Materials Science and Manufacturing Engineering Miscellaneous
- What are the upper and lower limits of the shaft represented by 60 f8 ?
Use the following data :
Diameter 60 lies in the diameter step of 50-80 mm. Fundamental tolerance unit, i, in μm = 0.45 D1/3 + 0.001 D, where D is the representative size in mm; Tolerance value for IT8 = 25i. Fundamental deviation for f shaft = –5.5 D 0.41
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D = √50 × 80 = 63.25 mm
i = 0.45(D)1 / 3 + .001 D μm = 1.856 × 10–6 m
Tolerance = 25i = 46.4 × 10–6 m
Fundamental deviation δ = –(5.5) D0.41 = –30.113 × 10–6 = 0.030 mm (absolute)
High limit = Basic size – Fundametal deivation
High limit = 60 – 0.030 = 59.97
Lower limit = High limit –Tolerance
Lower limit = 59.77 – 0.0464 = 59.924mmCorrect Option: A
D = √50 × 80 = 63.25 mm
i = 0.45(D)1 / 3 + .001 D μm = 1.856 × 10–6 m
Tolerance = 25i = 46.4 × 10–6 m
Fundamental deviation δ = –(5.5) D0.41 = –30.113 × 10–6 = 0.030 mm (absolute)
High limit = Basic size – Fundametal deivation
High limit = 60 – 0.030 = 59.97
Lower limit = High limit –Tolerance
Lower limit = 59.77 – 0.0464 = 59.924mm
- A hole is specified as 400.0000.050 mm. The mating shaft has a clearance fit with minimum clearance of 0.01 mm. The tolerance on the shaft is 0.04 mm. The maximum clearance in mm between the hole and the shaft is
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Minimum clearance = Minimum hole size – Maximum shaft size
∴ Maximum shaft size = 40.00 – .01 = 39.99 mm
Tolerance on shaft = 0.04 mm
Minimum shaft size = 39.99 – 0.04 = 39.95
Maximum clearance = Maximum hole size – Minimum shaft size
Maximum clearance = 40.05 – 39.95 = 0.10Correct Option: C
Minimum clearance = Minimum hole size – Maximum shaft size
∴ Maximum shaft size = 40.00 – .01 = 39.99 mm
Tolerance on shaft = 0.04 mm
Minimum shaft size = 39.99 – 0.04 = 39.95
Maximum clearance = Maximum hole size – Minimum shaft size
Maximum clearance = 40.05 – 39.95 = 0.10
- In order to have interference fit, it is essential that the lower limit of the shaft should be
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For making interference fit, seeing the figure it is essential that lower limit of shaft should be greater than upper limit of hole.
Correct Option: A
For making interference fit, seeing the figure it is essential that lower limit of shaft should be greater than upper limit of hole.
- GO and NO-GO plug gauges are to be designed for a hole 20.000+0.010+0.050 mm. Gauge tolerances can be taken as 10% of the hole tolerance. Following ISO system of gauge design, sizes of GO and NO-GO gauge will be respectively.
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Dimension of hole = 20.000+0.050+0.010
∴ Tolerance = 20.05 – 20.01 = 0.04 mmGauge tolerance = 0.04 × 10 = 0.004 mm 100
Size of NO – GO gauge = Maximum limit – gauge tolerance
= 20.05 – 0.004 = 20.046
Correct Option: D
Dimension of hole = 20.000+0.050+0.010
∴ Tolerance = 20.05 – 20.01 = 0.04 mmGauge tolerance = 0.04 × 10 = 0.004 mm 100
Size of NO – GO gauge = Maximum limit – gauge tolerance
= 20.05 – 0.004 = 20.046
- In an interchangeable assembly, shafts of size –25.000+0.040-0.0100 mm mate with holes of size 25.000+0.020-0.000 mm. The maximum possible clearance in the assembly will be
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Minimum shaft size = 25.000 – 0.04 = 24.96 mm Maximum hole size = (25.000 + 0.02) = 25.02 mm
∴ Maximum possible clearance = Maximum size of hole – minimum size of shaft
Maximum possible clearance = 25.02 – 24.96 = 0.060 mm = 60 micron.Correct Option: D
Minimum shaft size = 25.000 – 0.04 = 24.96 mm Maximum hole size = (25.000 + 0.02) = 25.02 mm
∴ Maximum possible clearance = Maximum size of hole – minimum size of shaft
Maximum possible clearance = 25.02 – 24.96 = 0.060 mm = 60 micron.
