THE STEADY-STATE VIBRATION RESPONSE OF A BAFFLED PLATE
SIMPLY-SUPPORTED ON ALL SIDES SUBJECTED TO HARMONIC PRESSURE WAVE
EXCITATION AT OBLIQUE INCIDENCE
Revision E
By Tom Irvine
December 22, 2014
Email: tom@vibrationdata.com
_____________________________________________________________________________________________
The baffled, simply-supported plate in Figure 1 is subjected to an oblique pressure wave on one
side. Only a side view along the length is shown because the pressure is assumed to be uniform
with width. This diagram and the corresponding pressure field equation are taken from
Reference 1.
Wavefront
Direction of
Propagation
Wavefront

w(x,y,t)

t
x
L
Figure 1.
 is the acoustic wavelength. t is the trace wavelength.
 t   / sin 
(1)
1
The governing differential equation from Reference 2 is
 4w
4w
4w 
 2z
D  4  2 2 2  4   h 2  P(x, t)
 x
x y
y 
t

(2)
The plate stiffness factor D is given by
D
Eh 3

12 1   2

(3)
where
E
is the modulus of elasticity

Poisson’s ratio
h
is the thickness

is the mass density (mass/volume)
P
is the applied pressure
The acoustic pressure is


2 x
P(x, t)  po cos  t 
 
t


(4)
The trace wave number k t is
kt 
2
t
P(x, t)  po cos  t  k t x  
(5)
(6)
The differential equation becomes
 4w
4w
4w 
2w
D  4  2 2 2  4   h 2  P(x, t)
 x
x y
y 
t

(7)
2
Let
w(x, y, t)  (x, y) u(t)
(8)
By substitution,
  4
 4
 4 
 2u
D  4  2 2 2  4  u  h  2  P(x, t)
 x
x y
y 
t

(9)
The homogeneous equation is
  4
 4
 4 
 2u
D  4  2 2 2  4  u  h  2  0
 x
x y
y 
t

(10)
  4
 4
 4 
h  2   D  4  2 2 2  4  u
 x
t
x y
y 

(11)
  4
 4
 4 

2

 c 2
 4
2 2
4 

x y
y 
 x
(12)
 2u
u
D

u
h
u  c2 u  0
(13)
  4
 4
 4 
D  4  2 2 2  4  u  c 2  0
 x
x y
y 

(14)
  4
 4
 4 
h

2

u c2
0
 4
2 2
4 

D

x

x

y

y


(15)
3
The mass-normalized mode shapes are
 mn 
2
 m x   n y 
sin 
 sin 

a b h
 a   b 

2
 m
 mn 

x
a b h  a

 m x   n y 
 cos 
 sin 


 a   b 
 
2 mn
2  m 
 m x   n y 

 sin 
 sin 

a b h  a 
 a   b 

2
 n   m x 
 n y 
 mn 

 sin 
 cos 

y
a b h  b   a 
 b 
(18)
(19)
2
2
y
(17)
2
2
x
(16)
2  n 
 m x   n y 

 sin 
 sin 

a b h  b 
 a   b 
 
2 mn
(20)
The natural frequencies are
c2  mn 2
 mn 
2
2
D   m   n  


 
 
 h   a   b  
2
D   m 
 n

 
 h  a 
 b




2
(21)
2



(22)
Let
2
mn
 m   n 
 
 

 a   b 
2
(23)
Thus
mn  mn 2
D
h
(24)
4
c 2  mn 2  mn 4
D
h
(25)
  4
 4
 4 
4
 4  2 2 2  4  u  mn   0
x y
y 
 x
(26)
Recall
 4w
4w
4w 
2w
D  4  2 2 2  4   h 2  P(x, t)
 x
x y
y 
t

(27)
Let


w(x, y, t)     mn (x, y)u mn (t)
(28)
n 1m1
By substitution,
 4   

 4w   

4w   
D  4     mn u mn   2 2 2     mn u mn   4     mn u mn  

 y 

 x 
x y  n 1 m1
 n 1 m1


 n 1 m1


 h

2   

u

  P(x, t)


mn
mn
t 2  n 1 m1

(29)
    4
   
    4

4
D     4  mn u mn   2    2 2  mn u mn      4  mn u mn  
 
 


  n 1 m1 x y
  n 1 m1 y

  n 1 m1 x
  

2
 h     mn 2 u mn   P(x, t)


t
 n 1 m1

(30)
5
  
   
   

