — The Foundation of Linear Static FEM

Category: Physics Fundamentals | 2026-03-25 | サイトマップ
NovaSolver Contributors
Table of Contents
  1. Equilibrium: The Core of Linear Static FEM
  2. Equilibrium Conditions: ΣF=0 and ΣM=0
  3. Free Body Diagrams in CAE Context
  4. Determinate vs Indeterminate Structures
  5. Truss Analysis: The Original FEM
  6. When Is Linear Static FEM Valid?
  7. Boundary Conditions and Constraint Equations
  8. Exploiting Symmetry in FEM Models
  9. Cross-Topics

1. Equilibrium: The Core of Linear Static FEM

Statics is the study of bodies in equilibrium — zero acceleration, zero net force, zero net moment. It sounds like the simplest branch of mechanics, but virtually every structural FEM analysis runs in "static" mode: the loads are applied slowly enough that inertia effects are negligible, and the solver finds the displacement field that puts every node in equilibrium.

Understanding statics deeply — including when it breaks down — is the single most important skill for correctly interpreting FEM results and deciding when a dynamic analysis is actually needed.

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When is linear static FEM actually valid, and when is it not? I feel like people just run static analysis by default without thinking about whether it's appropriate.

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You're right — it gets misused a lot. Linear static is valid when three things are all true: first, displacements are small (less than about 5–10% of the smallest structural dimension, or the geometry doesn't change significantly under load). Second, the material is linear elastic throughout — no yielding, no hyperelasticity, no creep. Third, the loading is quasi-static — applied so slowly that inertia forces are negligible. A good rule: if the loading frequency is below 10% of the structure's first natural frequency, static is fine. Fail any one of these three conditions and you need to upgrade: NLGEOM for large deformation, elastic-plastic material model for yielding, or a dynamic analysis for fast loading. The most common mistake I see is running static on impact events or seismic loads.

2. Equilibrium Conditions: ΣF=0 and ΣM=0

For a body in static equilibrium, the sum of all forces and the sum of all moments about any point must be zero:

$$\sum \mathbf{F} = 0: \quad \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$$ $$\sum \mathbf{M} = 0: \quad \sum M_x = 0, \quad \sum M_y = 0, \quad \sum M_z = 0$$

In 3D, these are six scalar equations of equilibrium. A free body in 3D has six degrees of freedom — it takes (at least) six constraints to fully restrain it. This is why correct boundary conditions in FEM require removing all six rigid body modes, and why "insufficient constraints" errors occur when boundary conditions are set up incorrectly.

Quick Verification: Reaction Force Check

After any FEM analysis, check that the sum of all reaction forces at fixed boundaries equals the total applied load. This is guaranteed by equilibrium, and if it doesn't hold, there's a modeling error. In Abaqus, request RF (reaction force) output and sum over all boundary nodes:

*NODE PRINT, NSET=FIXED_BC
RF,

The summed reaction force in each direction should equal the applied force in that direction.

3. Free Body Diagrams in CAE Context

A Free Body Diagram (FBD) isolates a component from its surroundings and shows all forces and moments acting on it. In FEM, the FBD mindset is critical for:

  • Correctly identifying applied loads: Is the load applied to a surface, a line, a point, or as a body force?
  • Setting up boundary conditions: What replaces the removed surroundings — a fixed wall, a spring, another component?
  • Interpreting reaction forces: The FBD tells you what reactions to expect; if FEM gives something else, find out why

FBD of a Bracket Under Horizontal Load

A bracket bolted to a wall with a vertical plate loaded 500 mm from the wall with a 10 kN horizontal force. The FBD of the bracket shows: 10 kN force at the load point, and at the wall attachment — vertical reaction, horizontal reaction, and a bending moment $M = 10 \text{ kN} \times 0.5 \text{ m} = 5 \text{ kN·m}$. This moment is what causes high stress at the bracket-wall interface and drives the bolt pattern design.

4. Determinate vs Indeterminate Structures

A statically determinate structure has reactions that can be found from equilibrium equations alone ($\sum F = 0$, $\sum M = 0$). Adding more constraints makes it statically indeterminate — the extra constraints are "redundant" and the load distribution among them depends on relative stiffness, not just equilibrium.

Structure TypeDegree of IndeterminacySolved ByExample
Simply supported beam0 (determinate)ΣF=0, ΣM=0Single-span simply supported bridge
Propped cantilever1Compatibility + equilibriumBeam on wall + column
Fixed-fixed beam2FEM or moment distributionContinuous beam over 3 spans
Frame (3D)3×members - 6FEMBuilding steel frames
General solid body∞ (indeterminate)FEM (continuum)Solid bracket, housing, casting

For highly indeterminate structures, FEM is the practical solution — the entire point of FEM is to handle arbitrary indeterminacy by treating the structure as a continuous body and finding the displacement field that satisfies equilibrium everywhere.

5. Truss Analysis: The Original FEM

A truss is a structure composed of straight two-force members (carrying only axial load — no bending) connected at joints. Truss analysis by the Method of Joints is essentially manual FEM: equilibrium at each pin joint gives two equations (ΣFx=0, ΣFy=0) that must be satisfied.

For a planar truss with $j$ joints and $m$ members and $r$ reactions, the determinacy check:

$$m + r = 2j \quad \text{(determinate)}, \quad m + r > 2j \quad \text{(indeterminate)}$$

The direct stiffness method — assembling the stiffness matrix for a truss — is computationally identical to FEM for bar elements. Students who understand truss analysis have grasped the core of FEM assembly.

