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BOUNDARY ELEMENT METHODS FOR VIBRATION PROBLEMS
by
Ashok D. Belegundu
Professor of Mechanical Engineering Penn State University
ASEE Fellow, Summer 2003
Colleague at NASA Goddard: Daniel S. Kaufman Code 542
MY RESEARCH IN LAST FEW YEARS
Noise reduction – passive approaches
V m
Helmholtz Resonator
Stiffeners
TVA/BBVA
Noisy Structure
Dynamic analysis
Acoustic analysis
2
MOST RECENT RESEARCH
Parallel optimization algorithms
05
101520253035
0 16 32 48 64
number of processors
Spee
dup
bracketing
intervalreductiontotal linesearchlinear speedup(reference)
3
OBJECTIVES AT NASA: SUMMER 2003 Tutorial ON Boundary Element Analysis (“BEM”) Computer codes (for tutorial and problem solving) Study the potential for BEM in Vibration at mid-frequency levels i.e. fill the ‘gap’ between FEA and SEA in view of:
Low freq – FEA is very good High freq -- SEA seems adequate (although
detailed response is not obtainable) Assess the state of the art in BEM for vibration analysis
4
BEM -- SIMPLE 1-D EXAMPLE
0)1(,0)0(,02
2
===+ uuxdx
ud
01
02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛+∫ dxx
dxudw
Integrate by parts twice:
0''1
0
1
02
21
0
1
0=++− ∫∫ dxxwdx
dxwduwuuw
Choose w such that )(2
2
ixxdx
wd−−= δ
‘Fundamental Solution’ :
)()()( ii xxxxstepxw −−−=
5
SIMPLE EXAMPLE – cont’d Then
∫+−=1
0
1
0
1
0'')( dxxwwuuwxu i (*)
boundary terms domain term
xi = ‘source ’, and w(x) depends on source point xi Choose source xi = 0+ = 0 + ε . Thus w(0) = 0 and remains zero as ε → 0. w(1) = -1, 1)1(' −=w , body force term = -1/3, and we get 3/1)1(' −=uWe can also recover the solution for all interior points from Eq. (*) : u(xi) = (xi – xi
3)/6
Note: the differential (equilibrium) is exactly satisfied within the domain in BEM – while boundary conditions are approximately satisfied, here, the boundary is only wo points, so BEM gives exact solution
6
AXIAL VIBRATION
),(2 txpucu =′′−&&
Harmonic loading: 22 /),( ctxpuku −=+′′
ρω /, Ecc
k ==
( ) 0/0
22 =++′′∫L
dxwcpuku
Integrate by parts twice,
∫∫ ++′′+′−′LL
LL dxcpwdxwkwuwuwu0
2
0
200
/)(
Fundamental solution w :
),(2ixxwkw δ−=+′′
),()(sin1ii xxstepxxk
kw −
−= ⇒
(*)/)(0
200 ∫+′−′=
LLL
i dxcpwwuwuxu
7
F Example
ρ A = 1 du/dx (L) = F/c2 u(0) =0
From Eq. (*), tip displacement, )(tan2 LkckFuL =
This is exact, since boundary conditions are exactly satisfied. Resonances occur when cos (kL) = 0, or kL = π/2, 3π/2, ... Eq. (*) then gives
{ })(sin)(cos)tan()( 2 iii xLkxLkLkckFxu −−−=
Timoshenko’s formula:
∑=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=,...5,3,1 2
2
222
4
12i
L
kL
iLcFu
π
Consider 3 modes in the expansion (FEA needs about 3 elements for each half sine wave) - red color in fig:
8
0 500 1000 15000
0.02
0.04
0.06
0.08
0.1
0.12
0 5 10 15 20 25-1.5
-1
-0.5
0
0.5
1
Mode 4 = blue color
9
BEAM VIBRATION
22
4
4
Acpvk
xv
ρ=−
∂∂
),(24
4
ixxwkxw
δ−=−∂∂
⇒
[ ] [ ]{ })(sin)(sinh2
1),( 2/3 iii xxkxxkk
xxstepw −−−−=
...)( =ixv
But now, also need to differentiate v wrt xi to get additional equation(s) for solution
10
STATIC PLATES BY DIRECT BEM
Dpw =∇ 4
Kirchoff theory for thin plates ( ) ( )
terms corner +
+−−=∇−∇ ∫∫∫
****4**4
SA
dSMMwVVwdAwwww θθ
Choose the ‘concentrated force’ fundamental solution w* such that
rrD
w
xxxw
*
ii
ln4
1 giveswhich
),()(
2
*4
π
δ
=
=∇
Thus
( )
( )∑
∫∫∫
=
−+
+−−+=
*K
1k wFFw
dSMMwVVwdAwpD
xwcSA
ii
**
*****
1)( θθ
where ci equals 0.5 if xi is on smooth S, and equals 1 if interior.
