NX Nastran DMAP Programmer's Guide - Kxcad.net

878
DDR2
Computes displacements due to mode acceleration
UNUSED1 Input-integer-default=-1. Unused.
UNUSED2 Input-integer-default=-1. Unused.
Remarks:
1. The solution matrix, UD1, may be used to also improve the data recovery output;
such as, stress, strain, etc.
2. USETD, UD, PD, MDD, OL, and LLL may not be purged.
3. DM may not be purged if support degrees-of-freedom exist.
4. For transient analysis, the velocities and accelerations (every second and third
column in UD) are unchanged in UD1.
5. DDR2 uses a static approximation for the effect of the higher modes.
Method:
The equivalent load vector is computed:
a
Pd { } { Pd } { Kdd} { ud } [ Bdd] u · { d}
[ Mdd] u ·· = –
–
– { d}
For a transient analysis problem { ud} , u are given explicitly. For
frequency response analysis,
· { d}
, and u ·· { d}
Eq. 4-5
Eq. 4-6
Eq. 4-7
where ω is the forcing frequency and { ud} is the complex response vector. ω comes
a
from PPF. The vector { Pd} is the sum of applied loads and inertia loads due to the
motion of the system approximated by its lower modes. The static solution using these
loads will provide a better answer for displacements.
If extra points are present, then:
2
2
u · { d}
= iω { ud }
u ·· { d}
ω 2
= – { ud }
a
{ Pd }
{ ud }
a
Pa ⎧ ⎫
→ ⎨----- ⎩P ⎬
e ⎭
a
ua ⎧ ⎫
→
⎨----- ⎩u ⎬
e ⎭
Eq. 4-8
Eq. 4-9