Solution VMT WS0607

Chair of Machine Tools Written Exam Task 1 P. 1/39

Esser Virtual Machine Tool Date: F07

SOLUTION

Machine Tool Design

Total score : 9.0 Points

1.1 Turning lathes

In the scetch below a turning lathe is shown with its basic components. Please fill in the correct names of the main components of the machine tool. (2.5 pts)

1.2 How can the design of a lathe help to reduce thermal errors using an inclined machine bed (slant bed) (please answer in one sentence)? (1 pt.)

The slant-bed design enables a quick removal (0.5 pts.) of the hot chips (0.25 pts.) and the cooling lubricant (0.25 pts.).

head stock (0.5 pts.) cross slide (0.5 pts.)

bed (0.5 pts.) longitudinal slide (0.5 pts.)

tail stock (0.5 pts.)



SOLUTION

1.3 Classification of manufacturing systems based on degree of automation

Please list the 5 degrees of automation in machine tools and state the typical features of automation. (3.75 pts)

Classification name Automation feature

1) _machine tool______ Generation of cutting motion / feed / process forces_

2) _NC-machine tool____ Automated control of machining sequences_______

3) _manufacturing centre_ Automated tool exchange with tool storage________

4) _manufacturing cell___ Aut. exchange of work piece with work piece storage

5) _transfer line, FMS __ Aut. flow of work piece and tool between machines_

Each classification name: 0.25 pts, each matching feature: 0.5 pts



SOLUTION

1.4 Force flux in machine tools

Scetch the force flux for a force load in direction of the y-axis. Start with defining the load. (1.75 pts.)

force load on cutter

0.25 pts. for each

feature of the scetch:

1. load

2. spindle bearing

3. Z-guide

4. Z-drive (no effect)

5. Y-guide (no effect)

6. Y-drive

7. table guide


R. Meidlinger Virtual Machine Tool Date: F07

SOLUTION

Construction and Design of Structural Components

Total score : 10,0 Points

3.1 Static and Dynamic Characteristics of Machine Tools

The static and dynamic behaviour of machine tools can be described by frequency response functions (FRFs) of compliance. Figure 3.1 shows a simplified milling machine tool in cross-bed-type. The naming of the components and the reference coordinate system can be taken from this picture.

The relative dynamic compliance between tool and workpiece is simulated in all three coordinate directions at the Tool-Center-Point (TCP). The FRF amplitudes can be taken from Figure 3.2 to 3.4.

Fig. 3.1 Simplified CAD-Model of the milling machine tool

Tool

Workpiece

Fixture

Table

Spindle-

carrier

Column

Machine bed

X

Y

Z



SOLUTION

0,00001

0,0001

0,001

0,01

0,1

1

0 50 100 150 200 250 300 350 400 450 500

Frequency [Hz]

Co

mp

lia

nc

e [

µm

/N]

Fig. 3.2 Relative Amplitude Response Plot GXX

0,00001

0,0001

0,001

0,01

0,1

1

0 50 100 150 200 250 300 350 400 450 500

Frequency [Hz]

Co

mp

lia

nc

e [

µm

/N]

Fig. 3.3 Relative Amplitude Response Plot GYY

0,06

0,037

0,03

70

105



SOLUTION

0,00001

0,0001

0,001

0,01

0,1

1

0 50 100 150 200 250 300 350 400 450 500

Frequency [Hz]

Co

mp

lia

nc

e [

µm

/N]

Fig. 3.4 Relative Amplitude Response Plot GZZ

3.1.1 Calculate the relative static stiffness values kXX, kYY and kZZ. (1 Point)

ii

iiG

1k = 0,25P

m

N7,16

06,0

1

G

1k

N

m

XX

XX µ===

µ also correct: [ ]

m

N2,18...4,15k XX µ

∈ 0.25P

m

N0,27

037,0

1

G

1k

N

m

YY

YY µ===


m

N6,28...0,25k YY µ

∈ 0.25P

m

N7,66

015,0

1

G

1k

N

m

ZZ

ZZ µ===


m

N3,83...6,55k ZZ µ

∈ 0.25P

3.1.2 Given is a mode shape of the machine tool at an eigenfrequency of 70 Hz

(Figure 3.5). Determine the relative compliance δrel in the main vibration direction. (1 Point)

0,015

105



SOLUTION

Fig. 3.5 Mode Shape of the Machine Tool at an Eigenfrequency of 70 Hz

Identification of the X-direction as the main vibration direction 0.25P

Identification of the amplitude peak at 70 Hertz within the GXX-Graph 0.25P

N

m3,0rel

µ=δ also correct: [ ]

N

m32,0...28,0rel

µ∈δ 0.5P

2.1.3 Given are two mode shapes at different eigenfrequencies (see below). Specify the mode shape that occurs at an eigenfrequency of 105 Hz. Use therefor the FRFs given in Figure 3.2 to 3.4 and justify your answer briefly. (1 Point)

Eigenfrequency 105 Hz ?

