(a) A frictionless pulley A as shown in the figure is supported by two bars AB and AC which are hinged at B and C to a vertical…

(a) Forces in bars AB and AC (10 marks)

Take the pulley at A as the free body. The frictionless pulley passes the cable round without changing its tension, so the cable pulls on it with T = 30 kN in both segments, one along AE (vertically downward) and one along AD (towards D). The geometry from the figure gives AB at 30° above the horizontal (towards B), AD at 30° below the horizontal (towards D) and AC at 60° below the horizontal (towards C). Assume both bar forces to be tensile (pulling on A towards B and C).

Cable forces on A: AD segment 30(-cos 30°, -sin 30°) = (-25.98, -15.0) kN, and AE segment (0, -30) kN.

ΣFx = 0: -25.98 - 0.866 F_AB - 0.5 F_AC = 0

ΣFy = 0: -15 - 30 + 0.5 F_AB - 0.866 F_AC = 0

The second equation gives F_AB = 90 + 1.732 F_AC. Substituting into the first, -25.98 - 0.866(90 + 1.732 F_AC) - 0.5 F_AC = 0, so -25.98 - 77.94 - 1.5 F_AC - 0.5 F_AC = 0, and F_AC = -51.96 kN. Then F_AB = 90 - 90 = 0.

Result: F_AC = 51.96 kN (30√3), compression, and F_AB = 0. The resultant of the two cable pulls, 51.96 kN, acts exactly along the line AC towards C, so bar AC alone resists it, and AB is a zero-force member.

(b) Bent cantilever ABCD (10 marks)

Members: AB horizontal 2 m (fixed at A), BC vertical 0.6 m, CD horizontal 0.3 m to the right, with the 12 kN load acting downward at D. The load is 2 + 0.3 = 2.3 m to the right of A.

Reactions at A: H_A = 0, V_A = 12 kN upward, and M_A = 12 × 2.3 = 27.6 kN·m (hogging, tension on the top of AB).

Shear force (kN): CD 12 (constant); BC 0 (the shear in a vertical member is the horizontal force, which is nil); AB 12 (constant). Axial force: BC carries 12 kN in tension; AB and CD carry none.

Bending moment (kN·m, hogging), from the free end: at D, 0; along CD it increases linearly to 12 × 0.3 = 3.6 at C; along BC it is constant at 3.6 (no horizontal force); along AB it increases linearly from 3.6 at B to 3.6 + 12 × 2 = 27.6 at A. The maximum is 27.6 kN·m at the support.

(c) Shape factor of a diamond section (10 marks)

The section is two triangles of base b and height d/2, joined at the mid-depth, with total area A = bd/2.

Elastic section modulus. The neutral axis passes through the centre, at half-depth. Each triangle has a moment of inertia about its base of b(d/2)³/12 = bd³/96, so

I = 2 × bd³/96 = bd³/48, and y_max = d/2, so Z = I/y_max = (bd³/48)/(d/2) = bd²/24.

Plastic section modulus. The plastic neutral axis divides the area into two equal halves; by symmetry it is also the horizontal centre line. Each half is a triangle of area A/2 = bd/4 whose centroid is (d/2)/3 = d/6 from the axis. So

Z_p = (A/2)(ȳ₁ + ȳ₂) = (bd/4)(d/6 + d/6) = bd²/12.

Shape factor S = Z_p/Z = (bd²/12)/(bd²/24) = 2.0.

(d) Loss of prestress in a pretensioned beam (10 marks)

Data: b = 250 mm, D = 300 mm; 12 wires of 6 mm diameter; initial stress 1200 MPa; wire centroid 100 mm above the soffit; E_s = 210 kN/mm², E_c = 35 kN/mm²; creep coefficient 1.6; residual shrinkage strain 3 × 10⁻⁴; relaxation 90 MPa.

Steel area A_p = 12 × π(6)²/4 = 339.3 mm². Initial force P = 1200 × 339.3 = 407.15 kN.

Section: A = 250 × 300 = 75 000 mm², I = 250 × 300³/12 = 5.625 × 10⁸ mm⁴, and the eccentricity e = 150 - 100 = 50 mm. Modular ratio m = 210/35 = 6.

Concrete stress at the level of the steel: f_c = P/A + Pe²/I = 407 150/75 000 + 407 150 × 50²/5.625 × 10⁸ = 5.43 + 1.81 = 7.24 N/mm² (self-weight neglected).

1. Elastic shortening: m f_c = 6 × 7.24 = 43.4 MPa.
2. Shrinkage: ε_sh E_s = 3 × 10⁻⁴ × 210 000 = 63.0 MPa.
3. Creep: φ m f_c = 1.6 × 6 × 7.24 = 69.5 MPa.
4. Relaxation of steel: 90 MPa (given).

Total loss = 43.4 + 63.0 + 69.5 + 90 = 265.9 MPa, which is 265.9/1200 = 22.2% of the initial stress. The effective prestress is about 934 MPa.

(e) Fillet weld of the flange to the web of a plate girder (10 marks)

Section: web 1500 × 15 mm, flanges 650 × 25 mm, factored shear V = 2800 kN.

Moment of inertia. Web: 15 × 1500³/12 = 4.219 × 10⁹ mm⁴. Each flange: area 650 × 25 = 16 250 mm², centroid 750 + 12.5 = 762.5 mm from the axis, so its contribution is 16 250 × 762.5² + 650 × 25³/12 = 9.448 × 10⁹ + 0.0008 × 10⁹ mm⁴. Total I = 4.219 × 10⁹ + 2 × 9.449 × 10⁹ = 2.312 × 10¹⁰ mm⁴.

First moment of one flange about the axis: Q = 16 250 × 762.5 = 1.239 × 10⁷ mm³.

Horizontal shear flow at the flange-web junction: q = VQ/I = 2800 × 10³ × 1.239 × 10⁷ / 2.312 × 10¹⁰ = 1501 N/mm (about 1.50 kN/mm).

Two fillet welds (one on each side of the web) share this, so the force per weld is 1501/2 = 750 N/mm.

Design strength of a fillet weld per unit length (IS 800:2007): P_w = 0.7 s f_u/(√3 γ_mw) = 0.7 s × 410/(1.732 × 1.25) = 132.6 s N/mm, with s the weld size in mm.

Required size: s = 750/132.6 = 5.66 mm.

Limits from IS 800 Table 21 and the maximum size rule: the thicker part joined is the 25 mm flange, so the minimum size is 6 mm (thickness 20-32 mm); the maximum is the thickness of the thinner part, 15 mm.

Provide a 6 mm continuous fillet weld on both sides of the web, shop welded. It satisfies the minimum (6 mm), the maximum (below 15 mm) and the requirement of 5.66 mm; the capacity is 2 × 132.6 × 6 = 1591 N/mm > 1501 N/mm, so it is safe.