2.810 Manufacturing Processes and Systems Quiz # 1 Oct 18, 2017

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1 NAME: SOLUTIONS COPY Manufacturing Processes and Systems Quiz # 1 Oct 18, 2017 Open book, open notes, calculators, computers with internet off. Please present your work clearly and state all assumptions. Put a box around your answer(s) for the problems with numerical answers. Problems: Points 1. How are these parts made? 25 pts 2. Injection Molding 25 pts 3. Profile cutting performance 15 pts 4. New Thermoforming Process 30 pts 5. Thermoforming Tooling 5 pts 100 pts

2 1. (25 pts) How are these parts made? Please look at the parts in front of you which are depicted below in the pictures and make 2 determinations: 1st what material is this part made from, and secondly what process or processes are used to make this part. Please explain what features about the part prompted you to give these answers. (Do not neglect giving your explanation. It is very important). If a part has multiple components address all of them. a) (6 pts max) Material: (+3) Aluminum/zinc/alloy. Low density. Non-ferrous material for die-casting. (We gave credit to all of these answers. Partial credit was also given to wrong materials but decent explanations. Little to no credit given for selecting a nonfeasible material). Process: (+3) Die casting. Ejector pin marks. Flash clearly visible. Material often diecasted because of its low melt temperature. (Note there are different types of casting that are definitely not the same! For full credit, die casting must be explicitly written). 2

3 b) (7 pts max) Material: (+3) Steel with zinc plating. Needs the strength for application and low springback. Zinc plating for the appearance. (A common answer here was aluminum. Aluminum would be too expensive for this application and have higher spring-back). Process: (+4) Steel wires are drawn (+1). Glued together with adhesive (+1). Sheet metal forming to create the crown and ends (+1). Cut into pieces (+1). (This was a very tricky part. The majority of students thought this was stamped and skipped over the full process. For the application to work, stamping or perforation would either be way too expensive or not high enough quality to ensure proper stapling). 3

4 c) (12 pts max) Material (body): (+3). Aluminum. Application doesn t require high strength and needs to be light weight. Aluminum is also easily extruded. Process: (+3) Extrusion. The constant cross-section suggests this process because of the low-cost. The shear marks are visible as the part was cut from the extruded stock. The body was then drilled/tapped. Material (end pieces): (+3) Stainless Steel. Common material for screws and simple hardware. Process: (+3) The material could be extruded into the rod stock. The threads need to be cold rolled. The hook and eye are then formed. (Some students thought the threads were machined. This would be far too expensive of a process for such a low-cost part). 4

5 2. (25 pts) Injection Molded Part Please consider the injection molded part below (there will be a copy of the part on your desk.) Section A-A a) Please draw the tooling required (core and cavity) to make this part. Also draw a top view of the cavity at section A-A. Draw on the picture where the fate is located. (15 pts max) (+2) The gate is located on the bottom of the cuvette, or shown at the top of the drawing above). (+4) The core dictates the geometry on the inside of the part. It is important to show the both views. (Points were deducted for poor detail on the core and cavity that may suggest not understanding the extent/function of the tool as well as the need to have it fully define the shape of the part). (+4) The cavity dictates the geometry on the outside of the part. Very important: no side pulls are needed in this orientation, which is why this tooling is correct. (Partial credit was given to other orientations of the mold, however if you had two symmetrical halves of the mold, you needed to accurately define the core and 5

6 cavity as well as the multiple side pulls that you then introduced. Failure to do so left you very susceptible to deductions). (+5) The cross-section of the cavity looks like this. (Points were deducted for not explicitly labeling the gate and cavity in this crosssection. It is critical to note that the cavity is the tool, not the empty space in the cross-section). I have attached a SolidWorks Part file at this link for you to download if you would like to see a 3D view of it or create your own sectional drawing to convince yourself of the geometry. 6

7 b) Please make a rough estimate of the cycle time to mold this part (sec). (6 pts max) (+1) Estimate the cooling time of the part using this thickness. t "##$ = h' π ' α ln 4 T./$0 T 23$$/.#$5 π T /6/"07#8 T 23$$/.#$5 Note that under traditional temperatures, this can be reduced to t "##$ = (h/2)' α (+1) Estimate the thickness of the part (constant) to be about 1mm. Note that if there were any larger thicknesses, that it would be the largest thickness not the average. >? ".A (+1) Note that the thermal diffusivity of plastic is α = 10 = 0.1..A B B (+1) Solve for the cooling time. t "##$ = (1mm/2)' = 2.5 seconds 0.1 mm' s (+1) Note that we are looking for the cycle time which includes fill time, ejection, and reset of the mold, not just cooling time. These can normally be ignored and the cycle time is roughly equal to the cooling time. (+1) However, in this particular situation, the cooling time is so short (seconds) that the other steps should not be ignored, since they account for several seconds themselves and would double or triple the cycle time. (There are a number of ways to calculate the other times. The key was to identify them and the magnitude similarity. Full credit was given for estimates, detailed calculations, as well as explanation). 7

