The cantilever beam shown is statically loaded by a concentrated force $\mathbf{F}.$ At time $t=0$ the force releases, so that $\mathbf{F\equiv0}$ for $t>0.$ Perform a transient dynamic analysis with damping included.

$E=2\times10^5\text{ MPa},$ $\nu=0.3,$ $\rho=7800\text{ kg/m$^3$}.$

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

Clamped at $x=0.$

Concentrated force $\mathbf{F}.$

$t=0$: $F_x=0\text{ N},$ $F_y=1\text{ N},$ $F_z=1\text{ N}$

$t>0$: $F_x=0\text{ N},$ $F_y=0\text{ N},$ $F_z=0\text{ N}$

• Hexahedra, BEAM56D2

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 Elastostatics. We exit with the binary file beam56d2.SOL containing the static solution on the first record and copy it to an auxiliary file 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 $\mathtt{TEND}=2T_1=0.235\text{ s},$ where $$T_1=\bar f_1^{-1}=\left[f_1\sqrt{1-\xi_1^2}\right]^{-1}$$ is the maximum damped period of the structure. The FE value must be substituted for $f_1,$ i.e., $f_1=8.55\text{ Hz}$—see example BEAM56D1.

In this example, we opt for the mode superposition method since the modal damping parameters $\xi_k$ are readily at hand. The time step $\mathtt{DT}=7.34375\times10^{-3}\text{ s}$ then merely applies to the output intervals, which are rounded-off to whole multiples of $\mathtt{DT}.$ It may be expected that the participation of the first bending mode will predominate in the overall response, thus we estimate $$w(\mathbf{x},t)\simeq\phi_1(\mathbf{x})e^{-\omega_1\xi_1 t}\sin(\omega_1 t+\varphi_1)$$ and similarly for displacements $u,$ $v$ in the $x$ and $y$ directions, respectively. It follows that $$\frac{w(\mathbf{x},n T_1)}{w(\mathbf{x},0)}=e^{-2n\pi\xi_1} \quad\Rightarrow\quad \frac{w(\mathbf{x},2T_1)}{w(\mathbf{x},0)}=0.2846.$$

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