turning assign
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8/8/2019 Turning Assign
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Turning: Engine Lathe
Turning is another of the basic machining processes. Information in this section is
organized according to the subcategory links in the menu bar to the left. Turning
produces solids of revolution which can be tightly toleranced because of the
specialized nature of the operation. Turning is performed on a machine called a lathe
in which the tool is stationary and the part is rotated. The figure below illustrates an
engine lathe. Lathes are designed solely for turning operations, so that precise control
of the cutting results in tight tolerances. The work piece is mounted on the chuck,
which rotates relative to the stationary tool.
http://www.efunda.com/processes/machining/turn_engine_lathe.cfmhttp://www.efunda.com/processes/machining/turn_elathe_chucks.cfm#turn_elathe_chucks_three_four_jaw_chuckhttp://www.efunda.com/processes/machining/turn_elathe_chucks.cfm#turn_elathe_chucks_three_four_jaw_chuckhttp://www.efunda.com/processes/machining/turn_elathe_chucks.cfm#turn_elathe_chucks_three_four_jaw_chuckhttp://www.efunda.com/processes/machining/turn_engine_lathe.cfm -
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Turning
Turning refers to cutting as shown below.
The term "facing" is used to describe removal of material from the flat end of a
cylindrical part, as shown below. Facing is often used to improve the finish of
surfaces that have been parted.
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Q1: Component Fig 1 to be manufactured from work piece of size 50 mm dia.
and 140mm length.
b. Operations Plan
Operations
1 Chucking of work piece and turning to length
2 Rough and finish turning from 50 dia. to 45 dia.3 Necking turning from 45 dia. to 35 dia.
4 Shoulder facing from 35 dia. to 10 mm dia.
5 Taper turning from end 35 dia. to end 10 dia.
6 Drill dept of 40 mm using 10 mm dia. drill bits
Q2: Reduce the diameter of raw material in a single pass into 45 mm in dia.
Calculate MMR if given: i. Spindle rotates 700 rpm
ii. Feed rate 0.5 mm/rev
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Time taken for single cut:
T = L / F x N
= 140 / 0.5 x 700
= 0.4 min
= 24 sec for MMR
Q3: Assume Plain milling; Calculate feed in mm/min machining cast iron
Given:
i. Cutter dia. = 75 mm
ii. Tooths = 60 helical
iii. Cutting speed = 60m/min
iv. Feed/tooth = 0.25 mm
Solution:
Feed rev = Feed/tooth x No. of teeth
RPM = 100 S / D
RPM = 100 (60) / (75)
= 25.45 rpm
Total feed/min = Feed/tool x No. of tooths x RPM
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= 0.25 x 60 x 25.45
= 382 mm/min
Q 4: The holding device which been used in above operation at Q1
i. Chucks
The chuck is integral to a lathe's functioning because
it fixtures the part to the spindle axis of the machine.
Below is shown a three-jaw chuck with jaws that are
all driven by the same chuck key. This arrangement
provides convenience in that parts can be mounted
and dismounted quickly.
ii. Tail Stock
The tail stock is used to provide a fixture at the end of the part opposite from the
chuck. It also can be used to support a long, thin part so that more radial cutting force
can be applied and higher rotational speeds can be attained without a "whipping"
instability effect. Drill bits can be fixtured in the tail stock to cut axial holes in a
turned part. These central holes are more accurate than a drill press or mill could
provide since the lathe is dedicated to operations in which an axially-symmetric part
is rotated about its central axis. The fixturing is more accurate since all fixturing is
based upon surfaces of revolution about the central axis, and the machining is more
rigidly supported for the same reason.
http://www.efunda.com/processes/machining/drill_bit_chuck.cfmhttp://www.efunda.com/processes/machining/drill_bit_chuck.cfm -
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Q 5: The holding device which been used in above operation at Q1
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