22-Mec-A4 Design and Manufacture of Machine Elements · December 2014

Question 3 of 6: Extended Taylor tool-life equation — speed, feed or depth of cut

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions divided into Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine elements); candidates answer two from Part A and two from Part B, and all questions carry equal value (25 % each). All six questions are worked here.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology — Ch. 10–12 (casting processes and design), Ch. 1 and 3 (structure, cold work, annealing), Ch. 21–22 (machining and tool life).
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design — Ch. 3 (stress, Mohr's circle, beam deflection), Ch. 4 (statically indeterminate shafts), Ch. 16 (clutches and brakes).
R. C. Hibbeler, Mechanics of Materials — Ch. 6–9 (bending, transverse shear, combined loading, stress transformation).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering — Ch. 7 (dislocations, strengthening, recovery and recrystallization).

it reports a single 5,200 N load and a dimension chain of 750 + 450 + 450 = 1,650 mm that "contradicts" the 2,400 mm overall. Q5 is solved against the drawing, not the caption. Likewise in Q4 the 3,000 lb load is read from the drawing as a horizontal force applied through the bolted plate on the neutral axis, and the section as a 8 in deep I-beam (¹⁄₂ + 3¹⁄₂ + 3¹⁄₂ + ¹⁄₂).

Question 3: Extended Taylor tool-life equation — speed, feed or depth of cut (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Taylor constant	$K$	100 (arbitrary units)
Speed exponent	$n_1$	0.10
Feed exponent	$n_2$	0.18
Depth-of-cut exponent	$n_3$	0.45
Initial cutting speed, feed, depth	$v_0, f_0, w_0$	1, 1, 1
Required production increase	—	30 % (one variable at a time)

Find. The tool life resulting from a 30 % increase in each of speed, feed and depth of cut taken separately, and hence a recommendation on the order in which the three should be raised.

Approach. Metal-removal rate is proportional to the product $v f w$, so raising any one of the three by 30 % raises production by 30 %; substitute each case into the extended Taylor equation in turn and compare the resulting tool lives against the baseline.

Establish the baseline tool life. With all three variables at unity the denominator is unity, so $$t_0 = \frac{K}{1^{1/n_1}\,1^{1/n_2}\,1^{1/n_3}} = K = 100 \text{ units.}$$ This is the reference against which the three proposals are judged.
Note the exponents that actually act. The equation is written with $1/n$ powers, so the sensitivity of tool life to each variable is set by the reciprocal of its exponent: $$\frac{1}{n_1} = 10, \qquad \frac{1}{n_2} = 5.556, \qquad \frac{1}{n_3} = 2.222 .$$ Everything in this question follows from those three numbers: cutting speed enters to the tenth power, feed to the 5.6th, and depth of cut only to the 2.2nd.
Case (a) — increase the cutting speed by 30 %. Setting $v = 1.30$ with $f = w = 1$, $$t_a = \frac{100}{1.30^{10}} = \frac{100}{13.786} = \boxed{7.25 \text{ units}}$$ a reduction of $(100-7.25)/100 = 92.7\,\%$ of the original tool life.
Case (b) — increase the feed by 30 %. Setting $f = 1.30$ with $v = w = 1$, $$t_b = \frac{100}{1.30^{5.556}} = \frac{100}{4.2949} = \boxed{23.28 \text{ units}}$$ a reduction of $76.7\,\%$.
Case (c) — increase the depth of cut by 30 %. Setting $w = 1.30$ with $v = f = 1$, $$t_c = \frac{100}{1.30^{2.222}} = \frac{100}{1.7914} = \boxed{55.82 \text{ units}}$$ a reduction of only $44.2\,\%$.

The three cases deliver identical production gains but wildly different tool lives. Increasing the depth of cut leaves the tool lasting more than seven and a half times as long as increasing the speed does.

Tool life after a 30 % production increase obtained three different ways.

(d) Recommendation

Increase the depth of cut first, the feed second, and the cutting speed last. The ordering is a direct reading of the exponents: because tool life varies as $w^{-2.222}$, $f^{-5.556}$ and $v^{-10}$, the variable with the smallest reciprocal exponent buys a given production increase at the smallest cost in tool life. This is the classic machining-economics result and it matches shop practice — take the deepest cut the setup will stand, then the heaviest feed the surface finish and tool strength allow, and only then raise the speed.

Two engineering qualifications belong with the recommendation. Depth of cut is limited by the stock available, by the rigidity of the workpiece, tool and machine (deep cuts raise the radial force and can provoke chatter or deflection), and by available spindle power. Feed is limited by the surface finish required — theoretical roughness rises roughly with the square of the feed — and by the strength of the tool nose, since feed increases the chip load directly. Cutting speed is nevertheless the variable most often raised in practice, precisely because it is the one least constrained by rigidity and finish; the exponents merely say that raising it is the most expensive way to buy production in tooling terms.

Case	Variable raised 30 %	Tool life $t$ (units)	Change from $t_0 = 100$
Baseline	—	100	—
(a)	Cutting speed $v$	7.25	−92.7 %
(b)	Feed $f$	23.28	−76.7 %
(c)	Depth of cut $w$	55.82	−44.2 %
(d)	Recommendation: raise depth of cut first, then feed, and cutting speed last.

Topic. Tool life in single-point turning; the extended Taylor equation and its use in machining economics.

Key relations. The classical Taylor equation is $v T^{\,n} = C$. The extended form used here, $t = K / (v^{1/n_1} f^{1/n_2} w^{1/n_3})$, adds feed and depth of cut as independent variables. Because production rate (metal-removal rate) is $\mathrm{MRR} \propto v f w$, a fractional increase $\lambda$ applied to any one variable gives the same production gain but multiplies tool life by $\lambda^{-1/n_i}$. The ratio of the resulting lives is therefore $\lambda^{(1/n_j - 1/n_i)}$, independent of $K$.

Why this works. The exponents encode how strongly each variable drives the wear mechanisms. Cutting speed dominates because nearly all the cutting work becomes heat, and tool-face temperature rises steeply with speed; diffusion and adhesive wear are thermally activated, so a modest temperature rise produces a large wear-rate rise. Feed raises the chip load and hence the mechanical and thermal loading on the tool, but less sharply than speed. Depth of cut mainly lengthens the engaged cutting edge, spreading the load and the heat over more edge length, so its effect on wear per unit length of edge is weakest of the three. Since $K$ cancels in every comparison, the answer depends only on the exponents — which is why the question can hand out arbitrary units and still have a definite answer.

Common pitfall. Inverting the exponents — using $v^{n_1}$ instead of $v^{1/n_1}$ — which reverses the ranking and produces the recommendation "raise the speed first". A second is arithmetic: $1.30^{10}$ is 13.79, and a mis-keyed power silently corrupts the whole comparison, so each should be re-evaluated. A third is to assume the three cases give different production increases; they do not, which is precisely what makes the comparison fair.

Source. Kalpakjian & Schmid, Manufacturing Engineering and Technology, Ch. 21.4 (tool life: wear and the Taylor equation) and Ch. 25 (machining economics).

Part B

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