User:Lzyvzl/Vibration Draft

Vibration refers to mechanical oscillations about an equilibrium point. The oscillations may be periodic such as the motion of a pendulum or random such as the movement of a tire on a gravel road.

Vibration is occasionally desirable. For example the motion of a tuning fork, the reed in a woodwind instrument or harmonica, or the cone of a loudspeaker is desirable vibration, necessary for the correct functioning of the various devices.

More often, vibration is undesirable, wasting energy and creating unwanted sound--noise. For example, the motions of engines, electric motors, or any mechanical device in operation are usually unwanted vibrations. Such vibrations can be caused by imbalances in the rotating parts, uneven friction, the meshing of gear teeth, parts that are dragging together, etc. Careful designs usually minimize unwanted vibrations.

The study of sound and vibration are closely related. Sound, pressure waves, are generated by vibrating structures (e.g. speaker cone) and pressure waves can generate vibration of structures (e.g. ear drum). Hence, when trying to reduce noise it is often a problem in trying to reduce vibration.

Types of vibration

Free vibration occurs when a mechanical system is set off with an initial input and then allowed to vibrate freely. Examples of this type of vibration are pulling a child back on a swing and then letting go or hitting a tuning fork and letting it ring. The mechanical system will then vibrate at one or more of its natural frequencies and damp down to zero.

Forced vibration is when an alternating force or motion is applied to a mechanical system. Examples of this type of vibration are a shaking washing machining due to an imbalance or the vibration of a building during an earthquake. In forced vibration the frequency of the vibration is dependent on the frequency content of the force or motion applied, but the magnitude of the vibration is strongly dependent on the behavior of the mechanical system.

Vibration analysis

The fundamentals of vibration analysis can be understood by studying the simple mass-spring-damper model. Indeed, even a complex structure such as an automobile body can be modeled as a summation of simple mass-spring-damper models. The mass-spring-damper model is an example of a simple harmonic oscillator and hence the mathematics used to describe its behavior is identical to other simple harmonic oscillators such as the RLC circuit.

Note: In this article the step by step mathematical derivations will not be included, but will focus on the major equations and concepts in vibration analysis. Please refer to the references at the end of the article for detailed derivations.

Free vibration without damping

To start, the investigation of the mass-spring-damper will assume the damping is negligible and that there is no external forces applied to the mass (i.e. free vibration).

The force applied to the mass by the spring is proportional to the amount the spring is stretched "x" (we will assume the spring is already compressed due to the weight of the mass). The proportionality constant, k, is the stiffness of the spring and has units of force/distance (e.g. lbf/in or N/m)

F_{\mathrm {s} }\ \ =\ \ -kx

The force generated by the mass is proportional to the acceleration of the mass per Newton’s second law of motion.

\Sigma \ F\ \ =\ \ ma\ \ =\ \ m{\ddot {x}}\ \ =\ \ m{\frac {d^{2}x}{dt^{2}}}

The sum of the forces on the mass then generates this ordinary differential equation:

m{\ddot {x}}+kx=0.

If we assume that we start the system to vibrate by stretching the spring by the distance of “A” and letting go, the solution to the above equation that describes the motion of the mass is:

x(t)=A\cos(2\pi f_{n}t)\!

This solution says that it will oscillate with an amplitude of “A” and with a frequency of $f_{n}$ , but what is $f_{n}$ ? $f_{n}$ is one of the most important quantities in vibration analysis and is called the undamped natural frequency.

$f_{n}$ is defined for the simple mass-spring system as:

f_{n}={1 \over {2\pi }}{\sqrt {k \over m}}\!

Note: Angular frequency $\omega$ ( $\omega =2\pi f$ ) with the units of radians/sec is often used in equations because it simplifies the equations, but is normally converted to “standard” frequency (units of Hz or equivalently cycles/second) when stating the frequency of a system.

