2 = 1 . 5 + 21 Xo = 2 " 2 ( xo ) x = ( 5 ; 5 ; 5 ; . . . . ) , 27 + 21 x6 6 " Xo = xo = - - 2 ( x ) = ( — 6 ; — 2 , — 2 , — 3 ; — 2 , 2 , 3 ; . . . . ) . 2 = 1 . 3 + 18,2 ( 20 ) = ( 6 ; 6 ; 6 ; . . . ) , 11 + ...

... 3 ( @M ) xo ' sin ( ot 301-302- - 03 ) 4 Z , 3 Z23 Z3 ( M ) 7 xo sin ( ot - 401 - 301⁄2 ) - 8 Z4 Z23 + + ( 6b ) 2 = ( @M ) 2 xo Z1 Z2 + ( @M ) * xo Xo 2 Z12 Z22 cos ( 2 wt - 01-02 ) + 3 ( @M ) xo 2 Z1 Z22 Z3 cos ( 2 wt - 01-202-03 ) cos ...

... xo 8 Z2 Z3 Z32 1 sin ( wt - 201302-203 ) 3 ( M ) Xo sin ( wt - 301-302- 4 Z , 3 Z2 Z3 ( WM ) 7 xo 3 ) - 8Z4Z23 sin ( ot 401 - 301⁄2 ) ( 6b ) 2 2 = ( @M ) 2 xo Z1 Z2 cos ( 2 wt - 01-02 ) ( 6c ) 73 = + — + + ( WM ) + xo 2 Z12 Z22 3 ( @M ) + ...

... 3 1 77. 77. + +f"(x0)(x-x0)” + ... + — f"(p)(x-xo) (8.2) 3! n! for some p between X and x0, and 1 f(x) = f(x0) + f"(x0)(x-xo)+ #f"(x)(x —wo)* 1 /// 3 1 77. n ++f"(x0)(x-x0)” + ..

... ( xo ) + 22 F ' ( xo ) + h h2 · F " ( xo ) + I 1.2 h3 1.2.3 F " ( xo ) + .. ( 1 ) hn hp + 1 + F " ( xo ) + · R . 1.2.3 . n I. 2 .3 ... n ( p + 1 ) p est un nombre entier quelconque . R est une quantité in- connue qu'il faut déterminer de ...

... 3 + 12 ( 6 L2 x 24 EI - 4 Ls + 24 ) ....... ( 115 ) Obviously , the maximum deflection occurs at the free end ... xo ) 3 EI WIL 1 3 EIL - xo ) 8 [ W ( 1 - xo ) 3 3 EI + ( L - l ) W ( l - xo ) 2 2 EI 2 W ( - xo ) 3 + 3 W ( L 6EI 1 2 ) 1 ...

... xo , μy = ( a + xo ) 2 + μlz , μlz ' = ( a + xo ) 3 + 3 ( a + xo ) μlz + μlz , μs = ( a + x ) ' + 6 ( a + xo ) 3 μ2 + 4 ( α + xo ) μ3 + μ4 . Substituting here in the place of μ , μ , μ their values from ( 7 ) , we shall obtain = ( a + b ) 3 ...

... xo ) x , differ in and only in the signs of their respective elements , the case of improper equivalence may be ... 3 + √5 2 3 + 5 2 xo = 2 + 2 2 = 1 . 2 ( x ) = ( 3 ; 3 ; 3 ; . . . . ) , , 2 ( x6 ) ,, = ( —3 ; — 3 ; — 3 ; . . . . ) . 2 ...

... xo - ( -2 ) and MoM 2 = 5 - xo , by [ 21 ] , for the abscissas of M1 , Mo , and M2 are respectively -2 , x 。, and 5 . And it is given that Hence P1Po 5 2 PoP2 Xo - ( -2 ) 5 = 5 - Xo 2 - Solving this equation , xo 3 . Similarly , draw ...

... 3 12 ( 6 L2x2 24 EI 4 L x3 + x1 ) ....... ( 115 ) == W Obviously , the maximum deflection occurs at the free end ... [ xo ) 3 W ( l - 20 ) 3 ( L — 7 ) W ( 1 − x0 ) 2 ) 3 EI + 2 EI = = R ( L xo ) 3 EI WI ( L 3 EIL 2 W ( - xo ) 3 + 3 W ...