English
Log in
Log in Simulate Now
English
Log in Simulate Now
Documentation
Help > OnScale Solve Validation Cases > Validation Case: Linear Analysis of Spherical Tank Under Internal Pressure

Validation Case: Linear Analysis of Spherical Tank Under Internal Pressure

Overview

This article is part of the series of FEA validation cases performed using OnScale Solve. The predicted stresses from the simulation results have been compared against analytical solutions.

Geometry

The geometry is available as an Onshape document. All planar surfaces have a symmetry boundary condition such that only one eighth of the tank is represented in the simulation.

Figure 1: Imported model in Solve

Material Properties

  • Structural Steel
    • Young’s Modulus = 200~\text{GPa}

Boundary Conditions

  • Symmetry Constraint
    • Applied to all planar faces (Face 3, 4 and 5)
  • Load
    • Pressure Load = 10~\text{kPa}
Figure 2: Cross section of tank with loading

Mesh

A second-order tetrahedral mesh is used for this study.

Reference Solution

The solution can be derived from [1]. The principal stresses can be computed using the first two equations, while the radial displacement can be computed using the third equation below.

The first and second principal stresses represent the tangential tensile stresses on the inner surface, while the third principal stress is the compressive stress on the inner surface that should match the applied internal pressure.

\sigma_{\text{1}} = \sigma_{\text{2}} = -\frac{P_{\text{i}} \cdot r_{\text{i}}^{3} \cdot\left(2 r^{3}+r_{\text{o}}^{3}\right)}{2 \cdot r^{3} \cdot\left(r_{\text{i}}^{3} r_{\text{o}}^{3}\right)}

\sigma_{\text{3}}=\frac{P_{\text{i}} \cdot r_{\text{i}}^{3} \cdot\left(r_{\text{o}}^{3}-r^{3}\right)}{r^{3} \cdot\left(r_{\text{i}}^{3}-r_{\text{o}}^{3}\right)}
u_{r}=\frac{3 \cdot P_{\text{i}} \cdot r_{\text{i}} \cdot r_{\text{o}}^{3}}{4 \cdot E \cdot\left(r_{\text{o}}^{3}-r_{\text{i}}^{3}\right)}

u_{r}=\frac{3 \cdot P_{\text{i}} \cdot r_{\text{i}} \cdot r_{\text{o}}^{3}}{4 \cdot E \cdot\left(r_{\text{o}}^{3}-r_{\text{i}}^{3}\right)}

Where:

r_{\text{i}}=\text{Internal~radius} = 2.007~\text{m}

r_{\text{o}} = \text{Outer~radius} = 3.264~\text{m}

E = \text{Young's~Modulus} = 200~\text{GPa}

We are evaluating the stresses at the internal radius; thus, r = r_{\text{i}}

Results Comparison

The tables below compares the values obtained using the analytical equation described above and the analysis performed using OnScale Solve. Furthermore, in OnScale Solve, the stress values are computed at the nodes (thus on the internal surface) by projecting the stress field from the element integration points.

# of VerticesAnalytical, \sigma_{\text{2}} (Pa)OnScale Solve, \sigma_{\text{2}} (Pa)Difference (%)
12779543.559569.060.27
25759543.559556.330.13
111189543.559545.150.02
233719543.559544.750.01
# of VerticesAnalytical, \sigma_{\text{3}} (Pa)OnScale Solve, \sigma_{\text{3}} (Pa)Difference (%)
1277-10000.00- 9918.81-0.82
2575-10000.00-9954.41-0.46
11118-10000.00-9983.65-0.16
23371-10000.00-9991.68-0.08
# of VerticesAnalytical, u_{\text{r}} (nm)OnScale Solve, u_{\text{r}} (nm)Difference (%)
127798.0696.66-1.43
257598.0696.87-1.21
1111898.0697.01-1.07
2337198.0697.09-0.99

References

[1]
R. Budynas, Roark’s formulas for stress and strain, 9th ed. McGraw-Hill Education, 2020.