SYSTEM.INITIALIZE: BLUEPRINT_UNFOLD
DWG TITLEPORTFOLIO BLUEPRINT
DRAWN BYDINESH KUMAR
SCALE1:1
REVISIONA.02
← Back to FEM Dashboard
CHAPTER 6Element Formulations

Euler-Bernoulli Beam Elements

1. Introduction and Structural Context

In structural and mechanical engineering, a beam is a slender structural member designed to support loads applied transverse (perpendicular) to its longitudinal centroidal axis. Unlike axial bar elements (which undergo tension or compression along their axes) and pinned truss elements (which transmit forces strictly along their member lines), beams undergo flexural deformation (bending). Bending is characterized by transverse displacement (deflection) and rotation of the cross-sections (slope).

To transition from 1D axial bar elements to 1D flexural beam elements, we must examine the nature of their strain fields:

For an axial bar, the strain is defined by the first derivative of the displacement field: $\epsilon_{xx} = \frac{du}{dx}$. This is a first-order differential relation, meaning that a linear interpolation of displacement yields a constant strain. This formulation requires only $C^0$ continuity, meaning that the displacement field itself must be continuous across element boundaries, but its derivative (strain and stress) can be discontinuous.
For a bending beam, the normal stress and strain vary linearly through the thickness of the member, and bending is governed by curvature. Curvature is mathematically related to the second derivative of the transverse deflection: $\kappa \approx \frac{d^2v}{dx^2}$. Because the strain energy in beam bending contains second-order derivatives, the deflection field must be $C^1$ continuous.

$C^0$ vs. $C^1$ Continuity: The Mathematical Necessity

In Finite Element Analysis, the selection of shape functions is governed by the order of derivatives in the variational statement (weak form). A field is $C^1$ continuous if both the function itself ($v(x)$) and its first derivative (slope $\theta(x) = \frac{dv}{dx}$) are continuous across element boundaries.

If we were to use a simple linear or quadratic Lagrangian interpolation (which is only $C^0$ continuous), the slope would be discontinuous at the nodes. This discontinuity means that the second derivative $\frac{d^2v}{dx^2}$ would become a Dirac delta function (infinite) at the nodes, implying infinite strain energy and infinite bending moments, which is physically impossible.

To resolve this issue and maintain a valid finite element formulation, we must ensure that both deflection $v$ and slope $\theta$ are continuous at the nodes. Thus, we define the slope $\theta = \frac{dv}{dx}$ as an active nodal degree of freedom. Consequently, each node of a 1D beam element has 2 degrees of freedom:

1.
Vertical deflection, $v_i$
2.
Slope/rotation, $\theta_i = \left(\frac{dv}{dx}\right)_i$

For a two-node element, this yields a total of 4 degrees of freedom per element, requiring cubic polynomials for shape interpolation.


2. Kinematic Assumptions of Euler-Bernoulli Beam Theory

Euler-Bernoulli beam theory (often referred to as classical beam theory) is built upon several simplifying kinematic assumptions originally formulated by Leonhard Euler and Daniel Bernoulli. These assumptions allow us to reduce a three-dimensional elasticity problem to a simple one-dimensional governing differential equation:

1.
Slenderness Ratio: The beam is assumed to be long and slender. The ratio of the length $L$ to the height/depth $h$ of the cross-section is typically greater than 10 or 15. If the beam is short and deep ($L/h < 10$), shear deformation becomes significant. In such cases, Timoshenko beam theory must be used, which includes independent shear deformation.
2.
Plane Sections Remain Plane (Bernoulli-Navier Assumption): Under the action of bending moments, any cross-section of the beam that was plane before bending remains plane after bending. Geometrically, this means that warping of the cross-section due to shear stresses is neglected.
3.
Normality Constraint: Cross-sections that were perpendicular (normal) to the longitudinal neutral axis before bending remain perpendicular to the neutral axis after bending. This directly implies that the transverse shear strain in the plane of bending is zero:
$$\gamma_{xz} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} = 0$$
4.
Small Deflections: The transverse displacement $v(x)$ is small compared to the cross-sectional dimensions of the beam, allowing us to approximate the curvature of the neutral axis as:
$$\kappa = \frac{\frac{d^2v}{dx^2}}{\left[1 + \left(\frac{dv}{dx}\right)^2\right]^{3/2}} \approx \frac{d^2v}{dx^2}$$
5.
Uniaxial Stress State: The transverse normal stresses $\sigma_{yy}$ and $\sigma_{zz}$ are assumed to be negligible compared to the axial bending stress $\sigma_{xx}$, allowing the use of the 1D Hooke's Law: $\sigma_{xx} = E \epsilon_{xx}$.

3. Displacement, Strain, and Stress Fields

Let the beam be aligned along the coordinate $x$-axis, with the $y$-axis representing the vertical direction. Let $v(x)$ represent the vertical deflection of the neutral axis ($y = 0$).

According to the normality constraint, when the beam bends, a cross-section at coordinate $x$ rotates by an angle $\theta(x) = \frac{dv}{dx}$. As a result of this rotation, any point in the cross-section at a vertical distance $y$ from the neutral axis experiences a longitudinal axial displacement $u(x, y)$ given by:

$$u(x, y) = -y \frac{dv(x)}{dx} = -y \theta(x)$$

The negative sign is crucial: for a positive slope (counter-clockwise rotation), points above the neutral axis ($y > 0$) move in the negative $x$-direction (compression), whereas points below the neutral axis ($y < 0$) move in the positive $x$-direction (tension).

