CELE Strength of Materials — Thin-Walled Pressure VesselsDetailed Explanation
Want to really understand Thin-Walled Pressure Vessels before tackling CELE Strength of Materials questions? This detailed explanation breaks down every key concept, shows you why it matters for the CELE 2026, and walks through the reasoning Professional Regulation Commission (PRC) — Board of Civil Engineering expects on high-difficulty questions.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Strength of Materials section sits under a "Core" weighting, and Thin-Walled Pressure Vessels is the 8th chapter in the 8-chapter CELE Strength of Materials rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Strength of Materials.
Thin-Walled Pressure Vessels - Detailed Explanation
Pressure vessels are among the most critical structural elements in civil and mechanical engineering practice. In the Philippine setting, they appear as water distribution pipelines, penstocks feeding hydroelectric plants (like those in Benguet and Bukidnon), LPG storage tanks, and boiler systems. The PRC Civil Engineer Licensure Examination consistently tests this topic because it bridges Strength of Materials directly with Hydraulics and structural design. This chapter develops the membrane stress theory for thin-walled cylindrical and spherical vessels from first principles, establishes the thin-wall validity criterion, introduces joint efficiency for welded and riveted seams, and guides you through the design procedure for minimum wall thickness. Mastery of these formulas — and the ability to avoid the classic exam pitfalls — will earn you reliable points on the board.
Concepts
The Thin-Wall Criterion
A pressure vessel is classified as thin-walled when the wall thickness t is small relative to the inner radius r — specifically when t/r ≤ 1/10, equivalently r/t ≥ 10. Under this condition, the normal stress through the thickness can be assumed to be zero (the radial stress is negligible), and the circumferential and longitudinal stresses are essentially uniform across the wall thickness. This simplification allows us to use simple free-body equilibrium rather than the full Lamé thick-cylinder equations. When t/r > 1/10, the radial variation of stress becomes significant and the Lamé equations must be used — but the board exam almost always deals with thin-walled cases, so always check the ratio first and state it explicitly in your solution. Use the inner radius (from the inner surface to the centreline) — not the outer radius — when checking the criterion.
Examples
This is a borderline case — on the exam, if t/r equals exactly 0.10, thin-wall formulas are still accepted. If t/r were 0.11 or higher, you must flag that Lamé equations are strictly required.
Scenario
A steel pipe has an inner diameter of 400 mm and a wall thickness of 20 mm. Check whether thin-wall formulas apply.
Solution
r = D/2 = 400/2 = 200 mm t/r = 20/200 = 0.10 Since t/r = 0.10 ≤ 0.10, the thin-wall criterion is exactly satisfied (borderline case). Thin-wall formulas may be applied.
Never blindly apply thin-wall formulas. A quick ratio check protects you from a fundamentally wrong approach — and examiners do include thick-wall vessels as distractors.
Scenario
A pressure vessel has r = 50 mm and t = 15 mm. Is it thin-walled?
Solution
t/r = 15/50 = 0.30 > 0.10 The vessel is NOT thin-walled. The Lamé thick-cylinder equations must be used.
Applications
- Determining whether a water distribution pipe qualifies for simplified stress analysis
- Checking penstocks in hydroelectric projects for thin-wall validity
- Preliminary screening of boiler drum designs
- Validating student solutions before applying hoop/longitudinal stress formulas
Misconceptions
- Using outer radius instead of inner radius in the t/r check — always use inner r
- Thinking t/r < 1/10 means 'much less than' — the criterion is ≤ 0.10, so 0.09 qualifies
- Applying thin-wall formulas when t/r = 0.15 just because it 'looks thin' — the check is mandatory
- Confusing the thin-wall criterion with the slenderness ratio used in column design
Related Concepts
- Lamé equations for thick-walled cylinders
- Plane stress state (σ_r ≈ 0 for thin walls)
- Principal stresses and Mohr's circle
- Hydraulic pressure (p = γh) as the source of internal pressure in penstocks and pipelines
Common Exam Questions
Example
A cylindrical tank has D = 600 mm and t = 25 mm. Verify if thin-wall theory is applicable. Answer: t/r = 25/300 = 0.083 < 0.10 — yes, thin-wall.
Approach
Compute t/r and compare with 0.10; state conclusion explicitly
Question Type
Verification/Check
Example
What is the maximum wall thickness for a thin-walled pipe of inner radius 250 mm? t_max = 0.10 × 250 = 25 mm.