8 c) Please make a rough estimate of the clamping force to mold this part. (4 pts max) (+1) We can estimate the force using the maximum pressure that the plastic material can withstand in the mold and the projected area. Force = Pressure x Area (projected) The maximum pressure is approximately 1000 bars according to Boothroyd s Table 8.5 in Design for Injection Molding bars = 1000 MPa = 1000 MN/m 2. (+2) The two dimensions in question are the length and width of the cross-section A-A. A good estimate is around 10mm or 1cm each. Area projected = 1.2cm x 1.2cm = 1.44 x 10 >T m ' (+1) Force = 100 MN m ' x 1.44 x 10>T m ' Force = 14.4 kn (The biggest mistake here was misinterpreting the projected area needed to keep the mold closed. This has to be from the perspective of the core and cavity that you drew in part a) since the material flowing in will cause them to separate. If students used a pressure drop formula rather than the Boothroyd estimate, it was equally important for them to correctly use the dimensions based on the geometry from part a). The answer could be calculated in either SI or English units depending on the conversion used. 8

9 3. (15 pts) Profile cutting performance Consider the profile milling of an aluminum bar stock 2cm X 4cm in cross sectional area and 10 m long. The tool has three teeth and is rotated at 400 rpm and is moved into the stock at 10 m/min (this is along the long axis of the bar stock). Please make rough estimates, the cross-sectional area of the materials removed is 2cm 2 : a. Estimate the material removal rate (cm 3 /minute). (5 pts max) (+2) The MRR is the volume removed in a given time frame. Here it would be the cross-sectional area multiplied by the linear feed into the stock. MRR = d(vol) dt (+2) Both of those values are given. = A dx dt = A v MRR = 2cm ' (1000 cm min ) (+1) Solve with the correct units. MRR = 2000 cm? min (Points were deducted for misinterpreting the feed and rotational velocity). Many people also did not convert meters to centimeters correctly. 9

10 b. Estimate the power needed to cut his aluminum stock (Watts). (5 pts max) (+2) The power can be estimated using the specific energy of removal for the material and the material removal rate. We calculated the MRR earlier. Power = μ B x MRR MRR = 2000 cm? 2000 x 10? = min 60 mm? s (+2) Using Table 21.2 from Kalpakjian (copied into the lecture notes), aluminum has a specific energy of removal of between 0.4 and 1.0 Ws/mm 3. We can use an average of 0.7 Ws/mm 3. (+1) Solve for power in the correct units. Power = 0.7 Ws 2000 x 10? x mm? 60 Power = W = 23.3 kw mm? s 10

11 c. Estimate the force on the tooth (Newtons). (5 pts max) (+2) Note that this is a profile mill and not all teeth are engaged with material so we can t just immediately go to the orthogonal cutting formulas. Instead, we need to use this formula for the force per tooth (slide 28 of the machining lecture). F " = f x d x μ B (This is a tricky question. Many people strictly went directly to the F = P / V or F = u x A equation, which is very simple to calculate using the two above results in parts a and b. However, that is for plunge milling. In this scenario, we are profile milling along an edge and so the most correct answer is estimating the depth of cut so that you didn t need to estimate (or wrongly assume) the number of teeth engaged). (+1) Let s estimate the depth of cut as 1 cm = 0.01 m. (Any reasonable estimate was accepted based on the above diagram. Partial credit was also given for those that mentioned the need for the depth of cut but didn t calculate it). (+1) Calculate the feed per tooth. f = f = v NΩ 10 m min (3 teeth = rev )(400 rev min ) m tooth (+1) Calculate the force per tooth using the correct units and comment on its large magnitude. F " = m tooth x 0.01 m x 0.7 Ws x (1000mm mm? F " = N = 58.1 kn m )? (Some people were thrown off or noted that this calculate force seemed very high. That is because of the extremely fast linear feed into the stock material of 10 m/min. Many people also had issues converting into Newtons. This stemmed mostly from not knowing how to break out Watts into fundamental units). 11

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13 4. (30 pts) Please consider a new thermoforming process innovation as shown below. Using a nonstick, porous, heated platen, the new device claims to greatly reduce the time to heat the plastic sheet prior to forming. In this process, the sheet is heated by conduction with a constant wall temperature. The heating step functions as follows (see the diagrams above): a) A plastic sheet at T = T room is loaded onto the nonstick porous platen at T = T platen. The sheet stays in this position until the top side (side A) of the sheet reaches T target. b) At this point, the platen and the sheet are rotated and the sheet is gently removed from the platen. In some cases, a slight air blast through the porous sheet may be needed to release the sheet. c) At this point the heated sheet is hanging free and the rest of the process is the same as in conventional thermoform. To prevent the sheet from overheating near the heated platen the nonstick surface also provides a heat transfer film coefficient with a resulting Bi -1 = The temperature of the sheet must not increase beyond Tdegrade Please analyze this process. Assume that the sheet is 2 mm thick. The properties of the sheet give a conventional thermal diffusivity as used in class, and the temperatures are: Troom = 20C, Tplaten = 300C, Ttarget = 180C and Tdegrade = 250C. You may assume 13

