General Machining Processes “Chip-forming”

(a) Classification of lathes

• Speed lathe: simple, 3 or 4 speeds, no feed or carriage; for wood turning, polishing, centring.
• Centre or engine lathe: general purpose, power feed, lead screw; most common in workshops.
• Bench lathe: small engine lathe mounted on a bench for light precision work.
• Tool room lathe: precision engine lathe with a wide speed range and many attachments.
• Capstan and turret lathes: hexagonal turret holds many tools for repetitive production of small parts.
• Automatic lathes: loading, feeding and tool change are automatic (cams); for mass production.
• Special-purpose lathes: wheel lathe, gap-bed lathe, T lathe, CNC lathe.

(b) Main parts of an engine lathe

 headstock           tailstock
 +------+  chuck       __ quill
 |gear  |==[O]--job--<|__|
 +------+   tool post      
 =====[ carriage ]============ bed (ways)
  legs   apron   lead screw & feed rod

• Bed: heavy cast iron base with precisely machined ways on which carriage and tailstock slide; supports all parts.
• Headstock: fixed at the left end; houses the spindle, gear box giving different speeds, and carries the chuck or face plate.
• Tailstock: on the right; supports long jobs with a dead centre and holds a drill or reamer; its body can be set over to turn a taper.
• Carriage: slides along the bed; consists of saddle, cross slide, compound rest, tool post and apron.
• Apron: attached to the carriage; holds the gears and clutch for hand and power feed and the half-nut for thread cutting.
• Feed rod and lead screw: the feed rod gives power feed for turning; the lead screw (Acme thread) moves the carriage for thread cutting.
• Cross slide and compound rest: cross slide moves the tool at right angles to the axis (facing, depth of cut); the compound rest swivels for taper turning.
• Tool post: clamps the cutting tool.

Work holding devices

Mounting and setting of tools

• The tool is clamped in the tool post (four-way turret post or quick-change post); the overhang is kept as small as possible for rigidity.
• The cutting tip is set exactly at the height of the job centre line. Methods: tail stock centre, a steel rule and the pointed centre, or trial cut. Above centre reduces clearance; below centre increases rake error and may cause the work to climb.
• Packing (shims) is used under the tool to raise it; the tool must be firmly clamped.
• The tool is set at right angles for turning and facing; for threading, the tool angle is checked with a thread gauge (centre gauge).
• The tool post is set to feed at the required depth by the cross slide dial, whose graduations show the movement (diametral reduction is twice the depth).

(a) Facing

Machining the end of the job flat and perpendicular to its axis. The tool is fed from the outside inward (or from the centre outward) with the cross slide, with the carriage locked.

  chuck |=========|
        |  job    |<-- tool moves across

(b) Straight turning

Reducing the diameter of a cylindrical job. The tool moves parallel to the axis by carriage feed. Roughing uses large depth of cut and feed; finishing uses small depth and feed with a fine nose radius.

(c) Parting off

Cutting a finished piece from the bar using a narrow blade-type parting tool fed radially by the cross slide. The tool is set on the centre line, used with low feed and coolant, and kept slightly over-sized in rigidity to avoid chatter.

(d) Knurling

Producing a diamond or straight cross-pattern on the surface for grip (handles, nuts). A knurling tool with two hardened rollers is pressed into the rotating job at low speed (about 1/3 of turning speed) with oil. The metal is displaced, not cut, so the diameter increases slightly.

(e) Boring

Enlarging an existing hole with a single-point boring bar held in the tool post. The job is rotated in a chuck and the tool feeds axially inside the hole. It gives accurate size and concentricity; the bar should be short and rigid.

Other operations: drilling and reaming (through the tailstock), taper turning, thread cutting, grooving, chamfering.

Taper turning is the production of a conical surface by gradually reducing the diameter along the length of the job.

