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5.5 Pressure Vessels, Welds, Fatigue, GD&T, and Design Safety

Key Takeaways

  • Thin-wall pressure-vessel cues: t < r/10; hoop stress σh = pr/t is twice the longitudinal stress σl = pr/2t.

  • Fatigue design uses the endurance limit Se and mean-stress lines: Goodman uses Su, Soderberg uses Sy, Gerber uses a parabola.

  • Stress concentration (Kt or Kf), surface finish, size, and loading modify the endurance limit before a fatigue check.

  • GD&T controls form, orientation, location, profile, and runout to limit manufacturing variation. Form controls do not require datums; use datum references where the selected control requires them.

  • Factor of safety n = strength/applied-stress reflects load uncertainty, consequence of failure, and inspection access.

Last updated: September 2026

Pressure vessels: identify the wall model

A pressure-vessel question begins by deciding whether the thin-wall approximation applies. The common FE cue is that wall thickness is small relative to radius — typically t < r/10. For a thin-walled cylinder of internal pressure p, mean radius r, and thickness t:

  • Hoop (circumferential) stress: σh = pr/t
  • Longitudinal (axial) stress: σl = pr/(2t)

Hoop stress is twice the longitudinal stress, which is why pressurized pipes split along their length, not around their circumference. For a thin sphere, both directions equal pr/(2t), making spheres the most material-efficient pressure container. When t is not small relative to r, the thin-wall formulas break down and thick-wall (Lamé) equations apply, giving a radial stress gradient through the wall.

Worked example. A thin cylinder has p = 2 MPa, inner radius r = 0.5 m, wall t = 10 mm. Hoop stress σh = pr/t = (2×10⁶)(0.5)/(0.010) = 100 MPa; longitudinal σl = 50 MPa. Compare σh to the allowable (Sy/n) to size the wall.

Fatigue: alternating and mean stress

Parts under cyclic loading can fail far below the static yield strength. Fatigue questions resolve the cyclic load into a mean stress σm = (σmax + σmin)/2 and an alternating stress σa = (σmax − σmin)/2. For steels, the endurance limit Se is the stress below which infinite life is expected; a rough starting estimate is Se ≈ 0.5·Su, then reduced by Marin factors for surface finish, size, loading type, temperature, and reliability, plus the fatigue stress-concentration factor Kf.

Mean stress shifts the safe alternating stress. The three FE mean-stress lines connect σa and σm:

CriterionEquation (failure when = 1)Reference strength
Soderberg (most conservative)σa/Se + σm/Sy = 1Yield Sy
Modified Goodman (common)σa/Se + σm/Su = 1Ultimate Su
Gerber (parabolic, least conservative)σa/Se + (σm/Su)² = 1Ultimate Su

Worked example. With Se = 200 MPa, Su = 500 MPa, σa = 100 MPa, σm = 150 MPa, the Goodman left side = 100/200 + 150/500 = 0.5 + 0.3 = 0.8 < 1, so the part survives, and the factor of safety is 1/0.8 = 1.25. Soderberg, using Sy < Su, would give a larger number and predict a smaller margin — that is why it is the conservative choice.

GD&T and the factor of safety

Geometric dimensioning and tolerancing (GD&T) is standardized by ASME Y14.5. It controls form, orientation, location, profile, and runout. Form controls such as flatness and circularity do not require a datum; orientation, location, and runout controls use the applicable datum references. A feature control frame gives the geometric symbol, tolerance and applicable modifiers, followed by datum references where required. Maximum Material Condition (MMC) allows a tolerance to grow (a bonus tolerance) as the feature departs from its maximum-material size — useful for clearance fits. FE questions usually test what a symbol controls and what a datum is, not a full tolerance stack-up.

The factor of safety is the unifying design-safety idea:

n = (material strength) / (applied stress)

Use Sy for a ductile-yielding check. Fatigue requires the applicable endurance or finite-life strength and any mean-stress correction; Su alone is not a fatigue limit. The number chosen reflects load uncertainty, material variability, consequence of failure (life-safety vs. nuisance), inspection access, and applicable codes/standards (e.g., ASME BPVC for pressure vessels, AWS for welds). A higher factor is justified when loads are poorly known or failure is catastrophic. Many FE problems stop at asking you to identify the governing failure mode (yield, fatigue, buckling, fracture) rather than complete a long calculation — recognize the mode, then apply the right strength and factor.

Welds, buckling, and combined-load checks

Welded joints deserve a closer look because the FE tests them often. A fillet weld carries load across its throat, the narrowest section of the triangular bead. For an equal-leg 45° fillet with leg size h, the throat is h/√2 ≈ 0.707h. For effective length L, A = 0.707hL, and direct shear is τ = F/A. When a load is applied eccentrically to a weld group, the weld sees a primary (direct) shear F/A plus a secondary shear from the moment, τ″ = Mc/J with J based on the effective throat area; the two are combined as vectors and the critical location depends on the vector sum. Complete-joint-penetration groove welds, by contrast, can develop load through the full effective joint thickness and are checked in tension or compression like the base metal.

Two more design-safety checks round out the topic. For an ideal slender elastic column, Euler buckling gives Pcr = π²EI/(KL)², using the least relevant flexural stiffness EI. Ideal effective-length factors are K = 1 for pinned-pinned, 0.5 for fixed-fixed, approximately 0.7 for fixed-pinned, and 2.0 for fixed-free. The end restraints and Euler assumptions must match the problem. A column can buckle far below its yield stress, so slenderness — not material strength — governs. Finally, combined loading (bending plus axial, or bending plus torsion on a shaft) is resolved with principal stresses and a yield theory; the distortion-energy (von Mises) stress σ′ = √(σ² + 3τ²) for the common bending-plus-torsion case is a frequent FE result.

The capacity used in a safety-factor check depends on the failure mode: yield strength for ductile yielding, an appropriate fracture criterion for brittle failure, an endurance/fatigue criterion for cyclic loading, and critical load for buckling. Select the mode before forming the capacity-to-demand ratio.

Thin-wall pressure-vessel quick recap

  • Hoop (circumferential) stress σ_h = pr/t is twice the longitudinal stress σ_l = pr/2t, so a cylinder fails along a longitudinal seam first.
  • A spherical vessel carries σ = pr/2t in every direction, half the cylinder hoop stress for the same p, r, t.
  • Thin-wall assumptions hold when r/t ≥ 10; below that, use thick-wall (Lamé) equations.
Test Your Knowledge

A thin-walled cylindrical tank has internal pressure p, radius r, and wall thickness t. How do the hoop and longitudinal stresses compare?

A

Longitudinal stress is twice the hoop stress

B

They are equal in a cylinder

C

Hoop stress is half the longitudinal stress

D

Hoop stress (pr/t) is twice the longitudinal stress (pr/2t)

Test Your Knowledge

A component sees σa = 100 MPa and σm = 200 MPa, with Se = 250 MPa and Su = 500 MPa. Using the modified Goodman criterion, is the part safe for infinite life?

A

No, the Goodman sum exceeds 1

B

Yes, the Goodman sum is 0.8 (< 1), so it is safe

C

Cannot tell without the yield strength

D

Yes, but only because Gerber is being used

Test Your Knowledge

On a feature control frame, what does the first item (the geometric symbol) followed by a datum reference letter establish?

A

The geometric characteristic being controlled and the reference (datum) it is measured from

B

The surface finish requirement

C

The nominal dimension of the feature

D

The material's yield strength

Sections you finish are checked off in the contents.