CELE Strength of Materials — Thin-Walled Pressure VesselsStudy Notes
Detailed study notes for CELE Strength of Materials — Thin-Walled Pressure Vessels. These are the kind of notes you would take if you were reviewing with someone who has already scored well on the CELE: organised by what Professional Regulation Commission (PRC) — Board of Civil Engineering tests first, followed by the nice-to-knows, and ending with the traps to avoid.
Exam context
On the CELE 2026, the Strength of Materials subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Thin-Walled Pressure Vessels lands at position 8th out of 8 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Strength of Materials on a typical CELE paper.
Thin-Walled Pressure Vessels - Study Notes
Pressure vessels are closed containers designed to hold fluids under pressure — a fundamental application in civil engineering practice across water supply systems, penstocks, boiler systems, and industrial storage tanks throughout the Philippines. When the wall thickness is small relative to the vessel's radius, simplified membrane stress formulas replace the complex thick-wall (Lamé) theory, making design analysis tractable and direct. This chapter develops the thin-wall criterion, derives hoop and longitudinal stresses in cylinders, establishes the stress state in spheres, and applies these principles to design and allowable-stress calculations. Mastery of these concepts is essential for civil engineers working on water distribution infrastructure, hydroelectric penstocks, and pressure-containing structures regulated under the NSCP 2015 and RA 544 (Plumbing Code of the Philippines).
Summary
Thin-walled pressure vessels represent a critical intersection of Strength of Materials, Hydraulics, and practical civil engineering design. The central theme is that when the wall thickness t is small relative to the inner radius r (specifically, t/r ≤ 1/10), the membrane stresses can be computed directly from equilibrium, yielding simple closed-form results. For cylinders, the hoop stress σ_h = pr/t (or pD/2t) dominates the longitudinal stress σ_l = pr/(2t), and this difference explains why cylinders fail along longitudinal seams. For spheres, the single stress σ = pr/(2t) is half the cylinder's hoop stress, making spheres far more efficient for high-pressure storage — a principle exploited in LPG tanks and gas storage worldwide. Joint efficiency factors η account for weld and seam weakness, reducing the effective allowable stress and requiring thicker walls in practice. The design methodology is straightforward: (1) verify the thin-wall criterion, (2) identify the governing stress (hoop for cylinders, axial for spheres), (3) include joint efficiency, (4) calculate the minimum thickness, and (5) verify both the thin-wall assumption and the stress limits in the final design. Filipino civil engineers encounter these principles in water distribution systems (MWSS, municipal water districts), hydroelectric penstocks (Cordillera, Mindanao), and industrial storage facilities. Mastery of this chapter is essential for the PRC Civil Engineer Licensure Examination and for safe, economical pressure-vessel design in professional practice.
Sections
A pressure vessel is classified as thin-walled when the wall thickness t is small compared to the radius r. The thin-wall criterion is: t/r ≤ 1/10 (equivalently, r/t ≥ 10) where r is the inner (gauge) radius and t is the wall thickness in consistent units. When this criterion is satisfied, the stress is approximately constant (uniform) through the wall thickness, justifying the use of simplified membrane stress formulas. If this condition is violated (t/r > 1/10), the vessel is classified as thick-walled and requires thick-wall (Lamé) theory, which involves exponential stress distributions that are beyond the scope of this chapter. Physical Reasoning: In a thin wall, the radial stress variation across t is negligible compared to the membrane (hoop and longitudinal) stresses. The assumption of constant stress through the thickness simplifies the equilibrium equations to a manageable level. This is why thin-wall analysis is so powerful in engineering practice — it trades accuracy at the wall surface for simplicity and speed in design. Always verify the thin-wall assumption at the end of any problem. If your calculated t violates t/r ≤ 1/10, you must either use thick-wall theory, increase the diameter, or reduce the pressure — the thin-wall formula is not valid. Practical Context in the Philippines: Water supply systems operated by the Metropolitan Waterworks and Sewerage System (MWSS) and municipal water districts frequently use thin-walled steel pipes with t/r ratios of 0.02 to 0.08, well within the thin-wall range. Similarly, penstocks for hydroelectric facilities in the Cordillera and Mindanao typically operate in this regime.
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1. The Thin-Wall Criterion and Validity
Examples
Checking thin-wall validity for a municipal water main
Problem
A water distribution pipe has an inner diameter D = 600 mm and wall thickness t = 8 mm. Verify that it qualifies as thin-walled.
Solution
Inner radius r = D/2 = 600/2 = 300 mm t/r = 8/300 = 0.0267 ≈ 0.027 Since 0.027 < 0.1, the thin-wall criterion is satisfied. ✓ Alternatively, r/t = 300/8 = 37.5 > 10, confirming thin-wall behavior.
Identifying a thick-walled vessel
Problem
A high-pressure industrial storage tank has D = 1000 mm and t = 150 mm. Is it thin-walled?
Solution
r = 500 mm t/r = 150/500 = 0.30 Since 0.30 > 0.1, this is a thick-walled vessel. Thin-wall formulas do NOT apply; Lamé theory must be used instead.