4
4
4
D     u mn 4  mn   2    u mn 2 2  mn      u mn 4  mn  
 
 


x
x y
y
  n 1 m1
  n 1 m1

  n 1 m1
  

d2
 h     mn 2 u mn   P(x, t)


dt
 n 1 m1

(31)
  
 4

  

4
4
d2
D    u mn  4  mn  2 2 2  mn  4  mn    h     mn 2 u mn   P(x, t)
 x




x y
y
dt


 n 1 m1

 n 1 m1
(32)
Note that
  4 mn
 4 mn  4 mn

2


4
x 2y 2
y 4
 x

4
 u  mn  mn

(33)
By substitution,
  

  

d2
D    mn 4 u mn  mn   h     mn 2 u mn   P(x, t)




dt
 n 1 m 1

 n 1 m 1

(34)
Multiply each term by  pq .
  

  

d2
D    mn 4 u mn  pq  mn   h     pq  mn 2 u mn    pq P(x, t)




dt
 n 1 m1

 n 1 m1

(35)
Integrate with respect to surface area. Note that the eigenvectors pq , mn are functions of
area, but the argument is omitted for brevity.
6
a b

D

0 0

  


a b
d2
4
   mn u mn  pq  mn  dxdy  0 0 h     pq  mn 2 u mn  dxdy
dt
n 1 m1

 n 1 m 1


a b

0 0
 pq P(x, t)dxdy
(36)
D

a b

0 0



a b  
d2
4

u


dxdy


h
u
 pq  mn  dxdy
   mn mn pq mn 


2 mn 0 0 


dt
n 1 m1

 n 1 m1


a b

0 0
 pq P(x, t)dxdy
(37)




   2
 d u mn


dxdy


h



pq
mn

2
0 0
 dt
n 1 m 1 
D   mn 4 u mn 
n 1 m 1
a b

a b

0 0



dxdy
0 0 pq mn 

a b
 pq P(x, t)dxdy
(38)
The eigenvectors are orthogonal such that
h 
a b

 pq  mn dxdy  0
if any of the indices (m, n, p, q) is unique with respect to the others
h 
a b
 pq  mn dxdy  1
if m=n=p=q
0 0

0 0
Thus
d 2 u mn
dt 2

a b
D mn 4
u mn     pq P(x, t)dxdy
0 0
h
(39)
7
mn 2  mn 4
d 2u mn
dt 2
D
h
(40)
 mn 2 u mn  
a b

0 0
mn P(x, t)dxdy
(41)
Add a modal damping term.
d 2u mn
dt 2
 2mn mn
a b
du mn
 mn 2 u mn    mn P(x, t)dxdy
0 0
dt
P(x, t)  po cos  t  k t x  
 mn 
d 2 u mn
dt
2
(43)
2
 m x   n y 
sin 
 sin 

a b h
 a   b 
 2mn mn
2po
du mn
 mn 2 u mn 
dt
a b h
(42)
(44)
a b
 m x   n y 
 sin 
 cos  t  k t x   dxdy
a   b 
0 0 sin 
(45)
d 2u mn
dt
2
 2mn mn
du mn
 mn 2 u mn  Lmn (t)
dt
(46)
The generalized force is
L mn (t) 
2po
a b
 m x   n y 
 sin 
 cos  t  k t x   dxdy
a   b 
  sin 
a b h 0 0
  n y  b  a  m x 
cos 
Lmn (t) 
  sin 
 cos  t  k t x    dx
n a b h   b  0  0
 a 
2bpo
(47)
(48)
8
L mn (t) 
2bpo
a
 m x 
cos  n   1  sin 
 cos  t  k t x    dx
0
n a b h
 a 
(49)
The phase angle is arbitrary for the steady-state analysis. Set  =0.
L mn (t) 
2bpo
a
 m x 
cos  n   1  sin 
 cos  t  k t x  dx
0
n a b h
 a 
(50)
Equation (50) is evaluated in Appendices A and B. There are three possible results as follows.
For k t 
m
,
a
Lmn (t) 
4po 1  cos  n  
(51)
4po 1  cos  n  
(52)
ab
cos  k t a / 2  sin  t    , for m  1,3,5,...
 ak 2  h
2 mn  t   1
  m 