� Yes � No 0.25P

If yes, justify your choice:

� Two amplitude peaks at 105 Hz

within the GYY and the GZZ-graph (see

FRPs)

� Vibration of this mode shape along

the Y- and the Z-direction

0.75P

Z

Y

X

Z

Y

X



SOLUTION

Eigenfrequency 105 Hz ?

� Yes � No

If yes, justify your choice:

No, because vibration directions of this

mode shape are along X and Z; no peak

at 105 Hz within the GXX-Graph

2.1.4 Explain briefly the different causes for rigid body modes on the one hand and flexural and torsional modes on the other hand. (1 Point)

Rigid Body Modes Flexural and Torsional Modes

Cause:

Relative deflections in bearings, drives and

joints

0.5P

Cause:

Eigenmodes/deformation of frame parts

0.5P

Z

Y

X



SOLUTION

of 39 Solution VMT WS0607.doc

3.2 Finite-Element-Analysis (FEA)

3.2.1 Explain the general idea the FEA is based on in two short sentences. (0.5 Points)

The Finite-Elements-Method/Analysis is based on the idea of dividing any continuum

into simple-shaped elements of a limited (finite) size. (0.75P)

For each of these elements, approximation statements are formulated, which describe

the required state variables by simple functions. (0.75P)

3.2.2 Specify the equation for the general dynamic calculation using the matrix notation. Name all of the variables. (1.5 Points)

[ ] { } [ ] { } [ ] { } { })t(FUKUCUM =⋅+⋅+⋅ &&&

[ ]M Mass Matrix

{ }U&& Acceleration Vector of the Nodes

[ ]C Damping Matrix

{ }U& Velocity Vector of the Nodes

[ ]K Global Stiffness Matrix

{ }U Dislocation Vector of the Nodes

{ })t(F Force and Moment Vector

3.2.3 How can the equation be simplified when performing a static calculation? (0.5 Points)

[ ] { } { }FUK =⋅ (0.5P)

3.2.4 How can the equation be simplified when calculating the linear eigenmodes and –frequencies? (0.5 Points)

[ ] { } [ ] { } 0UKUM =⋅+⋅ &&

(0.5P)

0.5P

0.25P

0.25P

0.25P

0.25P



SOLUTION


3.2.5 Which finite element type can be used for the approximation of the stiffness of ball screw spindles and linear guides, when examining complete machine tools? (0.5 Points)

Translational Springs (0.5P)

3.2.6 Which condition has to be fulfilled to be able to approximate structural components with ‚Thin Shell‘-Elements? Name a typical component. (1 Point)

Outer dimensions of the structure are large compared to the wall thickness. (0.5P)

Typical Components: machine bed, column, spindle carrier (0.5P)

3.2.7 Name two advantages and one disadvantage of using ‚Thin Shell‘-Elements instead of ‚Volume‘-Elements. (1.5 Points)

Advantages

� Number of nodes can be decreased significantly (decreasing of calculation time)

(0.5P)

� Optimization of the wall thicknesses can be conducted

(0.5P)

Disadvantage

� The effort to create mid-surface models of the parts is higher compared to volume

models.

(0.5P)


Zuber Virtual Machine Tool Date: F07

SOLUTION


Topic: Structural Optimisation

Total score : 12 Points

4.1 The mathematical formulation of a structural optimisation consists of the optimisation function (or optimisation goal) f(x) and a number of optimisation restrictions gi(x).

4.1.1 Name three possible optimisation goals which can be pursued in the structural optimisation of a machine tool!

Choose three from: minimum mass, maximum stiffness, maximum 1st

eigenfrequency, optimal damping, minimum stresses. Alternative answers may be possible.

