Beam Deflection with FEM: Euler-Bernoulli Beam Elements

From a truss bar to a beam

A truss bar only carries axial force and needs a single translational degree of freedom (DOF) at each node. A beam resists bending in addition to axial load, so each node needs a transverse displacement and a rotation DOF. The classical Euler-Bernoulli beam element assumes plane sections remain plane and perpendicular to the neutral axis (no shear deformation), giving a cubic (Hermite) displacement field between nodes — this exact element stiffness matrix is what the finite element method uses to model a beam under simply-supported, cantilever, or fixed-fixed boundary conditions.

Local beam stiffness (bending only, 4×4, DOFs: v1,θ1,v2,θ2):

  k_beam = (EI/L³) · [  12     6L   -12     6L  ]
                      [  6L    4L²  -6L     2L²  ]
                      [ -12   -6L    12    -6L  ]
                      [  6L    2L²  -6L     4L²  ]

  E = Young's modulus,  I = second moment of area
  L = element length,  v = transverse deflection,  θ = rotation

live demo · a beam settling into its deflected shape● LIVE

Why Hermite cubic shape functions

The governing equation for beam bending, EI·v'''' = q, is fourth order, which means the finite element solution must be continuous in both displacement and slope across element boundaries — so-called C¹ continuity. The cubic Hermite shape functions used in the standard beam element interpolate both displacement and slope simultaneously, guaranteeing this continuity automatically. Timoshenko beam elements extend the formulation to include transverse shear deformation, which becomes significant for short, deep beams where the span-to-depth ratio drops below about 10.

Assembling and reading the deflected shape

Just as with truss elements, each beam element's local stiffness is assembled into the global matrix K, boundary conditions are applied by partitioning free and supported degrees of freedom, and K_ff·D_f = F_f is solved for the unknown nodal displacements and rotations. Once D_f is known, each element's local displacements are recovered and used to reconstruct the deflected shape by interpolating the cubic Hermite functions within the element, along with the corresponding bending-moment and shear-force diagrams derived from successive derivatives of the deflection curve — exactly the interactive visualisation used in a browser-based beam simulator.

Boundary conditions change everything

The same beam, load and material give dramatically different results depending on how the ends are supported. A cantilever — fixed at one end, completely free at the other — produces the largest deflection and a bending moment that peaks at the fixed support. A simply supported beam allows rotation at both ends while preventing displacement, giving a smoother deflection curve. A fixed-fixed beam constrains both displacement and rotation at both ends, producing the smallest deflection of the three for a given load — the direct payoff of adding rotational stiffness at the supports in the global stiffness matrix.

Frequently asked questions

Why does a beam element need a rotation degree of freedom, unlike a truss bar?

A truss bar only stretches or compresses along its axis, so each node needs just a translational DOF. A beam resists bending as well, so its cross-section rotates as it deflects — each node therefore needs both a transverse displacement v and a rotation θ. The governing equation EI·v'''' = q is fourth order, which requires the displacement field to be continuous in both value and slope (C¹ continuity) across element boundaries, met by the cubic Hermite shape functions used in the standard beam element.

What is the difference between Euler-Bernoulli and Timoshenko beam elements?

The Euler-Bernoulli beam element assumes plane sections remain plane and perpendicular to the neutral axis, ignoring shear deformation entirely — accurate for slender beams. Timoshenko beam elements add transverse shear deformation to the formulation, which matters for short, deep beams where the span-to-depth ratio drops below about 10, since shear flexibility becomes a significant fraction of the total deflection in that regime.

How does a support condition (simply supported, cantilever, fixed-fixed) change the deflection?

Support conditions set which translational and rotational degrees of freedom are constrained to zero before the global system K_ff·D_f = F_f is solved. A cantilever fixes both displacement and rotation at one end and leaves the other completely free, producing the largest deflection for a given load. A simply supported beam fixes displacement at both ends but allows rotation, while a fixed-fixed beam constrains both displacement and rotation at both ends, producing the smallest deflection and a different bending-moment distribution along the span.

Try it live

Everything above runs in your browser — open Beam Deflection — Euler-Bernoulli Structural Simulator and choose simply supported, cantilever or fixed-fixed beams with point loads or a uniformly distributed load, then watch the deflected shape, bending-moment and shear-force diagrams update live. Nothing is installed, nothing is uploaded.

▶ Open Beam Deflection simulation