vibes assignments

1

MECH ENG 4105/7030 Advanced Vibrations Assignment 1 Low frequency analysis To be submitted to the Advanced Vibrations submission box on Level 2 of the Engineering

South building by 5pm on Tuesday 31 March 2015. 10 % of the total marks will be subtracted

for each day that the assignment is late. Note that a weekend counts as a 30 % penalty.

For this assignment it is suggested that you use Matlab to perform the calculations because

future assignments will build upon the work in this assignment. In addition to the results, you

will also need to submit a printed copy of the programs that you used to perform the

calculations, including comments, as an appendix.

Each of the plots that you present will need to include your student number in the title.

I have included some Matlab hints for this assignment on the back of this sheet. If you need

more assistance with Matlab, I suggest you examine the resources that have been placed on

MyUni in Advanced Vibrations under Course Material > Matlab Basics.

Task: A delicate piece of instrumentation is to be mounted on a rectangular steel panel, which

can be modelled as simply supported on all edges. The panel is 500 mm wide by 350 mm

high and 1.2 mm thick. The equipment is attached at a point (100 mm, 100 mm) from the

lower left corner, which is the origin of the coordinate system. The plate is driven by a force

located at (400 mm, 200 mm). Assume that the input force has a flat spectrum with amplitude

of 1 N from 10 to 500 Hz, with negligible phase variation across this frequency range.

The steel panel has the following properties: = 7850 kg/m3, = 0.3, E = 210 GPa, = 0.05.

1. Calculate the first 40 plate modes (i.e. (m,n) values) and their undamped resonance frequencies in Hz for transverse vibration and list them in ascending order in a table

together with their modal indices.

2. On a single graph, plot the displacement response in dB re 10-12 m at the equipment location from 10 to 500 Hz for

a) the first mode only; b) the first five modes only; c) the first 10 modes only; d) the first 20 modes only; and e) the first 40 modes.

3. Of the options considered, what is an appropriate number of modes to use in the analysis for this plate at 350 Hz, and why?

4. Of the options considered, what is an appropriate number of modes to use in the analysis for this plate at 500 Hz, and why?

5. Plot the vibration displacement distribution over the plate surface at 200 Hz. Which modes appear to dominate the vibrational response of the plate at this frequency?

From discussions with the manufacturer of the instrumentation, it is determined that the

instrumentation is particularly responsive to vibration at certain frequencies, and that the

current vibration amplitude at the equipment location is excessive at 350 Hz. It is desired to

reduce the vibration amplitude at the equipment location by 10 dB at 350 Hz.

6. What increase in damping ratio is required to achieve a 10 dB reduction in vibration amplitude?

7. What plate thickness is required to achieve a 10 dB reduction in vibration amplitude? 8. On a single graph, plot the displacement response in dB re 10-12 m at the equipment

location from 10 to 500 Hz using a suitable number of modes in your calculations (based

on consideration of the upper frequency) for the original system and the two systems

developed in Questions 6 and 7.

2

Hints

When calculating the first 40 plate modes, it is not possible to know from the start which 40

modes will be lowest in frequency. Hence you will need to calculate many more than 40

modes and then sort them to obtain the first 40 modes.

Matlab has a command to sort vectors, for example to sort a vector of values in a vector

x_val into ascending order you could use

[y_val,isort]=sort(x_val);

where y_val is the sorted values, and isort is the index showing where the sorted values

appeared in x_val. isort can then be used to sort associated vectors in the same order as

y_val, for example

m_sorted=m_val(isort);

n_sorted=n_val(isort);

omega_mn = sqrt(D/(rho*h))*y_val;

To use the above commands, the values that you wish to sort must be in a vector, rather than a

2-D array. This can be accomplished using the following structure of nested for loops:

i=1;

for m=1:whatever

for n=1:whatever

x_val(i)=(m*pi/Lx)^2+(n*pi/Ly)^2;

m_val(i)=m;

n_val(i)=n;

i=i+1;

end;

end;

Last hint. For the vibration distribution plot in Question 5, try using

surf(x,y,z)

where x is a vector of x coordinates, y is a vector of y coordinates and z is the displacement

over the grid formed by x and y, but beware of the required ordering of the matrix z.

SURF(x,y,Z) and SURF(x,y,Z,C), with two vector arguments replacing

the first two matrix arguments, must have length(x) = n and length(y)

= m where [m,n] = size(Z). In this case, the vertices of the surface

patches are the triples (x(j), y(i), Z(i,j)). Note that x corresponds

to the columns of Z and y corresponds to the rows.

For more details on the surf command, type help surf at the Matlab prompt.

To place your student number in the plot title, you can use the title command after the

plot or surf commands, i.e.

title(Displacement response of plate at 200 Hz (a1111111))

1

MECH ENG 4105/7030 Advanced Vibrations, Solutions to Assignment 1, 2015

The assignment is based on using the equations for the natural frequencies of a simply

supported plate and the displacement at a point on a simply supported plate given in

lectures. The natural frequency for the plate modes is given by

22

yxmn

L

n

L

m

h

D

where

23

112

EhD

and the displacement at a point on the plate is given by

N

n nnnn

innn

jM

xxFxX

122 2

)(),(

For a simply supported plate the mode shapes are given by

yxmn

L

yn

L

xmyx

sinsin,

and the modal mass is

4

yxmn

LhLM

An example Matlab program to perform the assignment tasks is shown in Appendix

A. Note that the example program is more substantial than I would expect students to

submit for the assignment, especially the plotting part. As I mentioned in lectures, the

program that is shown is probably not elegant from an experienced Matlab

programmer's perspective, but it does the job.

