CELE Strength of Materials — Thin-Walled Pressure VesselsExam Answer Templates

Marks

1

Topic

Thin-Wall Criterion

Difficulty

easy

Template Id

T1

Examiner Tip

One clean sentence with the correct inequality and defined variables earns the full mark. Do not waste time explaining Lamé theory here.

Model Answer

A pressure vessel is classified as thin-walled when the ratio of wall thickness t to inner radius r satisfies t/r ≤ 1/10 (equivalently, r/t ≥ 10). Under this condition, the stress is assumed uniform across the wall thickness.

Question Type

very_short_answer

Answer Structure

• Line 1: State the inequality t/r ≤ 1/10 (or r/t ≥ 10) with correct variable definitions [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing t/r ≤ 1/20 or any incorrect fraction — no mark.
• Defining r as outer radius instead of inner radius.
• Omitting variable definitions when the ratio alone is written.

Key Phrases To Include

• t/r ≤ 1/10
• inner radius
• uniform stress
• wall thickness

Marks

2

Topic

Cylindrical Pressure Vessels — Principal Stresses

Difficulty

easy

Template Id

T2

Examiner Tip

The ratio σ_h = 2σ_l is a key relationship examiners specifically look for. Always state it explicitly even in short answers.

Model Answer

A thin-walled cylinder under internal pressure p develops two principal membrane stresses: 1. Hoop (circumferential) stress: σ_h = pD/2t — acts tangentially around the circumference. 2. Longitudinal (axial) stress: σ_l = pD/4t — acts along the axis of the cylinder. The hoop stress is larger: σ_h = 2σ_l. Therefore, hoop stress governs the design of a cylindrical vessel.

Question Type

very_short_answer

Answer Structure

• Line 1: Name hoop stress with its formula σ_h = pD/2t [1 mark]
• Line 2: Name longitudinal stress with its formula σ_l = pD/4t AND state σ_h = 2σ_l relationship [1 mark]

Scoring Breakdown

Common Mark Deductions

• Giving only one stress formula — half mark only.
• Reversing the formulas (writing pD/4t for hoop) — full deduction.
• Stating longitudinal stress is larger — factual error, full deduction.

Key Phrases To Include

• hoop stress
• longitudinal stress
• σ_h = pD/2t
• σ_l = pD/4t
• σ_h = 2σ_l
• governs design

Marks

2

Topic

Cylindrical Pressure Vessels — Hoop Stress

Difficulty

easy

Template Id

T3

Examiner Tip

For 2-mark numerical problems, one mark is almost always reserved for the correct formula and one for the correct answer with unit. Never skip the formula line.

Model Answer

Given: D = 400 mm, t = 6 mm, p = 1.2 MPa Thin-wall check: t/r = 6/200 = 0.030 ≤ 0.10 ✓ (thin-wall formula applies) Hoop stress: σ_h = pD / (2t) σ_h = (1.2 × 400) / (2 × 6) σ_h = 480 / 12 σ_h = 40.0 MPa

Question Type

numerical

Answer Structure

• Line 1: Write all given data [0.5 mark]
• Line 2: Perform thin-wall check explicitly [0.5 mark]
• Line 3: Write the formula σ_h = pD/2t [0.5 mark]
• Line 4: Substitute and compute final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using σ_h = pD/4t (longitudinal formula for hoop) — incorrect formula, loses formula mark.
• Not converting or mixing up diameter and radius in the formula.
• Omitting the unit MPa in the final answer.
• Skipping the thin-wall check — minor deduction in some marking schemes.

Key Phrases To Include

• σ_h = pD/2t
• thin-wall check
• t/r ≤ 0.10
• MPa

Marks

2

Topic

Cylindrical Pressure Vessels — Failure Mode

Difficulty

medium

Template Id

T4

Examiner Tip

This is a reasoning question. Examiners reward students who connect the stress direction to the seam orientation. Use the words 'acts perpendicular to the longitudinal seam' for full credit.

Model Answer

A cylindrical pressure vessel under internal pressure develops two principal stresses: hoop (circumferential) stress σ_h = pD/2t acting perpendicular to the longitudinal axis, and longitudinal stress σ_l = pD/4t acting along the axis. Since σ_h = 2σ_l, the hoop stress is twice the longitudinal stress. A longitudinal seam is oriented parallel to the cylinder's axis, and the stress that acts to tear it apart is the hoop stress — the larger of the two. Therefore, the longitudinal seam is subjected to the greater stress and is more likely to fail first.

