Benchmark Elastic column compacted by self weight

Elastic column compacting by self-weight

In this benchmark, a static equilibrium for an elastic column compacting by its own weight is sought.

Analytic solution

Kinematics

$$ \varepsilon_{xx} = 0 \quad \varepsilon_{xy} = 0 \quad \varepsilon_{yy} = \ln(1 + u_{y,y}(y)) $$

Constitutive relations

$$ \begin{split} \sigma_{xx} &= \lambda \varepsilon_{yy} \\ \sigma_{xy} &= 0 \\ \sigma_{yy} &= (\lambda+2\mu)\varepsilon_{yy}. \end{split} $$

Mass balance

$$ \rho(y) = \frac{\rho_{0}}{1+\varepsilon_{yy}} $$

Momentum balance

$$ \begin{split} \sigma_{xx,x} + \sigma_{xy,y} &= 0 \\ \sigma_{yx,x} + \sigma_{yy,y} &= \rho g. \end{split} $$

Solution for strain

From the second equation for the momentum balance,

$$ (\lambda+2\mu)\frac{d\varepsilon_{yy}}{dy} = \frac{\rho_{0}g}{1+\varepsilon_{yy}} $$

Defining $M = \rho_{0}g/(\lambda+2\mu)$, we get a differential equation for $\varepsilon_{yy}$:

$$ \begin{equation} (1+\varepsilon_{yy})\varepsilon_{yy}' = \left( \frac{1}{2} (1+\varepsilon_{yy})^{2} \right)' = M \end{equation} $$

$$ \begin{equation} (1+\varepsilon_{yy})^{2} = 2My + C \end{equation} $$

Appling the boundary condition that $\sigma_{yy} = 0$ at $y=H$, we know $\varepsilon_{yy} = 0$ at $y=H$, too. So, we get

$$ \begin{equation} (1+\varepsilon_{yy})^{2} = 2My + (1-2MH) = 1 - 2M(H-y). \end{equation} $$

$$ \begin{equation} 1+\varepsilon_{yy} = \pm \sqrt{1 - 2M(H-y)}. \end{equation} $$

Finally, using the condition that $-1 < \varepsilon_{yy} < 0$ at $y=0$, we get

$$ \begin{equation} \varepsilon_{yy} = -1 + \sqrt{1-2M(H-y)} = -1 + \sqrt{1-\frac{2\rho_{0}g}{\lambda+2\mu}(H-y)}. \end{equation} $$

Solution for stress and density

$$ \begin{equation} \sigma_{yy} = (\lambda + 2\mu)\varepsilon_{yy} \end{equation} $$

$$ \begin{equation} \rho(y) = \frac{\rho_{0}}{1+\varepsilon_{yy}} \end{equation} $$

Solution for displacement

$$ \begin{equation} \varepsilon_{yy} = \ln \left( 1+ \frac{du_{y}}{dy}\right) = -1 + \sqrt{1-2M(H-y)}. \end{equation} $$

Solving this ODE, we get

$$ \begin{equation} \begin{split} u_{y}(y) = -y &+ \frac{1}{M} \Big[ \Big(-1 + \sqrt{1-2M(H-y)}\Big) e^{-1+\sqrt{1-2M(H-y)}} \\ &- \Big( -1 + \sqrt{1-2MH} \Big) e^{-1+\sqrt{1-2MH}} \Big]. \end{split} \end{equation} $$

Comparison with LHOGS solution

See the benchmark notebook