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Finite element analysis basics

Mechanical Engineering (B.E./B.Tech) · Design tools and practice · free study notes for exams, placements and interviews

In short

Finite element analysis replaces a real component — whose governing differential equations cannot be solved by hand — with a mesh of simple pieces (elements) joined at nodes, writes an approximate algebraic equation for each piece, assembles them into one large system [K]{u} = {F}, and solves it for displacements, from which strains and stresses follow. It is the standard way Indian product companies check strength, stiffness, vibration, heat and crash performance before cutting metal, because a simulation costs hours of CPU time while a prototype costs lakhs of rupees. It is examined as a full university subject (AICTE model curriculum carries Finite Element Methods/Analysis as a core or departmental elective in B.E./B.Tech Mechanical) with numericals on element stiffness matrices, shape functions and two- or three-element assemblies; note that FEM is not an explicitly listed topic in the current GATE ME syllabus, so weight it mainly for university exams, placement technical rounds and simulation-job interviews. Employers hiring on it include Tata Motors, Mahindra, Ashok Leyland, Bajaj, L&T, BHEL, HAL, ISRO, DRDO, Bosch, Cummins and the ER&D services firms (L&T Technology Services, Tata Technologies, Cyient, KPIT, Quest Global, Tata Elxsi).

Explained

What FEA is, and the eight steps of a solution

Element types, DOF and shape functions

Assembly, boundary conditions and loads

Analysis types

Meshing, element quality and convergence

Errors, validation, and Indian codes and practice

Key points

Formulas and examples

[K]{u} = {F} — K in N/mm (or N/m), u in mm, F in N. Π = ½{u}ᵀ[K]{u} − {u}ᵀ{F} minimised gives the same result.
Bar/truss element: [k] = (AE/L) × [[1, −1], [−1, 1]]; A = mm², E = N/mm² (MPa), L = mm ⇒ k in N/mm.
Linear bar shape functions: N₁ = 1 − x/L, N₂ = x/L; u(x) = N₁u₁ + N₂u₂; ε = (u₂ − u₁)/L; σ = Eε.
2D truss member in global axes (c = cos θ, s = sin θ): [k] = (AE/L) × [[c², cs, −c², −cs], [cs, s², −cs, −s²], [−c², −cs, c², cs], [−cs, −s², cs, s²]].
Euler–Bernoulli beam element (DOF v₁, θ₁, v₂, θ₂): [k] = (EI/L³) × [[12, 6L, −12, 6L], [6L, 4L², −6L, 2L²], [−12, −6L, 12, −6L], [6L, 2L², −6L, 4L²]]; EI in N·mm².
Element stiffness in general: [k] = ∫_V [B]ᵀ[D][B] dV; equivalent load {f} = ∫_V [N]ᵀ{q} dV + ∫_S [N]ᵀ{p} dS.
Plane stress [D] = E/(1 − ν²) × [[1, ν, 0], [ν, 1, 0], [0, 0, (1 − ν)/2]].
Plane strain [D] = E/[(1 + ν)(1 − 2ν)] × [[1 − ν, ν, 0], [ν, 1 − ν, 0], [0, 0, (1 − 2ν)/2]].
CST: area A = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|; [k] = t·A·[B]ᵀ[D][B] (t = thickness, [B] constant).
Mass matrices for a bar: consistent [m] = (ρAL/6)[[2, 1], [1, 2]]; lumped [m] = (ρAL/2)[[1, 0], [0, 1]]; ρ in kg/m³ (or tonne/mm³ in mm-N-tonne units).
Modal: ([K] − ω²[M]){φ} = 0; f = ω/2π in Hz. Buckling: ([K] + λ[K_g]){φ} = 0, P_cr = λP_applied.
Von Mises stress σ_v = √{½[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]}; compare with yield/FoS.
Thermal element: [k] = (kA/L)[[1, −1], [−1, 1]], k = conductivity W/m·K; thermal strain ε_th = αΔT.
Explicit stable time step Δt ≤ L_min/√(E/ρ) (L_min = smallest element edge).

Example 1 — stepped bar, two bar elements (classic exam question). A steel bar (E = 200 GPa = 200 000 MPa) is fixed at node 1. Element 1: A = 400 mm², L = 500 mm. Element 2: A = 200 mm², L = 500 mm. A 20 kN axial pull acts at node 3.

import numpy as np
E = 200e3                 # MPa = N/mm^2
A = [400.0, 200.0]        # mm^2
L = [500.0, 500.0]        # mm
K = np.zeros((3, 3))
for e in range(2):        # direct stiffness assembly
    k = (A[e]*E/L[e]) * np.array([[1.0, -1.0], [-1.0, 1.0]])
    K[e:e+2, e:e+2] += k
F = np.array([0.0, 0.0, 20e3])        # N, load at node 3
free = [1, 2]                         # node 0 (node 1) is fixed
u = np.zeros(3)
u[free] = np.linalg.solve(K[np.ix_(free, free)], F[free])
# u = [0.0, 0.125, 0.375] mm

Example 2 — interpolation inside an element. A 100 mm linear bar element has u₁ = 0.020 mm, u₂ = 0.060 mm, E = 200 GPa.

Example 3 — one beam element for a cantilever. L = 1 000 mm, E = 200 000 MPa, I = 1 × 10⁶ mm⁴, tip load P = 1 kN downward. Node 1 fixed, so only v₂ and θ₂ remain (rows/columns 3–4 of [k]):

Example 4 — mesh convergence at a hole. Thin plate, hole small compared with the width, remote tension 50 MPa; theory gives K_t = 3.0, so peak σ = 150 MPa. Successive Q8 meshes (quarter model, symmetry):

Elements around quarter-holePeak σ (MPa)Error vs 150 MPa
29636.0 %
412715.3 %
81416.0 %
161472.0 %
241490.7 %

The sequence rises monotonically towards the exact value — convergence from below. Accepting mesh 2 would have under-sized the part by a factor of more than 1.5 on stress.

Example 5 — first natural frequency, FE vs theory. Steel strip cantilever: L = 0.3 m, b = 25 mm, t = 3 mm, E = 200 GPa, ρ = 7 850 kg/m³.

AspectLinear elements (tet4, Q4, hex8)Quadratic elements (tet10, Q8, hex20)
Stress inside elementConstant / low orderLinear variation
Curved boundariesFacetedRepresented by mid-side nodes
Mesh needed for stressVery fineCoarser
Cost per elementLowHigher (more DOF, more Gauss points)
Typical useExplicit crash (hex8), quick stiffness checksStatic stress, fatigue, fracture

Where it is used

Interview questions

Common mistakes

Related topics

Mechanics of materials: stress, strain and deflection · CAD and 3D modelling · Vibrations and modal analysis · Machine design: failure theories and factor of safety · Heat transfer: conduction, convection and radiation · Computational fluid dynamics basics

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