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Lathe operations, taper turning and turning calculations

Short answer

Lathe operations include turning, facing, taper turning, threading, drilling. Taper methods: compound rest swivelled to half angle, tailstock offset for long tapers, taper attachment. Numerical: V = 94.2 m/min, MRR = 47.1 cm³/min, power = 1.96 kW, time = 3 min; compound rest 2.86°, tailstock offset 15 mm.

(a) List the main operations performed on a centre lathe and explain any three methods of taper turning. (b) A bar of 60 mm diameter is turned on a lathe at 500 rpm with a feed of 0.2 mm/rev and a depth of cut of 2.5 mm. The specific cutting energy of the material is 2.5 J/mm3^3. Determine the cutting speed, the material removal rate, the cutting power and the time to turn a length of 300 mm. Also, a taper is to be turned from 60 mm to 40 mm diameter over a length of 200 mm; find the compound rest setting and the tailstock offset if the whole bar length is 300 mm.

(a) Lathe operations and taper turning

The centre lathe holds the work between centres or in a chuck and the tool moves along or across the axis. Operations: plain (straight) turning, facing, step turning, chamfering, grooving, parting-off, taper turning, form turning, thread cutting, drilling, boring, reaming, knurling.

Methods of taper turning

  1. Compound rest: the top slide is swivelled to half the taper angle and fed by hand. Suitable for short, steep tapers (angle up to about 45 degree).
  2. Tailstock set-over: the tailstock is shifted sideways so that the work axis is inclined to the bed. Used for long gentle tapers; the centres are mismatched, which spoils the centre holes. Offset is given by S=D−d2⋅LlS = \frac{D-d}{2}\cdot\frac{L}{l} (L = bar length, l = length of taper).
  3. Taper turning attachment: a guide bar at the set angle moves the cross slide automatically while the carriage travels. Gives accurate, long tapers without disturbing the centres.
  4. Form tool: a tool shaped to the taper is plunged in; short tapers only.

(b) Numerical

Cutting speed

V=πDN1000=π×60×5001000=94.2 m/minV = \frac{\pi D N}{1000} = \frac{\pi\times60\times500}{1000} = 94.2\ \text{m/min}

Material removal rate

MRR=f d V=0.2×2.5×94.2×1000=47 124 mm3/min=785 mm3/sMRR = f\,d\,V = 0.2\times 2.5\times 94.2\times1000 = 47\,124\ \text{mm}^3/\text{min} = 785\ \text{mm}^3/\text{s}

Power

P=u×MRR=2.5×785.4=1963 W≈1.96 kWP = u\times MRR = 2.5\times 785.4 = 1963\ \text{W} \approx 1.96\ \text{kW}

Machining time

Tm=LfN=3000.2×500=3.0 minT_m = \frac{L}{fN} = \frac{300}{0.2\times500} = 3.0\ \text{min}

Taper

Half angle for the compound rest:

tan⁡θ=D−d2l=60−402×200=0.05  ⇒  θ=2.86∘\tan\theta = \frac{D-d}{2l} = \frac{60-40}{2\times200} = 0.05\;\Rightarrow\;\theta = 2.86^\circ

Tailstock offset for a 300 mm long bar:

S=D−d2⋅Ll=60−402×300200=15 mmS = \frac{D-d}{2}\cdot\frac{L}{l} = \frac{60-40}{2}\times\frac{300}{200} = 15\ \text{mm}

Answer: V=94.2V = 94.2 m/min, MRR=47.1MRR = 47.1 cm3^3/min, P=1.96P = 1.96 kW, time =3= 3 min; compound rest set at 2.86∘2.86^\circ; tailstock offset =15= 15 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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