Homework A Solutions

1)

    1. Min-Max

    20 + 30 + 15 = 65
    0.15 + 0.30 + 0.10 = 0.55 --> Tol = +/- 0.55

    Dimension A
    Min: 65.45, Max: 65.55

    2. Statistical

    3ó1 = 0.15 --> ó1 = 0.05

    3ó2 = 0.30 --> ó2 = 0.10

    3ó3 = 0.10 --> ó3 = 0.0333

    ó = (ó12 + ó22 + ó32)1/2

    ó = (0.052 + 0.102 + 0.03332)1/2

    ó = 0.117

    3ó = 0.350 --> +/-3ó = +/-0.350

    Min: 64.65, Max: 65.35


    3. Solutions are not equal. Statistical method is preferred in a manufacturing environment because it more closely resembles variations in dimension that occur in machining. Statistically, pieces will center about the median, but the tolerance will include about 99% of pieces produced.

2)

    Inner Fit (A) Outer Fit (B)
    1. Interference Fit (hole based)
    LN2: +0.005, -0.000
    LN3: +0.005, -0.000    		 Sliding Fit (shaft based)
    RC2: -0.004, -0.007
    RC4: -0.012, -.020
    2. Sliding Fit (hole based)
    RC2: -0.0025, -0.0045
    RC4: -0.008, -0.013    		 Interference Fit (shaft based)
    LN2: +0.014, +0.009
    LN3: +0.025, +0.020
    3. Case (1): Motor bearing, automobile wheel bearing
    Case (2): Camshaft of auto engine

    RC2 is a relatively free sliding fit, used for good positional accuracy. RC4 is a close running fit, usually used for accurate machinery.
    LN2 is a medium interference fit, for standard duty machines and positional accuracy. LN3 is a heavy interference fit, used for heavy operating machinery and loads.

3)

    óA = x
    óB = x/2
    3óTOTAL = 0.30 --> óTOTAL = 0.10

    óTOTAL = (óA2 + óB2 + óA2 ) ½ = 0.10
    óTOTAL = (2 x2 + ¼ x2)1/2
    óTOTAL = (9/4 x2)1/2 = 3/2 x
    --> x = 2/3 (0.10) = 0.0666
    --> óA = 0.0666
    --> óB = 0.0333

    Tolerance +/-3ó: Tol A = 0.20, Tol B = 0.10
    Parts A: 5 mm +/-0.20
    Parts B: 2 mm +/-0.10

4)

    Design 1. Simply, 105 +/- 0.60
    Min: 104.40, Max: 105.60
    Design 2. 35 + 20 + 50 = 105
    0.40 + 0.20 + 0.40 = 1.00
    --> 105 +/- 1.00
    Min: 104.00, Max 106.00

    2. Statistical

    Design 1. 3óA = 0.60 --> óA = 0.20
    óTOTAL = (óA2)1/2 = óA = 0.20
    3óTOTAL = 0.60
    Min: 104.40, Max: 105.60
    Design 2. 3óA = 0.40 --> óA = 0.1333
    3óB = 0.20 --> óB = 0.0666
    3óC = 0.40 --> óC = 0.1333
    óTOTAL = (óA2 + óB2 + óC2 )1/2 = 0.1999 --> 3ó = 0.60
    So, Min: 104.40, Max: 105.60

    3. Which is better in terms of cost and accuracy?

    Design 1 is both more accurate and much easier to machine and assemble. It is assumed then it would be less expensive to produce.