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摘要:
针对Wilson-
θ 方法高频耗散不能由频率无穷大时的谱半径ρ ∞精确控制问题,建立了ρ ∞和θ 之间的关系,形成了Wilson-ρ ∞方法,并将其数值性能与广义-α 方法进行比较。在Wilson-ρ ∞方法中,给定一个ρ ∞,存在2个θ 。对应不同θ 的Jacobi矩阵具有不同的本征根, Wilson-ρ ∞方法也因此具有不同的数值性能,根据谱半径特性给出推荐使用的θ 。同时,基于Wilson-ρ ∞方法的阻尼比和频率,构造单自由度受迫振动系统的模拟系统,初始条件与作用在其上的外力和原系统相同。可以看出,Wilson-ρ ∞方法结果与模拟系统解析解吻合,且稳态响应不存在累计幅值误差和相位误差。数值比较结果验证了所得结论。-
关键词:
- Wilson-θ方法 /
- 模拟系统 /
- 幅值衰减 /
- 周期延长 /
- 解析解
Abstract:The Wilson-
ρ ∞ technique is formed by establishing a relationship betweenρ ∞ andθ , as the high-frequency dissipation of the Wilson-θ method cannot be properly controlled byρ ∞ (spectral radius at infinite frequency). The numerical performances of this approach are compared with those of the Generalized-α method. In the Wilson-ρ ∞ method, there are two differentθ for a givenρ ∞. The characteristic roots of the Jacobi matrix corresponding to bothθ are different, and the corresponding Wilson-ρ ∞ method has different numerical performances. A betterθ is recommended according to the properties of the spectral radius. In addition, an analog system of a single degree-of-freedom forced vibration system is constructed with the dissipation and frequency of the Wilson-ρ ∞ method, and the initial conditions of which and the forces acting on the analog system are the same with those of the original system. It is evident that the steady state responses have no cumulative amplitude errors and phase errors, and the results of the Wilson-ρ ∞ method match the analytical solutions of the analog system.-
Key words:
- Wilson-θ method /
- analog system /
- amplitude decay /
- period elongation /
- analytical solution
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表 1 常用的$ {\rho }_{\mathrm{\infty }} $值及其对应θ
Table 1. Commonly used $ {\rho }_{\mathrm{\infty }} $ and corresponding θ
$ {\rho }_{\mathrm{\infty }} $ $ {\theta }_{1} $和分叉点$ {\tau }_{\text{b}} $ $ {\theta }_{2} $ 0.6 $ {\theta }_{1} $= 1.418259 , $ {\tau }_{\text{b}}=40.896\;625 $1.819322 0.7 $ {\theta }_{1} $= 1.409577 , $ {\tau }_{\text{b}}=19.537\;309 $2.441147 0.8 $ {\theta }_{1} $= 1.397065 , $ {\tau }_{\text{b}}=13.472\;628 $3.631715 0.9 $ {\theta }_{1} $= 1.382232 , $ {\tau }_{\text{b}}=10.601\;429 $7.156400 1.0 $ {\theta }_{1} $= 1.366025 , $ {\tau }_{\text{b}}=8.925\;094 $∞ -
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