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3850

3850. FEM model. Fcos( w t)=(m ・ a)cos( w t). A: Amplitude δ st : Static deformation ζ: Damping ratio(2%) ω n : Natural frequency (Support tube) ω: Frequency. If w = w n , A=25. Estimation of Acc. Data: Vertical @ATF(17:00 Feb. 10, 2004). Linear Spectrum. 2x10 -7 m/s 2.

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3850

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  1. 3850 FEM model Fcos(wt)=(m・a)cos(wt) A:Amplitude δst:Static deformation ζ:Damping ratio(2%) ωn:Natural frequency (Support tube) ω:Frequency If w=wn, A=25

  2. Estimation of Acc. Data: Vertical @ATF(17:00 Feb. 10, 2004) Linear Spectrum 2x10-7m/s2 Input Acc. = 2x10-7m/s2 Mass=90tons/9.8[m/s2] Self weight

  3. Calculation(1) QC-L: Tungsten 100mm QC-R: Tungsten 100mm CFRP: 3mm, 5mm, 10mm Df= 0Hz w= 0 – 1000Hz Fcos(wt)=(m・a)cos(wt) QC-L: Tungsten(100mm) QC-R: Tungsten(100mm) CFRP(3mm,5mm,10mm) Get amplitude difference between QC-L and QC-R.

  4. In case of 100mm-5mm(CFRP)-100mm, Df=0Hz 2nd mode 1st mode 5mm 100mm 100mm Amplitude: QC-L and QC-R Same phase Opposite phase Difference: QC-L and QC-R

  5. Amplitude: CFRP=3mm Df= 0Hz Amplitude: CFRP=10mm Amplitude Difference

  6. Calculation(2) QC-L: Tungsten 100mm QC-R: Tungsten 100mm EQC-R=0.97xEQC-L CFRP: 3mm, 5mm, 10mm Df= 1Hz w= 0 – 1000Hz Fcos(wt)=(m・a)cos(wt) QC-L: Tungsten(100mm) QC-R: Tungsten(100mm) EQC-R=0.97xEQC-L CFRP(3mm,5mm,10mm) Get amplitude difference between QC-L and QC-R.

  7. In case of 5mm(CFRP), Df=1Hz 2nd mode 1st mode Amplitude: QC-L and QC-R Same phase Opposite phase Difference: QC-L and QC-R

  8. Amplitude: CFRP=3mm Df= 1Hz Amplitude: CFRP=10mm Amplitude Difference

  9. Calculation(3) QC-L: Tungsten 100mm QC-R: Tungsten 100mm EQC-R=0.922xEQC-L CFRP: 3mm, 5mm, 10mm Df= 3Hz w= 0 – 1000Hz Fcos(wt)=(m・a)cos(wt) QC-L: Tungsten(100mm) QC-R: Tungsten(100mm) EQC-R=0.922xEQC-L CFRP(3mm,5mm,10mm) Get amplitude difference between QC-L and QC-R.

  10. In case of 5mm(CFRP), Df=3Hz 2nd mode 1st mode Amplitude: QC-L and QC-R Same phase Opposite phase Difference: QC-L and QC-R

  11. Amplitude: CFRP=3mm Df= 3Hz Amplitude: CFRP=10mm Amplitude Difference

  12. Calculation(4) QC-L: Tungsten 100mm QC-R: Tungsten 100mm EQC-R=0.872xEQC-L CFRP: 3mm, 5mm, 10mm Df= 5Hz w= 0 – 1000Hz Fcos(wt)=(m・a)cos(wt) QC-L: Tungsten(100mm) QC-R: Tungsten(100mm) EQC-R=0.870xEQC-L CFRP(3mm,5mm,10mm) Get amplitude difference between QC-L and QC-R.

  13. In case of 5mm(CFRP), Df=3Hz 1st mode 2nd mode Amplitude: QC-L and QC-R Same phase Opposite phase Difference: QC-L and QC-R

  14. Amplitude: CFRP=3mm Df= 5Hz Amplitude: CFRP=10mm Amplitude Difference

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