Approach
Set t/r = 0.10 as the maximum; solve for t_max = 0.10r
Question Type
Given ratio, find thickness
Key Points To Remember
- Thin-wall condition: t/r ≤ 0.10 (equivalently r/t ≥ 10)
- Use inner radius r (or inner diameter D = 2r) in all thin-wall formulas
- Radial stress is assumed zero in thin-wall analysis
- Stresses are uniform through the wall thickness
- Violating the criterion means Lamé equations are needed — state this if asked
- Always verify the criterion before applying the formulas and state the check in your solution
Hoop (Circumferential) and Longitudinal Stresses in Cylinders
When a closed cylindrical vessel is subjected to internal gauge pressure p, two principal membrane stresses develop on the wall: the hoop (circumferential or tangential) stress σ_h acting in the circumferential direction, and the longitudinal (axial) stress σ_l acting along the axis of the cylinder. These are derived from two separate free-body equilibrium cuts. Hoop stress — longitudinal cut: Slice the cylinder lengthwise through a diameter and isolate one half. The internal pressure acts on the projected rectangular area (D × L), and the two cut edges carry the hoop stress in the two wall pieces. Equilibrium gives: σ_h = pD / (2t) = pr / t Longitudinal stress — transverse cut: Slice the cylinder perpendicular to its axis and isolate one portion including an end cap. The pressure acts on the full circular end area (πD²/4), and the wall area at the cut is πDt. Equilibrium gives: σ_l = pD / (4t) = pr / (2t) The key relationship: σ_h = 2 × σ_l. Hoop stress is always twice the longitudinal stress in a cylinder. This is why cylinders fail by splitting along a longitudinal seam (hoop stress tears the longitudinal joint) rather than at the ends. Both stresses are tensile (positive) for internal pressure. On a stress element of the cylinder wall, σ_h acts tangentially and σ_l acts axially; the radial stress σ_r ≈ 0 (thin-wall assumption). This is a biaxial plane stress state.
Examples
Note the denominator: 2t for hoop, 4t for longitudinal. A common board error is using 4t for hoop and 2t for longitudinal — the ratio check (σ_h = 2σ_l) will catch this mistake immediately.
Scenario
A water pipe has inner diameter 600 mm and wall thickness 8 mm, carrying an internal pressure of 1.5 MPa. Find the hoop and longitudinal stresses.
Solution
Step 1 — Thin-wall check: t/r = 8/300 = 0.0267 ≤ 0.10 ✔ Step 2 — Hoop stress: σ_h = pD/(2t) = (1.5 × 600)/(2 × 8) = 900/16 = 56.25 MPa Step 3 — Longitudinal stress: σ_l = pD/(4t) = (1.5 × 600)/(4 × 8) = 900/32 = 28.13 MPa Verification: σ_h/σ_l = 56.25/28.13 ≈ 2.0 ✔
This example shows the link between Hydraulics (pressure head) and Strength of Materials. On the board exam, the pressure is often given as a water head — convert to MPa first using p = γh (with γ in MPa/m or kN/m² carefully converted).
Scenario
A penstock (steel pipe) carries water at a head of 200 m. The pipe has D = 1.0 m and t = 12 mm. Find the stresses. Take γ_water = 9.81 kN/m³.
Solution
Step 1 — Convert head to pressure: p = γh = 9.81 × 200 = 1962 kN/m² = 1.962 MPa Step 2 — Thin-wall check: t/r = 12/500 = 0.024 ≤ 0.10 ✔ Step 3 — Hoop stress: σ_h = pD/(2t) = (1.962 × 1000)/(2 × 12) = 1962/24 = 81.75 MPa Step 4 — Longitudinal stress: σ_l = pD/(4t) = 1962/48 = 40.875 MPa
Applications
- Water distribution pipelines in Philippine water districts (MWSS, MCWD service areas)
- Penstocks for NPC/PSALM hydroelectric plants (Agus River complex, Pantabangan)
- Industrial boiler drums and pressure piping
- Underground water mains subject to internal pressure from pump stations
- Compressed-air cylinders and pneumatic piping
Misconceptions
- Using σ_h = pD/(4t) — wrong; the 4 in the denominator belongs to σ_l
- Confusing 'hoop' with 'longitudinal' direction — hoop is circumferential (around the girth), longitudinal is along the axis
- Using absolute pressure instead of gauge pressure
- Forgetting that longitudinal stress arises from end caps — an open-ended cylinder has no σ_l
- Using outer diameter instead of inner diameter in the formulas
Related Concepts
- Biaxial stress state and principal stresses
- Mohr's circle for plane stress
- Maximum shear stress in the cylinder wall
- Joint efficiency and its effect on design
- Allowable stress design (ASD) approach
Common Exam Questions
Example
Given D = 800 mm, t = 10 mm, p = 2 MPa. Find σ_h and σ_l. Answer: σ_h = 80 MPa, σ_l = 40 MPa.
