Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

  1. In a machining operation, doubling the cutting speed reduces the tool life to 1/8th of the original value. The exponent n in Taylor's tool life equation VTn = C. is









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    Taylor’s tool life equation,
    VTn = C      ...(i)
    Where,   V = cutting speed and T = tool life
    When cutting speed is doubled and tool life reduces to 1/8, then

    2V
    1
    Tn = C       ...(ii)
    8

    Dividing equation (i) by equation (ii), we get
    VTn
    = 1
    2V
    T
    n
    8

    or   VTn = 2V
    T
    n = 2Vn2−3n
    8

    or   1 = 2⋅2−3n
    or   1 = 2−3n + 1
    or   2−3n + 1 = 20
    or   –3n + 1 = 0
    or   3n = 1
    or   n =
    1
    .
    3

    Correct Option: C

    Taylor’s tool life equation,
    VTn = C      ...(i)
    Where,   V = cutting speed and T = tool life
    When cutting speed is doubled and tool life reduces to 1/8, then

    2V
    1
    Tn = C       ...(ii)
    8

    Dividing equation (i) by equation (ii), we get
    VTn
    = 1
    2V
    T
    n
    8

    or   VTn = 2V
    T
    n = 2Vn2−3n
    8

    or   1 = 2⋅2−3n
    or   1 = 2−3n + 1
    or   2−3n + 1 = 20
    or   –3n + 1 = 0
    or   3n = 1
    or   n =
    1
    .
    3


  1. In an orthogonal cutting test on mild Steel, the following data were obtained
    Cutting speed    :  40 m/min
    Depth of cut      :  0.3 mm
    Tool rake angle  : +5º
    Chip thickness   : 1.5 mm
    Cutting force      : 900 N
    Thrust force       : 450 N
    Using Merchants analysis, the friction angle during the machining will be









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    μ =
    F
    =
    FCsinα + FTcosα
    NFCcosα − FTsinα

    =
    900sin5º + 450cos5º
    =
    526.72
    = 0.614
    900cos5º −450sin5º857.35

    ∴  λ = tan−1 = 31.5º
    Alternative method
    From Merchants' Circle

    Given:   FC = 900 N
    FT = 450 N
    Now
    tan(λ − α) =
    FT
    =
    450
    FC900

    ∴  (λ − α) = tan−1
    1
    2

    [where, α = Rake angle]
    Friction angle, λ = 26.5 + 5 = 31.50

    Correct Option: B

    μ =
    F
    =
    FCsinα + FTcosα
    NFCcosα − FTsinα

    =
    900sin5º + 450cos5º
    =
    526.72
    = 0.614
    900cos5º −450sin5º857.35

    ∴  λ = tan−1 = 31.5º
    Alternative method
    From Merchants' Circle

    Given:   FC = 900 N
    FT = 450 N
    Now
    tan(λ − α) =
    FT
    =
    450
    FC900

    ∴  (λ − α) = tan−1
    1
    2

    [where, α = Rake angle]
    Friction angle, λ = 26.5 + 5 = 31.50



  1. Through holes of 10 mm diameter are to be drilled in a steel plate of 20 mm thickness. Drill spindle speed is 300 rpm, feed 0.2 mm/rev and drill point angle is 120°. Assuming drill over travel of 2 mm, the time for producing a hole will be









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    Given: Diameter of hole,
    d =10 mm
    Thickness of steel plate,
    t = 20 mm

    α =
    120
    = 60º
    2

    Break through distance,
    A =
    d
    =
    10
    = 2.8867 mm
    2tanα2tan60º

    Total length of tool travel,
    l = t + A + 2
    = 20 + 2.8867 + 2 = 24.88
    Time for drilling
    =
    l
    =
    24.88
    = 0.4147 min
    fN0.2 × 300

    = 24.88 sec ≈ 25 sec.

    Correct Option: B

    Given: Diameter of hole,
    d =10 mm
    Thickness of steel plate,
    t = 20 mm

    α =
    120
    = 60º
    2

    Break through distance,
    A =
    d
    =
    10
    = 2.8867 mm
    2tanα2tan60º

    Total length of tool travel,
    l = t + A + 2
    = 20 + 2.8867 + 2 = 24.88
    Time for drilling
    =
    l
    =
    24.88
    = 0.4147 min
    fN0.2 × 300

    = 24.88 sec ≈ 25 sec.


Direction: A cylinder is turned on a lathe with orthogonal machining principle. Spindle rotates at 200 rpm. The axial feed rate is 0.25 mm per revolution. Depth of cut is 0.4 mm. The rake angle is 10°. In the analysis it is found that the shear angle is 27.75°.

  1. In the above problem, the coefficient of friction at the chip tool Interface obtained using Earnest and Merchant theory is









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    From Earnest and Merchants' theory
    2φ + λ – α = 90
    where μ = co-efficient of friction = tan λ
    ∴  λ = 90 + 10 – 2 × 27.75
    ∴  = 44.5º
    μ= tan (44.5)= 0.98

    Correct Option: D

    From Earnest and Merchants' theory
    2φ + λ – α = 90
    where μ = co-efficient of friction = tan λ
    ∴  λ = 90 + 10 – 2 × 27.75
    ∴  = 44.5º
    μ= tan (44.5)= 0.98



  1. The thickness of the produced chip is









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    Given, rake angle, α = 10º
    shear angle, φ = 27.75
    depth of cut, t1 = 0.4 mm
    Let t2 = thickness of chip produced

    Now,  
    t1
    =
    sinφ
    t2cos(φ − α)

    ⇒  t2 =
    t1cos(φ − α)
    sinφ

    =
    0.4cos(27.75 − 10)
    sin(27.75)

    =
    0.4 cos 17.75
    = 0.818 mm
    sin 27.75

    Correct Option: A

    Given, rake angle, α = 10º
    shear angle, φ = 27.75
    depth of cut, t1 = 0.4 mm
    Let t2 = thickness of chip produced

    Now,  
    t1
    =
    sinφ
    t2cos(φ − α)

    ⇒  t2 =
    t1cos(φ − α)
    sinφ

    =
    0.4cos(27.75 − 10)
    sin(27.75)

    =
    0.4 cos 17.75
    = 0.818 mm
    sin 27.75