ANALYTIC SOLUTIONS OF ELASTIC TUNNELING PROBLEMS Phần 4 doc

Section 4.2 Solution for a Deforming Hole in a Half-Plane 25

The expansion of the left-hand side of (4.24) is not as trivial, and will be

greatly simplified by multiplication with 1 − ασ. The right-hand side of the

modified equation can then be expanded as follows. We write

G

(ασ ) = 2µ(1 − ασ)h

g

m(ασ ) = ∞

k=−∞

Akσk. (4.31)

where we have assumed that the given displacements along the hole, multiplied

by 2µ(1−ασ), can be expanded into a Fourier series. Such an expansion should

be possible for all problems of practical significance. Note that the function

G

(ασ ) has same definition as in [42]. The logarithmic terms in the right-hand

side of (4.24) can be expanded with help of the formula given in (3.7):

log

1

1 − ζ = ∞

n=1

ζ n

n , |ζ | < 1, (4.32)

where the principle branch of the logarithm has been chosen. The modified

form of (4.25) resulting from the multiplication with 1 − ασ can be expanded

as follows (using (4.32) and the series in(4.31)):

(1 − ασ) h

g

m◦

(ασ ) = ∞

k=−∞

Ak ◦ σk (4.33)

where

Ak ◦ =







Ak +

h

Fx + i

h

Fy

2π

1 + (α2 − 1)k

k(k − 1)αk



k ≤ −2,

A−1 + α(

h

Fx + i

h

Fy )

4π

(2 − α2) − ακ(

h

Fx − i

h

Fy )

4π(1 + κ)

k = −1,

A0 + CF − α2(

h

Fx + i

h

Fy )

2π −

h

Fx − i

h

Fy

4π(1 + κ)

+ κ(

h

Fx − i

h

Fy )

4π(1 + κ) (1 − α2) k = 0,

A1 − αCF + ακ(

h

Fx + i

h

Fy )

2π

+

h

Fx − i

h

Fy

4πα(1 + κ)

+ α(κ − 1)(

h

Fx − i

h

Fy )

4π(1 + κ)

k = 1,

A2 − α2κ(

h

Fx + i

h

Fy )

4π

+

h

Fx − i

h

Fy

4π(1 + κ)

k = 2,

Ak − κ(

h

Fx + i

h

Fy )

2π

αk

k(k − 1)



k ≥ 3.

(4.34)