Lecture 8
Lecture annotation
Green’s tensor. Maxwell-Betti’s reciprocal theorem. Somigliana’s formulas. Fundamental
solution of the equations of elastostatics. Derivation of boundary integral equations of mixed
boundary problem of elasticity.
Introduction
It was shown in the preceding lecture, that BEM can be obtained using the integral (global,
weak) form of differential equations and boundary conditions. Such approach can be
understood as a special case of the weighed residue method , in which as weight functions the
fundamental solutions of the differential equation are chosen. From the point of the classical
mechanics, the BEM can be derived also by using Maxwell-Betti’s reciprocal theorem. In the
elasticity, such approach facilitates understanding the principles of the method, and therefore
it will be used in this lecture.
BEM consists in an application of the theory of FEM to the numerical solution of boundary
integrqal equations. In a conventional treatment of BEM, which is closely related to the
collocation method, the BEM leads to nonsymmetric system of linear algebraic equation for
unknown quantities at the nodes of elements into which the boundary of domain is divided. If
the BEM is combined with a suitable variational principle (or with a suitable energetic
method such as Galerkin’s method), a symmetric system of linear algebraic equations is
obtained, which can be reduced to the form corresponding to the condensed FEM (after
elimination of all inner degrees of freedom).
As a priority of the BEM to FEM, the lower number of unknowns, better accuracy of
approximation of higher derivatives, and an easy analysis of infinite domains are usually
considered. In the solid mechanics, BEM solutions have been found to be quite accurate in
problems such as the stress concentration or crack problems. There is a range of problems
where is useful to combine both methods. The symmetric variant of BEM simplifies this
combination.
Somigliana’s formulas
Earlier we have shown that the following system of equations is to be solved in any linear
elasticity problem:
 σ  F  0,
in , 

(1)
in ,  field equations
ε  T u

σ  Cε  ε0  in , 
ug
on u ,
(2)
 boundary conditions
t  σ n  p
on  . 
where ε0 stands for an initial deformation and the remaining terms have already been defined
see Lecture 4.
Combining the equations (1), the matrix notation of Lame’s equations follows
 C  T u  ε0   F  0 .
(3)
Remark 1. In the following, only an isotropic medium will be assumed. Then the tensor of elastic moduli C is
expressed e.g. using the shear modulus G and Poisson’s ratio , in component notation written as
1
Cijkl 
2G
ij  kl  G  il  jk  ik  jl  .
1  2
In the BEM, does not matter if is interpreted as a special case of the weighted residual method
or as a consequence of the Maxwell-Betti’s theorem, the fundamental solution U* satisfying
Lame’s equation
(4)
 C T U*  I  x   0 in  .
is of primary importance. ∞ denotes an infinite domain, I stands for the unit matrix and  is
the Dirac delta function, which is equal to zero everywhere except of the origin x = 0, where it
goes to infinity. Since
(5)
  (x)d  1 ,

the function I(x) represents three independent loading states caused, in sequence, by three
unit concentrated forces acting at the point x = 0 always in the direction of one of the
coordinate axes x1 = x, x2 = y, x3 = z. Fig. 1 illustrates the first loading state F1*  1 . The
*
*
*
corresponding displacements are u11
, u12
, u13
, and form the first row of the (33) matrix U*
The component notation of a general element of this matrix

ri rj 
1
(6)
uij* 
(3  4 ) ij 

16 (1  )Gr 
r r
expresses so-called Kelvin’s solution, where ij is the Kronecker symbol. The solution (6) is
sometimes referred to as Green’s tensor ij (Green’s function) or the fundamental solution of
the isotropic elasticity and it is derived in the Appendix 2. (back to Lecture 9-10)
Fig. 1: Unit loading state
Observe that in the local coordinate system in Fig.1, there is ri = xi, hence r xi = xi r = ri r .
These relations will be used for the computation of the stress state caused by the unit
concentrated force F1*  1 at a point determined by the radius vector r. Begin with Hooke’s
law into which substitute the strain-displacements relations:
 u u 
u
2G
 jk 
 jk l  G  j  k 
(7a)
 x

1  2
xl

x
k
j



and substitute for uj Kelvin’s solution uij
2

*
ijk
 uij* uik*
uil*
2G

 jk
G

 x
1  2
xl
 k x j

 

(7b)

rj
 ri
rk 
ri rj rk 
(1  2 )    jk   ik   ij   3
.
r
r 
r r r
 r



The notation  ijk
is to be understood in the following way:  ijk
is referred to the stress tensor
1

8 (1  )r 2
component  jk , which is generated by the unit concentrated force acting in the direction of xi
axis. Consider at the end point of the radius vector r a facet, with unit outer normal vector n,
see Fig. 2. This facet is a part of virtual surface. The traction vector acting upon this virtual
surface from the “outer part” reads as
*
tij*   ijk
nk 

1
8 (1  )r 2
 dr 
 rj
ri rj 
ri  
 (1  2 ) ij  3
  (1  2 )  ni  n j   ,
r r
r  
 dn 
r
(8)
Fig. 2: The surface tractions generated by unit concentrated force
where dr/dn = (r xl )nl = (rl/r) nl. The equation (8) defines a general element of (33) matrix
T*.
In the 2D problem, the matrices U* and T* are of (22) type. Here, we show their elements
assuming the case of the plane strain.
Kelvin’s solution has then the form

1
r r 
uij* 
(3  4 ) ln( 1 / r ) ij  i j  ,
(9)

