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Transient response, BEAM56D2

: Problem description: The cantilever beam shown is statically loaded by the concentrated force ${\bf F}$ . At time t=0 the force releases, so that ${\bf F}\equiv{\bf0}$ for t>0. Perform a transient dynamic analysis with damping included.

$\begin{figure} \centering\hspace{0pt} \epsfclipon\epsfxsize=6cm\epsffile{beam56d2.ps} \end{figure}$

Mesh: Four hexahedra--see appendix B.1.

Material properties: $E=2\times 10^5$ MPa, $\nu=0.3$ , $\rho=7800$ kg/m³.

Damping: Modal damping parameters $\xi_k=0.1$ for $k=1,2,\ldots$ .

Support: All the degrees of freedom fixed at x=0.


u=v=w=0 nodes: 1 2 3 4 21 22 23 24

Loading: Concentrated force




t=0: F_x=0 F_y=1 F_z=1 unit: N


t>0: F_x=0 F_y=0 F_z=0

Solution: It is necessary first to calculate the static deflection of the beam serving as an initial condition for ensuing free motion. This is carried out by using the procedures described in section II. We exit with the binary file beam56d2.SOL containing the static solution on the first record and copy it to an auxiliary file named beam56d2.1. Subsequently the force is set zero and the dynamic solver is executed starting with the initial displacement read-in directly from the auxiliary input file.

The displacement vectors are stored on disk at selected time intervals for further stress analysis. The computation is stopped at TEND=2T₁=0.235 sec, where

$\begin{displaymath}T_1=\bar f_1^{-1}=\left[ f_1\sqrt{1-\xi_1^2} \right]^{-1} \end{displaymath}$

is the maximum damped period of the structure. The FE value must be substituted for f₁, i.e. f₁=8.55 Hz--see example IV.1.1.

In this problem we opt for the mode superposition method since the modal damping parameters $\xi_k$ are readily at hand. The time step DT=7.34375 $\times 10^{-3}$ sec then merely applies to the output intervals, which are rounded-off to whole multiples of DT. It may be expected that the participation of the first bending mode will predominate in the overall response, thus we estimate

$\begin{displaymath}w({\bf x},t)\simeq \phi_1({\bf x}) e^{-\omega_1\xi_1 t}\sin(\omega_1t+\varphi_1) \end{displaymath}$

and similarly for displacements u, v in the x and y directions, respectively. It follows that

$\begin{displaymath}\frac{w({\bf x},nT_1)}{w({\bf x},0)}= e^{-2n\pi\xi_1} ~~\Rightarrow~~ \frac{w({\bf x},2T_1)}{w({\bf x},0)}= 0.2846 \end{displaymath}$

For example, the initial static displacement calculated at node 20 gives w(0)=0.937mm, therefore $w(2T_1)\simeq0.267$ mm, which compares well with the computed value 0.258mm.

$\begin{figure} \centering\hspace{0pt}\rotate{ \epsfclipon\epsfxsize=6cm\epsffile{bea56d2.ps}} \end{figure}$

Execute from prompt:
>rmd3 beam56d2.I1
>rpd3 beam56d2.I2
>srh3 beam56d2.I3
>hmot beam56d2.IM
>fefs beam56d2.I4
>heig beam56d2.IE
>cp beam56d2.SOL beam56d2.1
>hmod beam56d2.ID
>str3 beam56d2.I5

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