E27 Introduction to Manufacturing and Tolerancing Final Examination December 17, 2015 SOLUTIONS Instructions

Transcription

1 University of California, Berkeley Department of Mechanical Engineering E27 Introduction to Manufacturing and Tolerancing Fall 2015 Final Examination December 17, 2015 SOLUTIONS Instructions 1. Please do not open this exam paper until invited to do so. 2. This exam contains five questions. 3. Please attempt all questions. 4. The exam is closed book and closed notes but you may use five letter-size sides of notes. 5. Points available are shown next to each part of each question. 6. In parts of questions that rely on the answer to a previous part of the question, credit will be given for a correct approach even if the answer given to the previous part was incorrect. Thus it is in your interest to make an attempt at all parts of all questions. 7. The total number of points available is Please write answers in the blank spaces provided on this question paper. 9. There are two attachments at the back of this question paper (Sheet A and Sheet B), which you can detach to help you answer the questions. Please turn these in with your completed question paper but write all your answers within the spaces provided on the paper, and not on sheets A and B, which will not be graded. 10. Scratch paper is available for your working but will not be collected. 11. The exam will last three hours. 12. If you need clarification on what is being asked in any question, please raise your hand. 13. You may not communicate with other students in any way during the exam. 14. Good luck! Name: 1

2 Question 1: Subtractive and additive manufacturing [31 points] Consider the component of which a drawing is provided on Sheet A. First consider the case where this component is to be manufactured from aluminum alloy using subtractive techniques. The stock (starting) material is a 235 mm-diameter cylindrical bar. The machining process is as follows: 1. Face off the bar (to make a smooth surface on one end of the bar) 2. Turn a 140 mm length of the bar down to 230 mm diameter 3. Turn a mm-long portion of the bar down from 230 mm diameter to 100 mm diameter. 4. Bore the central 28 mm-diameter hole 5. Part off the component to separate it from the bar of stock material. a) Compute the volume of material removed during Step 3 alone (i.e., turning a portion of the bar down from 230 mm to 100 mm diameter). [4 points] Volume removed = length turned * (change in cross-sectional area) = (127.4 mm) * π/4 * [(230 mm) 2 (100 mm) 2 ] = mm 3 = 4.3 * 10 6 mm 3 (2 s.f.) b) Compute the machining energy consumed during Step 3, assuming the specific energy of machining the aluminum alloy to be 1.0 J/mm 3. [2 points] Energy consumed = (volume removed) * (specific energy) = 4.3 * 10 6 J (2 s.f.) c) Step 3 cannot be done in a single pass of the cutting tool because the reduction in diameter is too great. Instead, several passes are made to gradually reduce the diameter. Most of the cuts can be relatively fast and deep ( rough cuts ), while the final cut is slower and shallower to provide a smoother surface finish. During rough cutting, the cut depth is 4 mm and the feed rate is 400 mm/min. During the finishing cut, the cut depth is 1 mm and the feed rate is 200 mm/min. Compute the number of rough cuts required during Step 3. [3 points] Change in radius made during rough cuts = [( )/2] mm (1 mm) = 64 mm If each rough cut is 4 mm deep, 64/4 = 16 rough cuts are needed. 2

3 d) Compute the total time taken for Step 3, using the feed rates and cut depths given in part c) above. Include the time for all rough cuts and the finishing cut in your answer. [5 points] Time for one rough cut = mm / (400 mm/min) = min Time for finishing cut = mm / (200 mm/min) = min Total cutting time = (16* ) min = 5.73 min (3 s.f.) e) What is the volumetric material removal rate during the first rough cut in Step 3? [4 points] Take the effective diameter of the toolpath to be equal to the starting diameter of the bar less the cut depth: Volumetric removal rate = 400 mm/min * 4 mm * π * (228 4) mm = mm 3 /min = 1.13 * 10 6 mm 3 /min (3 s.f.) = 1.88 * 10 4 mm 3 /s (3 s.f.) Alternative approach is: R MR = 400 mm/min * π * ((230/2) 2 (222/2) 2 ) mm 2 = mm 3 /min Note that a constant feed rate is probably not optimal for such a large reduction in diameter. As the diameter reduces, the removal rate falls considerably, which wastes time, assuming that the removal rate during the first cut is the maximum feasible removal rate. f) What is the cutting power during the first rough cut in Step 3? [3 points] Cutting power = (removal rate) * (specific energy) = 1.88 * 10 4 mm/s * 1.0 J/mm 3 = 18.8 kw (3.s.f.) Step 3 might be considered overly time-consuming and wasteful of material. An alternative option is to use a new type of subtractive additive machine tool which combines turning capabilities with the ability to deposit material in powder form and fuse it to the component using heat from a laser. An illustration of this machine in action is below: 3