Key Points
- Thin-wall condition: t/r ≤ 1/10 (or equivalently r/t ≥ 10)
- Stress is essentially constant (uniform) through the wall thickness under this condition
- If t/r > 1/10, the vessel is thick-walled and requires Lamé theory — do not use thin-wall formulas
- Always verify the thin-wall assumption after calculating t in design problems
- Use inner radius and gauge pressure (not absolute pressure) in all formulas
A cylindrical pressure vessel under internal gauge pressure p develops two principal membrane stresses: one acting circumferentially (hoop stress) and one acting along the axis (longitudinal stress). These are derived by considering equilibrium of a free body obtained by cutting the vessel. HOOP (Circumferential) STRESS: Consider a longitudinal cut through the cylinder, exposing a half-cylinder cross-section. The internal pressure acting on the curved inner surface creates a net outward force that must be resisted by the tensile stress in the wall. Consider a cylindrical element of length L. The pressure force acting radially outward on the projected area is p × (D × L) = p·D·L. This force is resisted by the tensile stress σ_h acting on both cut surfaces, each of area (t × L). By equilibrium: p·D·L = 2·σ_h·t·L Simplifying: σ_h = p·D / (2t) = p·r / t where D = 2r is the inner diameter. LONGITUDINAL (AXIAL) STRESS: Consider a transverse cut perpendicular to the cylinder axis (imagine cutting a disk from the end). The pressure force on the circular end cap is p × (π·r²). This axial force must be resisted by the stress in the wall acting on the annular area π(r + t)² − π·r² ≈ 2π·r·t (for thin walls). By equilibrium: p·π·r² = σ_l · 2π·r·t Simplifying: σ_l = p·r / (2t) = p·D / (4t) CRITICAL RELATIONSHIP: σ_h = 2·σ_l The hoop stress is always twice the longitudinal stress. This is why cylindrical pressure vessels fail by splitting along a longitudinal seam — the hoop stress, being larger, reaches the yield or ultimate stress first and tears the vessel open along the weaker seam direction. Maximum Shear Stress: Using Mohr's circle or the stress tensor, the maximum in-plane shear (acting within the plane of the wall) is: τ_max(in-plane) = (σ_h − σ_l) / 2 = p·r / (4t) However, if we account for the radial stress (which is zero at the inner and outer surfaces and transitions in between), the absolute maximum shear stress is: τ_abs = σ_h / 2 = p·r / (2t) This distinction matters for failure criteria (Tresca vs. von Mises); for design, the hoop stress σ_h typically governs because it is the largest principal stress. Physical Interpretation: The hoop stress represents the tendency of the internal pressure to "hoop" the cylinder — to expand it radially. The longitudinal stress represents the end-cap effects — the ends of the cylinder push axially. In both cases, tension develops; the hoop dominates and is the design driver.
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2. Cylindrical Pressure Vessels — Hoop and Longitudinal Stress
Examples
Calculating hoop and longitudinal stress in a water distribution pipe
Problem
A water main has an inner diameter D = 600 mm, wall thickness t = 8 mm, and carries an internal gauge pressure of p = 1.5 MPa. Calculate the hoop and longitudinal stresses. Verify the thin-wall criterion.
Solution
Step 1: Verify thin-wall criterion. r = D/2 = 300 mm t/r = 8/300 = 0.0267 < 0.1 ✓ (thin-wall is valid) Step 2: Calculate hoop stress. σ_h = p·D / (2t) = (1.5 MPa × 600 mm) / (2 × 8 mm) σ_h = 900 / 16 = 56.25 MPa Alternatively: σ_h = p·r / t = (1.5 × 300) / 8 = 450 / 8 = 56.25 MPa ✓ Step 3: Calculate longitudinal stress. σ_l = p·D / (4t) = (1.5 × 600) / (4 × 8) = 900 / 32 = 28.13 MPa Alternatively: σ_l = p·r / (2t) = (1.5 × 300) / (2 × 8) = 450 / 16 = 28.13 MPa ✓ Step 4: Verify relationship. σ_h / σ_l = 56.25 / 28.13 = 2.0 ✓ Conclusion: The hoop stress (56.25 MPa) is twice the longitudinal stress (28.13 MPa), as expected. The hoop stress is the controlling stress for design.
Stress analysis in a penstock for hydroelectric power
Problem
A penstock in the Cordillera has an inner diameter D = 1.0 m, wall thickness t = 12 mm, and operates under a gauge pressure of p = 2.5 MPa (from water head and pumping). Find the hoop stress, longitudinal stress, and maximum shear stress.
Solution
Step 1: Check thin-wall. r = 500 mm, t/r = 12/500 = 0.024 < 0.1 ✓ Step 2: Hoop stress. σ_h = p·r / t = (2.5 MPa × 500 mm) / 12 mm = 1250 / 12 = 104.17 MPa Step 3: Longitudinal stress. σ_l = p·r / (2t) = (2.5 × 500) / (2 × 12) = 1250 / 24 = 52.08 MPa Step 4: Maximum in-plane shear. τ_max(in-plane) = (σ_h − σ_l) / 2 = (104.17 − 52.08) / 2 = 26.04 MPa Step 5: Absolute maximum shear. τ_abs = σ_h / 2 = 104.17 / 2 = 52.08 MPa Answer: σ_h = 104.17 MPa, σ_l = 52.08 MPa, τ_abs = 52.08 MPa. The hoop stress dominates and governs the design.
Maximum allowable pressure for an existing pipe
Problem
A cylindrical steel pipeline has D = 800 mm, t = 10 mm, and the allowable tensile stress is σ_allow = 90 MPa. What is the maximum gauge pressure it can safely carry?
Solution
The hoop stress governs (it is the largest). Setting σ_h = σ_allow: σ_h = p·D / (2t) = σ_allow p = (2t × σ_allow) / D = (2 × 10 mm × 90 MPa) / 800 mm p = 1800 / 800 = 2.25 MPa Answer: Maximum gauge pressure = 2.25 MPa. Note: If the problem asked for the maximum pressure based on longitudinal stress (unlikely, but for completeness): p_l = (4t × σ_allow) / D = (4 × 10 × 90) / 800 = 4.5 MPa The hoop-based pressure (2.25 MPa) is lower and therefore is the true limit.