(1.5 points)

4.1.2 Name three possible optimisation restrictions for a structural optimisation of a machine tool!

Choose three from: mass < maximum mass, stiffness > minimum stiffness, 1st

eigenfrequency > minimum value, damping > minimum value, costs < maximum costs, stresses < stress limit, maximum permittable design space, limits on wall thickness. Alternative answers may be possible.

(1.5 points)



SOLUTION


4.2 The frame of a forging press is subjected to high process forces, as shown in the sketch below.

Principle sketch of a C-type forming press

Process Forces

Maximum stresses

occur here

(it is sufficient to highlightone region!)

Principle sketch of a C-type forming press

Process Forces

Maximum stresses

occur here

(it is sufficient to highlightone region!)

4.2.1 In the sketch above, mark the region where the maximum mechanical stress in the machine frame due to the process forces has to be expected.

(1 point)

4.2.2 What type of structural optimisation should be used in order to minimise the maximum stress in the frame?

shape optimisation

(1 point)



SOLUTION


4.3 The spindle housing of a high-dynamic milling machine is made of fibre-reinforced plastic (FRP)

4.3.1 Name the two components which build up FRP, as used in machine tool design.

Fibres (carbon) and matrix (polymer)

(1 point)

4.3.2 Name two advantages of FRP over conventional engineering materials for machine tool structures.

Choose two from: low density, high stiffness, high strength, material can be “tailored” to load path by aligning the fibre directions

(1 point)

4.3.3 What other materials are usually used for machine tool frames? Name two.

Choose two from: cast iron (grey or spheroidic graphite), polymer concrete, steel

(1 point)



SOLUTION


4.3.4 Name two possible optimisation parameters that specifically take the material properties of FRP into account!

thickness of each laminate layer, fibre angles

(1 point)

4.4 The optimisation algorithms that are usually used in structural optimisations of machine tool components can be divided into three categories.

4.4.1 Name these three categories.

mathematical programming (MP)

optimality criteria (OC)

genetic or evolutionary strategies (GA or EA)

(1.5 points)

4.4.2 Which of the three algorithms utilises the gradient of the design goal with respect to the optimisation parameters in oder to find an optimal solution?

mathematical programming (MP)

(0.5 point)



SOLUTION


4.4.3 What is the main advantage of the gradient based approach?

Very fast, needs only a few iterations to find optimum (compared to other algorithms)

(0.5 points)

4.4.4 What is the main disadvantage of the gradient based approach?

very sensitive against local minima!

(0.5 points)


RWTH, University of Aachen Prof. Dr.-Ing. C. Brecher Virtual Machine Tool Date: F07

Responsible Assistant: W. Klein


Topic: Requirements on and calculation of guideway systems


5.1 Name two requirements and functions (two for each) of guideway systems in

machine tools (0,25P each). 1P

requirements: functions:

low wear connect relatively moving parts

low friction assure exact movement along one axis

high accuracy take up mass forces

high stiffness take up process loads

low costs

high damping

5.2 Explain how the effects listed below influence the life span and stiffness properties

of guideway systems (fill in: ↑ for increasing lifespan,.↓ for decreasing lifespan, - for

no influence) (0,25P per entry). 1,5P

effect influence on lifespan influence on stiffness

higher preload ↓ ↑

roller guideway instead of ball

guideway (same preload,

same guideway size)

- ↑

wear on carriage and rail

caused by abrasive chips

↓ ↓





5.3 Name two friction sources in rolling contact guideway systems (0,25P each). 0,5P

ball return

ball contact

sealing

5.4 Sketch qualitatively the physical principles of hydrodynamic, hydrostatic and ball

guideway systems and name the main components (0,5 P each). 1,5 P

oil storage tank

slide

hydrodynamicslideway

and plain bearingaerostaticslideway

and plain bearing

air pressuresystem

hydraulicsystem

hydrostaticslideway

and plain bearinglinear ball

guidance systemand ball bearing





5.5 Name three possible calculation objectives (with regards to the guideway system or

the machine) when calculating guideway systems (0,25P each). 0,75P

life span

load bearing capacity

stiffness (machine/ guideway system)

eigenfrequencies (machine)

5.6 Name three effects that cause guideway failure (0,25P each). 0,75P

overload

chips in the rolling elements‘ chains

wear of rolling elements and rail

low/no lubrication

5.7 Given is the displacement curve for a rolling contact guideway system (see figure).

Calculate the compression, tension and side stiffnesses at 40 kN load for a

guideway system of the sizes 45 and 65 each. Use the curves given below! 1,5P





tension load

compression load

side load

c45, tension= 40000 N/26 µm=1538 N/µm (+- 154 N/µm)

c65, tension= 40000 N/9 µm=4444 N/µm (+- 444 N/µm)

selection of displacements:

