Hi-Lok Grip, Edge Distance and Hole Condition on a Web Repair
A repair scheme calls for 3/16 in (4.8 mm) Hi-Lok fasteners through a 1.6 mm 2024-T3 web into an 11 mm aluminium block, giving a stack that measures 12.2 mm to 12.8 mm; twenty holes are involved. Stores holds pins with 12 mm, 13 mm and 14 mm grip lengths. Gauging shows most holes at 4.80-4.83 mm, one at 4.85 mm with 0.04 mm out-of-round, and the closest hole sits 9.6 mm from the web edge, centre to edge. Several countersinks gauge 0.15 mm below flush. The technician plans to use the 14 mm pins 'so there is plenty of thread for the collar' and to ream the out-of-round hole to 4.9 mm and fit the same 4.8 mm fastener.
1. Select the grip length and explain the failure modes of the alternatives, allowing for the 0.6 mm stack variation.
Read the worked answer
Rule: fit the shortest standard grip that is equal to or greater than the stack at that hole.
- 12 mm is below the 12.2 mm minimum stack - the pin will not protrude, the collar cannot engage the full thread and the joint cannot be clamped. Rejected.
- 13 mm is the correct choice for 12.2-12.8 mm: the collar engages full thread and seats on the material.
- 14 mm leaves 1.2-1.8 mm of unclamped shank. The collar runs out of thread before it seats, so the break-off torque is reached with the joint drawn up on the threads instead of clamped. The result looks like a good fastener but there is no preload, the pin threads sit in the shear plane, and the joint frets and fatigues.
Grip must be selected per hole, because a 0.6 mm variation spans most of a grip band. Verify on the first few holes with a trial installation and check collar seating and break-off before running the batch.
2. The technician wants to ream the out-of-round hole to 4.9 mm and fit the same 4.8 mm fastener. Evaluate this, including the edge-distance implications.
Read the worked answer
This is not acceptable.
- Edge distance: 9.6 mm / 4.8 mm = 2.0D, i.e. exactly the usual minimum with no margin. Enlarging the hole to 4.9 mm gives 9.6 / 4.9 = 1.96D, below the minimum, and increases tear-out risk. Pitch to adjacent fasteners must also be rechecked.
- Fit: a 4.8 mm pin in a 4.9 mm hole is a loose clearance fit where a transition or interference fit is required. The fastener will move, the fatigue benefit of the joint is lost, and the hole will elongate.
- The 4.85 mm hole with 0.04 mm out-of-round must first be compared with the drawing tolerance. If it is out of tolerance the options are, in order: use the specified oversize fastener (which itself needs its own edge-distance and pitch check and approved data), or obtain an engineering disposition (bushing, hole repair, or revised repair scheme).
- Never peen, fill or shim a hole to recover fit. Stop the job and get the data rather than 'making it fit'.
3. Several countersinks gauge 0.15 mm below flush. Is that serviceable, and what are the approved ways to deal with it?
Read the worked answer
Probably not serviceable - and it cannot be fixed with filler.
- Flush-head fasteners must sit within the drawing tolerance, typically flush to a small controlled recess (commonly around 0.05 mm, at most a defined value in the specification). 0.15 mm below flush is likely out of tolerance.
- Consequences: the head sits in a water and fluid trap that starts corrosion; local skin thinning at the hole creates a knife-edge with reduced bearing strength; the fastener may not clamp the sheet properly.
- Check the remaining material under the head against the knife-edge limit for every hole before installation - an out-of-tolerance countersink does not simply need a deeper cut.
- Approved rectification is normally to use the next larger head diameter where edge distance and spacing allow (approved data required), or to replace the section. Adding filler or sealant to disguise the recess is not a structural repair and hides the defect from the next inspection.
Record the gauging results with the job card so the disposition is traceable.