(a) Using limit state method, design a reinforced concrete circular column (with lateral ties) of height 3·0 m. It is subjected…

(a) Design of a tied circular RC column (15 marks)

Data: height 3.0 m, unfactored load 1150 kN, M25, Fe500, effective length factor 1.2.

Factored load: P_u = 1.5 × 1150 = 1725 kN. Effective length l_eff = 1.2 × 3.0 = 3.6 m.

Size of the column. Assume 1.5% longitudinal steel. For a short column with a small eccentricity (IS 456 cl. 39.3), P_u = 0.4 f_ck A_c + 0.67 f_y A_sc, with A_c = 0.985 A_g and A_sc = 0.015 A_g:

P_u = (0.4 × 25 × 0.985 + 0.67 × 500 × 0.015) A_g = 14.875 A_g, so A_g = 1725 × 10³/14.875 = 115 966 mm², giving D = 384 mm. Adopt D = 400 mm (A_g = 125 664 mm²).

Short-column check: l_eff/D = 3600/400 = 9 < 12, so the column is short and slenderness moments are not required.

Minimum eccentricity: e_min = l/500 + D/30 = 3000/500 + 400/30 = 6 + 13.3 = 19.3 mm, which is not more than 0.05D = 20 mm, so the axial-load formula for short columns applies.

Longitudinal steel: 1725 × 10³ = 0.4 × 25 × (125 664 - A_sc) + 0.67 × 500 × A_sc = 1 256 637 + 325 A_sc, so A_sc = 1441 mm².

Check against the limits: the minimum is 0.8% of A_g = 1005 mm² and the maximum is 6% (7540 mm²), so 1441 mm² is acceptable. A circular column needs at least 6 bars of 12 mm. Provide 6 bars of 18 mm diameter, A_sc = 6 × 254.5 = 1527 mm² (1.22%).

Capacity check: P_u = 0.4 × 25 × (125 664 - 1527) + 0.67 × 500 × 1527 = 1241 + 512 = 1753 kN > 1725 kN, so the design is safe.

Lateral ties. Diameter not less than one-quarter of the largest bar (18/4 = 4.5 mm) and not less than 6 mm; provide 8 mm ties. Pitch: not more than the least of the least lateral dimension (400 mm), 16 times the smallest longitudinal bar diameter (16 × 18 = 288 mm) and 300 mm, so provide 280 mm centre to centre. Clear cover 40 mm.

Details: column 400 mm diameter, 6-18φ longitudinal bars, 8φ circular ties at 280 mm c/c, 40 mm cover.

(b) Alloy strut: Euler load and maximum deflection (15 marks)

Data: pin-ended strut, L = 2.0 m, section 20 mm × 15 mm, E = 90 GN/m², yield stress 300 MN/m².

Least second moment of area: I = 20 × 15³/12 = 5625 mm⁴ = 5.625 × 10⁻⁹ m⁴, and the buckling occurs about this axis.

Euler load: P_E = π²EI/L² = π² × 90 × 10⁹ × 5.625 × 10⁻⁹/2.0² = 1249 N (about 1.25 kN).

When the bar has buckled, the load is P = P_E, and the bar has a central deflection δ, with a bending moment at mid-length M = Pδ (the buckled shape is a half sine wave, so the central deflection is the maximum). The extreme fibre stress is the sum of the direct and bending stresses:

σ_max = P/A + Pδ/Z

Area A = 20 × 15 = 300 mm² = 3 × 10⁻⁴ m². Section modulus about the weak axis Z = bd²/6 = 20 × 15²/6 = 750 mm³ = 7.5 × 10⁻⁷ m³.

Direct stress: P/A = 1249/3 × 10⁻⁴ = 4.16 MN/m².

Setting σ_max = 300 MN/m²: 300 = 4.16 + 1249 δ/(7.5 × 10⁻⁷) × 10⁻⁶, so the bending stress is 295.8 MN/m² and

δ = 295.8 × 10⁶ × 7.5 × 10⁻⁷/1249 = 0.1776 m ≈ 178 mm.

The maximum central deflection before the material yields is about 178 mm. (Because the load stays at P_E after buckling, the bar can deflect a long way before the extreme fibre reaches yield; the sinusoidal shape is used as the approximation.)

(c) Analysis of frame ABCD by moment distribution (20 marks)

Members: AB, 7 m (2I; A fixed) with 100 kN at 4 m from A; BC, 5 m (I; C fixed) with 80 kN at 2 m from B; BD, 5 m (2I; D fixed), unloaded. Joint B is rigid, and there is no sway. End moments are taken clockwise positive.

Fixed end moments:

M_AB = -Wab²/L² = -100 × 4 × 3²/7² = -73.47 kN·m; M_BA = +Wa²b/L² = 100 × 4² × 3/7² = +97.96 kN·m.

M_BC = -80 × 2 × 3²/5² = -57.60 kN·m; M_CB = +80 × 2² × 3/5² = +38.40 kN·m. BD carries no load, so its fixed end moments are zero.

Stiffness and distribution factors at B (K = 4EI/L, with far ends fixed):

K_BA = 4(2I)/7 = 1.143I, K_BC = 4I/5 = 0.800I, K_BD = 4(2I)/5 = 1.600I, sum = 3.543I.

DF_BA = 0.3226, DF_BC = 0.2258, DF_BD = 0.4516 (sum 1.000).

Distribution. Unbalanced moment at B = 97.96 - 57.60 = +40.36 kN·m; the balancing moment is -40.36, distributed as BA -13.02, BC -9.11, BD -18.23 kN·m. Carry-over one-half to the fixed far ends: AB -6.51, CB -4.56, DB -9.11 kN·m. As A, C and D are fixed, no further unbalanced moment arises and the process ends after one cycle.

• Joint B: FEM: BA +97.96, BC -57.60, BD 0; balance: -13.02, -9.11, -18.23.
• Far ends: FEM: AB -73.47, CB +38.40, DB 0; carry-over: -6.51, -4.56, -9.11.

Final end moments (kN·m): M_AB = -79.98, M_BA = +84.94, M_BC = -66.71, M_CB = +33.84, M_BD = -18.23, M_DB = -9.11. Check at B: 84.94 - 66.71 - 18.23 = 0.

Shear forces and reactions (from statics of each member with its end moments):

• AB: V_A = (100 × 3 - 84.94 + 79.98)/7 = 42.15 kN; V_B = 57.85 kN. Moment under the 100 kN load = 42.15 × 4 - 79.98 = 88.61 kN·m (sagging).
• BC: V_B = (80 × 3 - 33.84 + 66.71)/5 = 54.57 kN; V_C = 25.43 kN. Moment under the 80 kN load = 54.57 × 2 - 66.71 = 42.43 kN·m (sagging).
• BD: horizontal shear = (18.23 + 9.11)/5 = 5.47 kN; the moment changes sign at 18.23/27.34 × 5 = 3.33 m from B.
• Vertical reaction at D = 57.85 + 54.57 = 112.43 kN (axial force in the column); the totals check: 42.15 + 25.43 + 112.43 = 180 kN.

Bending moment diagram (kN·m): AB: -79.98 at A, +88.61 under the load, -84.94 at B; BC: -66.71 at B, +42.43 under the load, -33.84 at C; BD: 18.23 at B (right face in tension), zero at 3.33 m, 9.11 at D (left face in tension).

Shear force diagram (kN): AB: +42.15 from A to the load, -57.85 from the load to B; BC: +54.57 from B to the load, -25.43 to C; BD: constant 5.47.