Question Type

short_answer

Answer Structure

• Line 1: State both stress values and the relationship σ_h = 2σ_l [1 mark]
• Line 2: Explain that the hoop stress acts transverse to (and therefore tears) the longitudinal seam — the larger stress governs failure [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the vessel fails along a circumferential seam — factual error, full deduction.
• Naming hoop stress without explaining the orientation of the seam relative to the stress direction.
• Providing only formulas without any conceptual explanation.

Key Phrases To Include

• hoop stress
• σ_h = 2σ_l
• longitudinal seam
• acts perpendicular
• larger stress governs

Marks

3

Topic

Cylindrical Pressure Vessels — Both Stresses

Difficulty

easy

Template Id

T5

Examiner Tip

Always end with the ratio check σ_h / σ_l = 2 as a self-verification step. This shows examiners you understand the fundamental relationship and adds confidence to your answer.

Model Answer

Given: D = 900 mm, r = 450 mm, t = 10 mm, p = 1.8 MPa Thin-wall check: t/r = 10/450 = 0.022 ≤ 0.10 ✓ Hoop (circumferential) stress: σ_h = pD / (2t) = (1.8 × 900) / (2 × 10) = 1620 / 20 = 81.0 MPa Longitudinal (axial) stress: σ_l = pD / (4t) = (1.8 × 900) / (4 × 10) = 1620 / 40 = 40.5 MPa Check: σ_h / σ_l = 81.0 / 40.5 = 2.0 ✓ (hoop is twice longitudinal, as expected)

Question Type

numerical

Answer Structure

• Line 1: List all given data with units [0.5 mark]
• Line 2: Thin-wall check with result [0.5 mark]
• Line 3: Formula and computation for hoop stress with unit [1 mark]
• Line 4: Formula and computation for longitudinal stress with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Calculating only one stress — loses 1 mark.
• Omitting thin-wall check — 0.5 mark deduction.
• Arithmetic error in substitution — loses answer mark but retains formula mark.
• Swapping the denominator (using 4t for hoop and 2t for longitudinal) — loses both answer marks.

Key Phrases To Include

• σ_h = pD/2t
• σ_l = pD/4t
• thin-wall check
• 81.0 MPa
• 40.5 MPa
• σ_h = 2σ_l

Marks

3

Topic

Cylindrical Pressure Vessels — Derivation

Difficulty

medium

Template Id

T6

Examiner Tip

The most common error in derivations is using curved area instead of projected area for pressure. Examiners specifically check for the projected area = 2rL step.

Model Answer

Consider a thin-walled cylinder of inner radius r, wall thickness t, and length L under internal gauge pressure p. Step 1 — Cut plane: Make a longitudinal cut through a diameter, isolating a half-cylinder. This exposes the hoop stress σ_h acting on two rectangular cross-sectional areas (each t × L). Step 2 — Pressure resultant: The net upward force of internal pressure on the half-cylinder is the pressure times the projected area: F_pressure = p × (2r × L) = 2prL (acting upward on the projected rectangular area) Step 3 — Resisting force: The hoop stress in the two cut walls resists this force: F_resist = 2 × (σ_h × t × L) = 2σ_h tL Step 4 — Equilibrium (ΣFy = 0): 2σ_h tL = 2prL σ_h = pr/t = pD/(2t) ◀ Since t/r ≤ 1/10, the stress is essentially uniform across the thin wall, confirming the assumption.

Question Type

short_answer

Answer Structure

• Step 1: Describe the longitudinal cut plane and identify the surfaces on which σ_h acts [0.5 mark]
• Step 2: Correctly compute the pressure resultant force on the projected area 2rL [1 mark]
• Step 3: Write the resisting force expression involving σ_h, t, L [0.5 mark]
• Step 4: Apply equilibrium and simplify to σ_h = pr/t = pD/2t [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the curved surface area (πrL) instead of the projected area (2rL) for the pressure force.
• Forgetting the factor of 2 in the resisting force (only one wall instead of two).
• Not citing the thin-wall assumption to justify uniform stress.

