CELE Strength of Materials — Thin-Walled Pressure VesselsRevision Notes
Final-week revision notes for Thin-Walled Pressure Vessels. If you have already studied the full chapter, this page is your go-to refresher before sitting the CELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Civil Engineering tests in the Strength of Materials subtest.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Strength of Materials subtest is marked as "Core" in the official pattern, and Thin-Walled Pressure Vessels appears in position 8th of 8 in the CELE Strength of Materials review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Thin-Walled Pressure Vessels - Revision Notes
Pressure vessels are closed containers designed to hold fluids (liquids or gases) at pressures significantly different from ambient. In civil engineering practice, you encounter them as water supply pipelines, penstocks, storage tanks, boilers, and reservoirs. When the wall thickness t is small relative to the inner radius r (specifically t/r ≤ 1/10), stress distribution across the wall thickness is nearly uniform and can be determined by simple equilibrium — no need for advanced thick-wall (Lamé) theory. This chapter is a consistent board exam topic covering two principal geometries: cylindrical and spherical pressure vessels. Master the stress formulas, the thin-wall criterion, joint efficiency, and the design procedure, and you will answer these items confidently.
Sections
Formulas
Example
A pipe: D = 600 mm → r = 300 mm, t = 8 mm. Check: t/r = 8/300 = 0.027 ≤ 0.1 ✔ Thin-wall assumption valid.
Formula
t/r ≤ 1/10 (or equivalently r/t ≥ 10)
Variables
t = wall thickness (mm or m); r = inner radius (mm or m)
Application
Validity check before applying thin-wall stress formulas. If violated, use Lamé thick-cylinder equations instead.
Exam Tips
- The thin-wall check is often worth dedicated credit in a problem — never skip it.
- When the problem gives D (diameter), immediately compute r = D/2 in your scratch work to avoid mix-ups.
- If t/r is close to 0.1, state it and proceed — the problem almost certainly intends the thin-wall approach unless Lamé is explicitly requested.
Key Points
- A vessel is classified as thin-walled when t/r ≤ 1/10 (equivalently, r/t ≥ 10), where t is wall thickness and r is the INNER radius.
- Under this condition, stress is essentially uniform across the wall thickness — the membrane stress assumption holds.
- If t/r > 1/10, the vessel is thick-walled and requires Lamé's equations (outside scope of this chapter but know the boundary).
- Always verify the thin-wall assumption AFTER computing t; this step is frequently required in board problems.
- Inner radius r = D/2, where D is the inner diameter. Use inner dimensions throughout — NOT outer.
- The criterion ensures the simplified equilibrium approach gives results within about 5% of exact thick-wall theory.
Definitions
Term
Thin-Walled Pressure Vessel
Definition
A closed container holding internal pressure in which t/r ≤ 1/10, so wall stress is modeled as uniform (membrane stress) and computed from simple equilibrium.
Importance
Defines the applicability of the entire set of simplified formulas used in this chapter. Violating this criterion without switching to Lamé theory is a major error.
Term
Gauge Pressure (p)
Definition
Pressure measured above atmospheric pressure. In pressure vessel design, p always refers to gauge pressure — the net pressure pushing outward on the wall.
Importance
Board problems always use gauge pressure. Using absolute pressure when gauge is given will overestimate stresses.
Term
Membrane Stress
Definition
The in-plane tensile stress uniformly distributed through the wall thickness of a thin-walled vessel, with negligible radial (through-thickness) stress component.
Importance
Conceptual basis for all thin-wall formulas. Radial stress is negligible (≈ p at inner surface, 0 at outer — both small compared to hoop and longitudinal stresses).
Section Title
1. The Thin-Wall Criterion
Common Mistakes
- Using the outer radius or outer diameter instead of the inner radius/diameter in the formulas.
- Forgetting to perform the thin-wall check after computing the required thickness.
- Applying thin-wall formulas to a vessel where t/r > 0.1 — the error can be greater than 5%.
- Using absolute pressure instead of gauge pressure for p.
Formulas
Example
Pipe: D = 600 mm, t = 8 mm, p = 1.5 MPa. σ_h = (1.5 × 600)/(2 × 8) = 900/16 = 56.25 MPa
Formula
σ_h = pD/(2t) = pr/t
Variables
σ_h = hoop (circumferential) stress (MPa); p = internal gauge pressure (MPa); D = inner diameter (mm); r = inner radius (mm); t = wall thickness (mm)
Application
Governs design of cylindrical vessels — used to find required thickness or check stress against allowable. Governs longitudinal seams.
