TAILIEUCHUNG - Applied Structural and Mechanical Vibrations 2009 Part 9

Tham khảo tài liệu 'applied structural and mechanical vibrations 2009 part 9', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | The eigenfuncions are written Ớ A ------ Inrfnmr f sin ttỡ cos mớ where the constant A which a priori can depend on both n and m can be fixed by means of normalization. Different boundary conditions lead to more complicated calculations for example if our plate is simply supported at r R the boundary conditions to be imposed on the solution are from eq UJ M - 0 at r R and in polar coordinates the bending moment Mr is written explicitly M -D d2w i1 dw 1 d2w dr2 dr r1 de1 Things are even worse for a completely free plate in fact in this case the boundary conditions read eq _ 1 dM s Mr QT 0 r de where Mr is as above and the transverse shearing force Qr and the twisting moment Mre are given by ớ Q -DJvM dr _ d i 1 dm dr r de Rectangular plates Due to its importance in many fields of applied engineering let us now consider a uniform rectangular plate extending in the domain 0 X c a and 0 y b. The equation of motion of free vibrations is again which assuming a harmonic time dependence becomes eq for the function of the space variables u u x y . As in the preceding case this equation can be written as V2 72 v2 n 2 M 0 Copyright 2003 Taylor Francis Group LLC and we can express its solution as M Ml M2. Obviously it is now convenient to adopt a system of rectangular coordinates so that the Laplacian and biharmonic operators are written explicitly as The function u1 satisfies the equation V2 72 i 0 by separating the space variables and looking for a solution in the form Ui x y 1 x gi y we arrive at the two equations where a2 fl2 y2. Equations have the solutions 1 x A sin ax 4- B cos ax gỊ y c sin fly 4- D cos fly so that Ml x y Al sin ax sin fly Az sin ax cos fly 4- A3 cos ax sin fly A4 cosax cos fly The equation satisfied by the function u2 is V2 4- i7 2 M2 0 implying that its solution can be obtained from eq by replacing the trigonometric functions by hyperbolic functions. This means that we can write the

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