University of California, Berkeley Department of Mechanical Engineering. E27 Introduction to Manufacturing and Tolerancing.

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1 University of California, Berkeley Department of Mechanical Engineering E27 Introduction to Manufacturing and Tolerancing Fall 2015 Take-home midterm assignment Issued October 18, Due Wednesday October 28, 2015 at 11:59pm on bcourses You are welcome to access any resources you need when completing this assignment (including lecture notes, the Internet, reference texts, etc), and to ask the instructor and/or GSIs if you need clarification of what is being asked. We have now decided to allow collaboration and consultation between students. This assignment is designed to be challenging but will hopefully also be rewarding and informative. Question 1: Metal cutting (30 points) You are designing a process to lathe-turn a series of cylindrical mild steel components from an initial diameter of 150 mm down to a final diameter of 148 mm in one pass. The length of each component is 200 mm. (a) Assuming a feed of 0.2 mm/rev, what is the total distance travelled by the cutting tool across the material surface during the cutting of each component? [5 points] Your task is to decide between using a standard cutting tool made from high speed steel (HSS) and a cemented carbide (tungsten carbide/cobalt composite) tool, which is more expensive but whose lifetime degrades more gradually with increasing cutting speed. The running costs of the lathe are C run = $200/hour including all maintenance, energy, labor and overheads. Your objective is to minimize the total cost per part. At a cutting speed of V = 5 m/s, with a high speed steel cutting tool, it is found that 20 of the components can be cut before the tool fails and has to be replaced. With a cemented carbide tool at the same cutting speed, 80 components can be produced before the tool needs to be replaced. The HSS tool costs $10 and the cemented carbide tool costs $30. The Taylor equation, which links cutting speed V with lifetime T, is: Where n and C are material-dependent constants. The lifetime T is the actual amount of cutting time (rather than the number of components) for which the tool lasts. Reasonable values to assume for the exponent n in the Taylor equation when cutting mild steel are as follows: Tool material n High speed steel 0.10 Cemented carbide

2 (b) Write an expression for the tooling cost per component in terms of the price, P, of one tool, the cutting speed, V, the cutting distance per component, L cut, the Taylor exponent n, and the number of components, N 0, that can be cut with one tool at V = 5 m/s. (Hint: you do not need to include C in your expression as it cancels out.) [8 points] (c) Write an expression for the running cost per component in terms of the cutting distance per component, L cut, the cutting speed, V, and the hourly running cost, C run. [5 points] (d) Assume that the maximum range of cutting speeds that the lathe can deliver is 1 to 15 m/s. Make a plot which has cutting speed on the horizontal axis and the total cost per component on the vertical axis. Plot separate lines for HSS and cemented carbide tools. For each tool material, determine the cutting speed at which the cost per component is minimized. Which tool material offers the lower optimal cost? (We recommend using Matlab or Excel for this part of the question. Hint: use a log scale for the cost per part) [12 points] Question 2: Tolerancing and fit (8 points) (a) A hole is specified with a diameter (in inches) of In the table below are several possible toleranced diameters for a shaft that is to be inserted into the hole. For each dimension, circle the description that most accurately describes the kind of fit that would be obtained [4 points]: Toleranced dimension Type of fit Clearance Transition Interference Clearance Transition Interference Clearance Transition Interference Clearance Transition Interference (b) Expansion fit. A cylindrical steel component has an outside diameter of inches at room temperature and a collar into which it is to be mounted has an internal diameter of inches at room temperature. By how many Kelvin would the inch-diameter component need to be cooled to be able to be placed inside the collar with a clearance of 2

3 0.003 inches during fitting? Assume the thermal expansivity of steel to be /K. [4 points]. Question 3: Feasibility of fused deposition modeling with metals (62 points) In this question we will consider the energy inputs needed for fused deposition modeling of polymeric materials, and will then go on to consider the feasibility of directly printing metals. When printing a polymeric material (such as PLA, as in the Type A machines in Jacobs, or ABS, as in the Stratasys tools in the Machine Shop), the material is heated to around its melting point so that it enters a state in which it readily flows under load and can be driven through a nozzle. The process of forcing the material through a nozzle to make an output filament smaller in diameter than the input filament from the reel is called extrusion. Extrusion is a widely used industrial process at the macro-scale, for making plastics objects with constant cross-sections (e.g. curtain rails, desk edgings, drainage tubing, etc). In its molten form, a polymeric material actually has both viscous (flowing) and elastic (recoverable) behavior, and so is referred to as viscoelastic. The recoverable/elastic portion of the deformation is responsible for a phenomenon called swelling, whereby the final diameter of the deposited filament, D x, is slightly larger than the diameter of the nozzle in the die, D d. When the filament leaves the nozzle, stresses are released and the material expands. Figure 1: Extrusion of plastic filaments in FDM To design an effective print head, we need to know the load required to drive the filament through the nozzle at the required rate. A simple model for the flow of material through a circular nozzle is: 3