Q1. The output from the example program is shown below.

assign1_2015.m

Solutions to Advanced Vibrations Assignment 1, 2015

Calculates natural frequencies for a simply supported plate and

plots frequency response of plate at a point and vibrational response

at a frequency

AZ 23 March 2015

2

plate thickness, h (m) : 0.0012

plate damping ratio, zeta : 0.05

max number of plate modes to use for plate response calculations : 40

m = 1, n = 1, f = 35.8852 Hz

m = 2, n = 1, f = 71.2886 Hz

m = 1, n = 2, f = 108.1371 Hz

m = 3, n = 1, f = 130.2944 Hz

m = 2, n = 2, f = 143.5406 Hz

m = 3, n = 2, f = 202.5464 Hz

m = 4, n = 1, f = 212.9025 Hz

m = 1, n = 3, f = 228.5571 Hz

m = 2, n = 3, f = 263.9606 Hz

m = 4, n = 2, f = 285.1545 Hz

m = 5, n = 1, f = 319.1129 Hz

m = 3, n = 3, f = 322.9664 Hz

m = 5, n = 2, f = 391.3649 Hz

m = 1, n = 4, f = 397.1451 Hz

m = 4, n = 3, f = 405.5745 Hz

m = 2, n = 4, f = 432.5486 Hz

m = 6, n = 1, f = 448.9257 Hz

m = 3, n = 4, f = 491.5543 Hz

m = 5, n = 3, f = 511.7849 Hz

m = 6, n = 2, f = 521.1777 Hz

m = 4, n = 4, f = 574.1624 Hz

m = 7, n = 1, f = 602.3407 Hz

m = 1, n = 5, f = 613.901 Hz

m = 6, n = 3, f = 641.5976 Hz

m = 2, n = 5, f = 649.3045 Hz

m = 7, n = 2, f = 674.5927 Hz

m = 5, n = 4, f = 680.3729 Hz

m = 3, n = 5, f = 708.3103 Hz

m = 8, n = 1, f = 779.3581 Hz

m = 4, n = 5, f = 790.9184 Hz

m = 7, n = 3, f = 795.0127 Hz

m = 6, n = 4, f = 810.1856 Hz

m = 8, n = 2, f = 851.6101 Hz

m = 1, n = 6, f = 878.825 Hz

m = 5, n = 5, f = 897.1288 Hz

m = 2, n = 6, f = 914.2285 Hz

m = 7, n = 4, f = 963.6007 Hz

m = 8, n = 3, f = 972.0301 Hz

m = 3, n = 6, f = 973.2343 Hz

m = 9, n = 1, f = 979.9778 Hz

primary frequency of interest, pri_freq (Hz) : 350

x (dB) at output location @ 350 Hz = 116.6087 dB re 1e-12 m

x (m) at output location @ 350 Hz = 6.7676e-007 m

[3 marks total for correct modal indices, natural frequency values and units]

Q2. Vibration response at equipment location for h = 0.0012 m, = 0.05, for 1 mode, 5 modes, 10 modes, 20 modes, and 40 modes. [4 marks for adequately labelled

plot. Minus a mark for each of the following transgressions: unlabelled or

incorrectly labelled axis; no legend; data incorrect.]

3

Figure 1. Displacement response of plate. Example result for Question 2.

Q3. The appropriate number of modes is determined by examining the convergence

of the solution for the various numbers of modes used in the modal summation. If the

only frequency of interest is 350 Hz then, of the options considered, 40 modes should

be used to model the response for the given input and output locations, because the

modal summation series has converged at 40 modes (but not quite at 20 modes).

[2 marks]

Q4. 40 modes is sufficient to model the system accurately up to 500 Hz, because the

modal summation series has converged at this frequency. This can be confirmed by

comparing the response for 80 modes, which gives the same result as using 40 modes.

[2 marks]

Note: As a rough guide, modes with natural frequencies up to twice the frequency of

interest need to be included in calculations of vibrational response for reasonable

accuracy.

Q5. For damped structures with excitation at a frequency that does not correspond to

a natural frequency, the displacement response is generally complex. The plot shown

in Figure 2 is the absolute values of displacement over the surface of the plate.

0 50 100 150 200 250 300 350 400 450 50090

100

110

120

130

140

150

160

170Displacement response of plate

frequency (Hz)

dis

pla

cem

ent (d

B r

e 1

e-1

2 m

)

1 mode

5 modes

10 modes

20 modes

40 modes

4

Figure 2. Absolute vibration displacement of plate. Example result for Question 5 at

200 Hz.

[1 mark for adequately labelled plot of abs or other appropriate values]

The vibrational response of the plate at 200 Hz is not dominated by a single vibration

mode. Examination of the modal amplitudes at 200 Hz indicates that the (3,2), (4,1)

and (3,1) modes are the greatest contributors at this frequency.

[2 marks for explanation that not a single dominant mode, and listing some of

the greatest contributors]

Q6. For the plate with h = 0.0012 m and = 0.05, the displacement at 350 Hz at (x, y) = (0.1 m , 0.1 m) is found to be 6.77 10-7 m, or 116.6 dB re 1e-12 m. To achieve a

10 dB reduction the displacement needs to be reduced to below 2.14 10-7 m, or 106.6 dB re 1e-12 m, via rearrangement of

original

reduced

X

X10log2010

From trial and error, values of 0.185 (an increase of 0.135 or more) were found to achieve the desired reduction.

[2 marks for finding a value of damping that reduces the displacement by 10 dB]

Q7. From trial and error, increasing the plate thickness by 5.5 mm to 6.7 mm or more

was found to achieve the desired reduction. Note that there may be other plate

thicknesses that also achieve a 10 dB reduction in displacement.

[2 marks for finding a plate thickness that reduces the displacement by 10 dB]

00.1

0.20.3

0.40.5

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0

0.5

1x 10

-5

x (m)

Displacement response of plate at 200 Hz

y (m)

dis

pla

cem

ent (m

)

5

Note that a change in the plate thickness, h, requires the natural frequencies for the

plate to be re-calculated, and it is not sufficient to simply change the modal mass of

the plate, Mn.

Q8. At least 40 modes are needed in the calculations up to 500 Hz for the baseline,

increased damping and increased plate thickness cases.

Figure 3. Displacement response of plate for various modifications. Example result

for Question 8.

[2 marks for adequately labelled plot including baseline, increased damping, and

increased thickness]

An example Matlab program to plot the displacement response for the original and

modified plate systems is shown in Appendix A.