Key Phrases To Include

• longitudinal cut
• projected area = 2rL
• net pressure force = 2prL
• ΣFy = 0
• σ_h = pr/t = pD/2t

Marks

3

Topic

Spherical Pressure Vessels

Difficulty

easy

Template Id

T7

Examiner Tip

The most frequent error on sphere problems is applying the cylinder hoop formula. The denominator for a sphere is 4t, not 2t. State 'equal in all directions' to earn the concept mark.

Model Answer

Given: D = 2400 mm, r = 1200 mm, t = 12 mm, p = 2.0 MPa Thin-wall check: t/r = 12/1200 = 0.010 ≤ 0.10 ✓ For a spherical vessel, the membrane stress is equal in all directions: σ = pD / (4t) σ = (2.0 × 2400) / (4 × 12) σ = 4800 / 48 σ = 100.0 MPa The sphere develops a single uniform membrane stress of 100.0 MPa in all tangential directions.

Question Type

numerical

Answer Structure

• Line 1: State all given data, including unit conversions [0.5 mark]
• Line 2: Thin-wall check [0.5 mark]
• Line 3: Write the spherical vessel formula σ = pD/4t [1 mark]
• Line 4: Substitute and compute correct answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the cylinder hoop formula (pD/2t) for a sphere — incorrect formula, loses formula and answer marks.
• Not noting that the sphere has equal stress in all directions (omission of key concept).
• Using outer diameter instead of inner diameter.

Key Phrases To Include

• σ = pD/4t
• equal in all directions
• membrane stress
• thin-wall check
• 100.0 MPa

Marks

3

Topic

Cylindrical vs Spherical Vessels — Comparison

Difficulty

medium

Template Id

T8

Examiner Tip

Comparison problems require both a numerical answer AND a qualitative conclusion. Write 'The sphere requires half the wall thickness…' explicitly — examiners mark this sentence separately.

Model Answer

Given: D = 1000 mm, p = 3.0 MPa, σ_allow = 150 MPa, η = 1.0 Minimum wall thickness — Cylinder (hoop governs): t_cyl = pD / (2η·σ_allow) = (3.0 × 1000) / (2 × 1.0 × 150) = 3000 / 300 = 10.0 mm Minimum wall thickness — Sphere: t_sph = pD / (4η·σ_allow) = (3.0 × 1000) / (4 × 1.0 × 150) = 3000 / 600 = 5.0 mm Ratio: t_cyl / t_sph = 10.0 / 5.0 = 2.0 Conclusion: The sphere requires only HALF the wall thickness of the cylinder for the same pressure and allowable stress. The spherical vessel is twice as material-efficient, making it the preferred shape for high-pressure containment (e.g., LPG storage tanks).

Question Type

numerical

Answer Structure

• Line 1: State all given data [0.5 mark]
• Lines 2–3: Compute t_cyl with formula and result [1 mark]
• Lines 4–5: Compute t_sph with formula and result [1 mark]
• Line 6: State the ratio and conclusion on efficiency [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using the same formula for both shapes (applying cylinder formula to sphere).
• Not providing a comparison conclusion — loses the concept mark.
• Forgetting to round up to the next whole millimeter when specifically asked for design thickness.

Key Phrases To Include

• t_cyl = pD/2σ_allow
• t_sph = pD/4σ_allow
• ratio = 2
• sphere is more efficient
• half the wall thickness

Marks

5

Topic

Design for Allowable Stress with Joint Efficiency — Cylinder

Difficulty

medium

Template Id

T9

Examiner Tip

5-mark problems are solved step-by-step. Each step corresponds to roughly 1 mark. The thin-wall verification and the final stress check are separate steps that many students skip — these are easy marks. Never skip them.

Model Answer

Given: D = 1500 mm, p = 2.5 MPa, σ_allow = 120 MPa, η = 0.80 Step 1 — Identify the governing stress: For a cylinder, hoop stress (σ_h = pD/2t) is twice the longitudinal stress. The longitudinal seam is subjected to hoop stress. Therefore, hoop stress with longitudinal joint efficiency governs: Step 2 — Apply design formula: t = pD / (2 η σ_allow) t = (2.5 × 1500) / (2 × 0.80 × 120) t = 3750 / 192 t = 19.53 mm Step 3 — Round up to nearest whole mm: Use t = 20 mm (always round up for safety) Step 4 — Verify thin-wall criterion: r = D/2 = 750 mm t/r = 20/750 = 0.0267 ≤ 0.10 ✓ The thin-wall assumption is valid. Step 5 — Check actual stresses with t = 20 mm: σ_h (actual) = pD/(2t) = (2.5 × 1500)/(2 × 20) = 3750/40 = 93.75 MPa Allowable hoop stress at seam = η × σ_allow = 0.80 × 120 = 96.0 MPa 93.75 MPa < 96.0 MPa ✓ (design is safe) Answer: Minimum wall thickness = 20 mm.