4 Source: DMG Mori Suppose that instead of machining the component from a 235 mm-diameter bar, we start with a 105 mm-diameter bar, and follow the following process: 1. Face off the bar (to make a smooth surface on one end of the bar) 2. Turn a 140 mm length of the bar down from 105 mm diameter to 100 mm diameter 3. Deposit a cylindrical flange of material on to the 100 mm-diameter bar, building on top of the surface that was turned in Step 2 above. The flange is to be 230 mm in diameter and 12.6 mm thick. 4. Bore the central 28 mm-diameter hole 5. Part off the component to separate it from the bar of stock material. Consider the deposition process step (highlighted in bold above). For every unit volume of material that is deposited, a certain amount of energy must be supplied by the laser to fuse the metal powder. Let us call this quantity the specific deposition energy. g) What does the value of the specific deposition energy (in J/mm 3 ) need to be, in order for the total energy required to deposit the flange to be equal to the machining energy computed in part b) above? [3 points] Volume of material added by this additive process = (12.6 mm) * (π/4) * ( ) mm 2 = mm 3 For deposition energy to equal the removal energy computed for the subtractive approach in part (b) above, the specific deposition energy would need to equal: (volume removed in (a))/(volume added) * (specific energy for removal) = ( /424539) * 1.0 J/mm 3 = 10.1 J/mm 3 (3 s.f.) 4

5 h) What does the deposition rate of material (in mm 3 /s) need to be, in order for the total time required to deposit the flange to be equal to the machining time computed in part d) above? [3 points] (Deposition rate required for equal processing time) = (Volume added)/(machining time computed in (d)) = mm 3 /s * 5.73 min / 60 s/min) = 1234 mm 3 /s (3 s.f.) i) Comment on the two possible approaches considered in this question (purely subtractive, and hybrid subtractive additive). What considerations other than processing time, material usage and energy usage might be relevant? Are there any additional process steps that you would advocate inserting? [4 points] Comparative quality of material deposited: is the fused material fully solid or does it contain pores that might lower total strength by lowering the volume fraction of metal and/or promoting stress concentrations. Grain structure of the fused material may be non-optimal and unknown. Stock material on the other hand can be specified to have particular properties and subtractive processing can be designed to minimize thermal modification of material properties. Surface finish: laser-fused particles will have a much rougher surface finish than turned components. A final finishing cut on the deposited flange may be needed depending on the application. Capital costs of the tool. These hybrid systems are very new and will be far more expensive than standard lathes or even machining centers Running costs. Staff qualified to operate such a tool may be harder to find and thus command higher salaries than skilled machinists, since additive metal manufacturing is much newer than subtractive machining. In reality deposition rates of these hybrid machines are currently far lower than calculated in part (h). The makers of the tool, DMG Mori, quote a deposition rate of 1 kg/hour which, assuming this value is for stainless steel, equates to about 35 mm 3 /s. Students are not expected to know this to score full points on this question! 5

6 Question 2: Geometric Dimensioning and Tolerancing [24 points] Consider the component for which a dimensioned and toleranced drawing is supplied as Sheet A and reproduced below: a) Add an appropriately positioned feature control frame to the drawing above to specify that the total runout of the face marked with a single asterisk ( ) in the drawing should not exceed 0.1 mm when the component rotates about datum axis B. [4 points] b) Add an appropriately positioned feature control frame to the drawing above to specify that the flatness of the face marked with a single asterisk ( ) in the drawing should not be worse than 0.1 mm. [4 points] c) Add an appropriately positioned feature control frame to the drawing above to specify that the cylindrical feature marked with two asterisks ( ) must have cylindricity not exceeding 1.3 mm. [4 points] d) Now consider the accompanying Sheet B, which shows side views of three different components X, Y and Z, which are supposed to meet the tolerances specified on the drawing above and Sheet A, but have been imperfectly manufactured. For each component, consider the perpendicularity and concentricity tolerances as specified on the drawing above and Sheet A. Check the appropriate box to indicate whether or not the component satisfies each tolerance, and add a brief explanation of your reasoning. [12 points] 6

