(a) Find the maximum load P, which the bracket as shown can transmit. The bolt strength of 20 mm dia and 8·8 grade is assumed to…

(a) Maximum load on the eccentrically loaded bracket

The six bolts form two vertical rows 50 mm either side of the centre line (x = ±50 mm), with bolts at y = −50, 0 and +50 mm. The load P acts 400 mm from the centre line of the group, so e = 400 mm and the twisting moment is M = 400P kN·mm.

• Σr² = Σ(x² + y²) = 6 × 50² + 4 × 50² = 15 000 + 10 000 = 25 000 mm²
• Direct shear per bolt = P/6 = 0.1667P (vertical)
• The farthest bolts are the four corner bolts, r = √(50² + 50²) = 70.7 mm. Of these, the two in the row nearer to P are critical, because there the vertical component of the torsional shear adds to the direct shear.
• Torsional components at a corner bolt: horizontal M·y/Σr² = 400P × 50/25 000 = 0.8P, vertical M·x/Σr² = 400P × 50/25 000 = 0.8P
• Resultant on the critical bolt: R = P√(0.8² + (0.8 + 0.1667)²) = P√(0.640 + 0.934) = 1.255P

Setting R equal to the bolt strength of 82 kN (the lower of shear and bearing, limit state): 1.255P = 82, so P_max ≈ 65.3 kN.

(b) Section modulus for the two-span continuous beam

Use plastic collapse by virtual work. Factored load w_u = 1.75×25 = 43.75 kN/m. The simple supports A and C cannot develop moment, so the governing mechanism has plastic hinges at support B and at a sagging hinge at distance x from the simple support in one span. With the hinge deflection δ, external work = w_u Lδ/2 and internal work = M_p δ(L + x)/[x(L − x)]; minimising w_u gives x = (√2 − 1)L = 0.414L, not midspan. Equating gives M_p = w_u L²/11.66 = 43.75×6²/11.66 = 135.1 kN·m. Required plastic section modulus Z_p = M_p/f_y = 135.1×10⁶ N·mm / 245 N/mm² = 551.5×10³ mm³ = 551.5 cm³. Shape factor SF = Z_p/S, therefore the elastic section modulus is S = Z_p/SF = 551.5/1.144 = 482.1 cm³. Required elastic section modulus S ≈ 482 cm³; corresponding plastic modulus Z_p ≈ 551 cm³.

(c) Design of the cantilever stem

Geometry from the figure. The stem is 250 mm thick. The backfill rises 3 m above the front ground level, which is itself 1 m above the underside of the 400 mm base, so the stem cantilevers h = 3 + 1 − 0.4 = 3.6 m above the top of the base.

Earth pressure (Rankine, horizontal backfill). K_a = (1 − sin 30°)/(1 + sin 30°) = 1/3, so K_a γ = 6 kN/m² per m depth.

• Shear at the base of the stem: V = K_a γ h²/2 = 6 × 3.6²/2 = 38.9 kN/m
• Moment at the base of the stem: M = K_a γ h³/6 = 6 × 3.6³/6 = 46.7 kN·m/m
• Factored (γ_f = 1.5): M_u = 70.0 kN·m/m and V_u = 58.3 kN/m

Main steel (earth face, in tension). With 50 mm cover and 12 mm bars, d = 250 − 50 − 6 = 194 mm.

• M_u/bd² = 70.0 × 10⁶/(1000 × 194²) = 1.86 N/mm²
• From the given table, p_t = 0.455 + 0.6 × (0.484 − 0.455) = 0.472 %
• A_st = 0.472 × 1000 × 194/100 = 916 mm²/m
• Provide 12 mm bars @ 120 mm c/c on the earth face: A_st = 942 mm²/m, p_t = 0.486 %. The spacing is within 3d and 300 mm.
• Check: x_u = 0.87 × 500 × 942/(0.36 × 25 × 1000) = 45.5 mm, well below x_u,max = 0.46d = 89 mm, so the section is under-reinforced.

Shear check.

• τ_v = V_u/bd = 58.3 × 10³/(1000 × 194) = 0.30 N/mm²
• From the given table at p_t = 0.486 %, τ_c = 0.36 + (0.236/0.25) × 0.13 = 0.48 N/mm² (and IS 456 allows a factor k = 1.1 for a 250 mm slab)
• τ_v < τ_c, so no shear reinforcement is needed.

Distribution and front-face steel. Minimum steel 0.12 % of 1000 × 250 = 300 mm²/m. Provide 8 mm bars @ 300 mm c/c horizontally on both faces (335 mm²/m in total), and 8 mm bars @ 300 mm c/c vertically on the front face to hold the horizontal bars and control shrinkage cracking.

Detailing. The 12 mm main bars run the full height on the earth face (no curtailment), anchored into the base with a development length of L_d = 0.87 f_y φ/(4 τ_bd) = 0.87 × 500 × 12/(4 × 1.4 × 1.6) ≈ 583 mm, provided as a standard bend into the base slab. Clear cover is 50 mm on both faces.