Approach
Apply σ_h = pD/(2t) and σ_l = pD/(4t); always check thin-wall first
Question Type
Direct stress computation
Example
D = 500 mm, t = 8 mm, σ_allow = 120 MPa. p_max = 2(8)(120)/500 = 3.84 MPa.
Approach
Set σ_h = p_allow and solve for p = 2t σ_allow / D; hoop governs
Question Type
Find pressure given allowable stress
Example
Which stress governs the design of a cylindrical boiler? The hoop stress σ_h = pD/(2t), which is double the longitudinal stress.
Approach
State that hoop stress is larger; failure seam is longitudinal
Question Type
Identify governing stress / failure mode
Key Points To Remember
- σ_h = pD/(2t) = pr/t — hoop stress (circumferential, tangential)
- σ_l = pD/(4t) = pr/(2t) — longitudinal (axial) stress
- σ_h = 2σ_l always for a closed cylinder under internal pressure
- Hoop stress is the LARGER stress — it governs design of the longitudinal seam
- Longitudinal stress governs the circumferential (girth) seam
- Both stresses are tensile for internal pressure
- p must be gauge (internal) pressure — do not use absolute pressure unless gauge = absolute (vacuum outside)
Stresses in Spherical Pressure Vessels
A spherical vessel under internal pressure p is symmetric in all directions. Any great-circle cut through the sphere produces the same result: the pressure force on the projected circular area (πr²p) must be balanced by the membrane stress in the wall around the cut circumference (2πr t σ). Solving: σ = pD/(4t) = pr/(2t) This single membrane stress acts equally in all tangential directions — the sphere is in a state of equal biaxial tension. There is no distinction between 'hoop' and 'longitudinal' for a sphere. Critical comparison with a cylinder of the same D, t, and p: Sphere: σ = pD/(4t) Cylinder hoop: σ_h = pD/(2t) = 2 × (sphere stress) For the same pressure, radius, and thickness, a sphere is stressed at only HALF the level of a cylinder. Conversely, to carry the same pressure at the same stress level, a sphere needs only HALF the wall thickness of a cylinder. This efficiency is why high-pressure gas storage (LPG bullets, industrial gas spheres) uses spherical geometry, while lower-pressure applications (water tanks, fuel tanks) often use cylinders for ease of fabrication.
Examples
A cylinder of the same diameter and pressure would require t = pD/(2σ_allow) = 5400/240 = 22.5 mm — nearly double! This illustrates the efficiency advantage of the spherical shape.
Scenario
A spherical LPG storage tank has a diameter of 3 m and holds gas at 1.8 MPa internal pressure. The allowable stress is 120 MPa. Find the required wall thickness.
Solution
Step 1 — Required thickness: t = pD/(4σ_allow) = (1.8 × 3000)/(4 × 120) t = 5400/480 = 11.25 mm Use t = 12 mm (round up to next commercial size) Step 2 — Thin-wall check: t/r = 12/1500 = 0.008 ≤ 0.10 ✔ Step 3 — Actual stress: σ = pD/(4t) = (1.8 × 3000)/(4 × 12) = 5400/48 = 112.5 MPa < 120 MPa ✔
Rearrange the sphere formula to solve for p. This form appears frequently when the problem gives the geometry and asks for the working pressure.
Scenario
A spherical water tank has D = 4 m and t = 16 mm. Find the maximum allowable internal pressure if σ_allow = 100 MPa.
Solution
p_max = 4t σ_allow / D = (4 × 16 × 100) / 4000 = 6400/4000 = 1.6 MPa
Applications
- LPG spherical storage vessels at Petron, Shell, and Caltex terminals
- Pressurized water storage towers with spherical tanks
- Underwater pressure hulls for diving bells and submersibles
- Industrial high-pressure gas storage spheres
- Comparison problems on the board exam contrasting cylinder vs sphere efficiency
Misconceptions
- Using σ = pD/(2t) for a sphere — this is the CYLINDER hoop formula, not the sphere formula
- Thinking a sphere has two different stresses like a cylinder — it has one uniform membrane stress
- Confusing sphere formula with longitudinal cylinder formula (they are numerically equal, but apply to different geometries)
- Not rounding up the computed thickness to the nearest commercial plate size
Related Concepts
- Comparison of pressure vessel shapes (efficiency)
- Equal biaxial tension (σ_1 = σ_2) and Mohr's circle for sphere
- Thin-wall validity for spheres
- Design of LPG storage systems under DOLE/OSHC regulations in the Philippines
Common Exam Questions
Example
D = 2 m, t = 10 mm, p = 1.5 MPa. σ = (1.5 × 2000)/(4 × 10) = 75 MPa.
