FAA Aviation Mechanic General (AMG)Physics, Math and Fluid LinesHard
A tube has an outside diameter of 1.0 inch and a wall thickness of 0.049 inch. The tube material has an ultimate tensile strength of 40,000 psi. Using the thin-wall pressure vessel formula P = 2tS/D, what is the approximate internal burst pressure of the tube?
- A1,960 psi
- B490 psi
- C7,840 psi
- D3,920 psi
Show answer & explanationAnswer & explanation
Correct answer: D. 3,920 psi
Applying P = 2tS/D: P = (2 × 0.049 × 40,000) / 1.0 = 3,920 / 1.0 = 3,920 psi. This thin-wall hoop stress formula relates internal pressure, wall thickness, tensile strength, and diameter for a cylindrical pressure vessel like tubing.
Why the other options are wrong
- A. This results from using thickness once instead of the required factor of 2.
- B. This divides by an incorrect factor, understating burst pressure.
- C. This doubles the correct answer by an arithmetic error.
Thin-Wall Pressure Vessel (Hoop Stress) Formula
Relates internal pressure, tube wall thickness, tensile strength, and diameter: P = 2tS/D, used to estimate burst pressure of tubing under internal pressure.
- P = 2tS/D (P=burst pressure, t=wall thickness, S=tensile strength, D=diameter)
- Thicker walls and higher tensile strength increase burst pressure
- Larger diameter decreases burst pressure for the same wall thickness
Memory trick: 'Two times thickness times strength, over diameter' bursts the tube.