Example
Same pipe: σ_l = (1.5 × 600)/(4 × 8) = 900/32 = 28.13 MPa (= σ_h/2 ✔)
Formula
σ_l = pD/(4t) = pr/(2t)
Variables
σ_l = longitudinal (axial) stress (MPa); all other variables same as above
Application
Secondary stress in cylinder design. Governs circumferential (girth) seams. Used in computing shear stresses and failure on angled planes.
Example
56.25 MPa = 2 × 28.13 MPa ✔
Formula
σ_h = 2σ_l
Variables
σ_h = hoop stress; σ_l = longitudinal stress
Application
Quick verification check — if your computed σ_h is not exactly double σ_l, you made an arithmetic error.
Example
Pipe above: τ_max,in = (56.25 − 28.13)/2 = 14.06 MPa
Formula
τ_max,in-plane = (σ_h − σ_l)/2 = pr/(4t) = pD/(8t)
Variables
τ_max,in-plane = maximum in-plane shear stress (MPa)
Application
Mohr's circle of the biaxial stress state (σ_h, σ_l). Acts on planes at 45° to both principal planes.
Example
Pipe above: τ_abs = 56.25/2 = 28.13 MPa (equals σ_l — a useful cross-check)
Formula
τ_abs = σ_h/2 = pr/(2t) = pD/(4t)
Variables
τ_abs = absolute maximum shear stress (MPa), accounting for the zero radial stress face
Application
True maximum shear in the vessel wall — governs shear failure. Acts on a plane tilted 45° from the hoop plane toward the radial direction.
Exam Tips
- Memory aid: 'Hoop has 2 in the bottom, longitudinal has 4 in the bottom.' Or: 'Longitudinal force acts on a circle (area πr²), hoop force acts on a rectangle (area 2rt per unit length) — that's why the denominators differ.'
- When asked for the 'maximum stress,' always report σ_h (the larger one) unless the problem specifically asks for maximum shear.
- On a Mohr's circle sketch, place σ_h on the horizontal axis as the larger principal stress — the circle radius equals (σ_h − σ_l)/2 = τ_max,in-plane.
- If a problem asks about a diagonal crack at 45° to the axis, that failure is driven by the maximum shear stress.
Key Points
- A cylinder under internal pressure p develops TWO principal stresses: hoop (circumferential/tangential) stress σ_h and longitudinal (axial) stress σ_l.
- HOOP STRESS σ_h: derived from equilibrium of a longitudinal (lengthwise) free-body diagram — a half-cylinder slice cut along its length reveals the hoop forces.
- LONGITUDINAL STRESS σ_l: derived from equilibrium of a transverse (cross-sectional) free-body diagram — the end caps are pushed by pressure over area πr², resisted by the wall ring.
- Critical relationship: σ_h = 2 σ_l — hoop stress is ALWAYS twice the longitudinal stress in a closed-ended cylinder.
- Because σ_h > σ_l, cylinders under excessive pressure fail (burst) along a LONGITUDINAL seam — the hoop stress tears open that seam.
- Hoop stress controls the design of LONGITUDINAL welds/seams; longitudinal stress controls CIRCUMFERENTIAL welds/seams.
- Both stresses are TENSILE (for internal pressure), acting on orthogonal planes — they form a biaxial stress state.
- Radial stress at the inner wall ≈ −p (compressive) and at outer wall ≈ 0; for thin walls both are negligible compared to σ_h and σ_l.
- The in-plane shear (Mohr's circle of σ_h and σ_l) is τ_max,in-plane = (σ_h − σ_l)/2 = pr/(4t).
- The absolute maximum shear (including the zero radial face) is τ_abs = σ_h/2 = pr/(2t).
Definitions
Term
Hoop Stress (σ_h)
Definition
The circumferential tensile stress in the wall of a pressure vessel, acting tangentially (perpendicular to the axis and perpendicular to the radius). Resists the tendency of the cylinder to split open along its length.
Importance
Largest principal stress in a cylinder — governs the design and controls failure mode (longitudinal splitting). The denominator is 2t.
