To determine the required skin thickness for the fuselage of the supersonic passenger jet, we will use the concept of hoop stress (or circumferential stress) in cylindrical pressure vessels. The hoop stress is calculated using the formula:
\[ \sigma_h = \frac{P \cdot D}{2 \cdot t} \]
where:
- \(\sigma_h\) = hoop stress (Pa or N/m²)
- \(P\) = internal pressure (Pa or N/m²)
- \(D\) = external diameter of the cylinder (m)
- \(t\) = wall thickness of the cylinder (m)
Step 1: Calculate the required internal pressure at 24 km altitude.
At sea level, the pressure is approximately 101.3 kPa (101325 Pa). The cabin pressure should remain at least 72% of the sea-level value:
\[ P = 0.72 \times 101325 , \text{Pa} = 72915 , \text{Pa} \]
Step 2: Determine the external diameter of the fuselage.
The diameter of the fuselage is given as 4.5 m, so the external diameter \(D\) is:
\[ D = 4.5 , \text{m} \]
Step 3: Define the allowable stress with safety factor.
The maximum stress allowed in the fuselage skin is given as 75 MPa, and with a safety factor of 1.5, the allowable stress (\(\sigma_{\text{allowable}}\)) becomes:
\[ \sigma_{\text{allowable}} = \frac{75 \times 10^6 , \text{Pa}}{1.5} = 50 \times 10^6 , \text{Pa} \]
Step 4: Rearrange the hoop stress formula to solve for wall thickness \(t\).
Rearranging the hoop stress formula, we get:
\[ t = \frac{P \cdot D}{2 \cdot \sigma_h} \]
Substituting the known values:
\[ t = \frac{72915 , \text{Pa} \cdot 4.5 , \text{m}}{2 \cdot 50 \times 10^6 , \text{Pa}} \]
Now, calculating \(t\):
\[ t = \frac{72915 \cdot 4.5}{2 \cdot 50 \times 10^6} \]
\[ t = \frac{328117.5}{100000000} \]
\[ t = 0.003281175 , \text{m} = 3.28 , \text{mm} \]
Result:
The required skin thickness to ensure that the hoop stress does not exceed 75 MPa while accounting for a safety factor of 1.5 is approximately 3.28 mm.
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