(a) In Figure 2(a), AB is a rigid bar, CE is a steel rod of 20 mm diameter and DF is a copper rod of 20 mm diameter. If the end A…

(a) Stresses in the steel and copper rods and the reaction at A

Given: AC = 1 m, CD = 2 m and DB = 1 m, so C is 1 m, D is 3 m and B is 4 m from the hinge A. The steel rod CE (1 m) hangs from a support above the bar and the copper rod DF (1 m) stands on a support below it. Both have d = 20 mm, so A = (π/4)(20)² = 314.16 mm². E_s = 200 GN/m², E_c = 100 GN/m², and the load at B is 30 kN.

Nature of the forces: the load turns the rigid bar clockwise about A, so C and D both move down. C moving down stretches CE, so the steel rod is in tension. D moving down shortens DF, so the copper rod is in compression. Both rods pull or push the bar upward, P_s at C and P_c at D.

Equilibrium (moments about A): P_s × 1 + P_c × 3 = 30 × 4, so P_s + 3P_c = 120 … (1)

Compatibility: the bar is rigid and turns through a small angle θ, so δ_C = θ and δ_D = 3θ. Hence δ_c = 3δ_s, i.e. P_cL/(AE_c) = 3P_sL/(AE_s). L and A are equal, so P_c = 3(E_c/E_s)P_s = 1.5P_s … (2)

Solving (1) and (2): P_s + 4.5P_s = 120, so P_s = 21.82 kN and P_c = 32.73 kN.

Stresses: σ_s = 21 818/314.16 = 69.4 MN/m² (tensile) σ_c = 32 727/314.16 = 104.2 MN/m² (compressive)

Reaction at A (ΣF_y = 0, upward positive): R_A + 21.82 + 32.73 − 30 = 0, so R_A = −24.55 kN. The reaction at A is 24.55 kN acting downward, because the two rods together supply more upward force than the 30 kN load.

Check: δ_s = 21 818 × 1000/(314.16 × 2 × 10⁵) = 0.347 mm and δ_c = 32 727 × 1000/(314.16 × 1 × 10⁵) = 1.042 mm. Their ratio is 3, so compatibility holds, and B deflects by 4 × 0.347 = 1.39 mm.

(b) Porter governor: speed after lift, effort and power

Data: both arms l = 350 mm; the upper arms are pivoted on the axis and the lower arms on the sleeve at e = 25 mm from the axis; sleeve weight Mg = 250 N, ball weight mg = 25 N; r₁ = 125 mm at N₁ = 225 rpm. Friction is neglected.

Initial geometry: Upper-arm height h₁ = √(350² − 125²) = 326.92 mm, so tan α₁ = 125/326.92 = 0.3824. Lower-arm projection = √(350² − 100²) = 335.41 mm, so tan β₁ = 100/335.41 = 0.2981. q₁ = tan β/tan α = 0.780. Top pivot to sleeve = 326.92 + 335.41 = 662.33 mm.

Porter relation (instantaneous-centre method): mω²r = tan α [mg + (Mg/2)(1 + q)], which gives N² ∝ [m + (M/2)(1 + q)]/h.

(i) Speed after a 55 mm lift. The sleeve is now 662.33 − 55 = 607.33 mm below the top pivot. Solving √(350² − r²) + √(350² − (r − 25)²) = 607.33 gives r₂ = 185.9 mm, with h₂ = 296.52 mm and q₂ = 0.826. Then N₂²/N₁² = (h₁/h₂) × [25 + 125(1.826)]/[25 + 125(1.780)] = 1.1282, so N₂/N₁ = 1.0621. The ratio depends only on the geometry and loads, so with the given N₁ = 225 rpm, N₂ = 239 rpm.

(ii) Effort. The fractional speed change for the full lift is c = 1.0621 − 1 = 0.0621. With the sleeve held, the force needed at the sleeve rises to about 2c[Mg + 2mg/(1 + q)], and the effort is its mean: P = c[Mg + 2mg/(1 + q)] = 0.0621 × [250 + 50/1.780] = 0.0621 × 278.1 = 17.3 N. For equal arms (q = 1) this reduces to the familiar P = c(m + M)g.

(iii) Power of the governor is the work done at the sleeve: Power = P × lift = 17.3 × 0.055 = 0.95 N·m (J).

(c) Ferrous alloys: reasons for wide use and limiting characteristics

Ferrous alloys (steels and cast irons) are used in far greater tonnage than all other engineering metals together.

Reasons for extensive use:

1. Abundance. Iron ores are plentiful in the earth's crust, so raw material is widely available.
2. Economical production. Blast furnaces, basic oxygen and electric-arc steelmaking, rolling, casting and welding are cheap, mature processes, giving a very low cost per unit of strength.
3. Versatility. Carbon and alloy content and heat treatment tune properties over a wide range: ductile mild steel for car bodies, quenched-and-tempered steels for shafts and gears, stainless steels for corrosion resistance, and damping grey cast iron for machine beds.

Characteristics that limit their use:

1. High density (about 7.8 g/cm³), which penalises weight-critical uses; aircraft structures therefore favour aluminium and titanium alloys and composites.
2. Low electrical conductivity compared with copper and aluminium, so they are not chosen as conductors.
3. Susceptibility to corrosion (rusting) in ordinary atmospheres, which demands painting, galvanising or costlier stainless grades.

Ferrous alloys therefore dominate structural and machine uses where cost and strength decide, while non-ferrous metals take over where low weight, conductivity or corrosion resistance governs.