0 50 100 150 200 250 300 350 400 450 50080

90

100

110

120

130

140

150

160

170

180Displacement response of plate for various cases

frequency (Hz)

dis

pla

cem

ent (d

B r

e 1

e-1

2 m

)

baseline

increased damping

increased thickness

6

Appendix A - Example Matlab program to calculate plate response %assign1_2015.m

%

%Solutions to Advanced Vibrations Assignment 1, 2015

%

%Calculates natural frequencies for a simply supported plate and

%plots frequency response of plate at a point and vibrational

response

%at a frequency

%

%AZ 23 March 2015

help assign1_2015;

%set up geometric and physical constants

%steel

E=210e9; %Pa

nu=0.3;

rho=7850; %kg/m^3

%aluminium

%E=71.6e9; %Pa

%nu=0.34;

%rho=2700; %kg/m^3

Lx=0.5; %m

Ly=0.35; %m

h = input('plate thickness, h (m) : ');

zeta = input('plate damping ratio, zeta : ');

pl_modes = input('max number of plate modes to use for plate response

calculations : ');

D = E*h^3/(12*(1-nu^2));

x_in=0.4; %m

y_in=0.2; %m

x_out=0.1; %m

y_out=0.1; %m

F_in=1; %N

%calculate first 40 natural frequencies

i_max=40;

i=1;

for m=1:(i_max/2)

for n=1:(i_max/2)

x_val(i)=(m*pi/Lx)^2+(n*pi/Ly)^2;

m_val(i)=m;

n_val(i)=n;

i=i+1;

end;

end;

%sort eigenvalues into ascending order

[x_val,isort]=sort(x_val);

m_val=m_val(isort);

n_val=n_val(isort);

omega_mn = sqrt(D/(rho*h))*x_val;

for i=1:i_max

7

disp(['m = ',num2str(m_val(i)),', n = ',num2str(n_val(i)),...

', f = ',num2str(omega_mn(i)/(2*pi)),' Hz']);

end;

%calculate vibration response over frequency span 5 to 400 Hz

M_mn=rho*Lx*Ly*h/4;

modal_x=zeros(i_max,500);

for i=1:i_max

psi_in=sin(m_val(i)*pi*x_in/Lx)*sin(n_val(i)*pi*y_in/Ly);

psi_out=sin(m_val(i)*pi*x_out/Lx)*sin(n_val(i)*pi*y_out/Ly);

for f=10:500

omega=2*pi*f;

modal_x(i,f)=F_in*psi_in*psi_out/(M_mn*(omega_mn(i)^2+...

j*2*zeta*omega*omega_mn(i)-omega^2));

end;

end;

f=1:500;

x_ref=1e-12; %m

%calculate vibration distribution over plate at primary frequency of

interest in Hz

%using pl_modes modes

disp(' ');

pri_freq = input('primary frequency of interest, pri_freq (Hz) : ');

num_x_pts=30;

num_y_pts=20;

z=zeros(num_y_pts+1,num_x_pts+1);

for ii=1:num_x_pts+1

x=(ii-1)*Lx/num_x_pts;

for jj=1:num_y_pts+1

y=(jj-1)*Ly/num_y_pts;

for mn=1:pl_modes

psi_in=sin(m_val(mn)*pi*x_in/Lx)*sin(n_val(mn)*pi*y_in/Ly);

psi_out=sin(m_val(mn)*pi*x/Lx)*sin(n_val(mn)*pi*y/Ly);

z(jj,ii)=z(jj,ii)+F_in*psi_in*psi_out/(M_mn*(omega_mn(mn)^2+...

j*2*zeta*2*pi*pri_freq*omega_mn(mn)-(2*pi*pri_freq)^2));

end;

end;

end;

disp(' ');

disp(['x (dB) at output location @ ',num2str(pri_freq),' Hz =

',num2str(20*log10(abs(sum(modal_x(1:pl_modes,pri_freq)))/x_ref)),'

dB re 1e-12 m']);

disp(' ');

disp(['x (m) at output location @ ',num2str(pri_freq),' Hz =

',num2str(abs(sum(modal_x(1:pl_modes,pri_freq)))),' m']);

disp(' ');

%plot results

plot_type=0;

while plot_type~=3

plot_type=menu('data to plot','frequency response','vibrational

response',...

'quit plotting');

if (plot_type==1)

8

disp_type=0;

disp_type=menu('display type','linear','dB');

if disp_type==1

plot(f,abs(modal_x(1,:)),'r-',...

f,abs(sum(modal_x(1:5,:))),'g:',...

f,abs(sum(modal_x(1:10,:))),'k--',...

f,abs(sum(modal_x(1:20,:))),'b-.',...

f,abs(sum(modal_x(1:40,:))),'c-');

title('Displacement response of plate');

xlabel('frequency (Hz)');

ylabel('displacement (m)');

legend('1 mode','5 modes','10 modes','20 modes','40 modes');

end;

if disp_type==2

plot(f,20*log10(abs(modal_x(1,:))/x_ref),'r-',...

f,20*log10(abs(sum(modal_x(1:5,:)))/x_ref),'g:',...

f,20*log10(abs(sum(modal_x(1:10,:)))/x_ref),'k--',...

f,20*log10(abs(sum(modal_x(1:20,:)))/x_ref),'b-.',...

f,20*log10(abs(sum(modal_x(1:40,:)))/x_ref),'c-');

title('Displacement response of plate');


ylabel('displacement (dB re 1e-12 m)');

legend('1 mode','5 modes','10 modes','20 modes','40 modes');

end;

end;

if (plot_type==2)

for ii=1:num_x_pts+1

x(ii)=(ii-1)*Lx/num_x_pts;

end;

for ii=1:num_y_pts+1

y(ii)=(ii-1)*Ly/num_y_pts;

end;

surf(x,y,abs(z));

title(['Displacement response of plate at ',num2str(pri_freq),'

Hz']);

xlabel('x (m)');

ylabel('y (m)');

zlabel('displacement (m)');

end;

end;

Example Matlab program to plot various plate responses

The results produced by each run of the program must be saved to a mat file so that

they can be combined with other results. For example, after the plate response

calculations have been performed for the original plate, the following commands

would be used in Matlab: >>baseline_modal_x=modal_x;

>>save baseline baseline_modal_x f

Once an appropriate level of damping has been determined from trial and error, the

results would be saved via:

>>damped_modal_x=modal_x;

9

>>save damped damped_modal_x f

Similarly, once a suitable plate thickness has been determined, the results would be

saved using:

>>thick_modal_x=modal_x;

>>save thick thick_modal_x f

The following program would then be run to load the data and plot the results. %assign1_partb_2015.m

%

%Solutions to Advanced Vibrations Assignment 1, 2015

%

%Plots frequency response of plate at a point for various cases

%

%AZ 23 March 2015

help assign1_partb_2015;

x_ref=1e-12; %m

load baseline

load damped

load thick

%plot results

plot(f,20*log10(abs(sum(baseline_modal_x(1:40,:)))/x_ref),'r-',...

f,20*log10(abs(sum(damped_modal_x(1:40,:)))/x_ref),'g:',...

f,20*log10(abs(sum(thick_modal_x(1:40,:)))/x_ref),'k--

',...