Approach
Apply σ = pD/(4t); note the 4 in the denominator (unlike cylinder hoop which uses 2)
Question Type
Direct stress computation for sphere
Example
Show that a sphere and a cylinder (same D, t, p) have stress ratio of 1:2 (sphere:cylinder hoop).
Approach
Compute both stresses; sphere stress = cylinder longitudinal stress = half cylinder hoop stress
Question Type
Compare cylinder vs sphere
Example
D = 1.5 m, p = 2 MPa, σ_allow = 80 MPa. t = (2 × 1500)/(4 × 80) = 9.375 mm → use 10 mm.
Approach
Set σ = σ_allow and solve for t = pD/(4σ_allow)
Question Type
Minimum wall thickness of a sphere
Key Points To Remember
- σ = pD/(4t) = pr/(2t) — the single membrane stress in all tangential directions
- Sphere stress equals the LONGITUDINAL stress of a cylinder with the same D, t, and p
- Sphere stress is HALF the HOOP stress of a cylinder with the same D, t, and p
- Spheres are the most structurally efficient shape for internal pressure
- No preferential failure direction — the sphere tends to bulge uniformly
- For minimum wall thickness: t_sphere = pD/(4σ_allow) — cylinder needs twice this
- Apply thin-wall check t/r ≤ 0.10 to spheres as well
Joint Efficiency and Design for Allowable Stress
Real pressure vessels are fabricated by welding or riveting steel plates together. The joint (seam) is never as strong as the parent plate — welds may have inclusions, porosity, or heat-affected zones; riveted joints have reduced net area. The joint efficiency η (dimensionless, 0 < η ≤ 1) represents the ratio of the seam strength to the parent plate strength. Perfect welds: η = 1.0; good welds: η = 0.85–0.95; riveted joints: η = 0.60–0.80 (varies with rivet pattern). When sizing the wall, the allowable stress across the seam is reduced to η × σ_allow. The design formulas become: Cylinder (hoop governs longitudinal seam): t = pD / (2η σ_allow) Sphere or cylinder girth (circumferential) seam: t = pD / (4η σ_allow) For a cylinder, the LONGITUDINAL seam is subjected to hoop stress (the larger one) and the CIRCUMFERENTIAL (girth) seam is subjected to longitudinal stress (the smaller one). Therefore: • Longitudinal seam efficiency η_L governs the hoop stress check → t = pD/(2η_L σ_allow) • Circumferential seam efficiency η_C governs the longitudinal stress check → t = pD/(4η_C σ_allow) • Take the LARGER of the two computed thicknesses (the governing case) In many board problems, a single η is given without specifying which seam — apply it to the hoop stress formula (since hoop governs for cylinders). Design procedure summary: 1. Identify p, D, σ_allow, η 2. Compute t = pD/(2η σ_allow) for cylinder 3. Round t UP to the nearest whole millimeter or commercial plate size 4. Verify thin-wall: t/r ≤ 0.10 5. Compute actual stress to confirm it ≤ σ_allow
Examples
Joint efficiency η = 0.85 effectively reduces the allowable stress from 100 MPa to 85 MPa at the seam. Without η, the answer would be t = 12 mm — a non-conservative 17% underestimate.
Scenario
A cylindrical boiler 1.2 m in diameter must hold 2 MPa internal pressure. Allowable stress = 100 MPa, longitudinal joint efficiency = 0.85. Find the minimum wall thickness.
Solution
t = pD/(2η σ_allow) t = (2 × 1200)/(2 × 0.85 × 100) t = 2400/170 t = 14.12 mm Use t = 15 mm (rounded up) Thin-wall check: t/r = 15/600 = 0.025 ≤ 0.10 ✔
Even though η_C is smaller (0.75 vs 0.90), the circumferential seam case produces a smaller required thickness because it is governed by the smaller longitudinal stress. Always compute both and use the governing (larger) thickness.
Scenario
A cylindrical tank has a longitudinal joint efficiency of 0.90 and a circumferential joint efficiency of 0.75. D = 1.0 m, p = 1.5 MPa, σ_allow = 120 MPa. Find the governing wall thickness.