8 (1   )G 
r r
wherefrom

rj
 ri
1
rk 
ri rj rk 
*
(10)
 ijk

(1  2 )   jk   ik   ij   2

4 (1  )r 
r
r 
r r r
 r
 dr 
 rj
ri rj 
ri  
1
(11)
 (1  2 ) ij  2
  (1  2 )  ni  n j   .
4 (1  ) r  dn 
r r
r  
r
In the case of plane stress, Poisson’s ratio  is replaced in the preceding equations by the
constant    (1  ) .
tij*  
Remark 2. Note that Kelvin’s solution of Lame’s equation plays the same role as the fundamental solution 1/r,
or ln(1/r) respectively, in the case of Laplace’s and Poisson’s equations.
3
*
The transcription of the tensor of the third order  ijk
into the matrix * has to correspond to
the matrix notation of the tensor of the second order jk in the form of column matrix. In the
case of 3D problem, the (36) matrix * is set up such that the vector σ *i  =  i*11 ,  i*22 ,  i*33 ,
 i*23 ,  i*31 ,  i*12  enters its i-th row. In the case of 2D problem, we have (23) matrix *.
The matrices U*, *, and T* depend on the position of two points. Namely, each of their
elements is related to the initial and ending points of the radius vector r. In the local
coordinate system shown in Figs. 1 and 2, the initial point lies at the origin 0 and the ending
point has the coordinates of the point x. Though there is suitable to use the local coordinate
system for the BEM implementation, two-points character of matrices is not so apparent. The
more illuminating representation of two-points (two arguments) dependence is obtained in the
global coordinate system shown in Fig. 3. Here, the point at which the “cause” is introduced
(unit force Fi *  1 ) is denoted by x = (x1, x2, x3), and the point, where the response is
observed, is denoted by  = (1, 2, 3).
Start with the derivation of the boundary integral equations, which form the basis of the BEM.
Fig. 3 shows two states. The solid contour refers to a real state; the dash contour refers to a
fictitious state (the domain  was removed from ∞).
T
Fig. 3: Real and fictitious state
According to the Maxwell-Betti’s theorem, the work of real forces upon fictitious
displacements is equal to the work of fictitious forces upon real displacements. The latter
consists of two contributions– the work of fictitious tractions upon real displacements and the
*
work of fictitious stresses  ijk
upon the initial deformations (-  0 jk ).
In the component notation this theorem reads:
*
1 ui (x)   tij* (x, ξ ) g j (ξ )d(ξ )   tij* ( x, ξ )u j (ξ )d(ξ)    ijk
( x, ξ) 0 jk (ξ)d(ξ) 
u


(12)
  uij* (x, ξ )t j (ξ )d(ξ )   uij* ( x, ξ ) p j (ξ )d(ξ)   uij* ( x, ξ) Fj (ξ)d(ξ).
u


Wherefrom we express
ui (x)   uij* (x, ξ )t j (ξ)d(ξ)   tij* ( x, ξ)u j (ξ)d(ξ)  uiI ( x) ,
u

where
4
(13)
uiI (x)   uij* (x, ξ) p j (ξ)d(ξ)   tij* (x, ξ) g j (ξ)d(ξ) 

u
*
  uij* (x, ξ) Fj (ξ)d(ξ)    ijk
(x, ξ) 0 jk (ξ)d(ξ).

(14)

The notation in Eq. (13) represents three equations for unknown displacements ui (i = 1, 2, 3)
in the point x. The displacements uiI(x) are completely determined by the relation (14). If the
boundary tractions tj(x) on u (reaction of supports), and the boundary displacements uj() on
( –uj() are called distorsions) were known, the displacement field inside  would be also
determined.
The calculation of tj on u, and uj on , which is the first step of BEM analysis, will be
skipped for a moment, and the attention will be focused to the final stage of the analysis – the
calculation of the stress field. We start with the formula (7a), which will be supplemented
with the terms describing the influence of the initial deformation:
 u u   2G
u
2G

 ij (x) 
 ij m  G  i  j   
 ij  0 mm  2G 0ij  .
(15)


1  2
xm

 x j xi   1  2
For displacements, we substitute the expressions in Eqs. (13) and (14). The calculation is
rather lengthy and will not be presented in details here. We just present the final formula with
several useful instructions:
*
 ij (x)   dijk* (x, ξ)tk (ξ)d   sijk
( x, ξ)uk (ξ)d   ijI ( x) .
(16)
u

*
It can be easily verified that dijk
(x, ξ) =   (x, ξ) . The exchange of indices is explained by
the comparison of Eq. (7) with the relation
 uki* ukj* 
2G
ukl*
*
,
d ijk 
 ij
 G

(16a)
 x

1  2
xl

x
i 
 j
which is obtained by applying the relationship (13) in (15), and by considering the symmetry
of functions uij* = u *ji according to (6). The summing index j in Eq. (13) had to be changed to
k. The sign conversion relates to the transition from the local to global coordinates, where
rl xk = (l  xl ) xk = –lk. Interchanging the arguments x and we finally obtain a
symmetric relationship between both tensors
*
*
*
(17)
dijk
(x, ξ)   kij
(x, ξ)   kij
(ξ, x) .
*
kij
*
The formulation of the tensor sijk
is similar; substitute in the right-hand side of Eq. (16a) tlk*
instead of ulk* etc. Obtain
*
sijk

G
2 (1  )r 
 dr 
 rj
rk
ri 
ri rj rk 

(1  2 ) ij    ik   jk   

r
r
r
r r r

 dn 
 rj rk
r r 
   ni
 nj i k  
r r
 r r
(18)



r rj
(1  2 )   nk i  n j ik  ni kj   (1  4 )nk  ij  .
r r



In the case of 3D problem, substitute  = 2,  = 3,  = 5. In the case of 2D problem (plane
*
*
strain),  = 1,  = 2,  = 4. For the plane stress replace  in the expressions for dijk
and sijk
again by    (1  ) .
Likewise uiI, also the functions ijI(x) represent the response of domain ∞ to prescribed
actions. They can be computed by combining Eqs. (14) and (15). It results in
5
*
 ijI ( x)   dijk* (x, ξ ) pk (ξ)d   sijk
( x, ξ) g k (ξ)d 

u
*
*
  dijk
(x, ξ ) Fk (ξ )d    ijkl
( x, ξ ) 0 kl (ξ )d.