[350,350],[80,180],'b');

title('Displacement response of plate for various cases');


ylabel('displacement (dB re 1e-12 m)');

legend('baseline','increased damping','increased thickness');

MECH ENG 4105/7030 Advanced Vibrations Assignment 2 Statistical Energy Analysis To be submitted to the Advanced Vibrations submission box on Level 2 of the Engineering South building by 5pm on Tuesday 28 April 2015. 10 % of the total marks will be subtracted for each day that the assignment is late. Note that a weekend counts as a 30 % penalty. Each of the plots that you present will need to include your student number in the title. Question 1 Calculate the resonance frequencies of the steel plate from Assignment 1 (length 500 mm, width 350 mm, and thickness 1.2 mm with the properties: = 7850 kg/m3, = 0.3, E = 210 GPa, = 0.05) for the first 1000 modes. Sort the modes by frequency and order into one-third octave bands (using the preferred frequency bands and band limits), from the 31.5 Hz up to and including the 10 kHz one-third octave band. Calculate the band averaged modal density spectrum, n(f), from this data. Plot the modal density calculated by counting the number of modes in each frequency band with the results obtained using the appropriate analytical SEA expression for a flat plate. How many modes per band are required for the SEA analytical estimate of the modal density to be within 25% of the actual modal density for this plate? From which one-third octave band centre frequency (and above) is this criterion satisfied? How many modes are in this band? Hints To find the number of modes in a vector of natural frequency below a specified frequency value, you can use the find command together with the size command, for example: modes_below = size(find(omega_mn

between the subsystems assume the damping loss factors of the beam and plate are 0.3 and 0.15 respectively. Assume the aluminium properties are: = 2700 kg/m3, = 0.34, E = 71.6 GPa. Sketch a block diagram representing the power balance for the gearbox/beam/plate/room system, showing all potential power flows. Using SEA formulations, calculate the following in the 63 to 1000 Hz octave bands: The modal densities and number of modes in the beam, plate and acoustic space subsystems. The coupling loss factor for the beam-plate and plate-acoustic space connections.

The CLF for the beam-plate connection can be evaluated from

b

ppbp m

c4=

where p is the mass per unit area of the plate, cp is the bending wavespeed in the plate, and mb is the mass per unit length of the beam. You may assume that there is negligible direct coupling between the beam and the acoustic space.

The subsystem energies for the beam, plate and acoustic space. The vibration levels (in dB re 10-9 m/s) of the plate and beam and the sound pressure level

(in dB re 20 Pa) in the acoustic space. To do this you will need to build a 3 subsystem model representing the beam/plate/room problem and set up power balance equations, which you can then solve for system energies. A spreadsheet or Matlab solution will be essential.

For thin plates, the first critical frequency is given by

hccf

L

oc

255.0=

where cL is the plate longitudinal wavespeed, co the speed of sound in air, and h is the plate thickness. P is the plate perimeter and S the plate area (both for one side only).

[12 marks total]

2

1

MECH ENG 4020/7030 Advanced Vibrations, Assignment 2 Statistical Energy Analysis, 2015 solutions

Q1. Using the same calculation procedure as was used in Assignment 1, the first 1000

plate modes can be calculated for the simply supported steel plate.

The modal natural frequencies are then ordered into one-third octave bands, and the

number of modes in each band counted. This number is then divided by the one-third

octave band bandwidth in Hertz to give the model density n(f).

The SEA estimate for the modal density is given by

D

hLLn

yx

4)(

D

hLLnfn

yx

2)(2)(

The following properties for steel were used: E=210 GPa, =0.3, =7850 kg/m3.

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000

0.02

0.04

0.06

0.08

0.1

0.12Comparison of plate modal density for ss plate

one-third octave band centre frequency(Hz)

n(f

)

counted

analytical estimate

Figure 1. Modal density for a simply supported plate.

[Calculating modal density from the natural frequencies for the plate [1 mark],

calculating analytical SEA estimate [1 mark], and plotting values for comparison

with adequately labelled axes [1 mark].]

2

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000

20

40

60

80

100

120Comparison of plate modes in band for ss plate


delta N

counted

analytical estimate

Figure 2. Number of modes per band for a simply supported plate, up to 2500 Hz

only.

From a comparison of the simply supported plate modal densities calculated by

counting the number of modes in each band and those calculated using the SEA

analytical estimate, it can be seen that the that the lowest one-third octave band for

which the analytical SEA modal density is within 25% of the actual (or counted) modal density is the 80 Hz band. [1 mark]

The analytical SEA modal density is within 25% of the actual (or counted) modal density from the 1250 Hz one-third octave band and higher, i.e.

from 1250 Hz. [1 mark for band]

The 1250 Hz band contains 14 modes. [1 mark]

An example Matlab program to perform the calculations for Question 1 is shown in

Appendix A.

Q2. The values for compressional wave speeds for air and water can be found in a

reference such as Bies and Hansen, Engineering Noise Control 4th Edition Appendix

B. Assume ambient conditions.

For air at 20 C speed of sound = 343 m/s

For sea water at 13 C speed of sound = 1500 m/s used for calculations. (1530 m/s also OK.)

For steel E=210 GPa, =0.3, =7850 kg/m3

For air and water the wavenumber is given by

ock

25.1)()(75.0 countedSEA fnfn

3

As shown in the notes, for a rectangular plate the bending wavespeed is given by

4h

Dcb

and the wavenumber for bending waves is

bb

ck

A comparison of the wavenumber in different media is shown in Figure 3. The 1.2

mm steel plate will couple most effectively with air at the intersection of the

wavenumber plots for the two media, which is at 10 kHz. [0.5 mark for value for

air] For sea water the plots intersect at 200 kHz. [0.5 mark for value for sea water]

102

103

104

105

106

10-1

100

101

102

103

104

105

Comparison of wavenumbers for various media


wavenum

ber

(rad/m

)

air

sea water

1.2 mm steel plate

Figure 3. Wavenumbers for air, sea water and plate bending waves.

