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 P1P2 Tungsten Tube (t=100) CFRP tube

(2) 20t x 100w x 1440L (4) 20t x 100w x 695L2.5t 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 695L2.5t x 100w x 200L t=3mm 2: 40Hz 3: 161Hz 1: 26Hz 2: 43Hz 3: 184Hz 1: 30Hz Measurement FEM 1: 76Hz 2: 256Hz 3: 489Hz 1: 78Hz 2: 258Hz 3: 522Hz 1st ～ 3rd mode: Good agreement with FEM Support structure should be modeled in FEM.

Stiffness: 1: mm thick plate By connecting very weak structure, Deviation can be absorbed. Correlation can be given. Relative amplitude can be estimated. 1st: 99Hz 2nd: 548Hz 1st: 106Hz 2nd: 548Hz 1st: 88Hz 2nd: 126Hz Case-1 20mm thick plate Case-2

F 0 cos(  t)=(m ・ a)sin(  t) QC-L: Tungsten(100mm) QC-R: Tungsten(100mm) CFRP  0 – 1000Hz Calculation of relative amplitude 2m 3.85m 7m 8m QC-LQC-R CFRP thick. Difference of 1 st mode of resonant frequency between QC-R and QC-L. Get relative amplitude (QC-R)-(QC-L) (Model-A) (Model-B) F 0 cos(  t) 2m 7m 8m P1P2

Estimation of Input Acc. Data: Feb. 10, 2004) Linear Spectrum 2x10 -7 m/s 2 Input Acc. = 2x10 -7 m/s 2 Mass=90tons/9.8[m/s 2 ] Self weight Input data

1st mode 2nd mode Same phase Opposite phase Relative amplitude: QC-R and QC-L Amplitude: QC-R and QC-L t100mm t5mm In case of 100mm-5mm(CFRP)-100mm,  f=0Hz Relative amplitude between QC-R and QC-L QC-R QC-L Difference(nm) 1e-4 1

QC-L QC-R QC-LQC-R Other calculations Diff: Relative amplitude between QC-R and QC-L.

Results P1 P2 P1P2 Natural frequency: 76Hz, relative amp. : 0.1nm <1nm This 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 ～ 3nm In 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)

7m 8m 2m Right side: 70Hz L C R CFRP: Changed! Left side: 75Hz L R L R 1 st mode 2 nd mode 70Hz75Hz 1 st mode(CFRP: 1mm)2 nd mode Less than this thickness, correlation and opposite phase doesn't appear at 2 nd mode. Optimization of CFRP tube thickness 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)

Stiffness(CFRP tube) EI: Bending stiffness E: Young’s modulus I : Moment of Inertia CFRP tube(t=5mm) E=150GPa I=π ・ (d 1 4 -d 2 4 )/64 =π ・ ( )/64 =9.9×10 8 mm 4 Tungsten(t=100mm) E=415GPa I=π ・ (d 1 4 -d 2 4 )/64 =π ・ ( )/64 =1.4×10 10 mm 4 Ratio = CFRP:Tungsten = 1 : 39 CFRP Tube →Not efficient to increase natural frequency and decrease amplitude. ・ Natural frequency ・ Deformation

FRF(Frequency Response Function) X i : Output Acc. F j : Input force FFT FRF Input Output Input Output ○ Tests(Hammering test) FRF Table

A: 57Hz ○ Results (Taper flange, 12-M6) B: 129Hz C: 585Hz D: 1216Hz E: 1690Hz F: 2500Hz AB C D EF G-sensor Fixed

MAC(Modal Assurance Criteria) Modal assurance criteria quantitatively compare all the possible combinations of test and analysis mode shape pairs. ModeFreq. Damping(%) 130.4Hz Hz ３ 419Hz ４ 584Hz ５ 992Hz 8.02E-2 f2f2 fnfn f1f1 Damping ratio 5th 3rd 2nd 4th 5th １ st 4th 3rd 2nd 1st FEM Test 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 1st: 30Hz 3rd: 419Hz 2nd: 188Hz 5th: 992Hz 4th: 584Hz

F 0 sin(  t)=(m ・ a)sin(  t) F 0 : Input force(F 0 =ma) X: Amplitude X st : Static deformation ζ: Damping ratio(2%) ω n : Resonant frequency ω: Frequency If  =  n, X/X st =25 Harmonic analysis F 0 sin(  t): Excitation force  0 – 1000Hz: Sweep frequency F 0 sin(  t)

Model-1AModel1-BModel-2A Model- 2B Deformation(mm) Stress(MPa) Natural frequency(Hz) (Vertical) 1st mode nd Mode rd mode Harmonic Spectrum 1st mode nd Mode rd mode (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。 → f i : 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。