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Aloha Airlines Flight 243 (1988): Fatigue and Corrosion in Lap Joints

On April 28, 1988, Aloha Airlines Flight 243, a Boeing 737-200 on an inter-island flight in Hawaii, lost part of its upper fuselage at 24,000 ft (about 7,300 m) and suffered an explosive decompression. The airplane landed on nearby Maui, but one flight attendant was lost. The failure was in a fuselage skin lap joint, where the bond had come apart and many fatigue cracks had grown from rivet holes; the NTSB found the probable cause to be the maintenance program’s failure to detect this damage.

What happened

Date
April 28, 1988
Location
Near Maui, Hawaii, USA (flight from Hilo to Honolulu)
Airplane
Boeing 737-200 (line number 152), with 35,496 flight hours and 89,680 flight cycles
Casualties
Of 89 passengers and 6 crew, one flight attendant was swept overboard and presumed killed, and 8 people were seriously injured
Damage
About 18 ft (about 5.5 m) of cabin skin and structure aft of the cabin entrance door, above the floor line, separated
Investigation
U.S. National Transportation Safety Board (NTSB), report NTSB/AAR-89/03

Sequence of events and the investigation’s findings

Flight 243 was an inter-island flight from Hilo on the island of Hawaii to Honolulu. As it reached 24,000 ft, the upper fuselage skin and structure separated over about 18 ft aft of the cabin entrance door, causing an explosive decompression. The crew made an emergency descent and landed at Kahului Airport on Maui.

The NTSB determined that the probable cause was the failure of Aloha Airlines’ maintenance program to detect significant disbonding and fatigue damage, which ultimately led to the failure of the lap joint at the location called S-10L and the separation of the upper fuselage. Contributing factors included inadequate supervision of maintenance, the FAA’s failure to require inspection of all the lap joints proposed by Boeing in airworthiness directive AD 87-21-08, and the lack of a complete terminating action for problems with the early 737 cold-bonded lap joints (low bond durability, corrosion and premature fatigue cracking).

The chain of causes from the materials side

Aloha Flight 243: the chain of causes from the materials sidePoor bonding in manufactureLap joints bonded at room temperatureMoisture wicks into disbondsCorrosion and spreading disbondsRivets carry pressurization loadsKnife-edge countersinks raise stressPressurization cycles every flightFatigue cracks at many upper-row rivetsCracks link into a long crackThe S-10L lap joint failsUpper fuselage separates in flight
Aloha Flight 243: the chain of causes from the materials side

Early 737s (up to line number 291) bonded the fuselage skin lap joints with a scrim cloth impregnated with an adhesive that cures at room temperature (a cold bond). Manufacturing problems left randomly weak bonds and some areas not bonded at all, and made the joints susceptible to corrosion. Moisture wicking into disbonded areas caused further disbonding.

Once the bond failed, the rivets carried the pressurization loads. The rivet countersinks went through the full skin thickness, leaving a thin knife edge at the hole and a higher-than-usual stress concentration. Fatigue cracks therefore formed at many adjacent rivet holes (multiple site damage, MSD), and small cracks linked up into a long, critical crack. Visual inspection showed cracks growing from the holes of the top row of rivets.

Fatigue crack initiation and growth are driven by the number of flights (pressurization cycles), not flight hours, and Aloha’s airplanes flew about twice the cycle rate that Boeing had assumed. Lap joint corrosion had first been reported in 1970, and inspections of the fleet after the accident found swelling, bulging, popped rivets and blistered paint along lap joints.

On interpretations of the cause: a university student paper (Hashmi) suggests that chlorides in the marine operating environment probably caused pitting and drove the corrosion and corrosion fatigue. The NTSB’s probable cause is the maintenance program’s failure to detect the disbonding and fatigue damage.

The clues that revealed the cause

Open the guide with this case’s clues

Opening the guide with these clues ranks corrosion fatigue and fatigue at the top. Corrosion fatigue comes first because of the moisture and corrosion clues, but the NTSB’s probable cause is the failure to detect disbonding and fatigue damage; it does not identify corrosion fatigue as such.

Related failure modes: Fatigue / Corrosion fatigue

What changed afterwards: design, rules and inspection

Lessons

Check your understanding

The fracture surface of a part that failed by fatigue shows shell-like patterns visible to the naked eye (beach marks) and fine lines visible in an electron microscope (striations). Which statement about them is correct?

  1. ABoth form one line per load cycle, so counting them gives the number of cycles to failure
  2. BStriations roughly mark the crack advance in each cycle, while beach marks record history such as shutdowns
  3. CBoth form in the final fracture zone that broke all at once, gathering at the place farthest from the origin
Show answer and explanation

Answer: B. Striations roughly mark the crack advance in each cycle, while beach marks record history such as shutdowns

Striations are lines left as the crack tip opens and blunts, then closes and sharpens again in each cycle, and their spacing corresponds roughly to the crack growth per cycle, da/dN. The spacing is fine, at the µm level or below, and cannot be seen without an electron microscope. Beach marks are patterns that record positions of the crack front through shutdowns and restarts, changes in load level, differences in how corrosion progresses and so on, and each band contains many cycles. Both appear in the region where the fatigue crack advanced little by little, and the common center of the patterns shows the origin. The final fracture zone is where the remaining cross-section broke all at once, and it shows a ductile or brittle fracture surface.

