(a) A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in Figure 1(a). The crank is…

(a) Speed of the slider

r = 60 mm, l = 240 mm, ω = 10 rad/s, ∠AOB = 90°. AB = √(60² + 240²) = 60√17 mm, so cos∠ABO = 240/(60√17) = 4/√17.

The crank-pin velocity v_O = ωr = 10 × 60 = 600 mm/s is perpendicular to AO, i.e. along the rod OB. For a rigid rod the velocity components of O and B along OB are equal: v_B cos∠ABO = v_O ⟹ v_B = 600 × √17/4 = 150√17 mm/s.

Check (analytical): v = rω sinθ[1 + r cosθ/√(l² − r² sin²θ)] with cosθ = 1/√17, sinθ = 4/√17 gives 2400/√17 × 17/16 = 150√17 mm/s.

Speed of slider = 150√17 ≈ 618.5 mm/s ≈ 0.62 m/s (towards A for counterclockwise rotation in the drawn position).

(b) Factor of safety by three failure theories

σ₁ = 160 MPa, σ₂ = 0 (plane stress), σ₃ = −80 MPa, σ_e = 300 MPa.

(i) Maximum principal stress theory: FOS = σ_e/σ₁ = 300/160 = 1.875.

(ii) Maximum shear stress theory: τ_max = (σ₁ − σ₃)/2 = 120 MPa, allowable σ_e/2 = 150 MPa, FOS = 150/120 = 1.25.

(iii) Maximum distortion energy theory: σ_vm = √(σ₁² + σ₃² − σ₁σ₃) = √(160² + 80² + 160 × 80) = √44800 = 211.7 MPa, FOS = 300/211.7 = 1.417.

The maximum principal stress theory suits brittle materials and is unconservative here, while the shear stress theory (most conservative) and distortion energy theory (closest to test data) are used for ductile materials.

(c) Resultant of the pipe force system at A

Take A as origin with the axes of the figure (y upward, x and z as drawn, right-handed). From the figure, the 20 m run is along z, the 8 m run is downward and the 10 m run is parallel to x. At the free end (10, −8, 20) m the 600 N force is parallel to z and the 400 N force parallel to x.

Forces (N) and points of application (m): 900 N: (0, −900, 0) at (0, 0, 10) 200 N: (0, −200, 0) at (0, −4, 20) 300 N: (0, −300, 0) at (5, −8, 20) 500 N: (0, −500, 0), 400 N: (400, 0, 0), 600 N: (0, 0, 600), all at (10, −8, 20)

Resultant force: R = (400, −1900, 600) N, |R| = √(400² + 1900² + 600²) = 2032 N, direction cosines 0.197, −0.935, 0.295 (78.6°, 159.2°, 72.8° with x, y, z).

Moment at A, M = Σ r × F (N·m): 900 N: (9000, 0, 0), 200 N: (4000, 0, 0), 300 N: (6000, 0, −1500), 500 N: (10000, 0, −5000), 400 N: (0, 8000, 3200), 600 N: (−4800, −6000, 0). M_A = (24200, 2000, −3300) N·m, |M_A| ≈ 24.5 kN·m.

Equivalent system at A: R = 400i − 1900j + 600k N (2.03 kN) with couple M_A = 24200i + 2000j − 3300k N·m.

(d) Thermoplastics versus thermosetting plastics

Structure: thermoplastics have linear or branched chains held by secondary (van der Waals) bonds, whereas thermosets cure into a covalently cross-linked three-dimensional network.

Properties: thermoplastics are tougher, more ductile and impact resistant, with lower stiffness, lower heat resistance and creep at moderate temperature. Thermosets are harder, more rigid and brittle, dimensionally stable, with better heat, chemical and electrical resistance.

Response to heating: thermoplastics soften above the glass transition (and melt if semi-crystalline) and harden again on cooling, a reversible process allowing remoulding and recycling. Thermosets, once cured, do not soften on reheating, they char or decompose, so the change is irreversible and they cannot be remelted.

Processing: thermoplastics by injection moulding, extrusion, blow moulding and thermoforming. Thermosets by compression or transfer moulding, casting and resin transfer, with curing by heat or catalyst.

Applications: thermoplastics (polyethylene, polypropylene, PVC, polystyrene, nylon, polycarbonate) in packaging, pipes, bottles, films, gears and housings. Thermosets (Bakelite or phenolics, epoxy, melamine, urea-formaldehyde, polyester) in electrical switches and insulators, adhesives, laminates, composite matrices and cookware handles.

(e) Interference in the gear pair

m = 10 mm, z₁ = 14, z₂ = 54, addendum = m, φ = 20°. r₁ = 70 mm, r₂ = 270 mm, C = 340 mm, gear addendum radius r_a2 = 280 mm, gear base radius r_b2 = r₂ cosφ = 253.7 mm.

For no interference, the gear addendum circle must not go beyond the interference point on the pinion base circle: r_a2(max) = √(r_b2² + C² sin²φ) = √(64372 + 13523) = 279.1 mm. Since 280 mm exceeds 279.1 mm, interference occurs. (Equivalently, the minimum pinion teeth for G = 54/14 at 20° is 2/{G[√(1 + (1/G)(1/G + 2)sin²φ) − 1]} = 15.4, more than 14.)

Required pressure angle: set (r₂ + m)² = r₂² cos²φ + C² sin²φ, giving sin²φ = (2r₂m + m²)/(C² − r₂²) = 5500/42700 ⟹ φ = 21.03°.

The pressure angle must be increased by about 1.03°, to at least 21.03°, to eliminate interference.