Mechanical Engineering Design · Chapter 19
Mechanical Engineering Design · Part 4 — Special Topics

Finite-Element Analysis

A conceptual introduction to FEA for the designer: how the method discretizes a structure into elements and solves the assembled stiffness equations, the element library, mesh and convergence, load and boundary-condition application, symmetry modeling, and extensions to thermal, buckling, and vibration problems.

19–1The Finite-Element Method

Learning objectives

  • Translate a physical structure into elements, nodes, degrees of freedom, loads, and restraints.
  • Assemble and solve a transparent linear bar model while retaining units and sign conventions.
  • Select element families and mesh refinement measures appropriate to the response of interest.
  • Separate nominal from local stress, static from fatigue evidence, and strength from stiffness checks.
  • Verify equilibrium, boundary conditions, mesh convergence, and agreement with an independent model.
  • Document assumptions, data provenance, uncertainty, and the evidence still required for release.

From continuum to algebraic model

The finite-element method replaces a continuous field by interpolation over finite elements joined at nodes. Nodal degrees of freedom are the primary unknowns. Strain follows from displacement gradients, and constitutive data convert strain to stress. Thus a colored stress plot is a derived quantity, not direct evidence of strength or safety.

For small-displacement, linear-elastic statics the assembled model is \([K]\{u\}=\{F\}\). Geometry and modulus determine stiffness \([K]\); restraints prescribe selected components of \(\{u\}\); external loads populate \(\{F\}\). Material strength does not enter this equilibrium solution. It is compared later with an appropriate stress measure and failure criterion.

Approximation, error, and the quantity of interest

Discretization error arises because finite interpolation cannot exactly reproduce every curved boundary or field gradient. The \(h\)-method reduces element size; the \(p\)-method raises interpolation order. Convergence must be demonstrated for a named quantity—such as tip displacement, strain energy, or stress averaged over a specified path—not merely by obtaining a successful solve.

Point stress at a sharp re-entrant corner, point load, or ideal fixed edge may be singular and can increase without bound under refinement. That is nonconvergence of the reported quantity, not proof that the whole model is invalid. Replace the idealization with physical load spread or radius when known, or report a structural stress away from the singularity with a documented extraction rule.

Physical meaning

Finite elements enforce compatibility through shared nodal motion and equilibrium through assembled nodal forces. Refinement adds freedom for the numerical displacement field to bend and vary; it does not repair an incorrect load path, material law, contact assumption, or restraint.

19–2Element Geometries

Choose dimensionality from physics

Bars carry axial force; beams carry axial force, shear, bending, and often torsion; shells represent thin midsurfaces; plane-stress elements suit thin in-plane parts; plane-strain elements suit long, restrained sections; axisymmetric elements require geometry and loading symmetric about one axis. Three-dimensional solids are appropriate when through-thickness stress, complex contact, or genuinely three-dimensional geometry matters.

The simplest valid family usually gives the clearest audit trail. Slenderness, thickness-to-curvature ratio, expected stress state, connection eccentricity, and desired output govern selection. Mixing elements requires compatible degrees of freedom or a justified coupling; an indiscriminate rigid link can create artificial stiffness and local stress.

19–3The Finite-Element Solution Process

Element stiffness, assembly, and recovery

A two-node axial element has \(k=AE/L\) and local relation \(\{f_e\}=(AE/L)\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\{u_e\}\). Assembly adds overlapping nodal contributions. Essential boundary conditions remove rigid-body modes, the reduced equations yield unknown displacements, and back-substitution recovers reactions. Element strain is \(\epsilon=(u_j-u_i)/L\), followed by \(\sigma=E\epsilon\) for linear elasticity.

Bar-Element Stiffness & Global System
\[ k = \frac{A\,E}{L} \qquad\qquad [K]\{u\} = \{F\} \]Eqs. (19–1, 19–7)

Problem-solving strategy

  1. State the decision and response quantity before building geometry: displacement, stress, buckling factor, or frequency.
  2. Draw the physical free body and load path; define coordinates, positive directions, interfaces, and omitted features.
  3. Select dimensionality, element formulation, material law, loads, and restraints consistent with the real system.
  4. Solve a coarse model, audit reactions and deformation, then refine the response region and compare successive values.
  5. Corroborate with a hand model, test, alternate mesh, or benchmark and document acceptance criteria and unresolved risks.
EXAMPLE 1 Two-segment pull rod with an intermediate load
A straight steel pull rod is fixed at node 1. Element 1 has \(L_1=300\ \text{mm}\), \(A_1=400\ \text{mm}^2\); element 2 has \(L_2=200\ \text{mm}\), \(A_2=200\ \text{mm}^2\). With \(E=200{,}000\ \text{N/mm}^2\), loads of \(10\ \text{kN}\) and \(20\ \text{kN}\) act in the positive axial direction at nodes 2 and 3. Find displacements, element stresses, and the wall reaction.

