summary of support tube r&d nov. 11, ’04 kek h. yamaoka

Summary of Support tube R&D

Nov. 11, ’04

KEK H. Yamaoka

・ Components at IR region Supported by Tungsten tubes, CFRP tube・ For high luminosity 　 Ground motion, culture Noise

→ Analyses, Excitation tests

Study Items;

・ Consistency of analyses

・ How much is relative

amplitude?

|P1-P2| < 1nm(Criteria)・ Is necessary CFRP tube?→ How much thickness?

Introduction

4m

P1 P2

Tungsten Tube(t=100)

CFRP tube

(2) 20t x 100w x 1440L

(4) 20t x 100w x 695L 2.5t x 100w x 200L

(1) 20t x 100w x 695L

(3) 20t x 100w x 695L

Exciting Test

(5) φ80 x 10t x 200L(1/10 scale)

(6) φ80 x 10t (1/10 scale)t=3mm

Input exciting force(Excitation table or Impact hammer)　　　　↓ Measurement: natural frequencies Mode shape　　　　↓ Compare to the FEM results

20t x 100w x 695L 2.5t x 100w x 200L

t=3mm

2: 40Hz

3: 161Hz

1: 26Hz

2: 43Hz

3: 184Hz

1: 30HzMeasurement FEM

1

X

Y

Z

JUL 29 200318:01:57

DISPLACEMENT

STEP=1SUB =5FREQ=259.703DMX =38.732

1

X

Y

Z

JUL 29 200318:01:38

DISPLACEMENT


1: 76Hz

2: 256Hz

3: 489Hz

1

X

Y

Z

JUL 29 200318:06:24

DISPLACEMENT


1

2

34

56

7

8

910

1112

131415

1617

18

192021

(+Y): [ ] 521 右側ｺﾝﾌﾚ゚ｯｸｽ

: 1.0, : 振幅ﾄｳ゙ｪﾙ(g) X,Y,Z 方向遠

1

234

5678

9

101112

1314

1516

17

18 19

20

21

(+Y): [ ] 258 右側ｺﾝﾌﾚ゚ｯｸｽ


123

456

7891011121314

151617

18

192021

(+Y): [ ] 77.5 右側ｺﾝﾌﾚ゚ｯｸｽ


1: 78Hz

2: 258Hz

3: 522Hz

1st ～ 3rd mode: Good agreement with FEM Support structure should be modeled in FEM.

1 10 100 10001E-6

1E-5

1E-4

1E-3

0.01

0.1

1

10

100

Am

plit

ud

e(n

m)

Frequency(Hz)

1 10 100 10001E-6

1E-5

1E-4

1E-3

0.01

0.1

1

10

100

Am

plit

ud

e(n

m)

Frequency(Hz)

Stiffness: 1:512

2.5mm thick plate

By connecting very weak structure, Deviation can be absorbed. Correlation can be given.Relative amplitude can be estimated.

1st: 99Hz2nd: 548Hz

1st: 106Hz2nd: 548Hz

1st: 88Hz

2nd: 126Hz

Case-1

20mm thick plate

Case-2

1

X

Y

Z

F0cos(t)=(m ・ a)sin(t)

QC-L: Tungsten(100mm) QC-R: Tungsten(100mm)

CFRP

0 – 1000HzCalculation of relative amplitude

1

X

Y

Z

JUN 23 200409:31:56

ELEMENTS

REAL NUM

U

2m3.85m

7m8m

QC-L QC-R

3mm 5mm 10mm0Hz1Hz3Hz5Hz

CFRP thick.

Difference of 1st mode of resonantfrequency between QC-R and QC-L.

Get relative amplitude (QC-R)-(QC-L)

(Model-A)

(Model-B) F0cos(t)

2m7m8m

P1 P2

Estimation of Input Acc.

