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Chapter 8 · 3 hours

Material Removal Processes: “Abrasive and Non-Traditional”

Practice questions

Practice questions and answers

2 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 · 5 marks

Explain the standard marking system of a grinding wheel with an example, such as A 60 K 5 V. What are glazing and loading of a grinding wheel? How are truing and dressing done?

Answer

A grinding wheel is a bonded mass of abrasive grains. Each grain acts as a tiny cutting tool, and the wheel renews itself as the dull grains break out and fall away.

Standard marking, example A 60 K 5 V

  A     60     K      5      V
  |     |      |      |      |
abrasive grit  grade structure bond
Letter or numberMeaningHere
AbrasiveA = aluminium oxide, C = silicon carbide, B = CBN, D = diamondAluminium oxide: steels
Grain sizeGrit number (coarse 10-24, medium 30-60, fine 70-180, very fine 220+)60 = medium
GradeHardness, A (soft) to Z (hard); the strength of the bondK = medium
StructureSpacing of grains, 1 (dense) to 15 (open)5 = medium
BondV = vitrified, S = silicate, R = rubber, B = resinoid, E = shellacVitrified

Selection guide

  • Steel and soft metals use aluminium oxide; cast iron, carbide and non-ferrous use silicon carbide.
  • Hard materials need a soft grade so that dull grains drop out; soft materials need hard grades.
  • Fine grit for good finish; coarse grit for fast removal.
  • Wide contact area or soft material needs open structure.

Glazing and loading

  • Glazing: the abrasive grains become blunt and the wheel surface turns smooth and shiny because the grade is too hard. Cutting efficiency falls, heat and burning appear.
  • Loading: the spaces between the grains fill with chips of the work material (soft ductile metals such as aluminium or copper). The wheel cuts poorly.

Truing and dressing

  • Truing restores the wheel shape and makes it run concentric, using a single-point diamond or a diamond roll.
  • Dressing restores sharpness and exposes new grains by removing the glazed or loaded layer, using a dressing stick, star dresser (Huntington) or diamond. Often both are done together.
  • Practice · 4+4 marks

(a) Explain the working principle of electrical discharge machining (EDM) with a sketch. State its advantages and limitations. (b) Explain the principle of electrochemical machining (ECM). In ECM of pure iron (atomic weight 55.85, valency 2, density 7.86 g/cm3^3) a current of 1000 A is used with a current efficiency of 90 percent. Calculate the volumetric material removal rate and the tool feed rate if the cross-sectional area of the machined cavity is 400 mm2^2.

Answer

(a) EDM

In EDM, the work and the shaped tool (electrode) are both immersed in a dielectric fluid (kerosene, transformer oil or deionised water) and kept a small gap apart (0.01-0.5 mm). A pulsed DC supply makes a spark discharge across the gap; each spark generates temperature of 8000-12000 degree C in a tiny spot, melting and vaporising a small crater in the work. The dielectric flushes out the debris and re-insulates the gap. A servo system keeps the gap constant as the tool advances, and the tool shape is reproduced in the work.

   +---- pulse generator ----+
   |                         |
 tool (+/-)   spark gap    work (-/+)
   |__ dielectric fluid + flushing __|
          servo feed (down)

Advantages: machines any electrically conductive material regardless of hardness (hardened steel, carbide); complex cavities, dies and small deep holes; no cutting force, no burr; good accuracy.

Limitations: only conductive materials; slow metal removal; tool wear and recast (heat-affected) layer; high power; sharp inside corners not possible.

(b) ECM

In ECM, the work is the anode and the tool is the cathode. A low-voltage (10-20 V), high current (several thousand A/cm2^2) DC flows through a fast electrolyte (NaCl or NaNO3_3 solution) in a gap of about 0.1-0.5 mm. The work metal dissolves according to Faraday's law, and the electrolyte carries the hydroxide sludge away. The tool is not worn.

By Faraday's laws, mass removed per second:

m˙=η I Awz F\dot{m} = \frac{\eta\,I\,A_w}{z\,F}

Volumetric rate:

V˙=η I Awz F ρ\dot{V} = \frac{\eta\,I\,A_w}{z\,F\,\rho}

with Aw=55.85A_w = 55.85 g/mol, z=2z = 2, F=96 485F = 96\,485 C/mol, ρ=7.86\rho = 7.86 g/cm3^3, η=0.9\eta = 0.9, I=1000I = 1000 A.

V˙=0.9×1000×55.852×96 485×7.86=0.0331 cm3/s=33.1 mm3/s\dot{V} = \frac{0.9\times1000\times55.85}{2\times96\,485\times7.86} = 0.0331\ \text{cm}^3/\text{s} = 33.1\ \text{mm}^3/\text{s}

This equals 33.1×60=1988 mm3/min33.1\times60 = 1988\ \text{mm}^3/\text{min} (about 2.0 cm3^3/min).

Tool feed rate for the cavity cross-section of 400 mm2^2:

v=V˙A=33.1400=0.0828 mm/s=4.97 mm/minv = \frac{\dot{V}}{A} = \frac{33.1}{400} = 0.0828\ \text{mm/s} = 4.97\ \text{mm/min}

Answer: volumetric removal rate =33.1= 33.1 mm3^3/s (1.99 cm3^3/min); feed rate =4.97= 4.97 mm/min.

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

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