Proof-Test Strain and Acoustic-Emission Anomaly on a Titanium Helium Tank
A 320 mm internal diameter spherical helium pressurant tank for a satellite propulsion module is being proof-tested. The shell is annealed Ti-6Al-4V, 4.5 mm nominal wall, closed by two electron-beam girth welds with a measured weld-land eccentricity of 0.3 mm. Design pressure is 31 MPa, maximum expected operating pressure is 23 MPa, and the proof pressure is fixed at 1.25 times design, i.e. 38.75 MPa. Six biaxial strain rosettes are bonded around the shell and four acoustic-emission (AE) sensors monitor it. During the hold at 36 MPa, rosettes R3 and R5, both within 40 mm of the lower girth weld, read hoop strain 12% above the finite-element prediction, while the other four agree within 2%. Fourteen AE events above threshold were logged in the same region during the 34–36 MPa ramp. An ultrasonic thickness check at R3 gives 4.15 mm. The test conductor asks whether to continue to 38.75 MPa. The mass budget is closed and a replacement tank is a nine-month lead item.
1. Compute the membrane stress at design and proof pressure and state whether the planned proof level is defensible for this material (Fty = 830 MPa, Ftu = 900 MPa). Then rule on whether the test should proceed to 38.75 MPa.
Read the worked answer
Stress check (thin-wall sphere, sigma = pD/4t, mean D = 320 + 4.5 = 324.5 mm)
- Design, 31 MPa: 31 x 324.5 / (4 x 4.5) = 559 MPa = 0.67 Fty, 0.62 Ftu.
- Proof, 38.75 MPa: 38.75 x 324.5 / (4 x 4.5) = 699 MPa = 0.84 Fty, 0.78 Ftu.
- Local thin patch, 4.15 mm, at design pressure: 31 x 324.5 / (4 x 4.15) = 606 MPa = 0.73 Fty.
So the proof level itself is defensible: 1.25 x design keeps the membrane stress below yield (the usual rule is to stay under roughly 0.85–0.9 Fty so the tank is screened without gross permanent set), and even the thin patch keeps a 1.37 margin on yield at design pressure.
Ruling: hold at 36 MPa, vent, do not continue.
The logic is not about the stress number, which is fine, but about the evidence. A proof test is a deliberate over-stress; the decision to keep loading must rest on the response staying benign. Here it is not: a 12% non-axisymmetric strain deviation is outside the 2% scatter of the other four channels, and 14 AE events cluster in the same location during the same pressure band. Two independent measurements pointing at one spot is the pattern of a local geometric or flaw feature becoming active, not of instrument noise. Venting costs a day; driving a possible defect through the wall at 38.75 MPa costs the tank and nine months. Vent, re-analyse with the measured thickness and eccentricity, inspect, then re-proof with AE hold bands defined in advance.
2. Rosette R3 reads 12% high and the wall there is 4.15 mm against 4.5 mm nominal. How much of the strain deviation does the thickness shortfall explain, and what other mechanisms are credible?
Read the worked answer
Thickness contribution
For a thin shell the membrane stress varies as 1/t, so a wall at 4.15 mm is 4.15/4.5 = 0.922 of nominal, i.e. 7.8% thin, and the local stress and elastic strain rise by 1/0.922 - 1 = 8.5%.
That accounts for roughly 8.5 of the 12 percentage points. The residual, about 3.5%, is not large, but it is above the 2% channel-to-channel scatter and it sits exactly where a second effect is expected: the 0.3 mm weld-land eccentricity introduces a local bending component on top of the membrane field, and a rosette bonded on or beside a weld bead samples a thickness and a constraint that the nominal shell model does not contain. Rosette misalignment of a few degrees on a biaxial field also produces a small apparent scatter.
What this means
The strain reading is therefore largely explained by a benign geometric feature, not necessarily by a crack. That is why the immediate action is measurement, not rejection: ultrasonic thickness mapping of the whole band, penetrant or eddy-current scan of the weld land, and a re-run of the FE model with the measured local thickness and the eccentricity included. If the mapped thickness field reproduces the strain pattern and no flaw is found, the tank can be re-proofed with a revised hold band. If a flaw is found, the tank is rejected.
Note the thin patch is not a strength problem in itself: at design pressure it sees 606 MPa, still 1.37 below yield. It becomes a life problem only through low-cycle fatigue, which must be re-checked against the qualified cycle count.
3. What flaw size does a proof test at 38.75 MPa actually screen for? Use KIC = 75 MPa x sqrt(m) and a surface crack with geometry factor 1.12. Does that change the disposition?
Read the worked answer
Critical flaw size: for a surface crack, K = 1.12 x sigma x sqrt(pi x a), so a = (1/pi) x [KIC / (1.12 x sigma)]^2.
- At proof stress 699 MPa: 1.12 x 699 = 783; 75/783 = 0.0958; squared = 0.00918; divided by pi = 0.00292 m, i.e. a ≈ 2.9 mm.
- At MEOP stress, 23 x 324.5/18 = 415 MPa: 1.12 x 415 = 465; 75/465 = 0.161; squared = 0.0260; divided by pi = 0.00828 m, i.e. a ≈ 8.3 mm.
So the proof screens out any surface flaw deeper than about 2.9 mm, which is the size that would otherwise sit stable at MEOP and could grow to the 8.3 mm critical size during service. That is the whole value of the proof test, and it is also the reason not to abandon it.
Effect on disposition: this argues for fixing the anomaly and re-proofing rather than scrapping or skipping. Note the important caveat: the screening only works if the proof stress is actually reached across the whole shell. A locally thin or locally bending region reaches a different stress, so the 2.9 mm number applies to nominal wall. Once the thickness map is in hand, redo the calculation at the measured thickness (at 4.15 mm the proof stress is 758 MPa, so the screenable flaw there is about 2.5 mm) and confirm every point of the shell is screened at or below the qualified flaw size.