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)
Additionally, compute the longitudinal stress in the fuselage skin (in MPa or MegaPascal):

1 answer

To start, we need to determine the external pressure acting on the fuselage skin at the given altitude.

The atmospheric pressure decreases as we climb in altitude. At sea level, the standard atmospheric pressure is approximately 101.325 kPa.

To determine the pressure at 6,200 meters, we can use the barometric formula:

P = P0 * (1 - (L * h) / T0) ^ ((g * M) / (R * L))

Where:
P0 = sea level pressure (101.325 kPa)
L = temperature lapse rate (0.0065 K/m)
h = altitude (6,200 m)
T0 = sea level temperature (288.15 K)
g = acceleration due to gravity (9.80665 m/s^2)
M = molar mass of Earth's air (0.0289644 kg/mol)
R = ideal gas constant (8.31446261815324 J/(mol·K))

Plugging in the values:

P = 101.325 * (1 - (0.0065 * 6,200) / 288.15) ^ ((9.80665 * 0.0289644) / (8.31446261815324 * 0.0065))

Simplifying:

P = 101.325 * (1 - (0.0403 / 288.15)) ^ 3.48552

P = 101.325 * (1 - 0.000139827) ^ 3.48552

P = 101.325 * (0.999860173) ^ 3.48552

P ≈ 101.325 * 0.999343460

P ≈ 101.160 kPa

The external pressure acting on the fuselage skin at 6,200 meters altitude is approximately 101.160 kPa.

Now we can calculate the circumferential stress in the fuselage skin.

The circumference of the fuselage can be calculated using the formula:

C = 2 * π * r

Where:
r = radius of the fuselage (3.2 m)

C = 2 * π * 3.2

C ≈ 20.1 m

The circumference of the fuselage is approximately 20.1 meters.

The circumferential stress can be calculated using the formula:

σ_circ = (P * r) / t

Where:
P = pressure (101.160 kPa)
r = radius of the fuselage (3.2 m)
t = skin thickness (2.4 mm = 0.0024 m)

σ_circ = (101.160 * 10^3 * 3.2) / 0.0024

σ_circ ≈ 134,880,000 / 0.0024

σ_circ ≈ 56,200,000 N/m^2 or 56.2 MPa

The circumferential stress in the fuselage skin is approximately 56.2 MPa.

To calculate the longitudinal stress in the fuselage skin, we use the formula:

σ_long = P * r / (2 * t)

Using the same values as before:

σ_long = (101.160 * 10^3 * 3.2) / (2 * 0.0024)

σ_long ≈ 134,880,000 / 0.0048

σ_long ≈ 28,100,000 N/m^2 or 28.1 MPa

The longitudinal stress in the fuselage skin is approximately 28.1 MPa.