Engineering guide
Module ADR calculates the required minimum wall thicknesses and the design pressure for tanks for the carriage of dangerous goods to DIN EN 14025 in conjunction with ADR (road) and RID (rail). Annex A of EN 14025 adopts the governing dangerous-goods regulations, in particular the minimum wall thicknesses per ADR 6.8.2.1.18 and the equivalence formula for materials other than the reference mild steel.
To calculate the minimum wall thickness of a dangerous-goods tank to EN 14025, several criteria have to be checked in parallel: the pressure-based wall thickness for operation and test (with the design pressure derived from the vapour pressure at 50 °C, the filling/discharge pressure and the static and dynamic pressure components), the allowable stresses per EN 14025 and the ADR/RID limit of Min[0.5·Rm20; 0.75·Rp20], as well as the material-dependent equivalent minimum thickness relative to the reference mild steel. The least favourable value governs.
The module distinguishes fine-grain steel, other steels and aluminium, accounts for gravity discharge via the maximum filling height, and delivers the verifications for tank approval in road and rail transport.
Scope
- Minimum wall thickness
Typical workflow
Enter operating data and medium
The inputs are the design temperature, the filling/discharge pressure (0 for gravity discharge), the maximum filling height, the specific gravity of the filling medium, the vapour pressure at 50 °C, and the gross mass and tare weight of the tank.
Determine design and test pressure
From the vapour pressure, the static pressure component of the filling and the dynamic pressure from the travel accelerations, the module determines the design pressure per EN 14025 as well as the test pressure and the working pressure per clause 6.2 as the maximum of P<sub>B</sub> and MWP.
Establish allowable stresses
For the selected material (fine-grain steel, other steel or aluminium), the allowable stress per EN 14025 and the ADR/RID limit Min[0.5·Rm20; 0.75·Rp20] are determined from the tensile strength and proof stress at 20 °C.
Calculate pressure-based wall thicknesses
With the outside or inside diameter and the weld joint efficiency, the minimum thickness for the test condition per Eq. (A.3) and for the operating condition per Eq. (A.4) is calculated.
Check equivalent minimum thickness
The minimum thickness per ADR 6.8.2.1.18 (for mild steel 5 mm at D ≤ 1,800 mm, 6 mm at D > 1,800 mm) is converted to the actual material via the equivalence formula using tensile strength and elongation at fracture.
Report the governing minimum thickness
The required wall thickness is the maximum of the criteria; the module reports it as a Min(eT, eC, e1) comparison together with the as-built wall thickness.
Frequently asked questions
What is the difference between the EN 14025 calculation and the ADR minimum thickness?
EN 14025 yields the pressure-based wall thickness from the design and test pressures with the allowable stress of the material. Independently of that, ADR/RID 6.8.2.1.18 requires an absolute minimum thickness that protects the tank against mechanical damage (impact, overturning). Both criteria apply in parallel; even a tank operated without pressure must not fall below the ADR minimum thickness.
What does the equivalence formula with Rm and A stand for?
The ADR minimum thicknesses are defined for a reference mild steel with Rm = 370 N/mm² and 27 % elongation at fracture. If a stronger or tougher material is used, the wall thickness may be reduced via e1 = 21.4·e0/³√(Rm1·A1) — the product of tensile strength and elongation at fracture is a measure of the wall's energy absorption capacity on impact. The guaranteed minimum values of the material standard must be used, not acceptance-test values.
When may gravity discharge be assumed?
If the tank is filled and discharged without overpressure (filling/discharge pressure 0), the maximum filling height determines the static pressure component. For pressure discharge, the filling/discharge pressure must be applied instead. In addition, the vapour pressure at 50 °C or a boiling point above 35 °C decides which minimum design pressure per ADR applies for the tank code.
Why do ADR/RID limit the allowable stress to Min[0.5·Rm20; 0.75·Rp20]?
The limit ensures that even under overload in transport service the tank wall keeps a sufficient margin to the tensile strength and retains plastic reserves. It can be stricter than the allowable stress of the pure pressure vessel calculation and then governs the wall thickness.
Worked example
For a road tanker with a shell diameter above 1,800 mm, the equivalent minimum wall thickness per ADR 6.8.2.1.18 is to be determined — a worked example of the equivalence formula. The tank material has a guaranteed tensile strength of 490 N/mm² and a guaranteed elongation at fracture of 25 %.
Given
| Reference minimum thickness for mild steel e0 (D > 1,800 mm) | 6 mm |
|---|---|
| Tensile strength Rm1 (guaranteed minimum value) | 490 N/mm² |
| Elongation at fracture A1 (guaranteed minimum value) | 25 % |
Solution
Equivalence formula per ADR 6.8.2.1.18
The minimum thickness of the reference mild steel (Rm = 370 N/mm², A = 27 %) is converted to the actual material via the equivalence formula:
e1 = 21.4 · e0 / ³√(Rm1 · A1)
Inserting the values
Rm1 · A1 = 490 · 25 = 12,250
³√12,250 = 23.05
e1 = 21.4 · 6 / 23.05 = 5.57 mm
Assessment
The higher-strength, sufficiently ductile material permits a reduction of the ADR minimum thickness from 6 mm to 5.57 mm. The absolute lower limits of the ADR for the particular tank design (e.g. with or without protection against damage) as well as the pressure-based minimum thicknesses per EN 14025 Eq. (A.3)/(A.4) must be observed in addition — the largest value governs.
Result
| Equivalent minimum wall thickness e1 | 5.57 mm |
|---|
Technical content source: LV-Solved-Problems-Content 628d313247e7