430
32 On Boundary Value Problems of Inelastic Body Mechanics
The functions (32.50) in the polar coordinates (ρ, ϕ) correspond to stresses
σ r = −
q
π
ϕ 1 + ϕ 2 − 2θ 0 − (1 − ρ
2 )×
×
sin 2(α 1 − ϕ)
R 2
1
−
sin 2(α 2 − ϕ)
R 2
2
,
σ ϕ = −
q
π
ϕ 1 + ϕ 2 − 2θ 0 + (1 − ρ
2 )×
×
sin 2(α 1 − ϕ)
R 2
1
−
sin 2(α 2 − ϕ)
R 2
2
,
τ rϕ =
q(1 − ρ 2 )
π
ρ 2 − cos 2(α 1 − ϕ)
R 2
1
−
ρ 2 − cos 2(α 2 − ϕ)
R 2
2
.
(32.51)
Here it is indicated
ϕ 1 = arg ln
σ 2 − ζ
σ 1 − ζ
, ϕ 2 = arg ln
σ 4 − ζ
σ 3 − ζ
,
R
2
1 = (1 + ρ
2 )
2
− 4ρ
2 cos
2 (α 1 − ϕ),
R
2
2 = (1 + ρ
2 )
2
− 4ρ
2 cos
2 (α 2 − ϕ),
where α 1 , α 2 are the angles corresponding to the ends of the arc σ 1 σ 2 , and the beam
from which the polar angle ϕ is counted divides the arc σ 1 σ 2 in half.
The stresses inside the circle must be continuous everywhere, so the angles ϕ 1
and ϕ 2 must also be continuous functions of the point ζ . We satisfy this requirement,
assuming that ϕ 1 and ϕ 2 are the angles at which the arcs σ 1 σ 2 and σ 3 σ 4 are visible
from ζ , respectively (Fig. 32.15).
Keeping in mind that α 1 = −θ 0 , α 2 = θ 0 , we get
σ r = −
q
π
θ − 2θ 0 + 2(1 − ρ
2 ) sin 2θ 0
(1 + ρ 4 ) cos 2ϕ − 2ρ 2 cos 2θ 0
R 2
1 R 2
2
,
σ ϕ = −
q
π
θ − 2θ 0 − 2(1 − ρ 2 ) sin 2θ 0
(1 + ρ 4 ) cos 2ϕ − 2ρ 2 cos 2θ 0
R 2
1 R 2
2
,
τ rϕ =
2q(1 − ρ 4 )(1 − ρ 2 ) sin 2θ 0 sin 2ϕ
πR 2
1 R 2
2
,
(32.52)
with
θ = ϕ 1 + ϕ 2 =
θ 1 + θ 2
with ρ| cos ϕ| cos θ 0 ,
2π − θ 3 − θ 4 with ρ| cos ϕ| cos θ 0 ,
(32.53)