[1 mark for adequately labelled plot]

An example Matlab program to perform the calculations for Question 2 is shown in

Appendix A.

4

Q3. The physical arrangement of the motor, beam, plate and room system can be

considered to be arranged as shown in Figure 4. Note that in the solution of the

problem, it is assumed that there is no direct coupling between the beam and the

acoustic space.

Figure 4. Physical system arrangement of motor/beam/plate/room system.

This can be represented in a power balance block diagram as

Figure 5. Block diagram representation of power balance for

motor/beam/plate/room system. [1 mark]

The modal densities for the various subsystems are:

Beam bending

hc

L

c

Ln

Lb 4.32)(

where the beam bending wavespeed is given by

4A

EIcb

and L is the beam length and h is the beam depth or thickness.

motor

1. beam 2. plate 3. room

P2 in

P1 loss P2 loss P3 loss

P1 2

P2 1

P2 3

P3 2

F

room

plate

beam

5

Plate bending

D

hLLn

yx

4)(

Room

ooo c

P

c

A

c

Vn

1682)(

232

2

V is the room volume, A the room surface area, P the room perimeter. The room is

assumed to be cubic with equal length sides in each dimension.

The number of modes in each band for each subsystem is given by

)(nN

where is the frequency span of the analysis band.

The coupling loss factors for the connected subsystems are given as follows.

Beam-plate

bm

pcpbp

4

where p is the mass per unit area of the plate, pc is the bending wavespeed in the

plate, bm is the mass per unit length of the beam.

Plate-room

pM

radpAocopv

As shown in the assignment sheet, for thin plates, the critical frequency is given by

hLc

occf

255.0

and is equal to 2003 Hz in this instance.

21

ELc

where is the density of the plate, and h is the plate thickness. P is the plate perimeter and S the plate area.

6

The radiation efficiency of the plate is given in the assignment sheet as a function of

the plate parameter S

Ph and the frequency parameter

cf

f. The plate parameter is

0.01, and from the graph, the values for 10log10 in each of the octave bands are as

given in Table 1.

Table 1. Radiation efficiences for flat plate.

f (Hz) f/fc 10log10(rad) rad

63 0.0315 -22.5 0.0056

125 0.0624 -21.5 0.0071

250 0.1248 -20 0.0100

500 0.2496 -19 0.0126

1000 0.4993 -15 0.0316

[1 mark] table (preferably) or in code

The damping loss factor for the room is calculated from

V

Acoroom

4

where V is the room volume and A the room surface area.

Power input

The power input to the beam and room subsystems is zero, while the power input

from the motor to the plate is estimated by considering it as an isolated system.

Considering the following power and energy terms to be time and frequency band

averages,

din PP 22

7

22 EP d

2

2

22

nM

FE

Combining these equations gives

2

2

22

nM

FP in

where in this instance M is the mass of the plate subsystem and 2n the modal density of the plate subsystem.

P_2_in = 16.8 mW in each octave band.

[1 mark for calculating input power, either separately or within code]

Power Balance

The power balance equation for each of the subsystems can be written as follows:

Subsystem 1

1210 PP loss

2

2

1

1112110

n

E

n

EnE

Subsystem 2

232122 PPPP lossin

3

3

2

2223

1

1

2

2221222

n

E

n

En

n

E

n

EnEP in

Subsystem 3

3230 PP loss

2

2

3

3332330

n

E

n

EnE

Representing the power balance equations for the three subsystems in matrix form

gives

0

0

0

0

2

3

3

2

2

1

1

3323332

223223212221

1121121

inP

n

E

n

E

n

E

nn

nnn

nn

We can then solve for the subsystem energies using

PCn

E 11

8

and

nn

EE

for the individual subsystems, and utilising

2

11221

n

n and

3

22332

n

n

For vibratory subsystems, the subsystem energy level is related to the vibration levels

by

spvME 2

and rearranging gives

M

Ev

sp2

while for acoustic subsystems the acoustic pressure in the volume is related to the

subsystem energy level by

Vc

p

E

oo

sp

2

2

which can be rearranged to

V

cEp oo

sp

22

and the acoustic pressure is customarily represented in decibels as

2

2

10log10

ref

sp

p

p

SPL dB

where ref

p = 20 10-6 Pa

The vibration levels can also be expressed in decibels with a reference velocity of 10-9

m/s.

An example Matlab program to perform these calculations is shown in Appendix B.

The resulting subsystem energies and susbsystem responses are shown in the

following tables.

9

Table 2. Subsystem Modal Densities.

f (Hz) beam Plate Room

n() (modes.

s/rad)

N (modes)

n( ) (modes.

s/rad)

N (modes)

n( ) (modes.

s/rad)

N (modes)

63 0.0049 1.4 0.0475 13 0.0098 3

125 0.0035 1.9 0.0475 26 0.0238 13

250 0.0025 2.7 0.0475 53 0.0701 78

500 0.0017 3.9 0.0475 106 0.2352 523

1000 0.0012 5.5 0.0475 211 0.8558 3802

[1 mark] [1 mark] [1 mark]

Table 3. Subsystem Coupling Loss Factors.

f (Hz) beam-plate plate-room 63 36.2249 0.3701 e-3

125 25.7172 0.2348 e-3

250 18.1848 0.1659 e-3

500 12.8586 0.1044 e-3

1000 9.0924 0.1311 e-3

[1 mark] [1 mark]

Table 4. Subsystem Energies.

f (Hz) Ebeam (J) Eplate (J) Eroom (J)

63 0.2400 e-4 0.2348 e-3 0.1253 e-5

125 0.0901 e-4 0.1246 e-3 0.0847 e-5

250 0.0329 e-4 0.0648 e-3 0.0626 e-5

500 0.0119 e-4 0.0333 e-3 0.0407 e-5

1000 0.0043 e-4 0.0170 e-3 0.0522 e-5

[1 mark for solving for energy of each system at each frequency]

Table 5. Subsystem Responses.

f (Hz)

Beam Plate Acoustic Space

spv 2

(m/s)2

Vibration

velocity

level (dB

re 10-9

m/s)

spv 2

(m/s)2

Vibration

velocity

level (dB re

10-9

m/s)

spp2

(Pa2)

SPL (dB

re 20

Pa)

63 0.3766e-4 135.8 0.2647 e-5 124.2 0.0114 74.6

125 0.1414e-4 131.5 0.1405 e-5 121.5 0.0077 72.9

250 0.0517e-4 127.1 0.0730 e-5 118.6 0.0057 71.5

500 0.0187e-4 122.7 0.0375 e-5 115.7 0.0037 69.7

1000 0.0067e-4 118.2 0.0191 e-5 112.8 0.0048 70.8

[1 mark] [1 mark] [1 mark]

10

Appendix A - Example Matlab program for Questions 1 and 2 %assign2_q3_2015.m

%

%Solutions to Advanced Vibrations Assignment 2, 2015, question 3.

