Metal
Machining
ME 482 - Manufacturing Systems
Objectives
• Introduce cutting terminology and principles
• Review modern machining technologies and new methods
(papers)
• Introduce cutting parameters
• Develop cutting models
• Analyze a cutting example
ME 482 - Manufacturing Systems
Machining types
• Turning
• Drilling
• Milling
• Shaping
• Planing
• Broaching
ME 482 - Manufacturing Systems
Machining tools
• Single point
• Multiple point
ME 482 - Manufacturing Systems
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.
ME 482 - Manufacturing Systems
Machining surface finish
ME 482 - Manufacturing Systems
Machining terminology
Speed – 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 (a)
Flank – following edge of cutting tool
Relief angle – angle of tool’s following edge above part surface
ME 482 - Manufacturing Systems
Machining terminology (cont.)
ls
Chip thickness – thickness of machined chip (tc )
Orthogonal model
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 (f)
ME 482 - Manufacturing Systems
Cutting conditions
Note: - 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
ME 482 - Manufacturing Systems
Cutting geometry
Chip thickness ratio = r = to / tc
From the shear plane geometry:
r = ls sinf/[ls cos(f - a)]
which can be arranged to get
tan f = r cos a /[1 – r sin a]
ME 482 - Manufacturing Systems
Obviously, the
assumed failure
mode is shearing of
the work along the
shear plane.
Cutting geometry
Note from the triangles in (c) that the shear strain (g) can be
estimated as
g = AC/BD = (DC + AD)/BD = tan(f - a) + cot f
Thus, if know
r and a, can
determine f,
and given f
and a, can
determine g.
ME 482 - Manufacturing Systems
Cutting forces
Since R = R’ = R’’, we can get the force balance equations:
F = Fc sin a + Ft cos a
F = friction force; N = normal to chip force
N = Fc cos a - Ft sin a
Fc = cutting force; Ft = thrust force
Fs = Fc cos f - Ft sin f
Fs = shear force; Fn = normal to shear plane force
Fn = Fc sin f + Ft cos f
Forces are presented as function of
Fc and Ft because these can be
measured.
Friction angle = b
tan b = m = F/N
Shear plane stress:
t = Fs/As
where
As = to w/sin f
ME 482 - Manufacturing Systems
Cutting 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 ( b - a)/[cos ( f + b - a)]
Ft = Fs sin ( b - a)/[cos ( f + b - a)]
* The other forces can be determined from the equations on the previous
slide.
ME 482 - Manufacturing Systems
Merchant equations
The
Merchant
reln
is a function of a
Combining the equations from
the
previous
slides:
and b. Where did these variables
t = (Fc cos f - Ft sin f)/(t
f)
Merchant eqn
come
from?
ow/sin
The most likely shear angle will minimize the energy. Applying
Answer - Although the Merchant
dt/df = 0 gives:
eqn is not shown as a direct function
of a and b, these enter from
the reln
Merchant
equations for Fc and Ft from the
previous
slide!
What does the Merchant relation
indicate?
f = 45° + a/2 - b/2
If we increase the shear angle, we
- increase in friction angle decreases
decreaseshear
the toolangle
force and power
requirements!
- increase in rake angle increases shear angle
ME 482 - Manufacturing Systems
Cutting models
The orthogonal model for turning approximates the complex
shearing process:
to = feed (f)
w = depth of cut (d)
ME 482 - Manufacturing Systems
Cutting 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
ME 482 - Manufacturing Systems
Cutting energy
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”).
ME 482 - Manufacturing Systems
Machining 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 f = r cos a /[1 – r sin a]
tan f = 0.444 cos 10 /[1 – 0.444 sin 10] => f = 25.4°
Now calculate shear strain from g = tan(f - a) + cot f
g = tan(25.4 - 10) + cot 25.4 = 2.386 in/in
ME 482 - Manufacturing Systems
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 (shear stress):
Determine shear force from Fs = Fc cos f - Ft sin f
Fs = 350 cos 25.4 - 285 sin 25.4 = 194 lb
Determine shear plane area from As = to w/sin f
As = (0.02) (0.125)/sin 25.4= 0.00583 in2
The shear stress is
ME 482 - Manufacturing Systems
t = 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
ME 482 - Manufacturing Systems
answer!
Cutting temperatures
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:
Example in text calculates
DT = 0.4 U (v to
c)
DT = 936° total tool
where
temperature, given v =
200 ft/min, rc = 120 inDT = mean temperature rise (°F)
U = specific energy (in-lb/in3)
lb/(in3- °F) and K = 0.125
v = cutting speed (in/s)
in2/s
/K)0.333/(r
to = chip thickness before cut (in)
rc = 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!
ME 482 - Manufacturing Systems
Cutters
Toroid
ME 482 - Manufacturing Systems
Cutters
ME 482 - Manufacturing Systems
Machining
What did we learn?
ME 482 - Manufacturing Systems