Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

  1. Assertion (A) : The ratio of uncut chip thickness to actual chip thickness is always less than one and is termed as cutting ratio in orthogonal cutting.
    Reason (R) : The frictional force is very high due to the occurrence of sticking friction rather than sliding friction.









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    Sliding friction is very higher

    Correct Option: C

    Sliding friction is very higher


  1. Match List-I (Cutting tool materials) with ListII (Manufacturing methods) and select the correct answer using the codes given below :
    List-I List-II
    A. HSS1. Casting
    B. Satellite2. Forging
    C. Cemented carbide3. Rolling
    D. UCON4. Extrusion
    5. Powder metallurgy










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    HSS – Forging Stellite – Casting Cementite – Powder metallurgy UCON – Rolling

    Correct Option: D

    HSS – Forging Stellite – Casting Cementite – Powder metallurgy UCON – Rolling



  1. Match List-I with List-II and select the correct answer using the codes given below :
    List-I List-II
    A. Plane approach angle1.Tool face
    B. Rake angle2.Tool flank
    C. Clearances angle3.Tool face and flank
    D. Wedge angle4.Cutting edge
    5.Tool nose










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    Correct Option: C



  1. Consider the following statements: During the third stage of tool wear, rapid deterioration of tool edge takes place because
    1. Flank wear is only marginal
    2. Flank wear is large
    3. Temperature of the tool increases gradually
    4. Temperature of the tool increases drastically
    Which of the statements given above are correct?









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    Correct Option: B



  1. In a machining operation chip thickness ratio is 0.3 and the back rake angle of the tool is 10°. What is the value of the shear strain?









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    r = 0.3 =
    t1
    , α = 10°
    t2

    tanφ =
    rcosα
    1 - rsinα

    tanφ =
    0.3cos10°
    1 - 0.3sin10°

    φ = 17.3°
    Shear strain = tan (φ – α) + cot φ
    Shear strain = tan (17.3 – 10) + cot 17.3
    Shear strain = 3.34

    Correct Option: D

    r = 0.3 =
    t1
    , α = 10°
    t2

    tanφ =
    rcosα
    1 - rsinα

    tanφ =
    0.3cos10°
    1 - 0.3sin10°

    φ = 17.3°
    Shear strain = tan (φ – α) + cot φ
    Shear strain = tan (17.3 – 10) + cot 17.3
    Shear strain = 3.34