(19)

*
The underlined terms deserves a deeper analysis. Namely, the tensor  ikl
x j is singular at
the point x = , which does not allow to change the order of differentiation and integration,
that was used in remaining terms (the singularity is immediately obvious if Lame’s equations
*
(4) are re-written into Cauchy form  ijk
xk + ij(x – ) = 0, i,j = 1, 2, 3.). The interchange
of differentiation and integration leads to the occurrence of an additional term, which follows
from the relation
*
*
 ikl
 ikl

*


d



d


(20a)
ikl 0 kl
 x j 0kl
 x j  0kl d .
x j 
Focus on a plane problem; then  stands for a circle of radius  → 0. Apply Gauss’s
theorem and transform the second term in (20a) to the boundary integral. Consider in Eq (10)
that on the circle boundary there is r = , ri = i , hence ni = i /. Using formulas
 j
 j  

(20b)
 i  d  ij , 2 i  k l  0kl d  2  ij 0mm  2 0ij  ,
which can be easily proved by substitution i/ = cos, we get
*
 ikl

1 
1


*


d


 0 kl d  
(3  4 ) 0ij   2    ij 0 mm .
(20c)
ikl
0
kl



 x
x j 
4(1  ) 
2


j
Remark 3. Derive the second the equation (20b). The integrand details:
i  j  k  l
  n n n n   n n  n n   2n1n2 012  n2 n2 022  .
    0 kl i j k l 0 kl i j 1 1 011
Following combinations occur:
i  1, j  1: n14 011  2n13n2 012  n12 n22 022 ,
i  1, j  2 : n13n2 011  2n12 n22 012  n1 n23 022 ,
i  2, j  1: n13n2 011  2n12 n22 012  n1 n23 022 ,
i  2, j  2 : n12 n22 011  2n1 n23 012  n24 022 .
Integration along  leads to
2
2
1
  
2 n14 d  2   cos 4  d =2   1  2 cos 2  cos 2 2  d =2     ,

4
2 4
0
0
2
2
2 n24 d  2   sin 4  d =2  

0
0
2
1
  
1  2 cos 2  cos 2 2  d =2     ,

4
2 4
2
1

1  cos 4  d =
8
2
0
0
while the remaining integrals are equal to zero. Thus, it holds

i  1, j  1: 2   n14 011  2n13n2 012  n12 n22 022  d      011   022    011
2

2 n12 n22 d  2   sin 2  cos 2  d =2  

i  1, j  2 : 2   n13n2 011  2n12 n22 012  n1 n23 022  d    012 ,

i  2, j  1: 2   n13n2 011  2n12 n22 012  n1 n23 022  d    012 ,

i  2, j  2 : 2   n12 n22 011  2n1 n23 012  n24 022  d   


2
 011   022    022 .
Globally, the preceding results can written in the form of the second equation (20b) 
Remark 4. Derive the second term on the right-hand side of Eq. (20c). Using Gauss’s theorem get
6
*
 ikl

 0 kl d   
 ikl*  0kl  d     ikl*  0kl n j d  ,
  x
x j
j



because
0kl depends only on . From Eq. (10) we get for r = :
*ikl 0 kl n j  


 i

k
l
i  k  l
0 kl  
(1  2)    kl 0 kl  il 0 kl  ik 0 kl   2
4(1  ) 


  
 


nj


k n j
l n j
 i n j

  
1
0 mm 
0 ki 
0il   2n j i k l 0 kl  
(1  2)  
4(1  ) 


  
 


1
(1  2)  ni n j 0 mm  nk n j 0 ki  nl n j 0il   2n j ni nk nl 0 kl  .


4(1  ) 
It follows from the preceding relations
*
 ikl
1
(1  2 )  ni n j 0 mm  nk n j 0 ki  nl n j 0il   2n j ni nk nl  0kl  d  ,

 0 kl d  


  x
4

(1


)

j


wherefrom using (20b) we have
1
 4 (1  )  (1  2 ) ni n j 0mm  nk n j 0ki  nl n j 0il  2n j ni nk nl 0kl  d  



1 
1
1 
1



(1  2 )   ij 0 mm   kj 0 ki   lj 0il    ij 0 mm  2 0ij   
(3  4 ) 0ij   2  2   ij 0 mm  ,
4(1  ) 
2
4(1


)





which completes the proof.

With regard to Eq. (15) we finally obtain
*
*
  ijkl (x, ξ) 0kl (ξ)d    ijkl (x, ξ) 0kl (ξ)d 



(21)
G
 2 0ij (x)   ij 0 mm (x)  ,
4(1  ) 
where
*
*
  ikl
 *jkl 
2G
 mkl


 
 ij
G

 x

1  2
xm

x
j
i 


r r
G
r r 
r r r r

2(1  2 ) i j  kl  k l  ij   8 i j k l 
2 
2 (1   )r 
r r 
r r r r
r r
*
ijkl
(22)
r r

r r
r r
r r
 2  k j  il  l j  ik  k i  jl  l i  jk  
r r
r r
r r
r r

 (1  2 )( il jk   jl ik )  (1  4 ) ij kl .
The formulas (21) and (22) were derived assuming the conditions of plane strain. When
Poisson’s ratio is replaced by  =  / (1 +), they can be used also for plane stress
conditions.. Certain difficulties related to the local term in Eq. (21) can be avoided using two
different methods:
The first method ignores Eq.(16) and the derivative of displacements, standing in (15), are
calculated numerically using Eq. (13). The second method considers the fact that the local
terms drops out if the integration with respect to points  is preceded by the differentiation
with respect to the components of the vector x.

