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Chapter 5 · 4 hours

Drills and Drilling Processes

Practice questions

Practice questions and answers

3 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 6 marks

Draw and label a twist drill and explain its geometry. State typical values of point angle, helix angle, lip clearance angle and chisel edge angle.

Answer

A twist drill is a rotary cutting tool with two helical flutes and two cutting lips, used to produce round holes.

  shank      body (flutes)         point
 |<---->|<------------------->|<-------->
 ====== |  ///////////////  |  /\
 tang   |  helical flutes   | /  \  lips
                              \/  chisel edge

Main parts

  • Shank: straight (small drills, held in a chuck) or Morse taper (large drills, with a tang that drives and helps ejection).
  • Body: carries two helical flutes that form the cutting edges, carry chips out and let coolant in.
  • Land: raised strip following the flute; the margin on it guides the drill and keeps diameter.
  • Web: central core thickness between flutes; it increases toward the shank for strength.
  • Point: cone formed by two lips and the chisel edge.

Geometry angles

FeatureTypical valueRole
Point (included) angle118° for general steel and iron; 135° to 140° for hard materials; 90° for plastics, 130° for brassControls cutting action and thrust
Helix angle24° to 32° (about 30° general)Rake angle at the lip; chip removal
Lip clearance angle8° to 15°Prevents the heel rubbing behind the lip
Chisel edge angle120° to 135°Depends on lip clearance; 135° preferred
  • The two lips must be equal in length and angle; otherwise the hole will be oversize and the drill may break.
  • The chisel edge does not cut but pushes metal sideways, producing high thrust; web thinning reduces it.
  • Practice · 8 marks

A 20 mm diameter hole is to be drilled through a 50 mm thick mild steel plate with a twist drill of 118° point angle at a cutting speed of 25 m/min and feed of 0.25 mm/rev. Allow 2 mm for breakout (overtravel). Find (a) spindle speed, (b) drill approach, (c) feed rate, (d) time to drill the hole and (e) metal removal rate.

Answer

Given: D = 20 mm, t = 50 mm, v = 25 m/min, f = 0.25 mm/rev, point angle 118°, overtravel = 2 mm.

(a) Spindle speed

N=1000 vπD=1000×25π×20=397.9 rpmN = \frac{1000\,v}{\pi D} = \frac{1000 \times 25}{\pi \times 20} = 397.9\ \text{rpm}

(b) Drill approach (point allowance)

The drill must travel until its full diameter enters the plate.

A=D2cot⁡(118∘2)=202×tan⁡59∘=101.664=6.01 mmA = \frac{D}{2}\cot\left(\frac{118^\circ}{2}\right) = \frac{20}{2 \times \tan 59^\circ} = \frac{10}{1.664} = 6.01\ \text{mm}

(c) Feed rate

fm=f×N=0.25×397.9=99.5 mm/minf_m = f \times N = 0.25 \times 397.9 = 99.5\ \text{mm/min}

(d) Time

Total tool travel:

L=t+A+overtravel=50+6.01+2=58.01 mmL = t + A + \text{overtravel} = 50 + 6.01 + 2 = 58.01\ \text{mm} T=Lfm=58.0199.5=0.583 min≈35 sT = \frac{L}{f_m} = \frac{58.01}{99.5} = 0.583\ \text{min} \approx 35\ \text{s}

(e) Metal removal rate

MRR=π4D2×fm=π4(20)2×99.5=31,250 mm3/min\text{MRR} = \frac{\pi}{4}D^2 \times f_m = \frac{\pi}{4}(20)^2 \times 99.5 = 31{,}250\ \text{mm}^3/\text{min}
QuantityValue
N397.9 rpm
Approach6.01 mm
Feed rate99.5 mm/min
Time0.583 min
MRR31.25 cm³/min

Answer: N = 398 rpm, approach = 6.01 mm, feed rate = 99.5 mm/min, time = 0.583 min (35 s), MRR = 31.25 cm³/min.

  • Practice · 5 marks

Differentiate between reaming, honing and lapping. Write the purpose of counterboring and countersinking.

Answer

All three are finishing operations that bring a hole or surface to close tolerance and fine finish, but they differ in tool and action.

PointReamingHoningLapping
ToolMulti-tooth fluted reamerBonded abrasive stones in a hone headLoose abrasive paste between work and a soft lap
MotionRotation and feed in a drilled holeRotation plus reciprocation (crosshatch pattern)Rubbing movement of lap against work
Stock removed0.1 to 0.5 mm0.01 to 0.1 mm0.003 to 0.03 mm
CorrectsSize and finish; not hole position or straightnessSize, roundness, taper and straightnessFlatness, size, very fine finish (Ra < 0.1 µm)
SurfaceRa about 1 µmRa about 0.2 µmMirror finish
UseDowel holes, bearing seatsEngine cylinders, boresGauge blocks, valve seats, lenses

Counterboring

Enlarging the end of a hole to a larger diameter with a flat bottom so that a bolt head, nut or cap screw sits flush or below the surface. A pilot on the counterbore guides it in the original hole.

Countersinking

Making a conical enlargement at the mouth of a hole (usually 60°, 82° or 90°) so a flat-head screw or rivet lies flush. It is also used to chamfer and deburr holes and to make the centre-hole for lathe centres.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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