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)


\(\delta_{0} \textrm{[mm]}\) 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 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:

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


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

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


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:

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


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:

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


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

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:

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


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

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 \(g = 9810 \textrm{ mm s}^{-2}\):

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