Derivation for bridge natural frequency estimate in Eurocodes

Question

The Eurocodes gives the following equation for estimating a "simply supported bridge subject to bending only"*:

$$n_0 = \frac{17.75}{\sqrt{\delta_0}}$$

Where

• $n_0$ is the natural frequency in hertz
• $\delta_0$ is the deflection at mid-span under permanent actions in mm

The equation is seemly plucked from thin air, and there is no explanation as to where the constant 17.75 comes from. As an engineer I'm loath to use a formula I don't understand, but more than that it would be helpful to learn the fundamentals behind it so that I can see if it can be altered to work with other support conditions.

Can anyone provide a derivation / fundamental origin to this relationship?

*Full reference is: EN 1991-2:2003 6.4.4 [Note 8] (Equation 6.3), if that helps.

Answer 1

If we simplify the whole bridge into 2D thin beam with a constant section size, no internal damping and subject only to small vertical deflections, then the natural frequency is determined by simple harmonic motion:

$$ n_0 = \frac{1}{2 \pi} \sqrt{ \frac{ k } { m } } $$

Where $ n_0 $ is the natural frequency, $ k $ is the ratio between restorative force and deflection (the equivalent 'spring stiffness') and $m$ is the mass per unit length of the beam.

In a beam the restorative force is the internal shear caused by the deflected shape. As the force exhibited by a beam is proportional to the rate of change of shear, which is related to the stiffness ($EI$) and the rate of change of moment it can be shown (note: the deflection is proportional to the length of the beam) that:

$$ k = \alpha \frac{ EI } { L^4 } $$

Where $E$ is the Young's Modulus of the beam material, $I$ is the Second Moment of Inertia of the beam section, $L$ is the length of the beam and $\alpha$ is a constant determined by the support conditions and mode number of the response.

All of the literature I have seen expresses this in a way that more convenient for the frequency equation:

$$ k = \left( \frac{ K }{L^2} \right)^2 (EI) $$

Substituting back in,

$$ n_0 = \frac{ K }{ 2 \pi L^2 } \sqrt{ \frac{ EI } {m} } $$

Calculating the value of $K$ is quite involved, and there is an exact approach for simple solutions, and approximate methods including the free energy method and Raleigh Ritz. A few deviations for a simply supported beam can be found here.

It should be noted that this equation would have been enough, but as it requires a table for $K$ and the calculation of a value of $EI$ that represents the bridge as a homogenous beam, the authors of the Eurocode seem to have decided it would be better re-integrate the assumption that $k$ is constant along the beam.

To do this they have used the following relationship:

$$ \delta_0 = C \frac { w L^4 } { EI } $$

Where $\delta_0$ is the maximum deflection, $C$ is a constant dictated by the support conditions, $w$ is a constant uniformly distributed load across the length of the beam.

Under self-weight $w = gm$, where $g$ is acceleration due to gravity (9810 mm/s²; as deflection in this equation is given in mm).

Therefore (re-arranged:)

$$ \sqrt { \frac { EI } { m } } = L^2 \sqrt { 9810 } \frac { \sqrt { C } } { \sqrt { \delta_0 } } $$

And so:

$$ n_0 = \frac { 15.764 K \sqrt { C } } { \sqrt { \delta_0 } } $$

General values for $K$ and $C$ can be found in structural tables- for example here, and here, respectively.

For a simply supported beam:

$$ K = \pi ^ 2 \text{ and } C = \frac { 5 } { 384 } $$ $$ 15.764 K \sqrt { C } = 17.75 $$ $$ n_0 = \frac{ 17.75 } { \sqrt { \delta } } $$

Answer 2

Here's a possible answer.

I found this document (not sure of the exact source), which contains a related derivation:

In a simple harmonic motion problem, $$n_0=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$$ where $k$ is the elastic stiffness and $m$ is the mass undergoing vibration.

$$k=\frac{\text{load}}{\text{deflection}}=\frac{F}{\delta}$$ where $F$ is force and $\delta$ is the deflection. Thus, $$n_0=\frac{1}{2 \pi}\sqrt{\frac{F}{m\delta}}=\frac{1}{2 \pi}\sqrt{\frac{ma}{m \delta}}=\frac{1}{2 \pi}\sqrt{\frac{a}{\delta}}$$ But the deflection in your example is in millimeters, while it's in meters here, so I get about $$n_0=5.03 \sqrt{\frac{a}{\delta}}$$ If $a=12.4382$, we get your equation. But I'm not sure where this value comes from. It could be that another unit switch is needed, or it could be that this constant is only for a small subset of cases, where the acceleration is along those lines.

Answer 3

There is some more information on this in Ladislav Fryba's book "Dynamics of Railway Bridges" (1996). If you read chapter 4, you will see formula 4.53 on page 92:

$$f_1 = 17.753 v_{st}^{-1/2}$$

With $f_1$ being the first natural frequency in Hertz and $v_{st}$ the midspan deflection in mm. This is exacly the formula you are asking about.

This equation follows from the formula for the midspan deflection of a simply supported beam loaded by a uniformly distributed load μg

$$v_{st} = {5 \above 1pt 384} {\mu gl^4 \above 1pt EI} $$

which is substituted in

$$ f_j = {\lambda_j^4 \above 1pt l^4} ({EI \above 1pt \mu})^{1/2} $$

It yields $$ \lambda_1 = \pi $$

Substituting those equations into each other using g = 9.81 m/s^2 gives

$$ f_1 = {\pi \above 1pt 2} ({5 \above 1pt 384} g)^{1/2} v_{st}^{-1/2}$$

The numerical evaluation of this equation yields the desired equation.

Answer 4

Dynamics for engineers like myself, generally concerned with statics, can be fraught with easy to make errors, and misunderstandings. This formula is very useful for simply supported beams since it can be related quickly to the applied self weight loads and a proportion of live loading (generally 10%) without having to go into complications.

Also Cantilevers can use a similar constant (19.8 with udl, 15.8 with end point load). It all breaks down with continuous beams and frames.

I build in a natural frequency check with all beam designs to keep track of it. For timber structures for instance 8Hz is the target and for concrete floors/steel frames 4-6Hz - as a first pass.

There are also rough and ready methods for assessing dynamic responses around. I have to say dynamics still eludes and confuses me and always will! So I stay as simple as possible.