(a) A box culvert ABCD is shown in the above figure. By using member fixed end moments given above; calculate the final end…

(a) Members: walls AB and CD (I, 3 m), top slab BC (2I, 6 m), bottom slab AD (3I, 6 m). All four corners are rigid joints. Moments are clockwise-positive, as in the given FEMs.

Stiffness (K = I/L) and distribution factors:

• K_AB = K_CD = I/3; K_BC = 2I/6 = I/3; K_AD = 3I/6 = I/2.
• Joints A and D: DF(wall) = (1/3)/(1/3 + 1/2) = 0.4; DF(bottom slab) = 0.6.
• Joints B and C: DF(wall) = DF(top slab) = 0.5.
• Carry-over factor = ½ for every member (prismatic, far ends fixed against rotation during each cycle).

Structure and loads are symmetric about the vertical centre line, so there is no sway and each moment at C and D is the mirror (equal and opposite) of that at B and A. The distribution therefore only needs to follow joints A and B; the carry-over into AD comes from DA (= +½ × 45.9 = +22.95 in cycle 1), and into BC from CB.

Cycle 1: the unbalanced moment at A is −13.5 + 90 = +76.5, balanced by −30.6 (AB) and −45.9 (AD); at B it is +9 − 37.5 = −28.5, balanced by +14.25 on each side. Each subsequent correction is about half of the previous one, and the series converges to:

• M_AB = −49.26, M_AD = +49.26 kN·m
• M_BA = +18.95, M_BC = −18.95 kN·m
• M_CB = +18.95, M_CD = −18.95 kN·m
• M_DC = +49.26, M_DA = −49.26 kN·m

Check: the moments at every joint sum to zero; a slope-deflection solution gives the same values.

Sketch of the moments (drawn on the tension face):

• All four corners have tension on the outer face: 49.26 kN·m at A and D, 18.95 kN·m at B and C.
• Top slab mid-span: PL/4 − 18.95 = 75 − 18.95 = 56.05 kN·m, tension on the inner face.
• Bottom slab mid-span: wL²/8 − 49.26 = 30 × 6²/8 − 49.26 = 85.74 kN·m, tension on the inner face.
• Walls: the end moments exceed the span effect of the triangular pressure, so each wall stays in outer-face tension, with a minimum of about 15.7 kN·m roughly 1 m below the top.

(b) Design moment M_u = 1.5 × 200 = 300 kN·m. For M25 and Fe 500: f_ck = 25 N/mm², f_y = 500 N/mm², x_u,max = 0.46d = 209.3 mm.

Step 1, position of the neutral axis. With x_u = D_f = 125 mm, the flange alone resists

M = 0.36 f_ck b_f D_f (d − 0.42 D_f) = 0.36 × 25 × 1400 × 125 × (455 − 52.5) = 633.9 kN·m.

This exceeds 300 kN·m, so the neutral axis lies within the flange and the section is designed as a rectangle 1400 mm wide.

Step 2, limiting moment: M_u,lim = 0.133 f_ck b_f d² = 0.133 × 25 × 1400 × 455² = 963.7 kN·m > 300 kN·m, so a singly reinforced (under-reinforced) section suffices.

Step 3, steel area (IS 456, Annex G):

300 × 10⁶ = 0.87 × 500 × A_st × 455 × [1 − 500 A_st/(1400 × 455 × 25)]

Solving the quadratic gives A_st = 1596 mm².

Check: x_u = 0.87 f_y A_st/(0.36 f_ck b_f) = 0.87 × 500 × 1596/(0.36 × 25 × 1400) = 55.1 mm, which is below D_f = 125 mm and below x_u,max = 209.3 mm, confirming both assumptions.

Step 4, bars. Provide 2 bars of 25 mm + 2 bars of 20 mm (A_st = 1610 mm²) in one layer in the web.

• Width needed: 2 × 25 + 2 × 20 + 3 × 25 (clear gaps) + 2 × (25 cover + 8 stirrup) = 231 mm < 300 mm.
• With 1610 mm², x_u = 55.6 mm and M_R = 0.87 × 500 × 1610 × (455 − 0.42 × 55.6) = 302.5 kN·m ≥ 300 kN·m.
• A_st lies between the minimum 0.85 b_w d/f_y = 232 mm² and the maximum 0.04 b_w D = 6000 mm².

Provide 2 bars of 12 mm as hanger bars at the top of the web to hold the stirrups.

(c) Data: L = 10 m = 10 000 mm (simply supported); w = 100 kN/m = 100 N/mm; D = 500 mm, t = 20 mm, so the inner diameter d = 460 mm; E = 200 GPa = 2 × 10⁵ N/mm².

Moment of inertia of the tube:

I = π(D⁴ − d⁴)/64 = π(500⁴ − 460⁴)/64 = 8.70 × 10⁸ mm⁴

Maximum deflection (at mid-span, simply supported beam under UDL):

δ_max = 5wL⁴/(384EI) = 5 × 100 × (10⁴)⁴ / (384 × 2 × 10⁵ × 8.70 × 10⁸) = 74.8 mm

This is about span/134, far larger than normal serviceability limits (span/250 to span/325). The bending stress is also high: M = wL²/8 = 1250 kN·m and Z = I/250 = 3.48 × 10⁶ mm³, so σ = 359 N/mm², above the yield stress of ordinary structural steel (250 N/mm²). The elastic deflection therefore holds only for a higher-grade steel; otherwise the pipe is overloaded.