Solution
Case 1 — Longitudinal seam (hoop stress): t_1 = pD/(2η_L σ_allow) = (1.5 × 1000)/(2 × 0.90 × 120) = 1500/216 = 6.94 mm Case 2 — Circumferential seam (longitudinal stress): t_2 = pD/(4η_C σ_allow) = (1.5 × 1000)/(4 × 0.75 × 120) = 1500/360 = 4.17 mm Governing: t = max(6.94, 4.17) = 6.94 mm → Use t = 7 mm Thin-wall check: t/r = 7/500 = 0.014 ≤ 0.10 ✔
Applications
- Design of welded steel pipe for water supply and sewage systems
- Boiler drum sizing in power plants
- Riveted tank design (older infrastructure, heritage buildings with cast-iron cisterns)
- ASME Boiler and Pressure Vessel Code compliance (referenced in Philippine industrial regulations)
- Quality control: specifying weld quality (full penetration vs fillet) in terms of efficiency
Misconceptions
- Applying η to the pressure instead of the allowable stress — η modifies the strength side, not the load side
- Using η < 1 when a problem says 'seamless pipe' — η = 1.0 for seamless vessels
- Forgetting to round up — always go to the next higher millimeter or plate thickness
- Using η for both seams when the problem gives only one — apply the given η to the hoop stress check (conservative and standard practice)
Related Concepts
- Factor of safety (FS) and allowable stress design
- Weld quality and inspection (NDT, radiographic testing)
- Rivet patterns and net area calculations
- ASME Section VIII — Pressure Vessel Code (referenced in Philippine industry standards)
Common Exam Questions
Example
D = 900 mm, p = 1.2 MPa, σ_allow = 80 MPa, η = 0.80. t = (1.2 × 900)/(2 × 0.80 × 80) = 1080/128 = 8.44 mm → 9 mm.
Approach
Use t = pD/(2η σ_allow); substitute all given values; round up
Question Type
Find minimum thickness with joint efficiency
Example
D = 600 mm, t = 10 mm, σ_allow = 100 MPa, η = 0.85. p_max = 2(0.85)(100)(10)/600 = 1700/600 = 2.83 MPa.
Approach
Rearrange hoop formula: p = 2η σ_allow t / D
Question Type
Find maximum allowable pressure given t and η
Example
Is σ_h = 90 MPa safe if σ_allow = 100 MPa and η = 0.85? η σ_allow = 85 MPa < 90 MPa — NOT safe.
Approach
Actual stress in parent plate = pD/(2t); check against η σ_allow
Question Type
Effect of joint efficiency on stress
Key Points To Remember
- η ≤ 1.0; a value of 1.0 means a perfect, full-strength seam
- t_cylinder = pD/(2η σ_allow) — hoop stress governs, applied at the longitudinal seam
- t_sphere = pD/(4η σ_allow)
- Always round computed t UPWARD to ensure safety
- Verify thin-wall after computing t (since a thick wall from high pressure might violate the criterion)
- If two seam efficiencies are given, compute t for each and use the LARGER value
- σ_allow = F_y / FS or the value given directly — do not confuse with yield stress
Maximum Shear Stresses in Pressure Vessel Walls
The wall of a cylindrical vessel is in a biaxial stress state: σ_1 = σ_h (hoop), σ_2 = σ_l (longitudinal), σ_3 = 0 (radial, thin-wall assumption). Mohr's circle analysis gives three shear stress values: In-plane maximum shear (Mohr's circle of σ_1 and σ_2): τ_in-plane = (σ_h - σ_l)/2 = (pD/2t - pD/4t)/2 = pD/(8t) = pr/(4t) Absolute maximum shear (largest of the three Mohr's circles): τ_abs = σ_h/2 = pD/(4t) = pr/(2t) [since σ_3 = 0 and σ_1 = σ_h is the largest principal stress] For a sphere (σ_1 = σ_2 = σ = pD/4t, σ_3 = 0): τ_in-plane = 0 (equal biaxial tension, Mohr's circle collapses to a point in-plane) τ_abs = σ/2 = pD/(8t) These shear stress values are needed when checking for shear failure or when the problem asks about maximum shear. On the PRC board exam, the absolute maximum shear for a cylinder is a common follow-up question after finding the principal stresses.
Examples
The absolute maximum shear is twice the in-plane shear for a cylinder. It acts on a plane inclined 45° to the wall surface (out-of-plane). This is the relevant value if checking von Mises or Tresca yield criteria.
