2. Benchmark solution#

2.1. Calculation method#

The reference result was obtained by analytical calculation using the Airy function method.

  • Plane constraints:

      • \({\sigma }_{\mathrm{xx}}=(\mathrm{12.Py.}(x-L))/{\mathrm{2.H}}^{3}\)

\({\sigma }_{\mathrm{yy}}=0\)

\({\sigma }_{\mathrm{xy}}=\mathrm{6.P.}(({H}^{2}/4)-{y}^{2})/{\mathrm{2.H}}^{3}\)

  • Travel:

      • \(u=\frac{\mathrm{12P}}{{\mathrm{EhH}}^{3}}[y(\frac{{x}^{2}}{2}-\mathrm{Lx})-(1+\frac{\nu }{2})\frac{{y}^{3}}{3}]+\mathrm{Ay}+B\)

\(v=\frac{-\mathrm{12P}\nu }{{\mathrm{EhH}}^{3}}\frac{{y}^{2}}{2}(x-L)+\frac{\mathrm{12P}}{{\mathrm{EhH}}^{3}}[\frac{-{x}^{3}}{3}+\frac{{\mathrm{Lx}}^{2}}{2}+(1+\nu )\frac{{H}^{2}x}{4}]-\mathrm{Ax}+C\)

  • The constants \(A,B,C\) depend on the boundary conditions on the movements:

      • \(u(\mathrm{0,0})=v(\mathrm{0,0})=\frac{\partial v}{\partial x}(\mathrm{0,0})=0\)

\(u(\mathrm{0,}-\frac{H}{2})=v(\mathrm{0,}-\frac{H}{2})=u(\mathrm{0,}\frac{H}{2})=v(\mathrm{0,}\frac{H}{2})=0\)

2.2. Benchmark results#

Move from \(y\) to point \(x=L;y=0\): \(v=0.3413\cdot {10}^{-3}\) \(m\)

Constraint according to \(x\) at point: \(x=0;y=-H/2\) \({\sigma }_{\mathrm{xx}}=80.\cdot {10}^{6}\) \(\mathrm{Pa}\)

2.3. Uncertainties#

Analytical solution