FEM/BEM NOTES
Professor Peter Hunter
[email protected] Associate Professor Andrew Pullan
[email protected] Department of Engineering Science The University of Auckland New Zealand June 17, 2003
c Copyright 1997-2003 Department of Engineering Science The University of Auckland
Contents 1 Finite Element Basis Functions 1.1 Representing a One-Dimensional Field . 1.2 Linear Basis Functions . . . . . . . . . 1.3 Basis Functions as Weighting Functions 1.4 Quadratic Basis Functions . . . . . . . 1.5 Two- and Three-Dimensional Elements 1.6 Higher Order Continuity . . . . . . . . 1.7 Triangular Elements . . . . . . . . . . . 1.8 Curvilinear Coordinate Systems . . . . 1.9 CMISS Examples . . . . . . . . . . . .
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1 1 2 4 7 7 10 14 16 20
2 Steady-State Heat Conduction 2.1 One-Dimensional Steady-State Heat Conduction . . . . . 2.1.1 Integral equation . . . . . . . . . . . . . . . . . . 2.1.2 Integration by parts . . . . . . . . . . . . . . . . . 2.1.3 Finite element approximation . . . . . . . . . . . 2.1.4 Element integrals . . . . . . . . . . . . . . . . . . 2.1.5 Assembly . . . . . . . . . . . . . . . . . . . . . . 2.1.6 Boundary conditions . . . . . . . . . . . . . . . . 2.1.7 Solution . . . . . . . . . . . . . . . . . . . . . . . 2.1.8 Fluxes . . . . . . . . . . . . . . . . . . . . . . . . 2.2 An -Dependent Source Term . . . . . . . . . . . . . . . 2.3 The Galerkin Weight Function Revisited . . . . . . . . . . 2.4 Two and Three-Dimensional Steady-State Heat Conduction 2.5 Basis Functions - Element Discretisation . . . . . . . . . . 2.6 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Assemble Global Equations . . . . . . . . . . . . . . . . . 2.8 Gaussian Quadrature . . . . . . . . . . . . . . . . . . . . 2.9 CMISS Examples . . . . . . . . . . . . . . . . . . . . . .
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23 23 24 24 25 26 27 29 29 29 30 31 32 34 36 37 39 42
3 The Boundary Element Method 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 3.2 The Dirac-Delta Function and Fundamental Solutions 3.2.1 Dirac-Delta function . . . . . . . . . . . . . 3.2.2 Fundamental solutions . . . . . . . . . . . .
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43 43 43 43 45
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ii
CONTENTS
3.3 3.4 3.5 3.6 3.7 3.8
3.9 3.10 3.11 3.12
3.13 3.14 3.15 3.16
3.17 4
5
The Two-Dimensional Boundary Element Method . . . . . . . . . Numerical Solution Procedures for the Boundary Integral Equation Numerical Evaluation of Coefficient Integrals . . . . . . . . . . . The Three-Dimensional Boundary Element Method . . . . . . . . A Comparison of the FE and BE Methods . . . . . . . . . . . . . More on Numerical Integration . . . . . . . . . . . . . . . . . . . 3.8.1 Logarithmic quadrature and other special schemes . . . . 3.8.2 Special solutions . . . . . . . . . . . . . . . . . . . . . . The Boundary Element Method Applied to other Elliptic PDEs . . Solution of Matrix Equations . . . . . . . . . . . . . . . . . . . . Coupling the FE and BE techniques . . . . . . . . . . . . . . . . Other BEM techniques . . . . . . . . . . . . . . . . . . . . . . . 3.12.1 Trefftz method . . . . . . . . . . . . . . . . . . . . . . . 3.12.2 Regular BEM . . . . . . . . . . . . . . . . . . . . . . . . Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Axisymmetric Problems . . . . . . . . . . . . . . . . . . . . . . Infinite Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix: Common Fundamental Solutions . . . . . . . . . . . . 3.16.1 Two-Dimensional equations . . . . . . . . . . . . . . . . 3.16.2 Three-Dimensional equations . . . . . . . . . . . . . . . 3.16.3 Axisymmetric problems . . . . . . . . . . . . . . . . . . CMISS Examples . . . . . . . . . . . . . . . . . . . . . . . . . .
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48 53 55 57 58 60 60 61 61 61 62 64 64 64 65 67 69 72 72 72 73 73
Linear Elasticity 4.1 Introduction . . . . . . . . . . . . . . . . . . . 4.2 Truss Elements . . . . . . . . . . . . . . . . . 4.3 Beam Elements . . . . . . . . . . . . . . . . . 4.4 Plane Stress Elements . . . . . . . . . . . . . . 4.4.1 Notes on calculating nodal loads . . . . 4.5 Three-Dimensional Elasticity . . . . . . . . . . 4.5.1 Weighted Residual Integral Equation . 4.5.2 The Principle of Virtual Work . . . . . 4.5.3 The Finite Element Approximation . . 4.6 Linear Elasticity with Boundary Elements . . . 4.7 Fundamental Solutions . . . . . . . . . . . . . 4.8 Boundary Integral Equation . . . . . . . . . . . 4.9 Body Forces (and Domain Integrals in General) 4.10 CMISS Examples . . . . . . . . . . . . . . . .
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75 75 76 79 81 83 84 85 86 87 89 91 93 96 98
Transient Heat Conduction 5.1 Introduction . . . . . . . . . . . . . . . . . 5.2 Finite Differences . . . . . . . . . . . . . . 5.2.1 Explicit Transient Finite Differences 5.2.2 Von Neumann Stability Analysis . . 5.2.3 Higher Order Approximations . . .
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99 99 99 99 101 102
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ONTENTS C 5.3 5.4 5.5
iii
The Transient Advection-Diffusion Equation . . . . . . . . . . . . . . . . . . . . 103 Mass lumping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 CMISS Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
6 Modal Analysis 6.1 Introduction . . . . . . 6.2 Free Vibration Modes . 6.3 An Analytic Example . 6.4 Proportional Damping 6.5 CMISS Examples . . .
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7 Domain Integrals in the BEM 7.1 Achieving a Boundary Integral Formulation . . . . . . . . . 7.2 Removing Domain Integrals due to Inhomogeneous Terms . 7.2.1 The Galerkin Vector technique . . . . . . . . . . . . 7.2.2 The Monte Carlo method . . . . . . . . . . . . . . . 7.2.3 Complementary Function-Particular Integral method 7.3 Domain Integrals Involving the Dependent Variable . . . . . 7.3.1 The Perturbation Boundary Element Method . . . . 7.3.2 The Multiple Reciprocity Method . . . . . . . . . . 7.3.3 The Dual Reciprocity Boundary Element Method . .
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111 111 111 113 114 115
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117 117 118 118 119 120 120 121 122 124
8 The BEM for Parabolic PDES 8.1 Time-Stepping Methods . . . . . . . . . . . . . . . . . . . . 8.1.1 Coupled Finite Difference - Boundary Element Method 8.1.2 Direct Time-Integration Method . . . . . . . . . . . . 8.2 Laplace Transform Method . . . . . . . . . . . . . . . . . . . 8.3 The DR-BEM For Transient Problems . . . . . . . . . . . . . 8.4 The MRM for Transient Problems . . . . . . . . . . . . . . .
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135 135 135 137 138 139 140
Bibliography
143
Index
147
Chapter 1 Finite Element Basis Functions 1.1 Representing a One-Dimensional Field Consider the problem of finding a mathematical expression to represent a one-dimensional field e.g., measurements of temperature against distance along a bar, as shown in Figure 1.1a.
+
+
+ + +
+
+
+
+ ++ +
+ +
+ +
+ +
+
+
+
+
+ ++ +
+
(a)
+ +
+
(b)
along a bar. The points are the measured F IGURE 1.1: (a) Temperature distribution temperatures. (b) A least-squares polynomial fit to the data, showing the unacceptable oscillation between data points.
One approach would be to use a polynomial expression ! " $#%#%# and to estimate the values of the parameters , , and from a least-squares fit to the data. As the degree of the polynomial is increased the data points are fitted with increasing accuracy and polynomials provide a very convenient form of expression because they can be differentiated and integrated readily. For low degree polynomials this is a satisfactory approach, but if the polynomial order is increased further to improve the accuracy of fit a problem arises: the polynomial can be made to fit the data accurately, but it oscillates unacceptably between the data points, as shown in Figure 1.1b. To circumvent this, while retaining the advantages of low degree polynomials, we divide the bar into three subregions and use low order polynomials over each subregion - called elements. For later generality we also introduce a parameter & which is a measure of distance along the bar. is plotted as a function of this arclength in Figure 1.2a. Figure 1.2b shows three linear polynomials in & fitted by least-squares separately to the data in each element.
2
F INITE E LEMENT BASIS F UNCTIONS
+ + +
+ +
+ + +
+ + + +
+ + +
+ +
+ +
+
+
+ + +
+ + + +
+ +
&
+
+
&
(b)
(a)
'
'
F IGURE 1.2: (a) Temperature measurements replotted against arclength parameter . (b) The domain is divided into three subdomains, elements, and linear polynomials are independently fitted to the data in each subdomain.
1.2
Linear Basis Functions
A new problem has now arisen in Figure 1.2b: the piecewise linear polynomials are not continuous in across the boundaries between elements. One solution would be to constrain the parameters , , etc. to ensure continuity of across the element boundaries, but a better solution is to replace the parameters and in the first element with parameters )( and , which are the values of at the two ends of that element. We then define a linear variation between these two values by
+*,-.0/213*,4)()5*6
where * 879:*9;/( *,-?/@13* = *,-A* such that
+*,- = ()B*,C>() = B *,C
and refer to these expressions as the basis functions associated with the nodal parameters )( and . The basis functions = (>*, and = B*D are straight lines varying between 7 and / as shown in Figure 1.3. It is convenient always to associate the nodal quantity FE with element node G and to map the temperature HJI defined at global node K onto local node G of element L by using a connectivity matrix KM8GONLP i.e.,
EQMH IORSEUT VXW where KMGJNYLP = global node number of local node G of element L . This has the advantage that the
1.2 L INEAR BASIS F UNCTIONS
3
= ()B*,
= B *, 1/
/
7
/
*
7 F IGURE 1.3: Linear basis functions Z ( [P]\:^-_`[
and
Z + [U>\a[ .
* /
interpolation
*D- = ()B*D4)() = B*D4 holds for any element provided that >( and are correctly identified with their global counterparts, as shown in Figure 1.4. Thus, in the first element node global nodes:
element nodes:
Hk(
/
node j
H
)( cb cb cb cb cb cb cb 0 element
/
node i
1
*
H"
f > ( fg gf gf gf gf gf gf 0
node h
element j
1
*
HOl
)( ed ed ed ed ed ed ed 0
element i
1
*
F IGURE 1.4: The relationship between global nodes and element nodes.
*D- = ()B*D4)() = B *D4
with )(m?Hk( and ;H . In the second element is interpolated by
(1.1)
*D- = ()B*D4)() = B*D4 (1.2) with )(kMH and ;H , since the parameter H is shared between the first and second elements "
4
F INITE E LEMENT BASIS F UNCTIONS
the temperature field is implicitly continuous. Similarly, in the third element is interpolated by
(1.3) +*,- = ()B*,C>() = B *,C with )(noH and pH]l , with the parameter H being shared between the second and third " " elements. Figure 1.6 shows the temperature field defined by the three interpolations (1.1)–(1.3).
node
/
+ +
+
node i
node j +
element
+
+
/
+
+
+
element j
+
+
+
+
+
node h + +
element i
&
F IGURE 1.5: Temperature measurements fitted with nodal parameters and linear basis functions. The fitted temperature field is now continuous across element boundaries.
1.3
Basis Functions as Weighting Functions
It is useful to think of the basis functions as weighting functions on the nodal parameters. Thus, in element 1 at *Q7
87!qrs/2157D4)()t7u v )(
which is the value of at the left hand end of the element and has no dependence on
w / w / / )() / i )(> at *Q 2 / 1 h h)x h>x h h which depends on >( and , but is weighted more towards )( than w w / / /21 / )() / / )(> at *Q j jx jyx j j which depends equally on )( and w w i i i i / )(> at *Q 2 / 1 ) ) ( h h)x h>x h h
/ h / j i h
1.3 BASIS F UNCTIONS
W EIGHTING F UNCTIONS
AS
which depends on )( and
5
but is weighted more towards
than )( z/Pk.0/@1A/{/% {
at *Q$/
which is the value of at the right hand end of the region and has no dependence on )( . Moreover, these weighting functions can be considered as global functions, as shown in Figure 1.6, where the weighting function | E associated with global node G is constructed from the basis functions in the elements adjacent to that node.
|}( (a)
& | (b)
& | " (c)
& | l (d)
& F IGURE 1.6: (a) ~ ~~ (d) The weighting functions E associated with the global nodes \:^)~~~ , respectively. Notice the linear fall off in the elements adjacent to a node. Outside the immediately adjacent elements, the weighting functions are defined to be zero.
For example, | weights the global parameter H and the influence of H falls off linearly in the elements on either side of node 2. We now have a continuous piecewise parametric description of the temperature field +*, but in order to define + we need to define the relationship between and * for each element. A convenient way to do this is to define as an interpolation of the nodal values of . For example, in element 1
B*D- = (yB*DC)(> = B *DC
and similarly for the other two elements. The dependence of temperature on ,
(1.4)
+ , is therefore
6
F INITE E LEMENT BASIS F UNCTIONS
defined by the parametric expressions
= E2B*D4FE
+*,k OB*,k E
E
= E2B*DCFE
where summation is taken over all element nodes (in this case only j ) and the parameter * (the “element coordinate”) links temperature to physical position . *, provides the mapping between the mathematical space 7`9M*9
/ and the physical space )(}9M39M , as illustrated in Figure 1.7.
)( 7
*
/
*Q7#j
+ at Q* 7#j
)( *
)( 7
/
*Q7#j
+
7
)(
* [
F IGURE 1.7: Illustrating how and are related through the normalized element coordinate . The values of and are obtained from a linear interpolation of the nodal variables and then plotted as . The points at are emphasized.
[U
} [P
[@\3!~
1.4 Q UADRATIC BASIS F UNCTIONS
7
1.4 Quadratic Basis Functions The essential property of the basis functions defined above is that the basis function associated with a particular node takes the value of / when evaluated at that node and is zero at every other node in the element (only one other in the case of linear basis functions). This ensures the linear independence of the basis functions. It is also the key to establishing the form of the basis functions for higher order interpolation. For example, a quadratic variation of over an element requires three nodal parameters )( , and
"
+*,- = (>*,4)() = *,4 = " *,4 "
(1.5)
The quadratic basis functions are shown, with their mathematical expressions, in Figure 1.8. Notice that since = ()B*D must be zero at *M7# (node j ), = ()B*, must have a factor *157#! and since it is also zero at *$/ (node i ), another factor is B* 1A/P . Finally, since = ()B*D is / at *Q7 (node / ) we have = ()B*D-Mj2*1A/() = B *u(N¤* C = " B*u(N¤* " = lm*u( N* Fl where
= )( B*U( N¤* qrs/213*u(s)s/213* = B*U( N¤* qA*u()s/21a* (1.6) = B*U( N¤* qrs/213*u(sC* "= lJB*U( N¤* qA*u(z* Note that = ()B*U( N¤* = = ()+*u( = ()B* where = ()B*U( and = ()B* are the one-dimensional linear basis functions. Similarly, = B*u(N¤* = = B*u(s = ()B* q#¥#%# etc. These four bilinear basis functions are illustrated in Figure 1.9. = (
= node i
node
=
/
node j
"
*
* node h
7
*u(
=l
/
*U(
* 7
* 7
/
*U(
/
*U(
F IGURE 1.9: Two-dimensional bilinear basis functions.
Notice that = E2*u( N* is / at node G and zero at the other three nodes. This ensures that the temperature *U( N¤* receives a contribution from each nodal parameter FE weighted by = E2B*u(N¤* and that when +*u( N* is evaluated at node G it takes on the value E . As before the geometry of the element is defined in terms of the node positions FEN¤¦uE6 , Ga
1.5 T WO -
/!N%#%#¥#)NYh
AND
T HREE -D IMENSIONAL E LEMENTS
9
by
= E2*u( N* CFE
`M ¦M
E
= E2*u( N* 4¦UE E
which provide the mapping between the mathematical space *u( N* (where 7§9¨*U( N¤* 9©/ ) and the physical space BªN¤¦ . Higher order 2D basis functions can be similarly constructed from products of the appropriate 1D basis functions. For example, a six-noded (see Figure 1.10) quadratic-linear element (quadratic in *U( and linear in * ) would have
« ¬ = E2*u( N* 4FE EB*U(J1574#S!ys/21a* = B*U( N¤* kjU*U(>*u(O157#!)0/@1¯* =O"± B*U( N¤* kvh6*U(>0/21a*u(4* /
7
*
(1.7) (1.8) (1.9)
² h
7
= *u( N¤* kvh6*U()s/21a*u(s>s/21a* = lJ *u( N¤* kj2B*U(O1:/P>B*U(O1°7#! 4* = *u( N¤* kjU*U()B*u(O1°74#S!4* ¬
/ ³
j 7 #
i /
*U(
F IGURE 1.10: A -node quadratic-linear element (node numbers circled).