Term
Longitudinal Stress (σ_l)
Definition
The axial tensile stress in a closed-ended cylinder wall, developed because the internal pressure acts on the end caps and tries to pull the cylinder apart lengthwise.
Importance
One-half of hoop stress. Governs circumferential weld design. Absent in open-ended cylinders (e.g., guns, very long pipes with expansion joints).
Section Title
2. Cylindrical Pressure Vessels — Stress Analysis
Common Mistakes
- Using the hoop formula (pD/2t) for a SPHERE — the sphere formula has 4t in the denominator, not 2t.
- Forgetting the factor of 2 in the denominator for hoop: writing σ_h = pD/t instead of pD/2t.
- Confusing which stress governs which seam: HOOP governs LONGITUDINAL seams; LONGITUDINAL governs CIRCUMFERENTIAL seams.
- Calculating in-plane max shear as σ_h/2 instead of (σ_h − σ_l)/2.
- Mixing up τ_max,in-plane and τ_abs — the absolute maximum shear is TWICE the in-plane maximum for a cylinder.
Formulas
Example
Spherical tank: D = 3 m = 3000 mm, p = 1.8 MPa, σ_allow = 120 MPa. t = (1.8 × 3000)/(4 × 120) = 5400/480 = 11.25 mm → use t = 12 mm.
Formula
σ = pD/(4t) = pr/(2t)
Variables
σ = membrane stress in all tangential directions (MPa); p = internal gauge pressure (MPa); D = inner diameter (mm or m); r = inner radius; t = wall thickness
Application
Single formula for all orientations in a spherical vessel. Used to find required thickness or verify stress against allowable.
Example
For σ = 112.5 MPa (sphere at capacity), τ_abs = 112.5/2 = 56.25 MPa.
Formula
τ_abs,sphere = σ/2 = pD/(8t) = pr/(4t)
Variables
τ_abs,sphere = absolute maximum shear stress in sphere wall (MPa)
Application
Governs shear failure in spherical vessels. Acts on planes 45° to the wall tangent plane.
Exam Tips
- Association: 'Sphere = cylinder's longitudinal.' Both = pD/4t. This makes comparison problems very straightforward.
- For equal p, r, and t: t_sphere = t_cylinder/2 for the same allowable stress — sphere needs half the thickness. This is a common comparison question.
- If a problem mixes cylindrical and spherical parts (e.g., a tank with hemispherical end caps), apply the cylinder formula to the barrel and the sphere formula to the caps — they give DIFFERENT stresses for the same thickness.
Key Points
- By symmetry, every great-circle cross-section of a sphere gives an identical result — there is only ONE membrane stress acting equally in all directions in the wall.
- The sphere's membrane stress equals the cylinder's LONGITUDINAL stress (both use pD/4t) but is HALF the cylinder's hoop stress.
- This is why spheres are structurally more efficient for pressure containment: for the same p, r, and t, the sphere is stressed only half as much as the cylinder.
- High-pressure storage (LPG, compressed gas, water towers) uses spherical or nearly spherical shapes for this reason.
- No in-plane shear stress on any principal plane in a sphere (the two equal principal stresses give zero in-plane shear).
- Absolute maximum shear in the sphere wall: τ_abs = σ/2 = pD/(8t) — acting on planes tilted 45° toward the radial direction.
- Penstock (large-diameter pipe carrying water under head) design in civil engineering uses the cylindrical hoop formula with p = γH (fluid pressure at depth H).
Definitions
Term
Membrane Stress (Sphere)
Definition
The uniform biaxial tensile stress in the wall of a spherical vessel, equal in every tangential direction. Derived by cutting the sphere with any great-circle plane and applying pressure-area equilibrium to the resulting hemispherical free body.
Importance
Because the two principal stresses are equal, the in-plane Mohr's circle degenerates to a point — zero in-plane shear. This is unique to the sphere.
Section Title
3. Spherical Pressure Vessels — Stress Analysis
Common Mistakes
- Using the cylinder hoop formula (pD/2t) for a sphere — the sphere denominator is 4t.
- Thinking the sphere has two different stresses like a cylinder — it has one uniform membrane stress in all directions.
- Computing in-plane shear as (σ − σ)/2 = 0 but forgetting that the absolute maximum shear is still σ/2 due to the radial zero-stress face.
