19.2 Two-Dimensional Klyushnikov Model
287
The following loading path is considered: along the X axis from the point O to
the point M and then along a beam going from the point M. As a result, it is found
that tangential lines to the initial ellipsis drawn from the point M divide the plane
σ ∼ τ into 4 areas marked by Roman numbers in Fig. 19.3. If stress gains are such
that the additional loading trajectory is located in the area I , there is elastic loading.
If the additional loading vector is located in the area I I , gains of plastic strains are
defined by formulas of the strain theory of plasticity. In the case when the additional
loading path is located in the area I I I or I V , the ratios between stresses and strains
are obtained from formulas (19.1) in an extremely complicated form. The final result
is obtained by the author only for the case of orthogonal additional loading shown
in Fig. 19.3 as the vector τ . For this case,
τ
γ
= G
∗
=
G
1 +
3
2
G
1
E s
−
1
E t
.
(19.3)
The slip theory represented by the Batdorf–Budiansky model was not proved
experimentally, which the authors later admitted [3]. Despite the model predicts
some quality effects, their quantitative parameters were not rather satisfactory. In
Chap. 27 we show that this is not a reason to abandon slip theory in general.
19.2 Two-Dimensional Klyushnikov Model
As said above, the calculation of strain based on the Batdorf–Budiansky slip theory
is related to serious mathematical challenges so it is required to use approximated
calculations. At the same time, the hypotheses of the Batdorf–Budiansky model
are not that reliable to search for precise solutions. To simplify the primary
dependencies of slip theory, V.D. Klyushnikov proposed [16], [17] a model of a
two-dimensional medium for which e 33 = 0, σ 33 =
1
2
(σ 11 + σ 22 ). Then the stress
deviator components will be
σ
11 = −σ
22 =
1
2
(σ 11 − σ 22 ).
According to the non-compressibility condition e
p
11 + e
p
22 = 0, therefore we can
write
e
p
11 = −e
p
22 =
1
2
(e
p
11 − e
p
22 ).
It is deemed that the material may deform only by shear in planes perpendicular to
the plane x 1 x 2 . In some plane among these planes that make the angle ω with the
axis x 1 , the tangential stress will be
287
The following loading path is considered: along the X axis from the point O to
the point M and then along a beam going from the point M. As a result, it is found
that tangential lines to the initial ellipsis drawn from the point M divide the plane
σ ∼ τ into 4 areas marked by Roman numbers in Fig. 19.3. If stress gains are such
that the additional loading trajectory is located in the area I , there is elastic loading.
If the additional loading vector is located in the area I I , gains of plastic strains are
defined by formulas of the strain theory of plasticity. In the case when the additional
loading path is located in the area I I I or I V , the ratios between stresses and strains
are obtained from formulas (19.1) in an extremely complicated form. The final result
is obtained by the author only for the case of orthogonal additional loading shown
in Fig. 19.3 as the vector τ . For this case,
τ
γ
= G
∗
=
G
1 +
3
2
G
1
E s
−
1
E t
.
(19.3)
The slip theory represented by the Batdorf–Budiansky model was not proved
experimentally, which the authors later admitted [3]. Despite the model predicts
some quality effects, their quantitative parameters were not rather satisfactory. In
Chap. 27 we show that this is not a reason to abandon slip theory in general.
19.2 Two-Dimensional Klyushnikov Model
As said above, the calculation of strain based on the Batdorf–Budiansky slip theory
is related to serious mathematical challenges so it is required to use approximated
calculations. At the same time, the hypotheses of the Batdorf–Budiansky model
are not that reliable to search for precise solutions. To simplify the primary
dependencies of slip theory, V.D. Klyushnikov proposed [16], [17] a model of a
two-dimensional medium for which e 33 = 0, σ 33 =
1
2
(σ 11 + σ 22 ). Then the stress
deviator components will be
σ
11 = −σ
22 =
1
2
(σ 11 − σ 22 ).
According to the non-compressibility condition e
p
11 + e
p
22 = 0, therefore we can
write
e
p
11 = −e
p
22 =
1
2
(e
p
11 − e
p
22 ).
It is deemed that the material may deform only by shear in planes perpendicular to
the plane x 1 x 2 . In some plane among these planes that make the angle ω with the
axis x 1 , the tangential stress will be