and
Lmn (t) 
For k t 
ab
sin  k t a / 2  sin  t    , for m  2, 4, 6,...
 ak 2  h
2 mn  t   1
  m 

m
,
a
L mn (t) 
po ab
cos  n   1 sin  t 
n h 
(53)
Only the steady-state solution is needed. So define a participation factor mn and represent the
time varying term as a harmonic excitation function.
9
1
 mn p(t)
h
(54)
p(t)  po exp( jt)
(55)
Lmn (t) 
where
There are three participation factor cases.
For k t 
m
,
a
 mn 
4 1  cos  n  
 ak 2 
2
 mn  t   1
  m 

abh cos  k t a / 2  , for m  1,3,5,...
(56)
abh sin  k t a / 2  , for m  2, 4, 6,...
(57)
and
 mn 
For k t 
4 1  cos  n  
 ak 2 
2 mn  t   1
  m 

m
,
a
 mn 
1
abh cos  n   1
n
(58)
Define a joint acceptance function Jmn.
J mn 
1
mn
abh
mn  abh J mn
(59)
(60)
10
There are three joint acceptance cases.
For k t 
m
,
a
J mn 
4 1  cos  n  
 ak 2 
2
 mn  t   1
  m 

cos  k t a / 2  , for m  1,3,5,...
(61)
sin  k t a / 2  , for m  2, 4, 6,...
(62)
and
J mn 
For k t 
4 1  cos  n  
 ak 2 
2 mn  t   1
  m 

m
,
a
1
cos  n   1
n 
(63)
1
abh J mn p(t)
h
(64)
J mn 
Lmn (t) 
The equation of motion for the modal coordinates is
d 2 u mn
dt 2
 2mn mn
d 2 u mn
dt
2
du mn
1
 mn 2 u mn 
mn p(t)
dt
h
 2mn mn
du mn
1
 mn 2 u mn 
abh J mn p(t)
dt
h
(65)
(66)
11
d 2 u mn
dt
2
 2mn mn
du mn
ab
 mn 2 u mn 
J mn p(t)
dt
h
(67)
The temporal variable response to the applied force transposed to the frequency domain is


 mn
1 
 P  
U mn   
2
2
h  mn
   j2mn mn 


(68)


J mn
ab 
 P  
U mn   
2
2
h  mn
   j 2mn mn 


(69)




Recall


w(x, y, t)     mn (x, y)u mn  t 
(70)
n 1m1
Transpose to the frequency domain.
W(x, y, ) 
W(x, y, ) 
 
mn  mn  x, y 
1
P    
2
2
h
n 1m1 mn    j2mn mn
(71)
 
J mn  mn  x, y 
ab
P    
2
2
h
n 1m1 mn    j2mn mn
(72)




12
The frequency response function relating the displacement to the oblique pressure field is
mn  mn  x, y 
W(x, y, ) 1  
 
2
P  
h n 1m1 mn
 2  j2mn mn

W(x, y, )
ab

P  
h



n 1m1


J mn  mn  x, y 

2
mn
 2  j2mn mn
(73)
(74)
The bending moments are
 2
2 
M xx  x, y,   D  2   2  W  x, y, 
 x
y 

