(a) Analyse the continuous beam shown in the figure by slope-deflection method. Draw the Shear Force Diagram (SFD) and Bending…

(a) Continuous beam ABC by the slope-deflection method

Boundary conditions. A and C are fixed (θ_A = θ_C = 0), and B is a roller, so the only unknown is θ_B. There is no sinking of supports and EI is constant. End moments are taken clockwise-positive.

Fixed-end moments. For AB (L = 4 m, 4 kN at mid-span): FEM_AB = −PL/8 = −2.0 kN·m and FEM_BA = +2.0 kN·m. Span BC carries no load, so its FEMs are zero.

Slope-deflection equations (M = FEM + (2EI/L)(2θ_near + θ_far)):

• M_AB = −2 + 0.5EIθ_B
• M_BA = +2 + 1.0EIθ_B
• M_BC = (2/6)(2EIθ_B) = 0.667EIθ_B
• M_CB = 0.333EIθ_B

Joint condition at B. M_BA + M_BC = 0 gives 2 + 1.667EIθ_B = 0, so EIθ_B = −1.2 (anticlockwise).

End moments. M_AB = −2.6, M_BA = +0.8, M_BC = −0.8 and M_CB = −0.4 kN·m.

Reactions.

• Span AB: R_A = 2 − (M_AB + M_BA)/4 = 2 + 0.45 = 2.45 kN ↑; R_B(left) = 1.55 kN.
• Span BC: R_B(right) = −(M_BC + M_CB)/6 = +0.20 kN; R_C = 0.20 kN ↓.
• R_B = 1.55 + 0.20 = 1.75 kN ↑. Check: 2.45 + 1.75 − 0.20 = 4 kN ✓.

Bending moments (sagging +).

• A: −2.6 kN·m
• Under the load: −2.6 + 2.45 × 2 = +2.3 kN·m
• B: −0.8 kN·m
• C: +0.4 kN·m

Shear forces. +2.45 kN from A to the load, −1.55 kN from the load to B, and +0.20 kN over BC.

Points of contraflexure.

• −2.6 + 2.45x = 0 gives x = 1.06 m from A.
• 2.3 − 1.55(x − 2) = 0 gives x = 3.48 m from A (0.52 m left of B).
• In BC, −0.8 + 0.2x = 0 gives x = 4.0 m from B (2 m from C).

The hogging moment at A (2.6 kN·m) is the largest moment in the beam. The unloaded span BC bends only through the rotation of B, so C carries half the moment at B, with the opposite sign.

(b) Anchorage of 4–16 φ bars at the simple support

Development length (IS 456 cl. 26.2.1). τ_bd = 1.4 × 1.6 = 2.24 MPa for deformed bars. L_d = φ(0.87f_y)/(4τ_bd) = 16 × 435/(4 × 2.24) = 776.8 mm (≈ 48.5φ)

Moment of resistance with all bars at 0.87f_y. A_st = 4 × 201.1 = 804.2 mm².

• x_u = 0.87 × 500 × 804.2/(0.36 × 25 × 300) = 129.6 mm, which is less than x_u,max = 230 mm.
• M₁ = 0.87 × 500 × 804.2 × (500 − 0.42 × 129.6) = 155.9 kN·m

Check of cl. 26.2.3.3(c). The bar ends are confined by the compressive support reaction, so M₁/V may be increased by 30%: 1.3M₁/V + L₀ ≥ L_d, i.e. 1.3 × 155.9 × 10⁶/(350 × 10³) + L₀ ≥ 776.8, so 579.0 + L₀ ≥ 776.8. Required anchorage beyond the support centre: L₀ ≥ 197.8 mm.

Available anchorage. With 25 mm end cover, a straight bar gives only 150 − 25 = 125 mm beyond the centre, so the bars are bent up at 90°. The bend (internal radius 4φ) is worth 8φ = 128 mm, so L₀ = (125 − 5φ) + 8φ = 173 mm. A 100 mm vertical leg (36 mm beyond the standard 4φ) raises this to L₀ = 209 mm ≥ 197.8 mm ✓.

The bars also extend 275 mm into the support, more than L_d/3 = 259 mm (cl. 26.2.3.3(a)) ✓.

Result. Carry all four bars to 25 mm from the outer face of the support and finish them with a 90° bend and a 100 mm vertical leg (L₀ ≈ 209 mm).

(c) Tension capacity of 2 ISA 75 × 75 × 8 on the same side of the gusset

Each connected leg has six 20 mm bolts in one row (d₀ = 22 mm): gauge 40 mm from the heel and 35 mm to the toe, pitch 50 mm, end distance 35 mm, L_c = 250 mm. Tack bolting makes the two angles act together, so one angle is checked and the result doubled.

Areas per angle.

• A_g = (75 + 75 − 8) × 8 = 1136 mm²
• A_nc = (75 − 4 − 22) × 8 = 392 mm²
• A_go = (75 − 4) × 8 = 568 mm²

1. Gross yielding. T_dg = 2 × 1136 × 250/1.1 = 516.4 kN

2. Net-section rupture with shear lag.

• Shear lag width: b_s = w + w₁ − t = 75 + 40 − 8 = 107 mm.
• β = 1.4 − 0.076(75/8)(250/410)(107/250) = 1.4 − 0.186 = 1.214. This lies between 0.7 and f_uγ_m0/(f_yγ_m1) = 1.443, so it is accepted.
• T_dn per angle = 0.9 × 392 × 410/1.25 + 1.214 × 568 × 250/1.1 = 115.7 + 156.7 = 272.4 kN.
• For two angles, T_dn = 544.9 kN.
• By the simplified formula (six bolts, α = 0.8): T_dn = 0.8 × 2 × (1136 − 176) × 410/1.25 = 503.8 kN.

3. Block shear (per angle).

• A_vg = (35 + 250) × 8 = 2280 mm² and A_vn = (285 − 5.5 × 22) × 8 = 1312 mm².
• A_tg = 35 × 8 = 280 mm² and A_tn = (35 − 11) × 8 = 192 mm².
• T_db1 = 2280 × 250/(√3 × 1.1) + 0.9 × 192 × 410/1.25 = 299.2 + 56.7 = 355.9 kN
• T_db2 = 0.9 × 1312 × 410/(√3 × 1.25) + 280 × 250/1.1 = 223.6 + 63.6 = 287.2 kN
• For two angles, T_db = 574.5 kN.

4. Bolts (grade 4.6 assumed, single shear).

• Shear: V_dsb = (400/√3) × 245/1.25 = 45.3 kN.
• Bearing: k_b = 0.508, so V_dpb = 66.6 kN.
• For 12 bolts, capacity = 543.2 kN. L_j = 250 mm is less than 15d, so no long-joint reduction applies.

Design tensile capacity. The least of the member strengths is T_d = 516.4 kN, governed by gross-section yielding. If the simplified α-formula is used for rupture, rupture governs at 503.8 kN. Tack bolting lets the pair act as one member, and the shear-lag factor β credits the outstanding legs for the rupture strength.