(a) A rod shown in the figure below is subjected to a force of 95 kN. Determine the diameter d of the portion ②, if the stress…

(a) Stepped rod: diameter of portion ② and total elongation

Internal force. The rod is pulled by 95 kN at each end. A section cut through any portion must carry the same tension, so P = 95 kN in all four portions ① to ④.

Diameter of portion ②.

A₂ = P/σ_allow = 95 × 10³/115 = 826.1 mm²

d = √(4 × 826.1/π) = 32.4 mm (adopt 33 mm in practice)

Areas.

• A₁ = A₄ = π(50)²/4 = 1963.5 mm²
• A₂ = 826.1 mm²
• A₃ = π(85)²/4 = 5674.5 mm²

Deformation. δ = (P/E) Σ(Lᵢ/Aᵢ), with E = 205 × 10³ N/mm².

• Portion ①: L = 100 mm, A = 1963.5 mm², δ₁ = 0.0236 mm
• Portion ②: L = 50 mm, A = 826.1 mm², δ₂ = 0.0280 mm
• Portion ③: L = 100 mm, A = 5674.5 mm², δ₃ = 0.0082 mm
• Portion ④: L = 50 mm, A = 1963.5 mm², δ₄ = 0.0118 mm

Σ(L/A) = 0.1545 mm⁻¹, so δ = (95 × 10³/205 × 10³) × 0.1545 = 0.0716 mm (elongation)

If d = 33 mm is adopted, δ = 0.0707 mm. The slender portion ② contributes the largest share of the extension, even though it is the shortest.

(b) Moment of resistance: 250 × 500 beam, 3–16φ, M20/Fe500 (IS 456 LSM)

Effective depth.

d = 500 − 30 (cover) − 8 (stirrup) − 16/2 = 454 mm

Steel area.

A_st = 3 × π(16)²/4 = 603.2 mm²

This exceeds the minimum 0.85bd/f_y = 193 mm² and is below the maximum 0.04bD.

Neutral axis depth. From force equilibrium, 0.36 f_ck b x_u = 0.87 f_y A_st:

x_u = (0.87 × 500 × 603.2)/(0.36 × 20 × 250) = 145.8 mm

The limiting value for Fe500 is x_u,max = 0.46d = 208.8 mm. Since x_u is less than x_u,max, the section is under-reinforced and the steel yields first.

Moment of resistance.

M_u = 0.87 f_y A_st (d − 0.416 x_u) = 0.87 × 500 × 603.2 × (454 − 0.416 × 145.8)

M_u = 262.4 kN × 0.393 m = 103.2 kN·m

IS 456 Annex G gives the same result (103.3 kN·m): M_u = 0.87 f_y A_st d[1 − A_st f_y/(b d f_ck)].

Check. M_u,lim = 0.133 f_ck b d² = 137.1 kN·m, which is greater than 103.2 kN·m, confirming a ductile, under-reinforced section. The moment of resistance is ≈ 103 kN·m.

(c) Three-hinged parabolic arch under full UDL: zero bending moment

Arch axis. With the origin at the left springing A, L = 60 m and rise h = 10 m:

y = 4h x(L − x)/L² = x(60 − x)/90

Reactions.

• By symmetry, V_A = V_B = wL/2 = 15 × 60/2 = 450 kN.
• The moment about the crown hinge C of the left half is zero:
• V_A(L/2) − w(L/2)²/2 − H·h = 0
• H = wL²/(8h) = 15 × 60²/(8 × 10) = 675 kN

Moment at any section x.

M(x) = V_A x − w x²/2 − H·y

• The first two terms form the "beam moment" of a simply supported span: M_beam = w x(L − x)/2.
• The thrust term is H·y = [wL²/(8h)] × [4h x(L − x)/L²] = w x(L − x)/2.

So M(x) = w x(L − x)/2 − w x(L − x)/2 = 0 for every x.

