Design of heads and closures: flat, torispherical, ellipsoidal

Hemispherical, 2:1 ellipsoidal, torispherical, conical and flat heads: how each carries pressure, their code thickness formulas, standard proportions, depths and volumes, and how to choose between them.

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Why it matters

Every cylindrical vessel needs two closures, and the head type decides how thick they are, how much they cost to form, how much headroom they take and whether they can be opened. Choosing a flat cover where a dished head belongs can make the cover several times thicker than the shell; choosing a hemispherical head for a low-pressure tank wastes money on pressing. Head design is also where most students first meet discontinuity stresses.

Key ideas

Why curvature matters. A curved head carries pressure mainly by membrane tension, like the shell. A flat plate has no curvature, so it must carry the pressure by bending, which is far less efficient: its thickness grows with the square root of pressure and in proportion to diameter, and quickly becomes very large.

The common head types, from strongest to weakest shape:

  • Hemispherical: half a sphere. Membrane stress P·D/(4t) is half the cylinder's hoop stress, so it needs roughly half the shell thickness. Deepest and most expensive to form (pressed in segments); used for high-pressure vessels and large spheres.
  • 2:1 semi-ellipsoidal: depth (inside) = D/4, major-to-minor axis ratio 2. Its required thickness is almost exactly the shell thickness, so shell and head can be the same plate. The usual choice above roughly 1–1.5 MPa.
  • Torispherical (dished and flanged): a spherical crown of radius Rc joined to the cylinder by a toroidal knuckle of radius rk. Standard proportions are Rc ≈ D (crown radius not more than the outside diameter) and rk ≥ 6 % of the crown radius. Shallow and cheap to press, but the tight knuckle raises the stress, so it needs to be thicker than the shell — typically about 1.7–1.8 times for standard proportions. The most common head for vessels up to about 1–1.5 MPa.
  • Conical: used for hoppers, bottoms that must drain solids, and transitions between column sections. Stress increases with the half-apex angle α; above about 30° a knuckle or reinforcement at the junction is needed.
  • Flat: flat covers, blind flanges, manholes and small or low-pressure vessels; also where a removable cover is needed. Designed by plate-bending formulas with a code factor C that depends on how the plate is attached.

Discontinuity stresses. A cylinder and a head expand by different amounts under pressure; at the junction they must stay together, so local bending and shear appear. A hemispherical head and a cylinder of the same thickness expand differently; making the head thinner reduces the mismatch. In torispherical heads the knuckle develops a high compressive hoop stress, which is why the code formula has a stress-intensification factor and why very thin large torispherical heads can buckle in the knuckle.

Selecting a head. Up to moderate pressure: torispherical. Above it: 2:1 ellipsoidal. High pressure or very thick walls: hemispherical. Small openings and covers: flat. Headroom, column skirt height and the volume below the tangent line also influence the choice. Forming reduces thickness (especially in the knuckle), so the plate ordered must allow for forming thinning as well as the corrosion allowance.

Formulas

(ASME VIII Div. 1 forms, internal pressure; IS 2825 gives equivalent expressions. t excludes corrosion allowance and forming allowance.)

t = P·Di / (4·S·E − 0.4·P) (hemispherical head) t = P·Di / (2·S·E − 0.2·P) (2:1 semi-ellipsoidal head) t = P·L·M / (2·S·E − 0.2·P), M = ¼·(3 + √(L / r)) (torispherical head)

  • For standard proportions (L = Do, r = 0.06·L): M ≈ 1.77, giving t ≈ 0.885·P·L / (S·E − 0.1·P).

t = P·Di / (2·cos α·(S·E − 0.6·P)) (conical head or section, α ≤ 30°) t = d·√(C·P / (S·E)) (flat unstayed head)

  • P = design pressure (MPa); Di = inside diameter (mm); S = allowable stress (MPa, from code book); E = joint efficiency (–); L = inside crown radius (mm); r = inside knuckle radius (mm); M = stress-intensification factor (–); α = half-apex angle; d = effective diameter of the flat head (mm); C = attachment factor (–), from the code's sketches (roughly 0.17–0.33, take from your code book).

h = Di / 4 (inside depth of a 2:1 ellipsoidal head) V = π·Di³ / 24 (2:1 ellipsoidal head), V = π·Di³ / 12 (hemispherical head) h = Rc − √((Rc − rk)² − (Di/2 − rk)²) (inside depth of a torispherical head)

Worked examples

Example 1 (standard): comparing heads for one shell Given: P = 1.5 MPa, Di = 2000 mm, S = 130 MPa, E = 1.0 (seamless heads). Torispherical: L = 2000 mm, r = 120 mm (6 %).

