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Chapter 3 · 3 hours

Properties of Manufactured Products

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 · 3+5 marks

(a) Differentiate between unilateral and bilateral tolerance, and explain geometric tolerance with any four characteristic symbols. (b) A hole is dimensioned ϕ400+0.039\phi 40^{+0.039}_{0} mm. Three shafts are available: (i) ϕ40−0.050−0.025\phi 40^{-0.025}_{-0.050}, (ii) ϕ40+0.043+0.059\phi 40^{+0.059}_{+0.043}, (iii) ϕ40+0.026+0.042\phi 40^{+0.042}_{+0.026}. For each, find the maximum and minimum clearance (or interference) and state the type of fit.

Answer

(a) Unilateral and bilateral tolerance; geometric tolerance

PointUnilateralBilateral
VariationIn one direction only from basic sizeOn both sides of basic size
Example400+0.0440^{+0.04}_{0}40±0.0240 \pm 0.02
UseHole basis / shaft basis systems, easy fitsPositioning, centre distances

Geometric tolerance limits the form, orientation, location and run-out of a feature, not just its size. It is given in a feature control frame on the drawing. Characteristic symbols:

  • Straightness (a straight line) and flatness (a parallelogram) - form.
  • Parallelism (two parallel lines) and perpendicularity (a right-angle) - orientation.
  • Concentricity (two circles) and position - location.
  • Circular run-out (arrow) - run-out.

(b) Fit calculation

Hole: maximum size =40.039= 40.039 mm, minimum size =40.000= 40.000 mm.

Maximum clearance == hole max −- shaft min; minimum clearance == hole min −- shaft max (negative value = interference).

Shaft (i): shaft max =39.975= 39.975, min =39.950= 39.950

Cmax=40.039−39.950=0.089 mm,Cmin=40.000−39.975=0.025 mmC_{max} = 40.039 - 39.950 = 0.089\ \text{mm},\qquad C_{min} = 40.000 - 39.975 = 0.025\ \text{mm}

Both positive, so it is a clearance fit (running fit).

Shaft (ii): shaft max =40.059= 40.059, min =40.043= 40.043

Cmax=40.039−40.043=−0.004 mm,Cmin=40.000−40.059=−0.059 mmC_{max} = 40.039 - 40.043 = -0.004\ \text{mm},\qquad C_{min} = 40.000 - 40.059 = -0.059\ \text{mm}

Both negative, so maximum interference =0.059= 0.059 mm and minimum interference =0.004= 0.004 mm: an interference (press) fit.

Shaft (iii): shaft max =40.042= 40.042, min =40.026= 40.026

Cmax=40.039−40.026=0.013 mm,Cmin=40.000−40.042=−0.042 mmC_{max} = 40.039 - 40.026 = 0.013\ \text{mm},\qquad C_{min} = 40.000 - 40.042 = -0.042\ \text{mm}

The result can be clearance or interference, so it is a transition fit.

ShaftMax clearance (mm)Min clearance (mm)Fit
(i)0.0890.025Clearance
(ii)−0.004-0.004−0.059-0.059Interference
(iii)0.013−0.042-0.042Transition

Answer: (i) clearance fit, 0.089 to 0.025 mm; (ii) interference fit, 0.004 to 0.059 mm; (iii) transition fit, 0.013 mm clearance to 0.042 mm interference.

  • Practice · 5 marks

Define centre-line average roughness RaR_a and root-mean-square roughness RqR_q. The heights of 10 equally spaced points of a surface profile measured from the mean line within a sampling length are (in micrometre): +2.0, -1.5, +3.5, -2.5, +1.0, -3.0, +2.5, -1.0, +3.0, -4.0. Calculate RaR_a, RqR_q and the maximum peak-to-valley height. Name two factors that cause surface roughness in machining.

Answer

Surface texture has roughness (fine irregularities from the process), waviness (longer wavelength, from vibration or deflection) and lay (direction of the pattern).

