(a) (i) The recoil mechanism of a gun consists of a critically damped spring-damper system. The maximum permissible recoil…

(a)(i) Use the critically damped single-degree-of-freedom solution. The equation of motion is m d²x/dt² + c dx/dt + kx = 0. Critical damping makes the characteristic roots equal, so c = 2√(mk) and the repeated root is r = -ω, where ω = √(k/m). With zero initial displacement and initial recoil velocity v₀ = 10 m/s, x(t) = (A + B t)e^(-ωt). The initial conditions give A = 0 and B = v₀, so x(t) = 10 t e^(-ωt). Differentiate: dx/dt = 10 e^(-ωt)(1 - ωt). Maximum recoil occurs when dx/dt = 0, so t = 1/ω. Substituting gives x_max = 10/(ω e). Given x_max = 0.5 m, ω = 10/(0.5 e) = 20/e s⁻¹. Hence k = mω² = 500(20/e)² = 200000/e² N/m ≈ 2.71×10⁴ N/m, and c = 2mω = 2(500)(20/e) = 20000/e N s/m ≈ 7.36×10³ N s/m. Check: x(0)=0, v(0)=10 m/s, and the peak is exactly 0.5 m. The spring stiffness is the minimum value that keeps the peak at 0.5 m. Valid for linear spring and damper, zero initial displacement, and exact critical damping.

(a)(ii) Use specific shearing energy and the flywheel energy-fluctuation relation. Sheared area per hole is Aₛ = π d t = π(20 mm)(25 mm) = 500π mm². Energy per hole is Eₚ = 10 N·m/mm² × 500π mm² = 5000π J. For 10 holes/min, average power is P = 10Eₚ/60 = 2500π/3 W ≈ 2.62 kW. This is mean shaft power; the flywheel supplies the short peak demand during each punch. For the flywheel, C = (ω_max - ω_min)/ω_mean = 0.05. The fluctuation energy is ΔE = ½I(ω_max² - ω_min²). If ω_mean is the arithmetic mean of ω_max and ω_min, this becomes ΔE = Iω_mean²C. With rim speed v = ωR and rim mass m, I = mR², so ΔE = mv²C. Assuming the punch velocity is the given mean flywheel rim speed, because no drive ratio is specified, punching time is tₚ = 0.060 m / 25 m/s = 3/1250 s. Motor energy during punching at average power is Eₘ = P tₚ = (2500π/3)(3/1250) = 2π J. Thus the flywheel must supply ΔE = Eₚ - Eₘ = 4998π J. Required rim mass is m = ΔE/(v²C) = 4998π/(25² × 0.05) = 19992π/125 kg ≈ 502 kg. The result assumes the flywheel mass is concentrated at the rim. If motor energy during the stroke is neglected, m = 160π kg, differing by 0.04%.

[(b)] The figure printed on the original question paper for this part could not be recovered from the scan, so this part is not answered here. (b)(i) The shear force diagram cannot be drawn because the supplied description is a shaft design problem, not the beam AB loading, support conditions, point C, or rigid bar DEF required by the question. (b)(ii) The bending moment diagram cannot be drawn for the same reason; no beam loads, spans, or support reactions are available. (b)(iii) The maximum bending moment and its location cannot be found without the load diagram and support conditions. (b)(iv) The deflection at C cannot be found without the load diagram, the position of C, and the beam geometry. The given E = 200 GPa and I = 20×10⁶ mm⁴ cannot be used without those data. No alternative beam data are used.

(c) Defects in a steel article after hardening and their causes:

• Quench cracks: rapid cooling produces large thermal gradients; martensite formation adds volume expansion and high internal stress. High carbon, high hardenability, sharp corners, thick sections, too aggressive quench, insufficient preheat, or mechanical restraint can initiate cracks.
• Warping or distortion: uneven cooling of asymmetric or unequal sections, non-uniform martensitic transformation, residual stresses, improper support during quenching, or thermal and mechanical stresses can bend or twist the part.
• Soft spots or incomplete hardening: insufficient austenitizing temperature or holding time, slow quench, decarburized surface, alloy segregation, improper quench medium, or local overheating that coarsens the structure can leave regions below required hardness.
• Decarburization: exposure to oxidizing or decarburizing atmosphere at high temperature, overheating, scale formation, or prolonged furnace residence removes surface carbon and reduces surface hardness and fatigue life.
• Oxidation and scaling: high austenitizing temperature, prolonged heating, oxygen or sulfur in the atmosphere, poor furnace control, or inadequate protective atmosphere forms scale and pits.
• Coarse grain growth: overheating or excessive holding time in the austenite range enlarges grains, reducing toughness, hardenability, and fatigue resistance.
• Residual stresses: non-uniform martensitic transformation, prior cold working, improper preheat, inadequate stress relief, or uneven section cooling leaves internal stresses that can cause later cracking or distortion.
• Hardness variation: uneven austenitizing, uneven cooling, varying section thickness, decarburization, alloy segregation, or incomplete transformation gives different hardness in different zones.
• Surface roughness or pits: scale, slag, furnace contamination, improper handling, or quench medium contamination can roughen the surface and create stress raisers.
• Internal cracks: high hardenability, high carbon, restrained cooling, prior inclusions, insufficient preheat, or excessive section thickness can produce subsurface or internal cracking.
• Temper cracks, if tempered after hardening: slow cooling from tempering, high hardenability, impurities, or excessive section thickness can cause cracking during temper.
• Temper embrittlement, if tempered: slow cooling through tempering ranges, impurities such as phosphorus, high hardenability, or excessive tempering time can reduce impact toughness.