Question Type

numerical

Answer Structure

• Step 1: Identify that hoop stress governs (longitudinal joint efficiency applies to hoop/longitudinal seam) [1 mark]
• Step 2: Write design formula t = pD/(2ησ_allow) and substitute values [1 mark]
• Step 3: Compute t = 19.53 mm, round up to 20 mm with justification [1 mark]
• Step 4: Verify thin-wall: t/r = 20/750 = 0.027 ≤ 0.10 ✓ [1 mark]
• Step 5: Optional check — compute actual stress vs allowable at seam to confirm safety [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting joint efficiency η from the formula — loses formula and answer marks.
• Rounding down (using 19 mm) instead of rounding up — unsafe and loses rounding mark.
• Performing thin-wall check with the unrounded 19.53 mm instead of the chosen 20 mm.
• Not performing the final stress verification step.
• Using p as absolute pressure instead of gauge pressure if context specifies gauge.

Key Phrases To Include

• t = pD/(2ησ_allow)
• η = 0.80
• 19.53 mm
• round up to 20 mm
• thin-wall check t/r ≤ 0.10
• hoop governs
• 93.75 MPa < 96.0 MPa

Marks

5

Topic

Cylindrical Pressure Vessels — Penstock / Hydraulics Integration

Difficulty

hard

Template Id

T10

Examiner Tip

Penstock problems require a hydraulics step (p = ρgh) before any mechanics. This is a common board-exam integration between Hydraulics and Strength of Materials. Always convert head to pressure first, and always make a clear safety statement ('safe' or 'NOT safe').

Model Answer

Given: h = 250 m (water head), D = 800 mm, r = 400 mm, t = 9 mm, σ_allow = 100 MPa Step 1 — Convert head to gauge pressure: p = ρgh = (1000 kg/m³)(9.81 m/s²)(250 m) p = 2,452,500 Pa = 2.453 MPa Step 2 — Thin-wall check: t/r = 9/400 = 0.0225 ≤ 0.10 ✓ (thin-wall formulas apply) Step 3 — Hoop stress (governs in a cylinder): σ_h = pD / (2t) = (2.453 × 800) / (2 × 9) σ_h = 1962.0 / 18 σ_h = 109.0 MPa Step 4 — Safety check: σ_h = 109.0 MPa > σ_allow = 100 MPa ✗ Conclusion: The pipe is NOT safe. The hoop stress (109.0 MPa) exceeds the allowable stress (100 MPa) by 9%. The wall thickness must be increased. Minimum safe thickness: t_min = pD / (2 σ_allow) = (2.453 × 800) / (2 × 100) = 1962 / 200 = 9.81 mm → use t = 10 mm

Question Type

numerical

Answer Structure

• Step 1: Convert water head to pressure p = ρgh correctly in MPa [1 mark]
• Step 2: Perform thin-wall check with t/r and verify ≤ 0.10 [1 mark]
• Step 3: Apply σ_h = pD/2t and compute 109.0 MPa [1 mark]
• Step 4: Compare 109.0 MPa vs 100 MPa and state NOT safe [1 mark]
• Step 5: Compute minimum required thickness 9.81 mm → 10 mm [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert head to pressure (using h = 250 m directly in the formula).
• Using g = 9.8 m/s² — acceptable, gives p = 2.45 MPa; minor rounding difference.
• Concluding 'safe' instead of 'not safe' — critical error, loses judgment mark.
• Not recommending a revised wall thickness.

Key Phrases To Include

• p = ρgh
• 2.453 MPa
• σ_h = pD/2t
• 109.0 MPa
• NOT safe
• t_min = 10 mm

Marks

1

Topic

Pressure Concepts — Gauge Pressure

Difficulty

easy

Template Id

T11

Examiner Tip

For a 1-mark definition, one tight sentence with the formula and one sentence explaining the engineering reason earns full marks.