Key Points
- Hoop stress σ_h = p·D / (2t) = p·r / t — governs the design of cylindrical vessels
- Longitudinal stress σ_l = p·D / (4t) = p·r / (2t) — half the hoop stress
- Critical relationship: σ_h = 2·σ_l always holds for thin-walled cylinders
- Cylinders fail along longitudinal seams (hoop stress is the largest tensile stress)
- Maximum in-plane shear: τ = (σ_h − σ_l) / 2 = p·r / (4t)
- Absolute maximum shear (including radial): τ_abs = σ_h / 2 = p·r / (2t)
- Use gauge pressure p (not absolute) and inner diameter D or radius r
A spherical pressure vessel exhibits a fundamentally different stress state from a cylinder. By symmetry, every diametral plane through a sphere experiences the same stress distribution. When a sphere is cut by any great circle (a plane through the center), the internal pressure force must be balanced by the tensile stress in the wall, and this balance yields a single membrane stress acting uniformly in all directions within the wall. DERIVATION OF SPHERICAL STRESS: Consider a diametral cut through the sphere. The internal pressure acts over a projected circular area of π·r². This force must be resisted by tensile stress acting on the annular cross-section of the wall. For a thin wall, the annular area is approximately 2π·r·t. By equilibrium: p·π·r² = σ · 2π·r·t Simplifying: σ = p·r / (2t) = p·D / (4t) where D = 2r is the inner diameter. COMPARISON WITH CYLINDER: Notice that the spherical stress σ_sphere = p·r / (2t) is identical to the cylinder's longitudinal stress σ_l = p·r / (2t), and exactly HALF the cylinder's hoop stress σ_h = p·r / t. For the same diameter D, thickness t, and internal pressure p: σ_sphere = σ_cylinder,longitudinal = (1/2) × σ_cylinder,hoop This is the fundamental advantage of spheres: they experience lower stresses than cylinders under identical loading. A sphere requires only half the wall thickness of a cylinder to sustain the same pressure — or conversely, a sphere can hold twice the pressure with the same wall thickness. PRACTICAL IMPLICATIONS: This stress efficiency is why high-pressure storage tanks (liquefied petroleum gas, compressed air, nitrogen) are manufactured as spheres rather than cylinders. The reduced stress state translates directly to lower material cost, reduced wall thickness, and improved fatigue resistance. In the Philippines, LPG storage facilities and industrial gas storage typically employ spherical tanks for pressures above 1 MPa. MAXIMUM SHEAR IN SPHERE: Since the sphere has a single tensile stress σ in all directions (zero radial stress at the surfaces, transitioning through the wall), the maximum shear stress is: τ_max = σ / 2 = p·r / (4t) = p·D / (8t) This is half the maximum shear in a cylinder, confirming the overall stress advantage.
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3. Spherical Pressure Vessels
Examples
Designing a spherical LPG storage tank
Problem
An LPG storage tank is spherical with D = 3.0 m inner diameter. The design pressure is p = 1.8 MPa, and the allowable stress in the wall is σ_allow = 120 MPa. Calculate the required wall thickness and verify the thin-wall criterion.
Solution
Step 1: Calculate required thickness. σ = p·D / (4t) = σ_allow t = p·D / (4·σ_allow) = (1.8 MPa × 3000 mm) / (4 × 120 MPa) t = 5400 / 480 = 11.25 mm Use t = 12 mm (round up for fabrication). Step 2: Verify thin-wall criterion. r = D/2 = 1500 mm t/r = 12/1500 = 0.008 < 0.1 ✓ (thin-wall is valid) Answer: Required wall thickness = 12 mm. Comparison with cylinder: A cylinder of equal diameter and pressure would require: t_cyl = p·D / (2·σ_allow) = 5400 / 240 = 22.5 mm ≈ 23 mm The sphere uses half the wall thickness — a 50% material saving!
Comparing cylinder and sphere for water storage
Problem
A municipal water storage facility must hold 500 kPa (0.5 MPa) of pressure. Two designs are proposed: a cylinder (D = 2.0 m) and a sphere (D = 2.0 m). For both, the allowable stress is 100 MPa. Compare the required wall thicknesses.
Solution
Cylindrical design: t_cyl = p·D / (2·σ_allow) = (0.5 × 2000) / (2 × 100) = 1000 / 200 = 5.0 mm Spherical design: t_sph = p·D / (4·σ_allow) = (0.5 × 2000) / (4 × 100) = 1000 / 400 = 2.5 mm Answer: The cylinder requires 5.0 mm; the sphere requires 2.5 mm — exactly half. If both use 5.0 mm walls: - Cylinder can hold 0.5 MPa (as designed). - Sphere can hold 1.0 MPa (twice the design pressure). This demonstrates the sphere's superior efficiency.
Stress analysis in a spherical pressure vessel
Problem
A spherical tank (D = 2.5 m, t = 15 mm) holds gas at p = 1.2 MPa. Find the tensile stress and the maximum shear stress in the wall.
Solution
Step 1: Check thin-wall. r = 1250 mm, t/r = 15/1250 = 0.012 < 0.1 ✓ Step 2: Calculate tensile stress. σ = p·D / (4t) = (1.2 × 2500) / (4 × 15) = 3000 / 60 = 50 MPa Step 3: Calculate maximum shear. τ_max = σ / 2 = 50 / 2 = 25 MPa Answer: Tensile stress = 50 MPa; maximum shear = 25 MPa. For comparison, if this were a cylinder (same D, t, p): σ_h = p·D / (2t) = 3000 / 30 = 100 MPa (twice as much!) τ_max = 50 MPa (twice as much!)