Data: Vertical @ATF(17:00 Feb. 10, 2004)

(ﾛｸﾞ周波数0.1 1 10

1E- 6

1E- 7

1E- 8

1E- 9

Linear Spectrum

2x10-7m/s2

Input Acc. = 2x10-7m/s2

Mass=90tons/9.8[m/s2]

Self weight

Input data

1

X

Y

Z

APR 7 200414:19:12

DISPLACEMENT


U

1st mode 2nd mode

Same phase Opposite phase

Relative amplitude: QC-R and QC-L

Amplitude: QC-R and QC-L

t100mm t100mmt5mm

In case of 100mm-5mm(CFRP)-100mm, f=0Hz

Relative amplitude between QC-R and QC-L

QC-R QC-RQC-LQC-L

Dif

fere

nc

e(n

m)

1e-4

1

1

X

Y

Z

JUN 23 200409:31:56

ELEMENTS

REAL NUM

U

QC-L QC-R

1

X

Y

Z

QC-L QC-R

Other calculations

Diff: Relative amplitude between QC-R and QC-L.

Tungsten Tungstenf Mode 3mm 5mm 10mm 100mm 3mm 5mm 10mm 100mm

Freq.(Hz) 75.8 76.0 76.0 49.2 17.583 17.7 17.9 16.3Diff.(nm) 0.098 0.055 0.016 9.09E-05 0.065 0.026 0.008 2.60E-04Freq.(Hz) 77.5 78.6 81.4 113.0 20.2 21.8 25.0 44.4Diff.(nm) 0.213 0.206 0.192 0.080 2.816 2.398 1.763 0.325Freq.(Hz) 75.2 75.3 75.5 49.0 17.0 17.1 17.3 16.0Diff.(nm) 0.129 0.080 0.035 1.10E-03 0.823 0.480 0.234 3.60E-02Freq.(Hz) 77.0 78.2 80.9 112.6 19.8 21.4 24.7 43.3Diff.(nm) 0.174 0.177 0.177 0.081 2.357 2.186 1.696 0.334Freq.(Hz) 73.4 73.7 74.2 48.5 15.6 15.9 16.3 15.3Diff.(nm) 0.173 0.133 0.076 3.26E-03 2.339 1.496 0.752 1.30E-01Freq.(Hz) 76.9 77.8 80.4 111.9 19.5 20.9 24.1 41.0Diff.(nm) 0.094 0.116 0.143 0.082 1.547 1.726 1.532 0.349Freq.(Hz) 71.5 72.0 72.7 48.1 13.9 14.4 15.1 14.4Diff.(nm) 0.204 0.172 0.113 5.62E-03 3.813 2.631 1.396 2.84E-01Freq.(Hz) 76.8 77.6 80.0 111.0 19.3 20.6 23.6 38.4Diff.(nm) 0.056 0.078 0.114 0.084 1.077 1.343 1.343 0.359Freq.(Hz) 75.6 17.4Amp(p-p) 0.224 3.869

CFRP CFRP3-Point fixed(Both end+3.85m) 2-Point fixed(Both end)

1stCanti

1st

2nd5Hz

1st

2nd3Hz

0Hz1st

2nd

1st1Hz

2nd

Results

1

X

Y

Z

JUN 23 200409:31:56

ELEMENTS

REAL NUM

U

P1 P2

1

X

Y

Z

P1 P2

Natural frequency: 76Hz, relative amp. : 0.1nm <1nmThis is ideal configuration.In case of no CFRP tube(Cantilever): Amplitude=0.2nm <1nm　 CFRP tube is not necessary because of less than 1nm.　 However, it is difficult to amount on a very stiff base stand. So actual natural frequency must be lower than this value.

Natural frequency: 17Hz, relative amp. : 2 ～ 3nmIn case of no CFRP tube(Cantilever): Amplitude= 4nm

⇒ CFRP tube: - No efficient to reduce amplitude. - Deviation of natural frequency between two tubes can be absorbed.