 2
2 
M yy  x, y,   D  2   2  W  x, y, 
 y
x 

(75)
(76)
The bending stresses from Reference 3 are
E zˆ   2
2 



 W  x, y, 
1   2  x 2
y 2 
(77)
 yy  x, y,   
E zˆ   2
2 



 W  x, y, 
1   2  y 2
x 2 
(78)
xy  x, y,   

E zˆ   2
W  x, y,  


1    xy

(79)
 xx  x, y,   
ẑ is the distance from the centerline in the vertical axis
13
References
1. E. Richards & D. Mead, Noise and Acoustic Fatigue in Aeronautics, Wiley, New York,
1968.
2. T. Irvine, Natural Frequencies of Rectangular Plate Bending Modes, Revision B, Vibrationdata,
2011.
3. J.S. Rao, Dynamics of Plates, Narosa, New Delhi, 1999.
4. http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/plane_stress_principal.cfm
5. D. Segalman, C. Flucher, G. Reese, R Field; An Efficient Method for Calculating RMS von
Mises Stress in a Random Vibration Environment, Sandia Report: SAND98-0260, UC-705, 1998.
6. T. Irvine, Steady-State Vibration Response of a Plate Simply-Supported on All Sides Subjected to
a Uniform Pressure, Revision C, Vibrationdata, 2014.
APPENDIX A
Case 1:
L mn (t) 
kt 
m
a
2bpo
a

 m x 
cos  n   1 sin  t   sin 
 sin  k t x  dx
0
n a b h
 a 

a

 m x 
 cos  t   sin 
 cos  k t x  dx 
0
 a 

(A-1)
The definite integrals are evaluated using the wxMaxima software program, with the following
command:
14
ratsimp(trigexpand(integrate(A*sin(m*%pi*x/a)*sin(k*x)+B*sin(m*%pi*x/a)*cos(k*x),x,0,a)));
positive;
By substitution of the results,
L mn (t) 
2bpo 1  cos  n  
n a b h  2 m 2  a 2 k t 2 


 sin  t  a k cos  k a  sin  m   ma sin  k a  cos  m
2
t
t
t

 cos  t  a 2 k t sin  k t a  sin  m    ma cos  k t a  cos  m    ma 


(A-2)
Note that for all m values,
sin  m  0
Lmn (t) 
(A-3)
2bpo 1  cos  n  
n a b h  2 m 2  a 2 k t 2 


 sin  t   ma sin  k t a  cos  m

 cos  t    ma cos  k t a  cos  m    ma 
(A-4)
L mn (t) 
2po 1  cos  n   m ab
sin  t   sin  k t a  cos  m  
 2 m 2  a 2 k t 2  n h




 cos  t  cos  k t a  cos  m   1 
(A-5)
15
L mn (t) 
2po 1  cos  n  

ab
sin  t   sin  k t a  cos  m  
  ak 2  h
2 mn 1   t  
   m  

 cos  t  cos  k t a  cos  m   1 
(A-6)
Convert the trigonometric term to magnitude and phase format.
sin  t   sin  k t a  cos  m   cos  t  cos  k t a  cos  m   1 

cos  m  2 cos  k a  cos  m  1 sin t   
2
t
(A-7)
Note that for all m values,
cos  m    1
2
(A-8)
sin  t   sin  k t a  cos  m   cos  t  cos  k t a  cos  m   1

2  2 cos  k t a  cos  m  sin  t   
(A-9)
For odd m,
sin  t   sin  k t a  cos  m   cos  t  cos  k t a  cos  m   1
 2 1  cos  k t a  sin  t    , for m  1,3,5,...
(A-10)
16
sin  t   sin  k t a  cos  m  cos  t  cos  k t a  cos  m  1 
 4 cos 2  k t a / 2  sin  t    , for m  1,3,5,...
(A-11)
sin  t   sin  k t a  cos  m  cos  t  cos  k ta  cos  m 1
 2cos  k t a / 2  sin  t    , for m  1,3,5,...
(A-12)
Lmn (t) 
4po 1  cos  n  
ab
cos  k t a / 2  sin  t    , for m  1,3,5,...
 ak 2  h
2 mn  t   1
  m 

(A-13)
For even m,
sin  t   sin  k t a  cos  m   cos  t  cos  k t a  cos  m   1
 2 1  cos  k t a  sin  t    , for m  2, 4, 6,...
(A-14)
sin  t   sin  k t a  cos  m  cos  t  cos  k ta  cos  m  1 
 4sin 2  k t a / 2  sin  t    , for m  2, 4, 6,...
(A-15)
17
sin  t   sin  k t a  cos  m  cos  t  cos  k ta  cos  m 1
 2sin  k t a / 2  sin  t    , for m  2, 4, 6,...
(A-16)
Lmn (t) 
4po 1  cos  n  
ab
sin  k t a / 2  sin  t    , for m  2, 4, 6,...
 ak 2  h
2 mn  t   1
  m 