Model Answer

Gauge pressure is the pressure measured above atmospheric pressure (p_gauge = p_absolute − p_atm). In pressure vessel analysis, gauge pressure is used because the external atmosphere acts on the outside of the vessel wall. The net pressure loading that stresses the wall is therefore the internal pressure minus atmospheric, which is the gauge pressure.

Question Type

very_short_answer

Answer Structure

• Line 1: Define gauge pressure as excess above atmospheric [0.5 mark]
• Line 2: Explain that net wall loading = internal minus external (atmospheric), justifying use of gauge pressure [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Defining gauge pressure correctly but not explaining why it is used (missing second part).
• Confusing gauge pressure with vacuum pressure.

Key Phrases To Include

• gauge pressure
• above atmospheric
• net pressure
• p_absolute − p_atm

Marks

5

Topic

Shear Stresses in Cylindrical Vessels — Mohr's Circle

Difficulty

hard

Template Id

T12

Examiner Tip

The distinction between in-plane and absolute maximum shear is a high-level exam concept. The key insight is that σ_r = 0 at the outer wall, making it a third principal stress. τ_abs = σ_h/2 is always larger than the in-plane value for a cylinder. State this clearly.

Model Answer

Given: D = 600 mm, r = 300 mm, t = 8 mm, p = 1.6 MPa, τ_allow = 30 MPa Step 1 — Thin-wall check: t/r = 8/300 = 0.0267 ≤ 0.10 ✓ Step 2 — Principal stresses: σ_h = pD/(2t) = (1.6 × 600)/(2 × 8) = 960/16 = 60.0 MPa σ_l = pD/(4t) = (1.6 × 600)/(4 × 8) = 960/32 = 30.0 MPa Radial stress: σ_r ≈ 0 (thin-wall assumption — outer surface exposed to atmosphere) Step 3 — Maximum in-plane shear stress (from Mohr's circle of σ_h and σ_l): τ_in-plane = (σ_h − σ_l)/2 = (60.0 − 30.0)/2 = 15.0 MPa Step 4 — Absolute maximum shear stress (considering all three principal stresses: 60, 30, 0 MPa): τ_abs = (σ_max − σ_min)/2 = (60.0 − 0)/2 = 30.0 MPa Step 5 — Safety check: τ_abs = 30.0 MPa = τ_allow = 30.0 MPa ✓ (exactly at limit — acceptable but no margin) Note: In-plane shear (15.0 MPa) < 30 MPa ✓; absolute maximum shear (30.0 MPa) = 30 MPa ✓.

Question Type

numerical

Answer Structure

• Step 1: Thin-wall check [0.5 mark]
• Step 2: Compute σ_h = 60 MPa and σ_l = 30 MPa; state σ_r = 0 [1 mark]
• Step 3: Compute τ_in-plane = (σ_h − σ_l)/2 = 15.0 MPa [1 mark]
• Step 4: Compute τ_abs = (σ_h − 0)/2 = σ_h/2 = 30.0 MPa [1.5 marks]
• Step 5: Compare both shear stresses with τ_allow and state safety [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing only in-plane shear and ignoring absolute maximum shear — loses 1–1.5 marks.
• Not stating σ_r = 0 and not using it as the third principal stress for absolute max shear.
• Setting τ_abs = τ_in-plane — confusing the two definitions.
• Not performing the safety comparison.

Key Phrases To Include

• τ_in-plane = (σ_h − σ_l)/2
• τ_abs = σ_h/2
• σ_r = 0
• 15.0 MPa
• 30.0 MPa
• Mohr's circle

Marks

2

Topic

Spherical Vessels — Design and Efficiency

Difficulty

easy

Template Id

T13

Examiner Tip

Always write the formula with η in it for design-type questions, even if η = 1 in the specific problem — it shows complete knowledge of the design procedure.

Model Answer

Minimum wall thickness for a spherical vessel with joint efficiency η: t_sph = pD / (4η·σ_allow) A sphere requires thinner walls because by symmetry it develops only a single membrane stress σ = pD/4t in all directions, which is equal to the cylinder's longitudinal stress — and only HALF the cylinder's hoop stress (pD/2t). Since the cylinder's hoop stress governs its design (it is the larger stress), and the sphere's maximum stress is half that value, the sphere can withstand the same pressure with half the wall thickness for the same material and allowable stress.

