THE STEADY-STATE VIBRATION RESPONSE OF A BAFFLED PLATE
SIMPLY-SUPPORTED ON ALL SIDES SUBJECTED TO RANDOM PRESSURE PLANE
WAVE EXCITATION AT OBLIQUE INCIDENCE
Revision A
By Tom Irvine
December 27, 2014
Email: tom@vibrationdata.com
_____________________________________________________________________________________________
The random excitation paper is very similar to Reference 1 which covered harmonic excitation.
The baffled, simply-supported plate in Figure 1 is subjected to an oblique plane 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.
Direction of
Propagation x

 t
Wavefront
L

Figure 1.

is the acoustic wavelength.
 t
is the trace wavelength.
Wavefront w(x,y,t)
/ sin

(1)
1

The governing differential equation from Reference 2 is
D


 4 w
 x
4

2
 4 w
  y
2

 4 w
 y
4

   h
 2 z
 t
2

P(x, t) (2)
The plate stiffness factor D is given by
D

12
Eh
1

3
 
(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 mass-normalized mode shapes are
 mn

2
 a b h sin
 a
 sin
 b 
(4)

 x
 mn

2
 a b h
 m
 a
 cos
 a
 sin
 b 
(5)
 2
 x
2
 mn
 
2
 a b h
 m
  2 
 a
 sin  a
 sin
 b

(6)
2


 y
 mn

2
 a b h
 n b

 sin 
 mn
  mn
2
D
 h a
 cos
 b 
(7)
 2
 y
2
 mn
 
 mn

2
 a b h
 n
  2
 b
 sin
 a
 sin
 b

(8)
The natural frequencies are c
2   mn
2 

D h



 m a
  2

 n
 b
 2


2
(9)
 mn

D
 h

 m a
2
 n b
2


(10)
Let
 m
  2  n
 
 a


 b

2
(11)
Thus
(12) c
2   mn
2   mn
4
D
 h
(13)
The generalized force is
 mn

2
 a b h sin
 a
 sin
 b

(14)
Let

2
 abh
(15)
3

 mn
 sin m x a sin n y b
(16)
L mn
( t )
  
0 a sin m x a sin n y b p ( x , t ) dxdy (17)
L mn
( t )
 p o
 
0 a sin m a x sin n y b cos

 t
 k t x
 dxdy (18)
L mn
( t )
 p o
 
0 a sin m a x sin n b y cos

 t
 k t x
 dxdy (19)
L mn
( t )
 b n
 p o 
0 a sin m a x cos

 t
 k t x
 dx



 cos n y b b
0


(20)
L mn
( t )
 b

1
 cos n

L mn
( t )
 b

1
 cos n

L mn
( t )
 b

1
 cos n

 p o
 p o
 p o sin
 cos

0 a sin m x a cos

 t
 k t x
 dx (21)

0 a sin m x a cos

 t
 k t x
 dx (22)

0 a sin m a x sin k x dx t

0 a sin m x a cos
  dx
(23)
4

For k t
 m
 a
L mn
( t )

, n
 b 


1
2
 m cos
2  a
2 k t
 p
2 o 
 sin

 a
2 k t cos
   
 m
 a cos
   

 cos
  a

2 k t sin t m
  m
 a cos
   
 m
 a
 
(24)
Note that for all m values, sin
 

0 (25)
L mn
( t )
 n
 b 


1
2
 m cos
2  a
2 k t
 p
2 o   sin


 m
 a cos t m


 cos
  
 m
 a cos
   
 m
 a
 
(26)
L mn
( t )
 m
 ab

1 n
 

2 m

2 cos
 a
2 k t
2
 p  o
 sin

 t m


 cos


1
 
(27)
L mn
( t )
 p 
 o ab

1
2 m
2

 cos a
2 k t
2

 m  sin n
 t
 t m


 cos


1
 
(28)
5

L mn
( t )
 p o a b

1
 cos
 2 m
2

1


 ak
 m t


2


 m  sin n
 t
 t m


 cos


1
 
(29)
L mn
( t )
 p o a b

1
 cos
 2 mn

1


 ak
 m t


2



 sin
      
 cos
 t cos k t a cos m
 
1

(30)
V

2 sin

 k t a
2

 m
2


(31)
L mn
( t )
 p o a b

1

 2 mn


1


 cos ak
 m t


2



2 sin

 k t a
2

 m
2

 sin

 t
 

(32)
The phase angle ε will be regarded as arbitrary since only a steady-state solution is sought.
L mn
( t )

2 A p o a b

1
 cos
 2 mn 1



2

2
 m m

 m t


2
 sin

 k
2 t a

 m
2

 sin

 t
 

(33)
6

Note the following relationship for the modal wavelength
 m
.
 m
2 a

 m
 m
2 a m
(34)
(35)
L mn
( t )

2 A p o
 2 mn a b

1
 cos
1




 m t


2
 sin


2

 t m
 m
4

 m
2

 sin

 t
 

(36)
L mn
( t )

2 A p o a b

1
 cos
 2 mn 1




 m t


2
 sin
 m
2



 m t



 m
2 sin

 t
 

(37)
L mn
( t )

2 p o a b

1
 cos
 2 mn 1




 m t


2
 sin


 m
2



 m t

1



 sin

 t
 

(38)
For an arbitrary phase angle, the equation may be rewritten as for k t
 m
 a
,
L mn
( t )

2 A p o
 2 mn a b

1
 cos



 m t


2

1
 sin


 m
2



 t m 
1



 sin

 t
 

(39)
7

For k t
 m
 a
,
L mn
( t )
 p o a
2
 n b  cos

1
 sin
 
(40)
Again, only the steady-state solution is needed. So define a participation factor
 mn
and represent the time varying term as a harmonic excitation function.
L mn
( t )
  mn p ( t ) (41) where p(t)
 