Recognition strategy

The member is slender, coaxial, and linearly elastic, so two axial bar elements reproduce the piecewise-uniform exact solution. The intermediate nodal load makes assembly and force recovery visible.

Given information

  • Lengths, areas, modulus, and positive-\(x\) nodal loads are supplied; \(u_1=0\), while \(u_2,u_3\) and the reaction are unknown.

Model and interpretation

Use nodes at the support, area change, and free end. Small strain, perfect axial alignment, uniform material, and no self-weight are assumed. Tension and positive displacement point to the right.

Step-by-step solution
1. Compute element stiffnesses. \(k_1=A_1E/L_1=266{,}667 \text{N/mm}\) and \(k_2=A_2E/L_2=200{,}000\ \text{N/mm}\); these connect nodes 1–2 and 2–3.
2. Assemble and restrain. With \(u_1=0\), the reduced system is \(\begin{bmatrix}466667&-200000\\-200000&200000\end{bmatrix} \{u_2,u_3\}^{T}=\{10000,20000\}^{T}\), using newtons and millimetres.
3. Solve displacements. The second row gives \(u_3-u_2=0.1000\ \text{mm}\); substitution into the first gives \(u_2=0.1125\ \text{mm}\) and \(u_3=0.2125\ \text{mm}\).
4. Recover forces and stresses. \(N_1=k_1u_2=30.0\ \text{kN}\) and \(N_2=k_2(u_3-u_2)=20.0\ \text{kN}\), so \(\sigma_1=75.0\ \text{MPa}\), \(\sigma_2=100\ \text{MPa}\), and \(R_1=-30.0\ \text{kN}\).

Verification

Units: \(AE/L\) has units \((\text{mm}^2)(\text{N/mm}^2)/\text{mm}=\text{N/mm}\); stiffness times displacement therefore gives force, and \(\text{N/mm}^2=\text{MPa}\) for recovered stress.

Signs and direction: Both displacement differences are positive, hence both elements are in tension. The negative wall reaction opposes the two positive applied loads, exactly as the free-body convention requires.

Plausibility: Global equilibrium gives \(-30+10+20=0\ \text{kN}\). The thinner second segment carries less force yet has the larger stress, and its elongation \(0.1000\ \text{mm}\) equals \(NL/(AE)\).

Final result: \(\boxed{u_2=0.1125\ \text{mm},\ u_3=0.2125\ \text{mm}, \sigma_1=75.0\ \text{MPa},\ \sigma_2=100\ \text{MPa},\ R_1=-30.0\ \text{kN}}\).

Engineering interpretation

The solution quantifies stiffness response and nominal static stress; it does not establish allowable strength, fatigue life, or local stress at the area transition. Those require verified geometry, material data, and an appropriate concentration or detailed local model.

19–4Mesh Generation

Refine with a declared acceptance rule

Mesh density belongs where geometry, contact, loads, or response gradients demand it. Curved boundaries need enough elements to represent curvature; bending needs adequate through-thickness interpolation; contact needs compatible surface resolution. A practical relative-change measure is \(e_i=|q_i-q_{i-1}|/|q_i|\), where \(q\) is the predeclared response quantity.

Element Quality

Excessive skew, warpage, or aspect ratio can corrupt interpolation and integration. Quality limits are formulation- and solver-dependent, so use the solver manual rather than a universal invented cutoff. Review warnings, energy behavior, and contour smoothness, but never substitute a quality color map for convergence of the engineering response.

EXAMPLE 2Accepting a displacement mesh study
A linear-elastic bracket model predicts vertical tip displacements \(2.31\), \(2.46\), and \(2.49\ \text{mm}\) with characteristic local mesh sizes \(8\), \(4\), and \(2\ \text{mm}\). Reactions are \(9.98\), \(10.01\), and \(10.00\ \text{kN}\) for an applied \(10.00\ \text{kN}\) downward load. Decide whether the finest result satisfies a supplied successive-change criterion of \(2\%\).