Key Points
- Spherical stress: σ = p·D / (4t) = p·r / (2t) — acts uniformly in all directions
- Sphere stress = cylinder longitudinal stress = (1/2) × cylinder hoop stress
- For equal diameter, thickness, and pressure: sphere requires half the wall thickness of a cylinder
- Spheres are more efficient and are preferred for high-pressure storage (LPG, gases)
- Maximum shear in sphere: τ_max = p·r / (4t) — half that of a cylinder
- Use gauge pressure p and inner diameter D or radius r
- A sphere with the same t and D as a cylinder can safely hold twice the pressure
Real pressure vessels have welded or riveted seams that are potential weak points. A seam may have defects (incomplete penetration, porosity, slag inclusions), stress concentrations, or lower strength than the parent plate. To account for this, the design code introduces a joint efficiency factor η (a dimensionless number between 0 and 1) that reduces the effective allowable stress at the seam. JOINT EFFICIENCY FACTOR: The joint efficiency η depends on: - The type of joint (seamless vs. longitudinal seam vs. circumferential seam). - The inspection and quality level (radiographic inspection, ultrasonic testing). - The fabrication method (welded, riveted, etc.). Typical values: - η = 1.0: seamless tube or fully inspected weld. - η = 0.90 to 0.95: butt weld with partial inspection. - η = 0.85: butt weld with visual inspection only. - η = 0.70 to 0.80: riveted joint or poor-quality weld. APPLYING JOINT EFFICIENCY IN DESIGN: When sizing the wall thickness, the allowable stress is effectively reduced by the factor η. The design formula becomes: For a cylinder (hoop stress governs): t = p·D / (2·η·σ_allow) For a sphere: t = p·D / (4·η·σ_allow) The logic: if η = 0.85, the seam can sustain only 85% of the stress that the parent plate can. Therefore, we size the wall such that the working stress does not exceed η·σ_allow at the seam. WHICH SEAM GOVERNS? In a cylinder, two seams are critical: 1. **Longitudinal seam** — runs parallel to the axis and is subjected to hoop stress σ_h. This seam must resist the circumferential expansion of the pressure. 2. **Circumferential seam** — runs perpendicular to the axis (circumferential) and is subjected to longitudinal stress σ_l. Since σ_h = 2·σ_l, the **longitudinal seam** is the controlling design point. The design wall thickness is set to prevent hoop stress from exceeding η·σ_allow at the longitudinal seam: t = p·D / (2·η·σ_allow) REGULATORY CONTEXT (NSCP 2015 AND RA 544): The National Structural Code of the Philippines (NSCP 2015) references the ASME Boiler and Pressure Vessel Code (BPVC) and specifies inspection and certification requirements for pressure-containing structures. The Plumbing Code of the Philippines (RA 544) mandates that water supply and waste systems meet pressure vessel standards. Joint efficiency factors are specified based on the inspection and quality assurance level during construction.
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4. Joint Efficiency and Design for Allowable Stress
Examples
Designing a cylindrical boiler with joint efficiency
Problem
A cylindrical boiler (D = 1.2 m) must hold steam at p = 2.0 MPa. The allowable stress is σ_allow = 100 MPa. The longitudinal welded seam has an efficiency η = 0.85 due to visual inspection only. Calculate the minimum wall thickness. Use standard steel sheet sizes (0.5 mm increments for t ≥ 5 mm, 1 mm increments for t ≥ 10 mm).
Solution
Step 1: Calculate minimum thickness (hoop stress governs at longitudinal seam). t = p·D / (2·η·σ_allow) t = (2.0 MPa × 1200 mm) / (2 × 0.85 × 100 MPa) t = 2400 / 170 = 14.12 mm Round up to t = 14.5 mm (or 15 mm if only 1 mm increments are available). Step 2: Verify thin-wall criterion. r = 600 mm, t/r = 14.5/600 = 0.0242 < 0.1 ✓ Step 3: Verify stress at design thickness (t = 14.5 mm). σ_h = p·D / (2t) = (2.0 × 1200) / (2 × 14.5) = 82.76 MPa Stress at seam = η·σ_h = 0.85 × 82.76 = 70.35 MPa < 100 MPa ✓ Answer: Use t = 14.5 mm (or 15 mm). The hoop stress at the longitudinal seam is 70.35 MPa, within the allowable stress of 100 MPa.
Comparing joint efficiency scenarios for a water main
Problem
A cylindrical water pipe (D = 800 mm) operates at p = 1.0 MPa. The allowable stress is σ_allow = 90 MPa. Compare the required wall thickness for three scenarios: (a) seamless (η = 1.0), (b) butt weld with partial inspection (η = 0.90), and (c) butt weld with visual inspection only (η = 0.85).
Solution
Case (a) — Seamless (η = 1.0): t_a = p·D / (2·η·σ_allow) = (1.0 × 800) / (2 × 1.0 × 90) = 800 / 180 = 4.44 mm Use t = 4.5 or 5.0 mm. Case (b) — Partial inspection (η = 0.90): t_b = (1.0 × 800) / (2 × 0.90 × 90) = 800 / 162 = 4.94 mm Use t = 5.0 mm. Case (c) — Visual inspection only (η = 0.85): t_c = (1.0 × 800) / (2 × 0.85 × 90) = 800 / 153 = 5.23 mm Use t = 5.5 or 6.0 mm. Answer: Seamless: 4.5–5.0 mm Partial inspection: 5.0 mm Visual inspection: 5.5–6.0 mm The reduction in η increases the required thickness. Quality assurance and inspection are economically significant.