If you know the mass and stiffness of the system you can determine the frequency at which the system will vibrate once it is set in motion by an initial disturbance using the above formula. Every vibrating system has one or more natural frequencies that it will vibrate at once it is disturbed. This simple relation can be used to understand in general what will happen to a more complex system once we add mass or stiffness. For example, the above formula explains why when a car or truck is fully loaded the suspension will feel “softer” then unloaded because the mass has increased and hence reduced the natural frequency of the system.

What causes the system to vibrate under no force?

These formulas describe the resulting motion, but they do not explain why the system oscillates. The reason for the oscillation is due to the conservation of energy. In the above example we have extended the spring by a value of “A” and therefore have stored potential energy ( ${\tfrac {1}{2}}kx^{2}$ ) in the spring. Once we let go of the spring, the spring tries to return to its un-stretched state and in the process accelerates the mass. At the point where the spring has reached its un-stretched state it no longer has any energy stored, but the mass has reached its maximum speed and hence all the energy has been transformed into kinetic energy ( ${\tfrac {1}{2}}mv^{2}$ ). The mass then begins to decelerate because it is now compressing the spring and in the process transferring the kinetic energy back into potential. This transferring back and forth of the kinetic energy in the mass and the potential energy in the spring causes the mass to oscillate.

In our simple model the mass will continue to oscillate forever at the same magnitude, but in a real system there is always something that dissipates the energy, called damping, that makes the system eventually stop vibrating.

Free vibration with damping

File:Mass spring damper.png

Mass Spring Damper Model

We now add a viscous damper to the model that outputs a force that is proportional to the velocity of the mass. The damping is called viscous because it models the effects of an object in a fluid. The proportionality constant “c” is called the damping coefficient and has units of Force over velocity (lbf s/ in or N s/m).

F_{\mathrm {d} }\ \ =\ \ -cv\ \ =\ \ -c{\dot {x}}\ \ =\ \ -c{\frac {dx}{dt}}

By summing the forces on the mass we get the following ordinary differential equation:

m{\ddot {x}}+{c}{\dot {x}}+{k}x=0.

The solution to this equation depends on the amount of damping. If the damping is small enough the system will still vibrate, but will stop vibrating over time. This case is called underdamping--the case of most interest in vibration analysis. If we increase the damping to the point where the system no longer oscillates we reach the point of critical damping (if the damping is increased past critical damping the system is called overdamped). The value that the damping coefficient needs to reach for critical damping in the mass spring damper model is:

c_{c}=2{\sqrt {km}}

To characterize the amount of damping in a system a ratio called the damping ratio (also known as damping factor and % critical damping) is used. This damping ratio is just a ratio of the actual damping over the amount of damping required to reach critical damping. The formula for the damping ratio ( $\zeta$ ) of the mass spring damper model is:

\zeta ={c \over 2{\sqrt {km}}}.

For example, metal structures (e.g. airplane fuselage, engine crankshaft) will have damping factors less than 0.05 while automotive suspensions in the range of 0.2-0.3.

The solution to the underdamped system for the mass spring damper model is the following:

x(t)=Xe^{-\zeta \omega _{n}t}\cos({{\sqrt {1-\zeta ^{2}}}\omega _{n}t-\phi }),\ \ \omega _{n}=2\pi f_{n}

The value of “X”, the initial magnitude, and $\phi$ , the phase shift, are determined by the amount the spring is stretched. The formulas for these values can be found in the references.

The major points to note from the solution are the exponential term and the cosine function. The exponential term defines how quickly the system “damps” down – the larger the damping ratio, the quicker it damps to zero. The cosine function is the oscillating portion of the solution, but the frequency of the oscillations is different then the undamped case.

The frequency in this case is called the damped natural frequency, $f_{d}$ , and is related to the undamped natural frequency by the following formula:

f_{d}={\sqrt {1-\zeta ^{2}}}f_{n}

The damped natural frequency is less then the undamped natural frequency, but for many practical cases the damping ratio is relatively small and hence the difference is negligible. Therefore the damped and undamped description are often dropped when stating the natural frequency (e.g. with 0.1 damping ratio, the damped natural frequency is only 1% less then the undamped).