(+-10% dev.) 0,125P each (0,75P)

calculation of stiffness values:

(+-10% dev.) 0,125P each (0,75P)

c45, compression= 40000 N/17 µm=2353 N/µm (+- 235 N/µm)

c65, compression= 40000 N/9 µm=4444 N/µm (+- 444 N/µm)

c45, side= 40000 N/42 µm=952 N/µm (+- 95 N/µm)

c65, side= 40000 N/27 µm=1481 N/µm (+- 148 N/µm)





5.8 Given is a compliant machine tool column based on a guideway system (two

carriages per rail, see figure). The column has a weight of 1500 kg. Additionally, the

column is loaded with a vertical force of 40000 N in the middle axis of the column

(point P) so that all carriages are loaded equally. (gravity coefficent g=9.81 m/s²,

assume that each carriage takes the same amount of the gravity force!)

5.8.1 Using the stiffness value of exercise no. 7, calculate the overall

compression stiffnesses kcombined, 45 and kcombined, 65 of the column‘s

guideway system consisting of 4 carriages (assume kcarriage 45=2000 N/µm

and kcarriage 65=4000 N/µm in case of failing task 7!). 0,5 P

parallel springs: kcombined, 45= 4 * k45, compression= 4 * 2353 N/µm= 9412 N/µm

kcombined,65= 4 * k65, compression= 4 * 4444 N/µm= 17776 N/µm

(failed task 7:

kcombined, 45= 4 * k45, compression= 4 * 2000 N/µm= 8000 N/µm

kcombined,65= 4 * k65, compression= 4 * 4000 N/µm= 16000 N/µm)

(calculation: 0,25 P each)

x

y

F=40000 N

k carriage 45, 65

k column P

z





5.8.2 The column’s stiffness is given in three coordinate directions with respect to

point P:

,

,

,

60

60 /

2500

column x

column y

column z

k

k N µm

k

=

From the diagram in task 7 choose the minimum sized guideway system so that

the column’s overall compression displacement (column plus guideway system)

on point P is lower or equal than 27 µm (Assume that the gravity force and the

load F both act in point P! Pay respect to progressive stiffness curves in the

diagram!). 1,5 P

Displacement of point P at 54715 N: 21,9 µm (kcolumn, z)

Load per carriage: 54715 N/4=13678,8 N, allowable displacement: 5,1µm

Diagram ‚compression‘: check displacements at 13678,8 N (<5,1 µm)

� Minimum sized guideway system: 45

(displacement of column: 0,5 P,

calculated load per carriage 0,5P,

right selection of guideway system 0,5P)


Türker Yagmur Virtual Machine Tool Date: F07

SOLUTION


Topic: Bearings, Main spindles


5.1 In machine tools, different principles of guideways and bearings are used. Figure 1 shows 3 principles of guideways and bearings.

.

Figure 1: Different types of guideways and bearings

5.1.1 Name the principles of the shown guideways and bearings.

3*0.25 points = 0.75 points

5.1.2 Figure 2 illustrates a electromagnetic bearing. Name the different components.

Aerostatic bearing/guideway

Roller bearing/guideway

Hydrostatic bearing/guideway



SOLUTION


Figure 2: Electromagnetic bearing

5.1.3 Explain shortly, why the electromagnetic bearing is suitable for highest rotational speeds. Name also 2 advantages and 2 disadvantages of this principle.

Explanation:

The electromagnetic bearing comprises of non touching parts. Therefore, it has almost no friction torque and a lubrication is not required. (0.5 points)

Advantages:

- active machine elements

- low friction torques

Disadvantages:

- low maximum loads

- expensive periphery needed

Touch down bearing

Radial bearing

Axial bearing

Position sensor

4*0.25 points = 1 point



SOLUTION


- heat losses in the coil

4*0.25 points = 1 point

5.2 The different types of roller bearings differ in their properties. Therefore, it is not possible to use all types for all tasks.

5.2.1 Name two roller bearing types which are suitable to realize high rotational speeds. Explain why these types are suitable for high rotational speeds.