(42)
For k t
 m
 a
,
 mn

2 A a b

1
 cos
 2 mn



 m t


2

1
 sin


 m
2



 m t

1




(43)
For k t
 m
 a
,
 mn
 a b
2
 n
 cos

1

(44)
8

Define a joint acceptance function J mn .
J mn

 mn
 a b
 mn
(45)
1
J mn
(46) a b
For k t
 m
 a
,
J mn

2

1
 cos
 2 mn




 m t


2

1 sin


 m
2



 m t

1




(47)
The equation of motion for the modal coordinates is d
2 u mn dt
2

2
 mn
 mn du mn dt
  mn
2 u mn
  mn p ( t ) (48) d
2 u mn dt
2

2
 mn
 mn du mn dt
  mn
2 u mn

J mn p ( t ) (49)
The temporal variable response to the applied force transposed to the frequency domain is
U mn





 mn
2   2

 mn
 j 2
 mn
  mn



P (50)
9

U mn





 mn
2   2 a 
 b j
J
2 mn
 mn
  mn



P
 
(51)
Recall w(x, y, t)

 
   mn
(x, y)u mn
 
(52)
Transpose to the frequency domain.
W
 x , y ,
    n
 
 

1 m

1

 mn
2
 mn
  2

 mn
 j 2
( x ,
 mn y )
  mn
(53)
W
 x , y ,

   n
 
 

1 m

1

 mn
2 a
 b J
 2 mn 

 j 2 mn

( x , mn
 y )
 mn
(54)
The frequency response function relating the displacement to the oblique pressure field is
W
 x
P
, y ,
 
 
 n
 
 

1 m

1

 mn
2
 mn
  2

 mn
 j 2
( x ,
 mn y )
  mn
(55)
W

P x , y ,
 


 n
 
 

1 m

1

 mn
2 
J mn
 2


 mn j 2
( x , y )
 mn
  mn
(56)
The bending moments are
M xx
 x, y,

D


 2
 x
2
 2
 
 y
2

  
(57)
10

M yy
 x, y,

D


 2
 y
2
 2
 
 x
2

  
(58)
The bending stresses from Reference 3 are
 xx
 x, y,
1
  2



 x
2
2
 2
 
 y
2

  
(59)
 yy
 x, y,
1
  2



 2 y
2
 2
 
 x
2


  
(60)
 xy
 x, y,

1


 2
 

 zˆ is the distance from the centerline in the vertical axis
An example is given in Appendix B.
References
(61)
1.
T. Irvine, 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, Vibrationdata, 2014.
2.
T. Irvine, The Mean Square Force Due to Random Forces on a Rigid Beam, Revision B,
Vibrationdata, 2014.
3.
J.S. Rao, Dynamics of Plates, Narosa, New Delhi, 1999.
4.
T. Irvine, Steady-State Vibration Response of a Plate Simply-Supported on All Sides Subjected to a Uniform Pressure, Revision C, Vibrationdata, 2014.
5.
E. Richards & D. Mead, Noise and Acoustic Fatigue in Aeronautics, Wiley, New York,
1968.
11

12

APPENDIX A
Magnitude of Complex Trigonometric Term
V

 cos k a cos
 m

1

2 
 sin k a cos
 m

2
(A-1)
V
  cos
    
2   sin t
 
2 
2 cos
    
1 (A-2)
V
 
2 cos
   

2 (A-3)
V
  cos
 k t a
  m

 cos
 k t a
  m


2 (A-4)
V
 
2 cos
 k t a
  m


2 (A-5)
V

4 sin
2

 k t
2 a

 m
2


(A-6)
V

2 sin

 k t
2 a

 m
2


(A-7)
13

APPENDIX B
Example
Consider a rectangular plate with the following properties:
Boundary Conditions Simply Supported on All Sides
Aluminum Material
Thickness h = 0.125 inch
Length
Width a b
= 10 inch
= 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

= 0.03 for all modes
The normal modes and frequency response function analysis are performed via a Matlab script.
14

The normal modes results are:
Table B-1. Natural Frequency Results, Plate
Simply-Supported on all Sides fn (Hz) m n
302
656
1
2
1
1
855
1209
1246
1
2
3
2
2
1
1777
1799
2072
2131
2625
2721
3067
1
3
4
2
4
3
1
3
2
1
3
2
3
4
3134 5 1
3421 2 4
The fundamental mode shape is shown in Figure B-1. The corresponding joint acceptance function is shown in Figure B-2.
Now apply the sound pressure level from Mil-Std-1540C in Figure B-3 at an angle of
 
45

.
The resulting displacement and stress at the center of the plate are shown in Figures B-4 and
B-5, respectively.
15

Figure B-1.
16

Figure B-2.
17

Figure B-3.
18

Figure B-4. Center of Plate, 45 degrees Incidence
19

Figure B-5. Center of Plate, 45 degrees Incidence
20

APPENDIX C
Limiting Case, Uniform Pressure
Recall
For k t
 m
 a
 mn
,

2 A a b

1
 cos
 2 mn



 m t


2

1
 sin


 m
2



 m t

1




(C-1)
/ sin

(C-2)
The pressure field becomes a uniform and normal as
 
0 ,
 t
 
.
In this case,
 mn

2 A a b
 cos
 2 mn

1
 sin

 m
2

 (C-3)
The following equation is equivalent to (C-3) for m as an integer and ignoring the polarity shift.
 mn
 a b
 cos

1

 cos
 

1

(C-4)
 2 mn
Equation (C-4) is essentially similar to the result for uniform, normal pressure in Reference 4, allowance for a difference in the way that the mass density was accounted for.
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