2. Benchmark solution#
2.1. Calculation method#
The problem consists in analyzing the response of a mobile mass subjected to a non-zero initial speed and coming to shock against a wall that follows a buckling type of behavior law.
The equilibrium equation for the system is written as:
The resolution of takes place in several phases: before buckling of the wall, loading after buckling, unloading after buckling and free flight.
2.1.1. First phase#
Before buckling, the reaction force is equal: \({F}_{\mathrm{ext}}=-{K}_{1}x\)
The solution to the equation is in the following form: \(x(t)={A}_{1}\mathrm{sin}{\omega }_{1}t+{B}_{1}\mathrm{cos}{\omega }_{1}t\)
With: \({\omega }_{1}^{2}=\frac{K+{K}_{1}}{M}\)
Taking into account the initial conditions: \(x(0)=0\) and \(\frac{\partial x}{\partial t}(0)={v}_{0}\)
We identify: \({A}_{1}=\frac{{v}_{0}}{{\omega }_{1}}\) and \({B}_{1}=0\)
That is: \(x(t)=\frac{{v}_{0}}{{\omega }_{1}}\mathrm{sin}{\omega }_{1}t\)
The buckling limit threshold \({F}_{\mathrm{fl}}\) is reached at time \({t}_{\mathrm{fl}}\), such as: \(x({t}_{\mathrm{fl}})={d}_{\mathrm{fl}}=\frac{{F}_{\mathrm{fl}}}{{K}_{1}}\).
From where: \({t}_{\mathrm{fl}}=\frac{1}{{\omega }_{1}}\mathrm{Arc}\mathrm{sin}(\frac{{F}_{\mathrm{fl}}{\omega }_{1}}{{K}_{1}{v}_{0}})\)
and \(\frac{\partial x}{\partial t}({t}_{\mathrm{fl}})={v}_{\mathrm{fl}}\)
2.1.2. Second phase#
The next phase is the loading phase after buckling; the reaction force is constant as long as the speed remains positive. This reaction force is equal to: \({F}_{\mathrm{ext}}=-{F}_{\mathrm{seuil}}\) and the solution of the equation is written as: \(x(t)={A}_{2}\mathrm{sin}{\omega }_{0}t+{B}_{2}\mathrm{cos}{\omega }_{0}t-\frac{{F}_{\mathrm{seuil}}}{M{\omega }_{0}^{2}}\)
With: \({\omega }_{0}^{2}=\frac{K}{M}\)
Like \({\omega }_{0}^{2}\ll 1\), we can do a limited expansion of trigonometric functions.
That is: \(x(t)={A}_{21}{t}^{2}+{B}_{21}t+{C}_{21}\).
Taking into account the initial conditions \(x({t}_{\mathrm{fl}})={d}_{\mathrm{fl}}\) and \(\frac{\partial x}{\partial t}({t}_{\mathrm{fl}})={v}_{\mathrm{fl}}\), we obtain:
\({A}_{21}=(\frac{{v}_{\mathrm{fl}}{t}_{\mathrm{fl}}}{2}-\frac{{d}_{\mathrm{fl}}}{2}-\frac{{F}_{\mathrm{seuil}}^{2}}{4M}){\omega }_{0}^{2}-\frac{{F}_{\mathrm{seuil}}}{2M}\)
\({B}_{21}={v}_{\mathrm{fl}}+\frac{{F}_{\mathrm{seuil}}{t}_{\mathrm{fl}}}{M}+({d}_{\mathrm{fl}}{t}_{\mathrm{fl}}-\frac{{v}_{\mathrm{fl}}{t}_{\mathrm{fl}}^{2}}{2}){\omega }_{0}^{2}\)
\({C}_{21}={d}_{\mathrm{fl}}-\frac{{F}_{\mathrm{seuil}}{t}_{\mathrm{fl}}^{2}}{2M}-{v}_{\mathrm{fl}}{t}_{\mathrm{fl}}-\frac{{d}_{\mathrm{fl}}{t}_{\mathrm{fl}}^{2}}{2}{\omega }_{0}^{2}\)
The moment at which the speed is cancelled is: \({t}_{m}=-\frac{{B}_{21}}{2{A}_{21}}\)
By neglecting the terms in \({\omega }_{0}^{2}\), we deduce: \({t}_{m}={t}_{\mathrm{fl}}+\frac{M{v}_{\mathrm{fl}}}{{F}_{\mathrm{seuil}}}\),
and the maximum displacement is equal to: \(x({t}_{m})={d}_{m}\)
This makes it possible to obtain the cumulative plastic deformation: \({d}_{p}={d}_{m}-\frac{{F}_{\mathrm{seuil}}}{{K}_{2}}\)
2.1.3. Third phase#
This phase corresponds to unloading, the reaction force is equal to: \({F}_{\mathrm{ext}}=-{K}_{2}(x-{d}_{p})\)
The solution to the equation is written as: \(x(t)={A}_{3}\mathrm{sin}{\omega }_{2}t+{B}_{3}\mathrm{cos}{\omega }_{2}t+\frac{{K}_{2}{d}_{p}}{K+{K}_{2}}\)
With: \({\omega }_{2}^{2}=\frac{K+{K}_{2}}{M}\)
Taking into account initial conditions \(x({t}_{m})={d}_{m}\) and \(\frac{\partial x}{\partial t}({t}_{m})=0\),
we get: \({A}_{3}=({d}_{m}-\frac{{K}_{2}{d}_{p}}{K+{K}_{2}})\mathrm{sin}{\omega }_{2}{t}_{m}\) and \({B}_{3}=({d}_{m}-\frac{{K}_{2}{d}_{p}}{K+{K}_{2}})\mathrm{cos}{\omega }_{2}{t}_{m}\).
