Andalusia PAU Technology and Engineering II study materials
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Andalusia PAU Technology and Engineering II: Manufacturing and Production Systems: Machining, Forming, Joining and Metrology

Andalusia PAU Technology and Engineering II · Work through your exam, one topic at a time.

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Manufacturing and Production Systems: Machining, Forming, Joining and Metrology

This domain covers how a part is made and how it is verified. The PAU paper asks you to compute cutting parameters or machining time, to classify a process (conforming, forming, joining), and to interpret a measuring instrument's reading or a tolerance designation.

Machining

  • Turning produces cylindrical surfaces on a lathe; milling produces flat or profiled surfaces with a rotating cutter; drilling makes holes.
  • Cutting speed v = π·D·n/1000, where v is in m/min, D in mm and n in rpm; it depends on the tool and material, not on the part length.
  • Feed f in mm/rev is the advance per revolution; depth of cut a determines chip thickness.
  • Machining time per pass: t = L/(f·n), with L = machined length + approach allowance.

Forming and casting

  • Rolling, forging, extrusion and drawing are continuous or bulk forming; stamping and deep drawing work sheet metal.
  • Casting pours liquid metal into a mould; sand casting suits complex shapes, die casting suits high volumes of light alloys.
  • Sintering compacts and fuses metal powders for small precision parts.

Joining

  • Welding with a coated electrode, MIG/MAG (consumable wire plus shielding gas) and TIG (non-consumable tungsten electrode) all fuse the base metal.
  • Soldering joins with a low-melting filler without fusing the parts; adhesives and riveting avoid heat entirely.

Metrology

  • Vernier caliper: least count 0.05 or 0.02 mm; micrometer: 0.01 mm; dial gauge for runout.
  • ISO fits on a hole basis, e.g. 40 H7/g6: H7 is the hole tolerance, g6 the shaft, giving a clearance fit; H7/p6 would be an interference fit.
  • Roughness Ra is expressed in µm; a lower Ra means a finer finish and higher cost.

Worked example

A 60 mm diameter shaft 200 mm long is turned at v = 120 m/min with f = 0.2 mm/rev. Rotational speed n = 1000·v/(π·D) = 120 000/(π·60) = 636.6 rpm. Number of revolutions needed = 200/0.2 = 1 000. Time = 1 000/636.6 ≈ 1.57 min, about 94 s. If the diameter were doubled and the cutting speed kept constant, n would halve and the time would double.

Traps that cost marks

  • Using the radius instead of the diameter in v = π·D·n/1000.
  • Confusing mm/min with mm/rev, which changes the result by a factor of n.
  • Forgetting the approach and overtravel length when the question asks for real production time.
  • Reading a caliper to 0.01 mm when its least count is 0.05 mm, or misidentifying which element carries the H tolerance.