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009092

2026-06-12

VE0092 | Thin-Walled Toroidal Vessel

Description

A thin-walled toroidal vessel is loaded by means of internal pressure p according to the following figure. While neglecting self-weight, determine the von Mises stress both on the inner and outter radius at the test points 1 and 2.

Material	Isotropic Linear Elastic	Modulus of Elasticity	E	210000.000	MPa
Material	Isotropic Linear Elastic	Poisson's Ratio	ν	0.296	-
Geometry	Toroidal Vessel	Small Radius	r	150.000	mm
		Major Radius	R	500.000	mm
		Wall Thickness	t	2.000	mm
Load	Internal Pressure	Internal Pressure	p	0.500	MPa

Thin-Walled Toroidal Vessel

Thin-Walled Toroidal Vessel

Analytical Solution

The analytical solution is based on the theory of thin-walled vessels. This theory was introduced in VE0084 | Thin-Walled Spherical Vessel . The stress state of the thin-walled vessel is generally described by the Laplace equation:

\frac{σ_{1}}{r_{1}} + \frac{σ_{2}}{r_{2}} = \frac{p}{t}

Laplace equation

Where σ₁, σ₂ are stresses in meridian plane and the plane perpendicular to the meridian plane respectively and r₁, r₂ are the corresponding curvature radii. The mentioned stresses correspond to the principal stresses. The Laplace equation can be rewritten into the form of the normal forces per unit length in appropriate direction N_φ, N_θ:

\frac{N_{φ}}{r_{1}} + \frac{N_{θ}}{r_{2}} = p

Equilibrium equation in the form of normal forces

The radii r₁ and r₂ are in case of the toroidal vessel defined as:

r_{1} = r,

r_{2} = \frac{r_{0}}{\sin φ} = r + \frac{R}{\sin φ}

Radii definition

Where r₀ defines the radius of the parallel circle according to the following figure.

r_{0} = R + r \sin φ

Radius of the parallel circle

Thin-Walled Toroidal Vessel - Problem Sketch

Thin-Walled Toroidal Vessel - Problem Sketch

From the equilibrium of the inner outer forces, the normal force N_φ can be determined as follows:

N_{φ} = \frac{p r (r_{0} + R)}{2 r_{0}}

Meridional force per unit length

Substituting into the force equation, the normal force N_θ then follows:

N_{θ} = \frac{p r}{2}

Circumferential force per unit length

Based on above mentioned formulas for normal forces per unit length, the appropriate stresses can be defined:

σ_{φ} = \frac{p r (r_{0} + R)}{2 t r_{0}},

σ_{θ} = \frac{p r}{2 t}

Meridional and circumferential stress

The von Mises stress can be calculated according formula:

σ_{M i s e s} = \sqrt{σ_{φ}^{2} + σ_{θ}^{2} - σ_{φ} σ_{θ}}

Equivalent von Mises stress

Von Mises stress in the test points 1 and 2 then follows:

σ_{M i s e s 1} \approx 28.810 M P a,

σ_{M i s e s 2} \approx 39.369 M P a

Result stress at test points

RFEM Settings

Modeled in RFEM 6.14 and RFEM 5.39
Element size l_FE = 0.005 m
Isotropic linear elastic material is used
Large Deformation Analysis is used due to membrane shell element

Results

Quantity	Theory [MPa]	RFEM 6 [MPa]	Ratio [-]	RFEM 5 [MPa]	Ratio [-]
σ_{Mises 1}	28.810	28.812	1.000	28.812	1.000
σ_{Mises 2}	39.369	39.630	1.007	39.630	1.007

Remark: The stress is evaluated at the middle surface of the toroidal vessel.

Comparison of the von Mises stress course on the meridian circle -- theory and RFEM results, designation of test points 1 and 2

Comparison of the von Mises Stress Course on the Meridian Circle -- Theory and RFEM Results, Designation of Test Points 1 and 2

Von Mises stress distribution in RFEM

Von Mises stress distribution in RFEM

Model 006168 | Verification Example 000092 | 1

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Verification Example 000092 | 1

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