Numerical check.

• At the quarter point x = 15 m, y = 7.5 m: M = 450 × 15 − 15 × 15²/2 − 675 × 7.5 = 6750 − 1687.5 − 5062.5 = 0.
• At the crown, M = 0 by the hinge condition.

Interpretation. Under a uniform load the funicular (thrust) line is a parabola. Here it coincides with the arch axis, so the arch carries the load by pure axial compression. The thrust rises from H = 675 kN at the crown to √(675² + 450²) = 811 kN at the springings. Shear is also zero, so the arch is efficient and material-economical.

(d) Vertical deflection of C by Castigliano's theorem

Method. The load at C is treated as P (= 25 kN), so δ_C = ∂U/∂P = ∫ M(∂M/∂P) dx/EI.

Reactions. A carries 3 m of span to C, and the 10 kN/m acts over CB = 4 m.

• R_A = (4P + 10 × 4 × 2)/7 = (4P + 80)/7, so ∂R_A/∂P = 4/7 and R_A = 25.71 kN.
• R_B = (3P + 200)/7, so ∂R_B/∂P = 3/7 and R_B = 39.29 kN.

Segment AC (x from A, 0–3 m).

M = R_A x, and ∂M/∂P = 4x/7

∫ M(∂M/∂P) dx = 25.71 × (4/7) × 3³/3 = 132.24 kN·m³

Segment BC (x from B, 0–4 m).

M = R_B x − 10x²/2, and ∂M/∂P = 3x/7

∫ = (3/7)[39.29 × 4³/3 − 5 × 4⁴/4] = (3/7)(838.1 − 320) = 222.04 kN·m³

Total. EI δ_C = 132.24 + 222.04 = 354.3 kN·m³

EI = 210 × 10³ N/mm² × 150 × 10⁶ mm⁴ = 3.15 × 10¹³ N·mm² = 31 500 kN·m²

δ_C = 354.3/31 500 = 0.01125 m = 11.25 mm (downward)

Double integration of the elastic curve gives the same 354.3/EI.

(e) Fillet-weld design: ISA 75×75×8 to 10 mm gusset

Weld size check (IS 800:2007).

• The thicker part is 10 mm, so the minimum size is 3 mm.
• The maximum size at the rounded toe of the angle is ¾ × 8 = 6 mm.
• The 4 mm weld is acceptable.

Weld strength (shop weld, γ_mw = 1.25, f_u = 410 MPa for E250).

• Throat t_t = 0.7 × 4 = 2.8 mm
• f_wd = f_u/(√3 γ_mw) = 410/(1.732 × 1.25) = 189.4 N/mm²
• Strength per mm length: q = 2.8 × 189.4 = 530.2 N/mm

Design force. Use the larger of the tension and compression forces, P = 100 kN. A weld designed for 100 kN is automatically safe for the 90 kN compression, since fillet welds resist either sense of load in shear.

Balancing the welds about the centroid. The member's line of action passes through the centroid, C_z = 21.4 mm from the heel. Taking moments about each weld line:

• Heel (back) weld: P₁ = 100 × (75 − 21.4)/75 = 71.5 kN
• Toe weld: P₂ = 100 × 21.4/75 = 28.5 kN

Lengths.

• Heel weld effective length: L₁ = 71 467/530.2 = 134.8 mm. Adding 2s = 8 mm for end craters gives 142.8 mm, so provide 145 mm.
• Toe weld effective length: L₂ = 28 533/530.2 = 53.8 mm. With 8 mm added this is 61.8 mm, so provide 65 mm.
• Both effective lengths exceed the minimum of 4s = 16 mm.

Provide a 4 mm shop fillet weld, 145 mm along the heel and 65 mm along the toe of the connected leg. The effective capacity is (137 + 57) × 530.2 = 103 kN, which is at least 100 kN. The connection is therefore safe for both the 100 kN tension and the 90 kN compression.