11
A second equation is obtainable as:
( )( )
( )( )∑
∫∫∫
=
−−+
+−−−+=∂∂
+∂∂
*K
1k i
Si
Aii
wwFFw
dSMMwwVVwdAwpD
wcwc
**
*****
''
'''''1 θθηξ ηξ
where the ‘concentrated moment’ fundamental solution is given by *'w
)cos(ln2
1' απ
rrD
w * =
Expressions for V*, V’* etc are given in the literature. Note: Corners (eg. rectangular plates) require
special attention
Distributed load requires domain integration Here, a function s is defined such that , *2 ws =∇
rrr ∂∂
+∂∂
=∇1
2
22 and then converting to a
contour integral as ∫∫∫ ∂∂
=SA
dSrspdAwp )cos(* β
where β = angle between r and n. Similarly for . *'w
12
BEM WITH CONSTANT ELEMENTS : CONTOUR INTEGRATION
element kn
j
W(S) = Wj = constant for e
Gauss point r
source i
Thus, which is evaluated using
gauss quadrature
∫∫ =e
je
dSVwdSwV **
Linear elements will provide more accuracy
Finally: A x = b A is square, unsymmetric, dense, dim 2N+3K where N = no. of regular nodes, K = no of corner nodes Clamped Plate: x = [shear, moment] per unit length Simply-Supported Plate: x = [slope, shear]
13
DYNAMIC ANALYSIS OF PLATES
Dpw
Dhw =+∇ &&
ρ4 For general transient loading, apply Laplace transform and then use BEM. Here, we assume harmonic loading. Thus,
amplitudewewtw ti == )(,)(),( xxx ω
ck
Dpwkw ω
==−∇ ,24
Choose the ‘concentrated force’ fundamental solution w* such that
[ ])()(8
),()()1(
0)2(
0
2*4
rkiHrkHkiw
xxwkxw
*
ii
+=
−=−∇ δ
The ‘concentrated moment’ fundamental solution is given by
*'w
⎥⎥⎦
⎤
⎢⎢⎣
⎡ +)r+=
)(
()()cos('
12
1111
rkKc
kYcrkJciw * α
14
CORNER EFFECTS When source point is a corner, two concentrated moment fundamental solutions, and corresponding equations, need to be written. There are a total of three unknowns at each corner When a plate is loaded, corners tend to curl up – corner reactions keep these clamped
Details are omitted here for brevity
15
HANDLING DOMAIN (PRESSURE) LOADS IN DYNAMICS
∫∫∫∫AA
dAwpD
dAwpD
** '1,1
Approach 1: Create a domain mesh, and integrate (careful mesh design, and integration are necessary for accuracy) Approach 2: Express , where ai are determined through regression, and each of the known functions ψi can be written as . Then, the Divergence theorem allows us to convert the domain integral to a contour integral.
∑=m
iiawp1
*' ψ
is2∇
16
DIFFICULTIES IN BEM
1. Fundamental solutions are hard to derive for complicated differential equations
2. Domain integration requires special care
3. Numerical integration involving bessel functions need special care
17
BEM WITH MULTIPLE RECIPROCITY (BEM-MRM)
Dpwkw =−∇ 24
Choose the ‘concentrated force’ fundamental solution w* such that
),()( of Instead 2*4
ii xxwkxw δ−=−∇
Choose w* as the static fundamental solution :
rrD
w
xxxw
*
ii
ln4
1 giveswhich
),()(
2
*4
π
δ
=
=∇
We then have (ignoring corner terms):
( ) 1
)(
***** ∫∫∫
∫∫
+−−+
+=
Sssss
As
ii
dSMMwVVwdAwpD
xwc
θθ
A
*s
2 dAwwk
Let etcssssws s*
2*
34*
1*
24**
14 ,, =∇=∇=∇
and repeatedly use, with s = si , ( ) ( )∫∫∫ +−−=∇−∇
Ssss
A
dSMMwVVsdAwssw θθ ****4**4
18
BEM-MRM (Cont’d) to obtain an expression for the displacement
( )
( )∑
∫∫∫
=
−+
+−−+=
*K
1k wFFw
dSMMwVVwdAwpD
xwcSA
ii
**
*****
ˆˆ
ˆˆˆˆˆ1)( θθ
where
∑∑==
+=+=p
is
is
p
ii
is i
VkVVskww1
*2**
1
*2** ˆ,ˆ etc, along with a similar equation for for the slope.