G
 2 0ij (x)   ij 0 mm (x)  in Eq. (21). First of all
4(1  ) 
notice that, basing on Eqs. (14), (20a), and (20c), the expression 
 ikl*  0 kl d , and hence also the second
x j 
term in (20c), have a meaning of the contribution to the displacement derivative ui x j . If, in this sense, we
Remark 5. We account for the origin of the local term 
substitute the second term in Eq. (20c) into (15) and add the term expressing the influence of initial
deformations, we obtain
7
 1 
2G
1 
1

1



 ij
(3  4 ) 0 mm   2    mm 0 kk   2G 
(3  4 ) 0ij   2  2   ij 0 mm  
1  2
4(1  ) 
2
4(1


)








G
 2G

 2 0ij   ij 0 mm  ,

 ij 0 mm  2G 0ij   
1

2

4(1
 ) 


which is actually the second term in Eq. (21), (for 2D problem, mm =2!). The result can also be written in the
form
G
G
 2 0ij   ij  0 mm   

 2ik jl  ij kl   0kl ,
4(1  ) 
4(1  )
hence, with regard to Eq. (21), the tensor
*
 ijkl
can formally be written as
 (x, ξ) =  (x, ξ) – (x – )G(2ikjl + ijkl)/4(1 – ) )
*
ijkl
*
ijkl
*
*
The elements of tensors dijk
and sijk
can be stored into the matrices D* and S* (63 for 3D or
*
32 for 2D), similarly as we sorted the elements of tensor  ijk
into the matrix *. The only
*
difference is that the rows and columns are interchanged. The elements of the tensor  ijkl
are
stored into the square matrix (66) Σ (or (33) in the case of 2D problem).
The integral equations (13) and (16) are the basic relations used in the direct variant of the
BEM and there are called Somigliana’s formulas. For clarity, we write them in the matrix
notation:
u(x)   U* (x, ξ )t (ξ )d   T* ( x, ξ)u(ξ)d  u I ( x),
u

(23)
σ (x)   D* (x, ξ )t (ξ )d   S* (x, ξ )u(ξ )d  σ I (x).
u

Matrix expression for the column matrices uI(x), and I(x) is analogous:
u I (x)   U* (x, ξ )p(ξ )d   T* ( x, ξ )g(ξ)d 

u
  U (x, ξ )F(ξ )d   Σ (x, ξ ) ε0 (ξ )d,
*
*


σ I (x)   D* (x, ξ )p(ξ )d   S* (x, ξ )g(ξ )d 

(24)
u
  D (x, ξ )F(ξ )d   Σ (x, ξ ) ε0 (ξ )d.

*


Remark that with regard to the relation (17), it holds
D* (x, ξ)  Σ*T (x, ξ)  Σ*T (ξ, x).
(25)
Direct variant of BEM
Formulas for boundary point
Hitherto we have assumed that the point x, at which the displacements u are calculated using
the first Somigliana formula (23), is an interior point of the domain . If the point is moved to
the boundary , the equation takes the form
c(x)u(x)   U* (x, ξ)t (ξ)d   T* ( x, ξ)u(ξ)d  u I ( x)
(26)
u

and becomes the starting point for the calculation of unknown displacements u on the part of
the boundary , where loading is prescribed, and of unknown tractions t on the part of the
boundary u, where the displacement is prescribed.
The factor c(x) depends on the shape of the boundary in the neighbourhood of the point x. To
understand its meaning, it should be realized that the term c(x)ui(x) expresses the work of a
fictitious body force density Fi * , which is concentrated at the x, acts in the direction xi, and
has the resultant Fi * = 1 . The factor c(x) expresses a simple fact that in real (bounded) body
8
only a part of the resultant exerts its influence, i.e. the part of body forces which lie in the
domain .
The illustrating Fig. 4 restricts only to 2D case. In a small circular neighbourhood of the point
x, the uniformly distributed body force Fi * = 1/(2) fulfils the condition
1
1
*
2
 Fi d    2 d   2   1 .
Fig.4: Schematics explaining the calculation of the work of fictitious body force
The work of fictitious body force Fi * upon displacements ui is exerted only within the shaded
domain  defined by an angle. Provided that the displacement field is sufficiently smooth
in this domain, it may be considered to be constant in a small ( → 0) of the point x, which
gives
1
1  (x) 2
 (x)
*
 Fi ui d  ui (x)  2 d  ui (x)  2 2  ui (x) 2 ,
wherefrom
 ( x)
c ( x) 
.
(27)
2
According to this formula, the following values of the factor c(x), given in Tab. 1, correspond
to various nodes shown in Fig. 5
node i
1
2
3
4
5
c(xi) 1/(2) 1/2 1/(2) 1/4
1/2
Tab 1: Values of the factor c
6
1/4
In a similar fashion, values of the factor c can be obtained also for a three-dimensional
domain , where there is necessary to distinguish whether a boundary point lies on a smooth
surface, at the edge between two surfaces, or whether it is a node where three and more
surfaces intersect. Note that Eq. (26) remains valid also in interior points x, where, however,
we put c(x) = 2 / 2 = 1. Later, a specific algorithm will be demonstrated which allows to
avoid the calculation of the factor c.
Fig. 5: Boundary nodes of various kinds
9
Discretisation of BEM problem
In BEM, the boundary is divided into boundary elements. For a two-dimensional analysis, the
elements are normally straight lines (constant or linear element) or curves (quadratic or higher
order). For a three-dimensional analysis, the elements are usually quadrilaterals and triangles,
see Lecture 7. Thus, BEM, in contrast FEM, leads to the problem which is one less dimension
than the body being analyzed. Because the development of BEM was preceded by an intense
development of FEM, there has been possible to apply all findings from the FEM to the
geometrical description of BEM. This is particularly concerned the technique of isoparametric
elements.
Fig. 6 shows an example of a division of a boundary of two-dimensional domain  into
quadratic isoparametric elements. Empty circles refer to the division points, full circles refer
to internal nodes.
Boundary traction loading vector is approximated at an arbitrary point  by
t (ξ )  N p (ξ )rp ,
(28)
where Np is the matrix of basis (interpolating) functions and rp is the vector of nodal values of
boundary loading.
The vector of boundary displacements is approximated in a similar manner
(29)
u(ξ)  Nu (ξ)ru ,
with a similar meaning of terms used.
Fig. 6: Quadratic approximation on the boundary
Fig. 6 illustrates two quadratic basis functions. One of them belongs to the point of division of
boundary into elements (vertical hatching). Second of them belongs to internal nodes
(horizontal hatching). The degree of approximation of functions Np and Nu may differ. The
fact, that the stress field, and thus also the loading, is proportional, according to Eqs. (1) and
(2), to the first derivative of displacement field, suggests to use for the approximation of
displacements such polynomials, which are one order higher than those used to approximate
the boundary forces.
The ordering of vectors ru and rp depends on how the boundary is divided into elements. Let
on the part of the boundary u, where displacements are prescribed, be k nodes, and on the
part of the boundary , where loading is prescribed, be l nodes. If the prescribed values are
approximated in the same manner as unknown values, the following ordering of vectors is
obtained for 2D problem

 g
rp  t1x , t1 y ,
ru
1x
, g1 y ,
, tkx , tky pk 1, x , pk 1, y ,
, g kx , g ky uk 1 , vk 1 ,
, pk l , x , pk l , y
, uk l , vk l