%Solutions to Questions 1 and 2 in assign2_2015.m

%

%Solves SEA system for a beam/plate/cavity.

%

%A.Zander 1 June 2015

help assign2_q3_2015;

%Question 3


%beam and plate are aluminium

E=71.6e9; %Pa

nu=0.34;

rho=2700; %kg/m^3

%beam geometry and properties

L_beam=2.36; %m

width=0.01; %m

depth=0.01; %m

eta_beam=0.3;

%plate geometry and properties

Lx=2.36; %m

Ly=2.36; %m

h=0.0059; %m

eta_plate=0.15;

%room geometry and properties

Lm=2.5; %m

Ln=2.5; %m

Lp=2.5; %m

V=Lm*Ln*Lp; %m^3

alpha_bar=0.13;

c_o=343; %m/s

rho_o=1.21; %kg/m^3

%motor

F_in=20; %N^2

%calculate modal densities and number of modes for each subsystem in

octave bands

f_oct=[63 125 250 500 1000];

f_lo=[44 88 176 353 707];

f_hi=[88 176 353 707 1414];

%plate

D = E*h^3/(12*(1-nu^2));

for i=1:size(f_oct,2)

n_f_plate(i)=Lx*Ly/2*sqrt(rho*h/D);

num_modes_plate(i)=n_f_plate(i)*(f_hi(i)-f_lo(i));

end;

%beam

c_L=(E/rho)^0.5;


n_f_beam(i)=L_beam/((1/12)^0.25*(c_L*depth*2*pi*f_oct(i))^0.5);

11

num_modes_beam(i)=n_f_beam(i)*(f_hi(i)-f_lo(i));

end;

%room

A=2*(Lm*Ln+Ln*Lp+Lp*Lm);

P=4*(Lm+Ln+Lp);


n_f_room(i)=V*(2*pi*f_oct(i))^2/(pi*c_o^3)+A*2*pi*f_oct(i)/(4*c_o^2)+

P/(8*c_o);

num_modes_room(i)=n_f_room(i)*(f_hi(i)-f_lo(i));

end;

%also calculate loss factor for room

eta_room=c_o*A*alpha_bar./(4*2*pi*f_oct*V);

%calculate CLFs for connected subsystems

%beam plate CLF

m_b=width*depth*rho;

rho_p=rho*h;

c_b_plate = (2*pi*f_oct).^0.5.*(D/(rho*h)).^0.25;

eta_bp=4*rho_p*c_b_plate./(m_b*2*pi*f_oct);

%plate room CLF

c_L_plate=(E/(rho*(1-nu*nu)))^0.5;

f_critical=0.55*c_o*c_o/(c_L_plate*h);

plate_param=2*(Lx+Ly)*h/(Lx*Ly);

freq_param=f_oct/f_critical;

%reading values from graph supplied on assignment sheet

sigma_rad_dB=[-22.5 -21.5 -20 -19 -15];

sigma_rad=10.^(sigma_rad_dB/10);

eta_pv=rho_o*c_o./(h*rho*2*pi*f_oct).*sigma_rad;

%calculate input power

n_w_plate=n_f_plate/(2*pi);

P_2_in=F_in*pi*n_w_plate/(2*Lx*Ly*h*rho);

%calculate power balance matrices and solve for energies


P=[0;P_2_in(i);0];

eta_1=eta_beam;

eta_2=eta_plate;

eta_3=eta_room(i);

eta_12=eta_bp(i);

eta_23=eta_pv(i);

n_1=n_f_beam(i)/(2*pi);

n_2=n_f_plate(i)/(2*pi);

n_3=n_f_room(i)/(2*pi);

eta_21=eta_12*n_1/n_2;

eta_32=eta_23*n_2/n_3;

C=[(eta_1+eta_12)*n_1 -eta_12*n_1 0;...

-eta_21*n_2 (eta_2+eta_21+eta_23)*n_2 -eta_23*n_2;...

0 -eta_32*n_3 (eta_3+eta_32)*n_3];

modal_E=1/(2*pi*f_oct(i))*inv(C)*P;

E_beam(i)=modal_E(1)*n_1;

E_plate(i)=modal_E(2)*n_2;

E_room(i)=modal_E(3)*n_3;

end;

12

v_2_beam=E_beam./(L_beam*width*depth*rho);

v_beam=sqrt(v_2_beam);

v_2_plate=E_plate./(Lx*Ly*h*rho);

v_plate=sqrt(v_2_plate);

p_2_room=E_room*rho_o*c_o*c_o/V;

SPL=10*log10(p_2_room/(20e-6*20e-6));

Lv_beam=10*log10(v_2_beam/(1e-9*1e-9));

Lv_plate=10*log10(v_2_plate/(1e-9*1e-9));

Appendix B - Example Matlab program for Question 3

%assign2_q3_2015.m

%

%Solutions to Advanced Vibrations Assignment 2, 2015, question 3.

%Solutions to Questions 1 and 2 in assign2_2015.m

%

%Solves SEA system for a beam/plate/cavity.

