mayaview
Civil/Environmental
- Oct 20, 2014
- 4
Hello, everyone!
I am trying to using nonlinear spring elements in a FEM model, but first I want to run some test problems. I found nonlinear spring elements produced much smaller amplitude than analytical solution for SDF system. The problem is defined as following:
*Heading
**
** PARTS
**
*Part, name=Part-1
*Node
1, 0., 0., 0.
*Nset, nset=Part-1-RefPt_, internal
1,
*Nset, nset=_PickedSet2, internal
1,
*Nset, nset=_PickedSet3, internal
1,
*Element, type=MASS, elset=_PickedSet3_Inertia-1_
1, 1
*Mass, elset=_PickedSet3_Inertia-1_
1.,
*Spring,elset=Springs/Dashpots-1-spring, nonlinear, dependencies=1
2
0,0
288.0,0.18
6180.0,10
*Element, type=Spring1, elset=Springs/Dashpots-1-spring
2, 1
*End Part
**
**
** ASSEMBLY
**
*Assembly, name=Assembly
**
*Instance, name=Part-1-1, part=Part-1
*End Instance
**
*Nset, nset=Set-1, instance=Part-1-1
1,
*Nset, nset=_PickedSet6, internal, instance=Part-1-1
1,
*Nset, nset=_PickedSet7, internal, instance=Part-1-1
1,
*End Assembly
**
** BOUNDARY CONDITIONS
**
** Name: BC-1 Type: Displacement/Rotation
*Boundary
_PickedSet6, 1, 1
_PickedSet6, 6, 6
**
** PREDEFINED FIELDS
**
** Name: Predefined Field-1 Type: Velocity
*Initial Conditions, type=VELOCITY
_PickedSet7, 1, 0.
_PickedSet7, 2, 20.
** ----------------------------------------------------------------
**
** STEP: Step-1
**
*Step, name=Step-1, nlgeom=NO, inc=10000
*Dynamic,application=TRANSIENT FIDELITY
0.001,0.1,2e-08
**
** OUTPUT REQUESTS
**
*Restart, write, frequency=0
*Output, field, frequency=0
**
** HISTORY OUTPUT: H-Output-2
**
*Output, history, frequency=1
*Node Output, nset=Set-1
A2,V2,U2
*End Step
The spring is designed to have a stiffness 1600 for u < 0.18, and 600 for u>0.18.
The mass of the object is 1; the initial velocity of the object is 20.
The displacement vs acceleration figure is as following, the slope of the curve is the same as expected.
The displacement and velocity curve seem OK, too.
However, the amplitude should be 0.78 instead of 0.6 as in the result.
The amplitude can be calculated by the potential energy. The initial kinetic energy is 20^2/2=200. The potential energy at displacement 0.6 is 0.18^2*1600/2+(0.6^2-0.18^2)*600/2=125, which is much smaller than the initial kinetic energy.
I am trying to using nonlinear spring elements in a FEM model, but first I want to run some test problems. I found nonlinear spring elements produced much smaller amplitude than analytical solution for SDF system. The problem is defined as following:
*Heading
**
** PARTS
**
*Part, name=Part-1
*Node
1, 0., 0., 0.
*Nset, nset=Part-1-RefPt_, internal
1,
*Nset, nset=_PickedSet2, internal
1,
*Nset, nset=_PickedSet3, internal
1,
*Element, type=MASS, elset=_PickedSet3_Inertia-1_
1, 1
*Mass, elset=_PickedSet3_Inertia-1_
1.,
*Spring,elset=Springs/Dashpots-1-spring, nonlinear, dependencies=1
2
0,0
288.0,0.18
6180.0,10
*Element, type=Spring1, elset=Springs/Dashpots-1-spring
2, 1
*End Part
**
**
** ASSEMBLY
**
*Assembly, name=Assembly
**
*Instance, name=Part-1-1, part=Part-1
*End Instance
**
*Nset, nset=Set-1, instance=Part-1-1
1,
*Nset, nset=_PickedSet6, internal, instance=Part-1-1
1,
*Nset, nset=_PickedSet7, internal, instance=Part-1-1
1,
*End Assembly
**
** BOUNDARY CONDITIONS
**
** Name: BC-1 Type: Displacement/Rotation
*Boundary
_PickedSet6, 1, 1
_PickedSet6, 6, 6
**
** PREDEFINED FIELDS
**
** Name: Predefined Field-1 Type: Velocity
*Initial Conditions, type=VELOCITY
_PickedSet7, 1, 0.
_PickedSet7, 2, 20.
** ----------------------------------------------------------------
**
** STEP: Step-1
**
*Step, name=Step-1, nlgeom=NO, inc=10000
*Dynamic,application=TRANSIENT FIDELITY
0.001,0.1,2e-08
**
** OUTPUT REQUESTS
**
*Restart, write, frequency=0
*Output, field, frequency=0
**
** HISTORY OUTPUT: H-Output-2
**
*Output, history, frequency=1
*Node Output, nset=Set-1
A2,V2,U2
*End Step
The spring is designed to have a stiffness 1600 for u < 0.18, and 600 for u>0.18.
The mass of the object is 1; the initial velocity of the object is 20.
The displacement vs acceleration figure is as following, the slope of the curve is the same as expected.
The displacement and velocity curve seem OK, too.
However, the amplitude should be 0.78 instead of 0.6 as in the result.
The amplitude can be calculated by the potential energy. The initial kinetic energy is 20^2/2=200. The potential energy at displacement 0.6 is 0.18^2*1600/2+(0.6^2-0.18^2)*600/2=125, which is much smaller than the initial kinetic energy.