Calculator Inputs
Example Data Table
| Case | Model | ΔP (MPa) | Geometry | Thickness / Radii | Longitudinal Stress (MPa) |
|---|---|---|---|---|---|
| A | Thin | 2.0 | D = 200 mm | t = 8 mm | 12.5 |
| B | Thin | 10.0 | D = 100 mm | t = 5 mm | 50.0 |
| C | Thick | 30.0 | ri = 50 mm | ro = 75 mm | 24.0 |
Values are illustrative and assume closed-end vessels.
Formula Used
For a closed-end cylindrical vessel under net pressure ΔP = Pin − Pout, the longitudinal (axial) stress comes from balancing pressure force on the end cap against resisting wall area.
- Thin-walled approximation: σL = ΔP · D / (4t), where D is inner diameter and t is wall thickness.
- Thick-walled closed ends: σz = ΔP · ri2 / (ro2 − ri2), using inner and outer radii ri, ro.
- End-cap force: F = ΔP · A, with A = πr2.
How to Use This Calculator
- Select the model: thin-walled for small thickness, thick-walled otherwise.
- Enter internal pressure, and optional external pressure, with units.
- Provide geometry in the selected length unit.
- Optionally enter yield strength to estimate safety factor.
- Click Compute to display results above the form.
- Use the download buttons to export CSV or PDF reports.
Professional Notes on Longitudinal Stress
1) Why longitudinal stress matters
In closed-end cylinders, internal pressure loads the end caps and creates an axial force that must be carried by the shell. The resulting longitudinal stress influences weld sizing, flange selection, and fatigue performance near discontinuities. For thin cylinders, it is commonly about half the hoop stress, so both checks are essential.
2) Thin-wall validity check
The thin-wall model assumes the membrane stress is nearly uniform through thickness and that bending is negligible. A practical screening rule is t/r < 0.1. For example, D = 200 mm, t = 8 mm gives t/r = 0.08, supporting thin-wall use.
3) Thick-wall axial stress background
When thickness grows, the radial and hoop stresses vary across the wall, but the axial stress for closed ends remains uniform in simple theory. Using inner and outer radii, the calculator applies σz = ΔP·ri2 / (ro2 − ri2), which ties axial load to the metal area resisting it.
4) Role of external pressure
Many vessels operate with surrounding pressure: subsea housings, autoclaves, or vacuum jackets. The axial load depends on net pressure ΔP = Pin − Pout. A 2 MPa internal pressure with 0.5 MPa outside gives ΔP = 1.5 MPa, reducing stress and end-cap force proportionally.
5) Typical unit ranges and examples
Industrial cylinders often span 0.1–30 MPa depending on service. Example B in the table uses ΔP = 10 MPa, D = 100 mm, t = 5 mm: σL = 10×106·0.1/(4·0.005) = 50 MPa. Such transparent arithmetic helps peer review and QA.
6) Safety factor interpretation
If you enter yield strength, the tool reports SF = σy/σ. With σy = 250 MPa and σ = 50 MPa, the ratio is SF = 5. Design codes may require additional knockdowns for weld efficiency, corrosion allowance, and cyclic loading.
7) Practical data to document
A useful calculation record includes net pressure, geometry basis (inner diameter or radii), thickness definition, material strength basis, and the resulting end-cap force. The CSV and PDF exports standardize these fields so calculations are traceable, repeatable, and easy to attach to design dossiers or inspection files.
8) Common pitfalls and verification
Common errors include mixing inner and outer diameters, using thickness units inconsistently, and ignoring external pressure. Verify by checking proportionality: doubling ΔP should double stress and force; doubling thickness should halve thin-wall stress. For critical designs, validate with code-based methods and detailed analysis.
FAQs
1) What does “closed ends” mean here?
It means end caps or closures are present, so pressure acts on a finite end area and produces an axial force. Open-ended pipes do not develop the same longitudinal membrane stress from internal pressure.
2) When should I use the thin-walled option?
Use it when thickness is small compared with radius, commonly t/r < 0.1, and when stresses are expected to be membrane-dominated. Otherwise, choose the thick-walled option for better geometry fidelity.
3) Why is longitudinal stress smaller than hoop stress?
For thin cylinders, hoop stress balances pressure on a projected side area, while longitudinal stress balances pressure on the end area. The geometry leads to σL ≈ 0.5 σhoop under the thin-wall assumptions.
4) Does this include corrosion allowance?
No. Enter the effective remaining thickness you want to evaluate. If corrosion allowance applies, reduce the nominal wall thickness accordingly before computing stress, or compare multiple thickness scenarios to bound the design.
5) Can I use gauge pressure?
Yes, as long as internal and external pressures are consistent. If external pressure is atmospheric and you enter gauge internal pressure, set external pressure to zero. For absolute values, enter both as absolute pressures.
6) What does the end-cap force represent?
It is the axial force on a closure: F = ΔP·A. This value helps size bolts, flanges, and welds, and it provides a quick check that the longitudinal stress trend matches the axial load trend.
7) Why might my safety factor look too high or low?
Safety factor here is a simple yield-to-stress ratio. Real designs also account for code allowables, weld efficiency, temperature derating, fatigue, and stress concentrations. Treat it as a screening metric, not a code compliance statement.