NAME: Solutions (TA)

2.810 Manufacturing Processes and Systems Quiz # 1 Oct 17, 2018

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? 20 pts 2. Cutting Conditions 20 pts 3. Adiabatic Temp. Rise for wire drawing 20 pts 4. Mfg Energy Estimating 20 pts 5. Machining Process diagram 20 pts 100 pts

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1. (20 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(s) are used to make this part, 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) Sink Strainer (6 pts)

Material(s): (+3 pts) Stainless Steel (both parts). The material should be as cheap as possible an ideally does not rust easily due to its function (sitting in the bottom of a drain exposed to moisture). Some people overthought this part. We want a cheap material and one that is commonly used for sheet metal forming. Aluminum is too expensive. Process(es): (+1 pts) Steel wires are drawn down to the desired diameter (gauge). (+1 pts) Mesh is weaved together (in a 90-degree offset) and formed over a positive tool (notice the elongated cross-hatching on the sides of the indent).

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(+1 pts) Ring is stamped out of sheet metal and formed around the mesh edges. We were lenient on students saying that the steel wires were extruded instead of drawn. However, do note that question #3 is entirely about drawn wires, so this should have been a hint on this question if you were unsure about which. b) Zip Ties (6 pts)

Material(s): (+3 pts) Nylon (plastic). Based on the application, we want a material that s both high-strength and light-weight. We expect this to be injection molded, so our choice is narrowed to a plastic polymer. Process(es): (+3 pts) Injection molding. The gate is located on the head of the tie. We can see the parting line on both sides down along the entire zip die. There is also flash on the tail of the part. Most likely due to how cheap the part is and the aspect ratio, it is not worth risking a short-shot and so they error on the side of caution with the flash. Most people got this fully correct. A few students thought the teeth required side-pulls but that would contradict the parting line placement and be an unnecessary expense for such a cheap part/tool.

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c) Pencil (8 pts)

Material(s): (+1 pts) Wood for the body. We can see the color and grain structure. (+1 pts) Graphite for the tip/core. This needs to be something soft that can smear onto paper. (+1 pts) Rubber for the eraser. Often there are plasticizers added to allow for them to maintain their harness especially when used in abrasion. (+1 pts) Steel for the sheet metal clip. This part almost has no function besides to hold the pieces together and undergo crimping. Process(es): (+1 pts) Two wood halves (slats) are grooved out using a machine. We can see the parting line on the shaved pencils with two different grains/colors of wood. For such an inexpensive part with high quantities, it is not feasible to CNC machine or core out the body. Also, it is not possible to extrude or heat up wood and re-form it. (+1 pts) Forming/extrusion for the graphite writing core which is made from a slurry.

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(+1 pts) Extrusion for the erasers. This ensures that they maintain their low cost. They are tumbled in a large vat to round the edges. (+1 pts) Sheet metal forming for the metal clip. We can see the crimp marks when it was compressed by the tool for assembly with the wood body and the eraser. This was a very tricky question, but students did extremely well to think about what was possible given the material and function of the part. The eraser’s rounded edges threw some people off, saying it was injection molded. Think about how expensive it would be to machine the inside of a rounded tool just for a simple eraser.

The complete details of the pencil-making process can be found here: http://www.madehow.com/Volume-1/Pencil.html. It is fascinating to realize that the pencil is still manufactured nearly the same way it was centuries ago.

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2. (20 pts) Cutting Conditions Can the condition = (rake angle = shear angle) be met under reasonable cutting conditions? The easiest way to determine is to use the equations we have seen in class and relate them to either known strain or friction angle ranges that were demonstrated to be reasonable. (+3 pts) Use the equation that most easily relates the rake and shear angle 𝛾 = cot(∅) + tan(∅ − 𝛼) (+3 pts) Setting the rake angle and shear angle equal makes this equation become extremely simple. 𝛾 = cot(∅) + tan(0) tan(0) = 0 (+3 pts) Now we have an estimate for the shear under this constraint. 𝛾 = cot(∅) (+5 pts) From Slide 13 of the Machining lecture, the reasonable strain range is 1-10. (+4 pts) We can use the strain limits to assess the limits for the rake and shear angle. Any one example of calculating an angle from the strain rate range is satisfactory, but we’ve shown the boundaries here (not fully necessary). 1 = cot(∅) ∅ = 𝑐𝑜𝑡−1(1) = 45° 10 = cot(∅) ∅ = 𝑐𝑜𝑡−1(10) = 5.7°(𝑣𝑒𝑟𝑦𝑙𝑜𝑤) (+2 pts) Yes, there are reasonable cutting conditions within this range that satisfy the criteria as we have seen similar angles in lecture examples.