Bearing types:

Angular contact ball bearing (spindle bearing),

Deep groove ball bearing,


Explanation:

The called bearings consist of balls as sliding bodies. The small contact surface (point contact) provides a low friction between ball and inner/outer ring. Furthermore, a favourable lubrication can be realized with ball bearings. (0.5 points).

5.2.2 Associate relative to each other the different bearing types according to their properties respective the characteristics radial load accomondation and low

friction. Use for the classification „good“, „middle“, „bad“ the symbols „+“, „o“, „-“. Use following table.

Bearing type: Rad. load accomondation

Low friction

Angular contact ball bearing - +

Taper roller bearing o o

Double row cylindrical roller bearing + -



SOLUTION



5.3 Sketch a rigid-moveable bearing system with following characteristics: - At the rigid bearing side use a tandem back to back arrangement consisting of angular contact ball bearings.

- At the movable bearing side use a face to face arrangement consisting of angular contact ball bearings.

Hint: Sketch in the functional surfaces of the housing, i.e. the bearing seats. Consider how the moveable bearing is preloaded. Draw the pressure angles of the bearings and name the kind of arrangement (back to back (“O”) or face to face (“X”)). Make sure that the bearing rings are mounted in defined position on the spindle and in the housing. Be sure that the spindle cannot be pulled out of the housing!! Name all used components!

Use the following sketch as a help.

cover: 0.25 P

rigid bearing seat: 0.25 P flange

tandem back to back

arrangement: 0.5 P

face to face arrangement with

the correct preload: 0.5 P

lock nut: 0.25 P

correct pressure angles:

- rigid bearings: 0.25 P

- movable bearings: 0.25 P

moveable seat: 0.25 P

nut: 0.25 P

total: 3 P

sleeve: 0.25 P

cover: 0.25 P

rigid bearing seat: 0.25 P flange

tandem back to back

arrangement: 0.5 P

face to face arrangement with

the correct preload: 0.5 P

lock nut: 0.25 P

correct pressure angles:

- rigid bearings: 0.25 P

- movable bearings: 0.25 P

moveable seat: 0.25 P

nut: 0.25 P

total: 3 P

sleeve: 0.25 P



SOLUTION


5.4 During the formulation of the dynamic behaviour of a spindle-bearing-system, a linear differential equation system is developed with the prerequisite of smaller oscillation amplitudes.

5.4.1 Write down the inhomogeneous linear differential equation system for a non-damped system and name all matrices and vectors.

inhomogenious linear differential equaltion for a non-damped system:

(0.5 points)

Mass matrix

Acceleration vector

Stiffness matrix

Displacement vector

Force vector


[ ] { } [ ] { } { }FuKuM =⋅+⋅ &&

Chair of Machine Tools Written Exam Task ? P. 27/39

Gero Gerlach Virtual Machine Tool Date: F07

SOLUTION


Mechanical Feed Drive Components

Total score : 8.25 Points

The structure of a feed axis essentially contains an electric servomotor and the

mechanical transfer elements. All components in the force-flux between motor and tool

respectively workpiece are considered as feed drive components.

7.1.1 Please name two essential tasks of feed drives in machine tools. (0.5 P)

- Generation of relative motion between workpiece and tool for cutting process 0.25 P

- Generation of feed force and feed rate 0.25 P

7.1.2 Please enumerate 4 requirements on feed drives. (1.0 P)

- zero backlash, 0.25 P

- low friction, 0.25 P

- high acceleration capability, 0.25 P

- high stiffness, 0.25 P

- high geometric & kinematic accuracy of feed drive components 0.25 P



SOLUTION


7.2 A machine slide (S) carrying a workpiece (W) has to be driven by a feed drive consisting of a servo motor (M), a gear (G) and a rack-pinion-system (R/P).

