me 482 - manufacturing systems metal machining metal machining

ME 482 - Manufacturing SystemsME 482 - Manufacturing Systems

Metal

Machining

Metal

Machining


Objectives

• Introduce cutting terminology and principles

• Review modern machining technologies and new methods (papers)

• Introduce cutting parameters

• Develop cutting models

• Analyze a cutting example

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 types• Turning

• Drilling

• Milling

• Shaping

• Planing

• Broaching


Machining tools

• Single point

• Multiple point

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 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 finishMachining 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 edge

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

Flank – following edge of cutting tool

Relief angle – angle of tool’s following edge above part surface

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 edge

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

Flank – following edge of cutting tool

Relief 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 ()

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 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 geometryCutting geometry

Chip 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]

Chip 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 shearing of mode is shearing of the work along the the work along the

shear plane.shear plane.

Obviously, the Obviously, the assumed failure assumed failure

mode is shearing of mode is shearing of the work along the the work along the

shear plane.shear plane.


Cutting geometryCutting geometry

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

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

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

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

Thus, if know Thus, if know r and r and can can determine determine , , and given and given and and , can , can

determine determine ..

Thus, if know Thus, if know r and r and can can determine determine , , and given and given and and , can , can

determine determine ..


Cutting forcesCutting 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 force

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

Fn = Fc sin + Ft cos

Since 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 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.


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.


Merchant equationsMerchant equations

Combining the equations from the previous slides:

= (Fc cos - Ft sin tow/sin Merchant eqn

The most likely shear angle will minimize the energy. Applying d/d = 0 gives:

= 45° + Merchant reln

What does the Merchant relation indicate?

Combining the equations from the previous slides:

= (Fc cos - Ft sin tow/sin Merchant 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

- increase in friction angle decreases shear angle

- increase in rake angle increases shear angle

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

requirements!requirements!

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

requirements!requirements!

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

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

previous slide!previous slide!

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

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

previous slide!previous slide!


Cutting modelsThe orthogonal model for turning approximates the complex shearing process:

Cutting modelsThe orthogonal model for turning approximates the complex shearing process:

to = feed (f)

w = depth of cut (d)

to = feed (f)

w = depth of cut (d)


Cutting powerCutting power

Power is force times speed:

P = Fc v (ft-lb/min)

The cutting horsepower is

hpc = 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

Power is force times speed:

P = Fc v (ft-lb/min)

The cutting horsepower is

hpc = 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 energyCutting 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”).

Specific 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 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!


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.)


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!


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.)


Solution (cutting horsepower):

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

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


Solution (cutting horsepower):

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

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


Cutting temperaturesCutting 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!

In 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 calculates Example in text calculates TT = 936 = 936°° total tool total tool

temperature, given v = temperature, given v = 200 ft/min, 200 ft/min, cc = 120 in- = 120 in-

lb/(inlb/(in33- - °F°F) and K = 0.125 ) and K = 0.125 inin22/s/s


CuttersCutters

Toroid


CuttersCutters


MachiningMachining

What did we learn?What did we learn?

me 482 - manufacturing systems metal machining metal machining

Documents

cutting terminology

modern cutting tool

cutting parameters

cutting models

f depth of cut depth

surface slide

feed tool penetration

cutting example objectives