%

%A.Zander 1 June 2015

help assign2_q3_2015;

%Question 3


%beam and plate are aluminium

E=71.6e9; %Pa

nu=0.34;

rho=2700; %kg/m^3

%beam geometry and properties

L_beam=2.36; %m

width=0.01; %m

depth=0.01; %m

eta_beam=0.3;

%plate geometry and properties

Lx=2.36; %m

Ly=2.36; %m

h=0.0059; %m

eta_plate=0.15;

%room geometry and properties

Lm=2.5; %m

Ln=2.5; %m

Lp=2.5; %m

V=Lm*Ln*Lp; %m^3

alpha_bar=0.13;

c_o=343; %m/s

rho_o=1.21; %kg/m^3

%motor

F_in=20; %N^2

%calculate modal densities and number of modes for each subsystem in

octave bands

f_oct=[63 125 250 500 1000];

13

f_lo=[44 88 176 353 707];

f_hi=[88 176 353 707 1414];

%plate

D = E*h^3/(12*(1-nu^2));


n_f_plate(i)=Lx*Ly/2*sqrt(rho*h/D);

num_modes_plate(i)=n_f_plate(i)*(f_hi(i)-f_lo(i));

end;

%beam

c_L=(E/rho)^0.5;


n_f_beam(i)=L_beam/((1/12)^0.25*(c_L*depth*2*pi*f_oct(i))^0.5);

num_modes_beam(i)=n_f_beam(i)*(f_hi(i)-f_lo(i));

end;

%room

A=2*(Lm*Ln+Ln*Lp+Lp*Lm);

P=4*(Lm+Ln+Lp);


n_f_room(i)=V*(2*pi*f_oct(i))^2/(pi*c_o^3)+A*2*pi*f_oct(i)/(4*c_o^2)+

P/(8*c_o);

num_modes_room(i)=n_f_room(i)*(f_hi(i)-f_lo(i));

end;

%also calculate loss factor for room

eta_room=c_o*A*alpha_bar./(4*2*pi*f_oct*V);

%calculate CLFs for connected subsystems

%beam plate CLF

m_b=width*depth*rho;

rho_p=rho*h;

c_b_plate = (2*pi*f_oct).^0.5.*(D/(rho*h)).^0.25;

eta_bp=4*rho_p*c_b_plate./(m_b*2*pi*f_oct);

%plate room CLF

c_L_plate=(E/(rho*(1-nu*nu)))^0.5;

f_critical=0.55*c_o*c_o/(c_L_plate*h);

plate_param=2*(Lx+Ly)*h/(Lx*Ly);

freq_param=f_oct/f_critical;

%reading values from graph supplied on assignment sheet

sigma_rad_dB=[-22.5 -21.5 -20 -19 -15];

sigma_rad=10.^(sigma_rad_dB/10);

eta_pv=rho_o*c_o./(h*rho*2*pi*f_oct).*sigma_rad;

%calculate input power

n_w_plate=n_f_plate/(2*pi);

P_2_in=F_in*pi*n_w_plate/(2*Lx*Ly*h*rho);

%calculate power balance matrices and solve for energies


P=[0;P_2_in(i);0];

eta_1=eta_beam;

eta_2=eta_plate;

eta_3=eta_room(i);

eta_12=eta_bp(i);

eta_23=eta_pv(i);

n_1=n_f_beam(i)/(2*pi);

n_2=n_f_plate(i)/(2*pi);

14

n_3=n_f_room(i)/(2*pi);

eta_21=eta_12*n_1/n_2;

eta_32=eta_23*n_2/n_3;

C=[(eta_1+eta_12)*n_1 -eta_12*n_1 0;...

-eta_21*n_2 (eta_2+eta_21+eta_23)*n_2 -eta_23*n_2;...

0 -eta_32*n_3 (eta_3+eta_32)*n_3];

modal_E=1/(2*pi*f_oct(i))*inv(C)*P;

E_beam(i)=modal_E(1)*n_1;

E_plate(i)=modal_E(2)*n_2;

E_room(i)=modal_E(3)*n_3;

end;

v_2_beam=E_beam./(L_beam*width*depth*rho);

v_beam=sqrt(v_2_beam);

v_2_plate=E_plate./(Lx*Ly*h*rho);

v_plate=sqrt(v_2_plate);

p_2_room=E_room*rho_o*c_o*c_o/V;

SPL=10*log10(p_2_room/(20e-6*20e-6));

Lv_beam=10*log10(v_2_beam/(1e-9*1e-9));

Lv_plate=10*log10(v_2_plate/(1e-9*1e-9));

Advanced Vibrations 2015DSP & MCM Assignment

Due: Friday 4pm Week 13 (June 12)

1 The basics

q1.1. What is the frequency of sin(2pimt)? What is its period? [marks: 1]

q1.2. Consider a signal f (t) = sin(2pimt) + sin(2pint) with prime m and n(m 6= n). What is the period of f (t)? [marks: 1]

q1.3. Now consider f (t) for m = 0.4 and n = 0.5. What are the periodsof the two sinusoids that compose f (t)? What is the period of f (t)?

[marks: 2]

q1.4. Considering the above, what can we say about what happens to thelength of time needed to sample a signal the closer together the spac-ing of its sinusoids are? [marks: 1]

[Total marks: 5]

2 Sampling

Many signal analysers have adjustable settings for the number of linesdisplayed in their spectra. Typical values of lines are L {400, 800, 1600},corresponding to N = 2.56L points in the FFT. For this question assume theanalyser is set to display a spectrum up to 80 Hz.

q2.1. At what frequency is the analyser sampling the input signal? Why isthis larger than it strictly has to be? [marks: 2]

q2.2. With L = 1600 lines, what is the frequency resolution of the spectrum?[marks: 1]

q2.3. How long will it take to record a measurement with 16 averages andno overlap between averages? And with 50% overlap? [marks: 2]

[Total marks: 5]

1

3 FFT

Mathematically, the Fourier Transform can be compactly represented as

F =W f , (1)

where F and f are column vectors of length N andW is a square matrix ofsize N N with elements defined by

Wmn = exp(i2pi(m 1)(n 1)/N). (2)In Matlab:

q3.1. Define a column vector f of length N = 8 with random entries, con-struct matrixW , and calculate F using Eq. (1). [marks: 4]

q3.2. Calculate F using Matlabs fft function and ensure that the answersare equal (within numerical tolerance). [marks: 2]

q3.3. Now time the operations for N {8, 10, 12} (hint: use Matlabs tic andtoc functions), recording the time to constructW , the time to evaluateEq. 1, and the time to run fft. Tabulate the results. [marks: 3]

q3.4. Comment on the speed of the Fast Fourier Transform versus the matrixformulation, particularly as N increases. [marks: 1]

[Total marks: 10]

2

4 Power Spectral Density

As engineers, we rarely use fft on its own. Commands such as pwelchallow us to calculate the power spectrum and power spectral density in-cluding window functions and overlapping averages.