Three-dimensional basis functions are formed similarly, e.g., a trilinear element basis has eight nodes (see Figure 1.11) with basis functions
= >( *u( N* N* k.0/213*u()0/213* >z/213* = *u( N* N* " k.0/213*u(C* s/213 * " =]"± *u( N* N* " k.0/213*u()0 /213* 4* " =O´ *u( N* N* " k.0/213*u(C* * " " "
= B *U( N¤* N¤* -{*u(ys/213* )s/213* = lJ B*U( N¤* N¤* " -{*u(z* s/213 * " = B*U( N¤* N¤* " -{*u(ys /213* C" * =Oµ¬ B*U( N¤* N¤* " -{*u(z* * " " "
(1.10) (1.11) (1.12) (1.13)
10
F INITE E LEMENT BASIS F UNCTIONS
*" ·
¶ ²
i / j ¸
* h *u(
F IGURE 1.11: An -node trilinear element.
1.6
Higher Order Continuity
All the basis functions mentioned so far are Lagrange1 basis functions and provide continuity of across element boundaries but not higher order continuity. Sometimes it is desirable to use basis functions which also preserve continuity of the derivative of with respect to * across element boundaries. A convenient way to achieve this is by defining two additional nodal parameters w
D . The basis functions are chosen to ensure that !* x E w D w D , D !*}¹¹ ¼ !*Ox ( F½( and !*¾¹¹ ( !*Ox F½ ¹¹»º ¹¹º and since FE is shared between adjacent elements derivative continuity is ensured. Since the number of element parameters is 4 the basis functions must be cubic in * . To derive these cubic
Hermite2 basis functions let
+*,- tY*¾¿ * ¿!* " N D -®j6*@¿i!!* N !* 1 2
Joseph-Louis Lagrange (1736-1813). Charles Hermite (1822-1901).
1.6 H IGHER O RDER C ONTINUITY
11
and impose the constraints
87!- {>( 0/(O15jU Substituting , , and back into the original cubic then gives +*,-{)(>5F½( *¾8i6 13i6)(J1¿jU ½( 13F½ 4* 8F½( ¿F½ ®jU>(O1¿ju 4 * " or, rearranging,
+*,-à ¼( +*,4)():à (( B*D4F½( ®Ã ¼ B *,C tà ( *,4F½
(1.14)
where the four cubic Hermite basis functions are drawn in Figure 1.12. One further w step is required to make cubic Hermite basis functions useful in practice. The derivative
D !*Ox E
defined at node
G
is dependent upon the element * -coordinate in the two ad-
jacent elements. It is much more useful to define a global node derivative arclength and then use
w , w D w C& !*Jx E C&)x OI RSEUT VXW Ä !*)x E
w D C& x E
where
&
is
(1.15)
w C& where !*)x E is an element scale factor which scales the arclength derivative of global node K to the * -coordinate derivative of element node G . Thus DC& is constrained to be continuous D . A two- dimensional bicubic Hermite basis requires four across element boundaries rather than D*
derivatives per node
ªNOÅ *u ( ON Å * Å Å
and
Å*u( * Å Å
12
F INITE E LEMENT BASIS F UNCTIONS
à ¼( B*D-r/21°iu*! OtjU*!" /
slope r/
à (( B*D-v* * 1A/P
7
* /
/
à ¼ B *D-v*! P8i15jU*,
/
à ( *,-A*! P*1A/P 7
7
*
7
/
*
/
*
slope $/
F IGURE 1.12: Cubic Hermite basis functions.
The need for the second-order cross-derivative term can be explained as follows; If is cubic in *U(
* , then Å *u ( is quadratic in *U( and cubic in * , and Å * is cubic in *u( and quadratic Å side 1–3 Å of with * is specified by in * . Now consider the cubic variation w in Figure 1.13.w The the four nodal parameters )( , Å *Å x ( , " and Å Å * x " . But since Å Å *u( (the normal derivative) is also cubic in * along that side and is entirely independent of these four parameters, w four additional w parameters are required to specify this cubic. Two of these are specified by Å and Å , U * u ( x u * u ( x ( w w Å Å " and the remaining two by *Å UÅ ( Å * x ( and Å *uÅ ( Å * x " . and cubic in
1.6 H IGHER O RDER C ONTINUITY
*
w Å *u(ux Å "
w Å *u(ux Å (
13
node i
node
/
node h
F IGURE 1.13: Interpolation of nodal derivative
Æ
node j
Æ [ (
*u(
along side 1–3.
The bicubic interpolation of these nodal parameters is given by
¼( B *U(Fà ¼( B*U(Fà (( B*U(Fà ®Ã (( B*U(Fà ®Ã ¼( B*U(FÃ
¼( + * >()tà ¼ *u(Fà ¼( B* ¼ B* tà ¼ *u(Fà ¼ * Fl w" w ¼( B* Å tà ( *u(Fà ¼( * Å *u(ux ( *U(ux w Å w Å ¼ B* Å ( ¼ w Å *u(ux " tà *u(Fà * w Å Å *U(ux l (( B* Å tà ¼ *u(Fà (( * Å * x ( * x w Å w Å ¼ ( ¼ ( ®Ã ( B*U(Fà B* Å * x tà *u(Fà * Å * x w Å " Å w l ®Ã (( B*U(Fà (( B* *uÅ ( * x ®Ã ( B*u(sà (( * *uÅ ( * x w Å Å ( w Å Å ( ( ( ( à ( B*U(Fà B* *uÅ ( * x ®Ã B*u(sà * *uÅ ( * x ® Å Å " Å Å l
+*u( N¤* kà ®Ã ®Ã
(1.16)
14
F INITE E LEMENT BASIS F UNCTIONS
where
à ¼( B *D à (( B*D à ¼ B*D à ( B*D
2/ 15iu* ®jU* " *@*1:/P * Bi1¿jU*, * +*1A/P
(1.17)
are the one-dimensional cubic Hermite basis functions (see Figure 1.12). As in the one-dimensional case above, to preserve derivative continuity in physical x-coordinate space as well as in * -coordinate space the global node derivatives need to be specified with respect to physical arclength. There are now two arclengths to consider: &u( , measuring arclength along the *u( -coordinate, and & , measuring arclength along the * -coordinate. Thus
w w w &u( Å *u(ux Å &u(ux Å *u(6x E O I Ç R u E T È V W Ä w Å w Å w Å& E Å* x Å& x Å* x (1.18) E O I Ç R u E T È V W E Ä Å w w Å w C& w Å C& u ( *Å uÅ ( Å * x E Å &uÅ ( Å & x IORÇEuT VXW Ä !*u(ux E Ä !* x E w C&u( w 4& where !*u(ux E and !* x E are element scale factors which scale the arclength derivatives of global node K to the * -coordinate derivatives of element node G .
The bicubic Hermite basis is a powerful shape descriptor for curvilinear surfaces. Figure 1.14 shows a four element bicubic Hermite surface in 3D space where each node has the following twelve parameters
ªN Å &6 ( N Å & N u& Å ( & YN ¦N Å &6¦ ( N Å & ¦ N u& Å ( ¦ & N ÉN Å &uÉ ( N Å & É Å Å Å Å Å Å Å Å Å Å 1.7
and
Å&6( É & Å Å
Triangular Elements
Triangular elements cannot use the *u( and * coordinates defined above for tensor product elements (i.e., two- and three- dimensional elements whose basis functions are formed as the product of onedimensional basis functions). The natural coordinates for triangles are based on area ratios and are called Area Coordinates . Consider the ratio of the area formed from the points j , i and Ê:ªNY¦ in Figure 1.15 to the total area of the triangle
Ë (-
Area ÌAÊjuiÍ Area ÌM/Pj6iÍ
/ ¦ Î / j ¹¹ / ¦ ¹¹ K.¨À(>t¥(z¿ a second degree of freedom (in the form of another coordinate axis in Figure 2.3b) is added, the ) { and the residual is now also made orthogonal to = approximating vector is )( ) and hence to . Finally, in Figure 2.3c, a third ) degree )=3 of freedom (a third axis in Figure 2.3c) is permitted in the approximation >( = (k{ = ) v = with the result that the residual (now . For " " a 3D vector space we only need three also orthogonal to = ) is reduced to zero and " axes or basis vectors to represent the true vector , but in the infinite dimensional vector space associated w with û ü a spatially continuous field + we need to impose the equivalent orthogonality
condition
= !`7 x
for every basis function
=
used in the approximate representation of
B . The key point is that in this analogy the residual is made orthogonal to the current set of basis vectors - or, equivalently, in finite element analysis, to the set of basis functions used to represent the dependent variable. This ensures that the error or residual is minimal (in a least-squares sense) for the current number of degrees of freedom and that as the number of degrees of freedom is increased (or the mesh refined) the error decreases monotonically. 2.4
Two and Three-Dimensional Steady-State Heat Conduction
Extending Equation (2.1) to two or three spatial dimensions introduces some additional complexity which we examine here. Consider the three-dimensional steady-state heat equation with no source terms:
1ÑÅ Å
w
w w î ø Å Ö x 1 Å ¦ Aî @4Å m¦ x 1 Å É î BDÅ ÉJx v7 Å Å Å Å Å
2 .4 T WO
AND
T HREE -D IMENSIONAL S TEADY-S TATE H EAT C ONDUCTION
33
where î N îA@ and îCB are the thermal diffusivities along the , ¦ and É axes respectively. If the ø material is assumed to be isotropic, î î2 @ î B î , and the above equation can be written as
ø
1ED Ä
ÀîFDn-7
(2.13)
and, if î is spatially constant (in the case of a homogeneous material), this reduces to Laplace’s equation îFD ¨7 . Here we consider the solution of Equation (2.13) over the region G , subject to boundary conditions on H (see Figure 2.4). Solution region boundary: H
Solution region: G
F IGURE 2.4: The region I and the boundary J .
The weighted integral equation, corresponding to Equation (2.13), is
ûK
1?D Ä
ÈîFD
ý
®7
LG
(2.14)
The multi-dimensional equivalent of integration by parts is the Green-Gauss theorem:
ûK
MND Ä
DO PDQ
Ä
G:
DO
ûR
@Å G Å
LH
(2.15)
ý (see p553 in Advanced Engineering Mathematics” by E. Kreysig, 7th edition, Wiley, 1993). This is used (with ¯ , 1}î4 and assuming that î is constant) to reduce the derivative order from two to one as follows: ý ý ý ûK ûK ûR 1?D ÈîF D LG® îFn D D G¿1 îkÅ LH (2.16) Ä
Ä
ÅG
w , ý ý ø û D ý D 1 D î Dqx ! î ! ! !ô1 î ! . cf. Integration by parts is ø ø ø Using Equation (2.16) in Equation (2.14) gives the two-dimensional equivalent of Equation (2.6) û
34
S TEADY-S TATE H EAT C ONDUCTION
(but with no source term):
ý (2.17) îSDn D G: î Å G H Ä Å being given on another part of the subject to being given on one part of the boundary and Å ÅG boundary. The integrand on the LHS of (2.17) is evaluated using ý ý ý Dn D Å U T Å UT ÔÅ * í Å U* Tí Å *WV Å U*WV T (2.18) Ä Ä Ä Å Å Å Å Å Å ý = , as before, and the geometric terms Å * í are found from the where A = E6FE and Å UT inverse matrix *Å í Å U T ù ( Å U TX Å*í ûK
or, for a two-dimensional element,
2.5 Let G©
Å *u ( Å Å * Å
!#" Å *u¦ ( Å $ Å *¦ Å
Å *u ( Å ¦ Å *u( Å
ý
ûR
Å !#" ù ( / Å *¦ $ ¦ Å* Å u* ( Å * 1ÒÅ * Å Å Å Å
Å *¦ 1 Å * Å ¦ Å Å *u¦ ( 1:Å *u( Å *u( Å Å Å
!#" $
Basis Functions - Element Discretisation Z
Y G í , i.e., the solution region is the union of the individual elements. In each G í let ( í = EuFEÕ = z( >( = ®#¥#%#¢ =\[ [ and map each G í to the *u(N¤* plane. Figure 2.5 shows an
example of this mapping.
2 .5 BASIS F UNCTIONS - E LEMENT D ISCRETISATION
[ ^ Ý _
¸ I
I
"
I
^
l
¸ "
I
]
[( ^
]
l I
³^ ^
[(
^
[(
^
³ I
(
[ ^ ¸
_
]
35
[ ^ Þ
^
F IGURE 2.5: Mapping each I to the
[ ( Ûz[
I
(
[ ^
]
[( ^
n
plane in a 2`
] I
Þ
³
element plane.
For each element, the basis functions and their derivatives are:
= (k¨0/Â13*u(¢s/21a* = v *U(¢s/Â13* = ¨ " 0/Â13*u(z* = lÂv*U(z*
= Å *u( ( $ 1s/213* Å= ( Å * $1s/213*u( Å = Å *u( $/213* Å= ( Å * $12*u( Å = Å *u( " $12* Å= Å * " $/213*u( Å = Å *u( l {* Å= l Å * {*u( Å
(2.19) (2.20) (2.21) (2.22) (2.23) (2.24) (2.25) (2.26) (2.27) (2.28) (2.29)
36
S TEADY-S TATE H EAT C ONDUCTION
2.6
Integration
The equation is
i.e.,
ûK
îSDn Ä
D
ý
w ý î Å Å Å ¦ Å Å Å u has already been approximated by = E6FE ûK
:
G
ûR
ý -î Å G Å
ý
H
(2.30)
ý î ÅG Å
ûR
Å ¦ x G: (2.31) H Å ý andý is a weight function but what should this be = i.e., weight function is one of the basis chosen to be? For a Galerkin formulation choose functions used to approximate the dependent variable. This gives
w = E = = E = FE î Å Å Å ¦ Å ¦ x í Å Å Å Å where the stiffness matrix is E where r/!N%#%¥# #¢NYh ûK
ûR
î Å G = H Å and G$/6N%#%#%#>NYh and a :
G
(2.32) is the (element)
load vector. The names originated from earlier finite element applications and extension of spring systems, i.e., M a îC where î is the stiffness of spring and a is the force/load. This yields the system of equations E!FE ba . e.g., heat flow in a unit square (see Figure 2.6).
¦ B * /
7
/
*u(
F IGURE 2.6: Considering heat flow in a unit square.
2 .7 A SSEMBLE G LOBAL E QUATIONS
37
The first component (z( is calculated as
û(û(
(z(qî
ij î
¼ ¼
0/213¦ s/21a !£D¦
and similarly for the other components of the matrix. Note that if the element was not the unit square we would need to transform from BªNY¦ to B*U( N¤* coordinates. In this case we would have to include the Jacobian of the transformation and
=í
also use the chain rule to calculate Å . e.g., LV Å The system of E!FEÕca becomes
= = = Å E Å u* (E Å *u ( Å * E Å Å Å Å
= Å * Å * íE Å Å
Å * í . Å
" ( 1 ( 1 (( 1 "(( ! "" )( ! "" üedgf î 11 ( 1 " ¬( 1 ¬" 11 ( $ $ (Right Hand Side) (2.33) ¬( "( " ( ¬ " ¬ Fl" 1 ¬" 1 1 ¬ ¬ Note that the Galerkin formulation generates a symmetric stiffness matrix (this is true for self
adjoint operators which are the most common). Given that boundary conditions can be applied and it is possible to solve for unknown nodal temperatures or fluxes. However, typically there is more than one element and so the next step is required.
2.7 Assemble Global Equations Each element stiffness matrix must be assembled into a global stiffness matrix. For example, consider h elements (each of unit size) and nine nodes. Each element has the same element stiffness matrix as that given above. This is because each element is the same size, shape and interpolation.
1
" ( ¬( (
1 1
¬"
!" ( ( " 1 ( 1 1 " ( ( ( ( ( " " " ¬ ( " 1 1 ¬" 1 1( 1 "( " " " ¬ 1 ( 1 ¬(1 " ¬( " ( ( " ( " ( 1( ¬" ( 1 1 1 1 ¬ ( ( ( " ( ( """ ( " " ¬( " ¬( " 1 1 1 ¬" 1 1 ( 1 "( " 1 1 ( 1 "( 1 1 " ¬ " ¬ 1 1( 1 "( ¬ 1 " ¬1 " ¬ 1 ¬ ( ¬ ¬ 1 ¬ 1 ( ¬1 ¬ ¬ 1 ¬" 1 ( 1 "( ( 1 "(( 1 " ( " ¬( " 1 ( $ " ¬ 1 ¬1 " ¬ ¬ 1 ¬ ¬
!
)( """ " " " " " Fl " üedgf ± " " " " ¬µ´ $
(2.34)
38
S TEADY-S TATE H EAT C ONDUCTION
¦ ·
¶ i
é global node numbering
h
h
²
element numbering
/
j
/
j i
F IGURE 2.7: Assembling unit sized elements into a global stiffness matrix.