14 that during heating of the sheet heat loss from side A is minimal. In fact, assume; dt/dz = 0 at side A during heating on the platen, where z is the through the thickness direction. Please answer the following questions: a) How long will it take to heat the sheet (side A) to 180C? (10 pts max) (+2) First we need to recognize this is a heating situation (not cooling, so the numbers are reversed, although you achieve the same answer if you did not reverse) and calculate the dimensionless temperature. θ = T p$30/8 T 03qr/0 T p$30/8 T q##. = 300C 180C 300C 20C = 0.43 (+1) L is the thickness here (2mm) because dt/dz = 0. (+4) The key here is realizing that x/l = 0. This is the most important part of the problem. We have one-sided heating and so side A is actually in the center of the diagram that is drawn to supplement the Fourier charts. We know Bi -1 = 0.25 and can find the Fourier number on the upper left chart from the attached page. 14

15 (In order to use these charts for your answer, you have to accept this convention, therefore a solution with x/l equaling anything but 0 but using the chart must be incorrect). (+1) The Fourier number is 0.7 based on the chart analysis above. (+1) Since the Fourier number is just a dimensionless number relating the time, length, and thermal diffusivity, we can use that Fourier number to calculate the time to heat since the thermal diffusivity can be assumed and we know the thickness from earlier. F = t L ' t = FL' (+1) Solve for the time in the correct units. t = (0.7)(2mm)' = 28 seconds 0.1 mm' s (Partial credit was given if you used the conduction equation without the chart, although it is very difficult to solve both of these correctly because of the film thickness contribution which is not accounted for with that conduction equation. Most of the partial credit came from identifying the thickness and dimensionless temperature, which are the same for each method). b) How long will it take side B to reach 250C? (10 pts max) (+2) Like in part a, first we need to recognize this is a heating situation (not cooling, so the numbers are reversed, although you achieve the same answer if you did not reverse) and calculate the dimensionless temperature. 15

16 θ = T p$30/8 T 5/rq35/ T p$30/8 T q##. = 300C 250C 300C 20C = (+1) L again is unchanged (thickness = 2mm) since we are using the same assumption of dt/dz = 0 and the geometry is the same. (+4) The key here is realizing that x/l = 1. This is the most important part of the problem. We have one-sided heating and so side B is actually at the boundary of the diagram that is drawn to supplement the Fourier charts. We know Bi -1 = 0.25 and can find the Fourier number on the upper left chart from the attached page. (In order to use these charts for your answer, you have to accept this convention, therefore a solution with x/l equaling anything but 1 but using the chart must be incorrect). (+1) The Fourier number is 0.4 based on the chart analysis above. (+1) Since the Fourier number is just a dimensionless number relating the time, length, and thermal diffusivity, we can use that Fourier number to calculate the time to heat since the thermal diffusivity can be assumed and we know the thickness from earlier. F = t L ' t = FL' (+1) Solve for the time in the correct units. 16

17 t = (0.4)(2mm)' = 16 seconds 0.1 mm' s (Partial credit was given if you used the conduction equation without the chart, although it is very difficult to solve both of these correctly because of the film thickness contribution which is not accounted for with that conduction equation. Most of the partial credit came from identifying the thickness and dimensionless temperature, which are the same for each method). c) Does this process work as it is currently done? Why? Do you have any suggestion on how to further improve on this process? (10 pts max) (+3) This process currently does not work. (+2) It does not work because side B degrades approximately 12 seconds before side A is fully heated. (You must explicitly say if the process worked or not and also must give a reason why). (+5) Some ways that this process could be improved: Heating from both sides (changes the heat distribution) Using a lower platen temperature (and running the process longer) Using a thinner plastic sheet (thickness has a second power in the calculation) Using a more conductive material (changing the given Biot number) Achieving a higher film thickness (changing the given Biot number) (If you incorrectly interpreted this question, you were inclined to discuss the thermoforming process in this answer and not the heating. Partial, albeit minimal, credit was given for this as the process would fail during heating before the forming step, and commenting on heating was a much better answer). 17

18 5. (5 pts) Please consider the cross section of a thermoformed plastic cup as shown below. The measured thicknesses at several locations are shown. Was this cup formed on a male or female tool? Please provide a sequential drawing showing the forming process to support your answer and comment on why some sections are thicker than others. (+1) This part was clearly created on a female tool based on the thickness distribution. (+1) The part cooled first at the top of the mold. (+1) The material then touched the bottom and froze. (+1) Afterwards as the material continued to thin, it touches the side and then the corners last. (+1) See next page for the expected sequential drawings. (Points were deducted if you did not explicitly show that the corners touched last versus the sides and bottom. This normally required at least 2-3 drawings. Many students also did not draw sequential drawings which left them susceptible to deductions). 18

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