The vertical displacement of any point in the beam is assumed to depend only on $x$:

$$v(x, y) = v(x)$$

From the displacement field, we can compute the normal strain $\epsilon_{xx}$ using the strain-displacement relation:

$$\epsilon_{xx} = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\left(-y \frac{dv}{dx}\right) = -y \frac{d^2v}{dx^2}$$

Using Hooke's Law, the normal bending stress $\sigma_{xx}$ is:

$$\sigma_{xx} = E \epsilon_{xx} = -E y \frac{d^2v}{dx^2}$$

This demonstrates that the bending stress varies linearly through the depth of the beam, being zero at the neutral axis ($y=0$), reaching its maximum compressive value at the top fiber, and reaching its maximum tensile value at the bottom fiber.


4. Bending Moments, Shear Forces, and Governing Equations

The bending moment $M(x)$ at any section along the beam is the resultant couple of the internal normal stresses:

$$M(x) = \int_A -\sigma_{xx} y \, dA$$

Substituting the bending stress expression into this integral:

$$M(x) = \int_A -\left(-E y \frac{d^2v}{dx^2}\right) y \, dA = E \left( \int_A y^2 \, dA \right) \frac{d^2v}{dx^2}$$

The integral $\int_A y^2 \, dA$ is the second moment of area (commonly referred to as the area moment of inertia) $I$ about the neutral axis. Thus, we obtain the classical moment-curvature relation:

$$M(x) = E I \frac{d^2v}{dx^2}$$

By applying static equilibrium to a differential beam element, we can relate the bending moment to the transverse shear force $V(x)$:

$$V(x) = \frac{dM}{dx} = \frac{d}{dx}\left(EI \frac{d^2v}{dx^2}\right)$$

For a beam with a constant flexural rigidity $EI$:

$$V(x) = EI \frac{d^3v}{dx^3}$$

Similarly, the transverse distributed load $q(x)$ (force per unit length, positive upwards) is the derivative of the shear force:

$$q(x) = \frac{dV}{dx} = \frac{d^2M}{dx^2} = \frac{d^2}{dx^2}\left(EI \frac{d^2v}{dx^2}\right)$$

For constant $EI$, this yields the governing fourth-order ordinary differential equation (strong form) for beam deflection:

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

5. Physical Significance and Sign Conventions

To construct a consistent FEA formulation, we must define a clear sign convention. In standard structural mechanics, a sagging moment is considered positive (causing compression on the top fibers). However, in finite element analysis, it is more convenient to use a global coordinate sign convention:

Transverse deflection $v(x)$ is positive in the positive $y$-direction (upward).
Slope $\theta(x) = \frac{dv}{dx}$ is positive counter-clockwise.
Nodal forces $F_i$ are positive in the positive $y$-direction (upward).
Nodal moments $M_i$ are positive counter-clockwise.

The flexural rigidity $EI$ represents the beam's resistance to bending. It is the product of a material property ($E$, Young's Modulus) and a geometric property ($I$, Moment of Inertia). For a rectangular cross-section of width $b$ and height $h$, $I = \frac{bh^3}{12}$. This shows that the height of the beam has a cubic effect on bending resistance, explaining why structural members are oriented vertically (like I-beams) to maximize depth and minimize material volume.

Extended chapter reference

Euler-Bernoulli interpolation, bending stiffness, loads, shear, and moment

Beam elements interpolate both transverse displacement and rotation. Cubic Hermite functions provide the continuity required by bending theory and allow curvature, moment, and shear to be recovered from nodal DOFs.

Derive cubic Hermite shape functions for transverse displacement.
Construct bending stiffness and consistent distributed-load vectors.
Recover slope, curvature, bending moment, and shear force.
Recognize when Timoshenko theory is needed for deep beams.

Visual map

From formulation to verified result

Euler-Bernoulli Beam Elements finite element workflowA six-stage engineering workflow from model definition through verification.Euler-Bernoulli Beam ElementsModel-to-evidence sequence1
Beam idealization
2
Hermite interpolation
3
Curvature matrix
4
Bending stiffness
5
Boundary constraints
6
Moment recovery
Each stage produces evidence for the next. A reliable result preserves traceability from the physical idealization to the final engineering check.

Equation sheet

Governing relationships

Curvature
$$\kappa=\frac{d^2w}{dx^2}=[B_b]\{d\}^e$$

Second derivative of transverse displacement.

Moment
$$M(x)=EI\kappa(x)$$

Constitutive relation for elastic bending.

Bending stiffness
$$[k_b]^e=\int_0^L[B_b]^TEI[B_b],dx$$

Integrates curvature energy over the member.

Engineering comparison

Selection and interpretation table

TheoryKinematicsBest forPrimary caution
Euler-BernoulliPlane sections remain normalSlender beamsNeglects transverse shear
TimoshenkoIndependent rotationDeep beams and sandwichesLow-order shear locking
3D solidFull continuumJoints and local stressHigh mesh cost
ShellSurface kinematicsThin-walled beamsSection and drilling behavior

Verification and application

Checks before accepting the result

  • Rigid-body translation has zero curvature
  • End-force signs match convention
  • Moment diagram satisfies equilibrium
  • Mesh resolves load and section changes

Industry application

A machine base rail may be slender over most of its span but locally deep near bolted supports. A beam model efficiently predicts global deflection, while a shell or solid submodel resolves local bearing and weld stresses.