# Natural frequency of a simply supported bridge

EN 1992-2:2003 Clause 6.4.4 contains the formula

NOTE 8 For a simply supported bridge subjected to bending only, the
natural frequency may be estimated using the formula :

$$n_{0}\ \left\lbrack \text{Hz} \right\rbrack = \frac{17,75}{\sqrt{ \delta_{0}}}$$ (6.3)

where:

$$\delta_{0} \textrm{[mm]}$$ is the deflection at mid span due to permanent actions and is calculated, using a short term modulus for concrete bridges, in accordance with a loading period appropriate to the natural frequency of the bridge.

NR/GN/CIV/025 Clause 4.3.2.2 contains the same formula:

For simply supported members where the loading is mainly distributed uniformly $$n_{o}$$ may be taken as $$\frac{17.75}{\sqrt{\delta_{o}}}$$ where $$\delta_{o}$$ is the maximum deflection in mm of the member over the span $$L$$ under the loading defined for $$n_{o}$$ but with the concentrated load of 10kN represented by a uniformly distributed load of 20kN.

Some standards prefer to express this equation including a term for $$g$$, the acceleration due to gravity, such that the constant is unitless:

$n_{o} = 0.18 \sqrt { \frac{g}{\delta_{0}}}$

### Derivation

To derive this equation, we will use the Rayleigh-Ritz equation:

$T = V \label{a}\tag{Eq.1}$

where:

Symbol Description
$$V$$ is the maximum potential energy stored in the system
$$T$$ is the maximum kinetic energy stored in the system

For an Euler beam:

\begin{align} V & = \frac{1}{2}\int_{0}^{L}{\text{EI}\left\lbrack Y''\left( x \right) \right\rbrack^{2}\text{dx}} \label{b}\tag{Eq.2} \\ T & = \frac{1}{2}\int_{0}^{L}{m\omega^{2}\ \left\lbrack Y\left( x \right) \right\rbrack^{2}\text{dx}} \label{c}\tag{Eq.3} \end{align}

where:

Symbol Description
$$Y\left( x \right)$$ is the assumed mode shape of the beam (See NOTE 2)
$$\omega$$ is the angular frequency equal to $$2\pi n_{0}$$
$$m$$ is the mass per unit length of the beam
$$\text{EI}$$ is the bending stiffness of the beam

NOTE 1: These terms for $$V$$ and $$T$$ consider a unit amplitude as any amplitude terms will cancel when combined in $$\ref{a}$$.

NOTE 2: In order to obtain a good upper bound approximation of $$\omega$$ we only need assume a mode shape that is approximately correct, which lends itself to using numerical methods to iterate out the correct mode shape by minimising $$\omega^{2}$$. However, for this example we will use the exact mode shape:

$Y\left( x \right) = \sin{\pi\frac{x}{L}} \label{d}\tag{Eq.4}$

Therefore:

$Y''\left( x \right) = \operatorname{-sin}{\pi\frac{x}{L}} \label{e} \tag{Eq.5}$

Substituting the expressions for $$Y\left( x \right)$$ and $$Y''\left( x \right)$$ from $$\ref{d}$$ and $$\ref{e}$$ back into the equations for $$V$$ and $$T$$ in $$\ref{b}$$ and $$\ref{c}$$ we get:

\begin{align} V & = \frac{1}{2}\int_{0}^{L}{\text{EI}\left\lbrack \operatorname{-sin}{\pi\frac{x}{L}} \right\rbrack^{2}\text{dx}} \\ V & = \frac{\text{EI}\pi^{4}}{2L^{4}}\int_{0}^{L}{\operatorname{sin^{2}}{\pi\frac{x}{L}}\text{dx}} \\ V & = \frac{\text{EI}\pi^{4}}{4L^{4}}\label{f}\tag{Eq.6} \end{align}

and:

\begin{align} T & = \frac{1}{2}\int_{0}^{L}{m\omega^{2}\left\lbrack \sin{\pi\frac{x}{L}} \right\rbrack^{2}\text{dx}} \\ T & = \frac{m\omega^{2}}{2}\int_{0}^{L}{\operatorname{sin^{2}}{\pi\frac{x}{L}}\text{dx}} \\ T & = \frac{m\omega^{2}}{4} \label{g}\tag{Eq.7} \end{align}

Substituting the expressions for $$V$$ and $$T$$ in $$\ref{f}$$ and $$\ref{g}$$ back into the equation in $$\ref{a}$$:

\begin{align} \frac{m\omega^{2}}{4} & = \frac{\text{EI}\pi^{4}}{4L^{4}} \\ \omega^{2} & = \frac{\pi^{4}\text{EI}}{mL^{4}} \label{h}\tag{Eq.8} \end{align}

From data book formulae, the deflection $$\delta_{0}$$ of a simply supported beam under a uniformly distributed load $$mg$$ is given by the expression (see derivation):

$\delta_{0} = \frac{5mgL^{4}}{384EI}\label{i}\tag{Eq.9}$

Rearranging in terms of $$m$$:

$m = \frac{384EI\delta_{0}}{5\text{gL}^{4}}\label{j}\tag{Eq.10}$

Substituting the expression for $$m$$ in $$\ref{j}$$ back into the equation for $$\omega^{2}$$ in $$\ref{h}$$:

\begin{align} \omega^{2} & = \frac{5gL^{4}}{384EI\delta_{0}}\frac{\pi^{4}\text{EI}}{L^{4}} \\ \omega^{2} & = \frac{5g\pi^{4}}{384\delta_{0}} \\ 4\pi^{2}{n_{0}}^{2} & = \frac{5g\pi^{4}}{384\delta_{0}} \\ n_{0}^{2} & = \frac{5g\pi^{2}}{1536\delta_{0}} \\ n_{0} & = 0.18 \sqrt{\frac{g}{\delta_{0}}}\tag{Eq.11} \end{align}

where $$g = 9810 \textrm{ mm s}^{-2}$$:

$n_{0} \left\lbrack \text{Hz} \right\rbrack = \frac{17.75}{\sqrt{\delta_{0}\ \left\lbrack \text{mm} \right\rbrack}}\tag{Eq.12}$