Intro Predictive Maintenance 2 Part 5 pot

In most cases, this lower profile can be directly attributed to the difference in stiffness of the vertical plane when compared to the horizontal plane.. In this figure, the mass “M” is

broadband, narrowband, or component The unit of measurement is useful when theanalyst needs to know the total displacement or maximum energy produced by themachine’s vibration profile.

Technically, peak-to-peak values should be used in conjunction with actual displacement data, which are measured with a proximity or displacement transducer.Peak-to-peak terms should not be used for vibration data acquired using either relative vibration data from bearing caps or when using a velocity or accelerationtransducer The only exception is when vibration levels must be compared to vibra-tion-severity charts based on peak-to-peak values

shaft-Zero-to-peak Zero-to-peak (A), or simply peak, values are equal to one half of thepeak-to-peak value In general, relative vibration data acquired using a velocity trans-ducer are expressed in terms of peak

Root-mean-square Root-mean-square (RMS) is the statistical average value of theamplitude generated by a machine, one of its components, or a group of components.Referring to Figure 7–11, RMS is equal to 0.707 of the zero-to-peak value, A Nor-mally, RMS data are used in conjunction with relative vibration data acquired using

an accelerometer or expressed in terms of acceleration

must be taken because the vibration profile and energy levels generated by a machinemay vary depending on the location and orientation of the measurement.

7.5.1 Mass, Stiffness, and Damping

The three primary factors that determine the normal vibration energy levels and theresulting vibration profiles are mass, stiffness, and damping Every machine-train isdesigned with a dynamic support system that is based on the following: the mass ofthe dynamic component(s), specific support system stiffness, and a specific amount ofdamping

Mass

Mass is the property that describes how much material is present Dynamically, theproperty describes how an unrestricted body resists the application of an externalforce Simply stated, the greater the mass, the greater the force required to accelerate

it Mass is obtained by dividing the weight of a body (e.g., rotor assembly) by the

local acceleration of gravity, g.

The English system of units is complicated compared to the metric system In theEnglish system, the units of mass are pounds-mass (lbm) and the units of weight arepounds-force (lbf) By definition, a weight (i.e., force) of one lbf equals the force pro-

duced by one lbm under the acceleration of gravity Therefore, the constant, gc, which has the same numerical value as g (32.17) and units of lbm-ft/lbf-sec2, is used in thedefinition of weight:

(lbf/in) Machine-trains have three stiffness properties that must be considered invibration analysis: shaft stiffness, vertical stiffness, and horizontal stiffness.

Shaft Stiffness Most machine-trains used in industry have flexible shafts and

rela-tively long spans between bearing-support points As a result, these shafts tend to flex

in normal operation Three factors determine the amount of flex and mode shape thatthese shafts have in normal operation: shaft diameter, shaft material properties, andspan length A small-diameter shaft with a long span will obviously flex more thanone with a larger diameter or shorter span

Vertical Stiffness The rotor-bearing support structure of a machine typically has more

stiffness in the vertical plane than in the horizontal plane Generally, the structuralrigidity of a bearing-support structure is much greater in the vertical plane The fullweight of and the dynamic forces generated by the rotating element are fully sup-ported by a pedestal cross-section that provides maximum stiffness

In typical rotating machinery, the vibration profile generated by a normal machinecontains lower amplitudes in the vertical plane In most cases, this lower profile can

be directly attributed to the difference in stiffness of the vertical plane when compared

to the horizontal plane

Horizontal Stiffness Most bearing pedestals have more freedom in the horizontal

direction than in the vertical In most applications, the vertical height of the pedestal

is much greater than the horizontal cross-section As a result, the entire pedestal canflex in the horizontal plane as the machine rotates

This lower stiffness generally results in higher vibration levels in the horizontal plane.This is especially true when the machine is subjected to abnormal modes of operation

or when the machine is unbalanced or misaligned

Damping

Damping is a means of reducing velocity through resistance to motion, in particular

by forcing an object through a liquid or gas, or along another body Units of dampingare often given as pounds per inch per second (lbf/in/sec, which is also expressed aslbf-sec/in)

The boundary conditions established by the machine design determine the freedom ofmovement permitted within the machine-train A basic understanding of this concept

is essential for vibration analysis Free vibration refers to the vibration of a damped(as well as undamped) system of masses with motion entirely influenced by theirpotential energy Forced vibration occurs when motion is sustained or driven by anapplied periodic force in either damped or undamped systems The following sectionsdiscuss free and forced vibration for both damped and undamped systems

Free Vibration—Undamped To understand the interactions of mass and stiffness,

consider the case of undamped free vibration of a single mass that only moves

vertically, which is illustrated in Figure 7–12 In this figure, the mass “M” is ported by a spring that has a stiffness “K” (also referred to as the spring constant),

sup-which is defined as the number of pounds tension necessary to extend the spring one inch

The force created by the static deflection, Xi, of the spring supports the weight, W, of

the mass Also included in Figure 7–12 is the free-body diagram that illustrates thetwo forces acting on the mass These forces are the weight (also referred to as the

inertia force) and an equal, yet opposite force that results from the spring (referred to

as the spring force, Fs).

