Unformatted text preview: Name: SOLUTION EML 2023 –Computer Aided Design open book and notes; 50 min. Quiz 2A 14 April 2010 1. You are specifying the tolerance limits for a cylindrical fit whose nominal size is 3.00 inches. You wish to have a FN2 hole basis fit. Specify the size limits for the hole and the shaft. hole tolerance: 3.00 .0012 inches 0
shaft tolerance: 3.00.0029 inches .0022 2. You are specifying the tolerance limits for a cylindrical fit whose nominal size is 39 mm. You wish to have a sliding hole‐basis fit. Specify the size limits for the hole and the shaft using a preferred size. hole tolerance: 40 ‐ 40.025 mm shaft tolerance: 39.975 ‐ 39.991 mm Questions 3 through 10 refer to the part drawing. Units are in inches. 3. Sketch and label the datums on Figure 1. B C A Figure 1 4. What position tolerance is allowed for hole 2 if it is manufactured with a diameter of .26 inches? The MMC for hole 2 is 0.235 inches. At MMC the allowable tolerance is 0.012 inches. Since the hole was manufactured at 0.26 inches, the allowable tolerance is 0.037 inches. 5. If hole 1 is manufactured with a diameter of 0.250” and hole 2 is manufactured with a diameter of 0.240”, what is the closest allowable distance between the centers of these holes? Hole 1 will have an allowable tolerance of .027”. Hole 2 will have an allowable tolerance of .017”. The closest that the hole center points can be is 0.52 0.52 – 0.027/2 – 0.017/2 = 0.6851 inches. 6. What is the closest allowable distance between points P and Q? The closest allowable distance is .35” ‐ .015” = .335” . 7. What is the smallest allowable distance between points A and B? 1.5 – 0.3 = 1.47 inches 8. The actual location of hole 2 in a manufactured part is shown in Figure 2. What is the smallest allowable diameter for this hole? The center of the hole is a distance of .0052 .006 2 .0078 inches Figure 2 away from the perfect location. The diameter of the hole must be such that the position tolerance is 2 .0078” = .0156” . The allowable tolerance at MMC is given as .012”. The diameter of the hole will have to be .0156‐.012 = .0036” larger than the MMC diameter. thus the smallest allowable diameter is .235 + .0036 = .2386 inches. 9. What length of the center hole has been tapped with threads? 1 inch 10. Explain what the feature control from shown here means as it is applied to the tapered shaft. This is circular runout. The diameter at any cross section must not vary by more than .005 inches. ...
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This note was uploaded on 09/05/2011 for the course EML 2023 taught by Professor Mathews,crane during the Fall '09 term at University of Florida.
- Fall '09