structural vibration worked examples - taylor &...

An Introduction to Mechanical Engineering Part 2

© Hodder Education 2010

!!!! Structural vibration worked examples Natural Frequencies and Mode Shapes 1. Derive the equation of motion and hence find the natural frequencies for

the system shown in Figure Q1.

Solution Free-body diagram

θ2Lk ⋅ θLK ⋅θ

Since clockwise rotation has been chosen as the positive direction, when the bar moves down both springs will be put in compression and will exert upward reaction forces on the underside of the bar. For small angles, the change of length of the left-hand spring will be θ

2L .

Equation of motion Taking moments about the pivot in the same direction as the motion coordinate (i.e., clockwise), we get:

θ=θ−θ− 2

31.

2.

2mLLKLLLk



Rearranging gives:

043

1 22

2 =θ

++θ KLLkmL

or 043

=θ

++θ Kkm

From the coefficients of acceleration and displacement, the natural frequency is:

( )m

Kkm

Kk

n 443

3

4 +=+

=ω

2. A wheel (radius r, mass m, moment of inertia

about its centre I) can roll without slipping on a horizontal plane. It is restrained by a horizontal spring (stiffness k) attached at one end to the centre of the wheel and at the other end to a rigid vertical wall, as in Figure Q2. Derive the equation of motion and hence find the natural frequency for the system. What would the natural frequency be if there was no friction between the wheel and the plane?


r

θ

xG

k xG

F (friction)

G

C

Due to the friction at the point of contact with the horizontal plane, there is no slipping and the rotation of the wheel and the translation of its centre of mass are not independent motions, but are directly linked by the expression

θ= rXG (Q2.1)



For translation of the mass, GG xmFkx =−− (Q2.2)

For rotation about the centre of mass, θ=+ lFr (Q2.3)

Differentiate (Q2.1) twice to get an expression for θ , substitute into (Q2.3) to get an expression for F. Substitute this into (Q2.2) to give the equation of motion:

012 =+

+ GG kxx

rm

Imrkr

rm

kn +

=+

=ω∴ 2

2

2

1

If there was no friction between the wheel and the plane, F = 0. From equation (Q2.3) we see that 0=θ , so the wheel will not rotate at all. Equation (Q2.2) becomes

0=+ GG kxxm

mk

n =ω∴

3. Derive the equations of motion for the system in Figure Q3. Assume that

all displacements are small.

Solution Free-body diagrams

1m 3m2m

1x 2x 3x

( )211 xxk − ( )322 xxk −

For the left-hand spring, since the coordinates x1 and x2 are both positive when the masses move to the right, the change of length of spring k1 will be



the difference between to two displacements. The expression for the force in the spring can therefore be written as ( )211 xxk − . This force will be positive any time that x1 > x2. This would mean that the spring would be in compression and that the forces acting on it would be as shown below.

( )211 xxk −( )211 xxk −

This leads to the directions of the reaction forces shown in the free-body diagrams.

An equally correct expression for the force in the spring would be ( )121 xxk − . This force will be positive any time that x2 > x1, which would mean that the spring would be in tension and that the forces acting on it would be as shown below.

( )121 xxk −( )121 xxk −

That part of the free-body diagram would then be:

1m 2m( )121 xxk −

The two alternative force expressions are both correct, provided that the positive directions of the forces are identified correctly. Equations of motion Mass m1: ( ) 11211 xmxxk =−−

Mass m2: ( ) ( ) 22322211 xmxxkxxk =−−−+

Mass m3: ( ) 33322 xmxxk =−+

Rearranging gives:

=

−−+−

−+

000

0

0

000000

3

2

1

22

2211

11

3

2

1

3

2

1

xxx

kkkkkk

kk

xxx

mm

m



Note that the leading diagonal of the mass and stiffness matrices contain only positive terms and that both matrices are symmetric. 4. Find the natural frequencies and mode shapes for the system in Figure Q3

when k1 = 10 kN/m, k1 = 30 kN/m, m1 = m2 = 5 kg and m3 = 10 kg. Solution Put tXx ω= cos11 , etc. into the equations of motion from Question 2 and substitute the given mass and stiffness values.

