(a) Using moment area method, find the slopes and deflection at 'B' for the beam shown in the figure below. Point 'B' is a hinge…

(a) Slopes and deflection at hinge B (15 marks)

The hinge gives M_B = 0, so the beam is determinate. For segment BD, moments about B give R_D·2L − W·L = 0, so R_D = W/2. Then R_A = W/2 and the fixed-end moment is M_A = WL/2 (hogging). EI is constant.

BMD: A to B is a straight line from −WL/2 to 0. From B, M rises linearly to +WL/2 at C, then falls linearly to 0 at D.

Segment AB. A is fixed, so the tangent at A is horizontal.

• Area of the M/EI diagram = ½ × L × (−WL/2EI) = −WL²/4EI, so θ_B (left of the hinge) = WL²/4EI, clockwise.
• Deviation of B from the tangent at A = area × distance of its centroid from B = (−WL²/4EI)(2L/3) = −WL³/6EI, so δ_B = WL³/6EI, downward.

Segment BD. The slope just right of the hinge is different, so use the tangent at B.

• Area of the M/EI diagram over BD = ½ × 2L × WL/2EI = WL²/2EI. Its centroid is L from D, so the deviation of D from the tangent at B is t_DB = WL³/2EI.
• D is at support level and B is WL³/6EI below it. With θ positive anticlockwise, WL³/6EI = 2Lθ + WL³/2EI, so θ_B (right) = −WL²/6EI, i.e. WL²/6EI, clockwise.

The slopes on the two sides of the hinge differ by WL²/12EI, which the hinge allows.

(b) RC column design by limit state method (15 marks)

A_g = 250 × 400 = 100,000 mm². P_u = 700 kN (already factored).

Slenderness. The column is held in position at both ends and restrained against rotation at one end (IS 456 Table 28), so l_e = 0.8 × 3.2 = 2.56 m. l_e/b = 2560/250 = 10.2 < 12 and l_e/D = 6.4, so it is a short column.

Minimum eccentricity. e_min = l/500 + D/30 = 19.7 mm, taken as the 20 mm minimum. This equals 0.05D on the major axis, so the axial formula applies. On the minor axis 20 mm exceeds 0.05b (12.5 mm). A strain-compatibility check of the section with 4 bars at M_u = 700 × 0.02 = 14 kN·m gives a capacity of about 48 kN·m at P_u = 700 kN, so it is safe.

Steel. P_u = 0.4f_ck A_c + 0.67f_y A_sc. Concrete alone carries 0.4 × 25 × 100,000 = 1,000 kN > 700 kN, so minimum steel governs: 0.8% of A_g = 800 mm². Provide 4 bars of 16 mm (804 mm², 0.80%). Capacity = 1,000 + 0.67 × 415 × 804/1000 = 1,224 kN > 700 kN. Safe.

Ties. Diameter ≥ larger of 16/4 and 6 mm, so 6 mm. Pitch ≤ least of 250 mm (least lateral dimension), 16 × 16 = 256 mm and 300 mm, so 250 mm.

Details to draw. A 250 × 400 section with one 16 mm bar at each corner, 40 mm clear cover, 6 mm ties @ 250 mm c/c with 135° hooks, and a sectional elevation showing the tie spacing.

(c) Rectangular flume, uniform flow (20 marks)

Q = 0.100 m³/s, b = 0.60 m, y = 0.30 m, C = 56.

A = 0.60 × 0.30 = 0.18 m². P = b + 2y = 1.20 m. R = A/P = 0.15 m. V = Q/A = 0.100/0.18 = 0.556 m/s.

Chezy: V = C√(RS), so S = V²/(C²R) = 0.3086/(3136 × 0.15) = 6.56 × 10⁻⁴ (about 1 in 1,524).

Conveyance K = AC√R = 0.18 × 56 × √0.15 = 3.90 m³/s (check: K√S = 0.100 m³/s).

Froude number Fr = V/√(gy) = 0.556/√(9.81 × 0.30) = 0.32. Since Fr < 1, the uniform flow is subcritical (tranquil).