218
5 Plasticity
∂ φ
− E ∂ ((λ) =
σ
p
|σ
p |
E − E ∂ ((λ)
.
= 0.
(5.89)
As a conclusion the algorithmic tangent E a is thus finally expressed as
E
n
a = E − H 0 ((λ) E.
(5.90)
Note that, consequently, the algorithmic tangent degenerates to E a = 0 for λ > 0.
In one dimension the algorithmic tangent trivially coincides with its continuous
counterpart. It shall be noted, however, that this is at variance with the corresponding
result in two and three dimensions.
The algorithmic step-by-step update for the specific Prandtl model is summarized
in Table 5.5.
5.2.3 Specific Prandtl Model: Response Analysis
Prescribed Strain History: Zig-Zag
The response of the specific Prandtl model to a prescribed Zig-Zag strain history is
documented in Fig. 5.9a, b, c, d, e. (These shall be compared to the corresponding
response of the underlying, rigid-plastic, specific St. Venant model in Fig. 5.4a, b, c,
d, e.)
Figure 5.9a depicts the prescribed Zig-Zag strain history (t) with amplitude a =
5 and period T = 4 in the time interval t ∈ [0, t max = 10], whereby N = 100 time
steps with t = 0.1 are computed. Plastic time steps are emphasized by larger hollow
circles, whereas elastic time steps are indicated by smaller filled circles.
Figure 5.9b showcases the resulting stress history σ(t) that displays a trapezoidal
signal with ˙
σ(t) = E ˙
(t) in the elastic phases where |σ(t)| < σ y = 1 (E = 1, thus
the slopes in the elastic phases in Fig. 5.9a, b coincide), and ˙
σ(t) = 0 in the plastic
phases where |σ(t)| = σ y = 1.
The resulting σ = σ() diagram is highlighted in Fig. 5.9c. The expected
parallelogram-type format of the σ = σ() diagram is captured exactly, whereby
the slopes at = 0 and = ±5 obviously coincide with the elastic modulus E = 1.
Figure 5.9d demonstrates the plastic strain history p (t): during the plastic phases
p (t) evolves in parallel to the total strain with |˙ p (t)| = |˙ (t)| = 5, whereas p (t)
stays constant with | p (t)| = 4 (or as initial value p (t) = 0) during the elastic phases.
Finally, the plastic arc-length κ(t) in Fig. 5.9e follows linear in time from integrating ˙
κ(t) = |˙ p (t)| = 5 during the plastic phases and constant in time during the
elastic phases, thus κ max = [0.5 + 4 + 0.375] × 8 = 39 (for 4.875 plastic phases of
plastic arc-length 8 each).
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