T

T
,
.
The preceding component notation can be put into compact matrix form
 u rp 
 u ru 
rp   p  , ru   p  .
 ru 
 rp 
(30)
Applying the approximations (28) and (29) in Eq (26), we obtain its discretised form. Body
forces and initial deformations standing in the vector uI(x) complicate the algorithm of the
BEM. The corresponding integrals are evaluated numerically over cells into which the domain
 is divided. In contrast to FEM, introduced internal nodes do not increase the number of
10
unknowns. Corresponding terms occur on the right-hand side of resulting system of equations.
If these terms are zero, Eq. (26) obtains, by using the first formula in (24), an algorithmic
transparent form
 U* (x, ξ )N (ξ )d  u r   U* (x, ξ )N (ξ )d  p r 
p
p
 u
 p  
 p
 c(x)u(x)    T* (x, ξ)Nu (ξ )d  u ru    T* (x, ξ )N u (ξ )d  p ru .
 u

 

(31)
The scheme (31) enables us to set up a system of linear algebraic equations for unknown
components of vectors u rp and p ru . To do so, the point x has to pass all nodes of the boundary
 = u  In the case of 2D problem, the matrices U* and T* are of the type (22).
Therefore, at each node there are two unknowns (on u two components tx, ty, and on 
displacements u, v), which are found from two equations.
To exploit the partition of the vectors (30) into fields, we write in the same fashion also the
conditioned equations of BEM. They split into two matrix equations
uu
G u rp  up G p rp  uu Hu ru  up H p ru ,
(32)
pu
G u rp  pp G p rp  pu Hu ru  pp H p ru.
The first equation in (32) combines k relations (31) at the nodes on u, which is indicated by
the first left superscript of submatrices ijG and ijH. The second equation in (32) combines
l relations (31) at the nodes on ( the left superscript p. The second left superscript refers to
the partition of the vectors rp and ru according to Eq. (30). The factors c(x) enter the diagonal
of matrices uuH or ppH respectively, according whether the node x lies on the part of the
boundary u or  respectively.
Hence, every diagonal element consists of the sum of two terms. We describe a simple
algorithm how to determine these diagonal elements, which does not require finding the
values of the factor c(x). To this end, rewrite the system (32) in the compact form
Grp  Hru ,
(33)
which represents the most frequently form of the BEM equations when the body forces and
initial deformations are absent. In the case of 2D problem, the matrices G and H are of the
type [2(k + l), 2(k + l)]. Eq. (33) must also satisfy the conditions for the displacement of a
rigid body. If we prescribe at all point the same unit displacements u = v = 1, every
component of the vector ru is equal to unity ru = 1u and it holds
0  H1u .
(34)
It follows from Eq. (34), the diagonal element of the matrix H (and thus also the submatrices
uu
H and ppH) is determined by
(35)
hrr   hrs .
sr
It means that it equals to the negative sum of all remaining elements of r-th row of the matrix
H. By a suitable rearrangement of summands in the system (32), which separates known
quantities from unknowns, we arrive to the following form of BEM equations
 uu G,  up H   u rp   uu H,  up G   u ru   u rI 
(36)
 pu
  p    pu
 p    p .
pp
pp
 G ,  H   ru   H,  G   rp   rI 
The matrix of the system is nonsymmetric and regular, and the right-hand side of Eq. (36) is
known. Hence, using the Gauss elimination, the unknown elements of the vectors urp and pru
can be calculated.
After the determination of all boundary quantities, we can compute stresses in interior points
x of the domain . To this end, apply the approximations (28) and (29) in second
Somigliana’s formula in Eq. (23). Obtain
11
(37)
σ( x)    D* (x, ξ)N p (ξ)d  u rp    S* (x, ξ)Nu (ξ)d  p ru  σ I (x) ,
 u

 

where the last term follows (for body forces and initial deformations absent) from the second
formula (24)
(38)
σ I ( x)    D* (x, ξ)N p (ξ)d  p rp    S* (x, ξ)Nu (ξ)d  u ru .
 

 u

The ordering of elements in matrices *, D*, and S*, (for which the tensor notation formulas in
Eqs. (7),(10), (17), and (18) are available), must correspond to the ordering of elements of the
vector {. Thus, for 2D problem we have
σ   xx ,  yy ,  xy 
T
  11 ,  22 ,  12 
T
(39)
and therefore
*
*
*
*
 111

 s111

,  211
, s112
*
*
 111

,  1*22 ,  112



*
*
*
* 
*
*
*
D    1 22 ,  222  , S    s221 , s222  , Σ   *
.
*
*

,

,

,


211
222
212

*
*
*
*
 ,  
s , s 
 112
212 
 121
1 22 
(40)
Symmetric variant of BEM
The system of equations (1) and (2) describes the mixed boundary value problem of the
theory of elasticity, where stand up both the stress field , and the displacements field u. It
can be considered as a consequence of the general variational principle Hu-Washizu, which is
characterized by the functional
1

1  u u 
I (ε, u, σ )    Cijkl  ij  kl   ij  ij   ij  i  j   Fu
i i  dV 
 2
2  x j xi 


.
(52)
  pi ui dS    ij n j ( gi  ui )dS  I

u

  pi ui dS    ij n j ( gi  ui )dS .