Designing a spherical tank with joint efficiency
Problem
A spherical gas storage tank (D = 4.0 m) operates at p = 1.5 MPa. The allowable stress is σ_allow = 110 MPa, and the welded seams have η = 0.90. Find the minimum wall thickness and verify the thin-wall criterion.
Solution
Step 1: Calculate thickness. t = p·D / (4·η·σ_allow) = (1.5 × 4000) / (4 × 0.90 × 110) t = 6000 / 396 = 15.15 mm Round up: t = 15.5 or 16 mm. Step 2: Verify thin-wall. r = 2000 mm, t/r = 15.5/2000 = 0.00775 < 0.1 ✓ Step 3: Check stress (with t = 15.5 mm). σ = p·D / (4t) = (1.5 × 4000) / (4 × 15.5) = 96.77 MPa Stress at seam = η·σ = 0.90 × 96.77 = 87.09 MPa < 110 MPa ✓ Answer: Use t = 16 mm. The tensile stress at the seam is 87.09 MPa, within allowable.
Key Points
- Joint efficiency η accounts for seam weakness; it is a factor between 0 (no strength) and 1 (full strength of parent plate)
- Typical values: η = 1.0 (seamless), 0.90–0.95 (welded with inspection), 0.85 (butt weld, visual inspection), 0.70–0.80 (riveted or poor weld)
- Design thickness for cylinder: t = p·D / (2·η·σ_allow)
- Design thickness for sphere: t = p·D / (4·η·σ_allow)
- In cylinders, the longitudinal seam (subjected to hoop stress) is the controlling design point
- Always verify thin-wall criterion (t/r ≤ 1/10) after calculating t
- RA 544 and NSCP 2015 mandate inspection and quality standards; η is specified by code
- If η is not given in a problem, assume η = 1.0 (seamless or fully inspected)
This section consolidates the complete design workflow: (1) check or calculate the thin-wall criterion, (2) identify which stress governs (hoop for cylinders, single stress for spheres), (3) apply joint efficiency if applicable, (4) size the wall or verify existing dimensions, and (5) confirm all results. The approach mirrors real-world practice in water supply design, penstock engineering, and boiler fabrication.
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5. Design and Verification Problems
Examples
Complete design of a water supply penstock (cylinder)
Problem
Design a steel penstock to carry water from a reservoir to a power station 150 m below. The required discharge is Q = 0.5 m³/s, and the water velocity is v = 2.0 m/s. From hydraulics, the static head plus dynamic pressure creates a gauge pressure of p = 1.8 MPa. The allowable stress in the steel is σ_allow = 95 MPa, and the weld efficiency is η = 0.90. Assume a seamless tube is being selected from a catalog. Find the minimum inner diameter and wall thickness.
Solution
Step 1: Determine the inner diameter from flow rate and velocity. Q = (π·D²/4)·v 0.5 m³/s = (π·D²/4)·2.0 m/s D² = (4 × 0.5) / (π × 2.0) = 2 / (2π) = 0.3183 D = 0.564 m ≈ 565 mm For standardization, select D = 600 mm (common commercial size). Step 2: Check velocity with D = 600 mm. v = (4Q) / (π·D²) = (4 × 0.5) / (π × 0.6²) = 2.0 / (π × 0.36) = 1.77 m/s (acceptable) Step 3: Design the wall thickness (hoop stress governs). t = p·D / (2·η·σ_allow) = (1.8 × 600) / (2 × 0.90 × 95) t = 1080 / 171 = 6.32 mm Round up: t = 6.5 or 7.0 mm. Select t = 7.0 mm for safety and fabrication. Step 4: Verify thin-wall criterion. r = 300 mm, t/r = 7.0/300 = 0.0233 < 0.1 ✓ Step 5: Verify stresses. Hoop stress: σ_h = p·D / (2t) = (1.8 × 600) / (2 × 7.0) = 77.14 MPa Longitudinal stress: σ_l = p·D / (4t) = (1.8 × 600) / (4 × 7.0) = 38.57 MPa Stress at seam (hoop): η·σ_h = 0.90 × 77.14 = 69.43 MPa < 95 MPa ✓ Answer: Penstock specifications — Inner diameter: 600 mm, Wall thickness: 7.0 mm. Stresses: hoop = 77.14 MPa, longitudinal = 38.57 MPa. All criteria satisfied.
Verification of an existing municipal water tank (sphere)
Problem
An existing spherical water storage tank in a provincial town has D = 10 m, t = 18 mm, and operates at p = 0.8 MPa (from pumping head). The steel has σ_yield = 250 MPa. For a factor of safety of 2.5 against yield, what is the allowable stress? Does the tank satisfy this allowable stress? Check the thin-wall assumption.
Solution
Step 1: Calculate allowable stress. σ_allow = σ_yield / (Factor of Safety) = 250 / 2.5 = 100 MPa Step 2: Check thin-wall criterion. r = 5000 mm, t/r = 18/5000 = 0.0036 < 0.1 ✓ (thin-wall is valid) Step 3: Calculate actual stress in the tank. σ = p·D / (4t) = (0.8 × 10,000) / (4 × 18) = 8000 / 72 = 111.11 MPa Step 4: Compare with allowable. Actual stress (111.11 MPa) > Allowable stress (100 MPa) ✗ Conclusion: The tank is over-stressed by (111.11 − 100) / 100 = 11.1%. The tank does NOT satisfy the factor of safety of 2.5. To meet the requirement, either (a) reduce operating pressure to 0.72 MPa, (b) increase thickness to at least t = 20 mm, or (c) accept a lower factor of safety. For a large public tank, strengthening or depressurization is recommended.
