Example 16: Bearing capacity of foundation.

In this plane-strain example the Mohr-Coulomb plasticity law is used to calculate the bearing capacity of a foundation. The foundation is symmetric, so that only half of the problem is analyzed. The width of (half of) the foundation is $1.52 {\rm m}$. The size of the soil domain is taken to be $6 {\rm m}$ by $6 {\rm m}$. The Young's modulus of the soil is $207 \times \; 10^6 {\rm Nm^{-2}}$ and the Poisson ratio is $0.3$. For the Mohr-Coulomb law, both the yield rule angle and the flow rule angle are $20 \; {\rm degrees}$, and the cohesion is $0.069 \; 10^6 {\rm Nm^{-2}}$. At the bottom of the domain, all displacements are assumed to be fixed. At the left edge (the symmetry axis) and at the right edge, the horizontal displacement is fixed while the vertical displacement is free.

The classical solutions from Prandtl, Coulomb and Terzaghi give for the maximal average pressure $\sigma_{yy}$ at the foundation values in the range $0.98 \; 10^6 {\rm Nm^{-2}}$ up to $1.20 \; 10^6 {\rm Nm^{-2}}$.

To get the same order of accuracy as in the classical solutions, a coarse mesh with only sixteen quadratic elements is used. For optimal numerical stability, the elements are integrated fully (integration points in the nodes). The foundation is prescribed a downward velocity of $0.05 \frac{\rm m}{\rm s}$ in the calculation. With time steps of $10^{-3} {\rm s}$ the solution is advanced up to time $0.6 {\rm s}$. First the deformed mesh at time $1 {\rm s}$ is plotted below.

Secondly, the development for the average pressure at footing over time is plotted

The maximum value of average pressure is near $1.1 \; 10^6 {\rm Nm^{-2}}$.