u
The stationary point (, , u), satisfying the equation
 I (ε, u, σ)  0
(53)
is the solution of a given problem, because Eqs. (1) and (2) are the stationary conditions
(Euler’ equations) of the functional I (ε, u, σ ) .
During the discretisation, unknown boundary forces and unknown displacements are
approximated by polynomials as follows
t ( x)  u N p ( x ) u rp
on u
(54)
u(x)  p Nu (x) p ru
on   ,
where uNp and pNu are submatrices of the matrices Np and Nu, introduced in the relations (28)
and (29). Introduce the approximations in Eqs. (54) into Eq.(52) and obtain the modified
functional
(55)
I   I   ( u U  u U)T u rp  p PT p ru ,
where
u
U   u NTp udu , u U   u Ngdu , p P   p NTu pd .
(56)
u
u

The stationary conditions (Euler’ equations) for the functional I  are the equations (1) and the
generalized (global) boundary conditions
u
U  u U on u and p P  p T on   , where p T   p NTu td  .
(57)

The displacements field u(x) and the stress field (x) expressed by Eqs. (23) and (24) satisfy
the equations (1), however, do not satisfy the generalized boundary conditions (57). (Compare
with the boundary procedures, such as Trefftz’s method described in Leture5-6) Just these
conditions stand for the determination of unknown forces urp on u and unknown
12
displacements pru on  in the symmetric variant of BEM. To express these unknown, we
have to know a continuously distributed boundary forces t, which are obtained by combining
the relation t(x) = n(x) (x) with the second equation (23). This completes the calculation of
Somigliana’s formulas.
To emphasize the whole range of symmetries of Somigliana’s formulas, we rewrite them into
the form
u(x)   G uu (x, ξ )t (ξ )d   G up (x, ξ )u(ξ )d  u I (x), ( x   u )
u

t (x)   G pu (x, ξ )t (ξ )d   G pp (x, ξ)u(ξ)d  t I ( x),
u

σ (x)   G u (x, ξ )t (ξ )d   G  p (x, ξ )u(ξ )d  σ I (x),
u

( x   )
(58)
( x  )
Remark 6. Two-points (two-arguments) matrix kernels Ghk(x, ) in equations (58) and (59) are denoted by the
same letter as the numeric matrices in Eq. (32). The have, however, different meaning and are of different type.
To avoid confusion, the locations of indices is different.
The symbol − denotes the set of internal points infinitely close to the boundary . The first
two terms on the right-hand side express the response of domain ∞ to the actions of arbitrary
distributed forces t on the part of boundary u and displacements (–u) on . The last terms
express, in accordance with (23) and (24), the response of ∞ to the prescribed displacements
(–g) on u and to prescribed tractions p on . Here, we present the last terms only in a brief
form, in which the arguments x and  are not written down:
u I   G uu pd   G up gd   G uu Fd   G u  0 d,

u


p I   G pu pd   G pp gd   G pu Fd   G p  0 d,

u


(59)
σ I   G u pd   G p gd   G u Fd   G  0d.

u


As it was already explained at the outset, two-points matrix kernels stand for a fundamental
solution of a given problem. The function Ghk(x, ) expresses h-th influence at the point x
caused by a concentrated source at the point . An influence (relating to the first subscript)
can be the displacement (h = u), the boundary traction (h =t), and the stress (h = ). The index
k refers to a source associated such that the “source exerts work on an influence” i.e. unit
force (k = u), unit displacement (k = p), and unit volume (k = ).
Let us present two important properties of the kernel functions:
a) Symmetry with respect to x and  for h = k. if h ≠ k and x ≠ , then
G hk (x, ξ )  GTkh (ξ, x).
(60)
b) Let  be a surface in ∞. Then
T
(61)
  t (x)Guu (x, ξ)t(ξ)d(ξ)d(x)  0 for arbitrary t  0,
 

u
 
T
(x)G pp (x, ξ)u(ξ)d(ξ)d(x)  0
for arbitrary u  0
(62)
The left-hand side of inequalities apparently express double of the strain energy
corresponding to boundary tractions t or displacements u respectively. The inequality signs in
(62) refers to the rigid body displacement of the enclosed part of of ∞.
The transition to the symmetric variant of BEM is easy. It is sufficient to use the first two
formulas in Eq. (58) to (57) together with the approximations (54). It gives
K uu u rp  K up ( p ru )  u L,
(63)
K pu u rp  K pp ( p ru )  p L,
where
13
K uu  
u
u
K up  
u
u
u
K pp  
NTp (x)  G uu (x, ξ) u Ν p (ξ)d(ξ)d(x)  K Tuu ,
NTp (x)  G up (x, ξ) p Νu (ξ)d(ξ)d(x)  K Tpu ,

p

(64)
NTu (x)  G pp (x, ξ) p Νu (ξ)d(ξ)d(x)  K Tpp ,

u
L
u
u
p
L

NTp (x)[g (x)  u I (x)]d(x),
p
NTu (x)[p(x)  t I (x)]d(x).
(65)
Apparently, the method of deriving the system (63) can be understood as an application of
Galerkin’s method to Eq.(58). As we know, Galerkin’s method consists in multiplication of
the first equation by the matrix p NTu followed by the integration over u and considering that
on this part of boundary is u = g. That way we obtain the first equation (63). The construction
of the second equation is similar.
The system (63) is symmetric; the matrix Kuu is, with regard to (61), positive definite, and Kpp
is, with regard to (62), negative definite. Thus, it is possible to exclude from Eq. (63) the force
parameters urp on u and as a consequence, to reduce the system (63) to the form well-known
in the FEM:
Kr  R
(66)
where
1
1 u
(67)
K  K pp  KTup K uu
K up , R  p L  KTup K uu
L,
and r  p ru .
Basing on the preceding findings, it can be easily seen that the stiffness matrix K of the
domain  is symmetric and positive definite.
The double integration requires an increased consumption of computational time. On the other
side, it smoothes out singularities in the kernels of integral equations, so that it admits an
effective exploitation of so-called hypersingular fundamental solutions, which are otherwise
unemployable. The computation of the double integrals does not bring any special problems.
Numerical integration is in most cases quite satisfactory.
Formally, there exists a simpler algorithm leading to the final formulation in Eq. (66). It
requires performing two simple consecutive integrations, and it is rather sensitive to
truncation errors. Hence, it does not guarantee perfect symmetry of the resulting stiffness
matrix. We start with the system (33), extended for the equilibrium conditions of the body 
as a whole
Qrp  0 .
(68)
If in the case of 2D problem, only the equilibrium force conditions are used, the matrix Q will
have two rows; if the equilibrium moment condition is included, the matrix will have three
rows. The global conditions of equilibrium ensure that the relation between the force
parameters rp and displacement parameters ru obtained by the elimination of the extended
system
G, QT  rp   H 
(69)

      ru 
Q, 0   λ   0 
will be calculated with a required accuracy ( is a column matrix of Lagrange’s multipliers).
The solution of (69) written in the form
rp  Vru
is used in Clapeyron’s theorem (the work of applied forces is equal to the work of internal
forces)
T
T
'
 u td  ru K ru .