This yields the system of equations
( " ( 1 l 1 1 ( ¬ ( " " ¬ ¬( 1 ( 1 1 ( ¬"( ( 1 ¬" 1 " ( 1 " ( 1 " 1 ¬
!" " 1 (( 1 "(( ( " " 1 ¬" 1 "( 1 "( " " 1 1 " l 1 µ "( ( ( " 1 1 "1 ( " 1 ¬( 1 ( 1 "( 1 ( "" " 1 " ( l " ¬" 1 "( 1 "( "" 1 (( 1 ""(( " ( " ( 1 l "( ¬( 1 ¬" 1 "( 1 "( 1 ¬( 1 $ " " ¬ 1 " 1 ¬ 1 ¬ ¬
)( !#""" " " " " " Fl " ± " üedhf " " " " ¬ " ¬µ´ $
Note that the matrix is symmetric. It should also be clear that the matrix will be sparse if there is a larger number of elements. From this system of equations, boundary conditions can be applied and the equations solved. To solve, firstly boundary conditions are applied to reduce the size of the system. If at global node i , í is known, we can remove the i th equation and replace it with the known value of í . This is because the RHS at node i is known but the RHS equation is uncoupled from other equations so the equation can be removed. Therefore the size of the system is reduced. The final system to solve is only as big as the number of unknown values of u. As an example to illustrate this consider fixing the temperature ( ) at the left and right sides of ¶ the plate in Figure 2.7 and insulating the top (node ) and the bottom (node j ). This means that
2 .8 G AUSSIAN Q UADRATURE
39
there are only i unknown values of u at nodes (2,5 and 8), therefore there is a i ji matrix to solve. The RHS is known at these three nodes (see below). We can then solve the iQj i matrix and then multiply out the original matrix to find the unknown RHS values. ¶ The RHS is 7 at nodes j and because ûR it isý insulated. To find out what the RHS is at node we need to examine the RHS expression
7
Å G Å
$7
H
at node . This is zero as flux is always
at internal nodes. This can be explained in two ways.
¾( G
G
n
n
F IGURE 2.8: “Cancelling” of flux in internal nodes.
Correct way: H does not pass through node and each basis function that is not zero at on H Other way:
Å G Å
is zero
is opposite in neighbouring elements so it cancels (see Figure 2.8).
2.8 Gaussian Quadrature The element integrals arising from two- or three-dimensional problems can seldom be evaluated analytically. Numerical integration or quadrature is therefore required and the most efficient scheme for integrating the expressions that arise in the finite element method is Gauss-Legendre quadrature. Consider first the problem of integrating B*D between the limits 7 and / by the sum of weighted samples of B*D taken at points *U( N¤* N%#%#%#yN¤* (see Figure 2.3): Y (
û
¼
n*, !*Q
Y
í (lk
í B* í >7
Here í are the weights associated with sample points * í - called Gauss points - and is the k the approximation of the integral. We now choose the Gauss points and weights to exactly error in integrate a polynomial of degree jn m 1¨/ (since a general polynomial of degree jn m 1./ has jnm arbitrary coefficients and there are jnm unknown Gauss points and weights). m Mj we can exactly integrate a polynomial of degree 3: For example, with
40
S TEADY-S TATE H EAT C ONDUCTION
B*D
.... o
*U(
.... * Y F IGURE 2.9: Gaussian quadrature. q
[U is sampled at r
p
*
û(
B*D2!*Q
Let
(s*u(>
*
Gauss points
[ ( Ûz[ ~~~0[ ~ Y
*
¼ and choose *,-}tY*¾t*! O¿D*6" . Then û( û( û( û( û( *,Â!*Q !*¾t *k!*@t * !*@¿ * " !* (2.35) ¼ ¼ ¼ ¼ ¼ Since , , and are arbitrary coefficients, each integral on the RHS of 2.35 must be integrated exactly. Thus,
û( ¼
û(
D*Q?/
*-D*Q j/ ¼ û( * D*Q i/ ¼ û( * " D*Q h / ¼
k
(# /Ó k
(#»*u() k
k
#/ k
#»*
(2.37)
k
#»*
(2.38)
k
#»* "
(2.39)
k
(#»* ( k
k
(#»* ("
(2.36)
These four equations yield the solution for the two Gauss points and weights as follows:
2 .8 G AUSSIAN Q UADRATURE
41
From symmetry and Equation (2.36),
(k k
k
/ j#
Then, from (2.37),
* ?/@13*U( and, substituting in (2.38),
* ( s/Â13*u(s ij jU* ( 1¿jU*U() i/ 7N giving
*u(- ju/ t j v / i # Equation (2.39) is satisfied identically. Thus, the two Gauss points are given by
A similar calculation for a
k th
*U(k j/ 1 j v / * j/ v / j (q j/ i
iN N
(2.40)
k
degree polynomial using three Gauss points gives
*U(k j/ 1 xj/ w * j/ N * " j/ jl/ w
i N iN
k
k
k
(k / ¶ hé
(2.41)
" /¶
2 For two- or three-dimensional Gaussian quadrature the Gauss point positions are simply the values given above along each * í -coordinate with the weights scaled to sum to / e.g., for j x j Gauss
/
quadrature the h weights are all . The number of Gauss points chosen for each * í -direction is h governed by the complexity of the integrand in the element integral (2.8). In general two- and threedimensional problems the integral is not polynomial (owing to the
Å
Å L* Ví
terms which come from the
42
S TEADY-S TATE H EAT C ONDUCTION
Å * VLí ) and no attempt is made to achieve exact integration. The quadrature error must be balanced Å against the discretization error. For example, if the two-dimensional basis is cubic in the *u( -direction and linear in the * -direction, three Gauss points would be used in the *u( -direction and two in the * -direction. inverse of the matrix
2.9
CMISS Examples
1. To solve for the steady state temperature distribution inside a plate run CMISS example i/6/ 2. To solve for the steady state temperature distribution inside an annulus run CMISS example
i/Pj
3. To investigate the convergence of the steady state temperature distribution with mesh refinement run CMISS examples i/%h/ , i/%h,j , i/%hDi and i/%h6h .
Chapter 3 The Boundary Element Method 3.1 Introduction Having developed the basic ideas behind the finite element method, we now develop the basic ideas of the boundary element method. There are several key differences between these two methods, one of which involves the choice of weighting function (recall the Galerkin finite element method used as a weighting function one of the basis functions used to approximate the solution variable). Before launching into the boundary element method we must briefly develop some ideas that are central to the weighting function used in the boundary element method.
3.2 The Dirac-Delta Function and Fundamental Solutions Before one applies the boundary element method to a particular problem one must obtain a fundamental solution (which is similar to the idea of a particular solution in ordinary differential equations and is the weighting function). Fundamental solutions are tied to the Dirac1 Delta function and we deal with both here.
3.2.1 Dirac-Delta function What we do here is very non-rigorous. To gain an intuitive feel for this unusual function, consider the following sequence of force distributions applied to a large plate as shown in Figure 3.1
E
1
( 4 Ì E | E2k 7 4Í E ( þ % ;% % ;%
Paul A.M. Dirac (1902-1994) was awarded the Nobel Prize (with Erwin Schrodinger) in 1933 for his work in quantum mechanics. Dirac introduced the idea of the “Dirac Delta” intuitively, as we will do here, around 1926-27. It was rigorously defined as a so-called generalised function by Schwartz in 1950-51, and strictly speaking we should talk about the “Dirac Delta Distribution”.
44
T HE B OUNDARY E LEMENT M ETHOD
Each has the property that y
û
ù
y
| E2@!«r/
(i.e., the total force applied is unity)
but as G increases the area of force application decreases and the force/unit area increases.
{ z
{
(
{ {
z
(
F IGURE 3.1: Illustrations of unit force distributions
E.
As G gets larger we can easily see that the area of application of the force becomes smaller and smaller, the magnitude of the force increases but the total force applied remains unity. If we imagine letting G}| ~ we obtain an idealised “point” force of unit strength, given the symbol B , acting at = 0. Thus, in a nonrigorous sense we have Bk E â ó y | E@B the Dirac Delta“function”.
This is not a function that we are used to dealing with because we have 7 if 7 and “ 87D
~ ” i.e., the “function” is zero everywhere except at the origin, where it is infinite. ûy ûy BÂ!$/ since each | E2@!?/ . However, we have y y ù ù The Dirac delta “function” is not a function in the usual sense, and it is more correctly referred B to as the Dirac delta distribution. It also has the property that for any continuous function ûy (3.1) B +2D` Q À7D y
ù
ù
3 .2 T HE D IRAC -D ELTA F UNCTION
AND
F UNDAMENTAL S OLUTIONS
A rough proof of this is as follows ûy ûy ó +@!` E& â y | E2+@! y y
by definition of
ù
û ó G E& â y j @! ù E& â ó y G j B*D G j 87D
45
B
by definition of | EÂB
w / / by the Mean Value Theorem, where *O 1 G N ÖG x w / / 1 G N G x and as G| ~N)*| 7 since *O
The above result (Equation (3.1)) is often used as the defining property of the Dirac delta in more rigorous derivations. One does not usually talk about the values of the Dirac delta at a particular point, but rather its integral behaviour. Some properties of the Dirac delta are listed below ûy B*13Q BÂ!Q *D (3.2) y (Note:
+*13 d
where
ù
is the Dirac delta distribution centred at {* instead of `7 ) d B*1¯q ½ B*1
B*1
=
7 /
(3.3)
if *ÌP (i.e., the Dirac Delta function is the slope of the Heaviside2 if *ÍP
step function.)
*1aªNW1°¦-
B*13 1°¦
(3.4)
(i.e., the two dimensional Dirac delta is just a product of two one-dimensional Dirac deltas.)
3.2.2 Fundamental solutions We develop here the fundamental solution (also called the freespace Green’s3 function) for Laplace’s Equation in two variables. The fundamental solution of a particular equation is the weighting function that is used in the boundary element formulation of that equation. It is therefore important to be able to find the fundamental solution for a particular equation. Most of the common equations 2
Oliver Heaviside (1850-1925) was a British physicist, who pioneered the mathematical study of electrical circuits and helped develop vector analysis. 3 George Green (1793-1841) was a self-educated miller’s son. Most widely known for his integral theorem (the Green-Gauss theorem).
46
T HE B OUNDARY E LEMENT M ETHOD
have well-known fundamental solutions (see Appendix 3.16). We briefly illustrate here how to find a simple fundamental solution.
Å Å ¦ 7 in some domain G}@ . Å (analogous Å The fundamental solution for this equation to a particular solution in ODE work) is a solution of ý ý Å Å ¦ B*1aªWN 1°¦-v7 (3.5) Å Å in 2 (i.e., we solve the above without reference to the ý original domain G or original boundary conditions). The method is to try and find solution to Dn M7 in 2 which contains a singularity at the point *WN . This is not as difficult as it sounds. We expect the solution to be symmetric about the point *WN since *1aªWN 1°¦ is symmetric about this point. So we adopt a local polar coordinate system about the singular point B*WN . Consider solving the Laplace Equation
Let
×Õ *13 13¦ Then, from Section 1.8 we have
ý
w
ý
ý
×/ Å × × Å Ó× x × / Å Ø (3.6) Å Å ý Å is zero. Thus Equation (3.6) becomes For ×ÍA74N *13ªNW1°¦m7 and owing to symmetry, Å ý ÅØ w / Å × Å 7 × Å× Å× x D
This can be solved by straight (one-dimensional) integration. The solution is
ý
cgäà-×27
(3.7)
Note that this function is singular at ×Õ7 as required. To find and we make use of the integral property of the Delta function. From Equation (3.5) we must have
û
D
ý
r1
û
r1Õ/
where is any domain containing ×Õ7 . We choose a simple domain to allow us to evaluate the above integrals. If
(3.8)
is a small disk of
3 .2 T HE D IRAC -D ELTA F UNCTION
AND
F UNDAMENTAL S OLUTIONS
47
¦
B*Ns G
F IGURE 3.2: Domain used to evaluate fundamental solution coefficients.
radius
û
D
ý
ÍA7 L
centred at ×Õv7 (Figure 3.2) then from the Green-Gauss theorem
û
ý
ÅG Å û ý Å Å × jUÙ jU Ù
f
Å
is the surface of the disk
f
since
is a disk centred at ×7 so G and × are in the same direction
from Equation (3.7), and the fact that
is a disc of radius
Therefore, from Equation (3.8)
$1 jU/ Ù #
So we have
ý
r1 jU/ Ù àk×27
remains arbitrary but usually put equal to zero, so that the fundamental solution for the twodimensional Laplace Equation is
ý
r1 Uj / Ù à-×
w
jU/ Ù äà )×/ x
(3.9)
48
T HE B OUNDARY E LEMENT M ETHOD
where × *1a 1°¦ (singular at the point B*NW ). The fundamental solution for the three-dimensional Laplace Equation can be found by a similar technique. The result is
ý
h6Ù>/ ×
where × is now a distance measured in three-dimensions.
3.3
The Two-Dimensional Boundary Element Method
We are now at a point where we can develop the boundary element method for the solution of Dn `7 in a two-dimensional domain G . The basic steps are in fact quite similar to those used for the finite element method (refer Section 2.1). We firstly must form an integral equation from the Laplace Equation by using a weighted integral equation and then use the Green-Gauss theorem. From Section 2.4 we have seen that ý ý ûK ûK ý ûK Å G H1 n 7Q D ª# G: D ª#¡D G (3.10) Å This was the starting point for the finite element method. To derive the starting equation for the boundary element method we use the Green-Gauss theorem again on the second integral. This gives
ý ý Kû Å G H1 Dnª#¡D LG 7 Å (3.11) ý ûK ý ûK ý Kû Å G H1 ÓÅ G H« ¢ D G Å ýÅ = , one ý of the basis functions For the Galerkin FEM we chose , the weighting function, to be used to approximate . For the boundary element method we choose to be the fundamental ûK
solution of Laplace’s Equation derived in the previous section i.e.,
ý
?1 jP/ Ù äàk×
1°¦ (singular at the point B*NW =gG ). where × *1a Then from Equation (3.11), using the property of the Dirac delta
ûK
£D
ý
:$1
G
ûK
*1aªNW1°¦ G:$1@*NW
i.e., the domain integral has been replaced by a point value.
*NW4gG
(3.12)
3 .3 T HE T WO -D IMENSIONAL B OUNDARY E LEMENT M ETHOD
Thus Equation (3.11) becomes
*NW)
ý
ûK
Å G H§ Å
ûK
Å ý ÅG
B*NW4gG
LH
49
(3.13)
This equation contains only boundary integrals (and no domain integrals as in Finite Elements) and is referred to as a boundary integral equation. It relates the value of at some point inside
the solution domain to integral expressions involving and Å over the boundary of the solution G domain. Rather than having an expression relating the value Å of at some point inside the domain to boundary integrals, a more useful expression would be one relating the value of at some point on the boundary to boundary integrals. We derive such an expression below. The previous equation (Equation (3.13)) holds if B*NW 4gG (i.e., the singularity of Dirac Delta function is inside the domain). If +*Ns is outside G then
ûK
¢D
ý
®r1
LG
ûK
B *13ªNs13¦2 Gt7
since the integrand of the second integral is zero at every point except *NW and this point is outside the region of integration. The case which needs special consideration is when the singular is on the boundary of the domain G . This case also happens to be the most important point *NW for numerical work as we shall see. The integral expression we will ultimately obtain is simply Equation (3.13) with
+*Ns
replaced by
/ *NW . j
We can see this in a non-rigorous way as
follows. When B*NW was inside the domain, we integrated around the entire singularity of the Dirac Delta to get +*NW in Equation (3.13). When B*NW is on the boundary we only have half of the singularity contained inside the domain, so we integrate around one-half of the singularity to get
/ +*NW . Rigorous details of where this coefficient / j j
Let Ê
comes from are given below. ý ûK ¢D G in this case we denote the point B*NW¤¥G . In order to be able to evaluate
enlarge G to include a disk of radius about Ê (Figure 3.3). We call this enlarged region G ½ and let H ½ H ù ¦§ H ¦ . Now, since Ê is inside the enlarged region G ½ , Equation (3.13) holds for this enlarged domain i.e., ý RA¨&©«û ª&R © RA¨&¬© û ª'R © ý Å H (3.14) BÊQ) Å H
ÅG
We must now investigate this equation as â each of these in turn.
ó ¼ ¦M
ÅG
. There are h integrals to consider, and we look at
50
T HE B OUNDARY E LEMENT M ETHOD
G
½
H
¦ H
Ê
ù ¦
G
F IGURE 3.3: Illustration of enlarged domain when singular point is on the boundary.
