Время науки - The Times of Science

Ермолицкая А. Д. Ermolitskaya A. D. 2025 7 1 ( ) = [ 1 +∑ 2 ( ) =1 ] 4 ; 2 ( ) = [ 2 +∑ 2 ( ) =1 ] 4 ; Here 2 = 2 4 2 ; a 2 = ; 2 = 2 1 2 2 ; 1 2 = ; = 0 ; = 4 ; (11) Consider the case when the left end of the rod is supported. In this case it is necessary to satisfy the conditions (0) = 0; ′′ (0) = 0; (0) = 0; ′′ (0) = 0; (12) i.e. there is no deflection and bending moment at the left end. In this case { ′ (0) = 2 ; ′′′ (0) = 4 ; ′ (0) = 2 ; ′′′ (0) = 4 ; (13) where 2 , 4 , 2 , 4 – arbitrary constants. To find an analytical solution, we use the method of Laplace operational calculus. To do this, we introduce the operator and find the derivatives and substitute their values into (10). We obtain: { 4 + 2 2 − 2 − ( 4 + ) = 1 ( ) + 1 ( ), 4 + 2 2 − 2 + ( 4 + ) = 2 ( ) + 2 ( ), (14) where 1 ( ) = 2 2 + 4 + 2 2 − ( 2 2 + 4 ), 2 ( ) = 2 2 + 4 + 2 2 + ( 2 2 + 4 ). Hence we obtain a system for finding and : [( 4 + 2 2 − 2 ) 2 + ( 4 + ) 2 ] = = ( 4 + 2 2 − 2 )[ 1 ( ) + 1 ( )] + ( 4 + )[ 2 ( ) + 2 ( )]; [( 4 + 2 2 − 2 ) 2 + ( 4 + ) 2 ] = (15) = ( 4 + 2 2 − 2 )[ 2 ( ) + 2 ( )] − ( 4 + )[ 1 ( ) + 1 ( )]. Finding the roots of the characteristic equations 1 ( ) = (1 + ) 4 + 2 2 − 2 + ; 2 ( ) = (1 − ) 4 + 2 2 − 2 − ; (16) and transformed them, we write the roots as 1 = − − 1 , 2 = + 1 , 3 = − − 1 , 4 = + 1 ,