The reaction force is cancelled out when the displacement of the material point reaches the cumulative plastic deformation value \({d}_{p}\).
Like \(K\ll {K}_{2}\), we make the following approximation: \(\frac{{K}_{2}{d}_{p}}{K+{K}_{2}}\approx {d}_{p}\).
Thus, the moment \({t}_{d}\) which corresponds to the cancellation of the reaction force is such that:
\(x({t}_{d})={d}_{p}={A}_{3}\mathrm{sin}{\omega }_{2}{t}_{d}+{B}_{3}\mathrm{cos}{\omega }_{2}{t}_{d}+{d}_{p}\)
That is: \(\mathrm{sin}{\omega }_{2}{t}_{m}\mathrm{sin}{\omega }_{2}{t}_{d}+\mathrm{cos}{\omega }_{2}{t}_{m}\mathrm{cos}{\omega }_{2}{t}_{d}=\mathrm{cos}{\omega }_{2}({t}_{d}-{t}_{m})=0\).
From where: \({t}_{d}={t}_{m}+\frac{\pi }{2{\omega }_{2}}\).
2.1.4. Fourth phase#
The next phase corresponds to free flight phase \({F}_{\mathrm{ext}}=0\).
The solution to the equation is written as: \(x(t)={A}_{4}\mathrm{sin}{\omega }_{0}t+{B}_{4}\mathrm{cos}{\omega }_{0}t\)
The initial conditions are:
\(x({t}_{d})={d}_{p}\) and \(\frac{\partial x}{\partial t}({t}_{d})={v}_{d}={\omega }_{2}({x}_{m}-\frac{{K}_{2}{d}_{p}}{K+{K}_{2}})\mathrm{sin}{\omega }_{2}({t}_{m}-{t}_{d})\)
This results in:
\({A}_{4}={d}_{p}\mathrm{sin}{\omega }_{0}{t}_{d}+\frac{{v}_{d}}{{\omega }_{0}}\mathrm{cos}{\omega }_{0}{t}_{d}\)
\({B}_{4}={d}_{p}\mathrm{cos}{\omega }_{0}{t}_{d}-\frac{{v}_{d}}{{\omega }_{0}}\mathrm{sin}{\omega }_{0}{t}_{d}\)
Like \({\omega }_{0}^{2}\ll 1\), by doing a limited expansion of trigonometric functions up to order 2, the solution is in the following form:
\(x(t)={A}_{41}{t}^{2}+{B}_{41}t+{C}_{41}\)
With:
\({A}_{41}=\frac{{\omega }_{0}^{2}}{2}[{v}_{d}{t}_{d}-{d}_{p}(1-\frac{{\omega }_{0}^{2}{t}_{d}^{2}}{2})]\)
\({B}_{41}={d}_{p}{t}_{d}{\omega }_{0}^{2}+{v}_{d}(1-\frac{{\omega }_{0}^{2}{t}_{d}^{2}}{2})\)
\({C}_{41}={d}_{p}(1-\frac{{\omega }_{0}^{2}{t}_{d}^{2}}{2})-{v}_{d}{t}_{d}\)
By neglecting the terms in \({\omega }_{0}^{2}\), we get:
\(x(t)={v}_{d}t+{d}_{p}-{v}_{d}{t}_{d}\).
And we deduce the instant \({t}_{0}\) of transition to the initial position (\(x=0\)).
That is: \({t}_{0}={t}_{d}-\frac{{d}_{p}}{{v}_{d}}\).
2.1.5. C modeling reference solution#
The initial conditions chosen make it possible to have a kinetic energy of \(4J\), which should lead to a total compression of \(6.5m\) and a residual compression of \(3.1m\).
2.2. Reference quantities and results#
For A and B models
It is proposed to test the following quantities:
\({t}_{\mathrm{fl}}\): moment of onset of buckling;
\({d}_{p}\): cumulative plastic deformation;
\({t}_{0}\): moment of ironing at the initial position (after buckling and unloading).
Taking into account the numerical values of the input data, we obtain:
\({t}_{\mathrm{fl}}=\frac{\pi }{6}\) (expressed in seconds)
\({d}_{p}=3\) (expressed in meters)
\({t}_{0}=\frac{\pi }{6}+2\sqrt{3}+\frac{\pi +6}{\sqrt{2}}\) (expressed in seconds)
For C modeling
Maximum compression \(\mathit{max}(d)=6.5\) (expressed in meters)
Residual compression \({d}_{p}=3.1\) (expressed in meters)
Rebound speed after shock \({V}_{\mathit{rebond}}=-1.303840481\) (expressed in meters per second)
For modelling*D
The non-regression is tested without an analytical reference value.
2.3. Uncertainties about the solution#
The reference solution is analytical (to the second order).