Firstly consider
û
R ©
ý
ÅG Å
û
R ©
H
û
R ©
$1 jU/ Ù $1 jU/ Ù |
w
Å G 1 Uj / Ù äàk× x Å w / Å × 1 jUÙ àk× x Å Rû © H × / R û © H
1 jU/ Ù / BÊ,Ù
ý
H
by definition of
H
since
Å G¯® Å × Å Å
since ×
ó â R û © ý Å ¼ ¦M ÅG ó since â ¼ as â ó ¦M ¼ . ¦±°³²«´A¦M
H
¦
¦
by the mean value theorem for a surface with a unique tangent at Ê . Thus ý w / BÊ Rû © ó â ó ¼¦M Å G H } ¦â ¼ 1 jUÙ Ù x r1 Å By a similar process we obtain
on H
on H
8ÊQ j
(3.15)
w / ó cäâ ¼ 1 Uj Ù Å G ÀÊQCÙ à x 7 ¦M Å
(3.16)
3 .3 T HE T WO -D IMENSIONAL B OUNDARY E LEMENT M ETHOD
51
It only remains to consider the integrand over H . For “nice” integrals (which includes the ¦ ù integrals we are dealing with here) we have
ó äâ ¦
¼µ¶
û
RA¨©
(nice integrand) · ¸ ¹ºb» ¼ ½
(nice integrand) · ¸
since ¸\¾ ¿ÁÀ ¸ as ÂÄÃÅÆ¿MÇÉÈ . Note: If the integrand is too badly behaved we cannot always replace ¸\¾ ¿ by ¸ in the limit and one must deal with Cauchy Principal Values. (refer Section 4.8) Thus we have · ¸ ¹ºÒ» «¿ÂÄÃÄÊËÅ È µ¶ ½A¼ Ì&Í*ÎÏ ÎUÐÁÑ
¼½
Ó¿ÂÄÃÊËÅ È µ¶ ½A¼ ÌÍÔÎ Ñ Ï · ¸ º¹ ÎÐ
¼½ »
Î
Î Ï U ·L¸ ÎÐ4Ñ
(3.17)
· ¸
(3.18)
Ñ Ï Î Ð
Combining Equations (3.14)–(3.18) we get ÏÆÕÖ× Ø
¼½ Ï
Î
Ñ Î Ð
· ¸h»bÚ Ù
¼½
ÏuÕÖÛ×£Ø
Î Ï · ¸ ÎÐ Ñ
or ÚÙ
ÏÆÕÖ× Ø
¼½
Î Ï
Ñ Î Ð
· ¸h»
¼½
Î Ï U ·L¸ ÎÐ4Ñ
where » (i.e., singular point is on the boundary of the region). Ö ÕÜSÝWÞF×=ßhÎáà Note: The above is true if the point is at a smooth point (i.e., a point with a unique tangent) on Ö the boundary of . If happens to lie at some nonsmooth point e.g. a corner, then the coefficient à Ö Ú Ù is replaced by Úäâ ã where is the internal angle at (Figure 3.4). Ö â Ö â à
F IGURE 3.4: Illustration of internal angle å .
52
T HE B OUNDARY E LEMENT M ETHOD
Thus we get the boundary integral equation. æ
ÕMÖ×XÏuÕÖÛ×£Ø
¼½ Ï
Î
Ñ Î Ð U
· ¸g»
¼½
Î Ï · ¸ ÎUÐ Ñ
(3.19)
where »ç ÚäÙ ã ÂÄèélê Ñ
æ
ê2»ìë ö »
ÕMÖ×
ÕíÜ ôõ
çî
×ðïñØ ÕÞ Ùø
çóò
internalï angle Ú'ã õ÷
׫ï if Ößgà if ¸ and ¸ smooth at Öß Ö ¸ and ¸ not smooth at if Öß Ö
For three-dimensional problems, the boundary integral equation expression above is the same, with » Ñ
æ
ù ãÙ
ê2»ìë
ÕÖ×
»
ê ç6û × ï Ø Õú ×ï if Öüßgà if ¸ and ¸ smooth at Öüß Ö inner solid angle ¸ and ¸ not smooth at if ù ïã Öüß Ö
ÕÜ ö õ÷
ôõ
çî
×ï Ø ÕÞ Ùø
çò
and ÎÏ and the value of at a Ï Ï ÎUÐ point . Once the surface distributions of and ÎÏ are known, the value of at any point Ö Ï Ï Ö ÎÐ inside can be found since all surface integrals in Equation (3.19) are then known. The procedure à is thus to use Equation (3.19) to find the surface distributions of and ÎÏ and then (if required) Ï ÎÐ for the boundary data use Equation (3.19) to find the solution at any point . Thus we solve Ößýà first, and find the volume data as a separate step. Since Equation (3.19) only involves surface integrals, as opposed to volume integrals in a finite element formulation, the overall size of the problem has been reduced by one dimension (from volumes to surfaces). This can result in huge savings for problems with large volume to surface ratios (i.e., problems with large domains). Also the effort required to produce a volume mesh of a complex three-dimensional object is far greater than that required to produce a mesh of the surface. Thus the boundary element method offers some distinct advantages over the finite element method in certain situations. It also has some disadvantages when compared to the finite element method and these will be discussed in Section 3.6. We now turn our attention to solving the boundary integral equation given in Equation (3.19). Equation (3.19) involves only the surface distributions of
3þ .4 N UMERICAL S OLUTION P ROCEDURES
FOR THE
B OUNDARY I NTEGRAL E QUATION
53
3.4 Numerical Solution Procedures for the Boundary Integral Equation The first step is to discretise the surface ¸ into some set of elements (hence the name boundary elements).
¸ÿ»
ø
¸
(3.20)
(b)
(a)
F IGURE 3.5: Schematic illustration of a boundary element mesh (a) and a finite element mesh (b).
Then Equation (3.19) becomes
æ
Over each element ¸
ÕÖÛ×XÏÆÕÖ× Ø
ø
¼½
Ï
Î
Ñ Î Ð
· ¸g»
ø
Î Ï U ·L¸ ÎÐ Ñ
¼½
(3.21)
we introduce standard (finite element) basis functions Ï
»
Ï
and
ÎÏ ÎÐ
»
(3.22)
where are values of and on element ¸ and are values of and at node on Ï Ý Ï Ï Ý Ï â element ¸ . These basis functions for and can be any of the standard one-dimensional finite element Ï basis functions (although we are dealing with a two-dimensional problem, we only have to interpolate the functions over a one-dimensional element). In general the basis functions used for and Ï
do not have to be the same (typically they are) and these basis functions can even be different to the basis functions used for the geometry, but are generally taken to be the same (this is termed an isoparametric formulation).
54
T HE B OUNDARY E LEMENT M ETHOD
This gives æ
ÕMÖ×XÏÆÕÖ×NØ
ø
¼½
Ï
Î
·L¸h»
Ñ Î Ð U
ø
¼½
· ¸
(3.23)
Ñ
This equation holds for any point on the surface ¸ . We now generate one equation per node by Ö putting the point to be at each node in turn. If is at node , say, then we have Ö Ö æ Ï
Ø
ø
Ï
¼½
Î
Ñ Î Ð
· ¸g»
ø
¼½
· ¸
(3.24)
Ñ
where is the fundamental solution with the singularity at node (recall is çÒÚäÙ ã ÂÄèélê , where ê is theÑ distance from the singularity point). We can write Equation (3.24)Ñ in a more abbreviated form as æ
Ï
Ø
ø
Ï
»
ø
(3.25)
where
»
¼½
Î
Ñ Î Ð U
· ¸
and
»
¼½
· ¸
(3.26)
Ñ
Equation (3.25) is for node and if we have nodes, then we can generate equations. We can assemble these equations into the matrix system )
»)
)
(3.27)
(compare to the global finite element equations » ) where the vectors and are the vectors
th of nodal values of and . Note that the component of the matrix in general is not and Ï similarly for . At each node, we must specify either a value of or (or some combination of these) to have a Ï well-defined problem. We therefore have equations (the number of nodes) and have unknowns to find. We need to rearrange the above system of equations to get !#" "
»
(3.28)
where is the vector of unknowns. This can be solved using standard linear equation solvers, although specialist solvers are required if the problem is large (refer [todo : Section ???]). ! The matrices and (and hence ) are fully populated and not symmetric (compare to the finite element formulation where the global stiffness matrix is sparse and symmetric). The size of the and matrices are dependent on the number of surface nodes, while the matrix is dependent on the number of finite element nodes (which include nodes in the domain). As
3þ .5 N UMERICAL E VALUATION
C OEFFICIENT I NTEGRALS
OF
55
mentioned earlier, it depends on the surface to volume ratio as to which method will generate the smallest and quickest solution. The use of the fundamental solution as a weight function ensures that the and matrices are generally well conditioned (see Section 3.5 for more on this). In fact the matrix is diagonally ! dominant (at least for Laplace’s equation). The matrix is therefore also well conditioned and ) Equation (3.28) can be solved) reasonably easily. " The vector contains the unknown values of and on the boundary. Once this has been found, all boundary values of and are known. If a solution is then required at a point inside the domain, then we can use Equation (3.25) with the singular point located at the required solution Ö point i.e.,
»
ÏÆÕÖ×
ø
$
ç
ø
Ï
$
%
(3.29)
The right hand side of Equation (3.29) contains no unknowns and only involves evaluating the surface integrals using the fundamental solution with the singular point located at . Ö
3.5 Numerical Evaluation of Coefficient Integrals
We consider in detail here how one evaluates the and integrals for two-dimensional problems. These integrals typically must be evaluated numerically, and require far more work and effort than the analogous finite element integrals. Recall that
»¼½
Î
Ñ Î Ð
·L¸
and
» ¼½
·L¸
Ñ
where
Ñ
ê
»ç ÚäÙ ã ÂÄèAé;ê »
distance measured from node
In terms of a local coordinate we have Ü ø
» ¼
È
»
¼½
ÕÜC×
ÕÜC×
Ñ
ÕÜ ×'&)(QÕíÜ ×*&
Î Ñ
ÎUÐ
·
(3.30) Ü ø
ÕíÜ ×
&+(QÕÜC×,&
· Ü
» ¼
È
ÕÜ ×
Î
Î
Ñê
·Cê Õ Ü × · Ð
&+(QÕÜC×,&
· Ü
(3.31)
56
T HE B OUNDARY E LEMENT M ETHOD
The Jacobian
(QÕÜ ×
can be found by · ¸ · »
(QÕÜC×
·.» ·
Ü
» /
Ü
·î ·
ï Ü10
Ø
/
· ò ·
ï
(3.32)
Ü'0
·Cî ·Cò and can be found by straight differentiation of the · · Ü interpolation expression for î and ò Ü . ÕíÜ × ÕíÜ × The fundamental solution is
where
-
represents the arclength and
ê
Ñ ÕÜC×
»
»ç Ú'Ù ã ÂÄèé ë Õ
î
ÕÜC×
where î ò are the coordinates of node . Õ Ý ·C× ê To find we note that · Ð ·Cê ·
Õ
ê
çî
ÕÜ ×É×
»32 ê
× ïxØcÕ
5478 6
ò
ÕÜ ×
çò
× ï
(3.33)
Ð where 8 6 is a unit outward normal vector. To find a unit normal vector, we simply rotate the tangent ã ò.9 vector (given by î:9 ) by Ú in the appropriate direction and then normalise. Õ ÕíÜ ×¢Ý ÕíÜ ×É×
Thus every expression in the integrands of the and integrals can be found at any value of , and the integrals can therefore be evaluated numerically using some suitable quadrature schemes. Ü If node is well removed from element ¸ then standard Gaussian quadrature can be used to evaluate these integrals. However, if node is in ¸ (or close to it) we see that ê approaches 0 and the fundamental solution tends to ; . The integral still exists, but the integrand becomes singular. In such cases specialÑ care must be taken - either by using special quadrature schemes, large numbers of Gauss points or other special treatment. The integrals for which node lies in element ¸ are in general the largest in magnitude and lead to the diagonally dominant matrix equation. It is therefore important to ensure that these integrals are calculated as accurately as possible since these terms will have most influence on the solution. This is one of the disadvantages of the BEM - the fact that singular integrands must be accurately integrated. A relatively straightforward way to evaluate all the integrals is simply to use Gaussian quadrature with varying number of quadrature points, depending on how close or far the singular point is from the current element. This is not very elegant or efficient, but has the benefit that it is relatively easy to implement. For the case when node is contained in the current element one can use special quadrature schemes which are designed to integrate log-type functions. These are to be preferred when one is dealing with Laplace’s equation. However, these special log-type schemes cannot be so readily used on other types of fundamental solution so for a general purpose implementation, Gaussian quadrature is still the norm. It is possible to incorporate adaptive integration schemes that keep adding more quadrature points until some error estimate is small enough, or also to subdivide the current element into two or more smaller elements and evaluate the integral over each
3þ .6 T HE T HREE -D IMENSIONAL B OUNDARY E LEMENT M ETHOD
¸
57
¸
node ê
ê
node (a)
(b)
F IGURE 3.6: Illustration of the decrease in < as node = approaches element > .
subelement. It is also possible to evaluate the “worst” integrals by using simple solutions to the governing equation, and this technique is the norm for elasticity problems (Section 4.8). Details on each of these methods is given in Section 3.8. It should be noted that research still continues in an attempt to find more efficient ways of evaluating the boundary element integrals.
3.6 The Three-Dimensional Boundary Element Method The three-dimensional boundary element method is very similar to the two-dimensional boundary element method discussed above. As noted above, the three-dimensional boundary integral equation is the same as the two-dimensional equation (3.19), with and æ being defined as ÕMÖ× in Section 3.3. The numerical solution procedure also parallels that Ñ given in Section 3.4, and the
expressions given for and apply equally well to the three-dimensional case. The only real difference between the two procedures is how to numerically evaluate the terms in each integrand of these coefficient integrals.
As in Section 3.5 we illustrate how to evaluate each of the terms in the integrand of and .
58
T HE B OUNDARY E LEMENT M ETHOD
The relevant expressions are
»
¼½
ø »
¼ È
» ¼½
¼
ø
È
¼ È
Î
ÕÜ ÝsÜ × Ñ ê ï Î
ø »
ø
È
· ¸
Ñ Î Ð
ø
¼
Î
·Cê íÕ Ü ÝðÜ × · ï Ð ø
&?(QÕÜ
ø
ÝsÜ @× & ï
· ø · Ü Ü
(3.34) ï
· ¸
Ñ
ø
ÕÜ ÝsÜ × ï Ñ
ø ø · ø · ÕÜ sÝ Ü A × &?(QÕ Ü sÝ Ü ×,& Ü Ü ï ï ï
(3.35)
The fundamental solution is ø » ù ã Ù ø ÕÜ sÝ Ü × ê ï Ñ Õ Ü ÝsÜ × ø ø ï çî ø ø where ê » ë î ò çóò û çóû ÕÜ ÝsÜ × Õ ÕÜ sÝ Ü ×
× ï\ØcÕ ÕÜ sÝ Ü ×
× ïxØcÕ ÕÜ ÝsÜ × × ï ï ï ï ï · ê ·Cê »B2 ê C4D8 6 to find where î ò û are the coordinates of node . As before we use . Õ Ý Ý × · · 86 Ð The unit outward normal is found by normalising the cross product Ð of the two tangent vectors î ò û î ò û E ø E Î Î and (it relies on the user of any BEM code to » / Î ø Î ø Î ø » / Î Ý Ý Ý Ý ï ÎFÜ theÎÜ elements ÎFÜ 0 have been ÎFÜ ÎÜ ÎF Ü 0 ensure that defined set of element coordinates ø and ). ï E with ï E a consistent ï E E Ü Ü ø ø G F (where ø and The Jacobian is given by F are the two tangent vectors). ï (QÕÜ ÝsÜ × Note that this is differentï for the determinant in a two-dimensional finite element code in that ï ï case we are dealing with a two-dimensional surface in two-dimensional space, whereas here we have a (possibly curved) two-dimensional surface in three-dimensional space. The integrals are evaluated numerically using some suitable quadrature schemes (see Section 3.8) (typically a Gauss-type scheme in both the ø and directions). Ü Ü ï
3.7
A Comparison of the FE and BE Methods
We comment here on some of the major differences between the two methods. Depending on the application some of these differences can either be considered as advantageous or disadvantageous to a particular scheme. 1. FEM: An entire domain mesh is required. BEM: A mesh of the boundary only is required. Comment: Because of the reduction in size of the mesh, one often hears of people saying that the problem size has been reduced by one dimension. This is one of the major pluses of
3þ .7 A C OMPARISON
OF THE
FE
AND
BE M ETHODS
59
the BEM - construction of meshes for complicated objects, particularly in 3D, is a very time consuming exercise. 2. FEM: Entire domain solution is calculated as part of the solution. BEM: Solution on the boundary is calculated first, and then the solution at domain points (if required) are found as a separate step. Comment: There are many problems where the details of interest occur on the boundary, or are localised to a particular part of the domain, and hence an entire domain solution is not required. ) 3. FEM: Reactions on the boundary typically less accurate than the dependent variables. BEM: Both and of the same accuracy. 4. FEM: Differential Equation is being approximated. BEM: Only boundary conditions are being approximated. Comment: The use of the Green-Gauss theorem and a fundamental solution in the formulation means that the BEM involves no approximations of the differential Equation in the domain - only in its approximations of the boundary conditions. 5. FEM: Sparse symmetric matrix generated. BEM: Fully populated nonsymmetric matrices generated. Comment: The matrices are generally of different sizes due to the differences in size of the domain mesh compared to the surface mesh. There are problems where either method can give rise to the smaller system and quickest solution - it depends partly on the volume to surface ratio. For problems involving infinite or semi-infinite domains, BEM is to be favoured. 6. FEM: Element integrals easy to evaluate. BEM: Integrals are more difficult to evaluate, and some contain integrands that become singular. Comment: BEM integrals are far harder to evaluate. Also the integrals that are the most difficult (those containing singular integrands) have a significant effect on the accuracy of the solution, so these integrals need to be evaluated accurately. 7. FEM: Widely applicable. Handles nonlinear problems well. BEM: Cannot even handle all linear problems. Comment: A fundamental solution must be found (or at least an approximate one) before the BEM can be applied. There are many linear problems (e.g., virtually any nonhomogeneous equation) for which fundamental solutions are not known. There are certain areas in which the BEM is clearly superior, but it can be rather restrictive in its applicability. 8. FEM: Relatively easy to implement. BEM: Much more difficult to implement. Comment: The need to evaluate integrals involving singular integrands makes the BEM at least an order of magnitude more difficult to implement than a corresponding finite element procedure.