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It was critical for you to relate the analytical results to machining strain/friction bounds in some way (either through the slide example values or Kalpakjian tables or intuition). This problem could have been done in the reverse order (sub-steps are the same and the conclusion is the same), starting with the Merchant’s equation, then needing to estimate a reasonable angle of friction (and coefficient of friction) based on Slide 18 of the Machining lecture, then checking the magnitude of the strain. Either method received full credit, with point allocations to the analogous component. We could also double check how these ranges impact the angle of friction to ensure a realistic coefficient of friction using the Merchant’s equation. ∅ = 45° + 𝛼2 − 𝛽2

Before checking, we can first make the substitution. ∅ = 45° + ∅2 − 𝛽2

∅2 = 45° − 𝛽2

∅ = 90° − 𝛽 Based on our results of 5 and 45 degrees from earlier, the boundaries for the angle of friction will be 45 and 85 degrees. We can check the coefficient of friction in each case. 𝜇 = 𝑡𝑎𝑛(45) = 1 𝜇 = 𝑡𝑎𝑛(85) = 11.43(𝑣𝑒𝑟𝑦ℎ𝑖𝑔ℎ) The second coefficient of friction is extremely high which makes sense given the low shear angle. This is a situation where reasonable shear conditions may be satisfied but not reasonable friction conditions. If using the friction coefficient as your main source to determine reasonable conditions, note that on Slide 18 of the Machining lecture there was an example given with an angle of friction of 50 degrees and a 1.19 coefficient of friction.

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3. (20 pts) Adiabatic Temperature Rise for Wire Drawing

Please estimate the adiabatic temperature () rise for the copper alloy cylindrical wire drawing process shown in the diagram below. The wire is pulled through a die with a converging section that reduces the diameter of the wire from 2D to D = 2mm over a length of L = 1 cm. We know the force FA as well as the tensile stress 𝜎 = 90 MPa at point A, and velocity VA = 25 cm/sec at point A.

Fig. Die for wire drawing – not to scale a) (2 pts) Please estimate the force and velocity at point B and, (+1 pts) Assume incompressible and by mass conservation, 𝑚𝑖𝑛𝑠 = 𝑚𝑜𝑢𝑡𝑠

𝐴𝐴𝑉𝐴 = 𝐴𝐵𝑉𝐵 𝑉𝐵 = 𝐴𝐴𝐴𝐵 𝑉𝐴

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𝑉𝐵 = 𝜋(𝐷2)2𝜋(𝐷)2 𝑉𝐴

𝑉𝐵 = 𝑉𝐴4 = 6.25 𝑐𝑚𝑠𝑒𝑐 (+1 pts) Because there is no force acting on B, 𝐹𝐵 = 0

b) (18 pts) Estimate the adiabatic temperature rise () due to the plastic work in the die. (+5 pts) We know that the rate of work (plastic) must equal the rate of temperature rise. 𝑞ℎ𝑒𝑎𝑡𝑠 = 𝑊𝑝𝑙𝑎𝑠𝑡𝑖𝑐𝑠

(+2 pts) We use a familiar equation for the rate of temperature rise. 𝑞ℎ𝑒𝑎𝑡𝑠 = 𝜌𝐶 𝑑𝑇𝑑𝑡 = 𝜌𝐶 ∆𝑇∆𝑡

(+5 pts) Then from Slide 17 in the Injection Molding lecture (note that this is the most analogous example because in the slides we have rate of viscous work equaling rate of temperature rise) we know the rate of work is equal to power or force times velocity. Note how similar it is to the top line of that slide. The rate of plastic work is equal to the power divided by the volume. Power can be expanded into force times velocity. The volume can be expanded into the cross-sectional area (given with the diameter) times the length. The tensile stress is equal to F divided by A. Then we can expand the velocity into L divided by time. 𝑊𝑝𝑙𝑎𝑠𝑡𝑖𝑐𝑠 = 𝑃𝑉 = 𝐹𝐴𝑣𝐴𝑉 = 𝐹𝐴𝑣𝐴𝐴 ∗ 𝐿 = 𝐹𝐴𝑣𝐴𝜋𝐷24 ∗ 𝐿 = 𝜎𝐴𝑣𝐴𝐿 = 𝜎𝐴𝐿 ∗ 𝐿∆𝑡 = 𝜎𝐴∆𝑡 (+2 pts) Set the reduced expressions for the rates equal.

