Homework 5A - F23 Solution

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Feb 20, 2024

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ME250 | F23 | University of Michigan HW5A: Manufacturing Lectures (66 pts) - SOLUTIONS Due Tuesday, November 21st by 11:59 pm on Canvas This is an individual assignment, and your solution must be entirely prepared by you. Homework assignments must be completed on your own (unless they are team assignments), however you are encouraged to discuss the problems with your classmates. Upload a PDF of your solution to the Assignments tab on Canvas. Problem 1: Manufacturing Processes (19 points) A. Please identify the variable costs and fixed costs for each manufacturing method. Variable costs ($) Fixed costs ($) 3D Printing 5 0 Machining 3 2000 Injection Molding 0.25 20000 +0.5 points per correct variable cost +0.5 points per correct fixed cost 4 points total if all answers are correct B. Table below: Batch size 3D Printing ($/part) Machining ($/part) Injection Molding ($/part) 1 5 2003.00 20000.25 3 5 669.67 6666.92 10 5 203.00 2000.25 30 5 69.67 666.92 100 5 23.00 200.25 300 5 9.67 66.92 1000 5 5.00 20.25 3000 5 3.67 6.92 10000 5 3.20 2.25

Breakeven: 3D-Printing and machining: 1000 parts 3D-Printing and injection molding: 4211 parts (solve to find 4210.5 parts, MUST round up because you cannot have a fraction of a part) Machining and injection molding: 6546 parts (solve to find 6545.45 parts, MUST round up because you cannot leave a fraction of a part) +2.5 points for proper plot above with correct axes +2 points partial credit if 3 lines are not plotted on the same graph +0.5 points partial credit for each correctly drawn line (if not all three are correct) +0.5 points for each correct breakeven value ( 1.5 points total ) +0.25 points partial credit if they report 4210 or 4210.5 parts for 3D-printing and injection molding C. Table below: Batch size 300 3000 10000 3d printing 1500 15000 50000 Machining 2900 11000 32000 Injection molding 20075 20750 22500 For batch size 300, 3D printing is cheapest. For batch size 3000, machining is cheapest. For batch size 10000, injection molding is cheapest. +1 point for each correct process corresponding to each batch size ( 3 points total ) D. ● 3D printing cannot meet the order in time as it can only produce 500*8 = 4000 parts . ● Machining can produce 1500*(8-1) = 10500 parts . Therefore, it can easily meet the requirement

● Injection molding can produce 3000*(8-5) = 9,000 parts . Just enough to meet the requirement. +0.5 points for each correct value above ( 1.5 points total ) Costs – ● Machining cost of 9000 parts = $2,000 + 9000*($3) = $29,000 Profit = 9000*($10) - $29,000 = $61,000 ● Injection molding cost of 9000 parts = $20,000 + 9000*($0.25) = $22,500 Profit = 9000*($10) - $22,500 = $67,750 Based on these estimations, injection molding must be chosen. +1 point for each correct calculation of profit for both feasible options ( 2 points total ) +0.5 points for selection of injection molding as best option E. Lots of options (anything sensible). E.g., Availability of machines (maybe being used to process other orders). For the processes: achievable tolerances; surface finish, compatibility with the material used. +1 point for each reasonable answer ( 3 points total )

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Problem 2: Sand Casting (12 pts) A 71% steel by composition and 29% lead by composition alloy casting is made in a sand mold with a sand core. The density of steel is 7.82 g/cm 3 , density of lead is 11.30 g/cm 3 , density of sand is 1.60 g/cm 3 , and the chaplets are made of lead. The volume of the core is 0.020 m 3 , and the volume of the chaplets are 0.001 m 3 each. Also, each chaplet can support 80 N of force. How many chaplets are required above and below the sand core? ρ ? = ????𝑖?? ?? ????? = ρ ????? %???? + ρ ???? %???? = 7. 82 ? ?? 3 ( ) 0. 71 ( ) + 11. 30 ? ?? 3 ( ) 0. 29 ( ) = 8. 83 ? ?? 3 +1 point for equation +1 point for answer ρ ?ℎ?? = 11. 30 ? ?? 3 > 8. 83 ? ?? 3 = ρ ? ?? ?ℎ? ?????? ?ℎ?????? ?? ??? ?????? ?ℎ? ???????? F Chap = 80 N ? ? = ??𝑖?ℎ? ?? ???? = ρ ? ? ? ? = (1600 ?? ? 3 )(0. 020 ? 3 )(9. 81 ? ? 2 ) = 313. 92 ? +1 point for equation +1 point for answer ? ? = ???????? ????? ?? ???? = ρ ? − ρ ? ( ) ? ? ? = 8830 − 1600 ( ) ?? ? 3 0. 020 ? 3 ( ) 9. 81 ? ? 2 ( ) = 1418. 526 ? +1 point for equation +1 point for answer # ?ℎ?????? ?? ??? = ? ? 𝐹 ?ℎ?? = 1418.526 ? 80 ? = 17. 7 = 18 ?ℎ?????? +1 point for equation +1 point for answer

# ?????? ?ℎ?????? = ? ? +? 𝑇???ℎ?? ? ?ℎ?? 𝐹 ?ℎ?? = 313.92 ?+18 11300 ?? ? 3 ( ) 0.001? 3 ( ) 9.81 ? ? 2 ( ) 80 ? = 28. 86 = 29 ?ℎ?????? +1 point for W c +1 point for N topChap +1 point for W chap +1 point for answer