The plots to the side present how 0.1 and 0.3 damping ratios effect how the system will “ring” down over time. What is often done in practice is to experimentally measure the free vibration after an impact (for example by a hammer) and then determine the natural frequency of the system by measuring the rate of oscillation and the damping ratio by measuring the rate of decay. The natural frequency and damping ratio are not only important in free vibration, but also characterize how a system will behave under forced vibration.

Forced vibration with damping

In this section we will look at the behavior spring mass damper model when we add a harmonic force in the form below. A force of this type would, for example, be generated by a rotating imbalance.

F=F_{0}\cos {(2\pi ft)}\!

If we again sum the forces on the mass we get the following ordinary differential equation:

m{\ddot {x}}+{c}{\dot {x}}+{k}x=F_{0}\cos {(2\pi ft)}

The steady state solution of this problem can be written as:

x(t)=X\cos {(2\pi ft-\phi )}\!

The result states that the mass will oscillate at the same frequency, f, of the applied force, but with a phase shift.

The amplitude of the vibration “X” is defined by the following formula.

X={F_{0} \over k}{1 \over {\sqrt {(1-r^{2})^{2}+(2\zeta r)^{2}}}}

Where “r” is defined as the ratio of the frequency of the harmonic force over the undamped natural frequency of mass-spring-damper model.

r={\frac {f}{f_{n}}}

The phase shift , $\phi$ , is defined by following formula.

\phi =\arctan {\left({\frac {2\zeta r}{1-r^{2}}}\right)}

The plot of these functions, called the frequency response of the system, presents one of the most important features in forced vibration. In a lightly damped system when the forcing frequency nears the natural frequency ( $r\approx 1$ ) the amplitude of the vibration “blows up”. This phenomenon is called resonance. Subsequently the natural frequency of a system is often referred to as the resonance frequency.

If resonance occurs in a mechanical system it can be very harmful-- leading to eventual failure of the system. Consequently one of the major reasons for vibration analysis is to predict when resonance may occur and to determine what steps to take to prevent it from occurring. As the amplitude plot shows, adding damping can significantly reduce the magnitude of the vibration. Also, the magnitude can be reduced if the natural frequency can be shifted away from the forcing frequency by changing the stiffness or mass of the system. If the system cannot be changed, perhaps the forcing frequency can be shifted (for example, changing the speed of the machine generating the force).

The following are some other points in regards to the forced vibration shown in the frequency response plots.

At a given frequency ratio, the amplitude of the vibration, X, is directly proportional the amplitude of the force $F_{0}$ (e.g. If you double the force, the vibration doubles)
With little or no damping, the vibration is in phase with the forcing frequency when the frequency ratio (r ) < 1 and 180 degrees out of phase when the frequency ratio (r) >1
When r<<1 the amplitude is just the deflection of the spring under the static force $F_{0}$ . This deflection is called the static deflection $\delta _{st}$ . Hence, when r<<1 the effects of the damper and the mass are minimal.
When r>>1 the amplitude of the vibration is actually less than the static deflection $\delta _{s}t$ . In this region the force generated by the mass (F=ma) is dominating because the acceleration seen by the mass increases with the frequency. Since the deflection seen in the spring, X, is reduced in this region, the force transmitted by the spring (F=kx) to the base is reduced. Therefore the mass-spring-damper system is isolating the harmonic force from the mounting base—referred to as vibration isolation. Interestingly, more damping actually reduces the effects of vibration isolation when r>>1 because the damping force (F=cv) is also transmitted to the base.

Applying "Complex" Forces to the Mass-Spring-Damper Model

In a previous section only a simple harmonic force was applied to the model, but this can be extended considerably using two powerful mathematical tools. The first is the Fourier transform that takes a signal as a function of time (time domain) and break it down into its harmonic components as a function of frequency (frequency domain). For example, let us apply a force to the mass-spring-damper model that repeats the following cycle--a force equal to 1 Newton for 0.5 seconds and then no force for 0.5 seconds. This type of force has the shape of a 1 Hz square wave.

