2. Benchmark solution

The reference solution is analytical, and obtained as follows.

Next, we will note \(u={u}_{x}\). The equation in the relative coordinate system of \({P}_{2}\) is written as:

(2-1) \(m\ddot{u}+c\dot{u}+ku=-m\ddot{{u}_{e}}(t)\)

The load is applied using the AFFE_CHAR_SEISME command. The associated multiplier function FONC_MULT defines training acceleration as:

(2-2) \(\ddot{{u}_{e}}(t)={A}_{e}\mathrm{sin}({\omega }_{e}t)\)

The problem to be solved is therefore written as:

(2-3) \(m\ddot{u}+c\dot{u}+ku=-m{A}_{e}\mathrm{sin}({\omega }_{e}t)\)

Or, by defining the reduced quantities \({\omega }_{0}^{2}=\frac{k}{m}\) and \(\xi =\frac{c}{2m{\omega }_{0}}\):

(2-4) \(\ddot{u}+2{\omega }_{0}\xi \dot{u}+{\omega }_{0}^{2}u=-{A}_{e}\mathrm{sin}({\omega }_{e}t)\)

with the initial conditions:

(2-5) \(u(0)=0\)

(2-6) \(\dot{u}(0)=0\)

We are looking for a particular solution to () in the form:

(2-7) \(\stackrel{~}{u}(t)=D\mathrm{sin}({\omega }_{e}t+\varphi )\)

We ask:

(2-8) \(\stackrel{~}{U}(t)=D{e}^{i({\omega }_{e}t+\varphi )}\)

We then have \(\stackrel{~}{u}(t)=\mathrm{\Im }(\stackrel{~}{U}(t))\) (imaginary part).

Let the complex differential equation be:

(2-9) \(\ddot{\stackrel{~}{U}}+2{\omega }_{0}\xi \dot{\stackrel{~}{U}}+{\omega }_{0}^{2}\stackrel{~}{U}=-{A}_{e}{e}^{i({\omega }_{e}t)}\)

If \(\stackrel{~}{U}\) is a solution of (), then is a solution of (). By injecting the expression () into (), we find:

(2-10) \((-{\omega }_{\mathit{ex}}^{2}+2i{\omega }_{0}{\omega }_{e}\xi +{\omega }_{0}^{2})B{e}^{i\varphi }=-{A}_{e}\)

from where we can determine the parameters \(D\) and \(\varphi\) of the expression ():

(2-11) :math: `D=left|frac {{A} _ {e}} {{omega} _ {0} ^ {2} - {omega} _ {e} ^ {2} +2i {omega} +2i {omega} _ {omega} _ {0} {omega} _ {omega} _ {e}} {sqrt} {left ({omega} _ {0} ^ {2} - {omega} _ {e} ^ {2}right)} ^ {2}} + {left (2 {omega} _ {0} {omega} _ {0} {omega}} {omega}} _ {e}omega} _ {e}xiright)} ^ {2}}}} `

(2-12) \(\varphi =\mathit{arg}\left(-\frac{{A}_{e}}{{\omega }_{0}^{2}-{\omega }_{e}^{2}+2i{\omega }_{0}{\omega }_{e}\xi }\right)=\mathit{arg}\left({\omega }_{e}^{2}-{\omega }_{0}^{2}+2i{\omega }_{0}{\omega }_{e}\xi \right)=\mathit{atan}2\left(\frac{2{\omega }_{0}{\omega }_{e}\xi }{{\omega }_{e}^{2}-{\omega }_{0}^{2}}\right)\)

The general solution of the problem is then classically written as the sum of the particular solution and the homogeneous solution of a differential equation of order 2 with constant coefficients (see for example [1]:

(2-13) \(u(t)={e}^{-\xi {\omega }_{0}t}\left(A\mathrm{cos}({\omega }_{\xi }t)+B\mathrm{sin}({\omega }_{\xi }t)\right)+D\mathrm{sin}({\omega }_{e}t)\)

with:

(2-14) \({\omega }_{\xi }={\omega }_{0}\sqrt{1-{\xi }^{2}}\)

The initial conditions () and () give respectively:

\(A+D\mathrm{sin}(\varphi )=0\)

\({\omega }_{0}\left(-\xi A+B\sqrt{1-{\xi }^{2}}\right)+{\omega }_{e}D\mathrm{cos}(\varphi )=0\)

From which we can deduce:

(2-15) \(A=-D\mathrm{sin}(\varphi )\)

(2-16) \(B=-\frac{{\omega }_{e}D\mathrm{cos}\varphi }{{\omega }_{0}\sqrt{1-{\xi }^{2}}}+\frac{\xi D\mathrm{sin}(\varphi )}{\sqrt{1-{\xi }^{2}}}=-\frac{D}{\sqrt{1-{\xi }^{2}}}\left(\frac{{\omega }_{e}}{{\omega }_{0}}\mathrm{cos}(\varphi )+\xi \mathrm{sin}(\varphi )\right)\)

The analytical solution is therefore written as () where \(A\), \(B\),,, \(D\), \({\omega }_{\xi }\), and \(\varphi\) are given by (), (), (), (), () and () respectively.

2.1. Benchmark results

For this test case, an excitation pulse of the harmonic signal \({\omega }_{e}=2{s}^{-1}\), \(\xi =\mathrm{0,05}\) and \({A}_{e}=1m\mathrm{.}{s}^{-2}\) is chosen. Given the parameters of the problem, the natural frequency is \({\omega }_{0}=1{s}^{-1}\).

We test the values of the relative displacement \({u}_{x}\) at points \({P}_{2}\) and \({P}_{3}\) to \(t=10s\):

	Move to \(t=10s\)
Bow	\(\mathit{DX}\)
\(N2\) (dot \({P}_{2}\))	0.538736
\(N3\) (dot \({P}_{3}\))	0.538736

2.2. Uncertainty about the solution

Clarification on the integration of time in DYNA_NON_LINE