  1. Shell (for reference): t = P·R/(S·E − 0.6·P) = 1.5 × 1000 / (130 − 0.9) = 11.6 mm.
  2. Hemispherical: t = 1.5 × 2000 / (520 − 0.6) = 3000 / 519.4 = 5.78 mm.
  3. 2:1 ellipsoidal: t = 3000 / (260 − 0.3) = 3000 / 259.7 = 11.55 mm (same as the shell).
  4. Torispherical: M = ¼ × (3 + √(2000/120)) = ¼ × (3 + 4.08) = 1.77; t = 1.5 × 2000 × 1.77 / 259.7 = 20.5 mm.
  5. Hemispherical 5.8 mm, ellipsoidal 11.6 mm, torispherical 20.5 mm (before corrosion and forming allowances). At 1.5 MPa the ellipsoidal head is the sensible choice.

Example 2 (GATE level): why flat covers are kept small Given: a welded flat cover of effective diameter d = 600 mm, P = 1.0 MPa, S = 120 MPa, E = 1, attachment factor C = 0.33 (given; take from your code book in practice).

  1. Flat: t = d·√(C·P/(S·E)) = 600 × √(0.33 × 1.0 / 120) = 600 × 0.05244 = 31.5 mm.
  2. A 2:1 ellipsoidal head of the same diameter: t = 1.0 × 600 / (240 − 0.2) = 2.50 mm.
  3. The flat cover is about 12.6 times thicker. Doubling d to 1200 mm would double the flat cover to 63 mm, which is why flat heads are limited to small diameters, manways and low pressures.
  4. Depth and volume of a 2:1 ellipsoidal head for Di = 2000 mm: h = 2000/4 = 500 mm; V = π × 2³/24 = 1.047 m³.

Common mistakes

  • Using the cylinder formula for a hemispherical head (it needs about half) or for a torispherical head (it needs more, because of M).
  • Assuming torispherical heads are as strong as ellipsoidal ones because they look similar.
  • Ignoring forming thinning in the knuckle when ordering plate.
  • Using crown radius larger than the outside diameter or knuckle radius below 6 % — both outside code limits.
  • Treating flat heads with membrane formulas; they are governed by bending.
  • Confusing semi-axes with full axes when computing head volumes.

For GATE CH

Expect ranking questions (which head needs the least thickness for a given pressure), standard proportions (2:1 head depth D/4, torispherical crown and knuckle limits), quick thickness calculations with a given formula, and head volume or depth calculations used in tank-capacity problems. Practise the ellipsoidal and torispherical formulas and the volume expressions.

Quick check

  1. Rank hemispherical, 2:1 ellipsoidal and torispherical heads from thinnest to thickest for the same P and D.
  2. What is the inside depth of a 2:1 ellipsoidal head on a 1.6 m vessel?
  3. Why is a flat head so much thicker than a dished head?
  4. What are the usual crown and knuckle radii of a standard torispherical head?

Answers: 1. Hemispherical, 2:1 ellipsoidal, torispherical. 2. 0.4 m. 3. It carries pressure by bending instead of membrane tension. 4. Crown radius about equal to the outside diameter; knuckle radius at least 6 % of the crown radius.

Try answering each one aloud before you open it.

  1. 1.What is a flat head in process equipment design, and where is it typically used?Concept

    A flat head is a flat plate closing a shell, welded in or bolted as a blind flange. Having no curvature it carries pressure by bending, so its thickness is t = d·√(C·P/(S·E)) and grows in proportion to diameter. It is therefore used only for small diameters or low pressures: manhole and nozzle covers, blind flanges, small low-pressure vessels and atmospheric tank bottoms, or where a removable cover is needed.