14
Applying the approximations in Eqs. (28) and (29) , the preceding equation can be reduced to
the form
ruT  ΝTu (ξ)N p (ξ)d(ξ)Vru  ruT K 'ru ,

wherefrom
K' 
  Ν (ξ)N (ξ)d(ξ) V .

T
u
p
The matrix K’ is, in contrast to K in Eq. (66), positive semidefinite, because it contains the
degrees of freedom, which has the body as a rigid body.
The algorithms described above can be applied to the solution of complex structures by the
method of substructures. It is, of course, also possible to combine the traditional BEM
equations (33) with equations of the FEM. Such approach will no be entered into details here,
because it is less suitable from the point of the resulting algorithm.
The dual boundary integral equations in the fracture mechanics problems
One of the most important stages in the strength assessment of structural components is to
account of the effect of stress redistribution due to the presence of cracks. At the crack tip, the
largest concentration occurs, which stimulates further crack extension. The stress singularity
of r-1/2 within the neighbourhood of the notch with a vertex angle equal to 2 is a universal
property of linearly elastic bodies. The key point is the determination of the coefficient of the
singularity term – so called stress intensity factor K, which must be found by solving the
boundary value problem of the whole body. The stress intensity factor reflects an influence of
range of factors such that the crack length, the shape of body, and the character of loading,
and it does not depend on coordinates around the crack tip. The boundary integral equations
provide a very efficient instrument to determine the stress intensity factor. To formulate the
boundary integral equations for a cracked body, the following procedure becomes useful: The
given boundary value problem with prescribed loading on outer boundary is divided,
according to the principle of superposition, into two partial boundary problems, schematically
shown in Fig. 7.
Fig. 7: Schematic of the principle of superposition
The solution of the first partial boundary value problem provides a regular stress distribution
in the body without crack – hence, this solution does not contribute to the stress intensity
factor value in the original problem. In the second partial boundary value problem, there are
prescribed tractions on the crack faces, which are equal but opposite to those which would
prevail in the uncracked body in the place of a prospective crack. In such a way, the crack
faces in the original problem are ensured to be traction-free. Simultaneously it is obvious that
for the determination of stress intensity factor it is sufficient to solve only the second partial
boundary value problem. This procedure is particularly useful for unbounded bodies, in which
the initial stress distribution is known.
15
crack front
Fig. 8: Schematic of crack configuration
Remark 7. The typical linear dimension of a crack is very often much smaller then the distance between crack
and the outer boundary of finite body. Then the influence of the interaction of stress-free crack faces and the
outer boundary of the body can be neglected. It is thus seen that it is reasonable to formulate a stress intensity
factor problem for unbounded bodies.
For simplicity, we assume that the initial deformations are zero and the body forces are
absent. Consider first unbounded body with a crack cr   cr   cr , where  cr ,  cr stand for
crack faces, see Fig. 8. The boundary conditions are as follows:
p  ξ     p  ξ   , ξ cr .
(70)
cr
cr
The two crack surfaces are considered coplanar and are represented by a set of the same
points, they differ only by the orientation of unit normal, n+() =-n-(). Further consider that
the fundamental solution uij does not depend on the normal, while tij does depend on the
orientation of the normal. For points  cr the following symmetry relations follow:
uij  x, ξ    uij  x, ξ   , tij  x, ξ     tij  x, ξ  
cr
cr
cr
cr
(71)
Using the symmetry relations and boundary conditions we can rewrite the integral equation
(23) with the use of Eq. (24) to the form:
1 
(72)
u(x )   u(x  )      T* (x, ξ)u(ξ)d, x    cr ,
cr
cr 

2 
cr
or, in components form
1
ui (x  )   ui (x  )      tij (x, ξ)u j (ξ)d, x    cr

cr
cr 

2
cr
where u  ξ   u  ξ    u  ξ   is so called the crack opening displacement, and c(x) = 1/2.
cr
cr
The integral equation (72) cannot be used for the calculation of displacement on cr, because
it does not include the information about the boundary conditions, and thus leads to the
ambiguous formulation of the problem.
Consider now a finite cracked body with prescribed loading t() = p() on the outer boundary
e, and the crack surfaces are tractions-free, i.e. t() = 0 for  cr. In the components form,
the boundary integral equation has the following form for a smooth surface  = ecr



 e uij (x, ξ) p j (ξ)d   e tij (x, ξ)u j (ξ)d    tij (x, ξ)u j (ξ)d=


cr
1

ui (x), x  e ,

2
=
 1 u ( x  )  u ( x  )  , x    .


i
cr
cr
cr
 2  i

16
(73)
This equation is sufficient for the calculation of displacements on the outer boundary e and
on the crack surfaces cr if and only if the problem is symmetric with respect to the set of
tangential planes at all points  cr.
The boundary value problems for finite and infinite cracked bodies can be uniquely
formulated in terms of boundary integral equations with a strong singularity. The starting
point is the second Somigliana’s formula (23) written for x  cr. For the case of finite body
this formula results in
*
*
*
ti (x)  n j  x   e dijk
(x, ξ )tk (ξ)d  n j  x   e sijk
( x, ξ)uk (ξ)d  n j  x    sijk
( x, ξ) uk (ξ)d 
u

cr
(74)
*
*
n j (x)  e dijk
( x, ξ ) pk (ξ)d  n j  x   e sijk
( x, ξ) g k (ξ)d, x   cr .