The Fourier transform of the square wave generates a frequency spectrum that presents the magnitude of the harmonics that make up the square wave (the phase is also generated, but is typically of less concern and therefore is often not plotted). The Fourier transform can also be used to analyze non-periodic functions such as transients (e.g. impulses) and random functions. With the advent of the modern computer the Fourier transform is almost always computed using the Fast Fourier Transform (FFT) computer algorithm in combination with a window function.

In the case of our square wave force, the first component is actually a constant force of 0.5 Newtons and is represented by a value at "0" Hz in the frequency spectrum. The next component is a 1 Hz sine wave with an amplitude of 0.64. This is shown by the line at 1 Hz. The remaining components are at the odd frequencies and it takes an infinite amount of sine waves to generate the perfect square wave. Hence, the Fourier transform allows you to interpret the force as a sum of sinusoidal forces being applied instead of the more "complex" force (e.g. a square wave).

In the previous section, the vibration solution was given for a single harmonic force, but the Fourier transform will in general give multiple harmonic forces. The second mathematical tool, the principle of superposition, allows you to sum the solutions from multiple forces if the system is linear. In the case of the spring-mass-damper model, the system is linear if the spring force is proportional to the displacement and the damping is proportional to the velocity over the range of motion of interest. Hence, the solution to the problem with a square wave is summing the predicted vibration from each one of the harmonic forces found in the frequency spectrum of the square wave.

Frequency Response Model

We can view the solution of a vibration problem as an input/output relation--where the force is the input and the output is the vibration. If we represent the force and vibration in the frequency domain (magnitude and phase) we can write the following relation:

X(\omega )=H(\omega )*F(\omega )\ \ or\ \ H(\omega )={X(\omega ) \over F(\omega )}

$H(\omega )$ is called the frequency response function (also referred to the transfer function, but not technically as accurate) and has both a magnitude and phase component (if represented as a complex number, a real and imaginary component). The magnitude of the frequency response function (FRF) was presented earlier for the mass-spring-damper system.

|H(\omega )|=\left|{X(\omega ) \over F(\omega )}\right|={1 \over k}{1 \over {\sqrt {(1-r^{2})^{2}+(2\zeta r)^{2}}}},\ \ where\ \ r={\frac {f}{f_{n}}}={\frac {\omega }{\omega _{n}}}

The phase of the FRF was also presented earlier as:

\angle H(\omega )=\arctan {\left({\frac {2\zeta r}{1-r^{2}}}\right)}

For example, let us calculate the FRF for a mass-spring-damper system with a mass of 1 kg, spring stiffness of 1.93 N/mm and a damping ratio of 0.1. The values of the spring and mass give a natural frequency of 7 Hz for this specific system. If we apply the 1 Hz square wave from earlier we can calculate the predicted vibration of the mass. The figure illustrates the resulting vibration. It happens in this example that the fourth harmonic of the square wave falls at 7 Hz. The frequency response of the mass-spring-damper therefore outputs a high 7 Hz vibration even though the input force had a relatively low 7 Hz harmonic. This example highlights that the resulting vibration is dependent on both the forcing function and the system that the force is applied.

The figure also shows the time domain representation of the resulting vibration. This is done by performing an inverse Fourier Transform that converts frequency domain data to time domain. In practice, this is rarely done because the frequency spectrum provides all the necessary information.

The frequency response function (FRF) does not necessarily have to be calculated from the knowledge of the mass, damping, and stiffness of the system, but can be measured experimentally. For example, if you apply a known force and sweep the frequency and then measure the resulting vibration you can then calculate the frequency response function, and hence characterize the system. This technique is used in the field of experimental modal analysis to determine the vibration characteristics of a structure.

Multiple Degree of Freedom Systems and Mode Shapes

The simple mass-spring damper model is the foundation of vibration analysis, but what about more complex systems? The mass-spring-damper model described above is called a single degree of freedom (DOF) model since we have assumed the mass only moves up and down. In the case of more complex systems we need to discretize the system into more masses and allow them to move in more than one direction--adding degrees of freedom. The major concepts of multiple degrees of freedom (MDOF) can be understood by looking at just a 2 degree of freedom model as shown in the figure.