Methods

• Compound rest swivelled to half the taper angle (short, steep tapers).
• Tailstock set-over (long, gentle tapers between centres).
• Taper attachment (guide bar) on the back of the lathe (accurate, long tapers).
• Form tool (very short tapers) and combination of feeds in a CNC machine.

(a) Tailstock set-over

Given D = 60 mm, d = 40 mm, taper length l = 200 mm, total length L = 300 mm.

$$\begin{aligned} S &= \frac{D-d}{2}\times\frac{L}{l} \\ &= \frac{60-40}{2}\times\frac{300}{200} \\ &= 10 \times 1.5 = 15\ \text{mm} \end{aligned}$$

Here L is the total length of the job between centres. The tailstock is moved sideways by 15 mm (its direction decides which end is larger), and the carriage then feeds parallel to the bed.

(b) Compound-slide angle

Half taper angle $\alpha$:

$$\tan\alpha = \frac{D-d}{2l} = \frac{60-40}{2\times 200} = 0.05$$ $$\alpha = \tan^{-1}(0.05) = 2.86^\circ$$

The compound rest is swivelled to 2.86° from the axis (full included taper 5.72°). Taper = (D - d)/l = 20/200 = 1 in 10.

Answer: (a) tailstock set-over = 15 mm; (b) compound rest angle = 2.86°.

Given: D1 = 50 mm, D2 = 44 mm, length 300 mm, v = 30 m/min, f = 0.3 mm/rev, approach and over-run = 5 mm.

(a) Depth of cut

$$d = \frac{D_1 - D_2}{2} = \frac{50-44}{2} = 3\ \text{mm}$$

(b) Spindle speed

Cutting speed is based on the original diameter.

$$N = \frac{1000\,v}{\pi D_1} = \frac{1000\times 30}{\pi\times 50} = 191.0\ \text{rpm}$$

(c) Machining time

Tool travel $L = 300 + 5 = 305$ mm.

$$T = \frac{L}{f\,N} = \frac{305}{0.3\times 191.0} = 5.32\ \text{min}$$

(d) Metal removal rate

Use the mean diameter, $D_{avg} = (50+44)/2 = 47$ mm.

$$\begin{aligned} \text{MRR} &= \pi D_{avg}\, d\, f\, N \\ &= \pi \times 47 \times 3 \times 0.3 \times 191.0 \\ &= 25{,}380\ \text{mm}^3/\text{min} \end{aligned}$$

(Check: volume removed = $\frac{\pi}{4}(50^2-44^2)\times 300 = 132{,}866$ mm³; divided by the cutting time for 300 mm, 300/(0.3 x 191.0) = 5.24 min, gives 25,370 mm³/min, which agrees.)

Answer: d = 3 mm, N = 191 rpm, T = 5.32 min, MRR = 25.4 cm³/min.

Thread cutting on a lathe

The spindle and the lead screw are connected through a train of change gears so that for each revolution of the job the carriage moves one pitch of the thread.

1. Set the spindle speed low and the change gears (or quick-change box) for the pitch.
2. Fit a 60° V thread tool ground to the thread profile and set it on centre height and square to the job using the thread (centre) gauge.
3. Engage the half-nut on the lead screw at the right moment (use the thread dial for even-number threads).
4. Cut several passes by increasing the depth with the compound slide set at 29° (so the cutting is mainly on the leading flank), withdraw the tool at the end of each pass with the cross slide, and reverse the carriage to start.
5. Check the thread with a gauge or nut.

Change gear calculation

Train ratio is the ratio of job rotation to lead-screw rotation:

$$\frac{N_{spindle}}{N_{leadscrew}} = \frac{p_{lead\ screw}}{p_{thread}} \quad\Rightarrow\quad \frac{\text{Driver}}{\text{Driven}} = \frac{p_{thread}}{p_{lead\ screw}}$$ $$\frac{\text{Driver}}{\text{Driven}} = \frac{1.75}{6} = \frac{7}{24} = 0.2917$$

A simple train 7/24 is not possible (7 teeth too small). Choose a compound train by factorising:

$$\frac{7}{24} = \frac{7}{8}\times\frac{1}{3} = \frac{35}{40}\times\frac{20}{60}$$

So the first pair is gear on the spindle (driver) 35 teeth meshing with a driven gear of 40 teeth; the second pair has a driver of 20 teeth (on the same shaft as the 40) driving a gear of 60 teeth on the lead screw.

