Structural Analysis of Seastead Legs Under Side Wave Loading

Shape: NACA foil approximated as an ellipse.
Dimensions:
- Chord length (horizontal, front-to-back): 10 ft = 120 in.
- Maximum thickness (vertical): 4 ft = 48 in.
- Length (vertical span): 19 ft = 228 in.
- Wall thickness: 0.5 in.
Material: Marine aluminum with ultimate tensile strength \(\sigma_{ult} = 45,000\) psi.

Approximating the cross-section as a thin-walled ellipse:

Outer semi-axes: \(a_o = 60\) in, \(b_o = 24\) in.
Inner semi-axes: \(a_i = 59.5\) in, \(b_i = 23.5\) in.

Moment of inertia about the horizontal axis (parallel to chord):

\[ I_{xx} = \frac{\pi}{4} \left( a_o b_o^3 - a_i b_i^3 \right) \]

Calculation:

• \(b_o^3 = 24^3 = 13,824\) in³, \(a_o b_o^3 = 60 \times 13,824 = 829,440\) in⁴.
• \(b_i^3 = 23.5^3 = 12,977.875\) in³, \(a_i b_i^3 = 59.5 \times 12,977.875 = 772,183.5625\) in⁴.
• Difference: \(829,440 - 772,183.5625 = 57,256.4375\) in⁴.
• \(I_{xx} = \frac{\pi}{4} \times 57,256.4375 \approx 44,970\) in⁴.

Distance from neutral axis to outer fiber: \(c = b_o = 24\) in.

Section modulus: \(S = \frac{I_{xx}}{c} = \frac{44,970}{24} \approx 1,873.75\) in³.

The maximum bending moment before failure is:

\[ M_{allow} = \sigma_{ult} \times S = 45,000 \, \text{psi} \times 1,873.75 \, \text{in}^3 = 84,318,750 \, \text{lb-in} = 7,026,562.5 \, \text{ft-lb}. \]

Case A: Uniform load along the entire leg length (19 ft).

For a cantilever beam with uniform load \(q\) (lb/ft), maximum bending moment at the fixed end:

\[ M_{max} = \frac{q L^2}{2}, \quad L = 19 \, \text{ft}. \]

Setting \(M_{max} = M_{allow}\):
\[ q = \frac{2 M_{allow}}{L^2} = \frac{2 \times 7,026,562.5}{19^2} \approx 38,920 \, \text{lb/ft}. \]

Total force: \(F = q \times L = 38,920 \times 19 \approx 739,480 \, \text{lb}\) (330 metric tons).

Case B: Uniform load only on the submerged portion (9.5 ft).

Assuming the sideways wave force acts only on the submerged part (50% of leg), the maximum moment is:

\[ M_{max} = q \times d \times \left(L - \frac{d}{2}\right), \quad d = 9.5 \, \text{ft}, \, L = 19 \, \text{ft}. \]

Solving for \(q\):
\[ q = \frac{M_{allow}}{d \times (L - d/2)} = \frac{7,026,562.5}{9.5 \times 14.25} \approx 51,900 \, \text{lb/ft}. \]

Total force on submerged part: \(F_{sub} = q \times d = 51,900 \times 9.5 \approx 493,000 \, \text{lb}\) (221 metric tons).

Using a simplified drag-force formula for wave loading on a bluff body:

\[ F = \frac{1}{2} \rho C_d A v^2, \]

where:
- \(\rho = 1.94 \, \text{slugs/ft}^3\) (seawater density),
- \(C_d \approx 1.0\) (drag coefficient for a foil shape sideways),
- \(A =\) projected submerged area = \(d \times \text{thickness} = 9.5 \, \text{ft} \times 4 \, \text{ft} = 38 \, \text{ft}^2\),
- \(v =\) maximum water particle velocity.

For linear wave theory, the maximum horizontal velocity at the surface is approximately \(v_{max} \approx 0.5 H\) ft/s for a wave of height \(H\) (ft) and period \(T = 10\) s in deep water.

Thus:

\[ F = \frac{1}{2} \times 1.94 \times 1.0 \times 38 \times (0.5 H)^2 = 9.215 H^2 \, \text{lb}. \]

Setting \(F = 739,480 \, \text{lb}\) (Case A):
\[ H = \sqrt{\frac{739,480}{9.215}} \approx 283 \, \text{ft}. \]

For \(F = 493,000 \, \text{lb}\) (Case B):
\[ H = \sqrt{\frac{493,000}{9.215}} \approx 231 \, \text{ft}. \]

Note: These wave heights are extreme and exceed typical ocean wave conditions (even rogue waves rarely exceed 100 ft). This indicates the leg design is robust against wave-induced failure.