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Berechnung von Werkzeugmaschinen
3 Beispiele
Roberto Rossetti, CADFEM (Suisse) AG
grinding machine
milling machine
grinding machine
Simulation of 3 machine tool geometries
Hypothesis :
- Electric motors and their control loop are considered as infinitely rigid (the
equivalent stiffness values of the electric motors control are not considered in this
exercise)
- Contact surfaces between bodies are perfectly bonded (the bolts are stiff
enough to hold the parts together)
- The stiffness values for bearings, guideways, harmonic drives and ball screws
given by the suppliers are correct
- The geometrically simplified details have a negligible influence on the results
- As both displacements and rotations are small, a linear FEA is adequate
Goal :
- Get the global stiffness matrix and the influence coefficient matrix
- Get the coupling terms (off diagonal terms / crosstalk)
- Stress concentration regions (qualitative for better understanding)
Grinding machine 1 - Stiffness analysis
Shaft_A_axis
Z_axis
X_axis
Fixed and mobile
ballscrew support z
DesignModeler is very practical to perform such tasks
Simplification of the geometry
2 faces2 faces
5 faces
3 faces
Bearings, guideways , ball screws, “feet”
Kinematic connections like bearings or guideways are too complex to be
modelled efficiently using finite element analysis. This would need a very
large number of nodes and would end up with convergence issues. Thus,
it is best practice to model them with idealized joints (bushing).
Bushing : a bushing has 6 degrees of freedom, 3 translations and 3 rotations, all of which
can be characterized by their translational and rotational stiffness, respectively.
The 3 translational DOFs and the 3 rotations DOFs are : Ux, Uy, Uz, and φ, Θ, ψ. The
forces developed in the bushing are expressed as:
{F} are the forces, {T} are the torques, and [K] is the 6x6 stiffness matrix. Off diagonal
terms in the matrix are coupling terms between the DOFs. These are the so called
crosstalk terms.
z
y
x
U
U
U
KT
F
BUSHING
Bushing example : bearings
000000
000000
000000
00000
00000
00000
axial
radial
radial
K
K
K
K
y
z x
y
z x
Know how
Idealisation study at the ETHZ
Was is the best way to idealise guideways with FEM?
Specific elements available into ANSYS Workbench
Very good correlation between measurement and simulation
- 7 / 38-
Stiffness analysis
Mode Measurement [Hz] Simulation [Hz ] Delta abs(Delta)
1 326 314.53 3.52% 3.52%
2 432 425.78 1.44% 1.44%
3 461 460.76 0.05% 0.05%
4 464 509.65 -9.84% 9.84%
5 566 562.79 0.57% 0.57%
6 600 610.8 -1.80% 1.80%
7 775 776.69 -0.22% 0.22%
8 803 833.74 -3.83% 3.83%
9 927 956.68 -3.20% 3.20%
Mean value -1.48% 2.72%
The static structural analysis counts 3 steps :
Each step corresponds to a load case. Each load case
applies a unit force in one of the 3 directions of the
global coordinate system. A contact point between tool
and workpiece is defined (TCP). From the contact
point, a remote force to a surface of “tool side” and an
opposite remote force to a surface of “workpiece side”
are applied.
Stiffness analysis - load case
y
xz
tool side
workpiece side
Rigidity matrix
Influence coefficient matrix [m/N ] : (unknown)
Displacement vector [m] : (measured at the
remote point)
Force vector [N ] : (known)
Influence coefficient matrix ?
