Aeroelastic vibrations and stability of plates and shells by Sergey D. Algazin, Igor A. Kijko

By Sergey D. Algazin, Igor A. Kijko

Back-action of wind onto wings explanations vibrations, endangering the entire constitution. by means of cautious offerings of geometry, fabrics and damping, dangerous results on wind engines, planes, generators and autos might be shunned.

This booklet supplies an summary of aerodynamics and mechanics in the back of those difficulties and describes a number mechanical results. Numerical and analytical the way to examine and examine them are built and supplemented by means of Fortran code

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0000000E+00 5 Test problems | 27 y 1,0 0,5 x −1,0 −0,5 0,0 0,5 1,0 −0,5 −1,0 Fig. 2. 1. The first calculation was carried out for the velocity v = 0. As expected, the first eigenvalue is simple, while the other two are multiple. As the velocity is increased, the multiple eigenvalues are split (shifting to the right), but remain real-valued. Then, the first and second eigenvalues begin to approach each other (probably, at some instant they will completely merge to form a multiple eigenvalue, but such studies are impossible to perform, because the corresponding eigenvalue problem has a Jordan cell).

2. −1,0 1,0 0,5 −1,0 −0,5 0,0 0,5 1,0 −0,5 −1,0 Fig. 5. 2. 2, n = 4. This problem is computationally more challenging than the first one, and therefore it was solved on 13 × 25 and 15 × 31 grids. 2581 on the second. 390052). 5. Thus, for a nonround plate, the stability is not necessarily determined by the first eigenvalue. However, for θ = π /4, the stability is determined by the first eigenvalue. 369400) being the eigenvalue which determined the stability. 5) is qualitatively different: Re φ (0, y) has zeroes, because in this case the stability is determined by the second eigenvalue.

W2n )T we then have a system of linear equations: 2n ∑ B̄ j2 ,j1 wj1 = Rj2 + δj2 . j1 =0 18 | 4 Reduction to a problem on a disk Therefore, 2n wj1 = ∑ Cj1 ,j2 (Rj2 + δj2 ), C = B̄ −1 . 14) × ∑ Cj1 ,j2 (Rj2 + δj2 ) + Rn,M (ξ ; f , R, S). j2 =0 Here, fj = λ zj φj + Φj Φj = ???? ((vx Ur + vy Vr ) ????φ ????φ ???????????????? 1 + (vy Ur − Vr vx ) )???? , ????r r ????θ ????????????ζ =ζj j = 1, . . , M. Let ξ run through all interpolation nodes ξi , i = 1, 2, . . 15) ∑ Cj1 ,j2 ∑ H̄ i,j2 zi Hij j2 =0 i 2n 2n 1 λ R̄ i = Rn,M (ξi ; f , R, S) + ( − ) ∑ Bil ∑ Hj01 (ξl ) ∑ Cj1 ,j2 δj2 .

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