محاضرة بعنوان Milling Operations – TA 102 Workshop Practice

محاضرة بعنوان
Milling Operations – TA 102 Workshop Practice
By Prof. A. chANDRASHEKHAR

• Milling machines are used to produce parts having flat as well as
  curved shapes.
• Milling machines are capable of performing the usual flat,
  horizontal and vertical machining operations and can be used to
  do many other machining operations like gear teeth cutting,
  drilling, reaming, boring, slotting, tapping, keyway cutting, cam
  milling and so forth.
• In this process, the work piece is normally fed into a rotating
  cutting tool known as milling cutter. Equally spaced peripheral
  teeth on the cutter come in contact with the work piece
  intermittently and machine the work piece.- In conventional milling operation, metal is cut as the work piece
  is fed against a rotating multi point cutter.
• Different types of cutters are used in milling like end mill cutter,
  side and face cutter machine, form cutter and so forth.The Milling Machine
• Two types of milling machines are common. They are
  distinguished by the orientation of the axis of rotation of the
  spindle of the cutter.
• In horizontal milling machine, the axis of the spindle in
  horizontal and
• In vertical milling machine the axis of the spindle is vertical.Horizontal Knee and Column type milling
  machine
• Principal parts
  i. Base
  ii. Column
  iii. Knee
  iv. Saddle
  v. Table
  vi. Over arm
  vii. Spindle, and
  viii. arborVertical Knee and column type milling
  machine
• A horizontal milling
  machine can be converted in
  to a vertical milling machine
  by removing the arbor and
  attaching a vertical milling
  attachment to the spindle.The Milling Process
• Milling is a machining process for producing flat, countered and
  helical surfaces by means of multiple- cutting-edged rotating tools
  called milling cutters.
• There are two different methods of metal cutting in the milling
  operation
• up milling and
• down milling
  The difference between these two types of operations
  lies in the direction along which the work piece is fed in to the
  rotating milling cutter and the direction of rotation of cutter.UP Milling
• In up milling or conventional milling, the feed direction of
  the work piece is opposite to that of the cutter rotation.
• Each tooth of the cutter starts the cut with zero depth of
  cut, which gradually increases and reaches the maximum
  value as the tooth leaves the cut.
• The chip thickness at the start is zero and increases to the
  maximum at the end of the cut.
• The surface becomes slightly wavy, as the cut does not begin
  as soon as the cutter touches the work piece.UP Milling
• The resultant cutting force are directed upward and tend to lift the
  work piece upward from the table, and therefore, more secured
  clamping of the work piece is required.Down Milling
• Also called as Climb milling.
• The feed direction of the work piece is same as that of
  the cutter rotation.
• The maximum thickness of the chip at the start of the
  cut and decreases to zero thickness at the end of the cut.
• The resultant cutting forces in the down milling are
  directed downward in to the machine and tend to drag
  the work piece in to the cutter.
• This type of milling produces better surface finish and
  dimensional accuracy.DOWN MILLINGTypes of Milling CuttersOperating Conditions in Milling
• Cutting Speed is defined as the peripheral speed of the cutter.
  = πDN/1000 m/min
  Where,
  D = Diameter of the cutter in mm
  N = Rotational Speed of the cutter in rpm
• Both N and D refer to the milling cutter in milling
• The selection of the cutting speed depends on the properties of the
  material being cut, diameter and life of cutter, number of cutter teeth,
  feed, depth of cut as well as width of cut and coolant used.FEED
• Feed is defined as the movement of work piece relative to the
  cutter axis.
• In milling, the feed can be defined in three ways

1. Feed per tooth is defined as the distance advanced by the work
   piece between the time interval when two teeth come into cutting
   action.
   F
   1 = f / zN mm/tooth
   Where,
   f = feed rate in mm
   Z= number of teeth on the cutter periphery
   N = rpm of the cutter.2. Feed per cutter revolution F2 is defined as the distance
   advanced by the work piece in the time interval when
   cutter goes through one complete revolution.
   F
   2 = F1z mm/rev
2. Feed per minute is the distance advanced by the work
   piece in on minute f.
   f = F
   2N = F1zNDepth of cut

