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The maximum stress is
10
8× 10 :
Analytical
TEST08
Same as TEST01 but uses a special option to obtain a “quasi-static” calculation
(OPTI QUAS STAT FSYS beta). The code adds a linear damping represented by an
external force proportional to the mass for each degree of freedom:
Fqs
=− 4πβ
fsysmv
where m is the mass and v the velocity. In practice, only the product β fsys
is relevant.
The case β = 1 corresponds to critical damping for the frequency f
sys
. Here we use
β = 1 and f
sys
= 1 Hz, because in the large-strain case we saw in calculation TEST01
that the obtained period is 0.98 so the frequency is 1/ 0.98 ≈ 1.
The resulting displacement at 1.5 s is 0.221345, which is already very close to the
expected value of 0.221403 (i.e. exp(0.2) − 1) that would result from a large-strain
static analysis (see below).
Explanation: since ν = 0 there is no cross-section variation as long as the material
remains linear elastic (as assumed here). The Cauchy stress is therefore:
8 2 10
σ = F/ S0
= 1000 N / 2.5× 10 m = 4× 10 Pa .
10 11
This induces a strain (natural): e= σ / E = 4× 10 Pa / 2× 10 Pa = 0.2 . But we have
e= ln ( L/ L0
) so that L/ L0
= exp ( e)
and L= L exp ( e)
.
0
The elongation (or the displacement) is therefore: ∆ L= L− L0 = L0exp ( e)
− L0. Since
L
0
= 1 in this example, we get finally:
∆ L = exp(0.2) − 1 = 0.221403
7