Recognition strategy

This is a discretization acceptance decision for displacement, a generally well-behaved global response. Compare successive refinements and separately audit force balance; do not infer convergence from element count alone.

Given information

  • Three mesh sizes, three downward displacement magnitudes, three upward reaction magnitudes, load \(10.00\ \text{kN}\), and the project-supplied \(2\%\) criterion are explicit.

Model and interpretation

Take upward as positive, so displacements are negative and reactions positive. Geometry, linear material, loading, and constraints are unchanged; only local mesh size varies.

Step-by-step solution
1. Define the metric. Use \(e_i=|q_i-q_{i-1}|/|q_i|\), referenced to the newer solution, and apply it to the stated tip displacement rather than searching contours for a favorable value.
2. Evaluate the first change. From \(8\) to \(4\ \text{mm}\), \(e_2=|2.46-2.31|/2.46=0.0610=6.10\%\), so the medium mesh alone is not acceptable.
3. Evaluate the finest change. From \(4\) to \(2\ \text{mm}\), \(e_3=|2.49-2.46|/2.49=0.01205=1.20\%\), which is below the supplied \(2\%\) threshold.
4. Audit equilibrium and decide. The finest reaction error is \(|10.00-10.00|/10.00=0\%\). Accept \(2.49\ \text{mm}\) for this response, while recording the mesh sequence and criterion.

Verification

Units: Each convergence ratio divides millimetres by millimetres and is dimensionless; reaction error similarly divides kilonewtons by kilonewtons. The reported response remains a displacement in millimetres.

Signs and direction: The load and displacement are downward and therefore negative under the chosen convention; the positive support reaction opposes them. Magnitudes were used only inside absolute relative errors.

Plausibility: Displacement changes decrease from \(6.10\%\) to \(1.20\%\), a credible converging trend, while the finest reaction exactly balances the applied load. This does not certify a possibly singular peak stress.

Final result: \(\boxed{u_y\approx-2.49\ \text{mm};\ e_3=1.20\%<2\%,\ \text{accepted for tip displacement}}\).

Engineering interpretation

The accepted value is mesh-insensitive under the stated project rule, not universally exact. Release documentation should retain solver version, element formulation, mesh statistics, restraint and load definitions, and an independent stiffness or test comparison.

19–5Load Application

Nodal and Element Loads

Begin with a free-body diagram. Pressures, tractions, gravity, temperatures, bearing loads, and connector forces must reproduce the physical resultant and moment without creating unintended load paths. A concentrated nodal force may be adequate for remote displacement but creates unreliable local stress. Saint-Venant reasoning applies only sufficiently far from that idealized application.

19–6Boundary Conditions

Restraining the Model

Essential conditions prescribe motion; natural conditions prescribe traction. The model must suppress rigid-body motion without suppressing real compliance. Compare reactions, contact status, and deformed shape with the physical mounting. Remote constraints and rigid couplings need particular care because they may distribute load or add rotational stiffness not present in hardware.

19–7Modeling Techniques

Symmetry and Simplification

Symmetry is valid only when geometry, material, loading, restraints, and the response mode share that symmetry. On a symmetry cut, normal displacement and appropriate rotations are constrained; an antisymmetric load requires different conditions. Submodeling can resolve a local feature after the global model supplies boundary motion, but interpolation at the cut must lie outside the disturbed local region.

EXAMPLE 3Bracket stiffness correlation and redesign decision
A production-fixture bracket carries a service load of \(6.00 \text{kN}\). A verified linear FEA predicts \(0.740\ \text{mm}\) tip deflection and \(182\ \text{MPa}\) path-averaged von Mises stress. A prototype test at \(4.00\ \text{kN}\) measures \(0.520\ \text{mm}\). The specified stiffness limit is \(0.800\ \text{mm}\), and supplied material certification gives yield strength \(S_y=310\ \text{MPa}\) with required static factor \(n_d=1.50\). Assuming linear response, determine correlation, margins, and disposition.

Recognition strategy

This is model validation plus two distinct acceptance checks. Scale the test only under the stated linear assumption, compare displacement with the stiffness requirement, and compare computed stress—not load—with certified strength.

Given information

  • Service and test loads, predicted displacement and local path-averaged static stress, measured displacement, stiffness limit, certified yield strength, and required design factor are supplied.