Formulas
Example
Boiler: D = 1200 mm, p = 2 MPa, η = 0.85, σ_allow = 100 MPa. t = (2 × 1200)/(2 × 0.85 × 100) = 2400/170 = 14.12 mm → use t = 15 mm.
Formula
t_cyl = pD / (2 η σ_allow) [hoop governs — longitudinal seam]
Variables
t_cyl = required cylinder wall thickness (mm); p = gauge pressure (MPa); D = inner diameter (mm); η = joint efficiency of longitudinal seam (dimensionless); σ_allow = allowable tensile stress of plate material (MPa)
Application
Primary design formula for cylindrical vessels. Used when joint efficiency of the longitudinal seam is specified. Gives minimum t based on the critical hoop stress.
Example
If circumferential seam η = 0.60 for the same boiler: t = 2400/(4 × 0.60 × 100) = 2400/240 = 10 mm — less than 14.12 mm, so longitudinal seam still governs.
Formula
t_cyl = pD / (4 η σ_allow) [longitudinal stress — circumferential seam]
Variables
Same variables; η here is the efficiency of the circumferential seam
Application
Used when circumferential seam has a lower η than longitudinal. Check both; the larger t governs.
Example
Sphere: D = 3000 mm, p = 1.8 MPa, η = 1.0, σ_allow = 120 MPa. t = (1.8 × 3000)/(4 × 1.0 × 120) = 5400/480 = 11.25 mm → use 12 mm.
Formula
t_sph = pD / (4 η σ_allow)
Variables
Same variables; η = joint efficiency of the sphere's seam
Application
Design formula for spherical vessels with weld efficiency.
Example
Tank: D = 800 mm, t = 10 mm, η = 1.0, σ_allow = 90 MPa. p_max = (2 × 10 × 1.0 × 90)/800 = 1800/800 = 2.25 MPa.
Formula
p_max = 2t η σ_allow / D [cylinder, hoop controls]
Variables
p_max = maximum allowable internal pressure (MPa)
Application
Finding the maximum safe operating pressure of an existing vessel.
Exam Tips
- When the problem says 'joint efficiency = 0.85,' immediately note: this goes in the denominator, reducing the allowable capacity of the seam.
- If only one η is given, use it for the hoop (governing) formula — the problem intends for hoop to control.
- Board problems may give η as a percentage (e.g., 85%) — convert to decimal (0.85) before use.
- When solving for p_max of an existing vessel, use the smaller of the two computed pressures from hoop and longitudinal checks.
Key Points
- Real pressure vessels are fabricated from plates joined by welding or riveting. These joints (seams) are inherently weaker than the parent plate material.
- Joint efficiency η (eta) is a dimensionless factor, 0 < η ≤ 1.0 (never greater than 1.0), expressing the ratio of joint strength to parent-plate strength.
- Typical values: η = 1.0 (full-penetration butt weld, fully radiographed), η = 0.85 (spot-radiographed), η = 0.70 (no radiography). Check the problem statement.
- The joint efficiency REDUCES the effective allowable stress across the seam: effective allowable = η × σ_allow.
- For a cylinder: the longitudinal seam is stressed by σ_h (hoop). Use η of the longitudinal seam when computing t from the hoop formula.
- For a cylinder: the circumferential seam is stressed by σ_l (longitudinal). Use η of the circumferential seam when computing t from the longitudinal formula.
- In most problems, hoop governs (since σ_h = 2σ_l), so the longitudinal seam efficiency controls the required thickness.
- Design is complete only after: (1) computing t, (2) rounding UP to available plate size, and (3) verifying t/r ≤ 0.1 with the rounded value.
- Factor of safety (FS) is handled before η is applied: σ_allow = σ_ultimate / FS (or σ_yield / FS depending on the code basis). η then further reduces usable capacity.
Definitions
Term
Joint Efficiency (η)
Definition
The ratio of the strength of a welded or riveted joint to the strength of the solid (unjointed) parent plate. Ranges from 0 (no load capacity) to 1.0 (joint as strong as solid plate). Entered directly into the denominator of the design thickness formulas.
Importance
Failure to include η when given will result in an under-designed (unsafe) vessel — a critical design error. Always scan the problem for a stated η.
