1. Overview of Machining Technology2. Theory of Chip Formation in Metal

Machining3. Force Relationships and the Merchant

Equation4. Power and Energy Relationships in

Machining5. Cutting Temperature

A family of shaping operations, the common feature of which is removal of material from a starting workpart so the remaining part has the desired geometry

Machining – material removal by a sharp cutting tool, e.g., turning, milling, drilling

Abrasive processes – material removal by hard, abrasive particles, e.g., grinding

Nontraditional processes - various energy forms other than sharp cutting tool to remove material

Cutting action involves shear deformation of work material to form a chip As chip is removed, new surface is exposed

(a) A cross‑sectional view of the machining process, (b) tool with negative rake angle; compare with positive rake angle in (a).

Machining

Variety of work materials can be machined◦ Most frequently used to cut metals

Variety of part shapes and special geometric features possible, such as:◦ Screw threads◦ Accurate round holes◦ Very straight edges and surfaces

Good dimensional accuracy and surface finish

Wasteful of material ◦ Chips generated in machining are wasted

material, at least in the unit operation Time consuming

◦ A machining operation generally takes more time to shape a given part than alternative shaping processes, such as casting, powder metallurgy, or forming

Generally performed after other manufacturing processes, such as casting, forging, and bar drawing ◦ Other processes create the general shape of the

starting workpart◦ Machining provides the final shape, dimensions,

finish, and special geometric details that other processes cannot create

Speed is the relative movement between tool and w/p, which produces a cut

Feed is the relative movement between tool and w/p, which spreads the cut

Most Important Machining Operations:◦ Turning◦ Drilling◦ Milling

Other Machining Operations:◦ Shaping and Planing◦ Broaching◦ Sawing

Single point cutting tool removes material from a rotating workpiece to form a cylindrical shape

Three most common machining processes: (a) turning,

Turning

Used to create a round hole, usually by means of a rotating tool (drill bit) with two cutting edges

Drilling

Rotating multiple-cutting-edge tool is moved across work to cut a plane or straight surface

Two forms: peripheral (side) milling and face (end) milling

(c) peripheral milling, and (d) face milling.

Milling

1. Single-Point Tools◦ One dominant cutting edge◦ Point is usually rounded to form a nose radius◦ Turning uses single point tools

2. Multiple Cutting Edge Tools◦ More than one cutting edge◦ Motion relative to work achieved by rotating ◦ Drilling and milling use rotating multiple

cutting edge tools

(a) A single‑point tool showing rake face, flank, and tool point; and (b) a helical milling cutter, representative of tools with multiple cutting edges.

Cutting Tools

Three dimensions of a machining process: ◦ Cutting speed v – primary motion◦ Feed f – secondary motion◦ Depth of cut d – penetration of tool

below original work surface For certain operations, material

removal rate can be computed as RMR = v f d

where v = cutting speed; f = feed; d = depth of cut

Cutting Conditions for Turning

Speed, feed, and depth of cut in turning.

In production, several roughing cuts are usually taken on the part, followed by one or two finishing cuts

Roughing - removes large amounts of material from starting workpart◦ Creates shape close to desired geometry, but

leaves some material for finish cutting◦ High feeds and depths, low speeds

Finishing - completes part geometry◦ Final dimensions, tolerances, and finish◦ Low feeds and depths, high cutting speeds

A power‑driven machine that performs a machining operation, including grinding

Functions in machining:◦ Holds workpart◦ Positions tool relative to work◦ Provides power at speed, feed, and depth that

have been set The term is also applied to machines that

perform metal forming operations

where r = chip thickness ratio; to = thickness of the chip prior to chip formation; and tc = chip thickness after separation

Chip thickness after cut is always greater than before, so chip ratio always less than 1.0

c

o

tt

r

totc

Based on the geometric parameters of the orthogonal model, the shear plane angle can be determined as:

where r = chip ratio, and = rake angle

sincos

tanr

r

1

Figure 21.7 Shear strain during chip formation: (a) chip formation depicted as a series of parallel plates sliding relative to each other, (b) one of the plates isolated to show shear strain, and (c) shear strain triangle used to derive strain equation.

Shear Strain in Chip Formation

Shear strain in machining can be computed from the following equation, based on the preceding parallel plate model:

= tan( - ) + cot

where = shear strain, = shear plane angle, and = rake angle of cutting tool

Friction force F and Normal force to friction N Shear force Fs and Normal force to shear Fn

Forces in metal cutting: (a) forces acting on the chip in orthogonal cutting

Forces Acting on Chip

Vector addition of F and N = resultant R Vector addition of Fs and Fn = resultant R' Forces acting on the chip must be in balance:

◦ R' must be equal in magnitude to R ◦ R’ must be opposite in direction to R◦ R’ must be collinear with R

F, N, Fs, and Fn cannot be directly measured Forces acting on the tool that can be

measured:◦ Cutting force Fc and Thrust force Ft

©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e

Figure 21.10 Forces in metal cutting: (b) forces acting on the tool that can be measured

Cutting Force and Thrust Force

Coefficient of friction between tool and chip:

Friction angle related to coefficient of friction as follows:

NF

tan

F

F

N

Fn

Fs

R

F

F

N

R

F

F

N

R

Ft

Fc

F

F

N

R

Ft

Fc

F = Fc sin + Ft cos

N = Fc cos - Ft sin

F

F

N

Fn

Fs

R

F

Fn

Fs

R

F

Fn

Fs

R

Fc

Ft

F

Fn

Fs

R

Fc

Ft

Fs = Fc cos - Ft sin

Fn = Fc sin + Ft cos

Thus equations can be derived to relate the forces that cannot be measured to the forces that can be measured:

F = Fc sin + Ft cosN = Fc cos ‑ Ft sin

Fs = Fc cos ‑ Ft sinFn = Fc sin + Ft cos

Based on these calculated force, shear stress and coefficient of friction can be determined

Significance of Cutting forcesSignificance of Cutting forces In the set of following force balance equations:-

F = Fc sin + Ft cos F = friction force; N = normal to chip force

N = Fc cos - Ft sin Fc = cutting force; Ft = thrust force

Fs = Fc cos - Ft sin Fs = shear force; Fn = normal to shear plane force

Fn = Fc sin + Ft cos

In the set of following force balance equations:-

F = Fc sin + Ft cos F = friction force; N = normal to chip force

N = Fc cos - Ft sin Fc = cutting force; Ft = thrust force

Fs = Fc cos - Ft sin Fs = shear force; Fn = normal to shear plane force

Fn = Fc sin + Ft cos

Friction angle =

tan= = F/N

Shear plane stress:

= Fs/As

where

As = to w/sin

Friction angle =

tan= = F/N

Shear plane stress:

= Fs/As

where

As = to w/sin

Forces are presented as function of Forces are presented as function of FFcc and F and Ftt because these can be because these can be

measured.measured.

Forces are presented as function of Forces are presented as function of FFcc and F and Ftt because these can be because these can be

measured.measured.

Shear stress acting along the shear plane:

sinwt

A os

where As = area of the shear plane

Shear stress = shear strength of work material during cutting

s

s

AF

S

Cutting forces given shear strengthCutting forces given shear strength

Letting S = shear strength, we can derive the following equations for the cutting and thrust forces*:

Fs = S As

Fc = Fs cos (cos (

Ft = Fs sin (cos (

* The other forces can be determined from the equations on the previous slide.

Letting S = shear strength, we can derive the following equations for the cutting and thrust forces*:

Fs = S As

Fc = Fs cos (cos (

Ft = Fs sin (cos (

* The other forces can be determined from the equations on the previous slide.

Machining exampleMachining example

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower.

Solution (shear strain):

Determine r = 0.02/0.045 = 0.444

Determine shear plane angle from tan = r cos /[1 – r sin]

tan = 0.444 cos /[1 – 0.444 sin] => = 25.4°

Now calculate shear strain from = tan(-) + cot

= tan(25.4 - 10) + cot 25.4 = 2.386 in/in answer!

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower.

Solution (shear strain):

Determine r = 0.02/0.045 = 0.444

Determine shear plane angle from tan = r cos /[1 – r sin]

tan = 0.444 cos /[1 – 0.444 sin] => = 25.4°

Now calculate shear strain from = tan(-) + cot

= tan(25.4 - 10) + cot 25.4 = 2.386 in/in answer!

Machining example (cont.)Machining example (cont.)

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower.

Solution (shear stress):

Determine shear force from Fs = Fc cos - Ft sin

Fs = 350 cos 25.4 - 285 sin 25.4 = 194 lb

Determine shear plane area from As = to w/sin

As = (0.02) (0.125)/sin= 0.00583 in2

The shear stress is

= 194/0.00583 = 33,276 lb/in2

answer!

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower.

Solution (shear stress):

Determine shear force from Fs = Fc cos - Ft sin

Fs = 350 cos 25.4 - 285 sin 25.4 = 194 lb

Determine shear plane area from As = to w/sin

As = (0.02) (0.125)/sin= 0.00583 in2

The shear stress is

= 194/0.00583 = 33,276 lb/in2

answer!

Machining example (cont.)Machining example (cont.)

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower.

Solution (cutting horsepower):

Determine cutting hp from hpc = Fc v/33,000

hpc = (350) (200)/33,000 = 2.12 hp answer!

In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower.

Solution (cutting horsepower):

Determine cutting hp from hpc = Fc v/33,000

hpc = (350) (200)/33,000 = 2.12 hp answer!

Shear Plane Angle Ф = tan-1 [(r cos α )/(1 – r sin α)] Shear Strain = tan( - ) + cot Forces in Cutting:

F = Fc sin + Ft cosN = Fc cos ‑ Ft sin Fs = Fc cos ‑ Ft sinFn = Fc sin + Ft cos

Forces Fc and Ft in terms of Fs:

Fc = Fs cos (cos (

Ft = Fs sin (cos ( Merchant Relation:

Shear Stress:

= Fs/As

where As = to w/sin Cutting Power:

P = V Fc / 33,000 hp (V in ft /s and Fc in lb)

P = V Fc / 1000 kW (V in m /s and Fc in N)

PG = Pc / E