Model and interpretation

Positive load and displacement are downward; magnitudes are reported. Linear elasticity, repeatable fixture compliance, and the same load path are assumed. The test measurement includes any unremoved rig compliance.

Step-by-step solution
1. Scale the test response. The service-load test projection is \(\delta_{\text{test},6}=0.520(6.00/4.00)=0.780\ \text{mm}\), valid only if stiffness remains linear and contact does not change.
2. Quantify correlation. Relative to the projected test, \((0.740-0.780)/0.780=-0.0513\), so FEA underpredicts measured deflection by \(5.13\%\); investigate fixture and joint compliance before tuning properties.
3. Check stiffness. The test-correlated margin is \(0.800-0.780=0.020\ \text{mm}\), only \(2.50\%\) of the limit, whereas uncorrected FEA would imply a less conservative \(0.060\ \text{mm}\) margin.
4. Check static strength and decide. \(n_y=S_y/\sigma_{\rm vm} =310/182=1.70>1.50\). Static yield passes, but stiffness evidence is marginal; hold release pending uncertainty resolution or increase stiffness and retest.

Verification

Units: Load ratio and correlation error are dimensionless; multiplying the ratio by millimetres preserves displacement. Dividing certified strength in megapascals by computed stress in megapascals gives a dimensionless factor.

Signs and direction: Both physical deflections are downward. The negative correlation error means the model is stiffer than the tested assembly, so using uncorrected FEA would overstate the available stiffness margin.

Plausibility: Scaling \(0.520\ \text{mm}\) by the load ratio \(1.5\) gives \(0.780\ \text{mm}\), near both the FEA value and the limit. Yield passes because \(182(1.50)=273\ \text{MPa}<310\ \text{MPa}\).

Final result: \(\boxed{\delta_{\text{corr}}=0.780\ \text{mm},\ \text{FEA error}=-5.13\%, n_y=1.70;\ \text{hold for stiffness resolution}}\).

Engineering interpretation

Passing static yield does not erase a small stiffness margin, nor does one static test address fatigue, buckling, environment, manufacturing variation, or joint wear. A professional disposition records the discrepancy, uncertainty, model revision, and objective retest criterion.

19–8Thermal Stresses

Coupling Heat to Stress

A sequential analysis first solves temperature, then maps that field to the structural mesh. Free thermal strain is \(\epsilon_{th}=\alpha\Delta T\); stress arises only where expansion is constrained or gradients cause incompatible strain. Properties may depend on temperature, and mapped temperatures, reference temperature, contact conductance, and restraint stiffness require verification.

Thermal Strain & Fully Restrained Thermal Stress
\[ \epsilon_{th} = \alpha\,\Delta T \qquad\qquad \sigma = E\,\alpha\,\Delta T\ (\text{fully restrained}) \]Eq. (19–10 region)

19–9Critical Buckling Load

Buckling as an Eigenvalue Problem

Linear eigenvalue buckling supplies ideal load multipliers and mode shapes about a preload state. It commonly overpredicts real capacity because imperfections, residual stress, plasticity, contact, and load eccentricity are omitted. Use a nonlinear imperfection-sensitive analysis and applicable code or test evidence when buckling governs a released design.

Buckling Eigenvalue Problem
\[ \big([K] + \lambda\,[K_G]\big)\{u\} = 0, \qquad P_{cr} = \lambda\,P_{applied} \]Eq. (19–11 region)

19–10Vibration Analysis

Natural Frequencies and Mode Shapes

Modal analysis solves \(([K]-\omega^2[M])\{\phi\}=0\). An eigenvector gives relative shape, not operating amplitude. Forced response additionally needs excitation, damping, and enough modes for mass participation and response bandwidth. Density and attached mass are as important as stiffness; supports must represent operating conditions rather than a convenient laboratory idealization.

Free-Vibration Eigenvalue Problem
\[ \big([K] - \omega^2\,[M]\big)\{u\} = 0 \]Eq. (19–12 region)

19–11Summary

Chapter recap

  • Start with the free body and response decision; element choice follows physics, not software convenience.
  • Displacements are primary, while strain and stress are recovered and must be interpreted at valid locations.
  • Equilibrium, convergence, independent estimates, and test correlation provide complementary verification.
  • Static strength, stiffness, fatigue, buckling, thermal response, and vibration are separate acceptance questions.

Before moving on: you should be able to define a model and sign convention, assemble a bar system, judge a mesh sequence, identify a singular result, and state what code, vendor, or test evidence is still required before a professional release.