# Ø500 3500 1500 b1500 d4 x ø510 525 1500 m a 060 m a

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Chapter 6 / Exercise 3
Physics Laboratory Experiments
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Chapter 6 / Exercise 3
Physics Laboratory Experiments
Wilson Expert Verified
Geometric Dimensioning and Tolerancing for Mechanical Design Answer Guide 42 Figure 7-22 Zero positional tolerance conversion: problem 2.2. Convert the tolerance in Fig. 7-22 to the zero positional tolerances for the pin and the hole. Zero tolerance is not used when the tolerance applies at RFS, or when no bonus tolerance is available as in a tolerance specified for threads or press fit pins. Ø.996-1.006Ø1.000–1.006.004MABC.000MABC.004MABC.000MABCØ.998-1.004Ø.998–1.000HolePin
Geometric Dimensioning and Tolerancing for Mechanical Design Answer Guide 43 Figure 7-23 A hole specified at LMC: problem 3. 3. Calculate the minimum wall thickness between the inside diameter and datum feature B in Fig. 7-23. Datum feature B @ LMC Ø 2.480 I. D. @ LMC – Ø 2.020 Tolerance @ LMC – Ø .020.440 The wall thickness equals half of the differences in diameters or .220. (Calculating diameters and diving the final diameter by 2 minimizes errors.) Ø2.500 ± .020AØ2.000 ± .020.020LABL4.059 ±.003B
Geometric Dimensioning and Tolerancing for Mechanical Design Answer Guide 44 Figure 7-24 Boundary conditions: problem 4. 4. First calculate the virtual conditions and resultant conditions for the pin and hole. Then calculate the maximum and minimum distances for dimensions X and Y in Fig. 7-24. The Virtual Conditionof the PIN. The Virtual Conditionof the HOLE. VCp = MMC + Geo. Tol. VCh = MMC – Geo. Tol. VCp = 1.000 + .004 = 1.004 VCh = 1.000 – .004 = .996 VCp/2 = .502 VCh/2 =.498 Resultant Conditionof the PIN. Resultant Conditionof the HOLE. RCp= LMC – Geo. Tol. – Bonus RCh= LMC + Geo. Tol. + Bonus RCp=.998 – .004 – .002 = .992 RCh=1.006 +.004 +.006 =1.016 RCp/2 = .496 RCh/2 =.508 The maximum and minimum distances for dimension X: XMax= Dist. – RCp/2 – VCh/2 = XMin= Dist. VCp/2 – RCh/2 = XMax=3.000 – .496 – .498 = XMin= 3.000 – .502 – .508 XMax= 2.006 XMin=1.990The maximum and minimum distances for dimension Y: YMax=L @ MMC – Dist. – VCh/2 = YMin= L @ MMC – Dist. – RCh/2 = YMax=6.010 – 4.500 – .498 = YMin= 5.990 – 4.500 – .508 = AØ1.000-1.006.004MABCYØ.998-1.000.004MABCX3.0006.00B1.5001.000C.XX = ± .01ANGLES = ± 1°
Geometric Dimensioning and Tolerancing for Mechanical Design Answer Guide 45 YMax=1.012 YMin= .982
Geometric Dimensioning and Tolerancing for Mechanical Design Answer Guide 46 Chapter 8 Position, Location Chapter Review Page 153 1. The floating fastener formula is: T = H – F or H = F + T 2. T = The clearance hole Location Tolerance at MMC H = The Clearance Hole MMC diameterF = The Fastener’s MMC diameter, the nominal size3. The clearance hole LMC diameter formula isH @ LMC = (F +F head) / 2. 4. The fixed fastener is fixed by one or more of the members being fastened. 5. The formula for fixed fasteners is: t1+ t2= H – F or H = F + t1+ t2 6. The location tolerance for both the threaded hole and the clearance hole must come from the difference between the actual diameter of the clearance hole and the diameter of the fastener . 7. It is common to assign a larger portion of the tolerance to the threadedhole.
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