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10 Mathematical Structural Imperfections
10.9 Half-Plane with a System of Dislocations
Assume that infinitely many parallel rigid stripes of a similar length t and width δ
distanced from each other to the same distance a are introduced into the half-plane
y < 0. Using expression (10.40), the principle of superposition and formulas for
conversion of Muskhelishvili functions in the case of parallel transfer of coordinate
axes, we will obtain the functions and for the specified half-plane as
follows:
= −4Kδt
k=∞
k=−∞
z − ak
(z − ak + it) 2 (z − ak − it)
,
(z) = 4Kδt
k=∞
k=−∞
(3it − 2ak)(z − ak) 2 + (t 2 + akti)(z − ak) + akt 2
(z − ak + it) 3 (z − ak − it) 2
.
(10.41)
Using decomposition:
∞
k=1
1
z 2 − k 2 =
1
2z
π cot πz −
1
z
,
∞
k=−∞
1
(z − k) 2 =
π 2
sin
2 πz
;
∞
k=−∞
1
z − k
= π cot πz,
∞
k=−∞
1
(z − k) 3 =
π 3 cos πz
sin
3 πz
,
formulas (10.41) can be converted to
= −
Kδ
a
πi(cot πz 1 − cot πz 2 ) +
2π 2 t
a sin
2 πz 1
,
(z) =
Kδ
a
4π 3 t (it − az 1 ) cos πz 1
a 2 sin
3 πz 1
+
π 2 (t − z 1 ai)
a sin
2 πz 1
+
+
π 2 (t + z 2 ai)
a sin
2 πz 2
+ 2πi(cot πz 1 − cot πz 2 )
,
(10.42)
where z 1 =
1
a (z + it), z 2 =
1
a (z − it).
In this manner, Muskhelishvili functions for an elastic half-plane wedged by an
infinite number of parallel rectangular rigid stripes of the same width δ and length
t spaced with even pitch a are built. It is easy to determine stresses using the ratios
(10.42), (9.15). If we consider the half-plane as the basis of a pile foundation of a
building, and implemented absolutely rigid stripes as piles, the given solution allows
finding an additional loading that the single pile can take.
Indeed, with the known normal pressure on side edges of the pile σ y (±δ/2, y)
and the friction coefficients f between the pile and soil of basis, additional loading
that the single pile can take will be proportional to the value:
10 Mathematical Structural Imperfections
10.9 Half-Plane with a System of Dislocations
Assume that infinitely many parallel rigid stripes of a similar length t and width δ
distanced from each other to the same distance a are introduced into the half-plane
y < 0. Using expression (10.40), the principle of superposition and formulas for
conversion of Muskhelishvili functions in the case of parallel transfer of coordinate
axes, we will obtain the functions and for the specified half-plane as
follows:
= −4Kδt
k=∞
k=−∞
z − ak
(z − ak + it) 2 (z − ak − it)
,
(z) = 4Kδt
k=∞
k=−∞
(3it − 2ak)(z − ak) 2 + (t 2 + akti)(z − ak) + akt 2
(z − ak + it) 3 (z − ak − it) 2
.
(10.41)
Using decomposition:
∞
k=1
1
z 2 − k 2 =
1
2z
π cot πz −
1
z
,
∞
k=−∞
1
(z − k) 2 =
π 2
sin
2 πz
;
∞
k=−∞
1
z − k
= π cot πz,
∞
k=−∞
1
(z − k) 3 =
π 3 cos πz
sin
3 πz
,
formulas (10.41) can be converted to
= −
Kδ
a
πi(cot πz 1 − cot πz 2 ) +
2π 2 t
a sin
2 πz 1
,
(z) =
Kδ
a
4π 3 t (it − az 1 ) cos πz 1
a 2 sin
3 πz 1
+
π 2 (t − z 1 ai)
a sin
2 πz 1
+
+
π 2 (t + z 2 ai)
a sin
2 πz 2
+ 2πi(cot πz 1 − cot πz 2 )
,
(10.42)
where z 1 =
1
a (z + it), z 2 =
1
a (z − it).
In this manner, Muskhelishvili functions for an elastic half-plane wedged by an
infinite number of parallel rectangular rigid stripes of the same width δ and length
t spaced with even pitch a are built. It is easy to determine stresses using the ratios
(10.42), (9.15). If we consider the half-plane as the basis of a pile foundation of a
building, and implemented absolutely rigid stripes as piles, the given solution allows
finding an additional loading that the single pile can take.
Indeed, with the known normal pressure on side edges of the pile σ y (±δ/2, y)
and the friction coefficients f between the pile and soil of basis, additional loading
that the single pile can take will be proportional to the value: