1

MEEG 5113MEEG 5113

Set 1

2

3

4

5

Vibration AnalysisA simple model of a single degree-of-freedom system is a pendulum. The mass is concentrated at a single point, the cg, and the length of the pendulum is taken as the distance from the pivot point to the mass center (the length l in the figure to the right).

When the link is then disturbed from equilibrium, it will oscillate back and forth forever if there is no air drag or friction at the pivot.

Lacking these losses or energy dissipation methods, this system is said to conserve energy which means that the maximum kinetic energy and maximum potential energy are equal.

6

Vibration Analysis

m

W

L

x

y

O

For the pendulum shown below to be in dynamic equilibrium, the following must be true:

b bI M

where Ib = mass moment of inertia of the body about point b

b = any arbitrary point on the body = angular acceleration

As point b represents an arbitrary location, two unknowns can be eliminated by summing moments about the pivot point, O. Also note that the angle is measured from the reference position (vertically down in this case) and the positive direction for defines the positive direction for summing moments.

7

Vibration Analysis

m

W

L

O O

Oy

x

L sin

O

2

2

2

2

OOO IsinWLIIsinWLM

dtd

dtd

In order to proceed further two issues must be addressed. First, IO needs to be defined using the parallel axis theorem. This yields

2O dII mcg

Since the point mass is a sphere, Icg is a function of the radius of the sphere. However, for a point mass the radius is so small it can be taken as zero. Therefore, Icg = 0 and IO is simply md2 and the equation above becomes

22

2

mLsinWL

dtd

8

Vibration AnalysisSecond, the presence of the sine term makes the above equation a nonlinear, second order differential equation. However, if is limited to small values ( < 7o), sin can be replaced with to obtain

LLL

WL22

2 gmmg

mdtd

In this equation both acceleration and displacement are functions of time and (t) must be a function which repeats itself every second derivative with a negative coefficient. Harmonic and complex exponential functions meet this requirement. For simplicity the harmonic form will be used so that the angular displacement and acceleration can be written as

)()sinBcosA(

andsinBcosA)(

222

2

tttdtd

ttt

9

Vibration AnalysisBy equating the two equations for acceleration we obtain

sradgOR

tLgt

L

)()(2

This value of frequency is called the natural (characteristic) frequency of oscillation of the pendulum and depends only on the length of the pendulum and the local acceleration of gravity. If the pendulum is disturbed from equilibrium, it will oscillate with the frequency

The values of A and B from the equation for (t) are obtained from the initial conditions that were applied to the pendulum to begin the oscillation process.

10

Vibration AnalysisThere are four possibilities for those initial conditions which are

1. Initial displacement and initial velocity are both zero. While these initial conditions satisfy the equation, they produce a trivial solution as no motion will result.

2. Initial displacement is non-zero and the initial velocity is zero. These conditions are analogous to an archer preparing to take a shot at a stationary target. The bow is drawn back and held until aiming is complete. Then the arrow is released toward the target.

3. Initial displacement is zero and the initial velocity is non-zero. These conditions are analogous to a golfer hitting a tee shot. For a very small fraction of a second, the ball deflects without moving as the golf club impacts it.

4. Initial displacement and initial velocity are non-zero. These conditions are analogous to plucking a string on a guitar.

11

Vibration AnalysisA more commonly encounter model of a vibrating system is shown below. In this model, the mass undergoes translation in either the horizontal or vertical plane. In either case, the reference displacement location is the EQUILIBRIUM position. For the horizontal translation model, this represents the free length of the spring (no elongation of the spring). For the vertical translation model, this represents the location where the weight of the body is balanced by the force in the spring. If the free length of the spring is used for the vertical case, one must account for the energy needed to extend the spring to the distance where the spring force

balances the weight of the body. This approach is prone to error as this energy term is typically not handled correctly and this approach is not recommended.

12

Vibration AnalysisFor the translational system shown below, the first step is to draw a free body diagram of the mass. To do this, assume the body is moved slightly in the positive coordinate direction which will cause the spring to extend thus generating a restoring force on the body as shown in (a) below. As vibration for this model does not take place in the vertical direction, it is not necessary to include the weight of the body in this diagram. The inertia (effective) force is shown in (b). For dynamic equilibrium, this system must satisfy d’Alembert’s principle which states that

( ) ( ) ( ) ( )x x

kF kx t ma mx t x t x tm

Harmonic and complex exponential functions satisfy this expression.

13

Vibration AnalysisThe harmonic form is simplest to use for the undamped case which means that x(t) and ax(t) have the form

2

( ) cos sin

( ) ( ) cos sinx

ANDx t A t B t

a t x t A t B t

Substituting these definitions into the previous equation yields the result

2 2( ) ( ) kk kx t x tmm m

This result shows that if the mass is disturbed from either its equilibrium or steady-state condition it will oscillate with a natural or characteristic frequency that is solely dependent on k and m. And, in the absence of friction and air drag, the motion will continue forever.