M

m w

m s

Getriebe

J M , n Mn , n Mmax

Motor

Werkstück

Schlitten

Ritzel- Zahnstange

i G ,

,J GL

J GM

r R , J R

Führungen

r GLR

Figure 7.1: Structure of a rack-pinion-axis

The following data is available for the feed drive design:

Slide mass (including rack): mS = 400 kg

Workpiece mass: mW = 250 kg

Pinion wheel radius: rP = 50 mm

Pinion mass moment of inertia: JP = 0,019 kgm2

Motor mass moment of inertia: JM = 0,095 kgm2

Mass moment of inertia motor sided gear : JGM = 0,00003 kgm2

Mass moment of inertia load sided gear: JGL = 0,0005 kgm2

Gear wheel radius load sided rGL = 100 mm

Nominal motor speed: nMn = 600 min-1

Maximum motor speed: nMmax = 1400 min-1

motor gear rack-pinion-system

rP, JP

guidance

slide/ table

workpiece



SOLUTION


7.2.1 Please calculate the required gear ratio if a rapid table speed of vmax = vrts = 25 m/min is needed! (2.0 P)

Gear ratio: maxP

maxM

driven

driveup

n

n

n

ni == 0.5 P

Pinion speed: P

rtsP

r2

vn

⋅π= 0.5 P

Result: rts

maxMP

P

rts

maxM

maxP

maxMup

v

nr2

r2

v

n

n

ni

⋅⋅π=

⋅π

== 0.5 P

59,17minm25

min1400m05,02

minm25

min1400mm502i

1

1

1

1

up =⋅

⋅⋅π=

⋅

⋅⋅π=

−

−

−

−

0.5 P

7.2.2 Please choose a whole-numbered gear ratio to ensure the required rapid table speed. (0.75 P)

maxP

maxM

driven

driveupG

n

n

n

nii ==< => 59,1717iG <= 0.75 P

7.2.3 Please evaluate the total mass moment of inertia JRpd,tot reduced to the motor shaft! (2.5 P)

red,W,Sred,Pred,GLGMMtot,Rpd JJJJJJ ++++= 0.5 P

GL2

G

red,GL Ji

1J ⋅= 0.25 P

22

2red,GL kgm00000173,0kgm0005,017

1J =⋅= 0.25 P

22

2P2

G

red,P kgm0000657,0kgm019,017

1J

i

1J =⋅=⋅= 0.25 P

2

PWS2G

red,W,S r)mm(i

1J ⋅+⋅= 0.25 P

22

2

22

2red,W,S kgm0056228,0kgm625,117

1m05,0)kg250kg400(

17

1J =⋅=⋅+⋅=

0.5 P



SOLUTION


Result:

2

2tot,Rpd

kgm10072377,0

kgm)0056228,00000657,000000173,000003,0095,0(J

=

++++=

7.2.4 Please explain shortly the design possibilities to avoid backlash in rack-pinion-drives by a sketch and a description (Name the components). (0.75 P each) (1.5 P)

a) Double helical pinion, preloaded by spring force. (0.75 P)

The backlash is eliminated by the fact that two helical gear pinions comb with a toothed rack. The lower pinion is moved axially through elastic force on a spline shaft step. Through this, both pinions contact the opposite side of the toothed rack.



SOLUTION


b) Electrical preloading of two separate pinions: (0.75 P)

Electrically preloaded dual-pinion drives utilize two motors, each driving one pinion, which are loaded against each other by means of a special motor controller. The realised motor torque M is provided by the motor torques M1, M2. The single torque characteristic are staggered against each other to compensate the backlash of the drive-system. One pinion is considered the driving pinion, while the other is braking. During deceleration, the braking motor takes over control of the axis and the drive motor switches to preloading.


Andreas Schmidt Virtual Machine Tool Date: F07

SOLUTION


Simulation of Feed Drives


8.1 Rotary electrical motors are the usual choice as drives in a machine tool.

8.1.1 Name two of the three physical laws on which the functional principles of all electrical motors are based and describe their meanings in one sentence each.

Ampere’s law: The creation of a magnetic field via a current-carrying conductor.

Faraday’s law of induction: Induction of voltage in a coil via movement in a homogeneous magnetic field. When the electric circuit is closed, current flows in the coil. Alternatively to the movement, the strength of the magnetic field may vary.

Lorentz force law: A (Lorentz) force affects a current-carrying conductor in a magnetic field. Torque is produced by force couples on current-carrying pairs of conductors in the magnetic field or by direct, magnetic forces.

4 * 0,25 Points = 1,0 Points

8.1.2 Describe in 2-3 sentences the basic functional principle of a separately excited DC motor.

DC-excited coils in the stator generate a constant magnetic field (0,25). The armature coil in the rotor is also fed with DC current (0,25). The current-carrying conductors in the stator field experience (Lorentz-)forces, which result in a torque (0,25). The armature current is passed through graphite brushes (0,25) and a mechanical commutator (0,25), which changes the direction of current in the armature coils in a manner that keeps it constant under the stator poles (0,25). With increasing speed a growing counter voltage is induced into the armature circuit, which counteracts the armature voltage (0,25).