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I see that truss analysis and FEM give the same answer for a truss structure. Can I just use FEM for trusses without understanding the method of joints?

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You can get the numbers from FEM without knowing the method of joints, but you'd be missing important intuition. The method of joints teaches you which members are in tension and which in compression — and compression members can buckle while tension members can't. If you just look at a FEM stress plot, you might not notice that a critical compression diagonal is stressed beyond its Euler buckling load even though von Mises is "low." More importantly, hand-calculating a simple truss before building the FEM gives you an independent check. A 30% discrepancy between your hand calc and FEM result is a red flag — either your hand calc is wrong (find the mistake) or your FEM is wrong (find the modeling error). Using FEM as the only check has no check.

6. When Is Linear Static FEM Valid?

Three independent criteria must all be satisfied:

Criterion 1: Small Displacements (Geometric Linearity)

Displacements should be small compared to characteristic structural dimensions. A practical check: if max displacement/span < 0.05 (5%), geometric nonlinearity is generally negligible. For structures where geometry changes significantly under load — cables, membranes, shells under lateral pressure, slender columns near buckling — NLGEOM must be activated.

Criterion 2: Linear Elastic Material

All materials must remain below yield. A safety factor on static stress ≥ 1.5 relative to yield strength is typically required before assuming linear material behavior is adequate. For rubber, biological tissue, or materials under creep — nonlinear material models are required regardless of strain level.

Criterion 3: Quasi-Static Loading

The loading must be applied slowly enough that kinetic energy is negligible. The practical check:

$$\frac{f_{\text{load}}}{f_{n1}} \leq 0.1 \quad \Rightarrow \quad \text{static analysis adequate}$$

Where $f_{n1}$ is the structure's first natural frequency. For most civil structures (bridges, buildings) under normal loads, this criterion is easily met. For machinery, seismic events, and impact loads, it often fails.

Load TypeTypical FrequencyStatic Valid?Recommendation
Dead weight (gravity)0 Hz (static)YesLinear static FEM
Slow hydraulic press<0.1 HzUsually yesNonlinear static if plastic
Road bumps on vehicle2–15 HzCheck fn1Dynamic if fn1 < 150 Hz
Machinery vibration10–500 HzRarelyFrequency response or random vib
Blast/explosion100–10,000 HzNeverExplicit FEM
Seismic ground motion0.1–10 HzSometimesResponse spectrum or time history

7. Boundary Conditions and Constraint Equations

In FEM, boundary conditions replace the physical supports and connections that are removed when isolating the model:

  • Fixed (encastre): All 6 DOF constrained — models welded or bolted base
  • Pinned: 3 translational DOF fixed, 3 rotational free — models hinged support
  • Roller: Normal DOF fixed, tangential free — models bridge on a roller bearing
  • Symmetry: Normal translation and in-plane rotations constrained — halves or quarters the model
  • Spring support: Stiffness resists displacement — models foundation compliance

Over-Constraint: The Hidden FEM Error

Applying too many constraints (over-constraining) introduces artificial stiffness and wrong reactions. A common example: modeling a bolted joint by fully fixing all nodes at the bolt hole. In reality, the bolt allows in-plane rotation of the connected plate — fixing rotation creates a "built-in" effect that increases the apparent stiffness and reduces the actual stress at the critical location. Always model the actual physical constraint, not a convenient over-simplification.

8. Exploiting Symmetry in FEM Models

Symmetric structures under symmetric loads can be modeled at a fraction of full size. Types of symmetry and corresponding boundary conditions:

Symmetry TypeModel SizeBC on Symmetry PlaneExample
1 plane of symmetry1/2No normal displacement, no in-plane rotationSymmetric bracket
2 planes of symmetry1/4Apply to both planesRectangular plate with central hole
Axisymmetry2D cross-sectionAxisymmetric elements (CAX4)Pressure vessel, O-ring
Cyclic symmetry1/N sectorCyclic symmetry constraintTurbine blade, fan impeller

Symmetry is only valid when both geometry AND loads are symmetric. A symmetric structure under an asymmetric load must be modeled in full. A common shortcut error: symmetry axis passes through a weld or fillet — the actual geometry may not be perfectly symmetric due to as-built tolerances, requiring judgment on whether idealization is acceptable.

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I want to use cyclic symmetry for a 36-blade fan impeller in Abaqus. What's the sector angle, and are there any catches?

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Sector angle = 360°/36 = 10°. In Abaqus, you use *CYCLIC SYMMETRY MODEL with 36 sectors. Main catches: first, any loading must also have cyclic symmetry — a uniform centrifugal load does, but an asymmetric inlet distortion doesn't. Second, for modal analysis, you run separate analyses for each nodal diameter (0, 1, 2... up to N/2), because different nodal diameter modes don't couple in cyclic symmetry. Third, your blade-to-blade tie or contact interfaces need careful setup on the cyclic symmetry boundaries. The payoff is huge — instead of meshing all 36 blades (say 2 million elements), you mesh 1/36th (56k elements) and run 18× faster while getting the exact same answer for symmetric loading.

9. Cross-Topics

TopicConnectionLink
Newton's LawsStatics is F=ma with a=0Newton's Laws of Motion
Hooke's LawLinear statics requires Hooke's Law (σ=Eε) to hold throughoutSprings & Hooke's Law
Stress & StrainStatic equilibrium produces the stress and strain fields in FEMStress and Strain Basics
Fluid PressureHydrostatics is statics applied to fluid bodiesFluid Pressure and Buoyancy
Beam Deflection ToolInteractive beam deflection and moment diagramsBeam Deflection Calculator