Lathe and Turning Operations Questions and Answers

Answer: B

For a short taper in mass production the preferred method is the form tool.

A form tool is ground to the exact profile of the taper and is simply plunged in with the cross slide, so a short taper is produced in one movement with no setting of the compound or tailstock. That makes it the quickest method when many identical pieces are needed, provided the taper is short, since a long form tool would chatter.

The compound rest method is hand-fed and slow, the taper turning attachment is set up for longer tapers, and the tailstock offset method is used for long, gentle tapers on work between centres; none is suited to rapid repetition of a short taper.

Answer: C

The change gears are 60/80, as keyed.

The lead of a multi-start thread is the number of starts times the pitch, and the ratio of driver to driven gear (D_r over D_n) is the lead wanted over the lead of the lead screw:

\[ L = 4 \times 1.5 = 6\ \text{mm} \]

\[ \frac{D_{r}}{D_{n}} = \frac{6}{8} = \frac{3}{4} \]

60/80 reduces to 3/4. 80/100 is 4/5 and 80/120 is 2/3, so they would cut a lead of 6.4 mm and 5.33 mm.

Doubt: 45/60 also reduces to 3/4, so option A gives the same ratio. The official key gave C, and it is kept.

Answer: C

Knurling is done at one third of the turning speed.

Knurling presses hardened rolls into the work with a high radial force and the metal flows plastically to form the pattern. A slow spindle gives the rolls time to form the pattern, avoids double tracking and keeps the torque on the work and the rolls manageable. The Turner trade theory manual puts the speed at about one third of the turning speed.

Equal and double turning speed are far too fast. One fourth of turning speed is the option nearest to the keyed one, and some workshop guides quote a range of one third down to one fourth, but the standard rule taught for the trade is one third.

Answer: D

The spindle speed is 50 rpm.

Cutting speed is the surface speed of the work, \(v = \pi D N / 1000\) with \(v\) in m/min and \(D\) in mm. Rearranging:

\[ \begin{aligned} N &= \frac{1000\,v}{\pi D} = \frac{1000 \times 31.4}{3.14 \times 200} \\ &= \frac{31400}{628} = 50 \end{aligned} \]

Forgetting the factor 1000 that converts metres to millimetres, or using the radius instead of the diameter, produces the other answers: with the radius of 100 mm the result is 100 rpm.

Answer: C

The cross slide is used to supply feed to the tool.

The cross slide sits on the saddle and moves at right angles to the lathe axis, carrying the compound rest and the tool post with the tool. Moving it feeds the tool into the work for depth of cut and feeds it across the face of the work in facing.

The tool post is only the holder for the tool, so the feed is given to the tool through it, not to the tool post for its own sake. The chuck rotates with the spindle and is not fed, and the tailstock is moved along the bed by hand, not by the cross slide.

Answer: A

Setting the tool below centre height increases the front clearance angle.

The effective angles of a tool are measured from the line joining its point to the centre of the work. When the point is below centre, that line tilts, so the effective rake angle decreases and the effective front clearance angle increases by the same amount. (Setting it above centre does the opposite: rake increases and clearance decreases, and the tool may rub.)

So the clearance neither decreases, stays the same, nor becomes zero when the tool is set low.

Answer: B

The half included angle is \(\tan^{-1}\{(D - d)/2l\}\).

Across the length \(l\) of the taper, the diameter falls from \(D\) to \(d\), so the radius falls by \(\frac{D - d}{2}\). The half angle \(\alpha\) is the angle the tapered surface makes with the axis, in a right triangle with that fall in radius as the opposite side and \(l\) as the adjacent side:

\[ \tan\alpha = \frac{(D - d)/2}{l} = \frac{D - d}{2l} \]

\(\tan^{-1}\{(D - d)/l\}\) uses the change in diameter instead of radius, which overstates the angle. \(2(D - d)/l\) overstates it further, and \((2D - d)/l\) is not a difference of sizes at all.

Answer: C

The turning time is 10 min.

The reduction from 50 mm to 40 mm is done in one cut, so the tool travels the length of the rod once. Time is length divided by the feed per minute, which is feed per revolution times rpm:

\[ \begin{aligned} T &= \frac{L}{f N} = \frac{200}{0.1 \times 200} \\ &= \frac{200}{20} = 10\ \text{min} \end{aligned} \]

The diameters only fix the depth of cut (5 mm), which does not enter the time for a single pass. 20 min would be two passes, and 5 min and 4 min do not match the feed and speed given.

Answer: B

The machining time is 3.74 minutes.

First find the spindle speed for 30 m/min on the 50 mm rod. The feed, printed as 0.7 rev/min, is taken as 0.7 mm per revolution, the usual unit of turning feed:

\[ \begin{aligned} N &= \frac{1000 \times 30}{\pi \times 50} = 191\ \text{rpm} \\ T &= \frac{L}{f N} = \frac{500}{0.7 \times 191} \\ &= \frac{500}{133.7} = 3.74\ \text{min} \end{aligned} \]

The 45 mm finish diameter sets the depth of cut, not the time, since it is one pass. Basing \(N\) on 45 mm gives 212 rpm and about 3.4 minutes, which is not an option.

Answer: A

The keyed material removal rate is 7538.4 mm³/min.

The key uses the standard approximation MRR = cutting speed × feed × depth of cut, with the cutting speed taken at the original diameter. Feed per revolution is the axial speed divided by rpm, and depth of cut is half the reduction in diameter:

\[ \begin{aligned} f &= \frac{200}{500} = 0.4\ \text{mm/rev}, \quad d = 1\ \text{mm} \\ v &= \pi \times 12 \times 500 = 18850\ \text{mm/min} \\ \text{MRR} &= v f d \approx 7540\ \text{mm}^{3}/\text{min} \end{aligned} \]

With \(\pi\) taken as 3.141 this is 7538.4. Measured exactly, as the annulus removed per minute, \(\frac{\pi}{4}(12^2 - 10^2) \times 200\), the rate is lower, about 6910, because the cutting speed falls across the depth of cut; that value is not offered, so the paper follows the textbook formula.