=

ω−××−×−ω−×−

−ω−

000

101031030103510410

010510

3

2

1

244

4244

424

xxx

Its convenient to put 4

2

105ω=β to simplify the algebra. Hence:

( )( )( )iiiiii

xxx

ββ

βLLL

=

−−−−−

−−

000

2330341

011

3

2

1

The determinant of the coefficient matrix equated to zero gives the frequency equation, which is:

012132 22 =+− βββ

Although this is a cubic, there is no constant term, so that one of the roots will be 0=β , leaving only a quadratic to solve. The roots are:

01 =β 11.12 =β 39.52 =β

Giving: 01 =ω Hz 51.72 =ω Hz 51.163 =ω Hz

Mode shapes

From (i), ( ) 01 21 =−− XXβr ( )

−=

∴1

11

2

1rβ

XX

From (iii), ( ) 0233 22 =−+− XβX r ( )

−=

∴1

233

2

3rβ

XX

Combining the two results gives,



( )

( )

−

−=

r

r

β

β

XXX

233

11

1

3

2

1

Substituting the three values of rβ gives,

−=

=

89.3177.8

111

23

2

1

,13

2

1

ModeModeXXX

XXX

and

−

−=

386.01228.0

13

2

1

ModeXXX

As noted previously, one of the roots is 01 =β ( )Hz01 =ω and we see that the corresponding mode shape is 1:1:1:: 321 =XXX . This confirms that this is a rigid body mode where all three masses move together as if they were a single rigid body. 5. Derive the equations of motion for the system in Figure Q5. Assume that

all displacements and angles are small are small.



a

( )xak +θθ

x

( )xak +θ

AB

m

Solution Free-body diagrams Positive rotation of the bar moves end B upwards by aθ (angles are small) and positive displacement of the mass is downwards. The change of length of the spring is therefore the sum of the two displacements, that is, aθ + x. Equations of motion

For the bar: ( ) θ=+θ− Iaxak .

For the mass: ( ) xmxak =+θ−

Rearranging gives:

=

θ

+

θ

00

00 2

xkkakaka

xmI

Note that the leading diagonal of the mass and stiffness matrices contain only positive terms and that both matrices are symmetric.



6. Derive the frequency equation for flexural vibration of a uniform beam that

is free at both ends and find an expression for the mode shape function. Solution

At x = 0 and x = L, 02

2

2

2

=∂∂=

∂∂

xy

xy . Since ( ) ( ) txYtxy ω= cos, , 02

2

2

2

==dx

Yddx

Yd .

Since ( ) xCxCxCxCx λ+λ+λ+λ= coshsinhcossin 4321 (Q6.1)

At x = 0: 42

322

12

2

..0..0 CCCCxdYd λ++λ−=

and 432

212

2

2

.0..0. CCCCdx

Yd +λ++λ=

At x = L: 42

32

22

12

2

2

.cosh.sinh.cos.sin CxCxCLCLdx

Yd λλ+λλ+λλ−λλ−=

and 42

32

22

12

2

2

.sinh.cosh.sin.cos CLCLCLCLdx

Yd λλ+λλ+λλ+λλ−=

Assembling the equations into matrix format:

43 @CC

( )( )( )( )dQ

cQbQaQ

2.62.62.62.6

Expanding the determinant of the coefficient matrix and equating to zero gives the frequency equation:

( ) 01cosh.cos10 =−λλλ LL

Since a freefree beam can have two rigid-body modes, each with a frequency of zero, the frequency equation becomes:

( ) 01cosh.cos2 =−λλλ LL



There are no closed-form expressions for the roots of this equation, but numerical values are given in the book. The value of λ for each root is related

to the natural frequency by the equation A

EIr

ρλ=ω 2

Mode shapes The mode shape function for this beam is obtained by substituting each of the roots rλ in turn into equation (Q6.2) and solving for the coefficients C1 to C4. These can then be substituted into the general shape function (equation Q6.1) to give each mode shape.

