# Beams Natural Vibration Frequency

## Estimate structures 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

*f = (1 / (2 π)) (k / q)^{0.5} (2)*

*where *

*q = uniform distributed static dead load (N/m)*

*k = E I / L = stiffness*

*where *

*E = modulus of elasticity (N/m ^{2})*

*I = area moment of inertia (m^{4})*

*L = length (m)*

### General Rule for most Structures

*f = 18 / ( δ)^{0.5} (3)*

### 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 / F L)^{0.5} * (4)

*where *

*F = concentrated load (N)*

### 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} * (5)

#### Example - Natural Frequency of Beam

The natural frequency of an unloaded (only its own weight - dead load) *10 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 (weight = (26.2 kg/m) (9.81 m/s*can be calculated as

^{2}) = 257 N/m)*f = (π / 2) ((200 10^{9} N/m^{2}) (2140 10^{-8} m^{4}) / (257 N/m) (10 m)^{4})^{0.5} *

* = 2 Hz - *vibrations are likely to occur

The natural frequency of the same beam shortened to *6 m* can be calculated as

*f = (π / 2) ((200 10^{9} N/m^{2}) (2140 10^{-8} m^{4}) / (257 N/m) (6 m)^{4})^{0.5} *

* = 5.6 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} * (6)

### 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} * (7)

### 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} * (8)

### 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} * (9)