A modern jet flies at an altitude of 6,200 metres, during its climb to cruise altitude. For additional passenger comfort, the cabin pressure is maintained at 80% of the sea level value.

Given that the fuselage radius is 3.2 metres and that the skin thickness is 2.4 mm, compute the circumferential stress in the fuselage skin (in MPa or MegaPascal): (Assume ISA conditions)

1 answer

To calculate the circumferential stress in the fuselage skin of the jet, we need to follow these steps:

  1. Determine the pressure at altitude and the equivalent cabin pressure:

    • Standard atmospheric pressure at sea level: \( P_0 \approx 101325 , \text{Pa} \)
    • Cabin pressure maintained at 80% of sea level: \[ P_c = 0.8 \times P_0 = 0.8 \times 101325 , \text{Pa} = 81060 , \text{Pa} \]
    • The atmospheric pressure at an altitude of 6200 m can be found using the barometric formula or estimated values:
      • According to the ISA (International Standard Atmosphere), the pressure at 6200 m is approximately \( 54130 , \text{Pa} \).
  2. Determine the pressure difference acting on the fuselage:

    • The pressure difference \( \Delta P \) is calculated as: \[ \Delta P = P_c - P_a = 81060 , \text{Pa} - 54130 , \text{Pa} = 26930 , \text{Pa} \]
  3. Calculate the effective radius of the fuselage:

    • The outer radius \( R_o \) of the fuselage is calculated from the radius and skin thickness: \[ R_o = R + \text{skin thickness} = 3.2 , \text{m} + 0.0024 , \text{m} = 3.2024 , \text{m} \]
    • The inner radius \( R_i \) (if needed) can be considered for thin-walled pressure vessel approximations, but since we will use the outer radius for calculations of stress, we can use \( R_o \).
  4. Calculate the thickness of the fuselage skin:

    • The thickness \( t \) is given as: \[ t = 0.0024 , \text{m} \]
  5. Calculate the circumferential stress:

    • Using the formula for circumferential stress (\( \sigma_\theta \)) in a thin-walled cylindrical tube: \[ \sigma_\theta = \frac{R_o \Delta P}{t} \]
    • Plugging in the values: \[ \sigma_\theta = \frac{3.2024 , \text{m} \times 26930 , \text{Pa}}{0.0024 , \text{m}} \] \[ \sigma_\theta = \frac{86141.312 , \text{Pa m}}{0.0024} \] \[ \sigma_\theta \approx 35883880.08 , \text{Pa} \]
  6. Convert to MPa: \[ \sigma_\theta \approx 35.88 , \text{MPa} \]

Thus, the circumferential stress in the fuselage skin is approximately 35.88 MPa.