7 Component ID Perpendicularity tolerance Concentricity tolerance Component X Meets tolerance X Does not meet tolerance Explanation: Axis of manufactured feature deviates by as much as 10 mm from the component s center, so part of it lies outside a tolerance zone with diameter 10 mm and radius 5 mm. X Meets tolerance Does not meet tolerance Explanation: Axis of manufactured feature deviates by as much as 10 mm from center, so lies completely within a tolerance zone with diameter 30 mm and radius 15 mm Component Y Component Z Meets tolerance X Does not meet tolerance Explanation: The datum A established using the manufactured component is at an angle to the axis of the cylindrical feature being measured. Along the length of the feature, the manufactured feature s axis deviates by about 20 mm from a normal to datum A. This deviation places it outside the 10 mm-diameter tolerance zone. X Meets tolerance Does not meet tolerance Explanation: The cylindrical feature s axis appears to be perfectly perpendicular to datum plane A in the drawing. X Meets tolerance Does not meet tolerance Explanation: Both axes appear to be perfectly concentric in the drawing; the angle of datum plane A is irrelevant. X Meets tolerance Does not meet tolerance Explanation: The two axes are offset by about 10 mm which means that the cylindrical feature s axis lies within the diameter-30 mm tolerance zone specified about datum B Note that since only one view of each component is shown, it is conceivable that the concentricity tolerance is not met if the axes are not properly aligned in the out-of-page direction. Credit will be given for saying the concentricity tolerance is not (or may not be) met if this incompleteness of information is explicitly pointed out. Equally, for component Z, it is possible to argue that the perpendicularity tolerance may not be met if the circular feature tilted out of the page. 7

8 Question 3: Casting [20 points] Suppose that the component illustrated on Sheet A is to be manufactured from an aluminum zinc alloy by a casting process. a) Suggest a suitable casting process and describe the key steps in carrying out this process. Consider the size of the component and the tolerances specified. Make reference to: the specific mold material you would use, and its advantages and disadvantages; any secondary processing that your approach would involve. Four points for the molding process: 1 point for a reasonable process (method of inserting molten material) 1 point for a relevant description of the mold material 1 point for at least one valid advantage 1 point for at least one valid disadvantage Two points for explaining secondary processing needed or why it is not needed. [6 points] As we are not told the run size, it is hard to make a conclusive recommendation from among processes with very widely differing speed and set-up costs. However, given the large dimensional size, sand casting or some form of low pressure permanent mold process would seem the most likely candidates; however, well supported answers suggesting a die-casting or other approach will be accepted. Sand casting is one obvious choice. Aluminum zinc alloys are feasible to cast with a sand mold. A thorough answer on mold material will discuss the mechanism for holding the mold together, i.e. whether the sand is green, or held with a curing binder material. The use of core for the central hole could be mentioned. General advantages of a sand mold include its affordability for small runs, the reusability of green sand (but not of sand using polymeric binders). Disadvantages include the large tolerance (~ 0.5 mm or more) relative to permanent molds or machined casting dies. Surface finish is also an issue, with secondary machining for a smooth and flat finish being required much more often with this process than any other. As just mentioned, secondary processing will often take the form of machining to achieve flat surfaces or smooth and circular bores. For surfaces that do not need to meet tight tolerances, secondary processing might be taken to include the application of finishes (e.g. painting, powder coating, etc) although this was not discussed in class and is not expected to be mentioned. 8

