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Structural dynamics and vibration testing

Uniform beam bending frequencies

Bending natural frequency of a uniform Euler–Bernoulli beam from its flexural rigidity EI, mass per metre μ, length L and a dimensionless root λ set by the end supports.

Ref
SDY-11
Rev
v1
Field
Structural dynamics and vibration testing
Result
fn [Hz]

The bending frequencies of a uniform Euler–Bernoulli beam are fₙ = λₙ²√(EI/μ)/(2πL²). The dimensionless root λₙ is set by the supports: nπ for pinned–pinned, 1.8751 (first), 4.6941 (second) and so on for a cantilever, and 4.7300 (first) for clamped–clamped. Rotary inertia and shear deformation are neglected, so the formula overestimates frequencies of thick or short beams.

Bending natural frequency of a uniform Euler–Bernoulli beam from its flexural rigidity EI, mass per metre μ, length L and a dimensionless root λ set by the end supports.

Calculator

Solve for
Hz
112
N·m²
102,000
kg/m
0.120
m
0.25
  • Angular frequency, 2πf35.16 rad/s
  • Period of the vibration0.1787 s
  • Frequency with the length doubled1.399 Hz

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Natural frequency (Hz) current value 5.596 Hz
λnDimensionless root for the supportsKey driver
1.59229
LBeam length
0.224140
μMass per metre
1.7725
EIFlexural rigidity
1.2517.7
  • With a root of 1.875, a rigidity of 200 N·m², 2 kg/m and 1 m, the frequency is 5.596 Hz.
  • Frequency falls with the square of the length: a beam twice as long vibrates at one quarter of the frequency.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (Hz)Local stepChange in result
λn Dimensionless root for the supports1 to 12 1.592229.2227.6+10%+21 %
L Beam length0.2 to 5 m0.2238139.9139.7+10%−17.4 %
μ Mass per metre0.1 to 20 kg/m1.7725.0323.26+10%−4.65 %
EI Flexural rigidity10 to 2,000 N·m²1.25117.716.44+10%+4.88 %

Sweep of Beam length

Beam length (m)Natural frequency (Hz)
0.2139.9
0.88.744
1.42.855
21.399
2.60.8278
3.20.5465
3.80.3875
4.40.289
50.2238

Sweep one variable

Natural frequency (Hz)Hover, touch, or use the arrow keys on the chart
050100150 12345 Beam length (m) now 5.596

Variables and how each one contributes

fn

Natural frequency

Result · Hz

What the equation gives you. Everything else feeds it.

λn

Dimensionless root for the supports

· typical 1 to 12

The dimensionless root set by the end supports: nπ for pinned–pinned, 1.8751 for the first cantilever mode, 4.7300 for the first clamped–clamped mode. Enters squared, so the frequency rises with the square of the root. Raising it by 10 % (to 2.063) moves the result by +21 %, from 5.596 to 6.771 Hz.

EI

Flexural rigidity

N·m² · typical 10 to 2,000

The flexural rigidity, in N·m²: Young’s modulus times the second moment of area. A stiffer section raises the speed or frequency. Raising it by 10 % (to 220) moves the result by +4.88 %, from 5.596 to 5.869 Hz.

μ

Mass per metre

kg/m · typical 0.1 to 20

The mass per metre of the beam, in kg/m. A heavier beam lowers the speed or frequency. Raising it by 10 % (to 2.2) moves the result by −4.65 %, from 5.596 to 5.335 Hz.

L

Beam length

m · typical 0.2 to 5

The length, in m. A longer beam is slower to vibrate: the frequency falls with the square of the length. Raising it by 10 % (to 1.1) moves the result by −17.4 %, from 5.596 to 4.625 Hz.

Worked example with the values above

  1. Root squared: λₙ² = 1.875² = 3.516.
  2. Stiffness over mass: √(EI/μ) = √(200 / 2) = 10.
  3. fₙ = 3.516 × 10 / (2π × 1) = 5.5959 Hz.

Assumptions and limits

  • A uniform Euler–Bernoulli beam: slender, constant section, small deflections, no rotary inertia or shear deformation.
  • For a trial shape, it satisfies the geometric support conditions.

Common mistakes

  • Using the formula for thick or short beams, where shear and rotary inertia lower the frequencies.
  • Using the wrong dimensionless root for the supports.
  • Mixing the mass per metre with the total mass.

Statistics and uncertainty

Frequency scales with the square root of EI/μ and inversely with L², so a 2 % error in length gives a 4 % error in frequency. Section properties and the true support conditions are the dominant uncertainties in practice.

Where it is used

  • Benchmarking a finite-element beam model against a closed-form result.
  • Estimating bending frequencies and wave speeds for bars, shafts, pipes and cantilevers.

Maths you need

  • Second moment of area: the section property that sets bending stiffness
  • Mode shape: the pattern of motion at one natural frequency
  • Wavenumber: rad/m, 2π over wavelength

Related formulas

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