Cylindrical Reservoir with Walls of Constant Thickness

The reservoir undergoes the pressure of liquid, as shown on the picture. The bottom of reservoir is embedded into an absolutely rigid foundation.

In most cases in practice the thickness of reservoir wall h is small compared to both the radius R, and the depth of the reservoir d.
Taking into consideration the fact that the bottom of the reservoir does not experience any deformations, it is feasible to model the reservoir as the cylindrical wall whose bottom base is clamped.

Cylindrical Reservoir with Walls of Constant Thickness, the finite element model with applied loads and restraints

The finite element model with applied loads and restraints

Analytical solution of the problem takes the form:
,
where:
,
,
,
ρ – density of liquid,
g – gravitational acceleration (~ 9.8 m/s2).

Let us use the following data: depth of reservoir d = 1000 mm, radius R = 200 mm, thickness of the wall of the reservoir h = 3 mm, density of liquid ρ = 1000 kg/m3 .
Elastic properties are taken as: E = 2.1E+011 Pa, ν = 0.28.
Thus, w = 6.1366E-007 m (at z^ = 0.056 m).

After carrying out calculation with the help of AutoFEM, the following results are obtained:

Table 1.Parameters of finite element mesh

Finite Element Type	Number of Nodes	Number of Finite Elements
quadratic tetrahedron	4681	13428

Table 2.Result "Displacement, magnitude"*

Numerical Solution Displacement w*, m	Analytical Solution Displacement w, m	Error δ = 100%* \|w* - w\| / \|w\|
6.1491E-007	6.1366E-007	0.20

Cylindrical Reservoir with Walls of Constant Thickness, Result "Displacement, magnitude"

Conclusions:

The relative error of the numerical solution compared to the analytical solution for displacements was 0.20% for quadratic tetrahedral finite elements.

*The results of numerical tests depend on the finite element mesh and may differ slightly from those given in the table.