Term
Allowable Stress (σ_allow)
Definition
The maximum permissible stress in the wall material, equal to the material's strength (ultimate or yield, depending on code) divided by the factor of safety. Specified by the designer or given directly in board problems.
Importance
Defines the safe operating limit of the vessel. In Philippine practice, ASME Boiler and Pressure Vessel Code and applicable NSCP provisions guide allowable stress selection.
Section Title
4. Joint Efficiency and Design Thickness
Common Mistakes
- Forgetting to divide by η — treating the seam as if it were full-strength parent plate.
- Applying the longitudinal seam η to the circumferential seam formula and vice versa.
- Not rounding t UP to the next standard plate size — always round up in design, never down.
- Using the wrong formula when both seam efficiencies are given — must check both and use the larger (more conservative) required thickness.
Formulas
Example
p=2 MPa, D=500 mm, t=6 mm: σ_h=(2×500)/(2×6)=83.3 MPa; σ_l=41.7 MPa; ratio=2:1 ✔
Formula
CYLINDER: σ_h = pD/(2t), σ_l = pD/(4t), σ_h = 2σ_l
Variables
p [MPa], D [mm], t [mm] → stress [MPa]
Application
Complete stress state of a closed-ended thin-walled cylinder under internal pressure.
Example
p=2 MPa, D=500 mm, t=6 mm: σ=(2×500)/(4×6)=41.7 MPa (= σ_l of same cylinder ✔)
Formula
SPHERE: σ = pD/(4t)
Variables
p [MPa], D [mm], t [mm] → stress [MPa]
Application
Uniform membrane stress in all directions in a thin-walled spherical vessel.
Example
Cylinder: p=1.5 MPa, D=1000 mm, η=0.90, σ_allow=80 MPa → t=(1.5×1000)/(2×0.90×80)=1500/144=10.4 mm → use 11 mm
Formula
DESIGN: t_cyl = pD/(2ησ_allow) [hoop], t_sph = pD/(4ησ_allow)
Variables
η = joint efficiency (dimensionless, ≤ 1.0); σ_allow [MPa]
Application
Minimum required wall thickness accounting for joint efficiency.
Example
Water tower at H = 40 m head: p = 9.81 × 40 / 1000 = 0.392 MPa
Formula
HYDRAULIC HEAD: p (MPa) = γ_w × H / 1000 = 9.81H/1000 = 0.00981H
Variables
H = water head [m]; γ_w = 9.81 kN/m³; result in MPa
Application
Converting water pressure in hydraulics (head in meters) to MPa for use in pressure vessel stress formulas — critical for penstock and pipeline problems.
Exam Tips
- Create a two-column 'recipe card' in your mind: LEFT = Cylinder (hoop pD/2t, long pD/4t); RIGHT = Sphere (σ = pD/4t). The RIGHT formula equals the cylinder's LEFT-divided-by-2.
- When a problem asks to 'compare cylinder and sphere,' the answer almost always involves the 2:1 ratio of stresses or the 2:1 ratio of wall thicknesses.
- Absolute maximum shear is always half of the largest principal stress (for thin walls where σ_radial ≈ 0): τ_abs = σ_h/2 for cylinders, τ_abs = σ/2 for spheres.
- For penstock problems: first find p from γH, then apply hoop formula — a two-step process that trips up unprepared examinees.
Key Points
- The factor in the denominator tells you everything: 2t → hoop (cylinder); 4t → longitudinal (cylinder) or any direction (sphere).
- For the same pressure p, inner radius r, and thickness t: σ_h,cylinder = 2σ_l,cylinder = 2σ_sphere.
- For the same p, r, and allowable stress: t_cylinder = 2 × t_sphere — a sphere needs only HALF the wall thickness of a cylinder.
- Conversion reminder: 1 MPa = 1 N/mm² = 1000 kPa. Keep units consistent (p in MPa, D in mm → σ in MPa).
- Hydraulic pressure from a water column: p = γH = 9.81H kPa, where H is head in meters. Convert to MPa for use in formulas: p (MPa) = 0.00981H.
- In penstock design (civil engineering): p = γH (water pressure), and the hoop formula gives the governing tensile stress in the steel pipe wall.
- Design sequence: (1) Identify geometry (cylinder or sphere). (2) Identify p, D or r, σ_allow, η. (3) Apply governing formula. (4) Round UP. (5) Verify thin-wall. (6) State result.