The equations of motion of the 2DOF system are found to be:

m_{1}{\ddot {x_{1}}}+{(c_{1}+c_{2})}{\dot {x_{1}}}-{c_{2}}{\dot {x_{2}}}+{(k_{1}+k_{2})}x_{1}-{k_{2}}x_{2}=f_{1}

m_{2}{\ddot {x_{2}}}-{c_{2}}{\dot {x_{1}}}+{(c_{2}+c_{3})}{\dot {x_{2}}}-{c_{3}}{\dot {x_{3}}}-{k_{2}}x_{1}+{(k_{2}+k_{3})}x_{2}-{k_{3}}x_{3}=f_{2}\!

We can rewrite this in matrix format:

{\begin{bmatrix}m_{1}&0\\0&m_{2}\end{bmatrix}}{\begin{Bmatrix}{\ddot {x_{1}}}\\{\ddot {x_{2}}}\end{Bmatrix}}+{\begin{bmatrix}c_{1}+c_{2}&-c_{2}\\-c_{2}&c_{2}+c_{3}\end{bmatrix}}{\begin{Bmatrix}{\dot {x_{1}}}\\{\dot {x_{2}}}\end{Bmatrix}}+{\begin{bmatrix}k_{1}+k_{2}&-k_{2}\\-k_{2}&k_{2}+k_{3}\end{bmatrix}}{\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}={\begin{Bmatrix}f_{1}\\f_{2}\end{Bmatrix}}

A more compact form of this matrix equation can be written as:

{\begin{bmatrix}M\end{bmatrix}}{\begin{Bmatrix}{\ddot {x}}\end{Bmatrix}}+{\begin{bmatrix}C\end{bmatrix}}{\begin{Bmatrix}{\dot {x}}\end{Bmatrix}}+{\begin{bmatrix}K\end{bmatrix}}{\begin{Bmatrix}x\end{Bmatrix}}={\begin{Bmatrix}f\end{Bmatrix}}\!

Where ${\begin{bmatrix}M\end{bmatrix}}$ , ${\begin{bmatrix}C\end{bmatrix}}$ , and ${\begin{bmatrix}K\end{bmatrix}}$ are symmetric matrices referred respectively as the mass, damping, and stiffness matrices. The matrices are NxN square matrices where N is the number of degrees of freedom of the system.

In the following analysis we will consider the case where there is no damping and no applied forces (i.e. free vibration). The solution of a viscously damped system is somewhat more complicated and is shown in Maia, Silva. ^[1].

{\begin{bmatrix}M\end{bmatrix}}{\begin{Bmatrix}{\ddot {x}}\end{Bmatrix}}+{\begin{bmatrix}K\end{bmatrix}}{\begin{Bmatrix}x\end{Bmatrix}}=0

This differential equation can be solved by assuming the following type of solution:

{\begin{Bmatrix}x\end{Bmatrix}}={\begin{Bmatrix}X\end{Bmatrix}}e^{i\omega t}

Note: Using the exponential solution of ${\begin{Bmatrix}X\end{Bmatrix}}e^{i\omega t}$ is a mathematical trick used to solve linear differential equations. If we use Euler's formula and take only the real part of the solution it is the same cosine solution for the 1 DOF system. The exponential solution is only used because it easier to manipulate mathematically.

The equation then becomes:

{\begin{bmatrix}-\omega ^{2}{\begin{bmatrix}M\end{bmatrix}}+{\begin{bmatrix}K\end{bmatrix}}\end{bmatrix}}{\begin{Bmatrix}X\end{Bmatrix}}e^{i\omega t}=0

Since $e^{i\omega t}=0$ cannot equal zero the equation reduces to the following.