Definitions

RaR_a is the arithmetic mean of the absolute deviations of the profile from the mean line:

Ra=1n∑i=1n∣yi∣R_a = \frac{1}{n}\sum_{i=1}^{n}|y_i|

RqR_q (also RMS) is the square root of the mean of the squares of the deviations:

Rq=1n∑i=1nyi2R_q = \sqrt{\frac{1}{n}\sum_{i=1}^{n} y_i^2}

Check of the mean line

Sum of ordinates: 2.0−1.5+3.5−2.5+1.0−3.0+2.5−1.0+3.0−4.0=02.0-1.5+3.5-2.5+1.0-3.0+2.5-1.0+3.0-4.0 = 0, so the reference line is the mean line.

Calculation of RaR_a

∑∣yi∣=2.0+1.5+3.5+2.5+1.0+3.0+2.5+1.0+3.0+4.0=24.0\sum|y_i| = 2.0+1.5+3.5+2.5+1.0+3.0+2.5+1.0+3.0+4.0 = 24.0 Ra=24.010=2.4 μmR_a = \frac{24.0}{10} = 2.4\ \mu\text{m}

Calculation of RqR_q

∑yi2=4+2.25+12.25+6.25+1+9+6.25+1+9+16=67.0\sum y_i^2 = 4+2.25+12.25+6.25+1+9+6.25+1+9+16 = 67.0 Rq=67.010=6.7=2.59 μmR_q = \sqrt{\frac{67.0}{10}} = \sqrt{6.7} = 2.59\ \mu\text{m}

Maximum peak-to-valley height

Rmax=ymax−ymin=3.5−(−4.0)=7.5 μmR_{max} = y_{max} - y_{min} = 3.5 - (-4.0) = 7.5\ \mu\text{m}

Factors causing roughness in machining

  1. Feed marks: in turning the theoretical roughness is Rmax=f2/(8r)R_{max} = f^2/(8r) where ff is the feed and rr the nose radius.
  2. Built-up edge, tool wear, vibration (chatter) and machine tool rigidity.

Answer: Ra=2.4 μR_a = 2.4\ \mum, Rq=2.59 μR_q = 2.59\ \mum, Rmax=7.5 μR_{max} = 7.5\ \mum.

  • Practice · 5 marks

What are residual stresses? Explain how they are produced in manufacturing processes, their effects on the product, and the methods of reducing or removing them.

Answer

Residual stresses are stresses that remain in a body after all external loads and temperature gradients have been removed. They are self-balancing: tensile stress in one region is balanced by compressive stress in another.

Origin

Residual stress appears whenever different parts of a body deform plastically by different amounts, or cool at different rates.

  1. Non-uniform plastic deformation: in bending, the outer fibres are plastically stretched and when the load is removed the elastic core pushes them back, leaving compressive stress on the surface that was in tension. Rolling, drawing and shot peening also leave surface compression.
  2. Thermal gradients: in quenching, the surface cools first and contracts; later the core contracts against a rigid shell, leaving surface compression and core tension. In welding, the weld region shrinks against cold parent metal and is left in tension.
  3. Phase transformation: martensite formation causes volume expansion in some regions (hardening of steel).
  4. Machining and grinding: mechanical deformation under the tool and heat generated leave a thin surface layer in tension (grinding heat) or compression (sharp tools, low feed).

Effects

  • Distortion and warping when material is machined away or heated.
  • Compressive residual stress at the surface improves fatigue life and resists crack growth.
  • Tensile residual stress at the surface reduces fatigue strength and causes stress-corrosion cracking and cold cracks in welds.
  • Dimensional instability of precision parts with time.

Methods of reduction or removal

MethodPrinciple
Stress-relief annealingHeat below the transformation temperature (about 550-650 degree C for steel), hold, cool slowly
TemperingAfter hardening, relieves quench stresses
Stretching / light cold workSmall plastic deformation (about 1-2 percent) levels the stress
Vibratory stress reliefVibration at resonance redistributes stress
Shot peeningIntroduces beneficial surface compression
Controlled cooling and proper process designReduces thermal gradient

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

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