Scenario
For the water pipe in Example 1 (D = 600 mm, t = 8 mm, p = 1.5 MPa, σ_h = 56.25 MPa, σ_l = 28.13 MPa), find τ_in-plane and τ_abs.
Solution
In-plane shear: τ_in-plane = (σ_h - σ_l)/2 = (56.25 - 28.13)/2 = 28.12/2 = 14.06 MPa Alternatively: τ = pr/(4t) = (1.5 × 300)/(4 × 8) = 450/32 = 14.06 MPa ✔ Absolute maximum shear: τ_abs = σ_h/2 = 56.25/2 = 28.13 MPa Alternatively: τ_abs = pr/(2t) = (1.5 × 300)/(2 × 8) = 450/16 = 28.13 MPa ✔
Applications
- Checking shear stress against the shear allowable (typically 0.4 Fy to 0.6 Fy)
- Tresca (maximum shear stress) yield criterion for ductile metals
- Von Mises criterion calculations for pressure vessel design
- Identifying the orientation of potential shear failure planes
Misconceptions
- Stating τ_in-plane as the maximum shear — it is not; τ_abs (out-of-plane) is larger for a cylinder
- Using τ_abs = σ_l/2 — wrong; use the LARGER principal stress (σ_h) divided by 2
- For a sphere: saying τ_max = σ/2 in-plane — it is zero in-plane; σ/2 is the out-of-plane (absolute) maximum
Related Concepts
- Mohr's circle for three-dimensional stress
- Tresca and von Mises yield criteria
- Principal stresses and principal planes
- Shear failure in steel under combined stresses
Common Exam Questions
Example
σ_h = 80 MPa, σ_l = 40 MPa. τ_in-plane = 20 MPa, τ_abs = 40 MPa.
Approach
First find σ_h and σ_l; then τ_in-plane = (σ_h - σ_l)/2, τ_abs = σ_h/2
Question Type
Compute in-plane and absolute maximum shear
Key Points To Remember
- Cylinder: τ_in-plane = (σ_h - σ_l)/2 = pr/(4t) = pD/(8t)
- Cylinder: τ_abs = σ_h/2 = pr/(2t) = pD/(4t) (note: equals sphere membrane stress numerically)
- Sphere: τ_in-plane = 0 in the plane of the wall (equal biaxial tension)
- Sphere: τ_abs = σ/2 = pr/(4t) = pD/(8t)
- τ_abs always governs over τ_in-plane for design against shear
- For thin-wall vessels, the out-of-plane Mohr's circle (involving σ_3 = 0) gives the absolute maximum shear
Practice Problems
This problem links Hydraulics (pressure head conversion) with Strength of Materials. The penstock scenario is highly relevant to Philippine hydroelectric projects. Note that the longitudinal stress numerically equals the absolute maximum shear stress for a cylinder — this is not a coincidence (σ_l = pr/(2t) and τ_abs = pr/(2t)).
Problem
Problem 1 (Board-type, Cylindrical Pipe): A penstock made of steel has an inner diameter of 1.0 m and a wall thickness of 12 mm. It carries water at a pressure head of 250 m. (a) Verify the thin-wall criterion. (b) Find the hoop stress and longitudinal stress. (c) Find the absolute maximum shear stress. Take γ_water = 9.81 kN/m³.
Solution
(a) Thin-wall check: r = D/2 = 1000/2 = 500 mm t/r = 12/500 = 0.024 ≤ 0.10 ✔ Thin-wall formulas apply. (b) Convert head to pressure: p = γh = 9.81 kN/m³ × 250 m = 2452.5 kN/m² = 2.4525 MPa Hoop stress: σ_h = pD/(2t) = (2.4525 × 1000)/(2 × 12) = 2452.5/24 = 102.19 MPa Longitudinal stress: σ_l = pD/(4t) = 2452.5/48 = 51.09 MPa Check: σ_h ≈ 2 × σ_l = 2 × 51.09 = 102.19 MPa ✔ (c) Absolute maximum shear stress: τ_abs = σ_h/2 = 102.19/2 = 51.09 MPa
The factor of safety is applied first to get σ_allow, then joint efficiency η further reduces the effective allowable stress at the seam. Without η, the computed t would be 13.5 mm — 18% less than the safe value. Rounding up from 15.88 to 16 mm is essential for safety.