Definitions
Term
Penstock
Definition
A large-diameter steel pipe or conduit that conveys water under pressure from a reservoir to a hydroelectric turbine or treatment plant. Designed as a thin-walled cylindrical pressure vessel with p = γH.
Importance
A distinctly civil engineering application of hoop stress — a recurring board exam context. The high water head creates significant hoop stress in the steel wall.
Term
Biaxial Stress State
Definition
A state of stress in which two non-zero principal stresses act on mutually perpendicular planes, with zero stress on the third face (radial direction for thin walls). Mohr's circle of a biaxial state has radius = (σ₁ − σ₂)/2.
Importance
The cylinder wall is in biaxial tension (σ_h, σ_l); the sphere wall is in equal biaxial tension. Understanding this enables correct Mohr's circle analysis for shear stresses and failure.
Section Title
5. Summary of Formulas and Comparative Analysis
Common Mistakes
- Cylinder vs. sphere mix-up on the denominator factor (2t vs. 4t).
- Forgetting to convert water head H to MPa before substituting into the stress formula.
- Reporting the longitudinal stress as the answer when the problem asks for 'maximum stress' in a cylinder — always report hoop (σ_h) as maximum.
- Using inner diameter D in the formula but accidentally computing r from an outer diameter given in the problem.
- Rounding the required thickness DOWN instead of UP — always round up for design problems.
Connections
- HYDRAULICS CONNECTION: Internal pressure p in a buried pipe or penstock comes from fluid head: p = γH. Civil engineers computing penstock stresses must first apply hydraulics (Bernoulli, hydrostatics) to find H, then apply thin-wall hoop formula — a direct cross-subject link in the board exam.
- MECHANICS OF MATERIALS — MOHR'S CIRCLE: The biaxial stress state (σ_h, σ_l) of a cylinder is the direct application of Mohr's circle for plane stress taught in the stress transformation chapter. The principal stresses ARE σ_h and σ_l (no shear on those faces), and τ_max,in-plane is the circle's radius.
- STRUCTURAL DESIGN — NSCP 2015 / AISC 360: Steel plate allowable stress for pressure vessel walls is analogous to the allowable tension stress in NSCP structural members (0.6Fy for ASD). RA 544 (Civil Engineering Law of the Philippines) requires that structural components including pressure-retaining elements meet adopted design standards — reinforcing why σ_allow must be code-based.
- MATERIAL SCIENCE — DUCTILE FAILURE: The hoop stress being the critical stress in cylinders aligns with the concept that ductile materials under biaxial tension fail when a principal stress reaches the yield strength (Maximum Normal Stress theory) — connecting failure theories (covered in a later chapter) back to pressure vessel geometry.
- CONCRETE DESIGN — ACI 318 (Prestressed Concrete Pipes): Prestressed concrete pressure pipes use circumferential prestress (analogous to hoop stress) to precompress the concrete, counteracting the tensile hoop stress from internal water pressure. This connects thin-wall theory to ACI 318 prestressed design.
- GEOTECHNICAL / WATER RESOURCES: Water tanks and reservoirs used in Philippine waterworks (MWSS, LGU water utilities) are cylindrical or spherical vessels. The shell design uses exact hoop and longitudinal formulas covered here — a practical civil engineering context students encounter in their careers.
- THERMODYNAMICS / PLANT ENGINEERING: Steam boilers (cylindrical) — the classic thin-wall pressure vessel — apply these same formulas. Board exam items occasionally provide temperature-related expansion problems where the hoop stress from pressure is combined with thermal stress, linking to the thermal stress chapter.
Exam Strategy
Thin-walled pressure vessels typically appear as 1-3 items per board exam set, often as straightforward calculation items (find σ_h, find t, find p_max) or as conceptual comparison items (cylinder vs. sphere, which seam fails first). Your strategy: (1) IDENTIFY the shape immediately — cylinder or sphere. This determines which formula set to use. (2) LIST the given data: p (in MPa), D or r (in mm), t (if given), σ_allow, and η. (3) CHECK units — if water head H is given in meters, convert to MPa first using p = 0.00981H. (4) APPLY the correct formula — hoop for cylinder design (denominator 2t), membrane for sphere (denominator 4t). Include η in the denominator if given. (5) ROUND UP t if designing; compute exactly if finding stress or pressure. (6) VERIFY thin-wall: t/r ≤ 0.1 — state this explicitly to earn method marks. (7) VERIFY the 2:1 ratio (σ_h = 2σ_l) as an internal check when solving cylinder problems. Allot about 2 minutes per straightforward item, 4 minutes for multi-part problems. Errors almost always arise from: using wrong diameter (inner vs. outer), forgetting to include η, or using the sphere formula for a cylinder. A clear, labeled sketch of the free-body diagram (longitudinal half-cut for hoop, transverse cut for longitudinal) prevents formula mix-ups and earns partial marks.