9 b) Name and describe one defect that could occur in your chosen casting process and suggest a possible way of preventing this defect from occurring. [5 points] Possible defects that could be described are described in the lecture notes. Examples include: Flash material escapes between mold halves Misrun material solidifies before filling mold Solidification shrinkage orientation of the part in the mold (relative to gravity) may be crucial to ensure that the effects of solidification shrinkage are confined to regions where they do not adversely affect function and/or non-flat surfaces resulting from shrinkage can easily be machined away. Sand inclusions may result from incomplete mold compaction and/or pouring the molten metal too quickly Air entrainment could result from fast pouring and possibly from a riser that is too tall and without a taper such that air is sucked into the metal flow from the surrounding mold material. c) Suppose that this component, when cast, solidifies in 2000 seconds. Estimate the mold constant according to Chvorinov s Rule using an exponent of 2. You can refer to Sheet A for the dimensions of the component. Ensure that you give the correct units for the mold constant. [6 points] Volume of component, V = mm 3 Surface area of component (including inner bore), A = mm 2 From Chvorinov s Rule: C M = T TS ( V 2 A ) = 2000 s ( ) = 19.9 s (3 s. f. ) mm2 d) Briefly discuss how you would suggest positioning runners and any risers for casting this geometry. Consider which parts of the design are likely to solidify first. Include a sketch if this helps illustrate your answer. [3 points] While the effective design of runners and risers requires expertise, and possibly some iteration when casting a new component design, one suggested approach is shown below. Here, we connect the sprues to runners which enter the side of the flange, and risers vent the smaller-diameter end of the component. The component is oriented with its flange horizontal because its diameter is larger than the length of the component (the sketch is not to scale), so this orientation minimizes mold size and the amount of sand that has to be packed. The molten metal enters the component in its thinnest part which helps to guard against misrun: in this configuration, molten metal is passing through the narrow part of the 9

10 mold while it is at its hottest. The risers enable air venting so there is complete filling and visual feedback on when the mold is filled (metal appears in the risers). The sprues do not enter the flange directly downwards because the momentum of the fluid may dislodge/erode sand at the bottom of the sprue, so it is better if the bottom of the sprue is outside the actual component. Another plausible argument is that the flange should be furthest from the gate so that it fills first under gravity to prevent misrun. Other reasonably justified configurations will be accepted for full credit. If a different type of mold has been chosen, the design will likely be different. Orientation of part in mold must make it feasible to remove the pattern if one is being used. Question 4: Welding [15 points] Suppose that instead of machining or casting the component in Sheet A as one piece of material, it is proposed to fabricate it in two pieces of mild steel, as illustrated below, and then weld them together: 10

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12 a) What welding process would you recommend, and why? You can choose from oxyacetylene, manual metal arc, submerged arc, metal inert gas, tungsten inert gas, friction stir welding, and possibly others. Note that your selection should be well justified based on considerations such as the size and geometry of the components, the material (mild steel in this question), and the tolerances specified in Sheet A. [5 points] Again, multiple answers are possible. MIG, TIG, manual metal arc, and oxyacetylene could all be argued for using the pros and cons described in lecture notes. Submerged arc welding would not be suitable because the weld needs to be circular, is in a corner between two components. This weld is thus not suited to the geometry of a submerged arc welding setup which is really for long straight welds on large components. Friction stir welding is conceivable given the shape and materials, but the 100 mm diameter of the cylindrical feature may be too large for this approach to work effectively the range of velocities from the center to the edge of the component as it rotates may be too large to achieve sufficiently uniform heating. Electrical resistance (spot or seam) welding is probably not a good option because of the size of the components. Spot welding usually needs sheets not more than a couple of mm thick, to provide a suitable resistive path in which the energy can be locally dissipated as heat in the material. b) Briefly describe one advantage and one disadvantage of manufacturing this component as a welded assembly rather than by machining or casting. [2 points] Possible advantages include: Can use standard stock material (plate and bar), machine to the correct geometry, then weld. It avoids the large amount of material wastage that would come from turning down a 230 mm bar as in Q1. Compared to the subtractive/additive hybrid approach, the material properties of the stock plate and bar are likely to be well known whereas deposited and fused powder material will have less well characterized properties. Compared to casting, the fabrication effort and cost of welding may be less because a pattern is not needed (although a jig will be needed to hold the components in the correct relative orientation during welding). There is probably less manual intervention needed with welding as there is no sand mold to compact, no careful pouring to be done, and no sawing off of sprues/runners/risers and final machining of the component. For a simple geometry such as this, fabrication from a welded assembly seems very attractive. Possible disadvantages include: The weld joint may be a weak point in the assembly if it is not perfectly continuous. Moreover, there will be a heat-affected zone around the weld which may result in weakened material if not carefully controlled or followed by appropriate heat 12