Burst pressure calculation for a test
Problem
A cylindrical test vessel (D = 400 mm, t = 8 mm) is made of aluminum alloy with σ_ultimate = 310 MPa. The longitudinal seam (critical) has η = 0.80 due to the experimental nature of the weld. At what gauge pressure will the vessel fail by hoop stress rupture? Verify the thin-wall criterion.
Solution
Step 1: Check thin-wall. r = 200 mm, t/r = 8/200 = 0.04 < 0.1 ✓ Step 2: Calculate burst pressure (using ultimate stress and seam efficiency). At failure, the stress at the seam equals the ultimate stress: η·σ_h = σ_ultimate η·(p·D / 2t) = σ_ultimate p = (2t·σ_ultimate) / (η·D) p = (2 × 8 × 310) / (0.80 × 400) p = 4960 / 320 = 15.5 MPa Answer: The vessel will burst at a gauge pressure of approximately 15.5 MPa. This is a very high pressure, suitable only for specialized testing. Note that the seam efficiency (0.80) reduces the burst pressure by 20% compared to a seamless vessel (which would burst at 15.5 / 0.80 = 19.4 MPa).
Key Points
- Design workflow: 1) Check thin-wall, 2) Identify governing stress, 3) Apply joint efficiency, 4) Calculate or verify t, 5) Confirm thin-wall and stress limits
- For cylinders: hoop stress governs the design; use t = p·D / (2·η·σ_allow)
- For spheres: single stress governs; use t = p·D / (4·η·σ_allow)
- Always verify t/r ≤ 1/10 and σ ≤ η·σ_allow in the final design
- Use gauge (internal) pressure and inner diameter throughout
- Round up thickness to the nearest standard dimension or fabrication increment
Understanding these frequent errors helps avoid costly mistakes in PRC exams and professional practice. **PITFALL 1: Confusing Hoop and Longitudinal Stresses** The hoop stress σ_h = pD / (2t) is LARGER than the longitudinal stress σ_l = pD / (4t). The factor in the denominator distinguishes them: 2 for hoop, 4 for longitudinal. Reversing these is the most common error. Remember: "hoop = 2 × longitudinal" for cylinders. **PITFALL 2: Using Absolute Pressure Instead of Gauge Pressure** The formulas require gauge pressure (internal pressure minus atmospheric). Using absolute pressure (e.g., 1.8 MPa absolute = 0.7 MPa gauge at sea level in the Philippines) inflates the calculated stress. Most engineering problems state "gauge pressure" explicitly; if only "pressure" is given and p > 1 MPa, assume gauge unless context indicates otherwise. **PITFALL 3: Mixing Diameter and Radius** The formulas are expressed in both forms: σ_h = pD/(2t) or σ_h = pr/t. Mixing them (e.g., using D in one term and r in another) doubles or halves the answer. Always convert to consistent units (all in mm or all in m) and double-check: D = 2r. **PITFALL 4: Applying Thin-Wall Formulas to Thick-Walled Vessels** If t/r > 1/10, you must use thick-wall (Lamé) equations, not these membrane formulas. Always check t/r ≤ 1/10 before and after design. A common exam trap is to give dimensions that appear reasonable but violate the thin-wall criterion; students who apply the formulas anyway get the problem wrong. **PITFALL 5: Forgetting Joint Efficiency** If the problem specifies a weld or joint efficiency η, you must include it: t = pD / (2·η·σ_allow) for cylinders, not t = pD / (2·σ_allow). Omitting η understates the required thickness and fails the design. **PITFALL 6: Using Sphere Formulas on a Cylinder or Vice Versa** A cylinder has TWO stresses (hoop and longitudinal); a sphere has ONE. Using σ = pD/(4t) for a cylinder hoop calculation gives a value that is half the correct hoop stress, leading to under-design. Always confirm the vessel geometry. **PITFALL 7: Not Verifying the Final Design** After calculating t, students often stop without checking: (a) Thin-wall criterion: Is t/r ≤ 1/10 still satisfied? (b) Stress limit: Does the calculated or actual stress stay below η·σ_allow? (c) Physical sense: Is t reasonable for the diameter and pressure? (A 10 m tank with t = 2 mm is suspicious.) Best practice: Always perform a final verification step. **PITFALL 8: Confusion Between Stress Calculations and Allowable Stress** Given material strength (e.g., σ_yield = 250 MPa), you typically divide by a factor of safety to find σ_allow. Then you compare the working stress (from pressure formulas) to σ_allow, not directly to yield. Confusing these steps leads to under- or over-design.
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6. Common Pitfalls and Exam Mistakes
Examples
A classic exam mistake: confusing hoop and longitudinal
Problem
Students are given: D = 500 mm, t = 6 mm, p = 2 MPa. They calculate σ_h = p·D / (4t) instead of (2t). What is the error?
Solution
Incorrect calculation: σ_h = (2 × 500) / (4 × 6) = 1000 / 24 = 41.67 MPa Correct calculation: σ_h = (2 × 500) / (2 × 6) = 1000 / 12 = 83.33 MPa Error: The student used the longitudinal formula (dividing by 4t) for hoop stress. The incorrect answer is exactly half the correct answer — a dead giveaway. Always verify: hoop should be 2× longitudinal.
Pitfall: Thick-wall vessel with thin-wall formula
Problem
A vessel has D = 400 mm, t = 80 mm. A student applies σ_h = pD/(2t) with p = 1 MPa. Is the answer valid?
