Thư viện tri thức trực tuyến
Kho tài liệu với 50,000+ tài liệu học thuật
© 2023 Siêu thị PDF - Kho tài liệu học thuật hàng đầu Việt Nam
Tài liệu đang bị lỗi
File tài liệu này hiện đang bị hỏng, chúng tôi đang cố gắng khắc phục.
ANALYTIC SOLUTIONS OF ELASTIC TUNNELING PROBLEMS Phần 4 doc
Nội dung xem thử
Mô tả chi tiết
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)