Check: $\frac{35}{40}\times\frac{20}{60} = 0.875 \times 0.3333 = 0.2917 = \frac{1.75}{6}$.

 spindle gear 35 --> 40 | 20 --> 60 lead screw gear
          (driver)   (compound)    (driven)

Answer: compound train - drivers 35 and 20 teeth, driven 40 and 60 teeth.

Classification of shapers

• According to the ram drive: crank-and-slotted link, whitworth, or hydraulic shapers.
• According to the ram position and stroke: horizontal, vertical (slotter) and travelling-head shapers.
• According to the table: standard (plain) and universal shapers (table can tilt in several planes).
• According to the cutting stroke: push-cut (cutting on forward stroke) and draw-cut (on the return stroke).

Crank and slotted link mechanism

         ram  <-- connecting link
          |
    slotted link  (pivot O1 fixed, above)
         /|
        / |  sliding block on crank pin P
       /  |
 O2 (bull gear centre)  crank radius r

The bull gear rotates uniformly at O2. A sliding block fixed on the crank pin P slides in the slot of the link, which rocks about the fixed pivot O1 and, by a connecting link, drives the ram to and fro.

• When the crank pin moves through the larger arc (angle $\beta$ about O2, the far side of the circle from O1), the link swings slowly forward and the ram makes the cutting stroke.
• When the crank pin moves through the smaller arc ($\alpha$, near O1), the link swings quickly back and the ram makes the return stroke in less time.
• Since the bull gear speed is constant, the time ratio equals the angle ratio.

$$\frac{\text{time of cutting}}{\text{time of return}} = \frac{\beta}{\alpha} = \frac{360^\circ - \alpha}{\alpha}$$

In the extreme positions the link is tangent to the crank-pin circle, so in the right triangle O1-O2-P:

$$\cos\frac{\alpha}{2} = \frac{r}{d}$$

Numerical

$r = 100$ mm, $d = 250$ mm.

$$\begin{aligned} \cos\frac{\alpha}{2} &= \frac{100}{250} = 0.4 \\ \frac{\alpha}{2} &= 66.42^\circ,\quad \alpha = 132.84^\circ \\ \beta &= 360 - 132.84 = 227.16^\circ \\ \text{ratio} &= \frac{227.16}{132.84} = 1.71 \end{aligned}$$

Cutting stroke takes 1.71 times as long as the return stroke.

Answer: cutting : return time ratio = 1.71 : 1 (cutting angle 227.2°, return angle 132.8°).

Given: plate length 300 mm, width 160 mm; cutting speed v = 12 m/min = 12,000 mm/min; cutting : return time = 3 : 2; feed 1.5 mm per stroke.

Stroke length

$$L = 300 + 15 + 15 = 330\ \text{mm}$$

Double strokes per minute

Let k be the ratio of return time to cutting time: $k = 2/3$.

Time for the cutting stroke = $L/v = 330/12000 = 0.0275$ min.

Time for one double stroke:

$$t = 0.0275\,(1 + k) = 0.0275\times 1.6667 = 0.04583\ \text{min}$$ $$n = \frac{1}{t} = \frac{v}{L(1+k)} = \frac{12000}{330\times 1.6667} = 21.8\ \text{strokes/min}$$

Number of strokes needed

Width to be covered = 160 + 10 = 170 mm.

$$\text{strokes} = \frac{170}{1.5} = 113.3 \approx 114$$

Machining time

$$T = \frac{114}{21.8} = 5.23\ \text{min}$$

Answer: n = 21.8 double strokes per minute; machining time = 5.23 min.