From (Q6.2a), 022

12 =λ+λ− CC rr Hence, C4 = C2

From (Q6.2b), 033

13 =λ+λ− CC rr Hence, C3 = C21

From (Q6.2c),

0.cosh.sinh.cos.sin 42

32

22

12 =λλ+λλ+λλ−λλ− CLCLCLCL rrrrrrrr

Hence, 112 coshcossinsinh

CCLL

LLC r

rr

rr σ=λ−λλ−λ

=

Because equation (Q6.2) is a homogeneous set, it is not possible to define the value of C1 uniquely. If we arbitrarily assign it the value 1.0, the resulting mode shape expression is, ( ) ( )xxxxxY rrrrrr λ+λσ+λ+λ= coshcossinhsin



The mode shapes for the first four modes are shown below.

‐3

‐2

‐1

0

1

2

3

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Dis

plac

emen

t am

plit

ude

Y(x)

Axial position x/L

Y1 Y2 Y3 Y4

7. A 25 mm diameter shaft, 1.5 m long, is held by two roller bearings at one

end (giving a clamped boundary condition) and by a self-aligning ball bearing at the other end (giving a pinned boundary condition). Using the roots of the appropriate frequency equation given in Table 6.3 on page 398, find the first three critical speeds of the shaft.

Solution Recall that the critical speeds of a shaft are numerically identical to the natural frequencies in flexure of an equivalent beam with a circular cross-section.

The expression for the natural frequencies isA

EILLr

r ρ

λ=ω

2

, where the

roots of the frequency equation, Lrλ , are given in Table 6.3 in the book for a clamped-pinned beam. Taking E = 207 GN/m2 and ρ = 7800 kg/m3, along with the dimensions of the shaft, the expression for the natural frequencies is ( )231.14 Lrr λ=ω rad/s

= ( )27.136 Lrλ rev/min.

Results for the first three critical speeds are: 2107, 6981 and 14 245 rev/min.



Response of damped single-degree-of-freedom systems 8. If a heavily damped structure is given an initial displacement Z0 and then

released from rest, find the constants of integration and sketch the graph of z(t) against time.

Solution The general solution for a heavily damped system (equation 6.35) is

( ) tt eAeAtz 2121

λλ += in which 1λ and 2λ are both real and negative.

If ( ) 0Ztz = when t = 0, substitution into the general solution gives 210 AAZ += .

To apply the second boundary condition, first differentiate the general solution to give ( ) tt eAeAtz 21

2211λλ λ+λ=& .

Since the question specifies that ( ) 0=tz& when t = 0, we find that 22110 AA λ+λ= .

Solving for A1 and A2 gives:

12

201 λ−λ

λ=

ZA and

12

102 λ−λ

λ−=

ZA

In order to sketch the response, we will take as an example 10 =Z , 11 −=λ and 22 −=λ . In this case, we find that A1 = 2 and A2 = 1. If these values are put into the general solution, we get the following graph. Note that the overall response (the black curve) is consistent with the given initial conditions that

10 =Z and ( ) 0=tz& when 0=t .



‐1.5

‐1

‐0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3

Dis

plac

emen

t (m

)

Time (s)

Term 1

Term 2

Sum

9. A critically damped structure is subjected to an impulse such that it

acquires an instantaneous initial velocity,V0, while the displacement remains zero. Show that the subsequent displacement is given by ( ) tnteVtz ω−= 0& .

Solution The general solution for a critically-damped system (equation 6.38) is

( ) tt nn teAeAtz ω−ω− += 21

Since ( ) 0=tz when 0=t , 01 =A . Differentiating the remaining term in the response expression gives: ( ) [ ]tt

nnn eetAtz ω−ω− +ω−= 2&

Since ( ) 0Vtz =& when t = 0, A2 = V0 and ( ) tnteVtz ω−= 0

The plot below illustrates the time history of the response.



Dis

plac

emen

t

Time

V0

10. The rigid beam shown in Figure Q10 has a moment of inertia of 10 kgm2

about the pivot at A. End C is displaced downwards by 10 mm from its equilibrium position and then released from rest. Find the maximum upward displacement of C from its equilibrium position and the elapsed time at which this occurs.