60
T HE B OUNDARY E LEMENT M ETHOD
3.8
More on Numerical Integration
The BEM involves integrals whose integrands in generally become singular when the source point is contained in the element of integration. If one uses constant or linear interpolation for the geometry and dependent variable, then it is possible to obtain analytic expressions to most (if not all) of the integrals that will appear in the BEM (at least for two-dimensional problems). The expressions can become quite lengthy to write down and evaluate, but benefit from the fact that they will be exact. However, when one begins to use general curved elements and/or solve threedimensional problems then the integrals will not be available as analytic expressions. The basic tool for evaluation of these integrals is quadrature. As discussed in Section 2.8 a one-dimensional integral is approximated by a sum in which the integrand is evaluated at certain discrete points or abscissa ø ¼ ÈIH
ÕÜC×
·
ÜKJ
ø
H
ÕíÜ ×.L
where are the weights and are the abscissa. L Ü The weights and abscissa for the Gaussian quadrature scheme of order M are chosen so that the Ú above expression will exactly integrate any polynomial of degree Mcç or less. For the numerical Ù evaluation of two or three-dimensional integrals, a Gaussian scheme can be used of each variable of integration if the region of integration is rectangular. This is generally not the optimal choice for the weights and abscissae but it allows easy extension to higher order integration.
3.8.1
Logarithmic quadrature and other special schemes
Low order Gaussian schemes are generally sufficient for all FEM integrals, but that is not the case ø for BEM. For a two-dimensional BEM solution of Laplace’s equation, integrals of the form ¼
ÂÄèAé
ÕíÜ ×
ÕíÜ ×
· Ü
will be required. It is relatively common to use logarithmic schemes for this.
È H These are obtained by approximating the integral as ø ¼ È
Âèé
ÕÜ × H
ÕÜ ×
·
ÜNJ
ø H
ÕÜ ×OL
i.e., the log function has been factored out. In the same way as Gaussian quadrature schemes were developed in Section 2.8, log quadrature schemes can be developed which will exactly integrate polynomial functions . Tables of these ÕÜC× are given below H It is possible to develop similar quadrature schemes for use in the BEM solution of other PDEs, which use different fundamental solutions to the log function. The problem with this approach is the lack of generality - each new equation to be used requires its own special quadrature scheme.
3þ .9 T HE B OUNDARY E LEMENT M ETHOD A PPLIED
Ð
2
Abscissas = ê ç Ü L 0.112009 0.718539 0.602277 0.281461 Ð
3
Ü 0.063891 0.368997 0.766880
TO OTHER
E LLIPTIC PDE S
Weight Factors =
ç L 0.513405 0.391980 0.094615
Ð
4
Ü 0.041448 0.245275 0.556165 0.848982
61
L
ç L 0.383464 0.386875 0.190435 0.039225
TABLE 3.1: Abscissas and weight factors for Gaussian integration for integrands with a logarithmic singularity.
3.8.2 Special solutions Another approach, particularly useful if Cauchy principal values are to be found (see Section 4.8) is to use special solutions of the governing equation to find one or more of the more difficult integrals. For example » î is a solution to Laplaces’ equation (assuming the boundary conditions Ï are set correctly). Thus if one sets both and in Equation (3.27) at every node according to Ï the solution » î , one can then use this to solve for some entry in either the or matrix Ï (typically the diagonal entry since this is the most important and difficult to find). Further solutions to Laplaces equation (e.g., » î ç ò ) can be used to find the other matrix entries (or just used Ï ï ï to check the accuracy of the matrices).
3.9 The Boundary Element Method Applied to other Elliptic PDEs Helmholtz, modified Helmholtz (CMISS example) Poisson Equation (domain integral and MRM, DRM, Monte-carlo integration.
3.10 Solution of Matrix Equations !P"
» (refer (3.28)). Q The standard BEM approach results in a system of equations of the form ! As mentioned above the matrix is generally well conditioned, fully populated and nonsymmetric. For small problems, direct solution methods, based on LU factorisations, can be used. As the problem size increases, the time taken for the matrix solution begins to dominate the matrix assembly stage. This usually occurs when there is between RTS%S and S%S%S degrees of freedom, although it Ù is very dependent on the implementation of the BE method. The current technique of favour in the BE community for solution of large BEM matrix equations is a preconditioned Conjugate Gradient solver. Preconditioners are generally problem dependent - what works well for one problem may not be so good for another problem. The conjugate gradient technique is generally regarded as a solution technique for (sparse) symmetric matrix equations.
62
T HE B OUNDARY E LEMENT M ETHOD
àVU ¸
àVY
U
¸
¸XW
Y
F IGURE 3.7: Coupled finite element/boundary element solution domain.
3.11
Coupling the FE and BE techniques
There are undoubtably situations which favour FEM over BEM and vice versa. Often one problem can give rise to a model favouring one method in one region and the other method in another region, e.g., in a detailed analysis of stresses around a foundation one needs FEM close to the foundation to handle nonlinearities, but to handle the semi-infinite domain (well removed from the foundation), BEM is better. There has been a lot of research on coupling FE and BE procedures we will only talk about the basic ideas and use Laplace’s Equation to illustrate this. There are at least two possible methods. 1. Treat the BEM region as a finite element and combine with FEM 2. Treat the FEM region as an equivalent boundary element and combine with BEM Note that these are essentially equivalent - the use of one or the other depends on the problem, in the sense of which part is more dominant FEM or BEM) Consider the region shown in Figure 3.7, where àVY
»
» àVU ¸ » Y ¸ » U ¸ W=» X
The BEM matrices for
àVU
FEM region BEM region FEM boundary BEM boundary interface boundary
can be written as )
»
)
(3.36)
Î Ï where is a vector of the nodal values of and is a vector of the nodal values of Ï ÎUÐ The FEM matrices for can be written as) à Z
»[
(3.37)
3þ .11 C OUPLING
THE
FE
BE
AND
63
TECHNIQUES
where is the stiffness matrix and is the load vector. To apply method (i.e., treating BEM as an equivalent FEM region) we get (from EquaÙ tion (3.36)) ) ¾ ø (3.38) » \ If we recall what the elements of in Equation (3.37) contained, then we can convert in Equation (3.38) to an equivalent load vector by weighting the nodal values of by the appropriate basis functions, producing a matrix ] i.e., » ]B \ U Therefore Equation (3.38) becomes ) ¾ ø a` ] ^_ » ]B [ b » U
)
i.e.,
where
»[ U U
U
»3]c
¾ ø
an equivalent stiffness matrix obtain from BEM. Therefore we can assemble this together with original FEM matrix to produce an FEM-type . system for the entire region à U V Notes: 1.
is in general not symmetric and not sparse. This means that different matrix equation solvers must be used for solving the new combined FEM-type system (most solvers in FEM codes assume sparse and symmetric). Attempts have been made to “symmetricise” the U
matrix - of doubtful quality. (e.g., replace ) results).
U
by Ú Ù
^d U
çe gf
`
U
U
- often yields inaccurate
2. On ¸XW nodal values of ) and ) are unknown. One must make use of the following ) ) Î Î
W WU
8 U U
)
»
W
»
ç Î Î
( is continuous)
Z W
( is continuous, but ¸
8 Z Z
U
»
ç ¸ Z
)
) To apply method (i.e., to treat the FEM region as an equivalent BEM region) we firstly note that, as before, »h]B . Applying this to (3.37) yields »h]B an equivalent BEM system. This can be assembled into the existing BEM system (using compatability conditions) and use existing BEM matrix solvers. Notes: Ú
1. This approach does not require any matrix inversion and is hence easier (cheaper) to implement 2. Existing BEM solvers will not assume symmetric or sparse matrices therefore no new matrix solvers to be implemented
64
T HE B OUNDARY E LEMENT M ETHOD
3.12
Other BEM techniques
What we have mentioned to date is the so-called singular (direct) BEM. Given a BIE there are other ways of solving the Equation although these are not so widely used.
3.12.1
Trefftz method
Trefftz was the first person to perform a BEM calculation (in 1917 - calculated the value (numerical) of the contraction coefficient of a round jet issuing from an infinite tank - a nonlinear free surface problem). This method basically uses a “complete” set of solutions instead of a Fundamental Solution. e.g., Consider Laplaces Equation in a (bounded) domain à weighted residuals
Î Ï ·L¸ÿ» ¼ U j k Ñ ÎÐ
¼
i
j k
Ï
Î
Ñ Î Ð
· ¸
if 2 ï
»3S Ñ
The procedure is to express as a series of (complete) functions satisfying Laplace’s equation Ï with coefficients which need to be numerically determined through utilisation of the boundary conditions. Notes: 1. Doesn’t introduce singular functions so integrals are easy to evaluate 2. Must find a (complete) set of functions (If you just use usual approximations for matrix Ï system is not diagonally dominant so not so good) 3. Method is not so popular - Green’s functions more widely available that complete systems
3.12.2
Regular BEM
Consider the BIE for Laplace’s equation æ
¼
ÕMÖ×XÏuÕÖÛ×£Ø
jlk
Ï
Î
· ¸g»
Ñ Î Ð U
»
with
Î Ï · ¸ ÎUÐ Ñ
¼
j k
çbÚäÙ ã ÂÄèélê
Ñ The usual procedure is to put point at each solution variable node - creating an equation for each Ö node. This leads to singular integrands. Another possibility is to put point outside of the domain - this yields Ö à ¼
j k
Ï
Î
Ñ m n ÎÐ
· ¸h»
¼
j k
Î Ï ÎÐ m
· ¸
3þ .13 S YMMETRY
65
Following discretisation as before gives
ø
Ï
¼½
Î
Ñ m n ÎUÐ
· ¸g»
ø
¼½
· ¸
Ñnm
- an equation involving and at each surface node. Ï By placing the point (the singular point) at other distinct points outside one can generate Ö à as many equations as there are unknowns (or more if required). Notes: 1. This method does not involve singular integrands, so that integrals are inexpensive to calculate. 2. There is considerable choice for the location of the point . Often the set of Equations Ö generated are ill-conditioned unless chosen carefully. In practise is chosen along the unit Ö Ö outward normal of the surface at each solution variable node. The distance along each node is often found by experimentation - various research papers suggesting “ideal” distances (Patterson & Shiekh). 3. This method is not very popular. 4. The idea of placing the singularity point away from the solution variable node is often of Ö use in other situations e.g., Exterior Acoustic Problems. For an acoustic problem (governed by Helmholtz Equation 2 » S ) in an unbounded region the system of Equations pro Ïp Ø o ï Ï ï duced by the usual (singular) BEM approach is singular for certain “fictitious” frequencies (i.e., certain values of ). To overcome this further equations are generated (by placing the o singular point at various locations outside ). The system of equations are then overdeÖ à termined and are solved in a least squares sense.
3.13 Symmetry Consider the problem given in Figure 3.8 (the domain is outside the circle). Both the boundary conditions and the governing Equation exhibit symmetry about the vertical axis. i.e., putting î to ç î makes no difference to the problem formulation. Thus the solution q î û has the property Õ Ý × that q î û r » q ç¤î û î . This behaviour can be found in many problems and we can make Ú % t Õ Ý × Õ Ý . × s use of this as follows. The Boundary Element Equation is (with Mb» (i.e., M is even) constant elements)
ÚÙ Ï
Ø
ø Ï
¼½
Î
Ñ Î Ð
· ¸ÿ»
ø
¼½
Ñ
· ¸
»
Ù Ý*u*uvuNÝ
M
(3.39)
66
T HE B OUNDARY E LEMENT M ETHOD
û ¾yx{z
»w q
2
qb»\-
ï
î
ï
q
F IGURE 3.8: A problem exhibiting symmetry.
We have M Equations and M unknowns (allowing for the boundary conditions). From symmetry we know that (refer to Figure 3.9). Ï
»
ø¾
Ï:|,}
t
ñ»
(3.40)
Ù Ý*uvu*uNÝ
So we can write ÚÙ Ï
Ø
~
ø Ï
ö
ôõ ¼½ õ÷
for nodes » Ù Ý*uvu*uNÝ for nodes ñ» Ù Ý*uvu*u Ý If we define
Î t t
· ¸
Ñ Î Ð
Ø
½@O%¼ Ì
Î
· ¸ õ »
Ñ Î Ð
ø
~
õ
ö
ôõ ¼½ Ñ õ÷
t
· ¸ Ø
½,O%¼ Ì
Ñ
· ¸ õ
(3.41) õ
M are the same as the Equations . (The Equations for nodes \» t Ø Ù Ý*t u*u*u Ý ). The above Equations have only unknowns.
»
» ¼½
Î
¼½
Ñ Î Ð
· ¸
· ¸
Ø
½@y¼ Ì Ø
Ñ
½@O%¼ Ì
Î
Ñ Î Ð
· ¸
(3.42)
·L¸
(3.43)
Ñ
then we can write Equation (3.41) as ÚÙ Ï
Ø
~
ø
Ï
»
~
ø
\»
t
Ù Ý*u*u*u£Ý
and solve as before. (This procedure has halved the number of unknowns.)
(3.44)
3þ .14 A XISYMMETRIC P ROBLEMS
67
û
¸ }
ø¾
¸
î
¸ ø ¸
F IGURE 3.9: Illustration of a symmetric mesh. t ø to ¸ Note: Since 4» this means that the integrals over the elements ¸ will never } Ù Ý*u*u*u Ý contain a singularity arising from the fundamental solution, except possibly on the axis of symme~ try if linear or higher order elements are used. An alternative approach to the method above arises from the implied no flux across the û axis. This approach ignores the negative î axis and considers the half plane problem shown. However now the surface to be discretised extends to infinity in the positive and negative û directions and the resulting systems of equations produced is much larger. Further examples of how symmetry can be used (e.g., radial symmetry) are given in the next section.
3.14 Axisymmetric Problems ê û ) it If a three-dimensional problem exhibits radial or axial symmetry (i.e., ê ø û » ÏuÕ Ý Ý × ÏuÕ Ý Ý × is possible to reduce the two-dimensional integrals appearing in the standard boundary Equation ï to one-dimensional line integrals and thus substantially reduce the amount of computer time that would otherwise be required to solve the fully three-dimensional problem. The first step in such a procedure is to write the standard boundary integral equation in terms of cylindrical polars ê û Õ Ý LÝ × i.e., æ
ÕMÖ×XÏÆÕÖ×NØ
¼½
Ïg
¼ ï Î · Ñ m n È ÎÐ
*
¹º ê
· ¸h»
¼½
¼ ï · È Ñnm
*
¹º ê
· ¸
(3.45)
û û and ê are the polar coordinates of and respectively, and ¸ is the where ê Õ Ý Ý × Õ ±Ý*1Ý × Ö intersection m m of ¸ m and »3S semi-plane (Refer Figure 3.10). (n.b. is a point on the surface being integrated over.)
68
T HE B OUNDARY E LEMENT M ETHOD
û ¸
¸ à
ê F IGURE 3.10: Illustration of surface > for an axisymmetric problem.
û
ê ê
ï Ö ê ø
ê m
ê
ê
F IGURE 3.11: The distance from the source point ( ) to the point of interest ( ) in terms of cylindrical polar coordinates.
For three-dimensional problems governed by Laplace’s equation »
where ê m
is the distance from Ö
ê Ñnm . From Figure 3.11
to
ê ø c » ê ï ï Ø m ø ê » ë ê ï ï êe» ë ê ï m » ë
ê
Ú ç
ï
ê
ï ï
ê Ø
ê ê
:
m
Ø ç
ù ãÙ
ï
ç è%
Ú
ê ê m
Õ m
ç
è%
Õ èT
ç
*W×
m
Õ
ç m
*× Ø Õ
û
* m
ç6û
× Ø Õ ×
ï
û m
ç6û
W×Éï
(3.46)
3þ .15 I NFINITE R EGIONS
69
We define · ù Ù ã ¼ ï È n Ñ m » ÑXm
*
Ú
ã
Õ × Ø
where
»
(3.47)
Ø
is the complete elliptic integral of the first kind. Õ × is called the axisymmetric fundamental solution and is the Green’s function for a ring source as opposed to a point source. i.e., is a solution of Ñ m ' ÑXm (3.48) 2 êEçóê » S Ø Õ × ï Ñ m instead of and
(3.49) » S b ï Ø Ñ m $ where is the dirac delta centered at the point and êEçóê is the dirac delta centered on the Ö Õ × ring ê2»c m ê . Unlikem the two- and three-dimensional cases, the axisymmetric fundamental solution cannot be written as simply a function of the distance between two points and , but it also depends upon Ö the distance of these points to the axis of revolution. We also define 2
»
m
· ù Ù ã ¼ ï Î Ñ m n È ÎÐ
*
Î
(3.50)
Ñ m 1 ÎUÐ
For Laplace’s equation Equation (3.50) becomes »
m
ã Ù
Ø
Ú Ù ê
ê
m
ï
çê
ï
ØcÕ ç
û
çóû m
×
Õ ×
ç
Õ ×¢¡ Ðn£Õ
û ×Ø
m
çû ç ¤
Õ ×SÐ
z
Õ
× ¥ (3.51)
is the complete elliptic integral of the second kind. where Õ × Using Equation (3.47) and Equation (3.50) we can write Equation (3.45) as æ
ÕÖÛ×XÏÆÕÖ× Ø
¼½ Ï
Î
Ñ m X ÎÐ
· ¸h»
¼½
· ¸
(3.52)
Ñ'm
and the solution procedure for this Equation follows the same lines as the solution procedure given previously for the two-dimensional version of boundary element method.
