19.1 Batdorf–Budiansky Slip Theory
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Fig. 19.2
Batdorf–Budiansky model
components of the tensor e ij equal zero, except for e 12 =
1
2
γ nβ . We obtain
de
p
ij = (β 1i β 2j + β 2i β 1j )de 12 .
Having in mind that β 1i = n i , β 2i = β i , we can record as follows:
de
p
ij =
1
2
(n i β j + n j β i )F (τ nβ )dddβ.
By integration for plastic strain components, we obtain the following formulas:
e
p
ij =
1
2
dd
+π/2
−π/2
(n i β j + n j β i )F (τ nβ )dβ,
(19.1)
where tangential stress is calculated using the formula
τ nβ = σ ij n i β j .
(19.2)
The primary difficulty in practical calculation of strains using formulas (19.1)
is that the function F (τ nβ ) differs from zero only where τ nβ > τ s , whereas τ s is
the shear yield stress. Therefore, integrals in formulas (19.1) do not fall upon the
entire sphere of the singular radius but only some part of it. An open issue is also
the form of the function F (τ ). By considering the uniaxial elongation problem, the
paper by Batdorf and Budiansky assumes τ nβ = σ 11 n 1 β 1 , and the authors obtain the
following condition to define the integration area:
n 1 β 1 >
τ s
σ 11
.