{\begin{bmatrix}{\begin{bmatrix}K\end{bmatrix}}-\omega ^{2}{\begin{bmatrix}M\end{bmatrix}}\end{bmatrix}}{\begin{Bmatrix}X\end{Bmatrix}}=0

eigenvalue problem

This is referred to an eigenvalue problem in mathematics and can be be put in the standard format by pre-multiplying the equation by ${\begin{bmatrix}M\end{bmatrix}}^{-1}$

{\begin{bmatrix}{\begin{bmatrix}M\end{bmatrix}}^{-1}{\begin{bmatrix}K\end{bmatrix}}-\omega ^{2}{\begin{bmatrix}M\end{bmatrix}}^{-1}{\begin{bmatrix}M\end{bmatrix}}\end{bmatrix}}{\begin{Bmatrix}X\end{Bmatrix}}=0

or if we let ${\begin{bmatrix}M\end{bmatrix}}^{-1}{\begin{bmatrix}K\end{bmatrix}}={\begin{bmatrix}A\end{bmatrix}}$ and $\lambda =\omega ^{2}\,$

{\begin{bmatrix}{\begin{bmatrix}A\end{bmatrix}}-\lambda {\begin{bmatrix}I\end{bmatrix}}\end{bmatrix}}{\begin{Bmatrix}X\end{Bmatrix}}=0

The solution to the problem results in N eigenvalues (ie. $\omega _{1}^{2},\omega _{2}^{2},..\omega _{N}^{2}$ ), where N corresponds to the number of degrees of freedom. The eigenvalues provide the natural frequencies of the system. When these eigenvalues are substituted back into the original set of equations, the values of ${\begin{Bmatrix}X\end{Bmatrix}}$ that correspond to each eigenvalue are called the eigenvectors. These eigenvectors represent the mode shapes of the system. The solution of an eigenvalue problem can be quite cumbersome (especially for problems with many degrees of freedom), but fortunately most math analysis programs have eigenvalue routines.

The eigenvalues and eigenvectors are often written in the following matrix format and describe the modal model of the system:

{\begin{bmatrix}^{\diagdown }\omega _{r\diagdown }^{2}\end{bmatrix}}={\begin{bmatrix}\omega _{1}^{2}&\cdots &0\\\vdots &\ddots &\vdots \\0&\cdots &\omega _{N}^{2}\end{bmatrix}}

and

{\begin{bmatrix}\Psi \end{bmatrix}}={\begin{bmatrix}{\begin{Bmatrix}\psi _{1}\end{Bmatrix}}{\begin{Bmatrix}\psi _{2}\end{Bmatrix}}\cdots {\begin{Bmatrix}\psi _{N}\end{Bmatrix}}\end{bmatrix}}

A simple example using our 2 DOF model can help illustrate the concepts. Let both masses have a mass of 1 kg and the stiffness of all three springs equal 1000 N/m. The mass and stiffness matrix for this problem are then:

{\begin{bmatrix}M\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}

and

{\begin{bmatrix}K\end{bmatrix}}={\begin{bmatrix}2000&-1000\\-1000&2000\end{bmatrix}}

Then ${\begin{bmatrix}A\end{bmatrix}}={\begin{bmatrix}2000&-1000\\-1000&2000\end{bmatrix}}$ .

The eigenvalues for this problem given by an eigenvalue routine will be:

{\begin{bmatrix}^{\diagdown }\omega _{r\diagdown }^{2}\end{bmatrix}}={\begin{bmatrix}1000&0\\0&3000\end{bmatrix}}

The natural frequencies in the units of hertz are then (remembering $\omega =2\pi f$ ) $f_{1}=5.03Hz$ and $f_{2}=8.72Hz$ .

${\begin{bmatrix}\Psi \end{bmatrix}}={\begin{bmatrix}{\begin{Bmatrix}\psi _{1}\end{Bmatrix}}{\begin{Bmatrix}\psi _{2}\end{Bmatrix}}\end{bmatrix}}={\begin{bmatrix}{\begin{Bmatrix}-0.707\\-0.707\end{Bmatrix}}_{1}{\begin{Bmatrix}0.707\\-0.707\end{Bmatrix}}_{2}\end{bmatrix}}$

Since the system is a 2 DOF system, there are two modes with their respective natural frequencies and shapes. The mode shape vectors are not the absolute motion, but just describe relative motion of the degrees of freedom. In our case the first mode shape vector is saying that the masses are moving together in phase since they have the same value and sign. In the case of the second mode shape vector, each mass is moving in opposite direction at the same rate.