Solution Free-body diagram (AC= L)

θA B

θ3

2 &Lc ⋅ θLk ⋅

C

Equation of motion For positive (anti-clockwise) motion about the pivot at A,

( ) θ=⋅θ−⋅

θ− &&

AILkLLLc32

32



Rearranging, 09

4 22

=θ+θ+θ kLLcI A&

Or 05625020010 =θ+θ+θ &&

Hence, the undamped natural frequency, 75=ωn rad/s and damping ratio, 133.0=γ .

The damped natural frequency, 3.74=Ωn rad/s

Since γ < 1, the system is lightly-damped and the general solution (equation 6.42) is:

( ) ( )tBtBet nntn Ω+Ω=θ γω− sincos 21 (Q10.1)

Initial conditions: 0Θ=θ , 0=θ at t = 0

Hence, 01 Θ=B and 02 ΘΩγω

=n

nB

Here 67.65.101.0

0 −=−=Θ mrad

The resulting response is plotted below.

‐8

‐6

‐4

‐2

0

2

4

6

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

Ang

ular

dis

plac

emen

t (m

rad)

Time (s)

The response is a maximum when 0=θ& . Differentiating equation (Q10.1) and equating to zero leads to the condition that 0tan =Ω tn . The first non-zero root is π=Ω tn or t = 0.0423s. Substituting this time into equation (Q10.1) gives

369.4+=θ mrad and a corresponding tip displacement of 6.55 mm.



Note: An approximate value can often be found from inspection of the response time history. In this case, we see that the maximum upward displacement occurs half a cycle after the start. The period of damped

vibration is 0845.02 =Ω

π

n

s, so half a cycle is 0.0423 s as before. It happens

that this approach gives the exact result for this set of boundary conditions, but this is not always true. The default approach is to differentiate the displacement expression to get the velocity and then equate this to zero and solve for the time. This will always give the exact solution. 11. Figure Q11 shows a rocker arm with moment of inertia, IO, about the pivot

at O. Rubber blocks at ends A and B can each be modelled as a spring (stiffness, k) in parallel with a viscous damper (damping coefficient, c). The base of the block at A is attached to a rigid foundation. The base of the block at B is attached to a follower, which is driven by a cam that gives the follower a sinusoidal displacement of amplitude, Y, and frequency ω. Derive an expression for the steady-state amplitude of the displacement at A, assuming that the angular displacement of the bar is small.


θ&cakaθ +

θA BO

( ) ( )θ&& bycbθyk −+− This is an example of displacement excitation in which the follower attached to the bottom of the rubber block at B moves up and down with constant displacement amplitude, regardless of the rotational speed of the cam. The bottom of the block moves with displacement ( )ty and velocity ( )ty& and the top of the block moves with displacement ( )tbθ and velocity ( )tbθ& . In both cases, upward displacement and upward velocity are positive so that the change of length of the spring and relative velocity across the damper are ( )θ− by and ( )θ− && by respectively. These expressions imply that both spring and damper are being compressed so that the forces applied to end B of the



rocker by the spring damper model of the block are as shown in the free-body diagram. Equation of motion Taking moments of the forces about the pivot (anticlockwise positive):

( ) ( ) ( )( ) θ=θ−+θ−+θ+θ− &&&&&0Ibbycbykacaka

Rearranging, ( ) ( ) kbycbybakbacI +=θ++θ++θ 22220

&&&

Substitutions: ( ) ttYety ω= and ( ) ttet ω•Θ=θ

Hence,

( )( )( ) ( ) γ

+ω+ω−+ω+=Θ•

220

222 bactIbakcbtkb

The amplitude of the response is given by:

( ) ( )( )( ) ( )( ) γ

+ω+ω−+

ω+=Θ•2222

0222

22

bacIbak

cbkb

The amplitude of the displacement at A is •Θa

12. For an undamped system with n degrees of freedom, show that the

steady-state response in coordinate j due to a sinusoidal force of amplitude P and frequency ω applied in coordinate k is given by

( ) tPuu

txr r

krjrj ω

ω−ω= ∑

=sin

122

Solution Starting with the general equation of motion (equation 6.61), we uncouple the equations by making the substitution [ ] qx φ= (equation 6.62). Assuming that the modal vectors are scaled to unit modal mass and using their orthogonality properties, we get:

[ ] [ ] ( ) ( ) tftpqqI r =φ=

ω+ τ

O

O2

In this case, only coordinate k has any excitation, so the vector ( ) tp has only one entry, which is in row k. The modal space excitation vector is therefore,



( )

ωωω

=tPutPutPu

tf

kn

kr

k

sinsinsin1

The modal space equation for a typical mode r is: tPuqq krrrr ω=ω+ sin2 .