3 load cases :
fA
A
f
,
0
0
1
00
00
00
1
1
1
1
1
1
1
1
1
1
1
1
z
y
x
z
y
x
,
0
1
0
00
00
00
1
2
1
2
1
2
1
2
1
2
1
2
z
y
x
z
y
x
1
0
0
00
00
00
1
3
1
3
1
3
1
3
1
3
1
3
z
y
x
z
y
x
111
fA
1
3
1
2
1
1
1
3
1
2
1
1
1
3
1
2
1
1
1
zzz
yyy
xxx
A
,
0
0
1
00
00
00
2
1
2
1
2
1
2
1
2
1
2
1
z
y
x
z
y
x
,
0
1
0
00
00
00
2
2
2
2
2
2
2
2
2
2
2
2
z
y
x
z
y
x
1
0
0
00
00
00
2
3
2
3
2
3
2
3
2
3
2
3
z
y
x
z
y
x
222 fA
2
3
2
2
2
1
2
3
2
2
2
1
2
3
2
2
2
1
2
zzz
yyy
xxx
A
y
xz
(1)
(2)
y
xz
(1) stands for workpiece
(2) stands for tool
Results – rigidity matrix, influence coefficient
For any force applied to the machine-tool
The total deformation of the machine-tool is
The global influence coefficient matrix is given by
The global stiffness matrix is given by
2
3
1
3
2
2
1
2
2
1
1
1
2
3
1
3
2
2
1
2
2
1
1
1
2
3
1
3
2
2
1
2
2
1
1
1
zzzzzz
yyyyyy
xxxxxx
totA
21 ff
21
tot
12112112211 )( fAAfAfAfAfAtot
)( 21 AAAtot
1 tottot AK
Kf
Results – rigidity matrix, influence coefficient
matrig_param.txt :
\prep7
*dim, resultdisp1, array, 3, 3
*dim, resultdisp2, array, 3, 3
!read results step 1
set,1
*get,resultdisp1(1,1),node,node1,u,x
*get,resultdisp1(2,1),node,node1,u,y
*get,resultdisp1(3,1),node,node1,u,z
*get,resultdisp2(1,1),node,node2,u,x
*get,resultdisp2(2,1),node,node2,u,y
*get,resultdisp2(3,1),node,node2,u,z
!read results step 2
set,2
*get,resultdisp1(1,2),node,node1,u,x
*get,resultdisp1(2,2),node,node1,u,y
*get,resultdisp1(3,2),node,node1,u,z
*get,resultdisp2(1,2),node,node2,u,x
*get,resultdisp2(2,2),node,node2,u,y
*get,resultdisp2(3,2),node,node2,u,z
!read results step 3
set,3
*get,resultdisp1(1,3),node,node1,u,x
*get,resultdisp1(2,3),node,node1,u,y
*get,resultdisp1(3,3),node,node1,u,z
*get,resultdisp2(1,3),node,node2,u,x
*get,resultdisp2(2,3),node,node2,u,y
*get,resultdisp2(3,3),node,node2,u,z
a=sign(1,resultdisp1(1,1))*(resultdisp1(1,1)-resultdisp2(1,1))
b=sign(1,resultdisp1(2,2))*(resultdisp1(1,2)-resultdisp2(1,2))
c=sign(1,resultdisp1(3,3))*(resultdisp1(1,3)-resultdisp2(1,3))
d=sign(1,resultdisp1(1,1))*(resultdisp1(2,1)-resultdisp2(2,1))
e=sign(1,resultdisp1(2,2))*(resultdisp1(2,2)-resultdisp2(2,2))
f=sign(1,resultdisp1(3,3))*(resultdisp1(2,3)-resultdisp2(2,3))
g=sign(1,resultdisp1(1,1))*(resultdisp1(3,1)-resultdisp2(3,1))
h=sign(1,resultdisp1(2,2))*(resultdisp1(3,2)-resultdisp2(3,2))
i=sign(1,resultdisp1(3,3))*(resultdisp1(3,3)-resultdisp2(3,3))
a1=sign(1,resultdisp1(1,1))*resultdisp1(1,1)
b1=sign(1,resultdisp1(2,2))*resultdisp1(1,2)
c1=sign(1,resultdisp1(3,3))*resultdisp1(1,3)
d1=sign(1,resultdisp1(1,1))*resultdisp1(2,1)
e1=sign(1,resultdisp1(2,2))*resultdisp1(2,2)
f1=sign(1,resultdisp1(3,3))*resultdisp1(2,3)
g1=sign(1,resultdisp1(1,1))*resultdisp1(3,1)
h1=sign(1,resultdisp1(2,2))*resultdisp1(3,2)
i1=sign(1,resultdisp1(3,3))*resultdisp1(3,3)
a2=-sign(1,resultdisp1(1,1))*resultdisp2(1,1)
b2=-sign(1,resultdisp1(2,2))*resultdisp2(1,2)
c2=-sign(1,resultdisp1(3,3))*resultdisp2(1,3)
d2=-sign(1,resultdisp1(1,1))*resultdisp2(2,1)
e2=-sign(1,resultdisp1(2,2))*resultdisp2(2,2)
f2=-sign(1,resultdisp1(3,3))*resultdisp2(2,3)
g2=-sign(1,resultdisp1(1,1))*resultdisp2(3,1)
h2=-sign(1,resultdisp1(2,2))*resultdisp2(3,2)
i2=-sign(1,resultdisp1(3,3))*resultdisp2(3,3)
my_dispA_1=a*1000
my_dispA_2=b*1000
...
