Easy explanation of the equation of motion of beam vibration

beam vibration
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I will explain about vibration.

We take beam vibration as a typical example of a multi-degree-of-freedom system.

A multi-degree-of-freedom system can be thought of as having more than two degrees of freedom. You may need to analyze up to three degrees of freedom…

In addition to beam vibration, there are various patterns such as plates, but beams are the basis.

Even when considering complicated matters, I often use beams without fear of misunderstanding.

The concept of beams first appeared in the field of mechanics of materials when I entered university, but in my graduate school classes, it appeared as an applied concept in the world of vibration.

In textbooks, we will look at the derivation of the formula in detail, but we will omit the detailed calculations and focus on the interpretation of the results.

A lot of knowledge from material mechanics is used, so if you are interested in detailed derivations, please study material mechanics first. I used to use this formula as a matter of course, but I have often forgotten the detailed derivation. At first, I think it is sufficient to understand that it corresponds to a vibration system with one degree of freedom.

Beam vibration model

Consider the vibration model of a beam.

beam vibration

When it comes to the world of beams, it becomes specialized.

Meaning of constant

  • ρ: Density (kg/m 3 )
  • A: Cross-sectional area (m 2 )
  • E: Young’s modulus (N/mm 2 )
  • I: Moment of inertia (mm 4 )

Density and cross-sectional area remain the same.

Young’s modulus and moment of inertia are concepts that appear in mechanics of materials.

Any constant can be a design element.

However, it is difficult to change the density and Young’s modulus based on the intention of the designer, and the only way to do that is to change the cross-sectional area and moment of inertia.

  • Density and Young’s modulus: Determined naturally when the material is selected
  • Cross-sectional area and geometrical moment of inertia: cannot be determined unless the shape of the object is determined

Relationship between deflection, deflection angle, and moment

In material mechanics, there are concepts such as deflection, deflection angle, and moment.

DeflectionDeflection anglemomentshear force

the longitudinal direction of the beamx・Vertical directionywhenybecomes deflected.

Deflection is the amount of deformation when a force is applied to a beam .

Deflection isxIt changes depending on the value of. Since it targets vibration, it also depends on time \(t\).

Let’s use the following formula to intentionally express this.


Deflection angle, moment, and shear force are closely related to deflection .

  • Deflection angleΘis the deflectionyofxDifferentiated in the direction.
  • the deflection angleΘofxDifferentiated in the direction.
  • shear forceNis the momentM.ofxDifferentiated in the direction.

Let’s make an equation of motion based on the definition of this side.

Beam equation of motion

Even the equation of motion of the beam is basically the same.

$$ ρA\frac{\partial^2 y(x,t) }{\partial t^2 }+EI\frac{\partial^4 y(x,t) }{\partial x^4 } = f(x,t) $$

This formula is very important .

this is

$$ m\ddot{x}+kx=f $$

It has the same form as the expression.


The first term on the left side ism = ρA _Let’s think about this relationship.

This is because the equation of motion of the beam considers the following microsections of the beam.

Small interval like belowdxThe weight ofρ AdxIt becomes.

Weight of micro DX


The second term on the left side indicates the restoring force. This is a little more difficult.

When considering a beam, a minute intervaldxhas shear forceN( x , t )I think it will take.

shear force

The direction of shear force is directly from the mechanics of materials.

This minute intervaldxThe total force applied to

$$ \frac{\partial N}{\partial x}dx $$

It can be expressed as

Now, the momentM.and shear forceNcan be expressed as follows.

$$ M = -EI\frac{\partial^2 y(x,t)}{\partial x^2} $$

$$ N = \frac{\partial M}{\partial x} $$

Using these relationships

$$ -EI\frac{\partial^4 y(x,t)}{\partial x^4}dx $$

The force is a minute intervaldxwill join.

external force

The force applied to a beam of unit lengthf( x , t )Then, the infinitesimal intervaldxThe force applied tof( x , t ) dxis.

I just multiplied it.

For the two terms on the left hand side and for both terms on the right hand sidedxis included, so if you delete it, it becomes the equation of motion.


The equation of motion for beam vibration is shown in the figure below when compared with the equation of motion for vibration with one degree of freedom.

analogy (vibration)

It is exactly the same in that it consists of mass, restoring force, and external force.

natural frequency

To solve complex calculations such as beam motion equations, we generally rely on numerical calculation software such as MATLAB.

Here, let me introduce only the method of obtaining the natural frequency.

Using the Fourier transform, we separate the variables as follows:

$$ y(x,t)=φ(x)ψ(t) $$

It is a form that separates the term for position and the term for time.

$$ ρAφ(x)\frac{\partial^2 ψ(t)}{\partial t^2 }+EIψ(t)\frac{\partial^4 φ(x) }{\partial x^4 } = 0 $$

$$ \frac{1}{ψ(t)}\frac{\partial^2 ψ(t)}{\partial t^2 } = -\frac{ρA}{EI}\frac{1}{φ(x)}\frac{\partial^4 φ(x) }{\partial x^4 } = C $$

section on time

Let’s take a closer look at the section regarding time.

$$ \frac{1}{ψ(t)}\frac{\partial^2 ψ(t)}{\partial t^2 } = C $$

This solution is easy.

$$ ψ(t)=Ae^{iωt} $$

and takes the form of a Fourier transform.

This results in the same shape as an oscillating system with one degree of freedom.

t = 0This can be solved by determining the position and velocity under the initial conditions. (theoretically)

ωis the natural frequency.

Positional Section

Let’s take a closer look at the terms regarding position.

$$ -\frac{ρA}{EI}\frac{1}{φ(x)}\frac{\partial^4 φ(x) }{\partial x^4 } = C $$

This can be solved in the following form.

$$ φ(x)=B_1sin(λx)+B_2cos(λx)+B_3sinh(λx)+B_4cosh(λx) $$

For now, think of it like this.

λare called eigenmodes.

For beams, end conditions are required as boundary conditions.

  1. Fixed
  2. freedom

Conditions such as deflection and angle of deflection are determined.

Once the boundary conditions are determined,ωandλis determined by a 1:1 relationship.

Regarding eigenmodes, we will be talking about modal analysis, which is a slightly more specialized topic, so we will omit it this time.

For example, the eigenmode of a cantilever beam has the shape below.

Eigenmode (vibration)

A beam that vibrates in response to an external force at a specific frequency moves at that frequency in the form of a superposition of these eigenmodes.

Conceptually, it is similar to Fourier transform.


Few mechanical and electrical engineers at chemical plants have the luxury of having specialized knowledge regarding vibration.

If you want to study, the following books are useful.

created by Rinker
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I have explained the equation of motion of beam vibration by extracting only the essential part.

Knowledge of both material mechanics and vibration engineering is required.

Although there are limits to finding analytical solutions and we must rely on calculation software, this essence is very important for verification.

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