Machining

Machining removes material with cutting tools. Learn turning, milling and drilling, cutting speed/feed/depth, spindle speed, material removal rate, cutting force and power, tool life (Taylor), surface finish and machining time.

Most precision parts — shafts, gears, engine parts, moulds — are finished by machining. A hard tool shears off thin chips as the workpiece or tool rotates.

Choosing cutting speed, feed and depth of cut balances productivity, tool life, surface finish and power.

An analogy

Peeling a carrot with a peeler: the blade angle, how hard you press (depth) and how fast you move (speed/feed) decide how thick the peel is and how smooth the carrot ends up.

Main operations

Turning (lathe): the workpiece rotates, a single-point tool moves along it — shafts, pins. Milling: a rotating multi-tooth cutter moves over the workpiece — flats, slots, pockets. Drilling, boring, reaming and tapping make and finish holes.

Speed, feed and depth

Cutting speed depends on tool and work material (carbide in steel ≈ 150–250 m/min; HSS ≈ 25–35 m/min; aluminium much higher). Feed controls finish and chip thickness; depth of cut controls how much is removed per pass. Roughing uses large depth and feed; finishing uses small feed and higher speed.

Tool life and wear

Tools wear by abrasion, adhesion, diffusion and chipping. Taylor's equation shows speed dominates tool life: with n = 0.25, raising speed 33 % cuts life by ~70 %. Coatings (TiAlN, diamond), coolant and rigid setups extend life.

Mechanics and economics

Chip formation is concentrated shear along a shear plane (Merchant's theory). Forces cause deflection and chatter (self-excited vibration) — stability lobe diagrams find chatter-free speeds. Economic cutting speed minimises cost per part by balancing machining time against tool cost and change time.

Formulas

Worked example: Turning a shaft

Given: Ø50 mm steel, carbide at 150 m/min; f = 0.2 mm/rev, a_p = 2 mm, length 200 mm; k_c = 2000 N/mm²

Find: N, MRR, power and time.

Answer: 955 rpm, 60 cm³/min, 2 kW, ≈ 1.05 min per pass

Worked example: Taylor tool life

Given: V T^0.25 = 400

Find: Tool life at 150 and 200 m/min.

Answer: 51 min vs 16 min

Common mistakes

Applications

Key takeaways

Going further