(Model-A)

(Model-B)

1

X

Y

Z

APR 7 200414:19:34

DISPLACEMENT

STEP=1SUB =3FREQ=79.927DMX =.986417

U

1

X

Y

Z

APR 7 200414:19:12

DISPLACEMENT


U

7m

8m

2m

Right side: 70Hz

LC

RCFRP: Changed!

Left side: 75Hz

L R

L

R

1st mode

2nd mode

70Hz75Hz

1

X

Y

Z

APR 7 200414:09:13

DISPLACEMENT


U

1

X

Y

Z

APR 7 200414:08:52

DISPLACEMENT


U

1st mode(CFRP: 1mm) 2nd modeLess than this thickness, correlation and opposite phase doesn't appear at 2nd mode.

Optimization of CFRP tube thickness

1st mode 2nd modeCFRP(mm) Freq(Hz) Freq(Hz)

20 73.6 85.210 72.9 80.15 75.5 78.43 72.0 76.51 71.5 75.7

At least, thickness of CFRP: >3mm

Configuration of support system

・ Both-ends supported structure with CFRP tube connection 　→ Correlation is given to both-sides tubes in oscillating behavior.　　 Tungsten tube: 100mm thick, CFRP: 5mm thick Support position: Both ends and 3.85m from I.P.

・ Active vibration isolation system is necessary　→ Relative amplitude will be above 1nm. CFRP tube is not efficient to reduce amplitude less than 1nm.

・ It is necessary to design the stiff support base as possible → Natural frequency becomes high.　 → Amplitude is decreased proportional to frequency.

Conclusion CFRP tube(5mm)

Tungsten masks(100mm)

A

EI

lf ii

2

2 1

2

Stiffness(CFRP tube)

EI: Bending stiffness　 E: Young’s modulus　 I : Moment of Inertia

64

44IO dd

I

EI

pl

3CFRP tube(t=5mm) E=150GPa I=π ・ (d1

4-d24)/64

=π ・ (8004-7904)/64 =9.9×108mm4

Tungsten(t=100mm) E=415GPa I=π ・ (d1

4-d24)/64

=π ・ (8004-6004)/64 =1.4×1010mm4

Ratio = CFRP:Tungsten = 1 : 39

CFRP Tube→Not efficient to increase natural frequency and decrease amplitude.

・ Natural frequency

・ Deformation

fF

fXfH

j

iij

FRF(Frequency Response Function)

Xi: Output Acc.Fj: Input force

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

- 100

- 50

0

50

100

Out

put

Acc

.(m/s

2 )

Time(sec.)

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10- 5

0

5

10

15

20

25

30

35

40

45

50

Inpu

t Fo

rce(

N)

time(sec.)

10 100 1000

1E- 4

1E- 3

0.01

Fou

rier

Spe

ctru

m(N

)

Frequency(Hz)

10 100 1000

1E- 5

1E- 4

1E- 3

0.01

0.1

1

10

Fou

rier

Spe

ctru

m(m

/s2 )

Frequency(Hz)

Hz0.0 4E3

(

m/s

2̂ﾛｸﾞ振幅

1

0.1

1E- 2

10

Hz0.0 4E3

(

m/s

2̂ﾛｸﾞ振幅 1

0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

Hz0.0 4E3

(

m/s


0.1

10

100

frf- frf- frf-

frf- frf- frf-

frf- frf- frf-

FFT

FFT

FRF

Input

Output

Input

Output

○ Tests(Hammering test)

FRF Table

10 100 1000

1

10

100

FR

F((

m/s

2 )/N

)

Frequency(Hz)

P1 P2 P3 P4 P5 P6 P7 P8 P9

12

34

56

78

9

(+Y): 56.2 Hz (283右側


A: 57Hz

○ Results

(Taper flange, 12-M6)

123

45

6

78

9

(+Y): 129 Hz (- 11右側


B: 129Hz

C: 585Hz

1

2

34

5

6

7

8

9

(+Y): 1.21E3 Hz (196右側


D: 1216Hz

1

2

3

4

5

6

78

9

(+Y): 1.69E3 Hz (153右側


E: 1690Hz

1

23

4

56

78

9

(+Y): 2.5E3 Hz (143右側


F: 2500Hz

A B

C DE F

123456789

G-sensor

Fixed

*

''

*

2*

'

' FEr

FEr

testr

testr

FEr

testr

rrMAC

MAC(Modal Assurance Criteria)

Modal assurance criteria quantitatively compare all the possible combinations of test and analysis mode shape pairs.

