## Nbm Engineering Sample Drawings

41. True or false? All answers can be found in the text or in the figures in Chapter 5.

■ The clearance in a 'close-running fit' is smaller than that in a

'sliding fit'.

■ There is no out-of-roundness in a hole drilled by a new sharp drill.

■ The IT5 tolerance range is larger than the IT4 tolerance ranee.

■ One of the values of the 'H' or 'h' tolerance classes is always zero.

■ The tolerance class c 11 is the negative of class C11.

■ The GT symbol for symmetry is an 'equals' sign.

■ Reaming produces a hole with more variability than honing.

■ The relationship between the IT tolerance range and nominal size is linear.

■ The fit classes A to Z are a linear series.

■ Clearance fits are always associated with classes A to H (or a to h).

■ The shaft basis system of fits is superior to the hole basis system.

42. Explain the meaning of the following terms: limits, fits, universal tolerance, specific tolerance, IT tolerance grades, geometrical tolerance (Sections 4.5, 5.1, 5.3, 5.4 and 5.6).

43. The 9,lum out-of-roundness produced by the worn <|)10mm drill in Figure 5.2 corresponds to tolerance range IT7 (Section 5.3). If the diameter was different, the 9,lum error would correspond to a different class. Determine the IT class the 9,lum corresponds to for 05,010,015,020,025,030,035,040, 045 and 050mm holes. Plot the results in graphical form.

44. The table in Figure 4.13 gives errors in microns for 025mm holes produced by various manufacturing processes. Translate these into IT equivalents. (Note that the average roughness Ra cannot translate into an IT equivalent.) The taper, ovality and roundness are individual errors that contribute to an overall error. How do the Figure 4.13 individual errors sum to an overall error value which can be translated into an IT range? What is the overall IT value for each process?

45. Give the shaft and hole dimensions for the following tolerance cases. In each case, state the maximum, minimum and average clearance/interference values (Figures 5.11 and 5.12):

015 G7/h6; 0100,00 H7/n6; 37,5 h6/S7 □22,lF8/h7; 10,00 G7/h8; 025,4 H6/g6

46. Translate Figure 5.4 into a graph, i.e. plot a family of curves of IT1, IT2, IT3 ... IT11 against nominal size. Use the average of the nominal size range. This means that for the nominal size class of 120 to 180, you should use the average of 150 to plot the graphical values. Are the relationships linear, parabolic or what? Add to your graph the linear relationship ±(SIZE 0,5%) from Question 37. Comment on the various curves and their relationships. Is the Question 37 equation valid? 47. Add another column to the table in Figure 5.6 that states whether the process produces flat surfaces or cylindrical surfaces. Note that some processes can be used for either (e.g. lapping). From this, select two processes that can be used to machine the surfaces appropriate to each of the following fit classes. In each case select the process which gives minimum cost. Assume that the processing cost increases as the accuracy increases.

■ Flat: Loose-running fit - HI 1/cl 1 (Figure 5.11).

■ Cylindrical: Sliding fit - H7/g6 (Figure 5.11).

48. With reference to the third angle projection drawings in Figure Q48, reproduce each figure but include a geometric tolerance box for the following cases (Sections 5.5 and 5.6):

■ With respect to the 'tee' piece, face 'B' is to lie between two parallel planes 0,15mm apart that are perpendicular to datum face 'A' (perpendicularly GT).

■ With respect to the 'tee' piece, face 'C' is to lie between two planes 0,10mm apart that are parallel to datum face 'B' (parallelism GT).

■ With respect to the 'tee' piece, the top face 'D' is to lie between two parallel planes 0,02mm apart (flatness GT).

■ With respect to the 'turned' part, the axis of diameter 'D' is to lie in a cylindrical tolerance zone 0,1mm diameter (straightness GT).

■ With respect to the 'turned' part, the curved surface of diameter 'B' is to lie between two co-axial cylindrical surfaces 0,02mm radially apart (cylindricity GT).

■ With respect to the 'turned' part, the axis of diameter 'D' is to be contained in a cylindrical tolerance zone 0,05mm diameter which is co-axial with the axis of diameter 'B' (concentricity GT).

■ With respect to the 'turned' part, at the periphery of any cross-section, runout of surface 'B' is not to exceed 0,1mm when the part is rotated about diameter 'D' (runout GT).

■ With respect to the 'plate', the axis of hole 'B' is to be contained in a cylinder 0,06mm diameter inclined at 80° to the datum face 'D' (angularity GT).

■ With respect to the 'plate', the axis of hole 'A' is to be contained in a cylinder 0,03mm diameter which is parallel to the axis of hole 'C' (parallelism GT).

49. Figure 5.15 shows an assembly of two plates and a dowel. The upper plate will only assemble on the dowel if the top of the dowel is positioned correctly. This will depend on the dowel hole position and inclination (assuming the dowel is not bent!). In the example in Figure 5.15, the inclination tolerance zone value was <j)0,030 as an example. Calculate the required inclination tolerance zone value for the worst case situation, i.e., the upper plate hole and dowel diameters are at the worst extremes of their Hll and Cll ranges. Assume the upper plate hole is exactly on centre.

50. With reference to the movable jaw drawing in Figure 3.2, there are two functional 'bearing' surfaces. These are the 16mm wide tongue which locates in the body and the <j>15 x 7,5 central hole in which the bush (part 4) rotates. Using the fits tables in Figures 5.11 and/or 5.12, select fits appropriate to these two situations. From this, add these tolerances as well as general tolerances to your movable jaw detail drawing from Question 29.

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