Solution
Check thin-wall: r = 200 mm, t/r = 80/200 = 0.40 > 0.1 ✗ This is a THICK-WALLED vessel. The thin-wall formula is invalid. Calculation using thin-wall (wrong): σ_h = (1 × 400) / (2 × 80) = 400 / 160 = 2.5 MPa This answer is meaningless for a thick wall. The correct approach requires Lamé theory, which gives σ_h ≈ 1.8 MPa (lower than the thin-wall prediction because stress concentrates at the inner surface and the average is less). The thin-wall formula, when applied to thick walls, is unsafe and must be avoided.
Exam trap: Forgetting joint efficiency
Problem
A problem states: 'Design a cylindrical tank D = 1.5 m, p = 1.2 MPa, σ_allow = 80 MPa, η = 0.75 (riveted seam).' A student calculates t = pD / (2·σ_allow) = (1.2 × 1500) / (2 × 80) = 11.25 mm. Is this correct?
Solution
The student FORGOT the joint efficiency η. Incorrect: t = 11.25 mm (did not include η) Correct: t = pD / (2·η·σ_allow) = (1.2 × 1500) / (2 × 0.75 × 80) = 1800 / 120 = 15 mm The correct thickness is 33% larger (15 mm vs. 11.25 mm). The riveted seam (η = 0.75) is weaker and requires a thicker wall. An exam answer of 11.25 mm would be marked wrong because it ignores the seam weakness, leading to under-design.
Key Points
- Hoop stress > Longitudinal stress: σ_h = 2·σ_l (cylinders only)
- Use gauge pressure (not absolute) and inner diameter/radius in all formulas
- Always verify t/r ≤ 1/10 before and after design; if violated, use thick-wall theory
- Include joint efficiency η if specified; if not given, assume η = 1.0 (seamless)
- Cylinders have two stresses; spheres have one. Do not mix them up
- After calculating t, verify that the final stress stays within η·σ_allow
- Common exam trap: vessel dimensions that violate thin-wall but look 'normal' — always check the ratio
These questions represent the types and difficulty levels found on the PRC Civil Engineer Licensure Examination. Each is solved step-by-step following best practice.
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7. Practice Exam Questions and Solutions
Examples
EXAM QUESTION 1 — Design a cylindrical water main
Problem
A municipality plans to install a steel water main to supply 0.8 m³/s at a velocity of 1.5 m/s. The ground elevation change causes a gauge pressure of 1.6 MPa. The steel has σ_yield = 240 MPa, and a factor of safety of 2.0 is required. The seam is a butt weld with η = 0.85. (a) Find the inner diameter; (b) Calculate the required wall thickness; (c) Verify the thin-wall assumption; (d) Calculate the hoop and longitudinal stresses under design conditions.
Solution
SOLUTION: (a) Find inner diameter from flow rate. Q = (π·D²/4)·v 0.8 = (π·D²/4)·1.5 D² = (4 × 0.8) / (π × 1.5) = 3.2 / 4.712 = 0.6791 D = 0.824 m ≈ 825 mm Select standard size D = 800 mm (or 825 mm if available). Verify velocity: v = 4Q / (π·D²) = (4 × 0.8) / (π × 0.8²) = 1.59 m/s (acceptable) (b) Calculate allowable stress and wall thickness. σ_allow = σ_yield / FS = 240 / 2.0 = 120 MPa t = p·D / (2·η·σ_allow) = (1.6 × 800) / (2 × 0.85 × 120) t = 1280 / 204 = 6.27 mm Round up: t = 6.5 or 7.0 mm. Use t = 7.0 mm. (c) Verify thin-wall criterion. r = 400 mm, t/r = 7.0 / 400 = 0.0175 < 0.1 ✓ (thin-wall is valid) (d) Calculate stresses at design conditions. σ_h = p·D / (2t) = (1.6 × 800) / (2 × 7.0) = 91.43 MPa σ_l = p·D / (4t) = (1.6 × 800) / (4 × 7.0) = 45.71 MPa Verify ratio: σ_h / σ_l = 91.43 / 45.71 = 2.0 ✓ Stress at seam: η·σ_h = 0.85 × 91.43 = 77.71 MPa < 120 MPa ✓ ANSWER: (a) Inner diameter = 800 mm (or 825 mm depending on available sizes) (b) Wall thickness = 7.0 mm (c) Thin-wall criterion satisfied: t/r = 0.0175 < 0.1 (d) Hoop stress = 91.43 MPa, Longitudinal stress = 45.71 MPa. All stresses are within the allowable limit of 120 MPa.
EXAM QUESTION 2 — Compare cylinder and sphere for gas storage
Problem
A company must store compressed nitrogen at p = 2.0 MPa. Two designs are proposed: (A) a cylindrical tank (D = 2.0 m) and (B) a spherical tank (D = 2.0 m). For both, the allowable stress is σ_allow = 100 MPa (seamless construction, η = 1.0). (a) Calculate the wall thickness for each design. (b) Compare the material volumes (mass) needed for each design. (c) Which is more economical, and why?