For this undamped system, the steady-state response can be obtained by

making the substitution ( ) tQtq rr ω= sin to give, 22 ω−ω=

r

krr

PuQ

The response in coordinate is given by equation 6.65 as,

( ) ( ) tPuu

tqutxn

r r

krjrn

rrjrj ω

ω−ω== ∑∑

==sin

122

1

Approximate methods 13. Use Dunkerleys Method and Rayleighs Method to estimate the lowest

natural frequency of the torsional system shown in Figure Q13.

Solution Dunkerley’s Method There are three sub-systems in this case. Sub-system 1 with the left-hand inertia only:

Ik22

1 =ω

2kI



Sub-system 2 with the central inertia only:

k2k2I

The series combination of the torsional springs of stiffness 2k and k is given

by kkKequivalent

1211 += giving

32kKequivalent = . The natural frequency for sub-

system 2 is therefore:

Ik3

22 =ω

Sub-system 3 with the right-hand inertia only:

k2k kI

Here, the equivalent spring stiffness is given bykkkKequivalent

11211 ++= .

52kKequivalent = and the natural frequency for sub-system 3 is therefore:

Ik

522

3 =ω

Dunkerleys formula gives:

++=

ω 253

2111

2 kn

or Ik

lk

n 408.06

==ω

Rayleigh’s Method

[ ]23

22

21

2max 2

21 θ+θ+θω= IIIT

( ) ( )[ ]232

221

21max 2

21 θ−θ+θ−θ+θ= kkkU



Since the right-hand end of the system is free, the first mode shape will be such that θ3 > θ2 > θ1 and all inertias rotate in-phase with each other. Try θ3: θ2: θ1 = 1: 2: 3

This gives Ik

n 471.0=ω

The equivalent of a static deflection can be obtained by applying static torques to each disc proportional to its inertia. This leads to θ3: θ2: θ1 = 2: 5: 6

and to Ik

n 447.0=ω

For reference, the exact value isIk

n 445.0=ω

14. A shaft with universal joints at each end has a length of 6 m, a second

moment of area of 0.00025 m4 and a mass/unit length of 75 kg/m. It carries three discs, which can be regarded as point masses of 100, 150, and 200 kg located 1.2, 3.0 and 4.8 m from the left-hand end. Estimate the lowest critical speed using Dunkerleys and Rayleighs Methods. Take E = 207 GN/m2, ρ = 7800 kg/m2.

Solution Dunkerley’s Method This is a numerical example of the method presented in the book starting on page 438. The intermediate results for the sub-systems are as follows.

With only the 100 kg mass: 2521 s108076.2 −×=ω



With only the shaft mass: 2420 s10186.5 −×=ω

Combining these in Dunkerleys formula gives 27.24=ωn Hz minrev1456=

Rayleigh’s Method This is a numerical example of the method presented in the book, starting on page 442.

A good choice for the mode shape function is ( )LxxY π= sin since this is the

exact mode shape of the shaft before the discs are added.



63

42

0 2

2

10834.542

1 ×=π=

= ∫ L

Eldxdx

YdElUL

MAX

( )[ ] ( )[ ]222

0

3

1

222

3.2398.1265.112

21

21

ω=ω+ω=

ω+ρω= ∫ ∑=

L

pppMAX xYmdxxYAT

Equating gives minrev1491Hz85.24 ==ωn

15. Find a single-degree-of-freedom approximate model to analyse the motion

of the 1 kg mass in the system in Figure Q15 when it is vibrating near its lower natural frequency.

Solution The exact solution to this problem is presented as a worked example in the book starting, on page 427.