7 * 0,25 = 1,75 Points

8.1.3 Describe the idling speed n0 of the DC motor as a function of the armature voltage and the excitation.

Idling: T = 0 ⇒ IA = 0 ⇒ Φ⋅

=1

A0

k

Un

0,5 Points



SOLUTION


8.1.4 Describe the starting torque Tstart of the DC motor as a function of the armature voltage and the excitation.

Starting: n = 0 ⇒ Ui = 0 ⇒ 1 Astart

A

k UT

2 R= ⋅Φ ⋅

π

0,75 Points

8.2. The feedback control loop for the feed axes normally comprises “position control loop”, “velocity (rotational speed) control loop” and also “current control loop”.

8.2.1 Briefly discuss the reasons of employment of the three feedback loops.

Position control loop is to compare the actual position value with the reference position value. The difference between these two values (error signal) will be controlled (compensated, eliminated) in the position controller. Thus, the position controller is fairly important for the feed axes control to maintain the required positioning accuracy.

Velocity (rotational speed) control loop enhances the damping characteristic of the system. This means, this control loop enables the system to achieve the higher system dynamics.

Current control loop is normally enclosed in the feed (drive) amplifier and has the fastest cycle time (working frequency). The cycle time of this controller is generally under 250 sµ (= 4kHz). Main tasks of this control loop are to control

the acceleration of motors and to enhance the dynamic behaviour of the entire control system.

3*0,5 Points= 1,5 Points

8.2.2 What is the difference between open loop control and closed loop control?

open loop: calculation of output after one-time input, no feedback or adjustment, needs fixed, reliable system

closed loop: feedback of output towards the input, continuous comparison and control of the resulting error to reach target value exactly. Better against disturbances

2 * 0,5 Points = 1,0 Points



SOLUTION


8.3 Calculate the transfer function of the following block diagram.

+-

xinput xoutput

G1

G5

G2

G4G3

G6

+

Figure 3:block diagram of a fictitious control technical system

( )( )654

4321

input

output

GGG1

GGGG

x

x

+⋅−

⋅⋅−=

(3 different rules to concatenate subsystems have to be used)

3*0,25 + 0,75 = 1,5 Points

8.4 A PT2-System has three characteristic parameters, the gain constant K, the damping ratio D and the characteristic frequency ωn.

8.4.1 Which effects on the system behaviour do they have?

K controls the system´s amplification from input to output.

The Damping factor indicates the system’s tendency to oscillate.

The characteristic frequency ωn is the oscillation frequency of a system after an impulse excitation (without damping). (Alternative explanation: A forced excitation that matches the characteristic frequency leads to a resonance effect)

3*0.5 = 1.5 points



SOLUTION


8.4.2 Give the standard transfer function of a PT2-System and explain the following range of damping values: D = 0, 0 < D < 1, D = 1, D > 1.

( )2

nn

2

2

nPT

sD2s

KsG

2 ω+⋅ω⋅⋅+ω⋅

=

D = 0: Steady oscillation with the characteristic frequency ωn.

0 < D < 1: Step response of the system converges onto the gain constant K

(y∞ = K).

D = 1: Step response of the system converges onto the gain constant K without oscillation.

D > 1: Step response converges exponentially onto the gain constant K.

transfer function 0.5, range of damping values 4 * 0.25 = 1.5 Points



SOLUTION


Multi-Body Simulation


9.1 The Multi-Body-Simulation (MBS) with rigid bodies is able to compute the behaviour of mechanical systems with large motion amplitudes.

9.1.1 A multi-body simulation software usually offers kinematic, dynamic, static and inverse dynamic analysis. Please discribe the inverse-dynamic analysis. How is this type of analysis related to other types? What is defined as input and what is the output of this simulation?

Hybrid form of the dynamic- and kinematic-analysis. The motion of one or several bodies of the mechanical system is defined and is leading to a complete determination of the position, the velocity and the acceleration of the system. Afterward the algebraic equations of motion are solved by the use of the known positions, velocities and accelerations to determine the forces which cause these motions.

