BOUNDARY ELEMENT METHODS FOR VIBRATION PROBLEMS€¦ · VIBRATION PROBLEMS by Ashok D. Belegundu...
Transcript of BOUNDARY ELEMENT METHODS FOR VIBRATION PROBLEMS€¦ · VIBRATION PROBLEMS by Ashok D. Belegundu...
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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
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MY RESEARCH IN LAST FEW YEARS
Noise reduction – passive approaches
V m
Helmholtz Resonator
Stiffeners
TVA/BBVA
Noisy Structure
Dynamic analysis
Acoustic analysis
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MOST RECENT RESEARCH
Parallel optimization algorithms
05
101520253035
0 16 32 48 64
number of processors
Spee
dup
bracketing
intervalreductiontotal linesearchlinear speedup(reference)
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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
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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 −−−=
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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
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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
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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:
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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
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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
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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.
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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
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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]
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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
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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
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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∇
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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
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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
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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.
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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
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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
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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
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ANSYS RESULT
Circular plate concentrated
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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
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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
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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
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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
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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
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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.
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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)
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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 )