1 kN/m

1x

( ) tPtp ωsin11 =

1 kN/m 1 kN/m

2x

1 kg 2 kgk

1x

( ) tPtp ωsin11 =

m

Original system Approximate model

Model parameters Equating the maximum kinetic energy in the real and approximate systems,

[ ][ ]

21

22

21

22

21

221

2

.2.1

.2.121

21

XXX

m

XXmXTMAX

+=∴

+ω=ω=

Equating the maximum strain energy in the real and approximate systems,



( )[ ]( )[ ]

21

22

221

21

22

221

21

21

.1.1.1

.1.1.121

21

XXXXX

k

XXXXkXUMAX

+−+=∴

+−+==

A number of mode shape estimates can be made and the selection of model parameters based on the one that predicts the lowest natural frequency. Here are some examples.

Est. 1 Est. 2 Est. 3 Est. 4 Est. 5 X1 1 3 2 2 0.732

X2 2 4 3 1 1

k (kN/m)

6.00 2.89 3.50 1.50 3.00

m (kg) 9.00 4.56 5.50 1.50 4.73

ωn (Hz) 4.109 4.008 4.015 5.033 4.007 Estimate 5 predicts the lowest frequency and uses the exact mode shape. 16. Use the model from Question 15 to estimate the steady-state response of

each mass in the two-degree-of-freedom system due to a sinusoidal force of amplitude 10 N and frequency 3.8 Hz applied to the 1 kg mass.

Solution The equation of motion for the approximate model is

( )tpkxxm 111 =+

Put ( ) ttePtp ω= 11 and ttex ω1 . This gives:

mkP

X 21

1 ω−=

Using the model parameters from Estimate 5 in Q15 and the given force amplitude and frequency, the response X1 = 33.06 mm. The response X2 is in direct proportion to the estimated mode shape. That is,

mm17.45732.00.1

12 =×= XX .

For comparison, the exact values are X1 = 37.49 mm and X2 = 43.58 mm. As a comparison with the exact values from Figure 6.72 in the book, response predictions from the approximate model are shown below for excitation frequencies up to 10 Hz. The predictions are similar up to about 4.5 Hz, but are poor at higher frequencies due to the influence of the second mode, which is not included in the approximate single-degree-of-freedom model of course.



-30

-20

-10

0

10

20

30

0 2 4 6 8 10

Dis

pla

ce

me

nt

[mm

]

Frequency [Hz]

X1 [mm]

X2 [mm]

w1

w2

X1 (approx)

X2 (approx)

Vibration control techniques 17. Derive the displacement transmissibility for the system in

Figure Q17. Solution Free-body diagram

Equation of motion

( ) ( ) xmKxxycxyk =−−+−

( ) kyycxKkxcxm +=+++

Substitutions: ( ) ttYety ω= and ( ) ttXetx ω=

( ) ctmKkctk

YX

ω+ω−+ω+= 2

Transmissibility is the ratio of magnitudes, so,

( ) 2222

222

cmKk

ckYXTD

ω+ω−+

ω+==

( )xyk −

m

xKx

( )xyc && −



18. A compressor of mass 300 kg is to be installed on four isolators, two at each end. The centre of mass is 0.4 m from end A and 0.2 m from end B. In the end elevation, the isolators are located symmetrically with respect to the centre of mass.

Isolators are available with stiffnesses of 40, 70, 110, 180 and 290 kN/m and each has a maximum allowable static deflection of 10 mm. Select suitable isolators for the installation so that the isolation efficiency is at least 70% at the normal running speed of 870 rev/min. Estimate the actual isolation efficiency for each of the isolators you select. Neglect damping.

Solution From the quoted weight distribution, each isolator at end A will support 50 kg and each isolator at end B will support 100 kg. From equation 6.93 in the book,

MAX

MINMAXMAX T

mTk

+ω

=1

2

At A: mkN8.95=MAXk

Select isolators with stiffness 70 kN/m from the list. Static deflection is 7.01 mm, which is less than the allowable maximum of 10 mm. Actual transmissibility is 0.203 (Isolation efficiency = 79.7%). At B: mkN5.191=MAXk

Select isolators with stiffness 180 kN/m from the list. Static deflection is 5.45 mm, which is less than the allowable maximum of 10 mm. Actual transmissibility is 0.277 (Isolation efficiency = 72.3%).

structural vibration worked examples - taylor &...

Documents