Question 3 of 7: Metal Chip Types, Built-Up Edge, and the Orthogonal-Cutting Shear Angle

Question 3: Metal Chip Types, Built-Up Edge, and the Orthogonal-Cutting Shear Angle (20 marks: 7/7/6)

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.

(i) 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.

(ii) 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.

(iii) Orthogonal Cutting: Determining the Shear Angle

Given. Undeformed (uncut) chip thickness $t_0 = 0.0098$ in.; actual (deformed) chip thickness $t_c = 0.0169$ in.; rake angle $\alpha = 20^{\circ}$.

Find. The shear angle $\phi$ of the shear plane along which the chip is formed.

Approach. Compute the chip-thickness (chip-compression) ratio $r=t_0/t_c$, then apply Ernst & Merchant's orthogonal-cutting shear-angle relation, which follows from the geometry of the shear-plane/chip triangle.

1. Chip-thickness ratio. $r=\dfrac{t_0}{t_c}=\dfrac{0.0098}{0.0169}=\boxed{0.580}$.
2. Shear-angle relation. The shear angle satisfies $\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}$. Substituting $r=0.580$ and $\alpha=20^{\circ}$: $\tan\phi=\dfrac{0.580\cos20^{\circ}}{1-0.580\sin20^{\circ}}=\dfrac{0.580(0.9397)}{1-0.580(0.3420)}=\dfrac{0.5450}{0.8016}=0.6800$.
3. Solve for $\phi$. $\phi=\arctan(0.6800)=\boxed{34.2^{\circ}}$.