Problem 3: Manufacturing Processes (10 pts) Professor Austin-Breneman wants to upgrade his tractor to film a shot-for-shot remake of Fast and Furious V for a class video. The following pictures are a series of accessories he is adding to the tractor which were made using one of the manufacturing processes discussed in this course. Indicate which primary shaping process(es) were used to make the products and justify your answer. A. Lightweight plastic sunshade canopy (3 feet x 4 ft) (1 pt) Plastic Thermoforming, large plastic sheet stock and molded features, uniform thickness of features, cut edge. +0.5 pt for process, +0.5 for reasonable justification B. Metal rear brake drum (1 pt) Sand Casting with post-machining. Surface finish. Geometry. +0.5 pt for process, +0.5 pt for reasonable justification C. Metal hood (1 pt) Sheet metal Stamping. Made from sheet metal, molded shape with cutouts. +0.5 pt for process, +0.5 pt for reasonable justification D. Professor Austin-Breneman is looking to upgrade the seat for a more comfortable ride. He finds two options, the red metal one on the left, and the black plastic one on the right. What are the manufacturing processes used to create the two seats? Describe one manufacturing trade-off between the two different approaches? (4 pts) Stamping for the red metal part (also accepted die casting), injection molding for black seat. Injection molding can hold tighter tolerances, smoother surface finish and can create features like the ribs that stamping cannot. +1 pt for each process, +2 pt for justification E. Professor Austin-Breneman would like to create some custom tractor wheel rims like the ones shown below. He is considering sand casting and die casting as possible process choices. Suggest one process and give two reasons for your selection. (3 pts) For die casting: Better surface finish, quicker, better/tighter tolerancing, less incremental costs due to lower labor costs to break up molds/clean For sand casting: Material properties of cast iron, don’t have to pay the costs for metal die, better for lower production rates/smaller batch size

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+1 pt for process, +1 pt for each of the two reasons

Problem 4: Additive Manufacturing (9 pts) Please complete the chart below about additive manufacturing methods. List the full name of each process and place an “x” in the cells that describe each process. SLA – Stereolithography PolyJet Printing FDM – Fused Deposition Modeling SLS – Selective Laser Sintering DMLS – Direct Metal Laser Sintering SLM – Selective Laser Melting +0.5 points for each correct “x” and each name -0.5 points for each incorrect “x” NOTE: an additional x for multi-nozzle FDM is accurate)

Problem 5: Forging (16 pts) A cylindrical part is warm upset compressed by forging in an open die. Before forging, the diameter axially is 20mm and the height is 60mm. The coefficient of friction at the die-work interface is 0.20. The flow curve of the work material at forging temperature is defined by a strength coefficient of K is 530MPa, and a strain-hardening exponent of n is 0.23. Assume the volume of the part remained unchanged during the whole process and no barreling effects. A. The maximum strain of the part is 0.485. What is the maximum area of contact between the die and the part? ? 0 ℎ 0 = ? ??? ℎ ? [1] π? 0 2 ℎ 0 4 = ℎ ? ???? ε = ?? ℎ 0 ℎ ? ℎ 0 ℎ ? = ? ε ℎ 0 ? ε = ℎ ? Plug hf into [1] π? 0 2 ℎ 0 4 = ℎ 0 ? ε ???? h0 cancels out ? ??? = π? 0 2 ? ε ?????𝑖? 4 = π(0.2?) 2 ? 0.485 4 = 510 ?? 2 = 5. 10 ? 10 −4 ? 2 +1 point for the equation for h f +1 point for the equation ε ?????𝑖? +1 point for substituting hf into the equation ε ?????𝑖? +3 points for the correct answer A max B. What is the maximum force required to forge this part to its maximum area? (10pts) Df = .0255, hf = .0369 (1) 𝐹 = 𝐾 ? 𝑌 ? ? ??? (2) 𝐹 = 𝐾 ? (𝐾ε ? ) ? ??? (3) 𝐹 = (1 + 0.4µ? ??? ℎ ? )(𝐾ε ? )? ???

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Find final diameter and final height: ??𝑖??? = 4? ?𝑖??? π = 4(5.10 ? 10 −4 ? 2 ) π = 0. 0255 ? h±inal = ( π? 0 2 ℎ 0 4 ) / ( π? ?𝑖??? 2 4 ) = [(0. 02 ?) 2 * 0. 06? ] / [(0. 025?) 2 ] = 0. 0369 ? Find Kf: 𝐾? = (1 + 0.4µ? ??? ℎ ? ) = [1 + (0. 4 * 0. 2 * 0. 0255)/(0. 0369)] = 1. 055 Find Yf using given strain value: 𝑌? = 𝐾ε ? 𝑌? = (530 ?𝑃? ) (0. 485 0.23 ) = 449 ?𝑃? Plug values into equation 1: 𝐹??? = (1. 055) (449 ?𝑃?) 5. 1 ? 10 −4 ? 2 ( ) F = 0.241 MN = 241 kN +2 points for Eq 1 +2 points for substituting in the strain, Eq. 2 +2 points for substituting in the shape factor, Eq. 3 +4 points for the final answer. ( – 1pt for not including units)