FUNDAMENTAL SOLUTIONS 107
Here we shall consider the case of a point force P normal to the
interface at a depth c (see Figure 4.9).
Plane x
3
= 0
Observation point
Figure 4.9 Force normal to the boundary in the interior of
a
semi-infinite solid
Mindlin
3
found that the zero stress condition on x
3
= 0 is
produced by a combination of images at (0, 0,
— c)
of the following
types:
(a) a force in the x
3
direction at (0, 0, —c)
+ (b) a double force in the x
3
direction at (0, 0, -c)
+ (c) a centre of compression (three mutually perpendicular
double forces) at (0, 0,
—
c)
+ (d) a line of compression along the z axis from x
3
=
— c
to
x
3
= +c.
+ (e) a doublet at (0, 0, -c).
Superposing the displacement fields due to the original force and its
images we obtain the displacement field for the force normal to the
interface for the horizontal directions Xj and x
2
as
108
FUNDAMENTAL
SOLUTIONS
li*
-
ii*
Pr
(xi-c (3-4μ)(χ
3
-ο)
1
~
2
~16πα(1-μ)
r? r
4(l-n)(l-2^)
|
6cx
3
(x
3
+
c)·,
;
r^ ϊ
(4-77)
The
x
3
component
of
displacement
U
3
is
given by:
„,
P {3-4μ
t
8(l-/£)
a
-(3-4/i)
,
(x
3
-c)
3
^
16πσ(1-μ)
>Ί
t
(3-4/J)(X
3
+
C)
2
-2CX
3
6cx
3
(x
3
+
c)
2
+
13 + H
I
4
·'*'
These are the Mindlin fundamental solutions
for
the displacement
corresponding
to a
concentrated force
in
the positive
x
3
direction.
The corresponding stresses using cylindrical polar coordinates
(r
l9
0
l5
x
3
) corresponding to our Cartesian axes (x
1?
x
2
, x
3
) are given by
P /(l-2/z)(x
3
-c) (l-2/i)(x
3
+
7c)
°
rr
%π(-μ)
r r
|
4(1-/ι) (1-2μ) 3r
2
(x
3
-c)
|
r
2
(r
2
+
x
3
+
c)
rj
6c(l
-
2μ)
(x
3
+
c)
2
-
6c
2
(x
3
+
c)
-
3(3
-
4/j)r
2
(x
3
-
c)
_
30cr
2
x
3
(x
3
+
c)',
(4?9)
>
=
P(l-2g)/x
3
-c (3-4g)(z
+
c)-6c 4(1-|i)
σββ
8π(1-/ι),
r? r|
r
2
(r
2
+
x
3
+
c)
6c(x
3
+
c)
2
6c
2
(x
3
+
c)
.
+
3
5
- „ ' «
(4.80)
FUNDAMENTAL SOLUTIONS 109
8π(1-μ)
(l-2j/)(x
3
-c) (1-2μ)(χ
3
-<·) 3(x
3
-c)
3
r?
r?
3(3 - 4μ)χ
3
(χ
3
+
c)
2
- 3c(x
3
+
c)
(5x
3
- c)
30cx
3
(x
3
+ c)
3
3
(4.81)
Pr
σ* =
8π(1-μ) r r
1-2μ
+
1-2μ 3(x
3
-c)
2
r
3(3
—
4μ)χ
3
(χ
3
+
c) —
3c(3x
3
-f
c)
30cx
3
(x
3
+ c)
r2
(4.82)
The corresponding formulae for the situation of a concentrated force
parallel to the interface at point x
3
= +
c
are given by:
U* =
16πϋ(ί-μ)
3-4μ 1 x (3-4μ)χ
2
2cx
3
/ 7>x
-
+
-
+
-J +
-3
'2
r| V r^
|
4(1-μ)(1-2μ)
|
^
r
2
+ x
3
+ c
x
2
M^ + *3 +
C)
(4.83)
Px
t
x
2
/l 3-4μ 6cx
3
4(1-μ)(1-2μ)
2
16πσ(1-μ),Γ?
+
Ί rf
r
2
(r
2
+ x
3
+ c)
2
'
l
'
i/? =
Px
t
fx
3
—
c (3
—
4μ)(χ
3
— c)
6cx
3
(x
3
+c)
16πσ(1-μ) r
r
2
+
+
4(1-μ) (l-2/i)
^fo + *3 +
C)
(4.85)
110 FUNDAMENTAL SOLUTIONS
With stresses:
*?1
Px
t
Γ 1-2/i (1
8π(1-μ)|_ ^
+ _
-2μ)(5-4μ) 3x
2
3(3-4μ)χ
2
.3
'2
ri
4(1-μ) (1-2^ x
2
(3r
2
+ x
3
+ c)
r
2
(r
2
+ x
3
+c)
2
r|(r
2
+ x
3
+ c)
6c
5*1X3
+ —{3c-(3-2fi)(jc
3
+ c) + —3
(4.86)
1722
8TC(1
-
A*)
L
r
'
Px [1-2μ (1-2μ)(3-4μ) 3χ| 3(3-4μ)χ^
13 ' ^3 Z5
Λ
Α{-μ){-2μ)(
χ
xj (3r
2
+ x
3
+ c) j
|
r
2
(r
2
+x
3
+ c)
2
r{r
2
+x
3
+ c)
6c ( /, - w . 5x|x
3
+
—
c-(l-2/i)(x
3
+c)
+
-4^
(4.87)
φ
Pxi [ί-2μ 1-2μ 3(x
3
-c)
2
3(3 - 4μ) (χ
3
+ c)
2
733
8π(1-μ)|_ rj r| r* r
6c / ,« ~
w
x 5x
3
(x
3
+ c)
2
+ -5-
c
+ (l-2/x)(x
3
+ c) + ^-4 '-
^3 =
»■2
Px
Px
t
x
2
Γ 3(x
3
-c) 3(3-4/i)(x
3
+ c)
+
(4.88)
+
^£f
1
_
2/l
+
5
i^±f)
°Ϊ2 =
^2
P
8π(1-μ)
(l-2/z)(x
3
-c) (l-2^)(x
3
-c)
1 -1
r?
(4.89)
3x
2
(x
3
-c) 3(3-4/*)x
2
(x
3
+ c)
6c
^ ,.. ^ v 2 5x
2
x
3
(x
3
+ c)
y
. x
3
(x
3
+
c)
- (1 - 2/i)x
2
!-^V "
(4.90)
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