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23-Ind-B2 Manufacturing Processes · December 2015

Question 3 of 7: Turning Cutting Speed and Material Removal Rate, Metal Chip Types, and Built-Up Edge

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

Notes on this paper

National Exams — December 2015 — 98-Ind-B2 Manufacturing Processes. Closed book; candidates may use one of two calculators, the Casio or Sharp approved models. Any five of the seven questions constitute a complete paper; all questions are of equal value (20 marks each). Answers are written in point form but fully, with all calculations shown, as instructed. Complete answers to all seven questions follow.

Reference texts: Groover, Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 6th ed. — material selection, casting, metal-cutting theory, welding processes, and automation/numerical control; Montgomery, Introduction to Statistical Quality Control, 8th ed. — where quality-control concepts are referenced.

Question 3: Turning Cutting Speed and Material Removal Rate, Metal Chip Types, and Built-Up Edge (20 marks: 8/5/7)

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.

This entire question, including sub-part (i)'s calculation (identical numeric data: 6 in./½.

(i) Turning: Cutting Speed and Material Removal Rate

Given. 304 stainless steel rod, original diameter $D_o=0.500$ in., turned down to final diameter $D_f=0.480$ in.; spindle speed $N=400$ rpm; axial (feed) travel speed of the tool $v_f=8$ in./min; rod length $L=6$ in. (not needed for these two quantities).

Find. The cutting speed $v$ and the material removal rate $MRR$.

feed, $v_f$ $D_o=0.500$ in $D_f=0.480$ in Rotating workpiece, $N=400$ rpm Tool L = 6 in (uncut length)
Turning on a lathe: the rotating rod (original diameter $D_o$) is reduced to $D_f$ as the single-point tool feeds axially at $v_f$; the removed material is the thin annular ring between $D_o$ and $D_f$ swept out along the cut length.

Approach. Cutting speed is the surface (peripheral) speed of the rotating workpiece at the diameter being cut, $v=\pi D_o N$. Feed per revolution is the axial travel speed divided by the spindle speed, $f=v_f/N$, and the depth of cut is half the diameter reduction, $d=(D_o-D_f)/2$. The material removal rate is the volume of the removed annular ring per minute, which is the ring's cross-sectional area times the axial travel speed: $MRR=\tfrac{\pi}{4}(D_o^2-D_f^2)\,v_f=\pi D_{avg}\,d\,f\,N$, with $D_{avg}=(D_o+D_f)/2$.

  1. Cutting speed. $v=\pi D_o N=\pi(0.500\ \text{in})(400\ \text{rev/min})=628.3\ \text{in/min}=\boxed{52.4\ \text{ft/min}}$.
  2. Feed per revolution. $f=\dfrac{v_f}{N}=\dfrac{8\ \text{in/min}}{400\ \text{rev/min}}=\boxed{0.0200\ \text{in/rev}}$.
  3. Depth of cut. $d=\dfrac{D_o-D_f}{2}=\dfrac{0.500-0.480}{2}=\boxed{0.0100\ \text{in}}$.
  4. Material removal rate. $D_{avg}=(0.500+0.480)/2=0.490$ in, so $MRR=\pi D_{avg}\,d\,f\,N=\pi(0.490)(0.0100)(0.0200)(400)=\boxed{0.1232\ \text{in}^3/\text{min}}$. Check: $\tfrac{\pi}{4}(0.500^2-0.480^2)(8)=\tfrac{\pi}{4}(0.0196)(8)=0.1232$ in³/min.
Question 3(i) — final results
QuantityValue
Cutting speed, $v$ (at $D_o$, maximum)628.3 in/min (52.4 ft/min)
Feed, $f$0.0200 in/rev
Depth of cut, $d$0.0100 in
Material removal rate, $MRR$0.1232 in³/min
Convention note: the cutting speed is quoted at the original diameter $D_o$, which is where it is highest. At the machined diameter it is $\pi(0.480)(400)=603.2$ in/min, and the average is 615.8 in/min (51.3 ft/min). The simpler textbook shortcut $MRR=v\,f\,d$ with $v$ taken at $D_o$ gives 0.1257 in³/min, about 2% high, because it treats the whole ring as if it were at the outer diameter. The exact annular volume rate, 0.1232 in³/min, is the answer given above.

(ii) Types of Metal Chips and the Preferred Type

Metal-cutting chips are classified into four basic types by how the material deforms and separates ahead of the tool: (1) discontinuous chips, small segmented fragments produced when a brittle work material (or a ductile one cut at low speed with a small rake angle and high friction) fractures repeatedly ahead of the tool rather than flowing plastically; (2) continuous chips, a long unbroken ribbon formed when a ductile material is cut at high speed with a large rake angle and low friction, deforming plastically along the shear plane without fracturing; (3) continuous chips with a built-up edge (BUE), a continuous chip whose formation is disrupted by work material welding to and periodically breaking away from the tool's cutting edge; and (4) serrated (segmented) chips, semi-continuous chips with a saw-tooth profile of alternating zones of high and low shear strain. They form in metals with low thermal conductivity whose strength falls sharply with temperature (titanium alloys, nickel-base superalloys, austenitic stainless steels), because shear localizes into narrow bands.

The continuous chip (without BUE) is generally the best of the four: it gives a good, consistent surface finish and stable cutting forces, whereas discontinuous and serrated chips cause cyclic force fluctuations (and chatter) and a rougher finish, and a built-up edge degrades surface finish and dimensional accuracy and accelerates tool wear. The one practical drawback of a plain continuous chip is that the long ribbon can tangle around the tool, fixture or workpiece, which is why chip breakers are commonly added to curl and fracture an otherwise continuous chip into manageable lengths without sacrificing its favourable cutting mechanics.

(iii) Built-Up Edge: Effect and Control

A built-up edge (BUE) is a layer of work material that adheres to the tool's rake face at the cutting edge under the high pressure, temperature and friction of the cutting zone, welding itself to the tool. As successive layers accumulate the BUE grows until it becomes unstable and breaks away, with part of it carried off in the chip and part deposited onto the newly cut surface, then the cycle repeats.

Effects on the operation: BUE effectively (and unpredictably) changes the tool's actual rake angle and edge geometry, so cutting forces fluctuate; fragments deposited on the finished surface produce a rough, work-hardened surface finish and poor dimensional control; and the repeated welding/tearing cycle accelerates tool wear (particularly the crater and flank wear that determine tool life).

BUE can be eliminated or minimized by: increasing cutting speed (higher temperature reduces the material's tendency to adhere); increasing the tool's rake angle (reduces the contact pressure and friction on the rake face); reducing the depth of cut/feed; using an effective cutting fluid to reduce friction and carry away heat; and selecting a tool material/coating with lower chemical affinity for the work material, since BUE is fundamentally an adhesion (micro-welding) phenomenon between the two.

Part (ii) includes serrated chips; parts (ii) and (iii) carry 5 and 7 marks respectively.