Solution
SOLUTION: (a) Wall thicknesses. Cylinder: t_cyl = p·D / (2·σ_allow) = (2.0 × 2000) / (2 × 100) = 4000 / 200 = 20 mm Verify thin-wall: r = 1000 mm, t/r = 20/1000 = 0.02 < 0.1 ✓ Sphere: t_sph = p·D / (4·σ_allow) = (2.0 × 2000) / (4 × 100) = 4000 / 400 = 10 mm Verify thin-wall: r = 1000 mm, t/r = 10/1000 = 0.01 < 0.1 ✓ (b) Compare material volumes (surface area × thickness). Cylinder (assume length L = D = 2.0 m for a compact design): Surface area = π·D·L + 2·(π·D²/4) = π × 2 × 2 + π × 2² / 2 = 4π + 2π = 6π m² Volume = 6π × 0.02 = 0.377 m³ Sphere: Surface area = 4π·r² = 4π × 1² = 4π m² Volume = 4π × 0.01 = 0.126 m³ Ratio: V_cylinder / V_sphere = 0.377 / 0.126 = 2.99 ≈ 3.0 (c) Economic comparison. The sphere requires approximately one-third the volume of material (half the thickness, and the spherical surface area is less than a cylindrical shell with end caps). For steel at ~7800 kg/m³: Mass_cyl ≈ 0.377 × 7800 = 2940 kg Mass_sph ≈ 0.126 × 7800 = 980 kg The sphere uses about one-third the mass and thus one-third the cost of material. Fabrication of a sphere is slightly more complex, but the material savings typically outweigh the extra labor for high-pressure storage (p ≥ 1.5 MPa). ANSWER: (a) Cylinder: t = 20 mm; Sphere: t = 10 mm (sphere requires half the thickness) (b) Material volume for sphere ≈ 1/3 of cylinder; material mass for sphere ≈ 1/3 of cylinder (c) The sphere is more economical for high-pressure storage due to 2/3 savings in material cost. This is why LPG and compressed gas storage tanks are spherical.
EXAM QUESTION 3 — Verification of an existing penstock
Problem
An existing hydroelectric penstock (D = 1.2 m, t = 14 mm, steel with σ_yield = 250 MPa) has been in service for 20 years. A new hydro facility upstream increases the flow and water head. The new operating pressure will be p = 2.2 MPa (up from the original 1.8 MPa). A factor of safety of 2.5 is required against yield. Is the existing penstock adequate for the new pressure? If not, what options are available?
Solution
SOLUTION: Step 1: Calculate allowable stress. σ_allow = σ_yield / FS = 250 / 2.5 = 100 MPa Step 2: Verify thin-wall criterion (existing penstock). r = 600 mm, t/r = 14 / 600 = 0.0233 < 0.1 ✓ (thin-wall is valid) Step 3: Calculate current (actual) stress at new pressure p = 2.2 MPa. σ_h = p·D / (2t) = (2.2 × 1200) / (2 × 14) = 2640 / 28 = 94.29 MPa Step 4: Compare with allowable. Actual stress (94.29 MPa) < Allowable stress (100 MPa) ✓ Margin: 100 − 94.29 = 5.71 MPa (very tight — only 6% margin) Conclusion: The penstock is MARGINALLY ADEQUATE but with very little safety margin. Small uncertainties in material strength, weld quality, or corrosion could render it unsafe. Options: 1. Operate at the new pressure (94.29 MPa) with the understanding that there is minimal safety margin; recommend increased inspection. 2. Install a pressure-reducing valve to limit operating pressure to p_safe ≤ 2.0 MPa: p_safe = (σ_allow × 2t) / D = (100 × 28) / 1200 = 2.33 MPa Actually, at 2.0 MPa: σ = (2.0 × 1200) / 28 = 85.71 MPa (comfortable margin). 3. Replace the penstock section or wrap it with composite reinforcement. 4. Lower the design pressure and accept reduced flow. RECOMMENDATION: Option 1 (operate at 2.2 MPa) is technically acceptable but risky. Option 2 (operate at 2.0 MPa) is safer. For a critical infrastructure project, consult with the owner and consider the consequences of failure.
EXAM QUESTION 4 — Design with incomplete information
Problem
A cylindrical boiler is to be designed for p = 2.5 MPa. The allowable stress (for the parent plate) is σ_allow = 95 MPa. The problem does NOT specify the joint efficiency. (a) Assuming the boiler is seamless, find the wall thickness for a diameter of 1.5 m. (b) Assuming a butt weld with visual inspection (η = 0.85), find the new wall thickness. (c) Comment on the difference.
Solution
SOLUTION: (a) Seamless boiler (η = 1.0). t = p·D / (2·η·σ_allow) = (2.5 × 1500) / (2 × 1.0 × 95) t = 3750 / 190 = 19.74 mm Round up: t = 20 mm. Verify thin-wall: r = 750 mm, t/r = 20/750 = 0.0267 < 0.1 ✓ (b) Butt weld with visual inspection (η = 0.85). t = p·D / (2·η·σ_allow) = (2.5 × 1500) / (2 × 0.85 × 95) t = 3750 / 161.5 = 23.22 mm Round up: t = 23.5 or 24 mm. Use t = 24 mm. Verify thin-wall: r = 750 mm, t/r = 24/750 = 0.032 < 0.1 ✓ (c) Difference and interpretation. Increase in thickness: (24 − 20) / 20 = 20% (from 20 mm to 24 mm) The reduction in joint efficiency (from 1.0 to 0.85, a 15% reduction) translates to approximately a 20% increase in required thickness. This is because thickness appears linearly in the formula: when η decreases, t must increase proportionally to maintain the same stress at the seam. The thicker wall in case (b) compensates for the weaker weld. If a seamless or fully inspected weld is available, the cost savings (16% less material for case (a)) may justify the extra fabrication expense. ANSWER: (a) Seamless: t = 20 mm (b) Visual inspection weld: t = 24 mm (c) The 15% reduction in joint efficiency requires a 20% increase in thickness. This demonstrates the economic importance of weld quality and inspection in pressure vessel design.
Key Points
- Exam questions often combine multiple concepts: thin-wall verification, design, and comparison
- Read carefully: distinguish between "design" (find t) and "verify" (check if existing t is adequate)
- Show all steps and verify thin-wall criterion; partial credit is awarded for methodology
- Use consistent units (SI: MPa, mm, or Pa, m) throughout the solution
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