Results – rigidity matrix, influence coefficient
- 12 / 38-
T
Results – rigidity matrix, influence coefficient
Concept
Find the first eigenfrequencies and eigenmodes of the machine tool.
Generally speaking, for a given energy, the higher the eigenfrequency,
the lower the amplitude.
Goal is to
Identify the eigenmodes that might influence the machining process
Find constructive modifications that will change these modes and / or
increase the corresponding frequency.
- 13 / 38-
Modal analysis
Steps
Similar to a static analysis
Define point masses replacing non-idealised parts
The liaison to the ground is very important
23 Hz
28.4 Hz
35.3 Hz
50.7 Hz
48.1 Hz
58.1 Hz
62.3 Hz
77.8 Hz
68.5 Hz
97.6 Hz
flexible behavior
rigid behavior
Modal analysis – flexible vs. Rigid bodies
Variation of the influence coefficient matrix
The plates (1) and (2) are kept in the
model. By doing so, the top surfaces of
all feet of both plates remain in the
same plane. This will become clear
when the loads are defined.
(1)
(2)
New machine bed design
our proposal:
The existing vertical plates are aligned under the 12 feet (1), vertical metal sheets
are added (2), the geometry is partially closed (3) and the 5 feet thickness is
increased to 10 mm (4).
original design
(1) and (2)
(3)(4)
New machine bed design
Static Structural : Total deformation
Mass center displacement (µm)
Deformation Probe : -1.4 (-31%) 1.57 (-44%) -0.65 (-52%)
Deformation Probe 2 : 0.36 (-36%) 1.04 (-43 %) -0.39 (+238%)
New machine bed design - Results
Modal analysis
Original design New design
76 Hz
227 Hz
203 Hz
(+167%)
262 Hz
(+15%)
New machine bed design - Results
Concept
What is the influence of a temperature difference on the accuracy?
Let’s say the machine-tool is set up at 8am. Room temperature is 22°C
During the day, the room temperature rises to 23°C. The machine-tool will
change its size due to dilatation. What is the influence on accuracy?
- 20 / 38-
Temperature difference & Accuracy
Steps
3D parts are needed to replace bushings, for example so that the spindle
expansion can be taken into account + material data
As the temperature is uniform,
• there is no gradient
• no thermal coefficients are needed.
Definition of the temperature difference.
Important is the displacement difference between tool and workpiece
- 21 / 38-
Total deformation
x-dir. 2.1 μm
y-dir. -3.0 μm
z-dir. -0.1 μm
Temperature difference & Accuracy
The effect of material differences is illustrated in the figures above. The base plate
expanses more than the x and the z plates.
Temperature difference & Accuracy
3 translation
axes (x, y and
z axis)
Milling machine
Concept
The machine is excited over a frequency range by a given force (1000 N).
Amplitudes and phases are computed
Results accuracy depend greatly on the accuracy of the input force and on
the damping used
Very useful to determine how much an eigenmode will influence the
machining process and to get amplitudes between modes.
- 24 / 38-
Harmonic analysis
Steps
Similar to a static and modal analysis
Define frequency range and sampling.
- 25 / 38-
Harmonic analysis
Important is the displacement difference between tool and workpiece
MACHINE-TOOL WORKSHOP – 3 days
Table of content
WS01 : test bed guideways idealisation
WS02 : Static Structural Analysis
WS03 : Modal Analysis
WS04 : Parametric Static Structural – Stiffness
WS05 : Parametric Static Structural - gravity load
WS06 : Modal Analysis
WS07 : Thermal Dilatation
WS08 : Design Study - machine bed
WS09 : Static Analysis Milling Machine
WS10 : Modal Analysis Milling Machine
WS11 : Harmonic Analysis Milling Machine
Thanks!
CADFEM (Suisse) AG
Avenue de Cour 74
1007 Lausanne
Tél.: 021 601 70 80
info@cadfem.ch
www.cadfem.ch
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