• It is defined as the thickness of the layer of material
  removed in one pass of the work piece under cutter. It is
  expressed in mm.Width of cut
• The width of cut ω is the width of work piece surface contacting
  the cutter in a direction perpendicular to the feed.Material removal rate
• It is the volume of material removed in unit time.
  For milling
  MRR in mm3/min
  MRR = ω d f
  or MRR = ω d F
  2 N
  or MRR = ω d F
  1z NMachining time
• It is defined as the time required for one pass of width
  of cut ω for milling or machining a surface and is
  expressed in minutes.
   Where L is the length of cut includes length of job +
  length of approach + length of over travel distanceLength of cutProblem 1
• Determine the time required to mill a slot of 300 X 25 mm
  in a work piece of 300 mm length with a side and face milling
  cutter of 100 mm diameter, 25 mm wide and having 18
  teeth. The depth of cut is 5 mm, the feed per tooth is 0.1 mm
  and cutting speed is 30 m/min. assume approach and over
  travel distance of 50 mm.- Given
  L
  j = 300 mm, D = 100 mm, w = 25 mm,
  z = 18, d = 5 mm, F1 = 0.1 mm/tooth
  v = 30 m/min
  N = 1000 v / П D
  = 100 rpm
  Feed rate = feed per tooth X number of teeth on cutter X rpm
  f = F
  1zN = 0.1 X 18 X 100
  = 180 mm/min- L1 = L2 = d(D – d)
  = 22 mm
  Length of cut L = Lj + L1 + approach + Over travel
  = 300 + 22 + 50
  = 372 mm
  t = Length of cut/feed rate = L/ f = 372/180
  = 2.067 min
  = 2 min (app)Problem 2
• Determine the cutting time for cutting a 125 mm long
  keyway using HSS end-mill of 20 mm diameter having four
  cutting teeth. The depth of keyway is 4.5mm. Feed per tooth
  is 0.1mm and the cutting speed is 40 m/min. assume
  approach and over travel distances half of the diameter of the
  cutter and a depth of 4.5 mm can be cut in one pass.- Given
  Lj = 125 mm, D = 20 mm, z = 4
  d = 4.5 mm, F1 = 0.1 mm/tooth,
  v = 40 m/min
  N = 1000 v / ПD = 636 rpm
  Feed per minute = feed per tooth X no. of teeth X rpm
  = 0.1 X 4 X 636
  = 254.4 mm/min
  Length of cut L = 125 + 10 = 135 mm
  Cutting time = t = 135/254.4 = 0.53 min.Problem 3
• For a given milling operation, it was decided to switch from HSS
  cutter to carbide cutter, changing the cutting speed from 35
  m/min to 110 m/min. the other parameters of the cutting
  operation in the two cases are:
• carbide cutter HSS Cutter
  Cutter dia (mm) 150 125
  Feed (mm/tooth) 0.0425 0.0375
  Number of teeth 12 10- Calculate the following for each of the cutting tools
  a) Cutter rpm,
  b) Feed in mm/min,
  c) Time required to take 200-mm long cut including approach
  and over travel, and
  d) Percentage saving in time by changing from HSS to Carbide
  tool.Problem 3 Solution
• For HSS Cutter:
  A) N = 1000 X 35/ (П X 125) = 89 rpm
  B) f = F1zN = 0.0375 X 10 X 89 = 33.375 mm/min
  C) t = L/f = 200 / 33.375 = 6 min
• For Carbide Cutter:
  A) N = 233 rpm
  B) f = F1zN = 118.8 mm/min
  C) t = L/f = 1.68 min- Percentage saving in time = 6 – 1.68 /6 X100
  = 72 %Problem 4
• Write down a possible sequence of operations for
  manufacturing the component shown in figure. From the
  stock having a length of 140 mm and 60 mm diameter.

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