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
  1. A part shown in the figure is machined to the sizes given below
    P = 35.00 ± 0.08 mm
    Q = 12.00 + 0.02 mm
    R = 13.00+0.04-0.02 mm

    With 100% confidence, the resultant dimension W will have the specification








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    P = 35.00 ± 0.08 mm
    Q = 12.00 ± 0.02 mm
    R = 13.00+0.04-0.03 mm
    P = Q + W + R

    Now 35.00 ± 0.08 = 12.00 ± 0.02 + W + 13.00+0.04-0.03
    ⇒ W = 35.00 ± 0.08 - [ 12 ± 0.02 + 13.00+0.04-0.03 ]
    = (35 - 12 - 13)+0.08 - 0.02 - 0.04-0.08 + 0.02 + 0.03
    = 10.00+0.02-0.03 = 9.99 ± 0.03 mm

    Correct Option: B

    P = 35.00 ± 0.08 mm
    Q = 12.00 ± 0.02 mm
    R = 13.00+0.04-0.03 mm
    P = Q + W + R

    Now 35.00 ± 0.08 = 12.00 ± 0.02 + W + 13.00+0.04-0.03
    ⇒ W = 35.00 ± 0.08 - [ 12 ± 0.02 + 13.00+0.04-0.03 ]
    = (35 - 12 - 13)+0.08 - 0.02 - 0.04-0.08 + 0.02 + 0.03
    = 10.00+0.02-0.03 = 9.99 ± 0.03 mm

  1. The dimensional limits on a shaft of 25h7 are








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    Shaft Basic system

    D = √18 × 30 = 23.24
    i = 0.45 ³√D + 0.001(D)
    i = 0.45³√2324 + 0.001(23.24)
    i = 0.31 microns
    IT7 = 16i = 16 × 1.31= 20.96 microns.
    shaft Tolerance = 25 – (20.96 × 10–3) = 24.979mm.

    Correct Option: B

    Shaft Basic system

    D = √18 × 30 = 23.24
    i = 0.45 ³√D + 0.001(D)
    i = 0.45³√2324 + 0.001(23.24)
    i = 0.31 microns
    IT7 = 16i = 16 × 1.31= 20.96 microns.
    shaft Tolerance = 25 – (20.96 × 10–3) = 24.979mm.

  1. Allowance in limits and fits refers to








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    Allowances is always the difference between Maximum material limits.

    Correct Option: B

    Allowances is always the difference between Maximum material limits.

  1. 3 blocks of B1, B2 and B3 are to be inserted in a channel of width S maintaining a minimum gap of width T = 0.125 mm, as shown in figure. For P = 18.75 ± 0.08, Q = 25.00 ± 0.12, R = 28.125 ± 0.1 and S = 72.35 + X. The tolerance X is









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    Consider,P + Q + R + T(max.) = 18.83 + 25.12 + 28.225 + 0.125 = 72.3 = s
    Given s = 72.35 + X
    ∴ 72.35 + X = 72.3 or X = – 0.05

    Correct Option: C

    Consider,P + Q + R + T(max.) = 18.83 + 25.12 + 28.225 + 0.125 = 72.3 = s
    Given s = 72.35 + X
    ∴ 72.35 + X = 72.3 or X = – 0.05

  1. The fit on a hole-shaft pair system is specified as H7-h6, the type of fit is








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    NA

    Correct Option: A

    NA