(a) A submarine can be assumed to have cylindrical shape with rounded nose. Assuming its length to be 55 m and diameter 6·0 m…

(a) Power to overcome skin friction on the submarine (15 marks)

Treat the hull as a smooth cylinder in a turbulent boundary layer from the nose, and include skin friction only (no form or wave drag). The wetted area is the curved surface of the cylinder.

Re = VL/ν = 8.0 × 55/(1 × 10⁻⁶) = 4.4 × 10⁸. This is far above 5 × 10⁵, so the laminar length is negligible. For 10⁷ < Re < 10⁹ use the Prandtl–Schlichting formula:

C_D = 0.455/(log₁₀ Re)^2.58 = 0.455/(8.643)^2.58 = 0.00174.

A = πDL = π × 6.0 × 55 = 1,036.7 m².

F_D = C_D × ½ρV²A = 0.00174 × 0.5 × 1,030 × 8.0² × 1,036.7 ≈ 59,600 N.

P = F_D × V = 59,600 × 8.0 ≈ 4.77 × 10⁵ W, so power ≈ 477 kW.

(b) Beam with supports at B and E (15 marks)

From A: UDL 25 kN/m on AB (0 to 0.5 m, resultant 12.5 kN at 0.25 m); support B at 0.5 m; 300 kN·m anticlockwise couple at C (1.5 m); 50 kN at D (3.5 m); support E at 5.0 m; 40 kN at F (5.5 m).

(i) Reactions. Moments about B, anticlockwise positive: 12.5(0.25) − 50(3.0) − 40(5.0) + 4.5R_E + 300 = 0, so R_E = 10.42 kN (up). R_B = 12.5 + 50 + 40 − 10.42 = 92.08 kN (up).

(ii) V and M. V is positive when the net force on the left of the section is downward (figure convention), and M is positive when sagging. "L" and "R" mean just left and just right of the point.

• A: V = 0; M = 0.
• B: V = +12.5 (L), −79.58 (R); M = −3.125.
• C: V = −79.58; M = +76.46 (L), −223.54 (R).
• D: V = −79.58 (L), −29.58 (R); M = −64.38.
• E: V = −29.58 (L), −40.0 (R); M = −20.0.
• F: V = −40.0 (L), 0 (R); M = 0.

Check: M_C(L) = 92.08 × 1.0 − 12.5 × 1.25 = 76.46; the couple lowers it by 300. M_D = 10.42 × 1.5 − 40 × 2 = −64.38.

(iii) SFD. It is a straight inclined line over AB (UDL) and horizontal steps elsewhere, with jumps at B, D, E and F equal to the point loads and reactions.

(iv) BMD. It is a parabola over AB and straight lines elsewhere. It has a downward jump of 300 kN·m at C. The maximum sagging moment is 76.46 kN·m just left of C and the maximum hogging moment is 223.54 kN·m just right of C. M = 0 at x = 0.54 m (contraflexure).

(c) Beam design, IS 456 limit state (20 marks)

b = 300, D = 500, d = 500 − 40 = 460 mm, d′ = 40 mm, L = 5 m.

Loads. Self-weight = 0.3 × 0.5 × 25 = 3.75 kN/m, so w = 67.75 and w_u = 1.5 × 67.75 = 101.6 kN/m. M_u = w_uL²/8 = 317.6 kN·m. V_u = 254.1 kN.

Limiting moment. x_u,max = 0.48d = 220.8 mm. M_u,lim = 0.138 f_ck b d² = 175.2 kN·m. Since M_u > M_u,lim and the size is fixed, the beam is doubly reinforced.

Compression steel stress. ε_sc = 0.0035(1 − 40/220.8) = 0.00287. From the given curve (f_yd = 360.9), f_sc ≈ 353 N/mm².

Steel. M_u2 = 317.6 − 175.2 = 142.4 kN·m. A_sc = 142.4 × 10⁶/[(353 − 8.9) × 420] = 986 mm². A_st = 0.36 f_ck b x_u,max/0.87f_y + A_sc(f_sc − 8.9)/0.87f_y = 1,321 + 939 = 2,260 mm².

Provide: tension 4–25 φ + 1–20 φ = 2,278 mm² in one layer; compression 3–22 φ = 1,140 mm². Here x_u ≈ 199 mm < x_u,max and M_R ≈ 327 kN·m > 317.6 kN·m, so it is safe. The steel lies within 0.85bd/f_y and 4%.

Shear. At d from the support face (0.585 m from the support centre), V_u = 194.6 kN. τ_v = 1.41 N/mm² < τ_c,max = 2.8. With p_t = 1.65%, τ_c ≈ 0.73. V_us = (1.41 − 0.73) × 300 × 460 = 93.8 kN. For 2-legged 8 mm stirrups, s = 0.87 × 415 × 100.5 × 460/93,800 = 178 mm. Provide stirrups at 175 mm c/c for 1 m from each support and 300 mm c/c in the middle.

Deflection. f_s ≈ 239 N/mm², k_t ≈ 0.88, k_c ≈ 1.20, so allowable L/d = 20 × 0.88 × 1.2 = 21.2 against 10.9 actual. Safe.