3.15 Infinite Regions The boundary integral equations we have been using have been derived assuming the domain à is bounded (although this was never stated). However all concepts presented thus far are also
70
T HE B OUNDARY E LEMENT M ETHOD
¸ §
àV¦ îUÈ ¸
F IGURE 3.12: Derivation of infinite domain boundary integral equations.
valid for infinite regular (i.e., nice) regions provided the solution and its normal derivative behave appropriately as ¸gÀ ; . » S outside some surface ¸ . Consider the problem of solving 2 Ï ï ¸ is the centre of a circle (or sphere in three dimensions) of radius centred at some point îUÈ on ¸ and surrounding ¸ (see Figure 3.12). The boundary integral equations for the bounded domain can be written as àV¦ æ
$
ÕÖ×XÏÆÕÖ× Ø
¼½
If we let the radius
§
Î
À
Ï ;
Ñ Î Ð
$
· ¸
¼½ Ø
Ï
Î
· ¸g»
Ñ Î Ð
$
·L¸
Ñ
Ø
¼½
$
· ¸
(3.53)
Ñ
Equation (3.53) will only be valid for the points on ¸ if $ ¦
ÂÄÊ© ÃÅ ¨ ¼ ½ /
Î Ï
Ñ Î Ð
Ñ
$
ÕMÖ×XÏuÕÖ×NØ
¼½ Ï
$
çª
If this is satisfied, the boundary integral Equation for æ
¼½
Î
Ñ Î Ð
· ¸h»3S
(3.54)
à
0
will be as expected i.e.,
· ¸g»
¼½
$
Ñ
· ¸
(3.55)
3þ .15 I NFINITE R EGIONS
71
»
For three-dimensional problems with · ¸h»
·
Ñ
&?(«& §
·.¬ ¾ ø `
»33^ Î
Ñ
»33^
§
¾
ù ãÙ ê
where
&)(«&
»\b^
` §
ï
`
Ñ ï Î Ð § À where is the Jacobian and represents the asymptotic behaviour of the function as ø § ¾ &?(«& ÕM× ; . In this case Equation (3.53) will be satisfied if behaves at most as so that 7» § ¾ Ï Õ × . These are the regularity conditions at infinity and these ensure that each term in the § ¾ ø § Õ × ï (i.e., each term will À S as À ; ) integral Equation (3.53) behaves at most as § Õ × § For two-dimensional problems with we require to behave as so that » 3 Ä Â è é  è é § ¾ ø Õ Õ ×ð× Ï Õ × . For almost all well posedÑ infinite domain problems the solution behaves appropri QQ » Õ × ately at infinity.
72
T HE B OUNDARY E LEMENT M ETHOD
3.16
Appendix: Common Fundamental Solutions
3.16.1
Two-Dimensional equations
î ø î . Õ ï Ø ï × ï Laplace Equation
Here ê2»r®
Î ï Ïø î Î
Solution
Î ï Ïø î
Wave
Equation
Solution
Diffusion
Î
ï
»
Ø
Ø e
Î ï ï /ËÙ ÚäÙ ã  èé¯ ê
Helmholtz Equation Solution
ï »
Ï
Î ï Ï î
Ø
Î
Î ï Ï î
ùÙ
q²È ±
ï ï
È=»3S
0
Ø °
ïÏ
ê
ÈË»S
Øe
ï_³ Õ{° × Ï where q is the Hankel funtion. æ
Î ï Ï çÎ ï Ï È »bS / Î ï Ï î ø Ø î Ø Õµ´ð× ï Î æ is 0 Î:´ ï theÎ wave where ï ï speed.ï æ q çóê × » ç Úäã æ Õ æ ´ Ï çóê Õ ï´ï ï ×
Equation
Î ï Ïø î Ø Î ï where
Solution
» Ï
ç
o
Î ï Ï ç Ù ÎUÏ »3S î Îis the o δ ïï diffusivity. ê w± î ¹ / ç ù ï ù ã Ù Õ
o.´ 0
o¶´ð×¢·¸
¼»
Navier’s
Equation Solution
Î:º »S for a point load in direction ¿ . î Øe¾½ Î » ¹ w ¹ e ¹
»ç nÀ ã ÕÚ Ù
Ù çÂÁ
ï ×
ê
ï Ú êÆÅ ç Î ç Á êÆÅ êÆÅ Ç ç Á êÆÅ ÕÙ × ØÄ Ø Õ Ù × ÕÐ Ð × Î aЪtraction à for in direction where Á is Poisson’s ratio. /
ê
o
3.16.2 Here ê2» ®
Three-Dimensional equations î ø î î . Õ ï Ø ï Ø ïÈ × ï
0
w
3þ .17 CMISS E XAMPLES
Laplace
73
Î ï Ïø î
Equation Î
Solution
Î ï Ïø î Î
Solution Equation
Solution
ï »
Ï
ê Î ï Ï î
Ø
ù ãÙ
Î
Ø
Î ï Ï î
Ø
Î ï Ï î
ï ï
ï ï
Î
êËÊ ÌyÍ Õ
ï È
Î È ï çD ê
È »bS
Ø
Ø ° É
ï Ï
Øe
È=»3S
×
°
æ
Î ï Ï Î ï Ï ç / Î ï Ï î ø Ø î Ø î ï Î æ is Î È ï 0 ï theÎ wave where ï ï speed. ê ç : æ Ñ » ÐÏv´ ù ã Ï ê
Navier’s
Î
ù ãÙ
Helmholtz Equation
Wave
ï »
Ï
Î ï Ï î Ø
Î ï Ï Î:´ ï
»S
Ø Î
»
Equation
Î:º »ÓS for a isotropic homogenenous Kelvin î ØÒ¾½ Î solution for a point load in direction ¿ .
Solution
Ï
»
»
Ï
½
» w
½
»
ã'Õ Ù
/ Ä
ç
ù
ê Î ø Ï ½ ¾½ Ø çeÁ ê î ÕÙ × direction where Î Á Ù*Ô for a displacement in Õ o ratio and is the shear modulus. »
Á
»
Î
ê
î Î is Poisson’s 0 ï
3.16.3 Axisymmetric problems Laplace
For Ï
see Equation (3.47) and for
see Equation (3.51)
3.17 CMISS Examples 1. 2D steady-state heat conduction inside an annulus To determine the steady-state heat conÚ ù . duction inside an annulus run the CMISS example Ä
2. 3D steady-state heat conduction inside a sphere. To determine the steady-state heat conducÚTÀ tion inside a sphere run the CMISS example . Ä
3. CMISS comparison of 2-D FEM and BEM calculations To determine the CMISS comparison Ú ù Ú of 2-D FEM and BEM calculations run examples and . Ä Ä Ù 4. CMISS biopotential problems C4 and C5.
Chapter 4 Linear Elasticity 4.1 Introduction To analyse the stress in various elastic bodies we calculate the strain energy of the body in terms of nodal displacements and then minimize the strain energy with respect to these parameters - a technique known as the Rayleigh-Ritz. In fact, as we will show later, this leads to the same algebraic equations as would be obtained by the Galerkin method (now equivalent to virtual work) but the physical assumptions made (in neglecting certain strain energy terms) are exposed more clearly in the Rayleigh-Ritz method. We will first consider one-dimensional truss and beam elements, then two-dimensional plane stress and plane strain elements, and finally three-dimensional elasticity. In all cases the steps are: 1. Evaluate the components of strain in terms of nodal displacements, 2. Evaluate the components of stress from strain using the elastic material constants, 3. Evaluate the strain energy for each element by integrating the products of stress and strain components over the element volume, 4. Evaluate the potential energy from the sum of total strain energy for all elements together with the work done by applied boundary forces, 5. Apply the boundary conditions, e.g., by fixing nodal displacements, 6. Minimize the potential energy with respect to the unconstrained nodal displacements, 7. Solve the resulting system of equations for the unconstrained nodal displacements, 8. Evaluate the stresses and strains using the nodal displacements and element basis functions, 9. Evaluate the boundary reaction forces (or moments) at the nodes where displacement is constrained.
76
L INEAR E LASTICITY
4.2
Truss Elements
Consider the one-dimensional truss of undeformed length in Figure 3.1 with end points S S Õ Ý × and î ò and making an angle with the x-axis. Under the action of forces in the î - and ò Õ Ý × directions the right hand end of the truss displaces by in the î -direction and Ö in the ò -direction, Ï relative to the left hand end. ò
Ö
Õ×
Ø6ÏÝÙÔØ ÕØ×hÝÙu×
Ö
Ï
×
¿
î
F IGURE 4.1: A truss of initial length Ú is stretched to a new length Û . Displacements of the right hand end relative to the left hand end are Ü and Ý in the Þ - and ß - directions, respectively.
The new length is ¿ with axial strain ¿
w?»
ç Ù
ë ®
» »Óà
using
×
»
è%
and
Ù
»hsÃâá
Ö
Õµ×
Ø7 Ï¢× ï Ø ÕÙ*Ø
»
Ú × ï ØeÙ ï ØÕ ×gÏuØeÙ
ï Ø
Ö
Ú Ù Ø
, where
Ï
è%
.u
Ï Ø
×ï ç
Ö
× Ø7Ï ï Ø Ö
sÃâá
yu
Ñ
Ø
ï ç Ù
Ö
Ï ï Ø
ï ç Ù
ï
is defined to be positive in the anticlockwise direction.
Neglecting second order terms in the binomial expansion for small displacements and Ö is Ï wå»
Ù
èT
yu
Ï Ø
ðà á
®
yu
Ö
Õ Ù Øäã×
»
Ù Ø
ÚÙ
ã¤Ø
Õã ï ×
, the strain
(4.1)
æ
4 .2 T RUSS E LEMENTS
77
The strain energy associated with this uniaxial stretch is SE »
ÚÙ ¼ º
w·.ç
»
Ù è ¼ Ún é
wñ·î º
È
»
Ù ¼ é Úê È
è w
·î » ï
Ù è Ú5
w
ï
(4.2)
w is the stress in the truss (of cross-sectional area è ), linearly related to the strain w where » º via Young’s modulus . We now substitute for w from Equation (4.1) into Equation (4.2) and put ø and Ö ë » ç » Ö çeÖ ø , where ø Ö ø and Ö are the nodal displacements of the two Ï Ï Ï ÕÏ Ý × ÕÏ Ý × ends ofï the truss ï ï ï
SE »
ÚÙ
è
/
è%
yu
Ï ï
ç
Ï
ø Ø
ðà á
çäÖ ø Ö
.u
ï
ï
(4.3)
0
The potential energy is the combined strain energy from all trusses in the structure minus the work done on the structure by external forces. The Rayleigh-Ritz approach is to minimize this potential energy with respect to the nodal displacements once all displacement boundary conditions have been applied. For example, consider the system of three trusses shown in Figure 4.2. A force of S%S¯ì.í Ú Ù is applied in the î -direction at node . Node is a sliding joint and has zero displacement in the Ù y-direction only. Node is a pivot and therefore has zero displacement in both î - and ò - directions. Ä The problem is to find all nodal displacements and the stress in the three trusses.
node Ù S%î Ä
node
S%SDì.í
Ù Ú
S î Ä
Ä
Ù
node
Ú
Ä
F IGURE 4.2: A system of three trusses.
The strain in truss (joining nodes and ) is Ä Ù Ù ø ø Ö ø Ï èT S ðà á S » Ú Ä Ï Ä Ø Ä Ø Ú Ú The strain in truss (joining nodes and ) is Ù ø ç Ö ø ÕÏ Ï × %è 5ï%S sà áðï%S » ï Ø
ø Ö
ø Ö
ÚÙ
78
L INEAR E LASTICITY
Ú The strain in truss (joining nodes and ) is Ä
Ä
Ï
Since a force of PE »
ÚÙ
è
trusses
w
ï
ç
è% ç Õ
S Ä
» ×
Ú
Ï Ä
ï
acts at node in the î -direction, the potential energy is Ù
S%SDì.í
Ù
ï
S%S
Ù
ø » Ï
è
ÚÙ
Ú
ñòôó
Ä
ø Ï
ÚÙ Ø
ø¼õ ï Ö
Ø ó
Ú Ä
Ï
õ
ï
ï
ØcÕ
ø Ö
× ïö÷
ç Ù
S%S
Ï
ø
[Note that if the force was applied in the negative î -direction, the final term would be S%S ø ] Ø Ï Ù Minimizing the potential energy with respect to the three unknowns ø , Ö ø and gives Ï Ï ï PE è Î ø » (4.4) Ú Ä ø ÚÙ Ö ø õ Ú Ä ç S%S »bS Ï Ø ó Ù ÎÏ
è
è
»[R
» R G S truss) then ÿ Ù Equation (4.6) gives
G
¾ Ù È
Å
S
Å
ï
È G
Ú
ø:ó
PE ÎUÏ
¾
Î
If we choose
è
PE Î ø » Ö Î
ï
Ù
ï S1ìyü1þ Ù
»
,
ÄAÏ
Ø Ä
Ú Ä
Ö dø ù
Ø
õ
Ï
Ú
»bS
(4.5)
»3S Ä
(4.6)
ï
and c » G
Å»ÿR
ø » Ö
ÚÙ
Sûúýüþ
Equation (4.4) gives
Ö ¼ø õ
ÚÙ Ø
ó
Ï ø
ø Ï
è
»
Ä
Å (e.g., Ù ¾ ø Ù SCì.íÔÅ .
Ù
S%S
ÅQÅ
G
RTSÅ
Å
timber
»3S ï ù G
S
Ù ï
^dR
G
Ù
S
`
Equation (4.5) gives for two dimensions Ö
ø »ç
R
Ä
Ï
ø
ù Solving these last two equations gives ø » Å Å and Ö ø » ç R Å Å . Thus the strain in truss Ï ÄOu?Ä Ù uÙ ù Ú Ú Ú ¾ G ç ÚÙ R S È » S R is Ú Ä , in truss is çËS and in truss is zero. Õ ÄOu?Ä u Ä u ÙÙ Ù Ù uÙ × Ù Ú Ú G ¾ ¾ Ä G G » è wE\ » R S È Å SXì.üþ S S » ì.í (tensile), The tension in truss is è Ú º u Ä Ù Ù Ù Ù ÙÙ,Ô ï ï ç R RDì.í (compressive) and in truss is zero. The nodal reaction forces are shown in in truss is Ë u Ä Figure 4.3.
æ
4 .3 B EAM E LEMENTS
79
Ù
Ù
STSDì.í
S%Sðì.í
R ðì.í u R ðì.í u
F IGURE 4.3: Reaction forces for the truss system of Figure 4.2.
4.3 Beam Elements Simple beam theory ignores all but axial strain w and stress » w ( » Young’s modulus) º û along the beam (assumed here to be in the x-direction). The axial strain is given by wý» § , § where û is the lateral distance from the neutral axis in the plane of the bending and is the radius of curvature in that plane. The bending moment is given by
t
» ¼
º
'ûl· è
, where è is the beam
crossectional area. Thus e» º
t
» ¼
º
w?»