For this question, define the signal to be sampled to be

f (t) = sin(2pi15t) + 0.3 sin(2pi40t) + 0.01r, (3)

where r is normally distributed noise of unity amplitude (e.g., from Matlabsrandn function). Define sample frequency Fs = 100 Hz and an FFT lengthof N = 210.

q4.1. Create a time vector with sample frequency Fs containing M = Npoints and evaluate f (t) at these points. [marks: 2]

q4.2. Use pwelch to calculate and plot the power spectral density. Use arectangular window of size N. [marks: 1]

Hint: [Pxx,freq] = pwelch(x,ones(N,1),[],[],Fs);

q4.3. Repeat the steps above with a time vector containing M = 10N points.[marks: 1]

q4.4. Repeat again with M = 10N points and a Hanning window. [marks: 1]Hint: [Pxx,freq] = pwelch(x,hanning(N),[],[],Fs);

q4.5. Draw all three plots on one graph using a log-y-axis. [marks: 2]

q4.6. Repeat all three plots on another graph with N = 28. Ensure the y-axes are equal. Comment on the effects of changing N, M, and thetype of window. [marks: 3]

Finally, a demonstration of the meaning behind power spectral density. Ateach frequency bin, the PSD is the amount of energy at that frequency.

q4.7. Calculate the energy in the time signal using numerical integrationof f 2(t) over the range of one FFT period. [marks: 2]

q4.8. Then calculate the energy from the PSD by summing the response ateach frequency. [marks: 1]

q4.9. Are they equal? How does this affect the peaks of your graphs whenchanging N? [marks: 2]

[Total marks: 15]

3

5 Fourier interpolation and aliasing

We say that a signal can be reconstructed after sampling from it a finitenumber of points in time. Fourier theory allows us to write an equationof the reconstructed signal in terms of these points. Assume that a singleperiod of f (t) (constructed of a finite sum of sinusoids) is sampled at Mevenly-spaced points in time, then g(t) is the reconstructed signal given by

g(t) =2N

k=0

f (tk)sinc(p(t tk))sinc( 1q (t tk))

, (4)

where q is the period of f (t), M = 2N + 1 is the number of samples, andp = M/q is the sampling rate.

Consider sampling f (t) = sin(2pi0.4t) + sin(2pi0.5t) for three cases withM {5, 7, 11} points.q5.1. What is p and N in each case? [marks: 2]

q5.2. Use Matlab to implement Eq. (4) for each case, drawing three graphs,each showing: the original signal (solid line), the sampled points(points or circles), and the reconstructed signal (dashed). (Hint: onlyM = 11 works properly.) [marks: 4]

q5.3. What happens if too few points are used to sample a signal? [marks: 1]

Now investigate this phenomena when plotting the power spectral density.Consider:

f (t) = sin(2pi15t) + 0.3 sin(2pi40t) + 0.5 sin(2pi80t). (5)

q5.4. What are the frequencies of the three sinusoids in f ? [marks: 1]

q5.5. Calculate the PSD (using pwelch) of f (t) at a sample rate of Fs =200 Hz. (Use N = 210 FFT points and a rectangular window. Noaveraging is necessary.) Plot the spectrum with a log-y-axis. [marks: 2]

q5.6. Now reduce the sample frequency to 100 Hz and note the effect on thespectrum. Explain the shift in terms of the Nyquist frequency and thefrequency of the original sinusoid. [marks: 2]

q5.7. If we added a fourth sinusoid at 145 Hz, at what frequency would itappear when sampled at 200 Hz? And at 100 Hz? (You should now beable to calculate these without needing to plot the spectra in Matlab.)

[marks: 2]

[Total marks: 14]

4

6 MCM

The spectrum and time trace below were taken from the output shaft of adouble reduction gearbox of an extruder. The input speed is 992 RPM andthe output speed is about 40 RPM.

q6.1. Estimate the number of teeth on the output gear. [marks: 1]

q6.2. Can we conclude the fault type based on the current spectrum range?[marks: 1]

5. 5. Give the most likely machine fault and your brief reasoning for the following

spectrum:

a) Runspeed is 215 RPM. Machine is a ball mill.

b) Runspeed is 217.2 RPM. Machine is a cone crusher.

PREP - TL Extruder - 8"8x6-8" -08G GBOX OP DE HOR-ACC

Analyze Spectrum 01-Apr-03 15:05:45 (SST-Corrected) PK = .0406 LOAD = 100.0 RPM = 40. (.67 Hz)

2700 3000 3300 3600 3900 4200

0

0.01

0.02

0.03

0.04

0.05

Frequency in CPM

PK A

ccel

erat

ion

in G

-s

3504

.635

43.9

3465

.2

Analyze Waveform 01-Apr-03 15:05:45 PK = .0412 PK(+/-) = .0971/.0872 CRESTF= 3.27

5 6 7 8 9 10 11

-0.12

-0.09

-0.06

-0.03

0.00

0.03

0.06

0.090.12

Time in Seconds

Acc

eler

atio

n in

G-s

5

Give the most likely machine fault and your brief reasoning for the follow-ing spectra:

q6.3. Machine is a ball mill with runspeed 215 RPM: [marks: 2]


spectrum:





2700 3000 3300 3600 3900 4200

0

0.01

0.02

0.03

0.04

0.05

Frequency in CPM

PK A

ccel

erat

ion

in G

-s

3504

.635

43.9

3465

.2


5 6 7 8 9 10 11

-0.12

-0.09

-0.06

-0.03

0.00

0.03

0.06

0.090.12

Time in Seconds

Acc

eler

atio

n in

G-s

q6.4. Machine is a cone crusher with runspeed 217.2 RPM: [marks: 2]


spectrum:





2700 3000 3300 3600 3900 4200

0

0.01

0.02

0.03

0.04

0.05

Frequency in CPM

PK A

ccel

erat

ion

in G

-s

3504

.635

43.9

3465

.2


5 6 7 8 9 10 11

-0.12

-0.09

-0.06

-0.03

0.00

0.03

0.06

0.090.12

Time in SecondsA

ccel

erat

ion

in G

-s

[Total marks: 6]

6