Introduction 貯蔵弾性率等の説明 - unius-pa.com

1 1 Introduction Kouichi KAJISAKI unius PAT. ATTORNEYS OFFICE 1 unius unius 2 tan unius 3 =L1/L2 (MPa) (MPa) ( ) L2 L1 unius 4 1 1 2 1 1 2 1 1 1 t0 t1 (t0) (t1) unius 5 1 1 0 - 1 - 1 1 - 1 t0 t1 t2 t3 t4 t5 1 - 1 unius 6 1 1 0 - 1 - 1 1 - 1 t0 t1 t3 t4 t5 1 - 1 d /dt t2 (t0) (t1) (t2) 90 unius 7 1 1 0 - 1 - 1 1 - 1 t0 t1 t2 t3 t4 t5 1 - 1 (t0) (t1) (t2) (t) (t) G1+G2 i (t3) unius 8 unius 9 2 1 0 1 t0 t1 t2 t3 t4 t5 (t0) (t1) (t2) (90)

2 (180) (t3) (t4) (t5) 2sin( t) 2 1sin( t+ ) = 2sin( t) + 0cos( t) 0 4 t 360 0cos( t) 90 unius 10 1 t0 t1 t2 t3 t4 t5 (90) (180) (t)= 2sin( t)= 1cos sin( t) 2 90- 2= 1sin(90+ )= 1cos 0 (t)= 2sin( t) (t)/ (t) = 2/ 2 = 1cos / 2 G1 2 unius 11 0= 1sin (t)= 2cos( t) (t)/ (t) = 0/ 2 = 1sin / 2 G2 (t)= 0cos( t) = 1sin cos( t) 1 t0 t1 t2 t3 t4 t5 (90) (180) 2 0 2 unius 12 (tan ) 2 0 - 1 - 2 G2 G1 1sin / 2 1cos / 2 tan 0/ 2 2/ 2 2 2 0 0 0/ 2 0 =0 0/ 2 =35 0/ 2 =90 0/ 2 =56 unius 13 0 - 2 0 0 unius 14 PdU d PdUG d =[G 12/2-0]( =0 1) =G 1 1 unius 15 (t)= 1sin( t+ ) = 2sin( t) + 0cos( t) (t)= 2(cos( t)+i sin( t))= 2 e i t (t)/ (t)= 2/ 2+ 0/ 2 cos( t)

3 /sin( t) (t) (t) (t)= 2sin( t) (t) (t) 1* / 2 G1+i G2 = 0/ 2 = 1sin / 2 =G /(1+ 2 2) = 2/ 2= 1cos / 2 =G 2 2/(1+ 2 2) (t)= 1* (cos( t)+i sin( t))= 1* e i t 1* unius 16 G1 G2/G1 tan G2 G1+G2 i C C C tan unius 17