u
The equation (74) is amended by the integral equation
1
ui (x)   e uij (x, ξ ) p j (ξ )d   e tij (x, ξ )u j (ξ )d    tij (x, ξ ) u j (ξ )d, x   e .


cr
2
(75)
Remark 8. The integrals in Eq. (74) are hypersingular, it means they do not exist even in the sense of Cauchy’s
principal value and have to be considered in the sense of so called Hadamard’s finite part of hypersingular
integral, see Lecture 11, or they have to be regularized in a suitable way.
Eqs. (74) and (75) are to be solved for the unknown crack opening displacements ui and for
unknown displacements ui on the part of the outer boundary e , where loading is prescribed,
and for unknown tractions ti on the part of the outer boundary  ue , where displacements are
prescribed. These equations are known in literature as the dual boundary integral equations
(DBIE) and form the basis of the dual boundary element method (DBEM)- a general and
computationally efficient way of modelling crack problems in fracture mechanics. The
displacement and stress fields in the interior points of domain can then be calculated in
standard way using the integral relations given above.
For the case of infinite body with a crack, on whose surfaces the loading p is prescribed, there
is sufficient to apply only the equation (74), which takes the form:
*
pi (x)  n j  x    sijk
(x, ξ )uk (ξ)d  0, x   cr .
(76)
cr
This equation is further simplified for 2D problems with a straight crack. Consider a 2D
problem of isotropic elasticity in the plane x1x2, containing the crack cr lying in the plane
x1x3. The coordinates of the points x and  are from the interval -a, a. The unit normal
components ni(x+) and ni() on  cr then satisfy the relations ni(x+) = ni() =-i2. Let the
loading acts only in the direction of x2 axis, i.e. p2 =p. The relations (18) provide
G
1


s222

, s221
0
(77)
2 1       x 2
and, according to Eq.(76), the following boundary integral equation for the crack opening
displacement results
a
u2   
G
p  x 
d ,
(78)

2 1     a    x 2
which is a hypersingular integral equation of the first kind. It can be shown that the equation
(78) can be derived in an essentially simpler manner using the method of continuously
distributed dislocation, see e.g. [4].
Recommended literature for further reading:
[1] Baláš, J., Sládek J. and Sládek, V. Stress Analysis by Boundary Element Methods. Studies
in Applied Mechanics 23, Elsevier, Amsterdam,1989.
17
[2] Brebbia, C.A. and Dominguez, J. Boundary Elements, An Introductory Course, 2nd ed.
McGraw-Hill, New York, 1992.
[3] Aliabadi, M.H. and Rooke, D.P. Boundary element methods in fracture mechanics.
Applied Mechanics Review, 50, pp. 83-96, 1997.
[4] Hills, D.A., Kelly, P.A., Dai, D.N., Korsunsky, A.M., Solution of crack problems, Kluwer
Academic Press, Dordrecht,Boston, London, 1996.
Problems
1. Derive the formulas in Eq. (77).
2. Compute the displacements of the thick cylinder of the inner radius r1=10mm and of the
outer radius r2=25mm under an internal pressure p = 0.1 GPa, Young modulus E = 200 GPa
and Poisson’s ratio 0.25 using the boundary element solution. Assume that body forces are
absent.
Hint. This problem can be treated as a plane problem. The displacement at a point  inside the
body or on the boundary are expressed by Eqs (26) and (24)1 as
c(x)u(x)   U* (x, ξ)t (ξ)d   T* ( x, ξ)u(ξ)d   U* ( x, ξ)p(ξ)d   T* ( x, ξ)g(ξ)d (P2.1)
u


u
with the matrices U* and T* given in Eqs. (9) and (11). Due to symmetry, only one quarter of
the cylinder (0 /2) needs to be modelled. The prescribed boundary conditions are
i) v  0 and px  0 on side =0, r1  x  r2 ,
ii) pi  0, i  1, 2 on side r  r2 , 0     2,
iii) u  0 and p y  0 on side =  2, r1  y  r2 ,
iv) pi  ni p, i  1, 2 on side r  r1 , 0     2
Use quadratic elements for the boundary discretization. The geometry, displacements, and
tractions are modelled with this kind of element. Quadratic element shape functions are (see
also Fig.6)
1
1
N1      1 , N 2  1   1    , N 3   1   
2
2
The assumed displacements and tractions within i-th element are written in terms of the
element matrix shape function Niu  Nip  Ni , see also Eq.(30)
u(ξ)  Ni (ξ) p rui , t(ξ)  Ni (ξ) u rpi
where
1 2    1
0
Ni  
0
1 2    1

1   1   
0
0
1   1   
p
rui  u1i , v1i , u2i , v2i , u3i , v3i 
u
r  t , t , t , t , t , t
(P2.2)

1 2 1   
0
 (P2.3)
0
1 2 1    
T
i
p
i
1x
i
1y
i
2x
i
2y
i
3x
(P2.4)

T
i
3y
With (P2.2) and (P2.3) Eq. (P2.1) takes the form
M
N

 u r i     U* (x, ξ )N i (ξ )d  p r i  
*
i
U
(
x
,
ξ
)
N
(
ξ
)d

  

 
p
p


 i 1   i

i 1   ui
(P2.5)
M
N







 c(x)u(x)      T* (x, ξ ) N i (ξ )d  u rui       T* ( x, ξ ) N i (ξ )d  p rui .

  i 1   i


i 1   ui
where M is the number of elements into which the part u is divided and N is the number of
elements into which the part  is divided. For the part of the boundary i) we set
p
rui  u1i , 0, u2i , 0, u3i , 0 , u rpi  0, t1i y , 0, t2i y , 0, t3i y  , p rpi  u rui  0
T
T
Similarly, for the part of the boundary iii)
18
p
rui  0, v1i ,0, v2i ,0, v3i  , u rpi  t1ix ,0, t2i x ,0, t3i x ,0 , p rpi  u rui  0
T
T
On the part of the boundary ii) and iv) u=.
Finally, set up the system (32) and choose the simplest case when the whole boundary is
divided only in 4 elements such that each of the boundary parts i)-iv) is formed just by one
element.
19