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FUNDAMENTALS OF MACHINING
Or ORTHOGONAL MACHINING
Machining is the process of removing unwanted material from a workpiece
in the form of chips. If the workpiece is metal, the process is often called
metal cutting or metal removal.

What makes this process so unique and difficult to analyze?
BASIC MACHINING PROCESSES
BASIC MACHINING PROCESSES
TURNING PROCESS
TURNING PROCESS
1. For a turning operation, you have selected an HSS tool and turning a hot rolled free machining steel, Bhn
300. Your depth of cut will be 0.150 in. The diameter of the workpiece is 1.00 inches.
a. What speed and feed would you select for this job?
b. Using a speed of 105 sfpm and a feed of 0.015, calculate the spindle rpm for this operation.
c. Calculate the metal removal rate.
d. Calculate the cutting time for the operation with a length of cut of 4 in. and .10-in. allowance.
MILLING PROCESS
MILLING PROCESS
MILLING PROCESS
2. For a slab milling operation using a 5-in.-diameter, 11-tooth cutter. the feed per
tooth is 0.005 in./tooth with a cutting speed of 100 sfpm (HSS steel). Calculate the
rpm of the cutter and the feed rate (fm) of the table, then calculate the metal removal
rate, MRR, where the width of the block being machined is 2 in. and the depth of cut
is 0.25 in. Calculate the time to machine (Tm) a 6-in.-long block of metal with this
Setup. Suppose you switched to a coated-carbide tool, so you increase the cutting
speed to 400 sfpm. Now recalculate the machining time (Tm) with all the other
parameters the same.
DRILLING PROCESS

D = diameter of the
drill which rotates 2
cutting edges at rpm
Ns.
V = velocity of
outer edge of the lip of
the drill.
Ns = 12V/ΠD.
Tm = cutting time
= (L + A)/frNs
.

where fr is the feed rate in in. per rev. The allowance A = D/2.
The MRR =
which is approximately 3DVfr
3. The power required to machine metal is related to the cutting force (Fc) and the
cutting speed. For Problem 1, estimate cutting force Fc for this turning operation.
(Hint:You have to estimate a value of HPs for this material.)
4. In order to drill a hole in the material described in Problem 1 using an HSS drill, you
have to select a cutting speed and a feed rate. Using a speed of 105 sfpm for the HSS drill,
calculate the rpm for a -in.-diameter drill and the MRR if the feed rate is 0.008 inches per
revolution.
BROACHING PROCESS

The Tm for broaching is Tm = L /12V.
The MRR (per tooth) is 12tWV in.3/min
where V = cutting velocity in fpm,
W is the width of cut, t = rise per tooth.
Usually, 30 to 40% of the total energy goes into friction and 60 to 70% into the
shear process.
Doubling speed or depth of cut
In general, increasing the speed, the feed, or the depth of cut will increase the power
requirement. Doubling the speed doubles the horsepower directly. Doubling the feed
or the depth of cut doubles the cutting force Fc. In general, increasing the speed does
not increase the cutting force Fc, a surprising experimental result.
However, speed has a strong effect on tool life because most of the input energy is
converted into heat, which raises the temperature of the chip, the work, and the tool,
which effect the tool life.
The angle that the tool makes with respect to a vertical from the workpiece is called
the back rake angle a. A positive angle is shown in the schematic. The chip is formed by
shearing. The onset of shear occurs at a low boundary deformed by angle f with respect
to the horizontal.
In metal cutting, we observe that the onset of shear (to form the chip) is delayed by
increased hardness (so f increases directly with hardness).
If the work material has hard second-phase particles dispersed in it, they can act as
barriers to the shear front dislocations, which cannot penetrate the particle. The
dislocations create voids around the particles. If there are enough particles of the right
size and shape, the chip will fracture through the shear zone, forming segmented chips.
Free-machining steels, which have small percentages of hard second-phase particles
added to them, use this metallurgical phenomenon to break up the chips for easier chip
handling.
MERCHANT MODEL
The shear angle can be measured statically by instantaneously interrupting the cut
through the use of quick-stop devices. These devices disengage the cutting tool from
the workpiece while cutting is in progress, leaving the chip attached to the workpiece.
Optical and scanning electron microscopy is then used to observe the direction of
shear.
It is assumed that the resultant force R acting
on the back of the chip is equal and opposite to
the resultant force R acting on the shear plane.
The resultant R is composed of the friction
force F and the normal force N acting on the
tool–chip interface contact area.
The resultant force R’ is composed of a shear
force Fs and normal force Fn acting on the
shear plane area As
Since neither of these two sets of forces can
usually be measured, a third set is needed, which
can be measured using a dynamometer (force
transducer) mounted either in the workholder or
the tool holder.
The only symbol in this figure as
yet undefined is b, which is the
angle between the normal force
N and the resultant R. It is called
friction angle b and is used to
describe the friction coefficient
m on the tool–chip interface area,
which is defined as F/N so that
Unit power is sensitive to material properties (e.g., hardness), rake angle,
depth of cut, and feed, whereas ts is sensitive to material properties only.
CHATTER
LOBE DIAGRAM

The amplitude of chatter vibration may be more safely limited by temporary
reduction of the feed per tooth until a preferred speed and stable depth of cut have
been established.
SHAPING PROCESS
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311 chapter 20

