Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

Easy Questions
Moderate Questions
Difficult Questions

Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous

Direction: A φ 40 mm job is subjected to orthogonal turning by a + 10° rake angle tool at 500 rev/min. By direct measurement during the cutting operation, the shear angle was found equal to 25°.

  1. If the friction angle at the tool chip interface is 58° 10' and the cutting force components measured by a dynamometer are 600 N and 200 N, the power loss due to friction (in kNm/ min) is approximately








  1. View Hint View Answer Discuss in Forum

    r =
    t1
    		 =
    Vc
    		 =
    27.5
    		 = 0.437
    T2V62.8

    F
    		 =
    R sinβ
    		 , β = 58°10'
    FcR cos(β - α)

    F = 764 N
    loss of energy due to friction = F × Vc
    = 764 × 27.5 = 350.3 Watts

    Correct Option: D

    r =
    t1
    		 =
    Vc
    		 =
    27.5
    		 = 0.437
    T2V62.8

    F
    		 =
    R sinβ
    		 , β = 58°10'
    FcR cos(β - α)

    F = 764 N
    loss of energy due to friction = F × Vc
    = 764 × 27.5 = 350.3 Watts

  1. The velocity (in m/min) with which the chip flows on the tool face is








  1. View Hint View Answer Discuss in Forum

    D = 40 mm
    α = 10°
    N = 500 pm
    φ = 25°

    V =
    πDN
    		 =
    π × 40 × 500
    10001000

    V = 62.8 m/min
    Vc
    		 =
    sinφ
    Vscos(φ - α)

    Vc =
    62.8 × sin 25
    cos(25 - 10)

    Vc = 27.5 m/min

    Correct Option: D

    D = 40 mm
    α = 10°
    N = 500 pm
    φ = 25°

    V =
    πDN
    		 =
    π × 40 × 500
    10001000

    V = 62.8 m/min
    Vc
    		 =
    sinφ
    Vscos(φ - α)

    Vc =
    62.8 × sin 25
    cos(25 - 10)

    Vc = 27.5 m/min

Direction: A batch of 500 jobs of diameter 50 mm and length 100 mm is to be turned at 200 rev/min and feed 0.2 mm/rev.

  1. The number of tool changes required to machine the whole batch is








  1. View Hint View Answer Discuss in Forum

    Time/ piece =
    L
    		 =
    100
    		 = 2.5 min
    fN0.2 × 200

    No of Tool Changes =
    cutting time
    Tool life

    =
    674
    		 = 269.6 = 270
    2.5

    Correct Option: B

    Time/ piece =
    L
    		 =
    100
    		 = 2.5 min
    fN0.2 × 200

    No of Tool Changes =
    cutting time
    Tool life

    =
    674
    		 = 269.6 = 270
    2.5

  1. Applying Taylor's equation VT0.25 = 160, the tool








  1. View Hint View Answer Discuss in Forum

    No. of jobs = 500
    D = 50 mm, L = 100 mm
    N = 200 rpm, f = 0.2 mm/rev

    V =
    πDN
    		 = 31.4 m/min
    100

    T =
    160
    		1/0.25
    314

    T = 6 min

    Correct Option: C

    No. of jobs = 500
    D = 50 mm, L = 100 mm
    N = 200 rpm, f = 0.2 mm/rev

    V =
    πDN
    		 = 31.4 m/min
    100

    T =
    160
    		1/0.25
    314

    T = 6 min

  1. The rake angle of a cutting tool is 15°, shear angle 45° and cutting velocity 35 m/min. What is the velocity of chip along the tool face?








  1. View Hint View Answer Discuss in Forum

    α = 15°
    φ = 45°
    V = 35 m/min

    V
    		 =
    Vc
    		 =
    Vs
    cos(φ-α)sinφcos α

    V =
    V.sinφ
    		 =
    35sin45
    cos(φ-α)cos(45 - 10)

    Vc = 25.58m/min

    Correct Option: A

    α = 15°
    φ = 45°
    V = 35 m/min

    V
    		 =
    Vc
    		 =
    Vs
    cos(φ-α)sinφcos α

    V =
    V.sinφ
    		 =
    35sin45
    cos(φ-α)cos(45 - 10)

    Vc = 25.58m/min