# Beams Natural Vibration Frequency

Vibrations in a long floor span and a lightweight construction may be an issue if the strength and stability of the structure and human sensitivity is compromised. Vibrations in structures are activated by dynamic periodic forces - like wind, people, traffic and rotating machinery.

There are in general no problems with vibrations for normal floors with span/dept ratio less than 25. For lightweight structures with span above * 8 m (24 ft) * vibrations may occur. In general - as a rule of thumb - the natural frequency of a structure should be greater than * 4.5 Hz (1/s) *.

### Structures with Concentrated Mass

* f = (1 / (2 π)) (g / δ) ^{ 0.5 } (1) *

* where *

* f = natural frequency (Hz) *

* g = acceleration of gravity (9.81 m/s ^{2}) *

* δ = static dead load deflection estimated by elastic theory (m) *

** Note! ** - static dead load for a structure is load due to it's own weight or the weight of mass that is fixed to the structure.

### Structures with Distributed Mass

General rule for most structures

* f = a / (δ) ^{ 0.5 } (2) *

* a = numerical factor (in general 18) *

The numerical factor a can be calculated to * 15.75 * for a single lumped system but varies in general between * 16 and 20 * for similar systems. For practical solutions a factor of * 18 * is considered to give sufficient accuracy.

### Simply Supported Structure - Mass Concentrated in the Center

For a simply supported structure with the mass - or load due to gravitational force weight - acting in the center, the natural frequency can be estimated as

* f = (1 / (2 π)) (48 E I / M L ^{3} ) ^{ 0.5 } (3) *

* where *

* M = concentrated mass (kg) *

### Simply Supported Structure - Sagging with Distributed Mass

For a simply sagging supported structure with distributed mass - or load due to gravitational force - can be estimated as

* f = (π / 2) (E I / q L ^{4} ) ^{ 0.5 } (4) *

#### Example - Natural Frequency of Beam

The natural frequency of an unloaded (only its own weight - dead load) * 12 m * long DIN 1025 I 200 steel beam with Moment of Inertia * 2140 cm ^{4} (2140 10^{-8} m^{4} ) * and Modulus of Elasticity

*200 10*and mass

^{9}N/m^{2}*26.2 kg/m*can be calculated as

* f = (π / 2) ((200 10 ^{9} N/m^{2}) (2140 10^{-8} m^{4} ) / (26.2 kg/m) (12 m)^{4} ) ^{ 0.5 } *

* = 4.4 Hz - vibrations are likely to occur *

The natural frequency of the same beam shortened to * 10 m * can be calculated as

* f = (π / 2) ((200 10 ^{9} N/m^{2}) (2140 10^{-8} m^{4} ) / (26.2 kg/m) (10 m)^{4} ) ^{ 0.5 } *

* = 6.3 Hz - vibrations are not likely to occur *

### Simply Supported Structure - Contraflexure with Distributed Mass

For a simply contraflexure supported structure with distributed mass - or dead load due to gravitational force - can be estimated as

* f = 2 π (E I / q L ^{4} ) ^{ 0.5 } (5) *

### Cantilever with Mass Concentrated at the End

For a cantilever structure with the mass - or dead load due to gravitational force - concentrated at the end, the natural frequency can be estimated as

* f = (1 / (2 π)) (3 E I / F L ^{3} ) ^{ 0.5 } (6) *

### Cantilever with Distributed Mass

For a cantilever structure with distributed mass - or dead load due to gravitational force - the natural frequency can be estimated as

* f = 0.56 (E I / q L ^{4} ) ^{ 0.5 } (7) *

### Structure with Fixed Ends and Distributed Mass

For a structure with fixed ends and distributed mass - or dead load due to gravitational force - the natural frequency can be estimated as

* f = 3.56 (E I / q L ^{4} ) ^{ 0.5 } (8) *

## Related Topics

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Deflection and stress in beams and columns, moment of inertia, section modulus and technical information.

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