When there are many degrees of freedom, the best method of visualizing the mode shapes is by animating them. An example of animated mode shapes is shown in the figure below for a cantilevered I-beam. In this case, a Finite Element Analysis model was used to generate the mass and stiffness matrices and solve the eigenvalue problem. Even this relatively simple model has over a 100 degrees of freedom and hence as many natural frequencies and mode shapes. In general only the first few modes are important.

The Mode Shapes of a Cantilevered I-beam

1st lateral bending

1st torsional

1st vertical bending

2nd lateral bending

2nd torsional

2nd vertical bending

Multiple DOF problem converted to a single DOF problem

The eigenvectors have very important properties called orthoganility properties. These properties can be used to greatly simplify the solution of multi-degree of freedom models. It can be shown that the eigenvectors have the following properties:

{\begin{bmatrix}\Psi \end{bmatrix}}^{T}{\begin{bmatrix}M\end{bmatrix}}{\begin{bmatrix}\Psi \end{bmatrix}}={\begin{bmatrix}^{\diagdown }m_{r\diagdown }\end{bmatrix}}

{\begin{bmatrix}\Psi \end{bmatrix}}^{T}{\begin{bmatrix}K\end{bmatrix}}{\begin{bmatrix}\Psi \end{bmatrix}}={\begin{bmatrix}^{\diagdown }k_{r\diagdown }\end{bmatrix}}

${\begin{bmatrix}^{\diagdown }m_{r\diagdown }\end{bmatrix}}$ and ${\begin{bmatrix}^{\diagdown }k_{r\diagdown }\end{bmatrix}}$ are diagonal matrices that contain the modal mass and stiffness values for each one of the modes. (Note: Since the eigenvectors (mode shapes) can be arbitrarily scaled, the orthogonality properties are often used to scale the eigenvectors so the modal mass value for each mode is equal to 1. The modal mass matrix is therefore an identity matrix)

These properties can be used to greatly simplify the solution of multi-degree of freedom models by making the following the coordinate transformation.

{\begin{Bmatrix}x\end{Bmatrix}}={\begin{bmatrix}\Psi \end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}

If we use this coordinate transformation in our original free vibration differential equation we get the following equation.

{\begin{bmatrix}M\end{bmatrix}}{\begin{bmatrix}\Psi \end{bmatrix}}{\begin{Bmatrix}{\ddot {q}}\end{Bmatrix}}+{\begin{bmatrix}K\end{bmatrix}}{\begin{bmatrix}\Psi \end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}=0

We can take advantage of the orthogonality properties by premultiplying this equation by ${\begin{bmatrix}\Psi \end{bmatrix}}^{T}$

{\begin{bmatrix}\Psi \end{bmatrix}}^{T}{\begin{bmatrix}M\end{bmatrix}}{\begin{bmatrix}\Psi \end{bmatrix}}{\begin{Bmatrix}{\ddot {q}}\end{Bmatrix}}+{\begin{bmatrix}\Psi \end{bmatrix}}^{T}{\begin{bmatrix}K\end{bmatrix}}{\begin{bmatrix}\Psi \end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}=0

The orthogonality properties then simplify this equation to:

{\begin{bmatrix}^{\diagdown }m_{r\diagdown }\end{bmatrix}}{\begin{Bmatrix}{\ddot {q}}\end{Bmatrix}}+{\begin{bmatrix}^{\diagdown }k_{r\diagdown }\end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}=0

This matrix equation is the foundation of vibration analysis for multiple degree of freedom systems. In fact it can be viewed as N single DOF systems. A similar type of result can be derived for damped systems. ^[2]. The key is that the modal and stiffness matrices are diagonal matrices and therefore we have "decoupled" the equations. In other words, we have transformed our problem from a large unwieldy multiple degree of freedom problem into many single degree of freedom problems that can be solved using the same methods outlined above.

Instead of solving for x we are instead solving for q, referred to as the modal coordinates or modal participation factors.

