4.3 Generalized-Kelvin Model
141
a)
t
σ
(tmax = 100 × 0.1 σmax,min = ± 5.0)
b)
t
(tmax = 100 × 0.1 max,min = ± 10.0)
c)
σ
( max,min = ± 10.0 σmax,min = ± 5.0)
d)
t
v
(tmax = 100 × 0.1 v,max,min = ± 10.0)
e)
σ
( max,min = ± 10.0 σmax,min = ± 5.0)
f)
σ
( max,min = ± 10.0 σmax,min = ± 5.0)
Fig. 4.35 Response analysis of Standard-Linear-Solid Kelvin model with material data: η k = 1.0,
E k = 1.0 (τ k = 1.0), E 0 = 1.0 (E ∞ = E m = 0.5, η m = 0.25, τ m = 0.5). Prescribed Ramp stress
history with data: σ a = 5.0, a–d t 0 = 0.0, t 1 = 1.0, t 2 = 9.0, t 3 = 10.0; = 0.1, N = 100, e t 0 =
0.0, t 1 = 0.1, t 2 = 0.9, t 3 = 1.0; = 0.01, N = 100, f t 0 = 0.0, t 1 = 0.01, t 2 = 0.09, t 3 = 0.1;
= 0.001, N = 100
141
a)
t
σ
(tmax = 100 × 0.1 σmax,min = ± 5.0)
b)
t
(tmax = 100 × 0.1 max,min = ± 10.0)
c)
σ
( max,min = ± 10.0 σmax,min = ± 5.0)
d)
t
v
(tmax = 100 × 0.1 v,max,min = ± 10.0)
e)
σ
( max,min = ± 10.0 σmax,min = ± 5.0)
f)
σ
( max,min = ± 10.0 σmax,min = ± 5.0)
Fig. 4.35 Response analysis of Standard-Linear-Solid Kelvin model with material data: η k = 1.0,
E k = 1.0 (τ k = 1.0), E 0 = 1.0 (E ∞ = E m = 0.5, η m = 0.25, τ m = 0.5). Prescribed Ramp stress
history with data: σ a = 5.0, a–d t 0 = 0.0, t 1 = 1.0, t 2 = 9.0, t 3 = 10.0; = 0.1, N = 100, e t 0 =
0.0, t 1 = 0.1, t 2 = 0.9, t 3 = 1.0; = 0.01, N = 100, f t 0 = 0.0, t 1 = 0.01, t 2 = 0.09, t 3 = 0.1;
= 0.001, N = 100