&ûÁ· è
»
§
û ¼
§
û
ï
(4.7)
· è
»
§
(4.8) t
where
» ¼
û ï
· è
is the second moment of area of the beam cross-section. Thus,
§
»
and
Equation (4.7) becomes t º
The slope of the beam is
· L
·î
»
û
E»
(4.9)
and the rate of change of slope is the curvature
»
· » ·î
·
ï ·î
» L
ï
§Ù
(4.10)
80
L INEAR E LASTICITY
Thus the bending moment is t
»
·
ï ·î
» L
ï
and a force balance gives the shear force »ç ç
t
·
· · î Õ
ȍ
·î
99
L
(4.11)
99 L
(4.12) ×
and the normal force (per unit length of beam) ·.ç ·î
» ¹
»
· ï ·î
ç
99
Õ
L
(4.13) ×
ï This last equation is the equilibrium equation for the beam, balancing the loading forces ¹ with the axial stresses associated with beam flexure · ï ·Cî
ç
/
·
ï ·î
»e¹ L
(4.14)
ï ï 0 The elastic strain energy stored in a bent beam is the sum of flexural strain energy and shear strain energy, but this latter is ignored in the simple beam theory considered here. Thus, the (flexural) strain energy is
SE »
»
ÚÙ ¼ é ¼ È ÚÙ ¼ é È
@wñ· º
¼ Ï
§
û
·î »
Ñ
è
ï ·
ÚÙ ¼ é È
¼ w
· ï
·î »ÒÚ Ù ¼ é ÕL
È è
·Cî è
99
×ï
·î
where î is taken along the beam and è is the cross-sectional area of the beam. The external work associated with forces ¹ acting normal to the beam and moving through a transverse displacement L
is ¼ é
¹
È
PE »
L
·î . The potential energy is therefore
ÚÙ ¼ È
é
Õ L
99
×ï
·î ç ¼ È
é
¹ L
·î u
(4.15)
The finite element approximation for the transverse displacement must be able to represent L the second derivative 9 9 . A linear basis function has a zero second derivative and therefore cannot L represent the flexural strain. The natural choice of basis function for beam deflection is in fact cubic Hermite because the inter-element slope continuity of this basis ensures transmission of bending moment as well as shear force across element boundaries. ù The boundary conditions associated with the th order equilibrium Equation (4.14) or the equa-
æ
4 .4 P LANE S TRESS E LEMENTS
81
Ú tions arising from minimum potential energy Equation (4.15) (which contain the square of nd derivative terms) are more complex than the simple temperature or flux boundary conditions for the (second order) heat equation. Three possible combinations of boundary condition with their associated reactions are Boundary conditions Reactions
Simply supported zero displacement »bS t L zero moment » 9 9S»3S L (ii) Cantilever zero displacement »bS L 9 3 zero slope » » S L · (iii) Free end zero shear force ç » ç ·î Õ t » 9 9S3 » S zero moment
shear force ç slope » 9 FÕ L × shear force ç t moment
(i)
L
99
×
»3S
For two-dimensional problems, we define the displacement vector
ñò
w
w
ö÷
and stress vector *»
º ñò º º
slope
L
)
4.4 Plane Stress Elements w
displacement
»
Ï
Ö ¥
, strain vector }»
. The stress-strain relation for two-dimensional plane stress: ö÷
e» º
º
Ù
» º
w
ï
çeÁ Ù
»
Õ
çeÁ
ï Õ
Á
Ù Ø
Õ
w
w
ÁOw
Ø
Áyw
×
Ø
×
(4.16)
×
can be written in matrix form Ô»
where
»
gradients by
Ù
ç Á e
ï
ñò
Ù Á S
Á S
S
Ù Ù
S
çÂÁ
ö÷
. The strain components are given in terms of displacement
»
w
»
ÎÏ î
w
w
Î
Ö
»
Î Î
(4.17) ò
ÚÙ /
ÎÏ ò Ø Î Î
Î î
Ö 0
82
L INEAR E LASTICITY
The strain energy is SE »ÒÚ Ù ¼ »
ÚÙ ¼
Vf Ë·.ç
f
ÚÙ ¼ »
·.ç
»
w
Õ
ÚÙ ¼
º
w
Ø
ç Á e
Ù
º
w
ï
w
Ø w
ï Ø
º
Ú
·yç ×
ÁOw@w
ï Ø
Ø Õ Ù
The potential energy is PE »
çÂÁ ×
w
·.ç ï
)
external work »bÚ Ù ¼
SE ç
.f
·.ç ç ¼
fS· è
where represents the external loads (forces) acting on the elastic body. Following the steps outlined in Section 4.1 we approximate the displacement field finite element basis » , Öu» and calculate the strains Ö Ï Ï | |A: | | »
w?»
ÎÏ
î
w »
Î
Î
» ò
Î w »
Ö
ÚÙ
Î Î Î Î
)
with a
|
î Ï:|
| Ö
(4.19)
ò
ÎÏ ò Ø Î /
(4.18)
| Ö
Î
»
î
Î
Î
ÚÙ /
Î 0
Î
|
ò Ï:| Ø
| Ö
î
Î
| 0
or Î
Q»
w ñò
w
w
ñ
ò
Î S
ÚÙ Î
) S
î
Î
» ö÷
|
ò
Î Î From Equation (4.18) the potential energy is therefore ) )
PE »
ÚÙ ¼ ) »bÚ Ù
Õ
¼ ) f
) »
ÚÙ
f
×
f
Õ
×
f a ü·.ç
ç ¼
f
S·
| ÷
î
(4.20)
) )
) ç ¼ è
»3
Ï:| Ö ¥ |
·.ç ç ¼
u
)
ò
Î ÚÙ Î
|
ö
|
f
fF· è
· è
æ
4 .4 P LANE S TRESS E LEMENTS
83
»¼ f a·yç is the element stiffness matrix. We next minimize the potential energy with) respect to the nodal parameters
where
¼
· è
and Ö
»
» where
Ï:|
|
giving (4.21)
is a vector of nodal forces.
4.4.1 Notes on calculating nodal loads If a known stress acts normal to a given surface (e.g., a surface pressure), it mayø be applied by ¾ calculating equivalent nodal forces. For example, consider a uniform load ¹Nì.í}Å applied to the edge of the plane stress element in Figure 4.4a. The nodal load vector in Equation (4.21) has components ø H
¼ »
|
¹
|
·î »Â¹: ¼
·
|
È
(4.22) Ü
where is the normalized element coordinate along the side of length loaded by the constant Ü ¾ ø stress ¹pì.í*Å . If the element side has a linear basis, Equation (4.22) gives ø ø ø »Â¹5 ¼ È H
»Â¹5 ¼ H
ï
È
ø
ø ·
ï
· Ü
Ü
»e¹5 ¼ È »e¹5 ¼ È
ç
ÕÙ
·
ÜC×
Ü
»bÚ Ù
¹5
ø Ü
· Ü
»
ÚÙ
¹:
as shown in Figure 4.4b. If the element side has a quadratic basis, Equation (4.22) gives ø ø Ú ø e ø · e » ¹5 ¼ » ¹: ¼ / ÚÙ ç ç · »bÙ ¹5 Ü Ü ÕÙ Ü × Ü È È H 0 Ô ø ø Ú ù » ¹5 ¼ e · e » ¹: ¼ ç · » ¹5 Ü Ü Õ Ù ÜC× Ü È È ï H ï Ä ø ø Ú » ¹5 ¼ e · e » ¹: ¼ / ç ÚÙ · » Ù ¹5 Ü Ü Ü È È Ü È È H 0 Ô as shown in Figure 4.4c. A node common to two elements will receive contributions from both elements, as shown in Figure 4.4d.
84
L INEAR E LASTICITY
¹pì.í*Å
¾ ø ø
ø
¹5
ø
¹5
^
ï
ïÈ ø
(a)
¹5
ï
(b)
ø
éï ¹
ø ïÈ
¹5 È
ø
¹5
(c)
`
¹
¹
` ^ ` éï
¹¯^
ïÈ
éï ¹
^ ` éï
^ éï
(d)
F IGURE ø 4.4: A uniform boundary stress applied to the element ø ø side in (a) isø equivalent to nodal loads of Ú and Ú for the linear basis used in (b) and to Ú , Ú and Ú for the quadratic ïÈ basis usedï in (c). Two ï adjacent quadratic elements both contribute to a common node in (d), where the element length is now . éï
4.5
Three-Dimensional Elasticity
Consider a surface ¸ enclosing a volume of material of mass density ! . Conservation of linear à momentum over the domain results in the governing equilibrium equations à Ú b Å » S Û» (4.23) º Ø Ý Ù Ý Ý Ä where are the components of the stress tensor ( is the component of the traction or stress º º vector in the th direction which is acting on the face of a rectangle whose normal is in the th direction), and
is the body force/unit volume (e.g., "u»#!%$ ). Note that the notation
has been introduced to represent the derivative. Recall that the components of the linear (or small) strain tensor are )
w
»
ÚÙ
ÕÏ
)Å
Ø7Ï
Å
×
Û»
Ý
Ú
º
Å
»
Î:º î Î
(4.24)
Ù Ý ÝÄ
where is the displacement vector (i.e., is the difference between the final and initial positions of a material point in question). Note: we are assuming here that the displacement gradients are small compared to unity, which is appropriate for many materials in solid mechanics. However, for soft materials, such as rubber or biological tissue, then we need to use the exact finite strain tensor. The object of solving an elasticity problem is to find the distributions of stress and displacement in an elastic body, subject to a known set of body forces and prescribed stresses or displacements at the boundaries. In the general three-dimensional case, this means finding stress components Ô which arises from the conservation of angular momentum) and 3 displacements (» each º º Ï as a function of position in the body. Currently we have R unknowns ( stresses, strains and Ä Ù Ô Ô displacements), but only ï equations ( equilibrium equations and strain-displacement relations). Ä Ô To progress, we require an equation of state, i.e., a stress-strain relation or constitutive law. For a linear elastic material we may propose that the components of stress depend linearly on w . º
`
æ
4 .5 T HREE -D IMENSIONAL E LASTICITY
85
i.e., º
æ » »
w » ½
½
ù À where æ » are the components of a th order tensor, although symmetry of the strain and stress Ú ½ Ù tensors reduces the number of independent components to . Ù If the material is assumed to be isotropic (i.e., the material response is independent of orientation of the material element), then we end up with the generalized Hooke’s Law. º
» °
w »»
Ú'&
w
Ø
(4.25)
or inversely w
»
ÚÙ &
º
ç
Ú&
Ú& °
ÕÄ%°ÆØ
&
×
»» º
where , are Lam´es constants. & ° Note: , are related to Young’s modules and Poisson’s ratio Á by ° & 'Ú & » Õ{Ä%°ÛØ & × °ÆØ ÁQ»
Ú
& °
Õ °ÆØ × Providing that the displacements are continuous functions of position, then Equation (4.23), Equation (4.24) and Equation (4.25) are sufficient to determine the R unknown quantities. This Ù can often work with some smaller grouping or simplification of these equations, e.g., if all boundary conditions are expressed in terms of displacements, substituting Equation (4.24) into Equation (4.25) then into Equation (4.23) yields Navier’s equation of motion. & Ï
Å» »
& Ø Õ Û ° Ø
×XÏ
» Å»
b » S
Ø
Ý o
»
Ú Ù Ý ÝÄ
These equations can be solved for the unknown displacements. Then Equation (4.24) can be used Ä to determine the strains and Equation (4.25) to calculate the stresses.
4.5.1 Weighted Residual Integral Equation Using weighted residuals as before we can write
)
¼ k
Õº
Å
Ø
×XÏ
· à
»bS
(4.26)
is a (vector) weighting field. The are usually interpreted as a consistent set of where » ÕÏ × Ï virtual displacements (hence we use the notation instead of ). Ï L
86
L INEAR E LASTICITY
By the chain-rule Õº
»
Ï ×
Å
Å º
Ï
Øeº
Å
Ï
Therefore, the first term in the integrand of Equation (4.26) can be re-written ¼
Å º
k
Ï
·
¼ »
à
¼ 2
»
k
Õº
º
j k 4
where the domain integral involving “ 2
4
k
4
$· à
4 8
$
jlk
Õº
ç ¼ à
Ï ×
·
Î »
· ¸
k
Å
Ï
k
k
º
º
· à
ç ¼
· ¸ ç6¼
Å
Ï
Å
Ï ·
· à
(4.27) à
” has been transformed into a surface integral
î
º
à
Ï Ð
Î » ¼
·
using the divergence theorem 2
Å
k
»¼
¼
Ï ×
¼
or
Å
( k
·
» ¼ à
j k
( Ð
· ¸
*_) is the outward normal vector to the surface ¸ . where 8 » Ð Thus, combining Equation (4.26) and Equation (4.27) we have
¼ k
º
Ï
Å
·
»¼ à
k
»
¼
·
Ï
Ï
k
à¥Ø ·
à¥Ø
¼
º
jlk
¼
jlk
´
Ï
Ð Ï
· ¸
· ¸
(4.28)
E
where are the components of the internal stress vector ( ) and are related to the components of ´ the stress tensor ( ) by Cauchy’s formula º
E
» º
Ð
)
(4.29)
To arrive at this point, we have used weighted residuals to tie in with Chapter 2, however Equation (4.28) is more usually derived using the principle of virtual work (below). Note that the weighted integral Equation (4.28) is independent of the constitutive law of the material.
4.5.2
The Principle of Virtual Work
The governing equations for elastostatics can also be derived from a physically appealing ) argument. Let + be the external traction vector (i.e., force per unit surface area). For equilibrium, the work ) done by the external surface forces + » - ) , in moving through a virtual displacement » Ï
æ
4 .5 T HREE -D IMENSIONAL E LASTICITY
87
) E is equal to the work done by the stress vector » ) in moving through a compatible set of virtual ´ displacements . In mathematical terms, the principle of virtual work can be written
¼
-
Ï
j k
· ¸h»
¼
j k
Ï
´
·L¸ÿ»
¼
j k
º
Ð Ï
· ¸
(4.30)
using Cauchy’s formula (Equation (4.29)). The Green-Gauss theorem (Equation (2.15)) is now used to replace the right hand surface integral in Equation (4.30) by a volume integral, giving ¼
-
Ï
j k
· ¸g» ¼
^ k
Å º
Ï
غ
Ï
` Å
·
(4.31) à
Substituting the equilibrium relation (Equation (4.23)) into the first integrand on the right hand side, yields the virtual work equation ¼
º
k
Ï
Å
·
¼ »
à
k
Ï
·
àØ
¼
j k
-
Ï
· ¸
(4.32)
where the internal work done due to the stress field is equated to the work due to internal body forces and external surface forces. Note that Equation (4.32) is equivalent to Equation (4.28) via Equation (4.30). In practice, Equation (4.32) is in a more useful form than Equation (4.28), because the right hand side integrals can be expressed in terms of the known body forces and the applied boundary conditions (surface traction forces or stresses).
4.5.3 The Finite Element Approximation Let
and interpolate the virtual displacements from their nodal values. i.e., Ï 1 » / Ï 0/ ÕÏ 1 × » Î so Å 2/ (4.33) Ï î ÕÏ / × Î » 1 » Å » ÎFÜ ÕÏ / × î 0/ Î 5 Å ± » / ³ × , 6 ÕgÝ w × is the global node number of local node on element w , and Õ43
»#, à
à.-
where
Õ Ï / × the shorthand
Î 2/» has been introduced. 2/ ÎÜ Equation (4.32) gives Substituting this into
-
¼ k 7 8 º
Å»
2/
»
Å»
Î
ÎÜ î
»
· à
¹º Ï
3
5
±/
Å
³
Ñ
»
-
¼ k 7 8
2/
·
àØ
¼
j k87
-
2/
· ¸º¹ Ï
3
5 ±
Å
/ ³
Ñ
88
L INEAR E LASTICITY
and since the virtual displacements are arbitrary we get
¼ k
-
º
7
Å»
0/
Î
»
ÎFÜ î
·
» à
k
-
¼
7
·
2/
¼
à¥Ø
-
j k97
· ¸ º¹
2/
(4.34)
The next step is to express the stress components in terms of the virtual displacements º and their finite element approximation by substituting Equation (4.33) into Equation (4.24) (the strain-displacement relation) and in turn into Equation (4.25) (the generalized Hooke’s law). | which We first introduce the finite element approximation for the displacement field » Ï n | Ï gives w
»
ÚÙ
Î /
Î
î
Õ |nÏ
|
Î
× Ø Î
^
î
`
|
|nÏ
»
ÚÙ
Î /
ÎÜv½ Î 0
ÎFÜ* ½ î Ï
|
|
Î Ø
|
ÎÜv½ Î
ÎFÜ*½ î Ï
|
(4.35) 0
and w »»
»
» Å»
Ï
»
Î
|
ÎFÜ*»½ î Ï
|
ÎFÜ* ½ î Ï
ÎFÜ*½ Î
|»
Thus º
»
°
Î
ÎFÜ*»½ î Ï
|
Ú'&
|»
ÚÙ Î /
Ø
|
Î Ü*½ Î F ÎFÜ*½ Î which, due to symmetry of the stress tensor, simplifies to º
»
°
Î
ÎFÜ*» ½ î Ï
|
Ú&
|»
Ø
|
ÎÜv½ î Ï
|
ÎFÜv½ Î
ÎFÜ*½ î Ï
| 0
|
ÎFÜ*½ Î | ÎFÜ* ½ (4.36) » Å ÎFÜ*½ | ½ Ï î Ø î Î Î 0 where the summation index has been replaced with , but the parenthesis in implies that o ± there is no sum with respect to that particular index. ³ Substituting this expression into Equation (4.34) and simplifying, we get for each element
Ï
|
¼ k 7 8 /
ÎFÜv½ Î Å / °5 ± | ½ ³
Î
ÚÙ Î Ø
Å
° | ½
Î
ÎFÜ* ½ î 0 /
Å»
Ú&
»
ÎFÜ î Î
Ú'&
Ø
Å
| ½
Î
ÎFÜ*½ î 2 /
Å»
Î
ÎFÜ î
» 0
· à
» H
/
(4.37)
where denotes the right hand side terms in Equation (4.34). (Note that there has been some / carefulH manipulation of summation indices with the substitution of Equation (4.36) to arrive at Equation (4.37).) So for each element
/
|nÏ
|
» H
/
æ
4 .6 L INEAR E LASTICITY
WITH
B OUNDARY E LEMENTS
89
where ø
/
|
H
»
/
»
¼Á¼Á¼ È ø /
¼Á¼Á¼ È
°
2/
»
ÎFÜ* ½ F Î Ü î î Î Î (;ÕÜ
ø
Ú&
»
Ø
ÎFÜ*½ Î Ü î î Î Î
0
· ø· · ÝðÜ ðÝ Ü È × Ü Ü Ü ï ï
ø
Å Å» ( ÕÜ ; | ½2 /
ø
· ø· · ÝsÜ sÝ Ü È × Ü Ü Ü ï ï
È
¼Á¼ È Ø
:
-
2/ (
ï
È
(4.38)
ø · ø· ÕÜ sÝ Ü × Ü Ü ï ï
ø and : ø have been used to transform volume and surwhere the Jacobians (lÕíÜ ÝðÜ ÝðÜ È × ( ÕíÜ ÝsÜ × face integrals so that they can using -coordinates. (Note: without loss of ï be can beï calculated ï Ü generality, the above definition of assumes that ø are defined to lie in the surface ¸ .) ÝðÜ × / approximationÕíÜ leads So in summary, the finite element ï to element stiffness matrix components H that can be calculated from the known material parameters, the chosen interpolation functions, and the geometry of the material (note that the element stiffness components are independent of the unknown displacement parameters). Element stiffness components are then assembled into the global stiffness matrix in the usual manner (as described previously). Note that this is implicitly a Galerkin formulation, since the unknown displacement fields are interpolated using the same basis functions as those used to weight the integral equations.
