Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

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Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous

Direction: An orthogonal operation is carried out at 20 m/min cutting speed, using a cutting tool of rake angle 15 degree. The chip thickness is 0.4 mm and uncut chip thickness is 0.2 mm.

  1. The chip velocity (in m/min) is








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    Vc
    		 =
    T1
    		 = r
    VT2

    Vc = 0.5 × 20 = 10 m /min

    Correct Option: B

    Vc
    		 =
    T1
    		 = r
    VT2

    Vc = 0.5 × 20 = 10 m /min

  1. The shear angle (in degrees) is








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    V = 20m/min ∝ 15° t2 = 0.4 mm t1 = 0.2 mm r cos tan 1 rsin 0.2 cos15 0.4 tan 0.2 1 sin15 0.4

    tanφ =
    r cos∝
    1 - rsin∝

    tanφ =  
    0.2
    		 + cos 15
    0.4
    1 -
    0.2
    		 sin 15
    0.4

    φ = 29°

    Correct Option: C

    V = 20m/min ∝ 15° t2 = 0.4 mm t1 = 0.2 mm r cos tan 1 rsin 0.2 cos15 0.4 tan 0.2 1 sin15 0.4

    tanφ =
    r cos∝
    1 - rsin∝

    tanφ =  
    0.2
    		 + cos 15
    0.4
    1 -
    0.2
    		 sin 15
    0.4

    φ = 29°

Direction: Blind hole 10 mm diameter, 50 mm deep are being drilled in steel block, drilling spindle speed is 600 rpm, feed 0.2 mm/rev, point angle of drill is 120.

  1. During the above operation, the drill wears out after producing 200 holes.Taylor's tool life equation is of the form VT0.3 = C, where V = cutting speed in m/min and T= tool life in min. Taylor's constant C will be








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    VT0.3 =C
    No of holes = 200

    V =
    πDN
    		 = 18.84 min
    60

    Tool life = 200×0.442 = 88.33 min
    18.84 × 88.330.3 = 6 = 72.26.

    Correct Option: B

    VT0.3 =C
    No of holes = 200

    V =
    πDN
    		 = 18.84 min
    60

    Tool life = 200×0.442 = 88.33 min
    18.84 × 88.330.3 = 6 = 72.26.

  1. Machining time (in min) per hole will be








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    D = 10 mm Plate thickness = 50 mm N= 600 rpm feed = 0.2 mm/rev point Angle = 2∝ = 120°
    Approach = 0.3D = 0.3 × 10 = 3mm
    L = 3 + 50 = 53 mm

    Time/ hole =
    L
    		 =
    53
    		 = 0.442 min
    fN0.2 × 600

    Correct Option: C

    D = 10 mm Plate thickness = 50 mm N= 600 rpm feed = 0.2 mm/rev point Angle = 2∝ = 120°
    Approach = 0.3D = 0.3 × 10 = 3mm
    L = 3 + 50 = 53 mm

    Time/ hole =
    L
    		 =
    53
    		 = 0.442 min
    fN0.2 × 600

Direction: In an orthogonal cutting test, the following observations were made
Cutting force = 1200 N
Thrust force = 500 N
Tool rake angle = zero
Cutting speed = 1 m/s
Depth of cut = 0.8 mm
Chip thickness = 1.5 mm

  1. Friction angle during machining will be








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    Fc = 1200 N FT = 500 N ∝ = 0 VC = 1 m/s d = t1 = 0.8 mm t2 = 1.5 mm

    tanβ =  
    FT
    		 + tan ∝
    Fc
    1 -
    FT
    		 tan ∝
    Fc

    tanβ =  
    500
    		 + tan 0
    1200
    1 -
    500
    		 tan 0
    1200

    β = 22.62°
    {chip Reductioncocfficient}
    t1
    		 =
    Vc
    t2V

    Vc = 0.533 × 1
    Vc = 0.533m/s

    Correct Option: A

    Fc = 1200 N FT = 500 N ∝ = 0 VC = 1 m/s d = t1 = 0.8 mm t2 = 1.5 mm

    tanβ =  
    FT
    		 + tan ∝
    Fc
    1 -
    FT
    		 tan ∝
    Fc

    tanβ =  
    500
    		 + tan 0
    1200
    1 -
    500
    		 tan 0
    1200

    β = 22.62°
    {chip Reductioncocfficient}
    t1
    		 =
    Vc
    t2V

    Vc = 0.533 × 1
    Vc = 0.533m/s