  • 1. FUNDAMENTALS OF MACHINING Or ORTHOGONAL MACHINING
  • 2. Machining is the process of removing unwanted material from a workpiece in the form of chips. If the workpiece is metal, the process is often called metal cutting or metal removal. What makes this process so unique and difficult to analyze?
  • 3.
  • 6.
  • 9.
  • 10. 1. For a turning operation, you have selected an HSS tool and turning a hot rolled free machining steel, Bhn 300. Your depth of cut will be 0.150 in. The diameter of the workpiece is 1.00 inches. a. What speed and feed would you select for this job? b. Using a speed of 105 sfpm and a feed of 0.015, calculate the spindle rpm for this operation. c. Calculate the metal removal rate. d. Calculate the cutting time for the operation with a length of cut of 4 in. and .10-in. allowance.
  • 14. 2. For a slab milling operation using a 5-in.-diameter, 11-tooth cutter. the feed per tooth is 0.005 in./tooth with a cutting speed of 100 sfpm (HSS steel). Calculate the rpm of the cutter and the feed rate (fm) of the table, then calculate the metal removal rate, MRR, where the width of the block being machined is 2 in. and the depth of cut is 0.25 in. Calculate the time to machine (Tm) a 6-in.-long block of metal with this Setup. Suppose you switched to a coated-carbide tool, so you increase the cutting speed to 400 sfpm. Now recalculate the machining time (Tm) with all the other parameters the same.
  • 15. DRILLING PROCESS D = diameter of the drill which rotates 2 cutting edges at rpm Ns. V = velocity of outer edge of the lip of the drill. Ns = 12V/ΠD. Tm = cutting time = (L + A)/frNs . where fr is the feed rate in in. per rev. The allowance A = D/2. The MRR = which is approximately 3DVfr
  • 16.
  • 17. 3. The power required to machine metal is related to the cutting force (Fc) and the cutting speed. For Problem 1, estimate cutting force Fc for this turning operation. (Hint:You have to estimate a value of HPs for this material.)
  • 18. 4. In order to drill a hole in the material described in Problem 1 using an HSS drill, you have to select a cutting speed and a feed rate. Using a speed of 105 sfpm for the HSS drill, calculate the rpm for a -in.-diameter drill and the MRR if the feed rate is 0.008 inches per revolution.
  • 19. BROACHING PROCESS The Tm for broaching is Tm = L /12V. The MRR (per tooth) is 12tWV in.3/min where V = cutting velocity in fpm, W is the width of cut, t = rise per tooth.
  • 20.
  • 21.
  • 22.
  • 23.
  • 24.
  • 25. Usually, 30 to 40% of the total energy goes into friction and 60 to 70% into the shear process.
  • 26. Doubling speed or depth of cut In general, increasing the speed, the feed, or the depth of cut will increase the power requirement. Doubling the speed doubles the horsepower directly. Doubling the feed or the depth of cut doubles the cutting force Fc. In general, increasing the speed does not increase the cutting force Fc, a surprising experimental result. However, speed has a strong effect on tool life because most of the input energy is converted into heat, which raises the temperature of the chip, the work, and the tool, which effect the tool life.
  • 27.
  • 28. The angle that the tool makes with respect to a vertical from the workpiece is called the back rake angle a. A positive angle is shown in the schematic. The chip is formed by shearing. The onset of shear occurs at a low boundary deformed by angle f with respect to the horizontal.
  • 29. In metal cutting, we observe that the onset of shear (to form the chip) is delayed by increased hardness (so f increases directly with hardness).
  • 30. If the work material has hard second-phase particles dispersed in it, they can act as barriers to the shear front dislocations, which cannot penetrate the particle. The dislocations create voids around the particles. If there are enough particles of the right size and shape, the chip will fracture through the shear zone, forming segmented chips. Free-machining steels, which have small percentages of hard second-phase particles added to them, use this metallurgical phenomenon to break up the chips for easier chip handling.
  • 32. The shear angle can be measured statically by instantaneously interrupting the cut through the use of quick-stop devices. These devices disengage the cutting tool from the workpiece while cutting is in progress, leaving the chip attached to the workpiece. Optical and scanning electron microscopy is then used to observe the direction of shear.
  • 33. It is assumed that the resultant force R acting on the back of the chip is equal and opposite to the resultant force R acting on the shear plane. The resultant R is composed of the friction force F and the normal force N acting on the tool–chip interface contact area. The resultant force R’ is composed of a shear force Fs and normal force Fn acting on the shear plane area As Since neither of these two sets of forces can usually be measured, a third set is needed, which can be measured using a dynamometer (force transducer) mounted either in the workholder or the tool holder.
  • 34. The only symbol in this figure as yet undefined is b, which is the angle between the normal force N and the resultant R. It is called friction angle b and is used to describe the friction coefficient m on the tool–chip interface area, which is defined as F/N so that
  • 35.
  • 36. Unit power is sensitive to material properties (e.g., hardness), rake angle, depth of cut, and feed, whereas ts is sensitive to material properties only.
  • 37.
  • 38.
  • 40.
  • 41.
  • 42.
  • 43. LOBE DIAGRAM The amplitude of chatter vibration may be more safely limited by temporary reduction of the feed per tooth until a preferred speed and stable depth of cut have been established.