Solved Past Questions

A. Theory Answers (by chapter) concise

Ch 1 — Design methods & sections

• WSM vs LSM / why LSM is better: WSM is deterministic, governed by elastic theory with a single factor of safety on the material and does not give a realistic factor of safety; LSM is probabilistic, uses separate partial safety factors for loads (γ_f) and materials (γ_m), checks both strength and serviceability, and gives more economical sections. (See Ch 1 §1.5–1.6.)
• Characteristic strength/load: strength below which ≤ 5% of results fall; load with 95% probability of not being exceeded. Design: f_d = f/γ_m, F_d = Fγ_f.
• Balanced / under- / over-reinforced: defined by x_u vs x_u,max (= 0.53d/0.48d/0.46d for Fe250/415/500). Under-reinforced = ductile (preferred); over-reinforced = brittle (avoided). (Ch 1 §1.8, Ch 2 §2.4.)
• Idealized stress-strain / modular ratio: m = 280/(3σ_cbc). (Ch 1 §1.7.)
• Mechanical splices: parallel-thread couplers, taper-cut-thread couplers, cold-swaged sleeve with threaded end.

Ch 5/6 — Shear, torsion, bond, serviceability

• Behaviour of concrete under shear / bent-up bars: diagonal tension cracks form; stirrups and bent-up bars carry the shear beyond τ_c. (Ch 2 §2.9.)
• Anchorage vs flexural bond: anchorage bond resists pull-out along the embedded length; flexural (local) bond arises from the rate of change of bar force along the span. Code check L_d ≤ 1.3M₁/V_u + L_o. (Ch 2 §2.11.)
• Deflection & crack control: limit span/depth ratio (deflection); limit bar spacing/diameter and provide minimum steel (cracking); IS 456 surface crack-width formula. (Ch 5, IS 456 Annex.)
• Splicing: required to join bars where lengths are limited; lap, welded or mechanical; stagger laps, avoid at high-stress zones.

Ch 3/4/7–11 — Columns, footings, slabs, detailing, EQ

• Interaction diagram / short vs long column: see Ch 3 §3.4, §3.7.
• Mat foundation / why no shear steel in footings: footings are kept thick enough that concrete alone carries the shear (economical, avoids congestion). (Ch 4.)
• Ductility & ductile detailing (beam/column/joint): ability to deform inelastically without collapse; IS 13920 confining hoops, strong-column/weak-beam, joint confinement. (Ch 6 §6.3, §6.5.)
• Philosophy of EQ design: no damage in minor, repairable in moderate, no collapse in major shaking. (Ch 6 §6.2.)

B1. WSM — Moment of Resistance 2071 Shrawan, 2073 Shrawan, 2075 Chaitra, 2081 Bhadra

Beam 250 mm × 300 mm, 3−25 mm φ tension bars; σ_cbc = 7, σ_st = 140 N/mm².

1. A_st = 3×(π/4)×25² = 1471.9 mm²; m = 280/(3×7) = 13.33.
2. Critical NA: σ_cbc/(σ_st/m) = X_c/(d−X_c) ⇒ X_c = 120 mm.
3. Actual NA (b·x·x/2 = m·A_st(d−x)): 125x² + 19620x − 5.886×10⁶ = 0 ⇒ x = 152.3 mm.
4. x > X_c ⇒ over-reinforced. MOR = b·x·(σ_cbc/2)(d−x/3) = 250×152.3×3.5×(300−50.8) = 33.21 kN-m.

B2. WSM — Design of a Beam 2071 Chaitra

Design a 300 × 450 mm section carrying 30 kN/m over 3.6 m; M20, mild steel.

1. d = 450 − 50 = 400 mm; σ_cbc = 7, σ_st = 140 N/mm².
2. M = wL²/8 = 30×3.6²/8 = 48.6 kN-m.
3. ½σ_cbcb·x(d−x/3) = M ⇒ 350x² − 420000x + 48.6×10⁶ = 0 ⇒ x = 129.74 mm.
4. A_st = M/[σ_st(d−x/3)] = 48.6×10⁶/(140×356.75) = 973 mm² ⇒ 4−20 mm φ bars.

B3. LSM — MOR of a Singly-Reinforced Beam 2073 Shrawan, 2074/75/81

b = 250, d = 360 mm, M20, Fe415, 3−16 mm φ bars.

1. A_st = 602.88 mm²; x_u = 0.87f_yA_st/(0.36f_ckb) = 120.93 mm.
2. x_u,max = 0.48×360 = 172.8 mm ⇒ x_u < x_u,max (under-reinforced).
3. M_u = 0.87f_yA_std[1 − A_stf_y/(bd f_ck)] = 67.47 kN-m.

B4. LSM — Design of a Beam in Flexure (Doubly-Reinforced + Deflection) 2072 Chaitra

SS beam, effective span 5.5 m, overall 230 × 550 mm, superimposed load 50 kN/m (excl. self-weight); design for LS collapse in flexure and check deflection. Fe415, mild exposure, effective cover 50 mm. (worked exactly as in the manual)