The relationship between the weight of mass, M, and the static deflection of the spring

can be calculated using the following equation:

W = KXi

If the spring is displaced downward some distance, X0, from Xiand released, it will

oscillate up and down The force from the spring, Fs, can be written as follows, where

“a” is the acceleration of the mass:

It is common practice to replace acceleration, a, with the second derivative of

the displacement, X, of the mass with respect to time, t Making this substitution, the

equation that defines the motion of the mass can be expressed as:

Motion of the mass is known to be periodic Therefore, the displacement can bedescribed by the expression:

M g

d X

M g

d X

2 2

2

d X dt

X = Displacement at time t

X0= Initial displacement of the mass

w = Frequency of the oscillation (natural or resonant frequency)

t= Time

If this equation is differentiated and the result inserted into the equation that definesmotion, the natural frequency of the mass can be calculated The first derivative ofthe equation for motion yields the equation for velocity The second derivative of theequation yields acceleration

Inserting the expression for acceleration, or into the equation for Fsyields thefollowing:

Solving this expression for w yields the equation:

2 0

0

00

2

2 0

X=X0cos(wt)

Free Vibration—Damped A slight increase in system complexity results when a

damping element is added to the spring-mass system shown in Figure 7–13 This type

of damping is referred to as viscous damping Dynamically, this system is the same

as the undamped system illustrated in Figure 7–12, except for the damper, whichusually is an oil or air dashpot mechanism A damper is used to continuously decreasethe velocity and the resulting energy of a mass undergoing oscillatory motion.The system consists of the inertia force caused by the mass and the spring force, but

a new force is introduced This force is referred to as the damping force and is portional to the damping constant, or the coefficient of viscous damping, c The

pro-damping force is also proportional to the velocity of the body and, as it is applied, itopposes the motion at each instant

In Figure 7–13, the nonelongated length of the spring is “Lo” and the elongation caused

by the weight of the mass is expressed by “h.” Therefore, the weight of the mass is

Kh Part (a) of Figure 7–13 shows the mass in its position of stable equilibrium Part

(b) shows the mass displaced downward a distance X from the equilibrium position.Note that X is considered positive in the downward direction

Part (c) of Figure 7–13 is a free-body diagram of the mass, which has three forces

acting on it The weight (Mg/gc), which is directed downward, is always positive The

damping force which is the damping constant times velocity, acts opposite to

the direction of the velocity The spring force, K(X + h), acts in the direction opposite

to the displacement Using Newton’s equation of motion, where SF = Ma, the sum ofthe forces acting on the mass can be represented by the following equation, remem-bering that X is positive in the downward direction:

Dividing by

In order to look up the solution to the above equation in a differential equations table

(such as in CRC Handbook of Chemistry and Physics), it is necessary to change the form of this equation This can be accomplished by defining the relationships, cgc/M

= 2m and Kgc/M=w2, which converts the equation to the following form:

Note that for undamped free vibration, the damping constant, c, is zero and, therefore,

2 2

0

=

ww

d X dt

dX

2 2

2

2

= - m -w

d X dt

cg M

dX dt

Kg X M

d X dt

w Underdamping occurs when m is less than w.

The only condition that results in oscillatory motion and, therefore, represents amechanical vibration is underdamping The other two conditions result in periodicmotions When damping is less than critical (m < w), then the following equationapplies:

where:

Forced Vibration—Undamped The simple systems described in the preceding two

sections on free vibration are alike in that they are not forced to vibrate by any ing force or motion Their major contribution to the discussion of vibration funda-mentals is that they illustrate how a system’s natural or resonant frequency depends

excit-on the mass, stiffness, and damping characteristics

The mass-stiffness-damping system also can be disturbed by a periodic variation ofexternal forces applied to the mass at any frequency The system shown in Figure 7–12

is increased in complexity by adding an external force, F0, acting downward on themass

dX

2 2

2

d X dt

dX

2 2

2

2

= - m -w

In undamped forced vibration, the only difference in the equation for undamped free

vibration is that instead of the equation being equal to zero, it is equal to F0sin(wt):

Because the spring is not initially displaced and is “driven” by the function F0sin(wt),

a particular solution, X = X0sin(wt), is logical Substituting this solution into the aboveequation and performing mathematical manipulations yields the following equation

for X:

where:

X = Spring displacement at time, t

Xst = Static spring deflection under constant load, F0

Forced Vibration—Damped In a damped forced vibration system such as the one shown in Figure 7–14, the motion of the mass “M” has two parts: (1) the damped free

vibration at the damped natural frequency and (2) the steady-state harmonic motions

at the forcing frequency The damped natural frequency component decays quickly,but the steady-state harmonic associated with the external force remains as long asthe energy force is present

With damped forced vibration, the only difference in its equation and the equation for

damped free vibration is that it is equal to F0sin(wt) as shown below instead of beingequal to zero

With damped vibration, the damping constant, “c,” is not equal to zero and the tion of the equation becomes complex assuming the function, X = X0sin(wt - f) In

solu-M g

this equation, f is the phase angle, or the number of degrees that the external force,

F0sin(wt), is ahead of the displacement, X0sin(wt - f) Using vector concepts, the lowing equations apply, which can be solved because there are two equations and twounknowns:

fol-Solving these two equations for the unknowns X0and f:

g

c c

ˆ

ÊË

ˆ

¯

=-

21

w

ww

w

ww

w wtan

Vertical vector component:

Horizontal vector component:

dX dt

F0 Sin ( wt)

Figure 7–14 Damped forced vibration system.

For damped forced vibrations, three different frequencies have to be distinguished:the undamped natural frequency, the damped natural frequency,

and the frequency of maximum forced amplitude, sometimes

referred to as the resonant frequency.

7.5.2 Degrees of Freedom

In a mechanical system, the degrees of freedom indicate how many numbers arerequired to express its geometrical position at any instant In machine-trains, the rela-tionship of mass, stiffness, and damping is not the same in all directions As a result,the rotating or dynamic elements within the machine move more in one direction than

in another A clear understanding of the degrees of freedom is important because ithas a direct impact on the vibration amplitudes generated by a machine or processsystem

One Degree of Freedom

If the geometrical position of a mechanical system can be defined or expressed as asingle value, the machine is said to have one degree of freedom For example, theposition of a piston moving in a cylinder can be specified at any point in time by mea-suring the distance from the cylinder end

A single degree of freedom is not limited to simple mechanical systems such as thecylinder For example, a 12-cylinder gasoline engine with a rigid crankshaft and arigidly mounted cylinder block has only one degree of freedom The position of allits moving parts (i.e., pistons, rods, valves, cam shafts) can be expressed by a singlevalue In this instance, the value would be the angle of the crankshaft; however, whenmounted on flexible springs, this engine has multiple degrees of freedom In addition

to the movement of its internal parts in relationship to the crank, the entire engine cannow move in any direction As a result, the position of the engine and any of its inter-nal parts requires more than one value to plot its actual position in space

The definitions and relationships of mass, stiffness, and damping in the precedingsection assumed a single degree of freedom In other words, movement was limited

to a single plane Therefore, the formulas are applicable for all freedom mechanical systems

single-degree-of-The calculation for torque is a primary example of a single degree of freedom in amechanical system Figure 7–15 represents a disk with a moment of inertia, I, that is

attached to a shaft of torsional stiffness, k.

Torsional stiffness is defined as the externally applied torque, T, in inch-pounds needed

to turn the disk one radian (57.3 degrees) Torque can be represented by the ing equations:

In this example, three torques are acting on the disk: the spring torque, the dampingtorque (caused by the viscosity of the air), and the external torque The spring torque

is minus (-) kf where f is measured in radians The damping torque is minus (-) cf,

where “c” is the damping constant In this example, “c” is the damping torque on the

disk caused by an angular speed of rotation of one radian per second The external

torque is T0sin (wt)

or

Two Degrees of Freedom

The theory for a one-degree-of-freedom system is useful for determining resonant ornatural frequencies that occur in all machine-trains and process systems; however, fewmachines have only one degree of freedom Practically, most machines will have two

or more degrees of freedom This section provides a brief overview of the theoriesassociated with two degrees of freedom An undamped two-degree-of-freedom system

is illustrated in Figure 7–16

This diagram consists of two masses, M1and M2, that are suspended from springs, K1

and K2 The two masses are tied together, or coupled, by spring, K3, so that they are

ff˙˙

Figure 7–15 Torsional freedom system.

one-degree-of-forced to act together In this example, the movement of the two masses is limited tothe vertical plane and, therefore, horizontal movement can be ignored As in the single-degree-of-freedom examples, the absolute position of each mass is defined by its ver-tical position above or below the neutral, or reference, point Because there are two

coupled masses, two locations (i.e., one for M1and one for M2) are required to locatethe absolute position of the system

To calculate the free or natural modes of vibration, note that two distinct forces are

acting on mass, M1: the force of the main spring, K1, and that of the coupling spring,

K3 The main force acts upward and is defined as -K1X1 The shortening of the

cou-pling spring is equal to the difference in the vertical position of the two masses, X1

-X2 Therefore, the compressive force of the coupling spring is K3(X1- X2) The

com-pressed coupling spring pushes the top mass, M1, upward so that the force is negative

Because these are the only tangible forces acting on M1, the equation of motion forthe top mass can be written as:

If we assume that the masses, M1and M2, undergo harmonic motions with the samefrequency, w, and with different amplitudes, A1and A2, their behavior can be repre-sented as:

By substituting these into the differential equations, two equations for the amplituderatio, can be found:

sub-w4 w2 1 3

1

2 3 2

1 2 2 3 1 3

1 2 2

w

w

A A

2 2

2 3 3

=

w

-A A

K M

1 2

ww

for This means that there are two solutions for this example, which are of the form

A1sin(wt) and A2sin(wt) As with many such problems, the final answer is the position of the two solutions with the final amplitudes and frequencies determined bythe boundary conditions

super-Many Degrees of Freedom

When the number of degrees of freedom becomes greater than two, no critical newparameters enter into the problem The dynamics of all machines can be understood

by following the rules and guidelines established in the one- and freedom equations There are as many natural frequencies and modes of motion asthere are degrees of freedom

two-degree(s)-of-7.6 V IBRATION D ATA T YPES AND F ORMATS

There are several options regarding the types of vibration data that can be gatheredfor machine-trains and systems and the formats in which the data can be collected.Selection of type and format depends on the specific application There are two majordata-type classifications: time-domain and frequency-domain Each of these can befurther divided into steady-state and dynamic data formats In turn, each of these twoformats can be further divided into single-channel and multichannel

Actual time-domain vibration signatures are commonly referred to as time traces or time plots (see Figure 7–17) Theoretical vibration data are generally referred to as waveforms (see Figure 7–18).

Time-domain data are presented with amplitude as the vertical axis and elapsed time as the horizontal axis Time-domain profiles are the sum of all vibration com-ponents (i.e., frequencies, impacts, and other transients) that are present in themachine-train and its installed system Time traces include all frequency components,

A

A

1

2

but the individual components are more difficult to isolate than with frequency-domaindata.

The profile shown in Figure 7–17 illustrates two different data acquisition points, onemeasured vertically and one measured horizontally, on the same machine and taken

at the same time Because they were obtained concurrently, these data points can becompared to determine the operating dynamics of the machine

In this example, the data set contains an impact that occurred at 0.005 seconds Theimpact is clearly visible in both the vertical (top) and horizontal (bottom) data set

Figure 7–17 Typical time-domain signature.

Figure 7–18 Theoretical time-domain waveforms.

From these time traces, the vertical impact appears to be stronger than the horizontal.

In addition, the impact repeated at 0.015 and 0.025 seconds Two conclusions can bederived from this example: (1) the impact source is a vertical force, and (2) it impactsthe machine-train at an interval of 0.010 seconds, or frequency of 1/0.010 secondsequals 100 Hz

The waveform in Figure 7–18 illustrates theoretically the unique frequencies and sients that may be present in a machine’s signature Figure 7–18a illustrates the com-plexity of such a waveform by overlaying numerous frequencies The discretewaveforms that make up Figure 7–18a are displayed individually in Figures 7–18bthrough 7–18e Note that two of the frequencies (c and d) are identical but have a dif-ferent phase angle (f)

tran-With time-domain data, the analyst must manually separate the individual frequenciesand events that are contained in the complex waveform This effort is complicatedtremendously by the superposition of multiple frequencies Note that, rather than over-laying each of the discrete frequencies as illustrated theoretically in Figure 7–18a,actual time-domain data represents the sum of these frequencies as was illustrated inFigure 7–17

In order to analyze this type of plot, the analyst must manually change the time scale

to obtain discrete frequency curve data The time interval between the recurrences ofeach frequency can then be measured In this way, it is possible to isolate each of thefrequencies that make up the time-domain vibration signature

For routine monitoring of machine vibration, however, this approach is not cost tive The time required to manually isolate each of the frequency components andtransient events contained in the waveform is prohibitive; however, time-domain datahave a definite use in a total-plant predictive maintenance or reliability improvementprogram

effec-Machine-trains or process systems that have specific timing events (e.g., a matic or hydraulic cylinder) must be analyzed using the time-domain data format

pneu-In addition, time-domain data must be used for linear and reciprocating motionmachinery

Frequency-Domain

Most rotating machine-train failures result at or near a frequency component ated with the running speed Therefore, the ability to display and analyze the vibra-tion spectrum as components of frequency is extremely important

associ-The frequency-domain format eliminates the manual effort required to isolate the ponents that make up a time trace Frequency-domain techniques convert time-domaindata into discrete frequency components using a mathematical process called FastFourier Transform (FFT) Simply stated, FFT mathematically converts a time-based