(a) (i) A cylinder C weighing 1000 N is supported by two cylinders A and B weighing 500 N each as shown in the figure. Calculate…

(a)(i)

Reading the figure. The dimension arrows in the figure run across each full circle, so 1 m and 2 m are diameters. The radii are r_A = r_B = 0.5 m and r_C = 1.0 m, and the clear width between the walls is 3 m. This is the only consistent reading: two cylinders of 1 m radius sitting against both walls of a 3 m channel would overlap. All surfaces are taken as smooth.

Geometry. A touches the left wall and B the right wall, so their centres are at x = 0.5 m and x = 2.5 m, both at height 0.5 m. By symmetry the centre of C is at x = 1.5 m.

• Centre distance AC = r_A + r_C = 1.5 m, with horizontal projection 1.5 – 0.5 = 1.0 m.
• cos θ = 1.0/1.5 = 0.6667, so θ = 48.19° (line AC to the horizontal) and sin θ = 0.7454.

FBD of cylinder C. C carries its weight of 1000 N, the contact force P from A, and an equal force from B. Each contact force acts along the line of centres, at θ to the horizontal. ΣF_y = 0: 2P sin θ = 1000, so P = 1000/(2 × 0.7454) = 670.8 N ΣF_x = 0 is satisfied by symmetry.

FBD of cylinder A. A carries its weight of 500 N, the force P from C pressing down-left along CA, the wall reaction R (horizontal), and the floor reaction Q (vertical). ΣF_x = 0: R = P cos θ = 670.8 × 0.6667 = 447.2 N ΣF_y = 0: Q = 500 + P sin θ = 500 + 500 = 1000 N

Result: P = 670.8 N, Q = 1000 N and R = 447.2 N. As a check, the two floor reactions total 2000 N, which equals the total weight.

(a)(ii)

Given: σx = +80 MPa, σy = –40 MPa, τxy = 25 MPa, Syt = 250 MPa.

Principal stresses: σ₁,₂ = (σx + σy)/2 ± √[((σx – σy)/2)² + τxy²] = 20 ± √(60² + 25²) = 20 ± 65

σ₁ = 85 MPa (tensile) and σ₂ = –45 MPa (compressive). With σ₃ = 0, the principal stresses have opposite signs.

(I) Maximum shear stress (Tresca) theory Since σ₁ and σ₂ have opposite signs, τ_max = (σ₁ – σ₂)/2 = (85 + 45)/2 = 65 MPa. Yielding occurs when τ_max = Syt/2 = 125 MPa. FoS = 125/65 = 1.92

(II) Maximum distortion energy (von Mises) theory σ_e = √(σ₁² – σ₁σ₂ + σ₂²) = √(85² + 85 × 45 + 45²) = √13 075 = 114.35 MPa FoS = Syt/σ_e = 250/114.35 = 2.19

Interpretation. Tresca is the more conservative criterion, about 12% lower here. Distortion energy agrees better with tests on ductile metals, so 2.19 is the realistic margin and 1.92 is the safe design value.

(b)

Given: m = 7 kg (mg = 68.67 N) and M = 140 kg (Mg = 1373.4 N). Upper and lower arms are 180 mm, the lower-arm extension JE is 90 mm, the pivots are 25 mm from the axis, α = 40° and r = 190 mm.

Geometry (origin at the sleeve pivot level; heights measured upward):

• Knee J (upper and lower arms): radius 25 + 180 sin 40° = 140.70 mm, height 180 cos 40° = 137.89 mm. The upper pivot O is 275.78 mm above the sleeve pivot D.
• Ball E, on the 90 mm extension at r = 190 mm: radial offset 190 – 140.70 = 49.30 mm, so the rise above J is √(90² – 49.30²) = 75.30 mm. E is therefore 213.19 mm above D, and the extension makes 33.2° with the axis.
• Instantaneous centre I of the lower link DJE: D moves vertically and J moves perpendicular to OJ. I therefore lies where OJ produced meets the horizontal through D, at radius 25 + 2 × 115.70 = 256.40 mm.
• Lever arms about I: ID = 231.40 mm; height of E above I = 213.19 mm; horizontal distance from I to E = 256.40 – 190 = 66.40 mm.

Equilibrium of the lower link with its ball, moments about I. The upper arm is a two-force member and its force passes through I, so it drops out: F_c × 213.19 = mg × 66.40 + (S/2) × 231.40, where S = Mg without friction.

Equilibrium speed (no friction): F_c = (68.67 × 66.40 + 686.7 × 231.40)/213.19 = 766.77 N ω = √(F_c/(m r)) = √(766.77/(7 × 0.19)) = 24.01 rad/s N = 60ω/2π = 229.3 rpm

With sleeve friction f = 40 N. Friction opposes sleeve motion, so the effective sleeve load is Mg ± f.

• Rising (S = 1413.4 N): F_c = 788.48 N, ω = 24.35 rad/s, N₂ = 232.5 rpm
• Falling (S = 1333.4 N): F_c = 745.06 N, ω = 23.67 rad/s, N₁ = 226.0 rpm

Coefficient of insensitiveness = (N₂ – N₁)/N = (232.51 – 226.02)/229.29 = 0.0283 (2.83%)

Range of insensitiveness: between 226.0 and 232.5 rpm, a band of about 6.5 rpm. Within this band the sleeve does not move. The heavy central load keeps the insensitiveness small, which suits a Proell governor intended for close speed control.

(c)

A nanomaterial has at least one external dimension, or an internal structural unit, in the range of about 1–100 nm. In the standard scheme, nanomaterials are classified by how many dimensions lie outside the nanoscale, which is the number of directions in which electrons are free to move.

• 0-D: Dimensions in the nano range — All three (x, y and z each under 100 nm); Geometry — Spheres, clusters, cubes (dot-like); Examples — Metal nanoparticles, quantum dots, fullerene C₆₀
• 1-D: Dimensions in the nano range — Two; the length is large; Geometry — Wires, rods, tubes, fibres; Examples — Nanowires, nanorods, carbon nanotubes
• 2-D: Dimensions in the nano range — One (thickness); Geometry — Sheets, plates, films, layers; Examples — Graphene, MoS₂ nanosheets, thin films and coatings
• 3-D: Dimensions in the nano range — None as a whole, but built of nanoscale units; Geometry — Bulk solids made of nano-grains, nano-layers or dispersed nano-phases; Examples — Nanocrystalline metals, nanocomposites, bundles of nanowires or nanotubes

Significance. Confinement in more directions changes the electronic density of states, with quantum dots showing discrete levels, and raises the surface-to-volume ratio. This is why size-dependent optical, electrical and mechanical properties are strongest in 0-D and 1-D forms.