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𝜌𝐶 ∆𝑇∆𝑡 = 𝜎𝐴∆𝑡 (+6 pts) Eliminate the time component and solve for the temperature rise using the correct values for Copper and ensuring the correct units within the calculation. ∆𝑇 = 𝜎𝐴𝜌𝐶 = (90𝑀𝑃𝑎)(0.8 𝑔𝑐𝑚3) (0.4 𝐽𝑔𝐾)

∆𝑇 = 𝜎𝐴𝜌𝐶 = (9000 𝑁𝑐𝑚2)(0.8 𝑔𝑐𝑚3) (0.4 𝐽 = 𝑁 ∗ 100𝑐𝑚𝑔𝐾 )

∆𝑇 = 281𝐾 = 281℃ More than half the class just computed the adiabatic temperature rise due to cutting from Slide 33 of the Machining lecture. While this looks similar, this skipped a critical portion of the question, which was identifying that the rate of plastic work must equal the rate of temperature rise, like we did in the review for injection molding and viscous work. Considering so many students missed the crux of this question, a few more points were weighted towards the calculation and collecting the correct values for Copper.

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4. (20 pts) Mfg Primary Energy Estimate for an aluminum bracket for an aluminum bracket (You may assume that the efficiency of the electric grid is ⅓, give your answers as Primary energy)

Consider the part we looked at in class only now estimate the energy to make this part by: a) (10 pts) Powder Bed Fusion; Metal Laser shown in the diagram as red circles. This process is sometimes also called SLM for scanning laser melting. The powder aluminum alloy used to make this part has an embodied energy of 230MJ/kg and the support structure (waste) needed to make this part is equal to the weight of the part (444g). There are several measurements for this process in the “hockey stick” diagram. Please use the most recent data (2017) for a “single bed”. Give you answer in MJ. You can ignore the finishing operations. How does this compare with the machining estimate?

(+1 pts) Total mass of powder bed material used (twice the mass of the part): (2)(444𝑔) = 0.888𝑘𝑔 (+2 pts) Embodied energy in material (aluminum: 230 MJ/kg): 𝐸𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (230𝑀𝐽𝑘𝑔) (0.888𝑘𝑔) = 204𝑀𝐽 (+1 pts) The efficiency of the electric grid is 1/3 meaning, we must multiply by a factor of 3 in our electrical energy of the process.: (+2 pts) Electrical energy of powder bed is shown to be 1.2x10^9 J/kg based on the diagram (must include the grid efficiency factor) and the total mass is used again: 𝐸𝑚𝑎𝑛𝑢𝑓𝑎𝑐𝑡𝑢𝑟𝑖𝑛𝑔 = (1,200𝑀𝐽𝑘𝑔) (0.888𝑘𝑔)(3) = 3197𝑀𝐽 (+2 pts) Total primary energy is the sum of the embodied energy of the material and the electrical energy of the process.

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𝐸𝑡𝑜𝑡𝑎𝑙 = 204𝑀𝐽 + 3197𝑀𝐽 = 3401𝑀𝐽 (+2 pts) The total primary energy is higher than the 1825MJ estimate for machining due to printing both the support material and the part as well as a high estimate of powder bed electrical energy requirement.

Students lost points if they directly applied the machining calculations and did not think about how additive manufacturing processes both support and part material 2 points). We were lenient with this severe mistake because we didn’t cover the process of powder bed fusion in much detail. Several students also used the incorrect estimate based on the diagram and lost a small number of points (1 point).

b) (10 pts) For high volume production of this part consider a casting process and ignore the tooling. The melters (electric induction melters) shown in the “hockey stick” diagram are on the bottom right hand side. Assume the complete casting process requires about twice the MJ/kg required for the melters, and the scrap is 20%. For the cast parts you can use secondary aluminum. Give your answer in MJ. You can ignore the finishing operations. How does this compare with the machining estimate? (+1 pts) Total mass of casted material (includes 20% scrap): (1 + 0.2)(444𝑔) = 0.533𝑘𝑔 (+2 pts) Embodied energy in material (secondary aluminum: 25 MJ/kg): 𝐸𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (25𝑀𝐽𝑘𝑔) (0.533𝑘𝑔) = 13.3𝑀𝐽

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(+1 pts) Specific energy of casting. From the “hockey stick” diagram we can see the induction melters require approximately 2 MJ/kg. Given is the fact that a complete casting process requires twice as much as that, 4 MJ/kg. (+2 pts) Electrical energy of casting (note that you still need to include the grid efficiency factor): 𝐸𝑚𝑎𝑛𝑢𝑓𝑎𝑐𝑡𝑢𝑟𝑖𝑛𝑔 = (4𝑀𝐽𝑘𝑔) (0.533𝑘𝑔)(3) = 6.40𝑀𝐽 (+2 pts) Total primary energy is the sum of the embodied energy of the material and the electrical energy of the process. 𝐸𝑡𝑜𝑡𝑎𝑙 = 13.3𝑀𝐽 + 6.40𝑀𝐽 = 19.7𝑀𝐽 (+2 pts) The total primary energy is much lower than the 1825MJ estimate for machining. By inspection, this is mostly due to being able to use secondary aluminum and reduce the embodied energy by a factor of 10 as well as have a much lower energy associated with induction melting. Biggest areas of concern that caused point deductions (each usually 1-2 points):

- Not including the grid efficiency for the induction melting energy calculation - Not doubling the induction melting energy to approximate casting - Not correctly calculating the total mass or only performing the energy calculation

on only the scrap mass (see above diagrams showing how scrap and part mass equal the net volume that is cast). We were lenient with this because while we have done injection molding at length and it is analogous, we did not directly touch on casting.

