Chapter 10 · 3 hours

Measurement of Surface Finish

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 · 6 marks

Explain the effect of surface finish on fatigue life, wear, bearing properties, corrosion and fit. List the factors that affect surface finish.

Answer

Surface finish is the smoothness of the machined surface. Every machined surface has peaks and valleys; the valleys act as notches and the peaks carry the load.

Property	Effect of a rough surface
Fatigue life	Valleys act as stress raisers; cracks start at the roots of the tool marks, so fatigue strength falls. A fine finish (ground or polished) raises endurance limit
Wear	Peaks are sheared and broken in contact (high initial wear); a very smooth surface also does not hold oil. The best is a controlled medium finish
Bearing properties	Peaks carry the load on a small real area, giving high local pressure and quick wear of the bearing and journal; oil film breaks. Smooth surfaces give better load-carrying area and lubrication
Stress	High stress concentration at the valleys (notch effect), residual tensile stress may be left on a rough machined surface
Corrosion	Valleys retain moisture and corrosive agents; larger surface area makes corrosion faster. Smooth surface is more resistant
Fit	In a press fit, the peaks are flattened during assembly so the effective interference is smaller; in a running fit, wear of the peaks increases clearance and the fit loosens

Factors affecting surface finish

Cutting conditions: speed, feed (finish worsens as feed rises), depth of cut.
Tool geometry: nose radius, rake and clearance angles, sharpness and wear of the tool.
Tool and work material; formation of built-up edge.
Cutting fluid (lubrication and cooling).
Rigidity of machine, tool and work; vibration and chatter.
Machining process used: grinding and lapping give better finish than turning or milling.
Machine accuracy (spindle runout, slideway errors) and uniformity of the work material.

Practice · 6 marks

Define Ra, Rq (RMS) and Rz/Ry. The heights of ten equally spaced ordinates of a surface profile from the mean line are (in micrometres): +2.1, -1.8, +3.2, -2.4, +1.5, -3.0, +2.6, -1.2, +1.9, -2.8. Calculate Ra, Rq and the maximum peak-to-valley height. Convert Ra to micro-inches.

Answer

Definitions

Ra (centre-line average, CLA): the arithmetic mean of the absolute deviations of the profile from the mean line over the sampling length:

R_a = \frac{1}{n}\sum_{i=1}^{n}|y_i|

Rq (RMS): the root mean square deviation, $R_q = \sqrt{\dfrac{1}{n}\sum y_i^2}$ . It is about 1.11 times Ra for a sine-like profile.
Rz / Ry (peak to valley): the vertical distance between the highest peak and the lowest valley within the sampling length (Rz in ISO is the mean of five such heights in five sampling lengths).

Calculation

$i$	$y_i$ ( $\mu$ m)	$\lvert y_i\rvert$	$y_i^2$
1	+2.1	2.1	4.41
2	-1.8	1.8	3.24
3	+3.2	3.2	10.24
4	-2.4	2.4	5.76
5	+1.5	1.5	2.25
6	-3.0	3.0	9.00
7	+2.6	2.6	6.76
8	-1.2	1.2	1.44
9	+1.9	1.9	3.61
10	-2.8	2.8	7.84
Sum		22.5	54.55

R_a = \frac{22.5}{10} = 2.25\ \mu\text{m}

R_q = \sqrt{\frac{54.55}{10}} = \sqrt{5.455} = 2.34\ \mu\text{m}

Highest peak $= +3.2$ , deepest valley $= -3.0$ :

R_y = 3.2 - (-3.0) = 6.2\ \mu\text{m}

In micro-inches, 1 $\mu$ m $= 39.37\ \mu$ in: $R_a = 2.25 \times 39.37 = 88.6\ \mu$ in.

Answer: $R_a$ = 2.25 $\mu$ m (88.6 $\mu$ in), $R_q$ = 2.34 $\mu$ m, $R_y$ = 6.2 $\mu$ m.

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

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