13 treatment. Aesthetically, an obvious weld may be undesirable. Weld strength is highly dependent on the operator of the welding tool robotic welders are available, but are very expensive and for small job shops may not warrant the investment. Additionally robots may repeatably introduce welding defects if not perfectly programmed. There may be distortion of the components during welding and if alignment in the jig is not perfect, perpendicularity and concentricity may be difficult to achieve and hard to fix after the fact. Other well justified and valid comments will receive credit. c) Suppose that for a particular candidate welding technique, the heat transfer factor, f 1, is 0.8, the melting factor, f 2, is 0.6 and the specific energy of melting for mild steel is 10.5 J/mm 3. Assume that a weld with cross-sectional area 30 mm 2 is created all the way around the circumference of the smaller-diameter cylinder where it meets the flange. For a heat source of 1 kw, how long will the weld take to form? [4 points] Length of weld = π * 100 mm = mm Power available for melting = 1000 W * 0.8 * 0.6 = 480 W Volumetric rate of melting = (power available)/(specific energy of melting) = 480 W / (10.8 J/mm 3 ) = mm 3 /s Time for weld = (length of weld) * (cross-sectional area of weld) / (volumetric rate of melting) = mm * 30 mm 2 / mm 3 /s = 206 s (3 s.f.) = 3.44 minutes (3 s.f.) Now suppose that it is decided that the component can in fact be manufactured from a thermoplastic polymer, to reduce weight. d) Name and briefly describe one suitable welding technique for connecting thermoplastic components, explain how it works, and suggest what modifications, if any, might be needed to the design of the components to enable your chosen process to be used. [4 points] Friction stir welding, ultrasonic welding, and possibly microwave welding could be options. The key is to find a technique that will (a) localize melting at the interface between the components and (b) does not require electrical conductivity of the work (as polymers are almost all nonconductive). Thus, arc welding processes are ruled out and flame-based processes probably produce too diffuse a heat to prevent component distortion. If using ultrasonic welding, we would want energy directors at the interface, to locally increase stress and promote melting only where the weld is required. We may also want suitably designed features (e.g. a lip and recess at the edge of the cylindrical component) to hide from the end user material extruded from the weld. For microwave welding we would need to add an 13

14 electrically conductive antenna at the interface, which will be excited by the microwave radiation and generate heat to melt the material locally. Other well justified comments and modifications will receive credit. Question 5: Metrology [10 points, one point per valid method measurement pairing] For this question assume that you have access to any state-of-the-art metrology tool, both contact based and contact-free, as well as to traditional tools such as calipers, micrometers, hole gages, dial gages, etc. Suggest two different approaches to measuring each of the following, and briefly justify each in a few words: Measurement Method 1 Method 2 The total runout tolerance you labeled in Q2 part a) The cylindricity tolerance you labeled in Q2 part c) The internal diameter of the hole through the component on Sheet A The thickness of a sheet of paper CMM establish datum from measurements of the inside of the bore, scan the flange surface and interpret these measurements to deduce runout. CMM place component resting on its flange, and scan probe around the outside circumference at multiple axial positions. Fit minimum zone cylinders using software. CMM place component resting on its flange, and scan probe around the internal bore. In software, fit a circle to those measurements and extract diameter. CMM a precision of 1-2 microns is achievable with the most rigid systems, and paper is a few tens of microns thick, so a CMM would give some reasonable measurement of the thickness: place paper on a flat surface and scan stylus across it. Establish datum axis perhaps by mounting component on a rotating shaft with expanding jaws that clamp on the inside of the bore. A dial gage with its stylus oriented parallel to the shaft is held against the flange as the component is rotated. This is repeated for multiple radial positions of the dial gage. Rotate the component in V block with a dial gage resting on the top surface. Record maximum and minimum gage readings. Repeat at multiple axial positions. The full range of gage readings is an estimate (not exact) of cylindricity. Use internal micrometer, a hole gage, or if a large enough tool is not available, use calipers, being aware of the danger of undermeasuring the diameter use multiple measurements to improve confidence in the result. Micrometer: clamp paper between jaws. Resolution and repeatability of about a micron will ensure that paper can be effectively measured. 14

15 The surface roughness of a lens to be used in a microscope Optical interferometer has subwavelength resolution, and lenses need to have roughness less than about a quarter of a wavelength to prevent the light field from becoming too distorted. Stylus profilometer these can be obtained with a resolution of ~ 10 nm, so would be able to detect whether a surface is optically flat or smooth. CMM also accepted as an answer a stylus profilometer is effectively a 1D CMM with higher resolution designed for measuring microand nano-scale deflections. 15