4.6 Linear Elasticity with Boundary Elements Equation (4.28) is the starting point for the general finite element formulation (Section 4.5). In the above derivation, we have essentially used the Green-Gauss theorem once to move from Equation (4.26) to Equation (4.28) (as was done for the derivation of the FEM equation for Laplace’s equation). To continue, we firstly note that º
w
ÚÙ »
ÚÙ
»
» »
ÚÙ
where w
º
º
Ï
Ï
Ï
Å
Ø
º
Ø Ø
ÚÙ
Å
Ï
Å
ÚÙ
Å
ÚÙ
Å
º
º
Ï
Å
º
º
Ï
Ï
Å
are the virtual strains corresponding to the virtual displacements.
90
L INEAR E LASTICITY
Using the constitutive law for linearly elastic materials (Equation (4.25)) we have ¼ º k
Å
Ï
·
¼ »
à
k
»
º
à
w »» w
° k
w »» w »»
¼
° k
»c¼
·
¼
»
w
·
w
º
k
·
·
Ú'& àØ k
Ú& àØ k
w w
¼
à
w w
¼
·
· à
à
due to symmetry. Thus from the virtual work statement, Equation (4.28) and the above symmetry we have
¼ k
Ï
·
à¥Ø
¼
Ï
´
j k
·L¸ÿ»¼ k
Ï
·
à¥Ø
¼
´
jlk
Ï
· ¸
(4.39)
This is known as Betti’s second reciprical work theorem or the Maxwell-Betti reciprocity relationship between two different elastic problems (the starred and unstarred variables) established on the same domain. 3 » S ). Therefore Equation (4.39) can be written as Note that »ç Å (i.e., Å º Ø º
¼
k
^ º
Å
`
Ï
·
àØ
¼ k
Ï
·
» à
¼
j k
´
Ï
· ¸ ç ¼
j k
´
Ï
· ¸
(4.40)
represents the equilibrium state corresponding to the virtual displacements ). w Ý Ý´ Ï Note: What we have essentially done is use integration of parts to get Equation (4.28), then use it again to get Equation (4.39) above (after noting the reciprocity between and w ). º Since the body forces, , are known functions, the second domain integral on the left hand side of Equation (4.40) does not introduce any unknowns into the problem (more about this later). The first domain integral contains unknown displacements in and it is this integral we wish to à remove. We choose the virtual displacements such that
(
º
º
Å
Ø
w
»S
(4.41)
(or equivalently ç w »ëS ), where w is the th component of a unit vector in the th direction " Ø and w »3w ç . We can interpret this as the body force components which correspond to a Õ ÖÛ× positive unit point load applied at a point in each of the three orthogonal directions. Öüßhà Therefore
æ
4 .7 F UNDAMENTAL S OLUTIONS
¼
º
Å
k
Ï
·
91
» à
ç ¼
"
ç
=Õ k
w
Ö×
Ï
·
»ç à
Ï ÕMÖ×
w
i.e., the volume integral is replaced with a point value (as for Laplace’s equation). Therefore, Equation (4.40) becomes Ï
ÕMÖ×
w
»
¼
j k
´
· ¸ ç ¼
Ï
j k
´
Ï
· ¸
If each point load is taken to be independent then »
Ï
Ï »
´
´
Ï
ÕMÖËÝ × î ÕMÖËÝ ×
´
w
·
Ï
k
and
î
¼ Ø
à
(4.42)
Öüßgà
can be written as
(4.43)
w
(4.44)
where î and î represent the displacements and tractions in the th direction at î Ï ÕMÖËÝ × ´ ÕÖ¤Ý × corresponding to a unit point force acting in the th direction ( w ) applied at . Substituting these Ö into Equation (4.42) (and equating components in each w direction) yields
Ï
ÕÖÛ×
»c¼
j k
Ï
î î · ¸ î ç6¼ ÕÖËÝ . × ´ Õ × Õ × j k ´
î ÕÖ¤Ý S × Ï
î · ¸ î Õ × Õ × Ø
¼ k
Ï
î ÕÖËÝ ×
î · î Õ × à¥Õ ×
(4.45)
where (see later for ). Öüßgà Öüßg 1 ÎUà This is known as Somigliana’s identity for displacement.
4.7 Fundamental Solutions Recall from Equation (4.41) that º
satisfied
"
Å º
ØeËÕ
or equivalently
3 » w
Navier’s equation for the displacements Õ
Ï 1
Å» »
ç
Ï
"
ç
Õ
»bS
ÖÛ×
is Õ
Ø
ÖÛ×
w
Ù
ç
Ú Á
Ï
» Å»
Ø
b » S
Somigliana was an Italian Mathematician who published this result around 1894-1902.
(4.46)
92
L INEAR E LASTICITY Õ
where = shear Modulus. Thus satisfy Ï
Õ
Õ
Å» »
Ï
Ø
Ú ç
» Å»
Ï
Á
"
ç
Øe Õ
w
ÖÛ×
»S
(4.47)
Ù The solutions to the above equation in either two or three dimensions are known as Kelvin 2 ’s fundamental solutions and are given by "
Ï
ÕMÖËÝ
» ×
Ù ã
Õ
çÂÁ
ù ç
ê= Ø
(4.48)
× Ù Ô ÕÙ , for three-dimensions and for two-dimensional plane strain problems,
"
Ï
ÕÖ¤Ý
»
×
ç
Ànã
Ù
çeÁ
Õ
ÕÙ
; Õ{Ä
×
ù ç
Á
×
ÂÄèélêEçóêÆÅ êÆÅ >=
(4.49)
and ç
ê Ú Î ç ç Á ê¢Å çóêÆÅ (4.50) ´ ÕÖËÝ × ÕsÕ Ù × Ø@? × ÕÙ × Õ Ð Ð ×,¡ çÂÁ ê Î Ð U × â ÕÙ Ú A B Ú where » for two-dimensional plane strain and three-dimensional problems respec» ÝÄ â Ù Ý ? tively. " " " Here ê ê , the distance between load point ( ) and field point ( ), ê »Ôî ç î M Õ Ë Ö Ý × Ö Õ × ÕÖÛ× ê ê Î " » and Æê Å » . î ê " Î Õ × In addition the strains at an point due to a unit point load applied at in the th direction are Ö given by
"
»
w ¼»
ù
ã
"
ÕMÖËÝ
×
»
Ù
ç ã
À
Ù
çeÁ
â ÕÙ and the stresses are given by
º
»
"
ÕÖ¤Ý
×
»
ù ã
Ú
ç
ç Ù
çeÁ
Õ
ê
×
ê
Á
Ú ç
; ÕÙ
ç
; ÕÙ
êÆÅ êÆÅ
Ú Á
Á
×£Õ
× Õ
êÆÅ »
êÆÅ »
Ø
× â ÕÙ where and are defined above. ? â The plane strain expressions are valid for plane stress if
êÆÅ Ø
êÆÅ
Á
»
»
×
çóêÆÅ
çóêÆÅ
¼»
»
êÆÅ ê¢Å ê¢Å »'=
Ø@?
× ØC?
is replaced by
Á
ê¢Å êÆÅ êÆÅ »D=
»
Á Á
(This is a
Ù Ø differences. mathematical equivalence of plane stress and plane strain - there are obviously physical What the mathematical equivalence allows us to do is to use one program to solve both types of problems - all we have to do is modify the values of the elastic constants). Note that in three dimensions Ï 2
3 »
/ËÙ
ê
0
´
\ » Ó/Ù
ê
ï 0
Lord Kelvin (1824-1907) Scottish physicist who made great contributions to the science of thermodynamics
æ
4 .8 B OUNDARY I NTEGRAL E QUATION
93
and for two dimensions Ï
3 »
Õ
ÂÄèAé;ê ×
´
\ » /
ê
Ù u 0
Somigliana’s identity (Equation (4.45)) is a continuous representation of displacements at any point . Consequently, one can find the stress at any firstly by combining derivatives Öìߥà Öߥà of (4.45) to produce the strains and then substituting into Hooke’s law. Details can be found in Brebbia, Telles & Wrobel (1984b) pp 190–191, 255–258. This yields
º
¼½
ÕMÖ×
» ¼ Ø
k
¼»
Ï
Ï
»
"
ÕÖ¤Ý
×y´ "
ÕÖËÝ
×
»
" »
Õ
· ¸ ×
Õ
"
Õ
ç ¼½ ×
´
¼»
"
ÕMÖËÝ
×SÏ
»
"
Õ ×
· ¸ Õ
"
×
"
· ×
"
àÕ
×
Note: One can find internal stress via numerical differentiation as in FE/FD but these are not as accurate as the above expressions. Expressions for the new tensors » and ¼» are on page 191 in (Brebbia et al. 1984b). Ï ´
4.8 Boundary Integral Equation Just as we did for Laplace’s equation we need to consider the limiting case of Equation (4.45) as is moved to . (i.e., we need to find the equivalent of æ (in section 3) - called here æ .) Ö Îáà ÕÖ× ÕÖÛ× We use the same procedure as for Laplace’s equation but here things are not so easy. If we enlarge to 9 as shown. ÖßgÎUà à à
à
¸N¿ 9 ã
¸\¾ ¿ Ö à
F IGURE 4.5: Illustration of enlarged domain when singular point is on the boundary.
94
L INEAR E LASTICITY
Then Equation (4.45) can be written as
Ï
» ½nÌ&Í ¼ ½ Í
ÕMÖ×
"
Ï
ÕÖËÝ
}
"
"
· ¸
×.´ Õ ×
Õ
ç ½AÌ&Í ¼ ½ Í ×
"
´
ÕÖËÝ
}
"
×SÏ
Õ
×
¼ Ø F k E Ï
"
· ¸
Õ × "
ÕÖ¤Ý
×
"
Õ
"
· ×
(4.51)
àÕ ×
Ç We need to look at each integral in turn as S (i.e., À S from above). The only integral that ã ã presents a problem is the second integral. This can be written as ½AÌÍ ¼ ½ Í ´
}
"
ÕMÖËÝ
×SÏ
"
Õ
"
×
· ¸ Õ
¼½ Í »
×
"
´
ÕMÖËÝ
"
×SÏ
Õ
"
· ¸ ×
Õ
× ½A¼ Ì&Í
Ø
´
"
ÕÖ¤Ý
×SÏ
"
Õ
×
"
· ¸
(4.52)
Õ ×
The first integral on the right hand side can be written as ¼½ Í ´
"
ÕMÖËÝ
×SÏ
"
Õ ×
· ¸ Õ
"
×
» ¼½ Í G
´
"
ÕÖ¤Ý
× Ï Ã
È
"
Õ
HI
ç ×
Ï
by continuity of K
ÕÖÛ×
±
Ç
³
Ø7Ï
·L¸ î Õ × J
ÕMÖ×
¼½ Í ´
"
ÕMÖËÝ
×
"
· ¸
(4.53)
Õ ×
Let æ
As ã
Ç S
, ¸x¾ ¿4À
ÕMÖ×
»
Ø
¿MÃÄÇÉÅ È ¼½ Í ´
"
ÕMÖËÝ
×
What is a Cauchy Principle Value?
Consider LNMPO%QRS on TUV0RXWZY2[]\^Y`_aQb<M _]\4[dc
O
"
(4.54)
Õ ×
¸ and we write the second integral of Equation (4.52) as ¼ ½
where we interpret this in the Cauchy Principal Value3 sense. 3
· ¸
´
"
ÕMÖËÝ
×XÏ
"
Õ ×
·L¸ Õ
"
×
æ
4 .8 B OUNDARY I NTEGRAL E QUATION
Thus as æ
ÕÖÛ×LÏ
Ç ã
S
95
we get the boundary integral equation ¼½
ÕMÖ× Ø
´
"
ÕÖËÝ
"
×XÏ
Õ
× ¼½
»
(or, in brief (no body force), æ
Ï
Õ ×
´
"
Ï
¼½ Ø
"
· ¸
ÕMÖËÝ
Ï
"
· ¸
×.´ Õ ×
·L¸h»
¼½ Ï
´
"
Õ
¼
× Ø
Ï k
"
ÕMÖËÝ
×
"
Õ ×
· à
(4.55)
·L¸ ) where the integral on the left hand
side is interpreted in the Cauchy Principal sense. In practical applications æ and the principal value integral can be found indirectly from using Equation (4.55) to represent rigid-body movements. The numerical implementation of Equation (4.55) is similar to the numerical implementation of an elliptic equation (e.g., Laplace’s Equation). However, whereas with Laplace’s equation the unknowns were and ÎÏ (scalar quantities) here the unknowns are vector quantities. Thus it is Ï ÎÐ with matrices instead of indicial notation. more convenient to work i.e., use ) ø ø E Ï ´ » » ñò Ï ö÷hÝ ñò ´ ö÷ ï ï Ï È ´ È ) »
ñò
e UV
f]g*h
LiMjOQ`k*OlR
O
S k*O.n
e m O
ø Ï
È
ø Ï
Ï ï Ï ï« ï
Ï ø Ï ï È
Then
e
ø«ø Ï ø
Ï ï
E È È ö÷
Ï ïÈd È
»
Ý
ø«ø ø ï ø
´ ñò ´ ´ È
ø ´
ø
ï
´ È
´
ï« ï
´ È
ï
´ È ö÷ ï ´ ÈdÈ
UV m S k*OoRCpZqNr Osrtr U%m nXpZqNr Osrtr V
V U m Rupvq`_wY®aÔ }'¬^Ô (1.6)). ® element® (see Figure 1.9 ¼( Ï ¼ ¢ Ï Z Ï ¢ ¬Ö¬ ¢ ¼ }'¬^Ô ( ¼ }®*Ô }' ¬ }>® ¿ ¢ ¼ Ï Ï Ï
MM þ Ï
MM þ Ï Ñ Ñ Ñ Ñ M x x MM x M ® ® M ¥ ¥ ¥ ®Ö®.¢ } ¬ ¼ }>®aÔ } ¬ }® À ¢ Ï Z Ï ¢¥ Ï M Ï x Ñ ®
and . ã similarly Ï Z Ï ¢ ¬®.¢#  }¬ ¼ }'¬^Ô ( ¼ }'¬^Ô } ¬ }® ¢ ¥ ¼ ¥ Ï ¼ Á Ï and ( Ï Ï x x Ï x Ñ ® Ñ Ð Û ¬ ¢ ¼ }'¬^Ô }>® ( ¼ }>®a Ô } ¬ }® ¢ Ï ( ® ¥ ¥ Ï Ï ä Ï x x Ñ Ñ
and . and similarly Û ¬ ¢ }¬ ¼ }'¬^| ( Ô }® ¼ }>®a Ô } ¬ }®.¢ Ï ¿ ® Ï Ï Ñ Ñ
. and similarly ¥¤ ¬ ¬ ¬ ¬ ¬ ò ò ò ¬ Ä ¬ Ä ¬ ¬ ¬ ¬ ¬ ò ò ò ¬ Ä ¬ Ä ¢ ©ª à ¬ ¼ ¬Å ¬ ¬ ¬ ª ©
ò ò ò ¬ Ä ª« ¬ ¬ Ä ¬ ¬ ª« ¬ ¬ ¬ à p mass lumping
¬ ÅÄ ò ò ò ¬ Ä therefore u ²° ±± ²° ±± Ã
Å
Å
Ã
Ï Á
¯ ¬ ¢ ò ) The element mass is effectively lumped at the element vertices. Such a scheme has computa tional advantages when in Equation (5.26) because each component of the vector is obtained directly withoutS the need to solve a set of coupled equations. This explicit time integration scheme, however, is only conditionally stable (see (5.34)) and suffers from phase lag errors - see below. For evenly spaced elements the finite element scheme with mass lumping is equivalent to ¬ ® ¬ finite differences with central spatial differences. ¢ ÷ ¢ ò ÇC¢ ò In Figure 5.4, the finite element and finite differences (lumped f.e. mass matrix)Ï ¢ solutions of the ò ÇÆ Æ one-dimensional advection-diffusion equation (5.20) with ] , , ø are compared for the propogation and dispersion of an initial unit mass pulse at . The length of the solution domain is sufficient to avoid reflected end effects.® x ¼ Ô increases with time: The exact solution is a Gaussian distribution whose variance Æ Ñø ÷ ù ÔN¢ ] ñ Ñø ð ñ È Ü ñ + (5.38) ð ñ
The finite element solution, using the Crank-Nicolson-Galerkin technique, shows excellent amplitude and phase characteristics when compared with the exact solution. The finite difference, or lumped mass, solution also using centered time differences, reproduces the amplitude of the pulse very well but shows a slight phase lag.
5.5
CMISS Examples Ï
1. To solve for the transient heat flow in a plate run CMISS example ¥¥ 2. To investigate the stability of time integration schemes run CMISS examples ¥¥
Ð
Ï
and ¥¥
Ð%Ð
.
5 .5 CMISS E XAMPLES
109
Æ Ñø
¶
ù Ô ñ
¨eª¯« ª
¶
´
Ì Ê
¨[±Í(Ê