14

Vibration AnalysisAs shown below, if the displacement is a cosine function, the velocity will be 90o out of phase with it and the acceleration will be 180o out of phase.

x(t) = A cos t + B sin t v(t) = (-A sin t + B cos t)

a(t) = -2(A cos t + B sin t)

Normalized Displacement, Velocity, and Acceleration

-1

-0.5

0

0.5

1

Time

Ampl

itude

Disp.Vel.Accel.

For a given displacement, the velocity increases by the angular velocity and the acceleration by the square of the angular velocity. This means even a slight change in displacement can result in a significant velocity and even greater acceleration.

15

Vibration AnalysisThe third type of single degree-of-freedom system is a torsional system as shown below. In this case, the disk rotates in a horizontal plane about the shaft that acts like a torsional spring with a torsional spring constant of kT = GJ/L where G is the shear modulus of the material, J is the polar moment of the cross section, and L is the length of the shaft. For dynamic equilibrium, this system must satisfy d’Alembert’s principle which for this problem is

(t) = A cos t + B sin t

d/dt = (-A sin t + B cos t)

d2/dt2 = -2(A cos t + B sin t)

( ) ( ) ( ) ( )TT cg cg

cgIkM k t I I t t t

2 2( ) ( ) TT T

cgcg cg

kI

k kt tI I

which again yields

16

Vibration AnalysisAt times one encounters situations where more than one spring in used in an application. The physical arrangement of the springs relative to one another determines the effective or equivalent stiffness of the system. Shown below are three springs connected in series.

To determine the equivalent stiffness of this set of springs, a free body diagram is drawn of each spring. When this is done, it can be seen that each spring must support the same load F. For an extensional spring the load-deflection equation is F = k(x). For the above springs the equations are

1 2 31 2 3

1 2 31 2 3

1 1 1e

AND

F F Fx x xk k k

Fx x x x Fk k k k

17

Vibration AnalysisAs springs connected in series add as reciprocals, the magnitude of the equivalent stiffness is always controlled by the smallest value of k in the set and will be less than that smallest value. For the above spring set, let k1 = 10 lb/ft, k2 = 20 lb/ft, and k3 = 100 lb/ft. The equivalent spring stiffness is

- -

1 1 1 1 0.1 0.05 0.01 0.151ft/lb10 20 100

6.62252 lb/ft

e

e

ORk

k

When designing any system that has a series stiffness arrangement, one must be very careful to attempt to match the stiffnesses as closely as possible to avoid exceeding the maximum allowable deflection. This concern explains why the cylinder head gasket on most newer cars is now a stamped steel component. This allows the stiffness of the gasket to match that of the cylinder head and block as closely as possible.

18

Vibration AnalysisShown below are three springs connected in parallel.

To determine the equivalent stiffness of this set of springs, a free body diagram is drawn of each spring. When this is done, it can be seen that each spring must support the same deflection, x. For an extensional spring the load-deflection equation is F = k(x). For the above springs the equations are

1 1 2 2 3 3

1 2 31 2 3 e

AND

F k x F k x F k x

F F F F k k k x k x

19

Vibration AnalysisAs springs connected in parallel simply add together, the magnitude of the equivalent stiffness is always controlled by the largest value of k in the set and will be greater than that largest value. For the above spring set, let k1 = 10 lb/ft, k2 = 20 lb/ft, and k3 = 100 lb/ft. The equivalent spring stiffness is

10 20 100 130 lb/fte

k

When designing any system that has a parallel stiffness arrangement, one must be very careful to limit the equivalent stiffness to only the amount needed to carry the expected maximum load. This concern partially explains why makers of SUV’s are now claiming their vehicle rides like a regular passenger car. One benefit to using parallel springs is space saving. On automotive engines it is sometimes necessary to stiffen the valve springs without increasing the diameter of the spring or the diameter of the wire. This can be accomplished using two springs nested one inside the other. The overall stiffness increases, but the outside dimension does not.

20

Vibration AnalysisIn the preceding discussion, each element was an “ideal” element. The mass did not contribute stiffness to the system only mass. The spring did not contribute mass to the system only stiffness. A damper is considered to have no mass or stiffness and only dissipates energy. Several common items can be modeled as an “ideal” spring (i.e. they contribute little to the overall mass of the system).

