... ctgh bSX 1 / ( a + b ctgh bSX ) α R R = f ( SX , R , R ) R Ꭱ Ro = f ( SX , R . ) Ro = 1 SX = f ( R , Rg , R ) SX = Ar ctgh b SX = f ( Ro , R ) SX = ( Ar Ar ctgh ! 1 = = = R Ro = f ( R , R , R∞ ) Ꭱ f ( Ro , R ,, R∞ ) R Ro Ꭱ = f ( R1 ...
... ctgh bSX 1 / ( a + b ctgh bSX ) 1 a — Characteristic R = f ( SX , Rg , R ∞ ) Ꭱ Ꭱ Ro = f ( SX , R . ) Ro = SX = f ( R , Rg , R∞ ) SX :( SX = f ( Ro , R∞ ) SX = b ᄒ ( Ar ctgh bRo R co Ꭱ . ∞ = 2.303 = b a — Ar ctgh - Ar etgh R 1 Ꭱ ...
... ctgh € 1 2 2 ( 1 + N ctgh € ) Lp 2 € 2 E - ( 1 + Nctgh e ) 2 - ctgh e ( 1+ M ctgh € ) Lp c [ e ( 1 M We 2 1+ 2e√ / We / Lp ( NM ) ctgh e ( 2 € 2 / Lp ) ( 1 + ctgh e ) 2 - ctgh e [ e ( 1 + M ctgh e ) - M We / 2 ] 2 ( 2-35 ) In the case ...
... ctgh ( r− r ' B ( t ) , 2ħA ( t ) bo ( 2.42 ) b ( t ) = gg'sinh ( rr ' ) ( 2.43 ) c ( t ) q'2 + d ( t ) = ihbq'2 20'20 1 ctgh ( rr ' ) - -ln [ mn'sinh ( r – p ' ) ] + do , - 2 i ė ' i ( 2.44 ) c ( t ' ) = - + j'ctgh ( r - p ...
... ctgh) Q = Qc sinh(ctga+ctgh). /s (s = Dx), the resistance to tension–shear is obtained: From the first relationship, with Q s = Asfys and as = As Vsd = QzDx = aszfyd sina(ctga+ctgh) From the second, with Qc = rcbh = bsfc2sinh (see Fig ...
... ctgh V4a2-4 Ar tgh a- r b ctgh b , ( 42 ) worin nach Gl . ( 35 ) b = √ / a2 - 1. Ob der tgh oder der ctgh zu verwenden ist , hängt vom Argument ab . Area tangens hyperbolicus ist zu verwenden für -1≤x≤ + 1 , Area cotangens ...