(a) Two involute gears in mesh have pressure angle of 20° and module of 6 mm. The number of teeth on pinion is 28 and the larger…

(a) Contact ratio, angles of action and sliding-to-rolling velocity ratio

Data: m = 6 mm, φ = 20°, z₁ = 28 (pinion, driver), z₂ = 52, addendum = m on both.

Pitch radii: r₁ = mz₁/2 = 84 mm, r₂ = mz₂/2 = 156 mm. Addendum radii: r_a1 = 90 mm, r_a2 = 162 mm. Base radii: r_b1 = 84 cos20° = 78.934 mm, r_b2 = 156 cos20° = 146.592 mm.

Path of approach = √(r_a2² − r_b2²) − r₂ sinφ = 68.955 − 53.355 = 15.60 mm. Path of recess = √(r_a1² − r_b1²) − r₁ sinφ = 43.237 − 28.730 = 14.51 mm. Path of contact L = 15.60 + 14.51 = 30.11 mm.

(i) Contact ratio. Base pitch = πm cosφ = π × 6 × 0.93969 = 17.713 mm. Contact ratio = L/base pitch = 30.11/17.713 = 1.70, so on average 1.7 pairs of teeth are in contact.

(ii) Angle of action is the angle turned by each wheel while one pair is in contact, θ = L/r_b. Pinion: 30.11/78.934 = 0.3814 rad = 21.85°. Gear wheel: 30.11/146.592 = 0.2054 rad = 11.77°.

(iii) Sliding to rolling velocity. At a point at distance x from the pitch point along the line of action, sliding velocity = (ω₁ + ω₂)x and rolling (pitch-line) velocity = ω₁r₁ = ω₂r₂, so v_s/v_r = x(1/r₁ + 1/r₂) = x(1/84 + 1/156) = x × 0.014652 mm⁻¹. (I) Start of contact, x = 15.60 mm: v_s/v_r = 0.286. (II) Pitch point, x = 0: v_s/v_r = 0 (pure rolling). (III) End of contact, x = 14.51 mm: v_s/v_r = 0.266. With the pinion as driver, contact starts at the gear addendum circle; if the gear drives, the first and third values are interchanged.

(b) Minimum diameter of a shaft under bending and twisting

Data: M = 12 kNm = 12 × 10⁶ N·mm, T = 16 × 10⁶ N·mm, σ_y = 300 N/mm², FOS = 2. For a solid shaft σ_b = 32M/(πd³) and τ = 16T/(πd³). √(M² + T²) = √(12² + 16²) × 10⁶ = 20 × 10⁶ N·mm.

(i) Maximum principal stress theory. Allowable σ₁ = 300/2 = 150 N/mm². σ₁ = σ_b/2 + √((σ_b/2)² + τ²) = 16[M + √(M² + T²)]/(πd³) = 16 × 32 × 10⁶/(πd³) = 512 × 10⁶/(πd³). Setting σ₁ = 150: d³ = 512 × 10⁶/(π × 150) = 1.0865 × 10⁶ mm³, so d = 102.8 mm.

(ii) Maximum shear stress theory. Allowable τ_max = σ_y/(2 × FOS) = 75 N/mm². τ_max = √((σ_b/2)² + τ²) = 16√(M² + T²)/(πd³) = 320 × 10⁶/(πd³). Setting τ_max = 75: d³ = 320 × 10⁶/(π × 75) = 1.3581 × 10⁶ mm³, so d = 110.7 mm.

The shear stress theory is the more conservative one and gives the larger diameter, so a shaft of about 111 mm satisfies both.

(c) Short notes on ceramics and nano-materials

Ceramics: inorganic, non-metallic solids, usually oxides, carbides, nitrides or silicates, held by strong ionic and covalent bonds. They have high hardness, high compressive strength, high melting point, chemical inertness and electrical insulation, but are brittle, weak in tension and poor in thermal shock. Examples are alumina, silicon carbide, zirconia and porcelain, used for refractory linings, cutting tools, spark plugs, insulators and bio-implants. They are shaped by powder compaction, slip casting and sintering.

Nano-materials: materials with at least one dimension of 1–100 nm. The very high surface-area-to-volume ratio, quantum confinement and large grain-boundary fraction make their strength, catalytic, optical, electrical and magnetic behaviour differ from the bulk. Examples are nanoparticles, carbon nanotubes, graphene, quantum dots and nanocomposites, used in drug delivery, sensors, catalysts, coatings, electronics and energy storage. Concerns are toxicity, agglomeration and safe handling.