Mode Freq. 　Damping(%)　 1 30.4Hz 　　 1.68　 2 188Hz 　　 0.422　３ 419Hz 　　 0.303　４ 584Hz 　　 0.113　５ 992Hz 　　 8.02E-2

nf

ff

2

21

f2fnf1

2

Damping ratio

0.0 0.0

0.2 0.2

0.4 0.4

0.6 0.6

0.8 0.8

1.0 1.0

1

2

3

4

5

ﾃｽﾄ

5

4

3

2

1

FEM

5th3rd2nd

4th

5th１ st4th3rd2nd1st

FEMTest

MAC=1: Mode shape pairs is exactly match MAC=0: pairs that are completely independent

1st: 31Hz

3rd: 457Hz

5th: 1022Hz

4th: 534Hz

FEM

2nd: 192Hz

Comparison with FEM

Test

3D : Ref [ ] 988 表示ｺﾝﾌﾚ゚ｯｸｽ

: 1.0, :振幅ﾄｳ゙ｪﾙ(g) X,Y,Z :方向遠近

3D : Ref [ ] 32.5 表示ｺﾝﾌﾚ゚ｯｸｽ






1st: 30Hz

3rd: 419Hz

2nd: 188Hz

5th: 992Hz



4th: 584Hz

1

X

Y

Z

F0sin(t)=(m ・ a)sin(t)

F0: Input force(F0=ma)X: AmplitudeXst: Static deformationζ: Damping ratio(2%)ωn: Resonant frequencyω: Frequency If =n, X/Xst=25

Harmonic analysis

F0sin(t): Excitation force0 – 1000Hz: Sweep frequency

tFkxxcxm sin0

222 21

1

nnstX

X

F0sin(t)

Model-1A Model1-B Model-2AModel-

2B

Deformation(mm) 1.6 - 0.09 -

Stress(MPa) 23 - 5 -

Natural frequency(Hz)

(Vertical)

1st mode 17 15 71 152nd Mode 81 38 179 543rd mode 173 105 202 93

Harmonic response(nm) @QC1 8.0 8.0 0.2 6.0

Spectrum analysis(nm)

@QC1

1st mode 6.5 2.0 4.3 2.72nd Mode -1.7 1.1 0.2 0.23rd mode -0.4 0.1 1.9 0.002

(Model-1A) (Model-1B)

(Model-2A) (Model-2B)

現実に考えられる固有振動数のずれについて

Young’s modulus の違い　 Aluminum の場合種類によって　 6.9GPa – 7.3GPa の範囲がある。 → 　 fi: 3% different　 If 70Hz: 2Hz, 17Hz: 16.5Hz

寸法の誤差　寸法が多少ずれても断面 2 次モーメント（ I）に大きな差は生まれない。また、寸法を管理することができることから「寸法の誤差」による固有振動数のずれは小さいと考えられる。

組み立て及び設置誤差によるずれ　ネジ締めのトルク管理、寸法測定等で管理できるが完全にはできない。　　３％以内の誤差にできるのではないだろうか。（根拠はすこしあいまい）

全体では 5% ぐらいに抑えられるかもしれない。　 →　どのくらいの誤差に収められるか試験。

A

EI

lf ii

2

2 1

2

64

44IO dd

I

summary of support tube r&d nov. 11, ’04 kek h. yamaoka

Documents

w x 695l2

w x 1440l4

w x 200l1

w x 695l3

relative amplitude qcr

t x 200l110 scale6

fem results20t x

qclt100mmt100mmt5mmin