112
10 Mathematical Structural Imperfections
where the right part of the equation symbolically writes the sum of deductions of
the function N(z) in all special points a i of the area limited by the outline .
By equating the right parts of equations (10.25) and (10.26), we will obtain
1
2πi
+R
−R
N(ζ )dζ
ζ − z
=
1
2πi
C
N(ζ )dζ
ζ − z
− res N(a i ).
(10.27)
Using the first of the formulas (10.21), it can be easily ensured that for R → ∞, the
integral in the right part of formula (10.27) tends to zero. By making the passage to
the limit in formula (10.27), taking into account designations (10.22), we will obtain
ω 1 (z) = − res N(a i ).
(10.28)
From formula (10.28) after calculation and summing of deductions, let us find
ω 1 (z) = k
(z − 2iH ) cos α + H sin α
(z − iH ) 2
.
(10.29)
In a similar way, let us find
ω 2 (z) = k
H cos α + z sin α
(z − iH ) 2
.
(10.30)
10.6.4 Completion of Problem Solution
Substituting functions (10.29)–(10.30) and the respective derivatives into formulas
(10.23) gives Muskhelishvili functions for the half-plane on which boundary
distributed loads (10.21) are applied:
3 (z) = k
(3iH − z) cos α + i(z + iH ) sin α
(z − iH ) 2
,
3 (z) = k
(3iH z − 2H 2 − z 2 ) cos α − (iz 2 + 5H z) sin α
(z − iH ) 3
.
(10.31)
The components of stresses corresponding to these functions are designated by
the upper index “0”. Using the formulas of type (24.6) and dependencies (10.31),
we obtain