1. Helical spring – k = (d4G)/(8D3N) where d = wire diameter D = mean coil diameter G = shear modulus N = number of active coils

2. Rod (axial stiffness) – k = (AE)/L where A = cross sectional area E = Young’s modulus L = length

21

Vibration Analysis

3. Rod (torsional stiffness) – k = (GJ)/L where J = polar moment of the cross

section

4. Cantilever beam (tip load) – k = (3EI)/L3 where I = second moment of

the area about the neutral axis

5. Simply supported beam (center load) – k = (48EI)/L3

6. Clamped-clamped beam (center load) – k = (192EI)/L3

7. Clamped-pinned beam (center load) – k = (768EI)/7L3

22

Vibration AnalysisWhen the spring does contribute mass to the system, one must determine what the dynamic mass of the spring is, i.e. how much of the spring’s total mass contributes to the motion. In the figure below, it is easy to see that since one end of the spring is attached to ground that mass does not move at all. The other end is attached to the moving mass and it moves a great deal. Therefore, a portion of the spring’s mass will add to the dynamics. To determine how much mass to add to M, we first estimate the deflected shape of the spring and then use that same shape to compute the velocity of a small segment of the spring. This expression is used to determine the kinetic energyof the spring. From this result the dynamic mass can be found. An example for the figure to the right follows.

23

Vibration AnalysisFor the system shown below, the mass of the spring is considered large enough to contribute to the motion of the cart, i.e. it a “heavy” spring. The cart has a mass M, and the spring has a mass m, a free length l, and a stiffness k. For manner in which the spring is attached in the system, the static deflection shape of the spring is given by the equation y = (y/l) x. If the dynamic deflection shape of the spring is the same as the static deflection, the mass and velocity of the spring element dy are y my x AND dm dy

l l

The kinetic energy of the spring is then equal to

2

0

2 32

3

0

12

1 13 32 2

l

l

m yx dy

l l

m x y mx

l

T

T

mdyn = m/3

24

Vibration AnalysisTherefore,

3

/

3

Totalm

AND

krad s

mM

m M

This result indicates that if the mass of the spring cannot be ignored the natural frequency will be lowered. The amount it is reduced can be significant if m approaches M as it does in the valve train of an engine.In this case, the valve spring can fail due excessive engine speed if a careful analysis of the valve train has not been done. Such a failure results in considerable damage to the engine.

25

Vibration Analysis

At times it is desirable to alter the vibrational frequency of a system or design a system that has a specific fundamental frequency. One example of this situation is the design of a vibrating reed tachometer. A vibrating reed tachometer is a small handheld instrument which contains 5 to 10 cantilever beams each tuned to a specific frequency. Each reed is a cantilever beam with a weight added to the free end which serves to both tune the beam and act as an indicator.

26

Vibration Analysis

This yields the following velocity and kinetic energy expressions2 3

0

1 [3 ]2

x xy yl l

2 3

0

1 [3 ]2

x xy yl l

where y0 = maximum deflection

x = location along beam from left end

l = beam length

To design one reed, the dynamic mass of the cantilever beam must be determined. The first step is to establish the dynamic deflection of the beam. A good approximation is the static deflection curve for either a tip load or due to its own weight. For this example we will use

2

04 5 6

2

00

121 [9 6 ]2 4

l

l

T

T m y dx

x x xm y dxl l l

27

Vibration Analysis- OR -

2 2 2 2

0 0 0 0

33331 9 1 1 1 1[ 1 ]140 1402 4 5 7 2 2 2

beameff

mmlmlT y y y m y

What this result means is that a little less than ¼ of the mass of the beam is moving enough to contribute to vibration.

Now determine the amount of mass, M, which must be added to the tip of the cantilever beam to produce a fundamental frequency of 20 Hz if

b = 0.635 cm; h = 0.1016 cm; l = 8.89 cm;

= 0.07655 N/cm3; E = 200 x 109 N/m2

28

Vibration AnalysisFrom these values the following can be determined:

ml = 0.00475 kg; I = 0.0000555 x 10-8 m4; k = 473.96 N/m;

which allows the effective mass and then the added mass as follows:

3 2 2

12

33 3 473.96140 (4 )4004

0.0289

eff

eff

kf m

ml EIm Ml f

M kg

29

Vibration AnalysisFor the horizontal pendulum shown below, link AB is rigid, has a uniform cross section, but is not massless and it is desired to determine the frequency of vibration for the following conditions:

l = 0.3 m; a = 0.15 m; mrod = 10 kg; m = 7 kg; k = 2 kN/m

To proceed, the relationship between , a, and l must be established for small angle motion. For small angles, the chord length and arc length approach one another and the chord length represents the straight line motion at a and l keeping inmind that our reference position is the equilibrium or horizontal position. This yields the following equation of motion:

2

0A

kaI

30

Vibration AnalysisFor the resulting harmonic motion of the pendulum, the frequency of vibration is given by

21 Hz2 A

kafI

For this system, the mass moment of inertia is given by

IA = [7 + (1/3)(10)] (0.3)2 = 0.93 kg m2

This gives a frequency value of

22000 0.151 Hz

0.932

1.1 Hz

f

f

31