Quick Review Questions
A cylindrical tank has an inner diameter of 1.0 m and wall thickness of 20 mm. It carries an internal gauge pressure of 3.0 MPa. (a) Verify the thin-wall assumption. (b) Compute the hoop and longitudinal stresses.
Step 1 always: check t/r. Here t/r = 0.04, well within the 0.10 limit. Then apply the two cylinder formulas. The factor-of-2 relationship between σ_h and σ_l serves as a built-in check — if it does not hold exactly, recheck arithmetic.
A penstock (steel pipe) carries water at a head of 150 m. The pipe diameter is 800 mm and allowable stress is 120 MPa (η = 1.0). Find the required wall thickness.
Penstock problems require a hydraulics-to-mechanics conversion: first get p from γH (in kPa, then convert to MPa). Then apply the hoop design formula. The thin-wall check is mandatory. Rounding 4.91 mm up to 5 mm is correct design practice.
A spherical LPG storage tank has an inner diameter of 4.0 m, wall thickness of 16 mm, and allowable stress of 100 MPa. What is the maximum allowable gauge pressure?
For a sphere, invert the thickness formula to solve for p: p = 4tσ_allow/D (with η = 1.0 assumed since no seam efficiency is given). This is a direct application of the sphere formula. The tank can safely hold up to 1.60 MPa.
Why does a cylindrical vessel under internal pressure typically fail along a longitudinal seam rather than a circumferential seam?
This conceptual question is frequently asked in board exams. The key is connecting which stress acts across which seam: hoop stress acts perpendicular to the longitudinal seam (trying to split the cylinder lengthwise), while longitudinal stress acts perpendicular to the circumferential seam. Higher stress → more critical seam → governing failure mode.
A cylindrical pressure vessel has D = 600 mm, t = 10 mm, σ_allow = 80 MPa, and longitudinal seam efficiency η = 0.75. Find the maximum safe internal pressure.
Joint efficiency reduces the effective strength of the seam. The governing stress is hoop, acting across the longitudinal seam (η = 0.75). Substituting into the inverted design formula gives p_max. Always check both seams when efficiencies differ.
For a thin-walled cylinder with p = 2.0 MPa, r = 400 mm, t = 8 mm, find: (a) in-plane maximum shear stress; (b) absolute maximum shear stress.
Two different shear stresses! In-plane shear comes from the Mohr's circle of the biaxial stress pair (σ_h, σ_l) — its radius = (σ_h − σ_l)/2. Absolute maximum shear includes the out-of-plane (radial) direction where σ_3 ≈ 0 — the larger circle from σ_h to 0 gives τ_abs = σ_h/2. Note τ_abs = σ_l — a useful cross-check.
A cylindrical tank and a spherical tank have the same inner diameter D = 2 m, the same wall thickness t = 20 mm, and carry the same internal pressure p = 1.2 MPa. Compare their maximum stresses.
This comparison problem appears frequently in board exams. The sphere's single membrane stress = pD/4t equals the cylinder's longitudinal stress and is half the cylinder's hoop stress. The sphere is more efficient — it handles the same pressure at half the peak stress, or equivalently, needs only half the wall thickness for the same allowable stress.
Is the following statement TRUE or FALSE and why? 'The absolute maximum shear stress in a thin-walled cylindrical vessel equals the in-plane maximum shear stress.'
A classic concept question on 3D Mohr's circle interpretation. For a biaxial tensile stress state (both principal stresses positive), the absolute maximum shear is NOT the radius of the in-plane circle but the radius of the circle spanning from the maximum principal stress to zero (the radial direction). Always check all three Mohr's circles for the true τ_abs.
Ready to practise for the CELE 2026?
Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.