4 where (m 3 /s) is the volume flow rate of the material through the nozzle, (N/m 2 ) is the pressure at the entrance of the nozzle, and (m 5 /Ns) is a material- and geometry-dependent constant given by: 128 where is the die diameter, is the melt viscosity of the material being printed, and is the length of the nozzle opening. Assume that we are printing PLA at 190 C, and that the melt viscosity at this temperature is 100 Ns/m 2 (100 Pa.s) [1]. Also assume that the PLA is incompressible (i.e. that its volume remains constant over time). A consequence of this incompressibility assumption is that the volume flow rate is constant along the path of the material through the print head. Answer the following (we recommend using Matlab or Excel for the calculations and plotting): (a) Assume that a filament with diameter 1.75 mm is fed into the print head at a steady velocity of 2 mm/s. For values of nozzle diameter in the range 0.05 mm 1.0 mm, and a nozzle opening length of 0.3 mm, create a plot of the pressure,, at the nozzle entrance (vertical axis) against the nozzle diameter (horizontal axis). Remember to label your axes and include units. [8 points] (b) The load, F, that must be applied to the filament to drive it through the nozzle under the given conditions is given by. This model assumes that the only resistance to flow is within the nozzle (diameter ), and that there is negligible friction between the polymer and the portion of the die above the nozzle. Evaluate F using your pressures found in part (a), and create a plot of F against for 0.05 mm 1.0 mm. [5 points] (c) Assume that the specific heat capacity of PLA is 1800 J/kgK and its density is 1250 kg/m 3. The temperature of the 1.75 mm-diameter filament as it enters the print head is 20 C. Calculate q, the heat energy per unit volume, that needs to be input to the filament to raise it to its printing temperature. Note: we assume that the PLA is amorphous so that any latent heat of melting can be ignored. [5 points] (d) Create a third plot showing the ratio p/q against. The quantity p/q is the ratio of the mechanical work needed to drive a unit volume of PLA through the print head to the heat energy input needed to raise a unit volume of PLA to the print temperature. Note that mechanical work done by a load equals that load times the distance moved in the direction of the load, or, equivalently, a pressure times the volume swept out by that pressure. Thus, the pressure at the nozzle inlet, p, is equal to the mechanical work done on a unit volume of material to drive it through the nozzle. Comment on your results. How small does need to become before the mechanical work exceeds the thermal energy input? [8 points] 4

5 Now let us consider how FDM might be adapted for printing metals. Thus far, printing of metals has been restricted to expensive, large-scale industrial printers, many of which use lasers to fuse metal powders, or powders surrounded by binders. Consumer-scale printing has been limited to polymeric materials, which are good for many prototyping applications but do not offer the electrical or thermal conductivities or the mechanical strength that might be needed for other purposes. At the macro-scale, metal alloys are regularly extruded for example to produce aluminum alloy window and door frames. Here we ask whether micro-scale extrusion of metals could enable safer, more affordable and less energy-intensive metal 3D printing than is currently available. There has been at least one attempt to produce an FDM nozzle that can extrude a tin-based solder material 1. Some solders have the advantage that they are formulated to go through a pasty state as they are heated, enabling them to be extruded effectively as a viscous liquid which can fuse with the material below in much the same way as extruded polymers. To be useful for real engineering applications, though, we would need this process to work with more popular engineering alloys, e.g. those with aluminum as their main constituent. Most alloys have a much more sharply defined melting temperature, and so cannot be extruded in the molten state because the material would be far too fluid and would not hold its shape. Thus, we will look at hot extrusion of aluminum alloys at a significant proportion of, but still below, their melting temperature (Figure 2). Figure 2: Extrusion of metallic filaments in FDM In hot extrusion where the material remains solid throughout, the extrusion ratio, /, is the ratio of inlet to outlet material cross sections, and is a key parameter. Larger extrusion 1 5

6 ratios require larger loads to be applied, and the pressure at the inlet to the die can be approximated by: ln where is a material- and temperature-dependent constant. Let us assume that we are extruding an alloy called 1100 Aluminum at a temperature of 450 C, which, on an absolute temperature scale, is about 80% of the melting point of ~650 C. At this temperature (450 C), assume that 50 MPa. (e) Assume that an 1100 Aluminum filament with diameter 1.75 mm is fed into the print head. For values of die opening diameter in the range 0.05 mm 1.0 mm, and assuming that the opening length is short enough that the friction between the die and the extruded material is negligible, create a plot of the pressure,, at the die entrance (vertical axis) against the nozzle diameter (horizontal axis). Remember to label your axes and include units. [6 points] (f) The force F that must be applied to the input metal filament to drive it through the nozzle can be approximated by (note that the relevant diameter is the filament input diameter, not the nozzle input diameter as it is for the plastic extrusion model). Compute F for the full range of pressures found in part (e), showing them in a plot of F against. Do these forces seem practicable to achieve in a consumer-grade (e.g. desktop) system? If they seem too large, what variable(s) would you change to reduce the required load? [5 points] (g) Assume the specific heat capacity of our Al alloy is 900 J/kgK and its density is 2700 kg/m 3. Calculate q, the heat energy per unit volume, that needs to be input to the filament to raise it from room temperature of 20 C to its printing temperature of 450 C. [4 points] (h) Create a plot that shows the total energy input per unit volume of material, p+q, for both PLA and for 1100 Aluminum, against die diameter in the range 0.05 mm 1.0 mm, and assuming 1.75 mm in both cases. [10 points] Your graph will look something like this, but the shapes and relative magnitudes shown below are not accurate: 6

7 (i) Compare the values you plotted in part (h) above to the range of specific energies for machining for aluminum alloys that were given in lecture and in HW1. [5 points] (j) Suggest a few technical issues that have not been considered in this rudimentary analysis. For example, the model we used for metal extrusion does not consider extrusion rate is it realistic to think that the extrusion pressure will be insensitive to rate? Since the aluminum is extruded as a solid, how will it fuse to the part below? Would a post-printing sintering step be needed? Is a non-oxygen-containing chamber needed to prevent oxidation between the filaments as they are extruded? [6 points] Reference: [1] K. Hamad, M. Kaseem, and F. Deri, Melt Rheology of Poly(Lactic Acid)/Low Density Polyethylene Polymer Blends, Adv. Chem. Eng. Sci., vol. 01, no. 04, pp ,