It may be clearer to understand if we write ${\begin{Bmatrix}x\end{Bmatrix}}={\begin{bmatrix}\Psi \end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}$ as:

${\begin{Bmatrix}x_{n}\end{Bmatrix}}=q_{1}{\begin{Bmatrix}\psi \end{Bmatrix}}_{1}+q_{2}{\begin{Bmatrix}\psi \end{Bmatrix}}_{2}+q_{3}{\begin{Bmatrix}\psi \end{Bmatrix}}_{3}+\cdots +q_{N}{\begin{Bmatrix}\psi \end{Bmatrix}}_{N}$

Written in this form we can see that the vibration at each of the degrees of freedom is just a linear sum of the mode shapes. Furthermore, how much each mode "participates" in the final vibration is defined by q, its modal participation factor.

Forced vibration of Multiple DOF systems

The same techniques can be used to not only solve free vibration, but also the forced vibration problem.

{\begin{bmatrix}M\end{bmatrix}}{\begin{Bmatrix}{\ddot {x}}\end{Bmatrix}}+{\begin{bmatrix}K\end{bmatrix}}{\begin{Bmatrix}x\end{Bmatrix}}={\begin{Bmatrix}f\end{Bmatrix}}

If a harmonic force is applied with the following form:

{\begin{Bmatrix}f\end{Bmatrix}}={\begin{Bmatrix}F\end{Bmatrix}}e^{i\omega t}

This equation can be found to simplify with the orthogonality properties to:

{\begin{bmatrix}^{\diagdown }m_{r\diagdown }\end{bmatrix}}{\begin{Bmatrix}{\ddot {q}}\end{Bmatrix}}+{\begin{bmatrix}^{\diagdown }k_{r\diagdown }\end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}={\begin{bmatrix}\Psi \end{bmatrix}}^{T}{\begin{Bmatrix}F\end{Bmatrix}}

If we define a generalized force Q as:

Q={\begin{bmatrix}\Psi \end{bmatrix}}^{T}{\begin{Bmatrix}F\end{Bmatrix}}

The equation then has the following form.

{\begin{bmatrix}^{\diagdown }m_{r\diagdown }\end{bmatrix}}{\begin{Bmatrix}{\ddot {q}}\end{Bmatrix}}+{\begin{bmatrix}^{\diagdown }k_{r\diagdown }\end{bmatrix}}{\begin{Bmatrix}q\end{Bmatrix}}={\begin{Bmatrix}Q\end{Bmatrix}}

Again, this denotes N decoupled equations that are identical in form to the 1 DOF system that we have already solved. In

Vibration Measurement

Vibration measurements are taken for multiple purposes. For instance, the vibration of machines is often monitored to ensure they are working properly and not near failure. Vibration is also measured to determine how machines interact with their environment such as the vibration of a drill and its interaction with the human hand and arm. Vibration measurements are also taken to quantify the dynamic characteristics of a structure. For example, in the field of Experimental Modal Analysis a known input is applied to a structure and the vibration is recorded to determine how the structure responds.

The measurement of vibration can use many types of sensors that measure the motion of a structure as a function of time. The motion is typically measured with displacement, velocity, or acceleration (accelerometer) sensors. Via differentation or integration the vibration measurement can be transformed into any of these quantities.

v={\frac {dx}{dt}}\!

a={\frac {dv}{dt}}={\frac {d^{2}x}{dt^{2}}}\!

One of the common reasons is to quantify the vibration and its severity.

Bibliography

^ Maia,Silva Theoretical And Experimental Modal Analysis,Research Studies Press Ltd.,1997, ISBN 0471970670
^ Maia,Silva Theoretical And Experimental Modal Analysis,Research Studies Press Ltd.,1997, ISBN 0471970670

Rao, Singiresu, Mechanical Vibrations, Addison Wesley, 1990, ISBN 0-201-50156-2
Thompson, W.T., Theory of Vibrations, Nelson Thornes Ltd, 1996, ISBN 0-412-783908
Hartog, Den, Mechanical Vibrations, Dover Publications, 1985, ISBN 0-486-647854