  2. 2.Explain the design and application of torispherical heads in pressure vessels.Concept

    A torispherical (dished and flanged) head is a spherical crown of radius L joined to the shell by a toroidal knuckle of radius r; standard proportions are L about equal to the outside diameter and r at least 6 % of L. It is shallow and cheap to press, but the knuckle concentrates stress, so the code formula t = P·L·M/(2SE − 0.2P) includes M = ¼(3 + √(L/r)), about 1.77 for standard proportions, making it roughly 1.7–1.8 times the shell thickness. It is the usual head for vessels up to about 1–1.5 MPa.

  3. 3.Describe the characteristics of 2:1 ellipsoidal heads and their advantages in process equipment.Concept

    A 2:1 semi-ellipsoidal head has a major-to-minor axis ratio of 2, so its inside depth is D/4 and its inside volume is π·D³/24. Its required thickness, t = P·D/(2SE − 0.2P), is almost exactly the shell thickness, so the same plate can be used for shell and heads, and it is much thinner than a torispherical head of the same diameter. It is deeper and costlier to form than a torispherical head, so it is chosen above about 1–1.5 MPa.

  4. 4.Why are ellipsoidal heads preferred over flat heads in high-pressure applications?Application

    A curved head carries pressure by membrane tension, so its thickness grows linearly with P·D and stays close to the shell thickness. A flat head must carry it by bending; for a 600 mm cover at 1 MPa a flat plate needs about 31 mm while a 2:1 ellipsoidal head needs about 2.5 mm. At high pressure or large diameter a flat head becomes impractically thick and heavy.

  5. 5.What happens if a flat head is used in a high-pressure application?Application

    If a flat head is used in a high-pressure application, it may experience excessive deformation or even failure due to its inability to distribute stress effectively. Flat heads are not designed to handle high internal pressures, and using them in such conditions can lead to safety hazards and equipment damage.

  6. 6.How does the knuckle radius in a torispherical head affect its performance?Application

    The knuckle is where the crown curvature changes to the cylinder, and a sharp knuckle produces high local bending and a compressive hoop stress. A larger knuckle radius lowers L/r and hence the factor M = ¼(3 + √(L/r)), so the required thickness falls; with r = 6 % of L, M is about 1.77, while r = 10 % gives about 1.54. Codes set a minimum knuckle radius (6 % of the crown radius and at least three times the thickness); thin large heads with small knuckles can buckle there under internal pressure.

  7. 7.Calculate the thickness of a welded flat cover of effective diameter 600 mm at 1.0 MPa, allowable stress 120 MPa, E = 1 and attachment factor C = 0.33.Numerical

    Flat heads are designed for bending: t = d·√(C·P/(S·E)) = 600 × √(0.33 × 1.0/120) = 600 × 0.0524 = 31.5 mm, before corrosion allowance. A 2:1 ellipsoidal head of the same size would need only about 2.5 mm, which shows why flat heads are kept small. The factor C depends on the attachment detail and must be taken from the code.

  8. 8.For a torispherical head, how does the crown (dish) radius relate to the diameter of the vessel?Concept

    Codes limit the inside crown radius to not more than the outside diameter of the head, and a standard 'dished and flanged' head uses a crown radius about equal to the diameter with a knuckle radius of 6 % of the crown radius. A smaller crown radius with a larger knuckle (for example the 80 %/10 % proportions) gives a deeper, stronger head that is closer in thickness to an ellipsoidal head.

  9. 9.What are the implications of using a thicker head than necessary in a pressure vessel?Application

    Using a thicker head than necessary can lead to increased material costs and weight, which may not be economically efficient. It can also complicate the manufacturing process and may require additional support structures. However, it can provide a higher safety margin and potentially longer service life.

  10. 10.Determine the internal volume of a 2:1 ellipsoidal head on a 2 m inside-diameter vessel.Numerical

    The head is half an ellipsoid of revolution with semi-major axis a = D/2 = 1 m and semi-minor axis (depth) b = D/4 = 0.5 m. V = (2/3)·π·a²·b = (2/3) × π × 1² × 0.5 = 1.047 m³, the same as the shortcut π·D³/24 = π × 8/24 = 1.047 m³.

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