19
Advantages of BEM_MRM Simple fundamental solutions regardless of complexity of differential equation All domain integrals can be easily converted to contour integrals Claimed that number of terms, p, are small for convergence – is this true for all frequencies ? Integrals independent of k can be done once and stored within the frequency loop
20
COMPUTER CODE
Circular plate (done), Rectangular plate (in progress) Concentrated load, Constant boundary element Damping: E = E0 (1 + i η) Input: pressure as SPL (converted to point load) Outputs: Center displacement Vs frequency
[ ]
( ) πωω 2/,/w-
,sg'in onacceleratiinc. 1Hzat band,in freqlower & upperand
),(where
,)(1
2
2
nnnn
u
ub
nbb
fgacc(n)
accff
ffN
naccN
PSD
==
==
−=
= ∑
l
l
[ ]∑∞
=
=1
22 )(21
n
naccGRMS
Validation: with Timoshenko’s formulas and with Ansys
21
VALIDATION EXAMPLE
Clamped Circular Steel Plate, R=1m, η=.03 Concentrated load = 1 N
0 20 40 60 80 100 120 1400
0.5
1
1.5
2
2.5x 10
-5
frequency, Hz
Cen
ter D
ispl
acem
ent
Circular Plate
22
ANSYS RESULT
Circular plate concentrated
23
SAMPLE PROBLEM Clamped Circular Steel Plate 1m radius, 1 cm thick, η = .01 Pressure corresponds to about 50 Pa 22 1/3-octave bands, ranging from 16Hz to 2000 Hz
0 5 10 15 20 250
50
100
150
SP
L, d
B
SPL in 1/3 Octave Bands
24
CENTER DISPLACEMENT Vs HZ
0 500 1000 1500 2000 25000
1
2
3
4
5
6x 10
-3
frequency, Hz
Cen
ter D
ispl
acem
ent
Circular Plate
25
MEAN ACCELERATION IN G’S IN EACH
1/3RD OCTAVE BAND
0 5 10 15 20 250
10
20
30
40
50
60
70
80
Cen
ter A
ccel
erat
ion
in g
"s
Circular Plate
26
PSD IN EACH 1/3RD OCTAVE BAND
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
4
g2 /Hz
= in
t(g(f)
2 )/del
taf/d
elta
f
Circular Plate
27
DRAWBACK IN MASS MATRIX CALCULATIONS IN FINITE ELEMENTS
u = N q u = displacement field in element e q = nodal displacements of e N = shape functions – polynomials Mass matrix:
∫=e
dVNNm T
21
M is assembled from each m Choice of N critical in dynamics, since inertial force equals ω2 M , and solution is obtained from [K - ω2 M]-1 F Comment: for large ω , polynomial shape functions are inadequate Also, shear effects must be properly captured
28
A FEW KEY REFERENCES Book: “Boundary elements – an introductory course”, 2nd ed, C.A. Brebbia and J. Dominguez Paper: “Boundary integral equations for bending of thin plates”, Morris Stern Paper: “free and forced vibrations of plates by boundary elements”, C.P. providakis and D.E. Beskos, Computer methods in applied mechanics and engineering, 74, (1989), 23-250 Papers on BEM-MRM: see J. Sladek. Example: “Multiple reciprocity method for harmonic vibration of thin elastic plates”, V. Sladek, J. Sladek, M. Tanaka, Appl. Math. Modelling, 17, (1993), 468-475.
29
SUMMARY OF WORK COMPLETED (MAY 19TH – JULY 25TH, 2003)
Tutorial has been developed for BEM in vibration -- rods, beams, Laplace eq., plate bending Computer code for circular plate vibrations
clamped or simply-supported validated with Closed-form solutions and Ansys SPL input, PSD and and other metrics output
State of the art for vibration analysis has been assessed Rectangular (and other geometries) on-going Research issues identified (see next slide)
30
31
FUTURE WORK IN APPLYING BEM FOR PLATE VIBRATIONS
at MID-FREQUENCIES Accurate and efficient numerical integration of body force terms Develop BEM-MRM and compare with ‘standard’ BEM; more validation Include shear effects (BEM-MRM is attractive) Generalize geometries (any planar shape, thick plates, shells) Compare BEM, FEM, Experiment on a specific problem; Spectral element methods Other research: (shock loading, optimization )