FAA Aviation Mechanic Airframe (AMA)Airframe StructuresHard
A technician must determine if a single 3/16-inch diameter 2117-T4 (AD) solid rivet can safely carry a 750-pound single-shear load. The allowable shear stress for AD rivets is 30,000 psi. Using the shear area formula A = (π/4)d², what is the calculated allowable shear load, and is the rivet adequate?
- AExactly 750 pounds; marginally adequate
- BApproximately 375 pounds; not adequate
- CApproximately 588 pounds; not adequate
- DApproximately 828 pounds; adequate
Show answer & explanationAnswer & explanation
Correct answer: D. Approximately 828 pounds; adequate
Shear area A = (π/4)(0.1875)² = 0.0276 in². Allowable shear load = 30,000 psi × 0.0276 in² ≈ 828 pounds. Since 828 lb exceeds the 750-lb applied load, the rivet is adequate for this single-shear application.
Why the other options are wrong
- A. The calculation does not naturally produce exactly 750 pounds; this is a coincidental distractor value.
- B. This understates the shear area substantially, giving an unrealistically low load capacity.
- C. This value results from using an incorrect (smaller) diameter or area in the calculation.
Rivet Shear Strength Calculation
The allowable shear load a rivet can carry equals its allowable shear stress multiplied by its cross-sectional shear area (A = π/4 × d²).
- Shear area formula: A = (π/4)d²
- AD (2117-T4) rivet allowable shear stress ≈ 30,000 psi
- Compare calculated allowable load to actual applied load for adequacy
Memory trick: Shear load = Stress × Area — plug and check against the load.