Problem
Problem 2 (Board-type, Cylinder Design with Joint Efficiency): A cylindrical pressure vessel with inner diameter 1.5 m is subjected to an internal pressure of 1.8 MPa. The steel has a yield strength F_y = 250 MPa and a factor of safety FS = 2.5. The longitudinal welded joint has an efficiency of η = 0.85. Determine: (a) the allowable stress, (b) the minimum wall thickness, and (c) verify the thin-wall criterion.
Solution
(a) Allowable stress: σ_allow = F_y / FS = 250 / 2.5 = 100 MPa (b) Minimum wall thickness (hoop governs): t = pD / (2η σ_allow) t = (1.8 × 1500) / (2 × 0.85 × 100) t = 2700 / 170 t = 15.88 mm Use t = 16 mm (round up) (c) Thin-wall check: r = D/2 = 750 mm t/r = 16/750 = 0.0213 ≤ 0.10 ✔
For a sphere, the formula uses 4 in the denominator (not 2 as for cylinder hoop). Rounding up from 10.42 to 11 mm ensures that the actual stress (136.36 MPa) remains below the seam-reduced allowable (144 MPa). Always perform this back-check after rounding.
Problem
Problem 3 (Board-type, Spherical Tank): A spherical pressure tank holds nitrogen gas at 2.4 MPa internal pressure. The inner diameter is 2.5 m. The allowable tensile stress of the steel is 160 MPa and the weld efficiency is η = 0.90. Find: (a) the required minimum wall thickness, and (b) the actual membrane stress at the chosen thickness.
Solution
(a) Required thickness: t = pD / (4η σ_allow) t = (2.4 × 2500) / (4 × 0.90 × 160) t = 6000 / 576 t = 10.42 mm Use t = 11 mm Thin-wall check: t/r = 11/1250 = 0.0088 ≤ 0.10 ✔ (b) Actual membrane stress at t = 11 mm: σ = pD/(4t) = (2.4 × 2500)/(4 × 11) = 6000/44 = 136.36 MPa Check against seam: σ ≤ η σ_allow = 0.90 × 160 = 144 MPa 136.36 MPa < 144 MPa ✔ Safe.
When finding maximum pressure (rather than minimum thickness), the LOWER of the two computed pressures governs — the vessel fails whichever limit is reached first. Here the longitudinal seam (hoop stress) controls despite having the higher efficiency, because hoop stress is twice the longitudinal stress.
Problem
Problem 4 (Board-type, Maximum Allowable Pressure): A cylindrical steel tank has an inner diameter of 800 mm and a wall thickness of 10 mm. Two joint efficiencies are given: longitudinal seam η_L = 0.90, circumferential seam η_C = 0.80. The allowable stress is 90 MPa. Find the maximum allowable internal pressure.
Solution
Case 1 — Longitudinal seam (hoop stress controls): p_1 = 2η_L σ_allow t / D p_1 = (2 × 0.90 × 90 × 10) / 800 p_1 = 1620/800 = 2.025 MPa Case 2 — Circumferential seam (longitudinal stress controls): p_2 = 4η_C σ_allow t / D p_2 = (4 × 0.80 × 90 × 10) / 800 p_2 = 2880/800 = 3.60 MPa Governing (lower) pressure: p_max = min(2.025, 3.60) = 2.025 MPa ≈ 2.03 MPa Thin-wall check: t/r = 10/400 = 0.025 ≤ 0.10 ✔
This problem highlights the sphere's efficiency advantage. In practice, spheres are more expensive to fabricate than cylinders (complex forming and welding), so the material savings must be weighed against fabrication cost. Board exams frequently ask for this comparison — the answer is always that the sphere needs exactly half the thickness.
Problem
Problem 5 (Board-type, Comparison): A pressure vessel must store gas at p = 3 MPa. Two designs are proposed: (A) a cylinder with D = 1.2 m, and (B) a sphere with D = 1.2 m. For both, σ_allow = 150 MPa and η = 1.0. Find the required wall thickness for each and determine the material savings of the sphere over the cylinder (as a percentage reduction in plate area for equal vessel length).
Solution
Cylinder (Design A): t_A = pD/(2η σ_allow) = (3 × 1200)/(2 × 1.0 × 150) = 3600/300 = 12 mm Sphere (Design B): t_B = pD/(4η σ_allow) = (3 × 1200)/(4 × 1.0 × 150) = 3600/600 = 6 mm Material comparison: t_B/t_A = 6/12 = 0.50 The sphere requires 50% less wall thickness than the cylinder. Since the surface area formula is similar for equal diameter, this corresponds to a 50% saving in plate material. Verify thin-wall for both: Cylinder: t/r = 12/600 = 0.020 ≤ 0.10 ✔ Sphere: t/r = 6/600 = 0.010 ≤ 0.10 ✔
Exam Preparation Tips
- MEMORIZE THE FOUR KEY FORMULAS with their denominators: σ_h = pD/(2t), σ_l = pD/(4t), σ_sphere = pD/(4t), t_min_cylinder = pD/(2ησ_allow). Engrave these — the denominators 2 and 4 are where most errors occur.
