The Boundary Element Method - Mechanics of Structures: Variational and Computational Methods - page 560

all the forces δ(ξ ,x ) a j ,j = 1 , 2 , 3 for three-dimensional problems and j = 1 , 2 for two-

dimensional problems, can be written as

u i = u ji a j

(9.68)

In the rectangular coordinate system,

1

=

when i j

a i a j = δ =

ij

0 when i = j

Then u ji = u i a j . From Eq. (9.58b),

g i,kk −

g k,ik a j = (

1 1

u ji = g i a j ) ,kk − g k a j ) ,ik

− ν) (

2 ( 1 − ν) 2 ( 1

Since, from Eqs. (9.64) and (9.66), g i = ga i so that g i a j

= ga i a j = g δ

ij . Thus

1

u ji =

g ,kk δ −

− ν)

g ,ik δ (9.69)

ij kj

(

2 1

For three-dimensional problems, substitute Eq. (9.64) into Eq. (9.69) to obtain

1 1

u ji =

− ν)

g ,kk δ − g ,ik δ = [2 ( 1 − ν) r ,kk δ − r ,ik δ

kj ]

(9.70)

ij kj ij

2 ( 1 16 π G ( 1 − ν)

The derivatives of r are taken with reference to the coordinates of x i , i.e.,
r ,i = ∂

r

r ,j = ∂

r

(9.71a)

∂ x i ∂ x j

and

∂
∂

( x i − ξ ) r i

r

i

(

r ,i = ( x k − ξ )

2

=

2 =

(9.71b)

k

x i

x k − ξ k )

where r i = ( x i − ξ ) is the projection of r on the x i axis. From the relationships of Eq. (9.71),

i

r 2 r ∂

r 2 r 2 r i

∂ 1 r i

r i ∂

r −

x i = x i −

=

∂ ∂ ∂ x i r 3

Hence

3 3

2 r

r 2 r k

∂ − 2

r

r ,kk =

x k =

=

∂ r 3

k = 1 k = 1

Also

∂ r i

r

r ∂

2 r

∂ ∂ r ∂

∂

1
r 2

r i

r i ∂

r

r ,ik =

x k =

= =

x k −

∂ x i ∂ ∂ x k ∂ x i x k ∂ ∂ x k

i.e.,

1

−

r r ,i r ,k

when i = k

r ,ik =

1
r

( − ) =

1 r ,i r ,i when i k