1. d = 550 − 50 = 500 mm; assume M20. Self-weight = 25×0.23×0.55 = 3.16 kN/m → total = 53.16 kN/m → w_u = 1.5×53.16 = 79.74 kN/m.
2. M_u = w_ul²/8 = 79.74×5.5²/8 = 301.53 kN-m.
3. x_u,max = 0.48×500 = 240 mm; M_u,lim = 0.36f_ckb·x_u,max(d−0.42x_u,max) = 158.66 kN-m. M_u > M_u,lim ⇒ doubly-reinforced.
4. A_st1 = M_u,lim/[0.87f_y(d−0.42x_u,max)] = 1100.79 mm²; M_u2 = 142.87 kN-m; A_st2 = M_u2/[0.87f_y(d−d′)] = 879.36 mm². A_st = 1980.15 mm² → 5−25 mm φ (2453 mm²).
5. Compression steel: ε_sc = 0.0035(x_u,max−d′)/x_u,max = 0.00276 → f_sc = 351.8 (SP-16), f_cc = 0.446f_ck = 8.92. A_sc = M_u2/[(f_sc−f_cc)(d−d′)] = 925.96 mm² → 3−20 mm φ (942 mm²).
6. Deflection: f_s = 0.58f_y(A_st,req/A_st,prov) = 194.3 N/mm²; p_t = 2.13 → k₂ = 0.98; p_c = 0.82 → k₃ = 1.21. (L/d)_max = 1×0.98×1.21×20 = 23.7 > (L/d)_prov = 5500/500 = 11 ⇒ safe in deflection.

B5. Shear Reinforcement Design 2075 Chaitra, 2078/79/81

Beam 300 × 500 mm (d = 450), V_u = 200 kN, M20, Fe415, A_st = 4−20 mm.

1. τ_v = V_u/(bd) = 200000/(300×450) = 1.48 N/mm².
2. p_t = 100×1257/(300×450) = 0.93% ⇒ τ_c ≈ 0.60 N/mm² (IS 456 Table 19); τ_c,max = 2.8 > τ_v OK.
3. τ_c < τ_v ⇒ design stirrups. V_us = V_u − τ_cbd = 200000 − 81000 = 119000 N.
4. 2-legged 8 mm stirrups: A_sv = 100.5 mm². s_v = 0.87f_yA_svd/V_us = 361×100.5×450/119000 = 137 mm.
5. Provide 8 mm 2-legged stirrups @ 135 mm c/c (≤ 0.75d = 337 and 300 mm).

Variant — bent-up bars (2070 Chaitra, worked as in the manual): beam d = 450, b = 250, V_u = 125 kN, M20, 4−20 mm TOR with 2 bars bent up at 30°.

1. At the support (2 bars remain): A_st = 2×(π/4)×20² = 628 mm²; p_t = 0.56% → τ_c = 0.50 N/mm²; τ_v = 125000/(250×450) = 1.11 N/mm².
2. Shear resistance of the two bent-up bars = 2×0.87f_yA_svsin30° = 2×113.37 = 226.7 kN.
3. V′_us = V_u − τ_cbd = 125 − 56.25 = 68.75 kN. IS 456 caps the bent-up contribution at 50% ⇒ bent-up bars carry 34.375 kN and stirrups carry the other 34.375 kN.
4. Stirrups: s_v = 0.87f_yA_svd/V_us = 474.9 mm, but ≤ 0.75d = 337.5 and 300 ⇒ provide 8 mm 2-legged @ 300 mm c/c.

B6. Development Length 2076/79/81/82 (derivation)

L_d = φ·σ_s/(4τ_bd). For a 20 mm Fe415 bar in M20:

1. σ_s = 0.87×415 = 361 N/mm²; τ_bd = 1.2 (M20) × 1.6 (deformed) = 1.92 N/mm².
2. L_d = 20×361/(4×1.92) = 940 mm ≈ 47φ.
3. At a simply supported end also check L_d ≤ 1.3M₁/V_u + L_o.

B7. Column with Biaxial Bending 2072/74/75/76/78 …

Method (SP-16): given P_u, M_ux, M_uy, size b×D, grades.

1. Assume steel p = 100A_sc/(bD) (start ~1–2%); pick d′/D.
2. From SP-16 charts read the uniaxial capacities M_ux1, M_uy1 for the given P_u/(f_ckbD).
3. Compute P_uz = 0.45f_ckA_c + 0.75f_yA_sc → α_n (1.0 at P_u/P_uz = 0.2, 2.0 at 0.8).
4. Check (M_ux/M_ux1)^α_n + (M_uy/M_uy1)^α_n ≤ 1.0. Increase steel and repeat until satisfied; detail bars and ties. (Ch 3 §3.6.)

B8. Isolated Footing 2073/74/76/78/80/81/82

Column 400 × 400 mm, service P = 1200 kN, SBC = 200 kN/m², M20/Fe415.

1. Area = 1.1×1200/200 = 6.6 m² → 2.6 m × 2.6 m.
2. Net factored pressure p_u = 1.5×1200/6.76 = 266.3 kN/m².
3. Cantilever a = (2600 − 400)/2 = 1100 mm; M_u = p_u·B·a²/2 = 266.3×2.6×1.1²/2 = 418.9 kN-m.
4. Depth for flexure d = √(M_u/(0.138f_ckB)) = 242 mm; punching shear (perimeter d/2, τ_v ≤ 0.25√f_ck = 1.12) governs → take D ≈ 550, d ≈ 500 mm.
5. Find A_st from M_u (per m width), check one-way shear at d and development length. (Ch 4 §4.2.)

B9. Two-Way Slab 2073/75/78/79/80/81/82

Method (IS 456 Annex D): restrained panel l_y/l_x ≤ 2.

1. Assume depth from L/d (short span, basic 26× for continuous × modification); compute w_u = 1.5(DL + LL + FF).
2. Read moment coefficients α_x, α_y (Table 26) for the panel type & aspect ratio.
3. M_x = α_xw_ul_x², M_y = α_yw_ul_x² (mid-span & support).
4. Design steel both directions (short-span bars in the lower layer); provide corner torsion reinforcement = ¾ of mid-span steel over l_x/5 where corners are held down.
5. Check shear and deflection. (Ch 5 §5.3.)