Question Type

short_answer

Answer Structure

• Line 1: Correctly write t_sph = pD/(4ησ_allow) [1 mark]
• Line 2: Explain sphere's single stress = pD/4t = half of cylinder hoop stress pD/2t → thinner walls needed [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing t_sph = pD/(2ησ_allow) — copying the cylinder formula.
• Explaining correctly but omitting the formula.
• Vague explanation without referencing the specific stress values.

Key Phrases To Include

• t_sph = pD/4ησ_allow
• single membrane stress
• equal in all directions
• half of cylinder hoop stress
• more material-efficient

Marks

5

Topic

Design with Factor of Safety and Joint Efficiency — Cylinder

Difficulty

hard

Template Id

T14

Examiner Tip

When a problem gives ultimate stress and factor of safety separately, always compute σ_allow = σ_ult/FS as the very first step and box it. This prevents the common error of using σ_ult directly in the formula.

Model Answer

Given: D = 3000 mm, p = 0.45 MPa, η = 0.90, FS = 2.0, σ_ult = 400 MPa Step 1 — Determine allowable stress: σ_allow = σ_ult / FS = 400 / 2.0 = 200 MPa Step 2 — Identify governing stress: For a cylinder, hoop stress governs (σ_h = 2σ_l). The longitudinal joint (parallel to axis) carries hoop stress. Use hoop design equation with η: Step 3 — Apply design formula: t = pD / (2 η σ_allow) t = (0.45 × 3000) / (2 × 0.90 × 200) t = 1350 / 360 t = 3.75 mm Step 4 — Round up: Use t = 4 mm (round up to next whole mm for safety) Step 5 — Verify thin-wall assumption: r = D/2 = 1500 mm t/r = 4/1500 = 0.00267 ≤ 0.10 ✓ (well within thin-wall range) Step 6 — Verify design with chosen thickness: σ_h (actual) = pD/(2t) = (0.45 × 3000)/(2 × 4) = 1350/8 = 168.75 MPa Allowable at seam = η × σ_allow = 0.90 × 200 = 180 MPa 168.75 MPa < 180 MPa ✓ Design is adequate. Answer: Minimum wall thickness = 4 mm.

Question Type

numerical

Answer Structure

• Step 1: Compute σ_allow = σ_ult/FS = 400/2 = 200 MPa [1 mark]
• Step 2: State hoop governs; write t = pD/(2ησ_allow) [1 mark]
• Step 3: Substitute and compute t = 3.75 mm [1 mark]
• Step 4: Round up to t = 4 mm with justification [0.5 mark]
• Step 5: Thin-wall check t/r = 0.00267 ≤ 0.10 ✓ [0.5 mark]
• Step 6: Verify actual stress < allowable at seam [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to apply the factor of safety to find σ_allow before substituting.
• Using σ_allow = 400 MPa directly without dividing by FS.
• Omitting η from the design formula.
• Rounding down to 3 mm instead of up to 4 mm.

Key Phrases To Include

• σ_allow = σ_ult/FS
• t = pD/2ησ_allow
• 3.75 mm
• round up to 4 mm
• thin-wall ✓
• 168.75 MPa < 180 MPa

Marks

2

Topic

Maximum Allowable Pressure — Cylinder

Difficulty

easy

Template Id

T15

Examiner Tip

Maximum pressure questions are the reverse of thickness questions. Always use the hoop formula (the larger stress) for the most conservative (safe) answer. Using the longitudinal formula gives a higher — but unconservative — pressure limit.

Model Answer

Given: D = 500 mm, t = 8 mm, σ_allow = 80 MPa Thin-wall check: t/r = 8/250 = 0.032 ≤ 0.10 ✓ Hoop stress governs (larger stress in cylinder). Set σ_h = σ_allow: p_max = 2t·σ_allow / D p_max = (2 × 8 × 80) / 500 p_max = 1280 / 500 p_max = 2.56 MPa

Question Type

numerical

Answer Structure

• Line 1: State governing condition σ_h = σ_allow and rearrange: p = 2t·σ_allow/D [1 mark]
• Line 2: Substitute and compute p_max = 2.56 MPa with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using longitudinal formula p = 4t·σ_allow/D — this gives a non-conservative (higher) pressure, conceptual error.
• Correctly using hoop formula but computing p = 2tσ/r (using r instead of D without adjusting the factor).
• Omitting unit MPa on the answer.

Key Phrases To Include

• p = 2tσ_allow/D
• hoop governs
• 2.56 MPa