(A-17)
18
APPENDIX B
kt 
Case 2:
L mn (t) 
m
a
2bpo
a

 mx
cos  n   1 sin  t   sin 2 
0
n a b h
 a


 dx

a
 mx 
 mx  
 cos  t   sin 
 cos 
 dx 
0
 a 
 a  
(B-1)
The definite integrals were evaluated using wxMaxima with the following commands.
integrate((sin(m*%pi*x/a))^2,x,0,a);positive;
integrate((sin(m*%pi*x/a))*(cos(m*%pi*x/a)),x,0,a);positive;

 a 1  cos 2 (m)
 a 
L mn (t) 
cos  n   1    sin  t   

2m
n a b h
 2 

2bpo
L mn (t) 
 po

  cos  t 






(B-2)

ab
cos  n   1  m  sin  t   1  cos 2 (m) cos  t 
 mn h
L mn (t) 
2
po ab
cos  n   1 sin  t 
n h 
(B-3)
(B-4)
19
APPENDIX C
Example
Consider a rectangular plate with the following properties:
Boundary Conditions
Simply Supported on All Sides
Material
Aluminum
Thickness
h
=
0.125 inch
Length
a
=
10 inch
Width
b
=
8 inch
Elastic Modulus
E
=
10E+06 lbf/in^2
Mass per Volume
v
=
0.1 lbm / in^3 ( 0.000259 lbf sec^2/in^4 )
Mass per Area

=
0.0125 lbm / in^2 (3.24E-05 lbf sec^2/in^3 )
Viscous Damping Ratio

=
0.05
The normal modes and frequency response function analysis are performed via a Matlab script.
20
The normal modes results are:
Table C-1. Natural Frequency Results, Plate
Simply-Supported on all Sides
fn (Hz)
m
n
302
1
1
656
2
1
855
1
2
1209
2
2
1246
3
1
1777
1
3
1799
3
2
2072
4
1
2131
2
3
2625
4
2
2721
3
3
3067
1
4
3134
5
1
3421
2
4
Now apply a sound pressure wave with variable frequency at an angle of 90 degrees. This is
actually a special case called grazing incidence where the propagation vector is parallel to the
plate surface.
The fundamental mode shape is shown in Figure C-1. The resulting displacement and
transfer functions magnitudes are shown in Figures C-2 and C-3, respectively.
21
Figure C-1.
22
Figure C-2.
The maximum displacement response for oblique incidence at 90 degrees is:
max = 0.027 in/psi at 300.1 Hz
The same plate was subjected to uniform, normal pressure in Reference 6. The maximum
displacement for the uniform case was:
max = 0.138 in/psi at 300.1 Hz
The difference at this frequency between the two pressure fields is 14.2 dB.
Note that the acoustic wavelength at this frequency is about 45 inches. The bending mode
wavelength for the fundamental mode is 20 inches, twice the plate length.
23
Figure C-3.
The maximum von Mises stress response for the oblique case is:
max = 3209 (psi/psi) at
300.1 Hz
The maximum displacement for the uniform case was:
max = 16,580 in/psi at 300.1 Hz
The difference at this frequency is 14.3 dB.
24
APPENDIX D
Principal Stress
Principal Stresses
Stresses in Given
Coordinate System
The diagrams are taken from Reference 3.
The principle stresses are
2
x  y
 x  y 
   xy 2
1,2 
 

2
2


(D-1)
The angle at which the shear stress becomes zero is
tan 2 p 
2  xy
x  y
(D-2)
25
The von Mises stress  e is
2
2
e  1  1 2   2
(D-3)
The von Mises stress is used to predict yielding of materials under any loading condition from
results of simple uniaxial tensile tests. The von Mises stress satisfies the property that two stress
states with equal distortion energy have equal von Mises stress.
An alternate formula from Reference 4 is
2
2
 e   xx   yy   xx  yy  3 xy
2
(D-4)
26