Which statement correctly describes the curve of fatigue crack growth rate da/dN plotted against the stress intensity factor range ΔK on log-log axes?

  1. AThe middle region is a straight line given by the Paris law da/dN = C(ΔK)ᵐ, and below ΔK_th the crack hardly grows
  2. Bda/dN is constant regardless of ΔK, set only by the crack length and the material's yield stress
  3. CThe slope of the line decreases as ΔK grows, and crack growth stops when the maximum stress intensity factor approaches K_IC
Show answer and explanation

Answer: A. The middle region is a straight line given by the Paris law da/dN = C(ΔK)ᵐ, and below ΔK_th the crack hardly grows

The curve is usually divided into three regions. In the low region where ΔK approaches the threshold ΔK_th, the growth rate drops sharply, and below ΔK_th the crack effectively does not grow. The intermediate region is a straight line on log-log axes and can be described by the Paris law da/dN = C(ΔK)ᵐ, with an exponent m of roughly 2–4 for metals. In the high region where the maximum stress intensity factor K_max approaches the fracture toughness K_IC, the rate rises sharply and leads to unstable fracture, so the claim that growth stops is backwards. ΔK = YΔσ√(πa) depends on both the stress range and the crack length, so the rate is not determined by crack length alone.

Fatigue crack growth in a steel plate is described by the Paris law da/dN = C(ΔK)ᵐ (C = 1.0×10⁻¹², with da/dN in m/cycle, ΔK in MPa·√m, and m = 4), and ΔK = Δσ√(πa) (geometry factor 1). Under cycling with a stress range Δσ = 200 MPa, which value is closest to the number of cycles for the crack to grow from 1.0 mm to 5.0 mm? Assume that ΔK over this range (about 11–25 MPa·√m) is well above the threshold and within the range where the Paris law holds.

  1. AAbout 2.5×10⁵ cycles
  2. BAbout 5.1×10⁴ cycles
  3. CAbout 6.3×10⁴ cycles
Show answer and explanation

Answer: B. About 5.1×10⁴ cycles

For m = 4, da/dN = C(Δσ√π)⁴a², so separating variables and integrating gives N = (1/a₀ − 1/a_f)/[C(Δσ√π)⁴]. C(Δσ√π)⁴ = 1.0×10⁻¹²×200⁴×π² ≈ 1.58×10⁻² m⁻¹ and 1/a₀ − 1/a_f = 1000 − 200 = 800 m⁻¹, so N ≈ 800/(1.58×10⁻²) ≈ 5.1×10⁴ cycles. About 2.5×10⁵ cycles assumes the rate at the initial 1.0 mm, 1.58×10⁻⁸ m/cycle, continues to the end, ignoring that ΔK rises and the crack accelerates as it grows. About 6.3×10⁴ cycles drops the 1/a_f term. Most of the life is spent while the crack is still short, early on.

A steel part fails by fatigue after 1.0×10⁵ cycles at stress amplitude σ₁ and after 1.0×10⁶ cycles at σ₂. After it is used for 4.0×10⁴ cycles at σ₁, it continues to be used at σ₂. Which value is closest to the remaining life estimated with Miner's rule (failure when Σn/N = 1)?

  1. AAbout 6.0×10⁵ cycles
  2. BAbout 9.6×10⁵ cycles
  3. CAbout 6.0×10⁴ cycles
Show answer and explanation

Answer: A. About 6.0×10⁵ cycles

The damage at σ₁ is n₁/N₁ = 4.0×10⁴/1.0×10⁵ = 0.40, leaving 1 − 0.40 = 0.60. At σ₂, the damage per cycle is 1/N₂, so the remaining number of cycles is 0.60×1.0×10⁶ = 6.0×10⁵ cycles. About 9.6×10⁵ cycles subtracts the number of cycles itself, rather than the fraction of damage, from N₂, ignoring that the weight of one cycle differs with stress. About 6.0×10⁴ cycles multiplies the remaining fraction by the life N₁ at σ₁. Miner's rule is simple, but it cannot represent the effect of load sequence, and it is known that when the stress steps from high to low, the actual life tends to be shorter than estimated.

Related terms

Fatigue limit Fatigue striations Paris law Stress concentration factor Miner's rule

Main sources

  1. NTSB, Aircraft Accident Report NTSB/AAR-89/03: Aloha Airlines, Flight 243, Boeing 737-200, N73711, near Maui, Hawaii, April 28, 1988 (summary, archived NTSB site at UNT Libraries)
  2. FAA, Lessons Learned from Transport Airplane Accidents: Boeing 737-200 N73711
  3. N. Hashmi, Revisiting Aloha Airline Flight 243: Corrosion Engineer’s Stand point (paper citing pages of the NTSB report)

Dates, numbers of people, dimensions, temperatures and other values were checked against the official reports above and sources that cite them.

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