Ch5 Metal Machining

ME 482 - Manufacturing ME 482 - Manufacturing SystemsSystems

MetalMachining


Objectives

• Introduce cutting terminology and principles• Review modern machining technologies and new methods (papers)• Introduce cutting parameters• Develop cutting models• Analyze a cutting example


Machining types• Turning

• Drilling

• Milling

• Shaping

• Planing

• Broaching


Machining tools

• Single point

• Multiple point


Machining tool materials

Most modern cutting tool materials are a matrix of materials designed to be very hard. These materials will be covered in the next chapter.


Machining surface finish


Machining terminologySpeed – surface cutting speed (v)Feed – advance of tool through the part (f)Depth of cut – depth of tool into part (d)

Rake face – tool’s leading edgeRake angle – slant angle of tool’s leading edge ()

Flank – following edge of cutting toolRelief angle – angle of tool’s following edge above part surface


Machining terminology (cont.)

Chip thickness – thickness of machined chip (tc )

Depth of cut = to

Shear plane length – measured along shear plane chip (ls )

Chip width (not shown) – width of machined chip (w )

Shear angle – angle of shearing surface measured from tool direction ()

ls

Orthogonal model


Cutting conditionsNote: - Primary cutting due to speed

- Lateral motion of tool is feed- Tool penetration is depth of cut

The three together form the material removal rate (MRR):

MRR = v f d with units of (in/min)(in/rev)(in) = in3/min/rev (or vol/min-rev)

Types of cuts: Roughing: feeds of 0.015 – 0.05 in/rev depths of 0.1 – 0.75 in

Finishing: feeds of 0.005 – 0.015 in/rev depths of 0.03 – 0.075 in


Cutting geometryChip thickness ratio = r = to / tc

From the shear plane geometry:

r = ls sin/[ls cos(-)]

which can be arranged to get

tan = r cos /[1 – r sin]

Obviously, the Obviously, the assumed failure assumed failure

mode is mode is shearing of the shearing of the work along the work along the shear plane.shear plane.


Cutting geometryNote from the triangles in (c) that the shear strain () can be estimated as

= AC/BD = (DC + AD)/BD = tan(-) + cot

Thus, if Thus, if know r and know r and

can can determine determine

, and , and given given

and and , can , can determine determine

..


Cutting forcesSince R = R’ = R’’, we can get the 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 forceFs = Fc cos - Ft sin Fs = shear force; Fn = normal to

shear plane forceFn = Fc sin + Ft cos

Friction angle = tan= = F/N

Shear plane stress: = Fs/As

where As = to w/sin

Forces are presented as Forces are presented as function of Ffunction of Fcc and F and Ftt because these can be because these can be measured.measured.


Cutting forces given shear strengthLetting 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.


Merchant equationsCombining the equations from the previous slides:

= (Fc cos - Ft sin tow/sinMerchant eqn

The most likely shear angle will minimize the energy. Applying d/d = 0 gives: = 45° + Merchant reln

What does the Merchant relation indicate?- increase in friction angle decreases shear angle- increase in rake angle increases shear angle

If we increase the shear If we increase the shear angle, we decrease the angle, we decrease the tool force and power tool force and power

requirements!requirements!

The Merchant reln is a The Merchant reln is a function of function of and and . Where . Where did these variables come did these variables come from? from?

AnswerAnswer - - Although the Although the Merchant eqn is not shown Merchant eqn is not shown as a direct function of as a direct function of and and , these enter from , these enter from the equations for Fthe equations for Fc c and Fand Ft t

from the previous slide!from the previous slide!


Cutting modelsThe orthogonal model for turning approximates the complex shearing process: to = feed (f)

w = depth of cut (d)


Cutting powerPower is force times speed:

P = Fc v (ft-lb/min)The cutting horsepower ishpc = Fc v/33,000 (hp)The unit horsepower is

hpu = hpc/MRR units?

Due to efficiency losses (E about 90%), the gross hp is

hpg = hpc/E


Cutting energySpecific energy is

U = Fc v/(v tow) = Fc /(tow) (in-lb/in3)The table shown contains power and specific energy ratings for several work materials at a chip thickness of 0.01 in. For other chip thicknesses, apply the figure to get a correction factor multiply U by correction factor for thickness different than 0.01”).


Machining exampleIn 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 = 0.444Determine 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/inanswer!


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 lbDetermine 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.)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!


Cutting temperaturesIn machining 98% of the cutting energy is converted into heat. This energy flows into the work part, chip, and tool. Cook determined an experimental equation for predicting the temperature rise at the tool-chip interface during machining:

T = 0.4 U (v to/K)0.333/(c)where

T = mean temperature rise (°F)U = specific energy (in-lb/in3)v = cutting speed (in/s) to = chip thickness before cut (in) c= volumetric specific heat of the work

material (in-lb/(in3-°F)) K = thermal diffusivity of the work material (in2/s)Note - To get total temperature at tool-chip interface, must add in ambient temperature!

Example in text Example in text calculates calculates TT = = 936936°° total tool total tool temperature, temperature, given v = 200 given v = 200

ft/min, ft/min, cc = 120 = 120 in-lb/(inin-lb/(in33- - °F°F) ) and K = 0.125 and K = 0.125

inin22/s/s


Cutters

Toroid


Cutters


Machining

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Ch5 Metal Machining

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