Euler–Bernoulli beam theory

Category: Structural Analysis | Integrated 2026-04-06
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Euler-Bernoulli Beam Theory

Euler: Theoretical Foundations

The Most Basic Beam Theory

🧑‍🎓

Professor, what is the "Euler-Bernoulli beam"? Euler also appeared in buckling theory, right?


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It's the bending theory of beams established by Leonhard Euler and Daniel Bernoulli in the 18th century. It is the most fundamental theory in structural mechanics and also the starting point for beam elements in FEM.


Basic Assumptions

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What assumptions does it make?


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Three Basic Assumptions:


1. Plane Sections Remain Plane Assumption — Cross-sections that were plane before deformation remain plane after deformation.

2. Orthogonality Assumption — Cross-sections remain orthogonal to the beam axis (neutral axis) after deformation.

3. Small Deformation — Deformations are sufficiently small.


🧑‍🎓

Assumption 2 seems important. If the cross-section doesn't tilt, that means...


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Correct. It means shear deformation is ignored. The beam deforms only due to bending, and the tilt of the cross-section due to shear is zero. This is the most significant feature of the Euler-Bernoulli beam and also its greatest limitation.


Governing Equation

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Please tell me the differential equation for bending.


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A fourth-order ordinary differential equation concerning deflection $w(x)$:


$$ EI \frac{d^4 w}{dx^4} = q(x) $$

Here, $EI$ is the bending stiffness, and $q(x)$ is the distributed load.


🧑‍🎓

A fourth-order differential equation! Does solving it require four integrations?


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Yes. The four integration constants are determined by four boundary conditions. Two conditions are needed at each end (displacement and rotation, or shear force and moment).


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Physical meaning of each derivative:


DerivativePhysical QuantityFormula
$w$Deflection
$w' = dw/dx$Rotation angle $\theta$
$w'' = d^2w/dx^2$Curvature $\kappa = M/(EI)$$M = EI w''$
$w''' = d^3w/dx^3$Shear Force$V = -EI w'''$
$w'''' = d^4w/dx^4$Distributed Load$q = EI w''''$
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Differentiating deflection four times returns the load. That's elegant.


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Memorizing this relationship is very useful for verifying FEM results. It forms a chain: deflection → rotation angle → curvature → moment → shear force → load.


Effect of Ignoring Shear Deformation

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How much error occurs when shear deformation is ignored?


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The error is smaller when the beam's span-to-depth ratio ($L/h$) is larger:


$L/h$Contribution of Shear DeformationAccuracy of Euler-Bernoulli
> 20< 1%Sufficiently accurate
10 to 201 to 5%Practically acceptable
5 to 105 to 20%Caution required
< 5> 20%Inaccurate. Should use Timoshenko beam
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So differences start to appear around $L/h < 10$.


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Typical steel beams (H-shaped steel) have $L/h = 15 \sim 25$, so Euler-Bernoulli is sufficient. However, sandwich panels or short connecting beams ($L/h < 5$) require the Timoshenko beam theory.


Beam Elements in FEM

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What does the Euler-Bernoulli beam element look like in FEM?


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A 2-node beam element, with 3 degrees of freedom per node (in 2D): deflection $w$, rotation angle $\theta$, and axial displacement $u$.


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An important feature is that the shape functions are Hermite polynomials (cubic polynomials). Unlike standard FEM elements (Lagrange polynomials), rotation angles are also nodal variables. This allows the accurate solution of the fourth-order bending differential equation with just 2 nodes.


🧑‍🎓

Two nodes can solve a fourth-order equation! Is one element exact?


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For a uniform cross-section and constant distributed load, a single element yields the exact solution. This is a major advantage of the Euler-Bernoulli beam element. Nodes need to be placed at points where concentrated loads act, but otherwise, a coarse mesh is sufficient.


Summary

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Let me organize the Euler-Bernoulli beam theory.


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Key points:


  • Classical beam theory ignoring shear deformation — Cross-sections always remain orthogonal to the neutral axis.
  • $EI w'''' = q$ — Fourth-order ordinary differential equation.
  • Sufficiently accurate for $L/h > 10$ — Applicable to slender beams.
  • Uses Hermite interpolation in FEM — Exact solution for uniform beams with 2 nodes.
  • Use Timoshenko beam for $L/h < 5$ — Shear deformation cannot be ignored.


🎓

The deflection formula for a cantilever beam $\delta = PL^3/(3EI)$ solved in strength of materials classes is precisely the solution of this theory. FEM beam elements discretize this theory; the principle is the same.


Coffee Break Trivia

The Birth of Euler Beam Theory

The Euler-Bernoulli beam theory originates from Leonhard Euler's 1744 work "De Curvis Elasticis." The assumption that "cross-sections remain plane and orthogonal to the beam axis after deformation" was initially controversial, but for slender beams (slenderness ratio L/h>10), it still provides accuracy within 1% error today.

Computational Methods for Euler

Beam Element Stiffness Matrix

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Please tell me the stiffness matrix for the Euler-Bernoulli beam element.


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For 2 nodes, with 2 bending degrees of freedom per node ($w_i, \theta_i$), a 4×4 stiffness matrix:


$$ [K_b] = \frac{EI}{L^3} \begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix} $$

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Where do these numbers $12, 6L, 4L^2$ come from?


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They are obtained by differentiating the Hermite shape functions to create the $B$ matrix and integrating $\int_0^L B^T EI B \, dx$. Since Hermite shape functions are cubic polynomials, curvature (second derivative) becomes linear, allowing exact integration.


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Adding axial force degrees of freedom ($u_i$) gives an independent 2×2 stiffness:


$$ [K_a] = \frac{EA}{L} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$

This is combined with the bending $[K_b]$ to form a 6×6 (2D) or 12×12 (3D) beam element stiffness matrix.


Equivalent Nodal Loads

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How is a distributed load applied to a beam element?


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Convert the distributed load $q$ into equivalent nodal loads. For a uniformly distributed load:


$$ \{F_{eq}\} = \frac{qL}{12} \begin{Bmatrix} 6 \\ L \\ 6 \\ -L \end{Bmatrix} $$

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A force of $qL/2$ and moments of $\pm qL^2/12$ at each node... that's the fixed-end moment for a fixed-fixed beam, right!


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Perfect understanding. Equivalent nodal loads are equal to the fixed-end reactions with the sign reversed. Knowing this correspondence allows intuitive verification of whether equivalent nodal loads are calculated correctly.


Element Names by Solver

ElementNastranAbaqusAnsys
2-node Beam (EB)CBARB23 (2D), B33 (3D)BEAM3 (2D), BEAM4
2-node Beam (Timoshenko)CBEAMB21 (2D), B31 (3D)BEAM188/189
🧑‍🎓

Are Nastran's CBAR and CBEAM different?


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CBAR is the Euler-

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