- ALWAYS STATE AND VERIFY THE THIN-WALL CRITERION first in every solution. Examiners award method marks for this check even if subsequent arithmetic has minor errors.
- UNIT CONSISTENCY: p in MPa (N/mm²), D and t in mm gives stress directly in MPa. Alternatively, p in kN/m² and D, t in m gives stress in kN/m² = kPa. Never mix units mid-calculation.
- PRESSURE HEAD CONVERSION: When a water problem gives head h in meters, convert using p = γh = (9.81 kN/m³)(h m) = 9.81h kPa = 0.00981h MPa. For h = 100 m: p = 0.981 MPa ≈ 1.0 MPa (a handy check value).
- REMEMBER THE RATIO σ_h = 2σ_l AS A SELF-CHECK: After computing both stresses, verify this ratio. If it does not hold, you mixed up the formulas.
- JOINT EFFICIENCY: η multiplies σ_allow, not p. The formula t = pD/(2ησ_allow) is equivalent to saying the effective allowable at the seam is ησ_allow. Do not put η in the numerator.
- SPHERE vs CYLINDER: A sphere has half the stress of a cylinder under the same conditions. The sphere formula pD/(4t) numerically matches the CYLINDER LONGITUDINAL formula — do not mistake one for the other. Context (geometry) determines which to use.
- ROUND UP t TO THE NEXT WHOLE MILLIMETER: Never round down wall thickness — a thinner wall means higher stress and possible failure. Commercial plates come in discrete thicknesses (6, 8, 10, 12, 16, 20 mm, etc.) — rounding to the next available commercial size is engineering practice.
- BACK-CHECK AFTER ROUNDING: After choosing a rounded-up t, compute the actual stress and confirm it is ≤ σ_allow (or ≤ ησ_allow). This is the last line of defense and earns full solution marks.
- BOARD EXAM STRATEGY: In a 4-choice MCQ, eliminate options by checking the denominator in the stress formula. If an option gives σ = pD/(4t) for a CYLINDER, it corresponds to longitudinal stress, not hoop — cross it out if hoop is asked.
- GAUGE PRESSURE vs ABSOLUTE: Board problems almost always state 'internal pressure' which is gauge pressure. If the problem mentions absolute pressure and atmospheric pressure separately, subtract atmospheric to get gauge before using the formulas.
- RA 544 (Civil Engineering Law) and the PRC Code of Ethics require engineers to design conservatively. This means using η < 1 when any seam exists, using FS ≥ 1 (never equal to 1 in practice), and always rounding wall thickness upward — these professional habits must be reflected in your exam solutions.
In summary
Thin-walled pressure vessel analysis is a foundational topic in the PRC Civil Engineer Licensure Examination's Strength of Materials component. The entire subject reduces to four elegant formulas derived from equilibrium — no advanced elasticity theory required — but the board exam tests whether you know the precise differences: 2t vs 4t in the denominator, cylinder vs sphere geometry, hoop vs longitudinal direction, and the modifying role of joint efficiency. The most important relationships to internalize are: (1) hoop stress is always twice the longitudinal stress in a cylinder; (2) the sphere formula equals the cylinder's longitudinal formula numerically, but a sphere is stressed at only half the level of a cylinder under the same conditions; (3) joint efficiency η reduces the effective strength at the seam and must be incorporated in design; and (4) the thin-wall criterion t/r ≤ 0.10 must be verified for every problem before applying any formula. In Philippine civil engineering practice, these concepts underpin the design of water distribution mains, penstocks for NPC hydroelectric facilities, LPG storage tanks, and industrial boilers — all regulated under the Mechanical Engineering and Electrical Engineering Codes alongside RA 544 (Civil Engineering Law). A civil engineer who deeply understands these principles not only passes the board exam but is equipped to contribute safely to the nation's water supply and energy infrastructure. Review the formulas daily, practice the five worked problems in this chapter, avoid the documented pitfalls, and approach every board problem with the systematic six-step procedure outlined above — thin-wall check, formula selection, computation, efficiency adjustment, rounding up, and back-verification.
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.