3 * 0,5 Punkte = 1,5 Punkte

9.1.2 There are two general types of connecting elements in the MBS, that can be used to assemble single bodies into a multi-body system. Please give the name of both element types and characterize the type of mathematical equation that result from the element definition

Kinematic joints – algebraic non-linear equations

Force elements (Spring-Damper, etc) – differential equations


9.1.3 The figure shows the schematic assembly of a feed drive for machine tools with electric-servo-motor and a ballscrew-nut-system. Please give the missing names of the joint types, that are used to build a kinematic model of the drive system. The two linear guides are modelled with one constraint. Give the number of kinematic degrees of freedom.



SOLUTION


motor

table

linear guide

transmissiongearing

synchronous belt drive

fixed bearingball screw spindle

spindle nut -rail-shoe

machine bed

motor

table

linear guide

transmissiongearing

synchronous belt drive

fixed bearingball screw spindle

spindle nut -rail-shoe

machine bed

x

y

z

x

y

z

Figure 1 Feed drive with ballscrew-nut system

Fixed bearing between machine bed and spindle revolute joint (1DOF)

Spindle nut between spindle and table screw joint (1DOF)

Linear guide between table and machine bed translational joint (1DOF)


9.1.4 Please give the kinematic constraint equations for the joints in the feed drive system. Remember that a screw is described by the pitch h. If there is no constraint equation for a specific direction mark it as a function of time f(t).

Fixed bearing

=

ϕ∆

ϕ∆

ϕ∆

∆

∆

∆

0

0

)t(f

0

0

0

z

y

x

z

y

x

Linear guide

=

ϕ∆

ϕ∆

ϕ∆

∆

∆

∆

0

0

0

0

0

)t(f

z

y

x

z

y

x



SOLUTION


Spindle nut

π⋅ϕ∆

=

ϕ∆

ϕ∆

ϕ∆

∆

∆

∆

0

0

)t(f

0

0

2/h

z

y

x x

z

y

x

Example: Ball joint

=

ϕ∆

ϕ∆

ϕ∆

∆

∆

∆

)t(f

)t(f

)t(f

0

0

0

z

y

x

z

y

x


9.2 Multi-Body-Simulations can be performed both with rigid and flexible bodies.

9.2.1 In which case is the use of flexible bodies necessary? State two effects onto the simulation results, if rigid instead of flexible bodies are used?

The use of flexible bodies is necessary if deformation of individual machine parts have significant impact onto the machine tool’s dynamic behaviour. Otherwise, neither the eigen-frequencies nor the weak points of the structure can be determined correctly.

0,5 points + 2 times 0,25 points --> 0,5 points

9.2.2 If you have to compute the dynamic behaviour of the system shown in Figure 1, it can be calculated with Finite-Element-Software Packages or with Multi-Body-Simulation. State in which domains, time- or frequency domain, each simulation is running in?

Multi-Body-Simulation: Time domain; FE-Simulation: Frequency domain. 2 times 0,25 points --> 0,5 points

9.2.3 The static and dynamic flexible properties of the table in Figure 1 should be incorportated in the multi-body-computation. Therefor a Finite-Element model of the table was built up. Which computations have to be performed for the integration of flexible bodies into the multi-body-simulation model?

1.) Normal mode computation with constraints on each degree of freedom that transfers forces in the multi-body-simulation-modell.



SOLUTION


2.) Calculation of static correction modes

2 x 0,5 points �� 1,0 points

9.2.4 With the use of flexible bodies in MBS, more realistic results with respect to the dynamic behaviour and calculated mode shapes can be achieved. In this context the MAC-value is often of interest. For what stands the acronym MAC? For what is the MAC-value used? Which meaning do high values of the MAC-value have?

1.) MAC : Modal Assurance Criterion

2.) The MAC-value is used to compare measured and calculates mode shapes

3.) High MAC-values mean a high level of consistency between measured and simulated mode shapes.

3 x 0,5 points �� 1,5 points

9.2.5 Nowadays more and more machine tools use linear direct drives as feed drive systems. What is the problem when modelling these kind of feed drives compared to normal ball screw drives? Explain how this problem can be treated by the Multi-Body-Simulation?

Linear direct drives do not have mechanical stiffness properties like mechanical components like e.g. ball screw drives. Linear direct drives’ stiffness and damping is determined solely by the drive’s feedback control system. The simulation of this control system can only be performed in the Computer Aided Control Engineering (CACE) environment (MATLAB/Simulink) with the simulation of the control loop.

2 times 0,5 points --> 1,0 points

Solution VMT WS0607

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