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5. (20pts) Machining Process Diagram Recall the Process Window diagram for injection molding in terms of temperature and pressure, now please draw your version of the Process Window for Machining (Face Milling) in terms of MRR and cutting force on the diagram below. Do this for a given material and indicate what happens if you exceed these limits. Please label your boundaries and provide enough description so we can follow your arguments.

In general, this answer was graded based on how much correct information you gave about this relationship. (+2 pts) Correctly give the force equation with its components for face milling 𝐹 = 𝑓 ∗ 𝑑 ∗ 𝑢𝑠 (+2 pts) Correcting give the MRR equation with its components for face milling 𝑀𝑅𝑅 = 𝑣𝑤𝑑 (+2 pts) Unlike the injection molding process window where T and P are both independent, the cutting force is largely the independent variable in this question, while the MRR is dependent, hence why they are drawn on their respective axes. This suggests that there is some relationship between them and that they might be coupled. (+5 pts) A large portion of the points were given to rearranging the two equations to relate the two parameters and give some indication of the shape and slope.

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𝑀𝑅𝑅 = (𝑓𝑛𝑁)𝑤𝑑 𝑀𝑅𝑅 = (𝑓𝑑)(𝑤𝑛𝑁) From here you could set the (fd) from the MRR equation equal to the F equation and solve, 𝑀𝑅𝑅 =(𝑤𝑛𝑁𝑢𝑠 ) 𝐹

As we can see, there is a positive relationship between MRR and F. The slope also is dictated by the specific energy of removal of the material and the spindle speed, the number of teeth, and the width of the tool, all of which are given and could either be high or low (which creates the diagonal boundaries of the window). (+2 pts) (Label A on picture) There is an upper y-axis boundary which is the power of the spindle. This is given by the equation below and should represent itself as a straight horizontal line as both the power of the spindle and the specific energy of removal are constants for a given mill and material. The motor would stall above this line. 𝑀𝑅𝑅𝑚𝑎𝑥 = 𝑃𝑚𝑎𝑥𝑢𝑠

(+2 pts) (Label B on picture) There is an upper x-axis boundary which is the maximum cutting force that the tool can withstand before breaking (based on either feed or depth of cut). This boundary could also be from the part bending or distorting, or chatter at the interface (stick-slip phenomenon). 𝐹𝑚𝑎𝑥 = 𝑓 ∗ 𝑑 ∗ 𝑢𝑠 (+2 pts) (Label C on picture) There is a lower x-axis boundary which is the minimum cutting force needed to cut the material and shows up as a straight vertical line. 𝐹𝑚𝑖𝑛 = 𝑓 ∗ 𝑑 ∗ 𝑢𝑚𝑖𝑛 (+2 pts) (Label D on picture) There is a lower y-axis boundary which is possible but uneconomical since MRR is so low. The machine would have an unnecessarily high power rating if the user was constantly operating in this range and they should seek a lower power mill.

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(+1 pts) Some other interesting areas that you could put on this chart that we have covered in class:

- High-speed machining (high MRR and moderate/high F) - Surface finishing (high MRR and low F) - Temperature degradation of the tool/material (high MRR and high F):

We know that the change in temperature is a function of the velocity and feed, which show up in these MRR and F equations. There is also a 1/3 power to this relationship, so it most likely is a straight line or cuts off the top corner of highest MRR and F. ∆𝑇 = 0.4 𝐻𝜌𝐶 (𝑣𝑓𝛼 )0.33 You could have also gotten to that relationship between MRR and F through an easier method without using the individual MRR and F expressions (which is like the rearranging that I do in my quiz review slides for machining). 𝑃 = 𝐹𝑉 𝑃 = 𝑢𝑠𝑀𝑅𝑅 𝑢𝑠𝑀𝑅𝑅 = 𝐹𝑉 𝑀𝑅𝑅 = 𝑉𝑢𝑠 𝐹

Several students (probably being short on time) jumped straight to a shape that looked like the injection molding processing window and then labeled the four sides using their intuition. With the above rubric, this received approximately half credit, assuming those four sides were correctly labeled. Many students also correctly wrote down the equations from an analytical approach but then didn’t apply intuition to how that would affect the shape of the window and so lost 2 points for the shape. Overall, while these were the general guidelines, more credit was given to interesting and correct relationships.